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Some Important Lie Symmetries in Both General Relativity and
Teleparallel Theory of Gravitation
By
Suhail Khan
A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENT FOR THE DEGREE OF DOCTOR OF
PHILOSOPHY IN MATHEMATICS
Supervised by Dr. Ghulam Shabbir
Faculty of Engineering Sciences,
Ghulam Ishaq Khan Institute of Engineering Sciences and Technology Topi, District: Swabi, Khyber Pakhtunkhwa, Pakistan.
2011
ii
Author’s Declaration
I Suhail Khan S/O Haji Nawab declare that the work in this dissertation was carried out in
accordance with the regulations of Ghulam Ishaq Khan Institute of Engineering Sciences and
Technology Topi, Swabi. The work is original except where indicated by special reference in the
text and no part of the dissertation has been submitted for any other degree either in Pakistan or
overseas. Most of the work presented in this dissertation has been published in reputed journals.
Suhail Khan
ES0503
Faculty of Engineering Sciences,
GIK Institute, Topi, Swabi,
Khyber Pakhtunkhwa, Pakistan.
August, 2011.
iii
Certificate
It is certified that the research work presented in this thesis, entitled “Some
Important Lie Symmetries in Both General Relativity and Teleparallel
Theory of Gravitation” was conducted by Mr. Suhail Khan under the supervision
of Dr. Ghulam Shabbir.
(External Examiner) (External Examiner) Prof. Dr. Khalida Inayat Noor, Prof. Dr. Rasheed A. Muneer Professor, Professor, COMSATS Institute Islamabad. NUST, Islamabad. (Supervisor) (Dean) Dr. Ghulam Shabbir, Prof. Dr. Syed Ikram A. Tirmizi Associate Professor, Faculty of Engineering Sciences. Faculty of Engineering Sciences.
iv
Dedication
I DEDICATE THIS THESIS TO
MY LOVING SON
SAAD AHMAD
AND
MY LOVING ANGEL
SILAH SUHAIL
v
Table of Contents Table of Contents v
Abstract viii
Acknowledgments ix
1. Preliminaries 1.1 Introduction................................................................................................................................... 1
1.2 Basics of General Relativity ......................................................................................................... 3
1.2.1 Manifolds .............................................................................................................................. 3
1.2.2 Tangent Spaces ..................................................................................................................... 5
1.2.3 Tensors .................................................................................................................................. 6
1.2.4 Covariant Derivative ............................................................................................................. 8
1.2.5 Parallel Transport and Lie Derivative ................................................................................... 9
1.2.6 Riemann Curvature Tensor ................................................................................................. 10
1.2.7 Symmetries in General Relativity ....................................................................................... 11
1.2.8 Space-Times and Tetrad...................................................................................................... 12
1.2.9 Classification of Tangent Spaces ........................................................................................ 14
1.3 Basics of Teleparallel Theory of Gravitation.............................................................................. 14
1.3.1 Tetrad in Teleparallel Theory ............................................................................................. 14
1.3.2 Space-Time Structure.......................................................................................................... 15
1.3.3 Teleparallel Lie Derivative ................................................................................................. 17
1.4 Literature Review of Conformal, Homothetic, Killing and Self Similar Vector Fields in General Relativity .................................................................................................................................... 18
1.5 Literature Review of Teleparallel Killing and Homothetic vector fields in Teleparallel Theory of Gravitation.............................................................................................................................. 21
2. Teleparallel Killing Vector Fields in Bianchi Types I, II, VIII and
IX Space-Times 2.1 Introduction................................................................................................................................. 23
vi
2.2 Teleparallel Killing Vector Fields in Bianchi Type I Space-Times............................................ 24
2.3 Teleparallel Killing Vector Fields in Bianchi Type II Space-Times .......................................... 53
2.4 Teleparallel Killing Vector Fields in Bianchi Types VIII and IX Space-Times......................... 69
2.5 Summary of the Chapter ............................................................................................................. 85
3. Teleparallel Killing Vector Fields in Kantowski-Sachs, Bianchi
Type III, Static Cylindrically Symmetric and Spatially
Homogeneous Rotating Space-Times 3.1 Introduction................................................................................................................................. 87
3.2 Teleparallel Killing Vector Fields in Kantowski-Sachs and Bianchi Type III Space-Times...... 88
3.3 Teleparallel Killing Vector Fields in Static Cylindrically Symmetric Space-Times ................ 102
3.4 Teleparallel Killing Vector Fields in Spatially Homogeneous Rotating Space-Times............. 133
3.5 Summary of the Chapter ........................................................................................................... 147
4. Teleparallel Proper Homothetic Vector Fields in Bianchi Type I,
Non Static Plane Symmetric and Static Cylindrically Symmetric
Space-Times 4.1 Introduction............................................................................................................................... 149
4.2 Teleparallel Proper Homothetic Vector Fields in Bianchi Type I Space-Times....................... 150
4.3 Teleparallel Proper Homothetic Vector Fields in Non Static Plane Symmetric Space-Times . 180
4.4 Teleparallel Proper Homothetic Vector Fields in Static Cylindrically Symmetric Space-Times200
4.5 Summary of the Chapter ........................................................................................................... 231
5. Proper Conformal Vector Fields in Non Conformally Flat Non
Static Cylindrically Symmetric, Kantowski-Sachs and Bianchi
Type III Space-Times 5.1 Introduction............................................................................................................................... 234
5.2 Proper Conformal Vector Fields in Non Conformally Flat Non Static Cylindrically Symmetric Space-Times ............................................................................................................................. 235
5.3 Proper Conformal Vector Fields in Non Conformally Flat Kantowski-Sachs and Bianchi type III Space-Times ........................................................................................................................ 263
vii
5.4 Summary of the Chapter ........................................................................................................... 278
6. Self Similar Vector Fields in Kantowski-Sachs, Bianchi Type III,
Static Plane Symmetric and Static Spherically Symmetric Space-
Times 6.1 Introduction............................................................................................................................... 280
6.2 Self Similar Vector Fields in Kantowski-Sachs and Bianchi Type III Space-Times................ 281
6.3 Self Similar Vector Fields in Static Plane Symmetric Space-Times ........................................ 286
6.4 Self Similar Vector Fields in Static Spherically Symmetric Space-Times ............................... 292
6.5 Self Similar Vector Fields in Static cylindrically Symmetric Space-Times ............................. 301
6.6 Summary of the Chapter ........................................................................................................... 310
References………………………………………………………………………………………...312
viii
Abstract
The aim of this thesis is to study some important Lie symmetries of well known space-times in
general relativity and teleparallel theory of gravitation. In teleparallel theory we investigated
teleparallel Killing vector fields for Bianchi types I, II, III, VIII, IX, static cylindrically
symmetric, Kantowski-Sachs and spatially homogeneous rotating space-times and proper
teleparallel homothetic vector fields for Bianchi type I, static cylindrically symmetric and non
static plane symmetric space-times. In general relativity conformal vector fields are studied for
non conformally flat Kantowski-Sachs, Bianchi type III and non static cylindrically symmetric
space-times. We also study self-similar vector fields in general relativity for Kantowski-Sachs,
Bianchi type III, static plane symmetric and static spherically symmetric space-times.
In teleparallel Killing vector fields for Bianchi types I, II, III, VIII, IX, static cylindrically
symmetric, Kantowski-Sachs and spatially homogeneous rotating space-times we have shown
that the number of teleparallel Killing vector fields for the above space-time are either same in
number or more to the Killing vector fields in general relativity. Interestingly, in the presence of
torsion it turns out that in some cases of Bianchi type I and static cylindrically symmetric space
times the dimension of teleparallel Killing vector fields is ten.
In proper teleparallel homothetic vector fields for Bianchi type I, static cylindrically symmetric
and non static plane symmetric space-times we have shown that the number of teleparallel proper
homothetic vector fields is always one which is the same in number to proper homothetic vector
field in general relativity. In general the dimension of the teleparallel homothetic vector fields is
always greater to the homothetic vector fields in general relativity.
In general relativity we have considered Kantowski-Sachs, Bianchi type III and non static
cylindrically symmetric space-times for proper conformal vector fields. We have shown that the
above space-times possess proper conformal vector fields for a special choice of metric
functions.
In general relativity we have investigated self similar vector fields for Kantowski-Sachs, Bianchi
type III, static plane symmetric, static spherically symmetric and static cylindrically symmetric
space-times. All the above space-times admit self-similar vector fields for a special choice of the
metric functions. Here, we have also discussed self similarity of first, second, zeroth and infinite
kind for both the tilted and non-tilted.
ix
ACKNOWLEDGMENTS
I have really no words to express the deepest sense of gratitude to Almighty Allah, who enabled
me to complete this thesis successfully. I wish to express my sincere gratitude to my respectable
and praiseworthy supervisor Dr. Ghulam Shabbir for giving me an opportunity to work under
his dynamic supervision. His motivation, continuous guidance, patience and valuable advices
have been fruitful in the completion of this uphill task.
I must acknowledge the valuable discussions with Dr. Muhammad Jamil Amir who’s guidance
enabled me to work in the field of teleparallel theory of gravitation. I am also indebted to my
father Haji Nawab for his encouragement, moral and financial support throughout my studies. I
should thank my wife for her patience and forbearance. Without her help it would have been
impossible for me to continue my studies. I am also thankful to my son Saad Ahmad and my
daughter Silah Suhail who spent most of their time without me. I am deeply indebted to my
mother, sisters, mother in law and brother Tahir Khan who always prayed for my brilliant
success. The name of my father in law Mr. Faramosh Khan needs to be specially mentioned here
for his moral and financial support he extended to me during my studies. Without his financial
support for my family it would not be possible for me to complete this task.
My thanks are due to some of my friends Abdul Mateen, Dr. Rana M. Ramzan, Dr. Amjad Ali,
Alamgeer Khan, Fazal Wahab, M. Tufail and Adam Khan who cooperated with me during my
studies.
Finally, I profoundly appreciate the financial support of Higher Education Commission of
Pakistan whose generous support made my dreams true. By awarding indigenous 5000
scholarship to me HEC has lightened my financial burden which allowed me to achieve a higher
goal. The transparency and generosity of HEC has inspired me to give back my best to the
community.
I would also thank Director Higher Education (Colleges) Khyber Pakhtunkhwa for granting me
study leave.
Suhail Khan
1
Chapter 1
Preliminaries
1.1 Introduction
Since long, man is trying to unfold mysteries of the universe and this quest for
exploring secretes of universe is increasing day by day. With all the modern
technology in hands man is still unable to resolve completely the mystery of gravity.
It is fact that no one can explain gravity at the most elementary level. In history,
gravity remained one of the most attractive topics to study. The common people
understand gravity just as a force to help them in staying on surface of earth. The
significance of gravity is much wider because it is responsible for the paths followed
by the planets around the Sun.
Among the ancient philosophers, Aristotle had a great insight of massive bodies. He
believed that universe was made up of five elements: earth, water, air, fire and ether.
He called ether the heavenly element and the rest of four as earthly elements. In his
law of terrestrial motion, he states that “all terrestrial bodies tend to go to their natural
state of rest”[1]. According to him anything taken from earth will alternately fall to
the earth, which is the natural state of rest for that body. Galileo Galilei has observed
the solar system through his telescope and concluded that Sun is the center of the
solar system and all the other planets revolve around it. He also discovered that
objects of different weights fall to the surface of earth with the same speed. Newton
was another notable figure who worked on gravity. He says that gravity travels
instantaneously throughout space. He thought that gravity is a force between the two
objects. According to Newton’s law of gravitation “The force of gravity is
proportional to the product of the two masses and inversely proportional to the square
of the distance between them”. The gravitational phenomena observed in the solar
2
system do not perfectly match with Newton’s law especially for Mercury’s orbit
where deviation of Mercury from precession is exceptional.
It was Albert Einstein who made the picture clear that in our universe everything is
relative. In 1905 he presented his theory of special relativity which he based on two
principles: (i) all inertial frames are physically equivalent and (ii) The speed of light
in vacuum is constant for all inertial observers. Since this theory is limited to deal
with linear motion, it has given the name special relativity. After spending 10 more
years Einstein proposed his second theory of relativity known as general theory of
relativity. The basic approach of this theory is to describe gravitational interaction
among massive objects by geometrizing the space-time.
It is the major aim of theoretical physics to procure a stable theory which unifies
general theory of gravity with the laws of quantum mechanics. One attempt to obtain
such a theory was made by Einstein in 1928. He used the mathematical structure of
distant or absolute parallelism also called teleparallelism. At each point of tangent
space of the four dimensional space-time he introduced a tetrad field. The tetrad field
was then used to compare the direction of the tangent vectors at different locations of
the space-time structure. This teleparallel theory involves sixteen components at each
point for the specification of four tetrad vectors while there are only ten components
for the specification of symmetric metric tensor. Einstein erroneously supposed that
the additional six degrees of freedom guaranteed by the tetrad is representing
electromagnetic field [2]. The attempt made by Einstein to unify general relativity
and electromagnetism failed but the concept introduced by him is still significant.
Today we study teleparallel theory as an independent theory of gravitation without
unifying if with the laws of quantum mechanics. Therefore, gravity can be studied
alternately in general relativity or teleparallel theory of gravitation.
Gravitation describes two equivalent descriptions because of the gravitational
property called universality, i.e. in nature every object feels gravity the same (at least
at classical level). This point can be described in both teleparallel and general theory
of relativity. In general relativity theory the role of gravitation is to give the same
3
acceleration to two objects of different masses. Because of this property, gravitation
is described through curvature of the space-time. On the other hand torsion in
teleparallel theory acts as a force on objects. In teleparallel theory of gravitation there
are no geodesics, but there are force equations [3].
To understand our universe we need to know the physical and geometrical features of
our space-times. To understand these features different theories have been developed
but general relativity seems more relativistic theory at least at classical level. General
relativity theory is governed by Einstein’s field equations. These field equations are
highly non linear. Therefore, to solve Einstein field equations certain symmetry
restrictions are required. Symmetries are therefore widely studied in general
relativity. In this chapter a brief introduction to the basic definitions involved in
general relativity and teleparallel theory of gravitation are given. All the stuff for
basic definitions involved in general relativity and teleparallel theory of gravitation
have been taken from [4-7]. The layout of this chapter is as follows: In section (1.2) a
brief description about manifolds is given. Notions of tangent spaces, tensors,
covariant derivative, parallel transport, Lie derivative and Riemann curvature tensor
are given. A brief introduction to symmetries in general relativity is given which
involves definition of a space-time, tetrad formalism and tangent space classification.
In section (1.3) some basic definitions in teleparallel theory, space-time structure and
Lie derivative are given. In section (1.4) literature review of conformal, homothetic,
Killing and self similar vector fields in general theory of relativity is also given. In the
last section (1.5) a brief literature review of teleparallel Killing and homothetic vector
fields are given in teleparallel theory of gravitation.
1.2 Basics of General Relativity 1.2.1 Manifolds
The aim of this section is to give a precise definition and some properties of
manifolds. A manifold is a set which is composed of the pieces that look like open
4
subsets of mℜ and these pieces can be combined mutually smoothly. One can define
manifold as [5]:
An m-dimensional, ,∞C real manifold M is a set together with a collection of
subsets λU satisfying the following properties:
(1) Each Mp∈ lies in at least one ,λU i.e., λU covers M .
(2) For each ,λ there is a bijective map ),(: λλλλ ϕϕ UU → where )( λλϕ U is an
open subset of mℜ .
(3) If any two sets λU and µU overlap, φµλ ≠∩UU (where φ denotes the empty
set), we can consider the map 1−λµ ϕϕ o (where o denotes composition) which takes
points in mUU ℜ⊆∩ ][ µλλϕ to points in .][ mUU ℜ⊆∩ µλµϕ
A schematic diagram showing the defining notions of a manifold is shown below in
the figure 1.2.1.
Figure 1.2.1
We require these subsets of mℜ to be open and this map to be ,∞C i.e., infinitely
continuously differentiable. Each map ),( λλ ϕU is known as a chart or a coordinate
5
system. If the domain of a chart is the whole M it is called global. The collection of
those charts whose union constitutes the whole manifold is known as an atlas of the
manifold. Basically an atlas gives differentiability structure on .M Since a manifold
has given a structure by coordinate system; we can now define smoothness and
differentiability of maps on manifolds. For such purposes suppose 1M and 2M are
manifolds and let ),( 11λλ ϕU and ),( 22
µµ ϕU denote the chart maps. A map
21: MMh → is said to be ∞C if for each λ and µ , the map 112 )( −λµ ϕϕ oo h taking
mV ℜ⊆1λ to mV ℜ⊆2
µ is ∞C in the sense used in advanced calculus. If 21: MMh →
is ,∞C bijective and has ∞C inverse, then h is called a diffeomorphism and ,1M 2M
are said to be diffeomorphic. Diffeomorphic manifolds have identical manifold
structure [5]. A one parameter group of diffeomorphism tψ over a manifold M is a
∞C map from MM →×ℜ such that for a fixed ℜ∈t MMt →:ψ is a
diffeomorphism and for all ,, ℜ∈st we have .stst +=ψψψ o Now we are interested
to define Hausdorff, compact and connected spaces. A manifold is said to be
Hausdorff if it satisfies the Hausdorff separation axiom: whenever sr, are two
distinct points in M , there exist open sets sr WW , in M such that φ=∩ sr WW and
., sr WsWr ∈∈ A topological space is said to be compact if each open cover of the
space has finite sub cover. M (as a topological space) is said to be connected if the
only subsets of M which are both open and closed are the empty set φ and the entire
set M itself [5].
1.2.2 Tangent Spaces
In this section it will be shown that at each point of an m-dimensional smooth
manifold M there is a well defined family of finite dimensional real vector spaces. In
,mℜ there is a one-to-one correspondence between directional derivatives and
vectors. A vector ),...,( 1 mwww = defines the directional derivative operator
6
∑ ∂∂ν
νν )/( xw and vice versa [5]. To characterize directional derivatives, its linearity
and ‘Leibnitz rule’ behavior is checked when acting on functions. On a manifold M
consider the collection Φ of ∞C maps from M into .ℜ A tangent vector at a point
Mp∈ is defined to be a map ℜ→Φ:η which obeys the Leibnitz rule and is linear
[5]. By linearity of tangent vector w we mean that ),()()( khkh ηβηαβαη +=+
for all .,and, ℜ∈Φ∈ βαkh The Leibnitz rule is defined as
).()()()()( hpkkphkh ηηη += A collection of tangent vectors at a point p denoted
by pW has a natural structure of a vector space called the tangent space. This tangent
space obeys the addition and scalar multiplication laws respectively as:
)()())(( 2121 hhh ηηηη +=+ and ).())(( hh ηαηα = An important property of the
tangent space is given in the following theorem:
Theorem:
Let M be an m-dimensional manifold. Let Mp∈ and let pW denote the tangent
space at .p Then mWp = dim [5].
1.2.3 Tensors
To define a tensor precisely we consider a finite dimensional vector space W and its
dual space .*W A tensor ,T of type ),( sl over W is a multilinear map [5]
.......: ** ℜ→××××× 434214434421sl
WWWWT This definition of tensor simply mean that for l
dual vectors and s ordinary vectors, T yields a number. A type )1,0( tensor is exactly
a dual vector and a type )0,1( tensor is an element of **W where **W is nothing but
an ordinary vector because we identify **W as .W The collection ),( slΦ which
represents tensors of type ),,( sl has a natural structure of vector space with the well
known rules of addition and scalar multiplication [5].
7
In general relativity we often apply the two operations of contraction and outer
product to the tensors. These two operations are explained in the following lines.
Contraction of tensors of ),( sl type is a map )1,1(),(: −−Φ→Φ slslO defined as
follows:
If T is a tensor of ),( lk type then ,,...),...,..,(...,1
*∑=
=m
wwTTOσ
σσ where *σw is the
dual basis of W which has basis σw and these vectors are inserted into the ith and
the jth slots of .T The tensor OT obtained here is independent of the choice of basis
µw so contraction operation is well defined [5].
The second operation on tensors is the outer product. Given any tensor 1T of ),( sl
type and another tensor 2T of ),( sl ′′ type, a new tensor can be constructed of type
),( ssll ′+′+ which is called the outer product of 1T and 2T denoted as 21 TT ⊗ by the
following simple rule. Given ll ′+ dual vectors **1 ,..., llww ′+ and ss ′+ vectors
,,...,1 ssww ′+ we define 21 TT ⊗ acting on these vectors to be the product of
),...,;,...,( 1**1
1 sl wwwwT and ).,...,;,...,( 1
**12 sss
lll wwwwT ′++′++ Construction of tensors
on one hand is to take outer products of vectors and dual vectors. A tensor which can
be expressed as such an outer product is called simple [5].
A metric is an important tensor constructed on a manifold. A metric tensor is defined
as a non-degenerate, second order symmetric type )2,0( tensor. By symmetric we
mean that ,αββα gg = where βαg denotes the components of the metric tensor
∑ ⊗=βα
βααβ
,.dxdxgg The metric at each point Mp∈ is a multilinear map from
.ℜ→× pp WW Basically, a metric is an inner product on the tangent space at each
point. Other notation which is used for the metric tensor is ,2ds thus in term of 2ds
we may write ∑=νµ
νµµν
,
2 .dxdxgds Plus and minus signs appearing in the metric
tensor is called the signature of the metric. Different types of metrics can be found
8
like positive definite metrics where the signature is ).,,,( ++++ Riemannian metrics
have positive definite signature. Space-time metric has signature ),,,( +++− and such
metrics are called Lorentzian [5]. From here onward we follow the Einstein’s
summation convention in which summation is understood. For example, µµ xa in a
four dimensional space-time simply means .3
0∑=µ
µµ xa
1.2.4 Covariant Derivative
A derivative operator ∇ on M is called covariant derivative if it map each smooth
tensor field of ),( sl type to a smooth tensor field of )1,( +sl type and satisfy the five
properties given as [5]:
1) Linearity: For all ),(, slKH Φ∈ and ,, ℜ∈µλ
.)( ......
......
......
......
11
11
11
11
lk
lk
lk
lk bb
aaebb
aaebb
aabb
aae KHKH ∇+∇=+∇ µλµλ
2) Leibnitz rule: For all ),,(),,( slKslH ′′Φ∈Φ∈
).()()( ......
......
......
......
......
......
11
11
11
11
11
11
lk
lk
lk
lk
lk
lk dd
ccebb
aadd
ccbb
aaedd
ccbb
aae KHKHKH ′
′′
′′
′ ∇+∇=∇
3) Commutativity with Contraction: For all ),,( slH Φ∈
.)( ............
............
11
11
lk
lk bcb
acadbcb
acad HH ∇=∇
This means that both the contraction and covariant differentiation operations
commute each other.
4) Consistency with the notion of tangent vectors as directional derivatives on
scalar fields: For all Φ∈H where Φ is the collection of ∞C maps and all
,pa W∈η .)( ,; a
aa
aa
aa
a HHHHH ηηηηη ===∇=
5) Torsion Free: For all ,Φ∈H .HH abba ∇∇=∇∇
9
1.2.5 Parallel Transport and Lie derivative
A connection µ∇ can be used for defining notion of parallel transport of a vector or
tensor along a curve Ω with tangent .aτ Mathematically, if a vector νw given at
each point on the curve Ω satisfies ,0=∇ νµ
µτ w then it is said to be transported
parallely along the curve [5].
We now turn our attention towards Lie derivative. Let M be a manifold and tψ be a
one parameter group of diffeomorphism. For every Mp∈ a vector field K over M
determines a unique curve )(tpΩ such that pp =Ω )0( and pK is the tangent vector
to curve. Along a curve )(tpΩ the local coordinates µs are the solutions of the
system of ordinary differential equations )( µνν
sxdt
ds= with initial value
).()0( pxs νν = To introduce a new type of differentiation we consider the map tψ
dragging each p with coordinates νx along the curve )(tpΩ through the point p
into the image )(tr pΩ= with coordinates ).(tsν When the parameter t is
sufficiently small, the map tψ is one-to-one and represents a map called pull-back
map Tt*ψ for any tensor .T The Lie derivative of T with respect to K is defined as
,lim *
0 ⎥⎦⎤
⎢⎣⎡ −
=→ t
TTTL t
tK
ψ where T
t*ψ and T are of the same type tensors and are
evaluated at the same point .p Generally, for a smooth tensor field T of type ),( sl
the components of TLK
becomes [5]
ss
sjli
jj
sjli
sjli
ll
sjli
ii
sjli
sjli
sl
sl
bbbaaa
bbbaaa
bbbaaaa
bbbaa
abbb
aaabbb
aabb
aabb
aa
K
KT
KTKTKT
KTKTKTTL
;............
;............
;............
;............
;............
;............
;......
......
11
11
11
11
11
11
11
11
11
11
...
......
...
µµ
µµ
µµµ
µ
µµ
µµµ
µ
++
+++−−
−−−=
10
The Lie derivative also satisfies the following useful properties [4]. In the following
h is some smooth function, H and K are smooth vector fields, ℜ∈ba, and ,S T
are smooth tensor fields.
(1) ,)( SLTTLSSTL HHH ⊗+⊗=⊗
(2) ,)( SLbTLaSbTaL HHH +=+
(3) ,TLTL HH ρρ = (4) ,TLbTLaTL KHbKaH +=+
(5) ],[ KHKLH = (6) ),(hHhLH =
where ρ denotes the contraction operation. The Lie derivative plays a vital role in
gravitational theories because we describe symmetries through it.
1.2.6 Riemann Curvature Tensor
We shall now define Riemann curvature tensor associated with metric tensor. Let α∇
be a derivative operator and αw be a dual vector field then the following relation
holds ,δγβαδ
γαβγβα wRww =∇∇−∇∇ where γβαδR is called the Riemann
curvature tensor and the relation itself is known as Ricci identity [78]. In terms of
christoffel symbols the Riemann curvature tensor is given as
.,,δαµ
µγβ
δβµ
µγα
δαγβ
δβγαγβα
δ ΓΓ−ΓΓ+Γ−Γ=R Riemann curvature tensor can be
decompose into two parts known as trace and trace free parts. The trace of Riemann
curvature tensor known as Ricci tensor can be obtained by contracting the first and
third indices of the Riemann tensor as .βγαγ
βα RR = The Ricci scalar R is given as
[5] .αδδα
αα RgRR == The Weyl tensor δγβαC known as the trace free part of the
Riemann curvature tensor for manifolds of dimension 3≥m can be obtained as [5]
( ) .)2)(1(
22
2][][][ βδγααδγββδγααβγδαβγδ ggR
mmRgRg
mCR
−−−−
−+= (1.2.1)
11
A space-time metric is said to be conformally flat if all the components of Weyl
tensor become zero and is said to be flat if all the components of Riemann curvature
tensor become zero.
1.2.7 Symmetries in General Relativity
Symmetry of the space-time is a local smooth diffeomorphism which preserves some
geometrical feature of M [4]. A diffeomorphism ψ will be symmetry of a tensor T
if the tensor remains unchanged when pulled back under ψ [4]. In general relativity
the symmetries of metric are important. A metric remains invariant when pulled back
through ψ i.e. .* ababt gg =ψ This type of diffeomorphism is known as isometry [5].
A vector field µK generates a one parameter family of isometries, called a Killing
vector field. Any vector field K on a manifold can be decomposed as [4]
ababba BhK +=21
; (1.2.2)
where abKbaab gLhh == )( and )( baab BB −= are symmetric and skew symmetric
tensors on ,M respectively. The vector field K is said to be conformal if the
associated tψ called the local diffeomorphism, with K maintain the metric structure
through conformal factor η i.e. ,ggt ηψ =∗ where ℜ→U:η is a smooth conformal
function on some open subset U of .M This is equivalent to [4]
,abab gh η= (1.2.3)
or
.,,, abcbac
cacb
ccababK
gKgKgKggL η=++≡ (1.2.4)
If η is a smooth conformal function on some open subset U of M then K is called
conformal vector field. If η becomes constant on ,M then K is called homothetic
vector field (proper homothetic vector field when 0≠η ) while if 0=η it becomes
12
Killing vector field. Also if the vector field K is not a homothetic vector field then it
is called proper conformal vector field. Another kind of symmetry which we will be
discussing in these thesis is self similar vector fields. A vector field K is said to be
self similar if it satisfies the two conditions [8] that
aaKuuL α= (1.2.5)
,2 ababKhhL δ= (1.2.6)
where au is the four velocity of the fluid satisfying 1±=aauu and baabab uugh ±=
is the projection tensor and ., ℜ∈δα If ,0≠δ the ratio ,/δα which is scale
independent, characterizes similarity transformation and known as similarity index. If
the above ratio gives unity, K turns out to be a homothetic vector field, which is
known as first kind self similarity. If 0=α and ,0≠δ similarity is called of the
zeroth kind. If the ratio is neither zero nor one, it is referred to as self-similarity of the
second kind. Self similarity is called of infinite kind when 0≠α and .0=δ If both
,0== αδ K turns out to be a Killing vector fields [9]. A self similar vector field K
can be tilted or non tilted to the four velocity vector .au When au is time like then
1−=aauu and .baabab uugh += In such case the self similar vector field K will be
parallel, orthogonal or tilted (that is neither parallel nor orthogonal) to the time like
vector field au when ,)(u
uFK∂∂
= x
xfK∂∂
= )( or x
xu
uK∂∂
+∂∂
+= )( βα
respectively [9]. The above theory also valid for space-like vector field .au
1.2.8 Space-Times and Tetrad
A space-time is a set ),( gM where M is a four dimensional smooth, connected,
compact and Hausdorff manifold and g is a lorentzian metric with signature
),,,,( +++− which is symmetric and non degenerate. A tetrad can be constructed at
each point of a manifold by taking a system of four linearly independent vectors.
13
Different types of tetrads can be formed over a point Mp∈ in the tangent space. For
instance, one is an orthonormal tetrad ),,,( zyxt and the other is called a real null
tetrad ).,,,( yxml The inner product for orthonormal tetrad members satisfy [4]
1====− aa
aa
aa
aa zzyyxxtt with all other products as zero. Also the inner
product of the members of real null tetrad ),,,( yxml satisfy 0== aa
aa mmll and
1=== aa
aa
aa yyxxml with all other products as zero. The null vectors l and m for
real null tetrad are given as )(2
1 aaa tzl += and ).(2
1 aaa tzm −= Given a tetrad
we can check it to be an orthonormal or real null tetrad if the completeness relation
for metric abg at p holds i.e. a tetrad will be an orthonormal if it satisfies
,zzyyxxttg abababaab +++−= (1.2.7)
and it will be a real null tetrad if it satisfies
.2 )( babaabbabababaab yyxxmlmlyyxxmlg +++=++= (1.2.8)
A complex null tetrad ),,,( ssml may also be introduce at a point ,Mp∈ where, l
and m are as defined in the real null tetrad and s and its conjugate s are defined
from the real null tetrad by )(2
1 aaa iyxs += and ).(2
1 aaa iyxs −= The inner
product among ,l ,m s and s are 1=aaml and 1=a
a ss with all other product as
zero. It is clear that as and as are complex null vectors. In this type of tetrad the
completeness relation at Mp∈ is given as [6]
.22 )()( babababababaab sssslmmlssmlg +++=+= (1.2.9)
14
1.2.9 Classification of tangent spaces
In Minkowski space-time a non zero vector w is called time like, space like or null
according as ),( wuη is negative, positive or zero [4], where η is defined as
.),( baab wuwu ηη = A one dimensional subspace of Minkowski space is called time
like, space like or null if it is spanned by a time like, space like or null vector
respectively. A set of two dimensional subspaces (2-spaces) of the tangent space
MTp is called space like, time like, null if it contains no null directions, exactly two
null directions, exactly one null direction respectively [6]. It is obvious to see that
these are the only possibilities for the 2-spaces because for a real null tetrad
),,,( yxml at p and for two independent vectors u and w at the point, if ),( wu
represents a 2-space then ),( nl is time-like, ),( yx is space-like and
),(),,(),,( xmylxl and ),( ym are null [6]. The set of those vectors which are
orthogonal to each member of the 2-space is called the orthogonal complement of that
set. Therefore, ),( xl and ),( yl are orthogonal complements as are ),( yx and ),( ml
also ),( xm and ),( ym . Similarly a three dimensional subspace (3-space) of
Minkowski space is called time like, space like or null if its orthogonal complement is
space like, time like or a null 1-space.
1.3 Basics of Teleparallel Theory of Gravitation 1.3.1 Tetrad in Teleparallel Theory
In teleparallel theory, a tetrad field represents the gravitational field. A tetrad field S
is basically a map ,: MTMS p→ where M is Minkowski space and MTp is the
tangent space at the point .p A tetrad and its dual are defined respectively as [7]
νν ∂= aa SS .ν
ν dxSS bb = (1.3.1)
15
Tetrad field νaS and its inverse field denoted by ν
aS satisfies the relations
,µν
µν δ=a
a SS ,abb
a SS δνν = (1.3.2)
where abδ is the kronecker delta. It is important to mention that the above equations
hold for non-trivial tetrad field. A trivial tetrad field and its dual can be written as:
,ννδ ∂= aae .ν
νδ dxe bb = (1.3.3)
If we choose a trivial tetrad field then all the torsion components will vanish. To work
in the teleparallel theory of gravity one needs to select a non trivial tetrad field.
1.3.2 Space-time structure
The Riemannian metric can be generated from the tetrad field as [7]
.νµνµ η baba SSg = (1.3.4)
where abη is the Minkowski metric given by ).1,1,1,1(diag −=abη A weitzenböck
connection can be defined through a non trivial tetrad field as [10]
.µνθθ
µνa
a SS ∂=Γ (1.3.5)
Unlike christoffel symbol weitzenböck connection is not symmetric in its lower two
indices and therefore generates torsion in the space-times. The teleparallel covariant
derivative ρ∇ of a covariant tensor of rank 2 in terms of weitzenböck connection is
defined as [7]
,, θνθµρθµ
θρνρνµµνρ AAAA Γ−Γ−=∇ (1.3.6)
where comma denotes the partial derivative and θρνΓ are weitzenböck connections
defined as above. The covariant derivative of the tetrad field is given by
., θθρνρννρ
aa SSS Γ−=∇ (1.3.7)
16
Now using equation (1.3.5) in equation (1.4.7) we get ,0=∇ νρaS which means that
in teleparallel theory tetrads are parallely transported. The weitzenböck and
christoffel symbols have the relation
,0νµ
θµν
θµν
θ
N+Γ=Γ (1.3.8)
where
],[21
νµθ
µθ
ννθ
µνµθ TTTN −+= (1.3.9)
is a tensor quantity called the contortion tensor and νµθ0Γ is the christoffel symbols
defined as
).(21
,,,0
σνµνµσµνσσθ
µν
θ
gggg −+=Γ (1.3.10)
As discussed above weitzenböck connections are not symmetric with respect to its
lower two indices, therefore their difference give us the torsion in the space-times.
Mathematically, torsion can be defined as [11]
,θµν
θνµµν
θ Γ−Γ=T (1.3.11)
which is anti symmetric with respect to its lower indices. The Riemann curvature
tensor in terms of weitzenböck connection in teleparallel theory is given as [12]
.,,λµσ
θνλ
λνσ
θµλ
θνµσ
θµσνσµν
θ ΓΓ−ΓΓ+Γ−Γ=R (1.3.12)
Now using equation (1.3.8) in equation (1.3.12) we have [3]
,00 =+= θµνσ
θµνσ
θµνσ GRR (1.3.13)
where θµνσ0R represents Riemann curvature tensor in general relativity and
,- λνσ
θµλ
λµσ
θνλ
θµσ
νθνσ
µθµνσ NNNNNNG +−∇∇= (1.3.14)
17
is the tensor quantity based on weitzenböck connection only. From (1.3.13) it is clear
that in teleparallel theory curvature of the space-time vanishes identically. In
teleparallel theory torsion is responsible for the gravitational interaction.
1.3.3 Teleparallel Lie Derivative
We have already discussed Lie derivative along a vector field in section (1.2.5). Lie
derivative plays an important role in both general and teleparallel theories because we
study symmetries of the space-times through it. The teleparallel version of the Lie
derivative was introduced in [13]. The teleparallel Lie derivative of a second rank
covariant tensor along a vector field K is given as
).(,,, νρσ
σλνλσ
ρσν
ρν
νλλν
ρνν
νρλρλ TETEKKEKEKEELK
T ++++= (1.3.15)
Similarly, for a second rank contravariant tensor teleparallel Lie derivative is given as
[13]
).(,,, νσρσλ
νσλρσν
νρλν
νλρνν
νρλρλ TETEKKEKEKEEL
K
T +−+−= (1.3.16)
A vector field K is said to be teleparallel homothetic vector field if it satisfies [72-
74]
,2)(,,, ρλνρσ
σλνλσ
ρσν
ρν
νλλν
ρνν
νρλρλ η gTgTgKKgKgKggLK
T =++++= (1.3.17)
where η is constant on .M The teleparallel vector field K is called proper
teleparallel homothetic vector field when 0≠η while if 0=η it becomes teleparallel
Killing vector fields.
18
1.4 Literature Review of Conformal, Homothetic, Killing and Self Similar Vector Fields in General Relativity
In general relativity the above mentioned symmetries can be used to study the laws of
conservation of energy, momentum and angular momentum [14]. These symmetry
restrictions not only give us the laws of conservation but provide some physical and
geometrical information about the space-time. Some geometrical and physical
features through symmetries are the study of cosmological voids, cosmological
perturbations, gravitational collapse, primordial black holes, star formation and
cosmic censorship [15]. In literature Killing, homothetic, conformal and self similar
symmetries of the space-times have been studied by different researchers. Bokhari et
al [16] classified static spherically symmetric space-times according to its Killing
vector fields. They explored that static spherically symmetric space-times admit
minimum 4 linearly independent Killing vector fields. Qadir and Ziad [17, 18]
worked to investigate the Killing isometry for static cylindrically symmetric space-
times and non static spherically symmetric space-times. They completely classified
these space-times according to their metrics and isometries. They showed that the
number of Killing vector fields for static cylindrically symmetric space-times can be
,3 ,4 ,5 ,6 7 or .10 In [19] the authors completely classified non static plane
symmetric space-times according to their Killing vector fields and metrics. Authors
obtained classes of metrics that admit Killing vector fields of dimension ,3 ,4 ,5 ,6
7 and .10 In [20] it is found that the dimension of the homothetic vector fields for
space-times admitting m Killing vector fields is .1+m In [21] authors explored that
the maximum possible dimension of homothetic vector fields is .11 In [22] authors
investigated spherically symmetric space-times according to its homothetic vector
fields. They have given complete classification and it is explored that this space-time
admit homothetic vector fields of dimension ,4 ,5 ,6 8 and .11 They showed that
19
the space-time becomes flat when admit 11 homothetic vector fields. In [23, 24]
authors completely classified static plane symmetric space-times and static
cylindrically symmetric space-times according to homothetic motions. In [23] they
showed that space-time admits ,5 ,6 8 and 11 homothetic vector fields. In [24]
authors completely classified static cylindrically symmetric space-times according to
its homothetic vector fields and metrics. They showed that there exist ,4 ,5 7 and 11
homotheties. They argued that some of the metrics admitting homothetic vector fields
were left in the classification of Hall and Steele [21] because some classes of metrics
admit homothetic vector fields which do not have global topological structure. They
also extended some local homotheties globally. On the same topic authors in [25-27]
have discovered homothetic vector fields and their metrics in Bianchi type I, static
cylindrically symmetric and non static plane symmetric space-times.
In [28, 29] the authors introduced the method of finding a conformal change in the
metric on a manifold. The conformal change is taken such that the Lie algebra of
conformal vector fields on the manifold with respect to one metric becomes Killing or
homothetic algebra of the other metric. According to the authors of these two papers,
the Lie algebra of conformal vector fields can be reduced to homothetic algebra under
some assumptions if the space-time is not conformally flat. The theorems stated in the
above two papers were not precise. Afterward in 1990, Hall [30] gave some counter
examples for the theorems given in [28, 29] by considering non conformally flat
space-times admitting proper homothetic vector fields with isolated zero. In [31] the
authors discussed conformal vector fields in a space-time and found out the maximum
dimension of conformal vector fields for non conformally flat space-times. In this
paper the authors discussed deficiencies in the theorems of [28, 29] and also corrected
them. They found that the maximum dimension of the conformal vector fields for non
conformally flat space-times is seven and for conformally flat space-times it is 15.
Kramer et al [32] considered rigidly rotating, stationary axisymmetric, perfect fluid
space-times admitting a proper conformal vector field under certain assumption of
Lie algebra and showed that no such solutions of Einstein field equations exist. By
20
considering static spherically symmetric space-times [33], authors obtained
conformal vector fields in non conformally flat and conformally flat cases. The
showed that non conformally flat static spherically symmetric space-times admit at
the most two proper conformal motions. They also listed all eleven proper
conformation vector fields for conformally flat static spherically symmetric space-
times. Hall [34] established some restrictions on existence of conformal and
homothetic vector fields in spaces admitting Killing vector fields. G. Shabbir et al
[35-39] explored conformal vector fields in Bianchi type I, static cylindrically
symmetric, non static spherically symmetric, Bianchi types VIII and IX and spatially
homogeneous rotating space-times by using direct integration method. Using the
same method we investigated proper conformal vector fields in non conformally flat
Kantowski-Sachs [40], Bianchi type III [41] and non static cylindrically symmetric
space-times [42]. We showed that all the above space-times possess proper conformal
vector fields for special choice of the metric functions.
Much work has been done to find self-similar solutions of the space-times. Coley [43]
in his paper discussed the self similarity differences between generalized similarity
and first kind self-similarity. He argued about the suitability of generalization of
homothety as self-similarity. He also discussed various mathematical and physical
properties of space-times admitting self-similarity. In his work Coley introduced the
governing equations for perfect fluid cosmological models and a set of integrability
conditions have derived for the existence of proper self-similar vector fields. In [8]
authors discussed some important symmetries of Bianchi type I space-times. They
determined the self similarities of Bianchi I metrics without any constraint on the type
of the fluid. They showed that Bianchi type I space-times admit self similarity of first
and zeroth kind. They explored that Bianchi type I space-times which admit self
similar vector field of second kind with co-moving fluid are only Kasner type space-
times. H. Maeda et al [44] classify all spherically symmetric space-times admitting
self-similar vector fields of the second, zeroth or infinite kind. They studied the cases
in which the self-similar vector field is not only tilted but also parallel or orthogonal
21
to the fluid flow. M. Sharif and S. Aziz [45, 46] published their work on self
similarity solutions for spherically symmetric and cylindrically symmetric space-
times, respectively. In [45] they explored some properties of the self similar solutions
of the first kind for spherically symmetric space-times and also checked the
singularities of these solutions. In [46] they studied the cylindrically symmetric
solutions which admit self-similar vector fields of second, zeroth and infinite kinds,
for the tilted fluid case, parallel and orthogonal cases. They showed that the parallel
case gives contradiction both in perfect fluid and dust cases and the orthogonal
perfect fluid case yields a vacuum solution while the orthogonal dust case gives
contradiction. Using an algebraic and direct integration method we investigated
proper self similar vector fields of first, second zeroth and infinite kind for Bianchi
type III [47], static plane symmetric [48], static spherically symmetric[49] and static
cylindrically symmetric space-times [80]. We showed that all the above space-times
possess proper self similar vector fields for special choice of the metric functions.
1.5 Literature Review of Teleparallel Killing and
Homothetic Vector Fields in Teleparallel
Theory of Gravitation
Although general relativity is a powerful theory of gravity, it has some controversial
and unresolved problems. One of the problems is the energy and momentum
localization. This theory has no consistent formalism to localize energy and
momentum. For this reason the problem of localization of energy was also considered
in teleparallel theory of gravitation [50-55]. As a second step in this theory
teleparallel versions of the exact solutions of general relativity have been studied by
many authors [56-63]. In teleparallel theory symmetries of the metric tensor were
ignored. As a pioneer work Sharif and Jamil [13] introduced the teleparallel version
of the Lie derivative for Killing vector fields and used those equations to find the
22
teleparallel Killing vector fields in Einstein universe. After that Sharif and Bushra
[64] explored teleparallel Killing vector fields for spherically symmetric static space-
times. The above authors did not classify the space-times according to their
teleparallel Killing vector fields. We started our work in this field by classifying
space-times according to their teleparallel Killing vector fields by using direct
integration method. We classify Bianchi types I, II, III, VIII, IX, Kantowski-Sachs,
static cylindrically symmetric, spatially homogeneous rotating and non static
cylindrically symmetric space-times according to their teleparallel Killing vector
fields [65-71]. Later we extended our work to the classification of Bianchi type I,
static cylindrically symmetric and non static plane symmetric space-times by
teleparallel proper homothetic vector fields [72-74] which are pioneer papers in this
field. In the above papers we have also established a brief comparison between
teleparallel Killing vector fields and Killing vector fields in general relativity.
23
Chapter 2
Teleparallel Killing Vector Fields in
Bianchi Types I, II, VIII and IX Space-
Times
2.1. Introduction
This chapter is devoted to investigate teleparallel Killing vector fields in Bianchi
types I, II, VIII and IX space-times by using direct integration technique. This chapter
is organized as follows: In section (2.2) teleparallel Killing vector fields of Bianchi
type I space-times are investigated. In section (2.3) teleparallel Killing vector fields in
Bianchi type II space-times in the context of teleparallel theory have been explored.
In the next section (2.4) teleparallel Killing vector fields of Bianchi type VIII and IX
space-times are explored in the context of teleparallel theory of gravitation. A brief
comparison of Killing vector fields in both the theories is also given. Last section
(2.5) of the chapter is dedicated to a detailed summary of the work.
24
2.2. Teleparallel Killing Vector Fields in Bianchi
Type I Space-Times
Bianchi type-I space-time is a spatially homogeneous space-time. In general relativity
it admits an abelian Lie algebra of isometries ,3G acting on space like hyper surfaces
generated by the space like Killing vector fields [78] .,,zyx ∂∂
∂∂
∂∂ The line element in
the usual coordinates ),,,( zyxt (labeled by ),,,,( 3210 xxxx respectively) is given by
[75, 78]
.2)(22)(22)(222 dzedyedxedtds tCtBtA +++−= (2.2.1)
where BA, and C are functions of t only. The tetrad components and its inverse
can be obtained by using the relation (1.3.4) as [65]
),,,,1diag( )()()( tCtBtAa eeeS =µ ).,,,1diag( )()()( tCtBtAa eeeS −−−=µ (2.2.2)
Using equation (1.3.5), the corresponding non-vanishing Weitzenböck connections
are obtained as
,101 •=Γ A ,20
2 •=Γ B ,303 •=Γ C (2.2.3)
where dot denotes the derivative with respect to .t The non vanishing torsion
components by using (1.3.11) are
.,, 303
033
202
022
101
011 ••• =−==−==−= CTTBTTATT (2.2.4)
Now using (2.2.1) and (2.2.4) in (1.3.17) we get the teleparallel Killing equations as:
03,3
2,2
1,1
0,0 ==== XXXX (2.2.5)
,01,2)(2
2,1)(2 =+ XeXe tBtA (2.2.6)
,01,3)(2
3,1)(2 =+ XeXe tCtA (2.2.7)
25
,02,3)(2
3,2)(2 =+ XeXe tCtB (2.2.8)
,01)(20,
1)(21,
0 =−− • XAeXeX tAtA (2.2.9)
,02)(20,
2)(22,
0 =−− • XBeXeX tBtB (2.2.10)
.03)(20,
3)(23,
0 =−− • XCeXeX tCtC (2.2.11)
Now integrating equation (2.2.5), we get
).,,(),,,(),,,(),,,(
4332
2110
yxtPXzxtPXzytPXzyxPX
==
== (2.2.12)
where ),,,(1 zyxP ),,,(2 zytP ),,(3 zxtP and ),,(4 yxtP are functions of integration
which are to be determined. In order to find solution for equations (2.2.5) to (2.2.11)
we will consider each possible form of the metric for Bianchi type I space-times and
then solve each possibility in turn. Following are the possible cases for the metric
where the above space-times admit teleparallel Killing vector fields:
(I) )(),(),( tCCtBBtAA === and .,, CBCABA ≠≠≠
(II)(a) ),(),( tBBtAA == and .tan tconsC =
(II)(b) ),(),( tCCtAA == and .tan tconsB =
(II)(c) ),(),( tCCtBB == and .tan tconsA =
(III)(a) )(),(),( tCCtBBtAA === and ).()( tCtB =
(III)(b) )(),(),( tCCtBBtAA === and ).()( tCtA =
(III)(c) )(),(),( tCCtBBtAA === and ).()( tBtA =
(IV) )(),(),( tCCtBBtAA === and ).()()( tCtBtA ==
(V)(a) )(),(,tan tCCtBBtconsA === and ).()( tCtB =
(V)(b) )(,tan),( tCCtconsBtAA === and ).()( tCtA =
(V)(c) tconsCtBBtAA tan),(),( === and ).()( tBtA =
26
(VI)(a) )(tAA = and .tan tconsCB ==
(VI)(b) )(tBB = and .tan tconsCA ==
(VI)(c) )(tCC = and .tan tconsBA ==
We will discuss each possibility in turn.
Case (I):
In this case we have ),(tAA = ),(tBB = ),(tCC = ,BA ≠ CA ≠ and .CB ≠ Now
substituting equation (2.2.12) in equation (2.2.6), we get
.0),,(),,( 2)(23)(2 =+ zytPezxtPe ytA
xtB (2.2.13)
Differentiating equation (2.2.13) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (2.2.13) we get
⇒−= − ),(),,( 1)(2)(22 ztEezytP tAtBy ),,(),(),,( 31)(2)(22 ztEztEeyzytP tAtB +−= − where
),(3 ztE is a function of integration. Now refreshing the system of equations (2.2.12)
we get
).,,(),,(),(),,(),(),,,(
43212
31)(2)(2110
yxtPXztEztxEXztEztEeyXzyxPX tAtB
=+=
+−== −
(2.2.14)
Considering equation (2.2.7) and using equation (2.2.14) we get
.0)],(),([),,( 31)(2)(2)(24)(2 =+−+ − ztEztEeyeyxtPe zztAtBtA
xtC (2.2.15)
Differentiating equation (2.2.15) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (2.2.15) we get
.0),(),(),( 3)(21)(24)(2 =+− ztEeztEeyytEe ztA
ztBtC Differentiating this equation with
respect to y twice, we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting
back the above value in equation (2.2.15) and solving, we get
27
)()(),( 31)(2)(21 tKtKezztE tBtC += − and ),()(),( 42)(2)(23 tKtKezztE tAtC +−= − where
),(1 tK ),(2 tK )(3 tK and )(4 tK are functions of integration. Substituting back the
above information in equation (2.2.14) we get
).,()()(),,()()(
),()()()(),,,(
5213
231)(2)(22
42)(2)(23)(2)(21)(2)(21
10
ytEtxKtKyxXztEtxKtKezxX
tKtKeztKeytKezyXzyxPX
tBtC
tAtCtAtBtAtC
++=
++=
+−−−=
=
−
−−−
(2.2.16)
Considering equation (2.2.8) and using equation (2.2.16) we get
.0),(),()(2 2)(25)(21)(2 =++ ztEeytEetKxe ztB
ytCtC (2.2.17)
Differentiating the above equation with respect to x we get .0)(1 =tK Substituting
the above value in (2.2.17) and differentiating the resulting equation with respect to
,y we get ),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are
functions of integration. Substituting the above values back in (2.2.17) we get
),()(),( 75)(2)(22 tKtKezztE tBtC +−= − where )(7 tK is a function of integration.
Refreshing the system of equations (2.2.16) we get
).()()(),()()(),()()(
),,,(
652375)(2)(232
42)(2)(23)(2)(21
10
tKtKytxKXtKtKeztxKXtKtKeztKeyX
zyxPX
tBtC
tAtCtAtB
++=+−=
+−−=
=
−
−− (2.2.18)
Considering equation (2.2.9) and using equation (2.2.18) we get
.0)()(),,()()(
)()]()(2[)()()]()(2[4)(214)(22)(2
2)(23)(23)(2
=−+−+
−++−
tKetAzyxPtKetKez
tKetAtCztKeytKetAtBytA
txttA
ttC
tCttt
tBtBtt (2.2.19)
Differentiating (2.2.19) with respect to ,x we get
),,(),(),,(0),,( 7611 zyEzyxEzyxPzyxPxx +=⇒= where ),(6 zyE and ),(7 zyE
are functions of integration. Substituting the above value back in (2.2.19) and
differentiating twice with respect to ,y we get
28
),()(),(0),( 9866 zKzKyzyEzyEyy +=⇒= where )(8 zK and )(9 zK are functions
of integration. Substituting back the above value in (2.2.19) and differentiating with
respect to ,y we get .0)()()()]()(2[ 83)(23)(2 =++− zKtKetKetAtB ttBtB
tt
Differentiating this equation with respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzK z
Substituting back this value in the above equation and solving we get
.,)( 2)(2)(
2)()(2)(
13 ℜ∈+−= −−− ∫ cecdteectK tBtAtAtBtA Now substituting all the above
information in (2.2.19) and solve after differentiating twice with respect to ,z we get
.,,)(0)( 434399 ℜ∈+=⇒= ccczczKzK zz Substituting back in equation (2.2.19) and
solve after differentiating with respect to ,z we get
.,)( 5)(2)(
5)()(2)(
32 ℜ∈+−= −−− ∫ cecdteectK tCtAtAtCtA Now substituting all the above
information in equation (2.2.19) and solving we get
.,)( 6)(
6)()(
44 ℜ∈+= −−− ∫ cecdteectK tAtAtA Refreshing the system of equation
(2.2.18) with the help of above information, we get
),()(
),()(
,
),,(
65)(2)(5
)()(2)(3
3
75)(2)(2)(2)(2
)()(2)(1
2
)(6
)()(4
)(5
)()(3
)(2
)()(1
1
7431
0
tKtKyexcdteexcX
tKtKzeexcdteexcX
ecdteecezcdteezceycdteeycX
zyExcxzcxycX
tCtAtAtCtA
tBtCtBtAtAtBtA
tAtAtAtAtAtAtAtAtA
+++−=
+−+−=
++−+−=
+++=
−−−
−−−−
−−−−−−−−−
∫∫
∫∫∫(2.2.20)
where .,,,,, 654321 ℜ∈cccccc Now considering equation (2.2.10) and using equation
(2.2.20) we get
.0)()()()()()]()(2[
)]()([)]()([),(27)(27)(25)(25)(2
)(2
)()(1
71
=−−+−+
−−−++ ∫ −
tKetBtKetKeztKetBtCz
etBtAcxdteetBtAxczyExctB
tttB
ttCtC
tt
tAtt
tAtAtty (2.2.21)
Differentiating equation (2.2.21) with respect to ,x we get
.0)]()([)]()([2 )(2
)()(11 =−−−+ ∫ − tA
tttAtA
tt etBtAcdteetBtAcc Solving this equation
and remember that in this case ,0)()( ≠− tBtA tt we get ,01 =c which on back
29
substitution give us .02 =c Substituting the above information in equation (2.2.21)
and differentiating with respect to ,y we get
),()(),(0),( 111077 zKzyKzyEzyEyy +=⇒= where )(10 zK and )(11 zK are
functions of integration. Substituting the above value in (2.2.21) and differentiating
with respect to z twice, we get .,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzK zz
Substituting back this value in equation (2.2.21) and differentiating with respect to ,z
we get .0)()()]()(2[ 5)(25)(27 =+−+ tKetKetBtCc t
tCtCtt Solving this equation we get
.,)( 9)(2)(
9)()(2)(
75 ℜ∈+−= −−− ∫ cecdteectK tCtBtBtCtB Now substituting all the above
information in equation (2.2.21) we get ⇒=−− 0)()()( 7)(27)(28 tKetKetBc t
tBtBt
.,)( 10)(
10)()(
87 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations
(2.2.20) we get
),(
,
,
),(
6)(2)(9
)()(2)(7
)(2)(5
)()(2)(3
3
)(10
)()(8
)(9
)()(7
2
)(6
)()(4
)(5
)()(3
1
118743
0
tKeyc
dteeycexcdteexcX
ecdteecezcdteezcX
ecdteecezcdteezcX
zKycyzcxcxzcX
tCtB
tBtCtBtCtAtAtCtA
tBtBtBtBtBtB
tAtAtAtAtAtA
++
−+−=
++−=
++−=
++++=
−
−−−−−
−−−−−−
−−−−−−
∫∫∫∫∫∫
(2.2.22)
where .,,,,,,, 109876543 ℜ∈cccccccc Now considering equation (2.2.11) and using
equation (2.2.22) we get
.0)()()(
))()(())()(())(
)(())()(()(22
6)(2)(26
)(9
)()(7
)(
5)()(
311
73
=−−
−−−+−
−−+++
∫∫
−
−
tKetCetK
etCtBycdteetCtBycetC
tAcxdteetCtAxczKycxc
tCt
tCt
tBtt
tBtBtt
tAt
ttAtA
ttz
(2.2.23)
Differentiating equation (2.2.23) with respect to ,x we get
.0)]()([)]()([2 )(5
)()(33 =−−−+ ∫ − tA
tttAtA
tt etCtAcdteetCtAcc Solving this equation
and remember that in this case ,0)()( ≠− tCtA tt we get ,03 =c which on back
30
substitution gives us .05 =c Substituting the above information in equation (2.2.23)
and differentiating with respect to ,y we get
.0)]()([)]()([2 )(9
)()(77 =−−−+ ∫ − tB
tttBtB
tt etCtBcdteetCtBcc Solving this equation
and remember that in this case ,0)()( ≠− tCtB tt we get ,07 =c which on back
substitution gives us .09 =c Now substituting all the above values in equation
(2.2.23) and differentiating with respect to ,z we get
.,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzK zz Substituting the above value in
equation (2.2.23) we get ⇒=−− 0)()()( 6)(26)(211 tKetKetCc t
tCtCt
.,)( 13)(
13)()(
116 ℜ∈+= −−− ∫ cecdteectK tCtCtC Refreshing the system of equations
(2.2.22) we get
,,
,,)(
13)()(
113)(
10)()(
82
)(6
)()(4
1121184
0
tCtCtCtBtBtB
tAtAtA
ecdteecXecdteecX
ecdteecXczcycxcX−−−−−−
−−−
+=+=
+=+++=
∫∫∫
(2.2.24)
where .,,,,,, 13121110864 ℜ∈ccccccc The line element for Bianchi type I space-times is
given in equation (2.2.1). The above space-time admits seven linearly independent
teleparallel Killing vector fields which are ,t∂∂ ,)(
xe tA
∂∂− ,)(
ye tB
∂∂− ,)(
ze tC
∂∂−
,)(1
xtG
tx
∂∂
+∂∂
ytG
ty
∂∂
+∂∂ )(2 and ,)(3
ztG
tz
∂∂
+∂∂ where ,)( )()(1 ∫ −−= dteetG tAtA
∫ −−= dteetG tBtB )()(2 )( and .)( )()(3 ∫ −−= dteetG tCtC Killing vector fields in general
relativity are ,x∂∂
y∂∂ and .
z∂∂ It is evident that teleparallel Killing vector fields are
different and more in number to the Killing vector fields in general relativity.
Case (II)(a):
In this case we have tCtBBtAA tancos),(),( === and ).()( tBtA ≠ Now
substituting equation (2.2.12) in equation (2.2.6), we get
31
.0),,(),,( 2)(23)(2 =+ zytPezxtPe ytA
xtB (2.2.25)
Differentiating equation (2.2.25) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (2.2.25) we get
⇒−= − ),(),,( 1)(2)(22 ztEezytP tAtBy ),,(),(),,( 31)(2)(22 ztEztEeyzytP tAtB +−= − where
),(3 ztE is a function of integration. Now refreshing the system of equations (2.2.12)
we get
).,,(),,(),(),,(),(),,,(
43212
31)(2)(2110
yxtPXztEztxEXztEztEeyXzyxPX tAtB
=+=
+−== −
(2.2.26)
Considering equation (2.2.7) and using equation (2.2.26) we get
.0)],(),([),,( 31)(2)(2)(24 =+−+ − ztEztEeyeyxtP zztAtBtA
x (2.2.27)
Differentiating equation (2.2.27) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (2.2.27) we get
.0),(),(),( 3)(21)(24 =+− ztEeztEeyytE ztA
ztB Differentiating this equation with
respect to y twice we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting
back the above value in equation (2.2.27) and solving, we get
)()(),( 31)(21 tKtKezztE tB += − and ),()(),( 42)(23 tKtKezztE tA +−= − where ),(1 tK
),(2 tK )(3 tK and )(4 tK are functions of integration. Substituting back the above
information in equation (2.2.26) we get
).,()()(),,()()(
),()()()(),,,(
5213
231)(22
42)(23)(2)(21)(21
10
ytEtxKtKyxXztEtxKtKezxX
tKtKeztKeytKezyXzyxPX
tB
tAtAtBtA
++=
++=
+−−−=
=
−
−−−
(2.2.28)
Considering equation (2.2.8) and using equation (2.2.28) we get
32
.0),(),()(2 2)(251 =++ ztEeytEtxK ztB
y (2.2.29)
Differentiating the above equation with respect to x we get .0)(1 =tK Substituting
the above value in (2.2.29) and differentiating the resulting equation with respect to
,y we get ),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are
functions of integration. Substituting the above values back in (2.2.29) we get
),()(),( 75)(22 tKtKezztE tB +−= − where )(7 tK is a function of integration.
Refreshing the system of equations (2.2.28) we get
).()()(),()()(),()()(
),,,(
652375)(232
42)(23)(2)(21
10
tKtKytxKXtKtKeztxKXtKtKeztKeyX
zyxPX
tB
tAtAtB
++=+−=
+−−=
=
−
−− (2.2.30)
Considering equation (2.2.9) and using equation (2.2.30) we get
.0)()(),,()()(
)()()()()]()(2[4)(214)(22
23)(23)(2
=−+−+
−+−
tKetAzyxPtKetKz
tKtAztKeytKetAtBytA
txttA
t
tttBtB
tt (2.2.31)
Differentiating (2.2.31) with respect to ,x we get
),,(),(),,(0),,( 7611 zyEzyxEzyxPzyxPxx +=⇒= where ),(6 zyE and ),(7 zyE
are functions of integration. Substituting the above value back in (2.2.31) and
differentiating twice with respect to ,y we get
),()(),(0),( 9866 zKzKyzyEzyEyy +=⇒= where )(8 zK and )(9 zK are functions
of integration. Substituting back the above value in (2.2.31) and differentiating with
respect to ,y we get .0)()()()]()(2[ 83)(23)(2 =++− zKtKetKetAtB ttBtB
tt
Differentiating this equation with respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzK z
Substituting back this value in the above equation and solving we get
.,)( 2)(2)(
2)()(2)(
13 ℜ∈+−= −−− ∫ cecdteectK tBtAtAtBtA Now substituting all the above
information in (2.2.31) and solve after differentiating twice with respect to ,z we get
.,,)(0)( 434399 ℜ∈+=⇒= ccczczKzK zz Substituting back in equation (2.2.31) and
33
solve after differentiating with respect to ,z we get
.,)( 5)(
5)()(
32 ℜ∈+−= ∫ − cecdteectK tAtAtA Now substituting all the above
information in equation (2.2.31) and solving we get
.,)( 6)(
6)()(
44 ℜ∈+= −−− ∫ cecdteectK tAtAtA Refreshing the system of equation
(2.2.30) with the help of above information, we get
),()(
),()(
,
),,(
65)(5
)()(3
3
75)(2)(2)(2
)()(2)(1
2
)(6
)()(4
)(5
)()(3
)(2
)()(1
1
7431
0
tKtKyexcdteexcX
tKtKzeexcdteexcX
ecdteecezcdteezceycdteeycX
zyExcxzcxycX
tAtAtA
tBtBtAtAtBtA
tAtAtAtAtAtAtAtAtA
+++−=
+−+−=
++−+−=
+++=
∫∫
∫∫∫
−
−−−−
−−−−−−−−−
(2.2.32)
where .,,,,, 654321 ℜ∈cccccc Now considering equation (2.2.10) and using equation
(2.2.32) we get
.0)()()()()()(
)]()([)]()([),(27)(27)(255
)(2
)()(1
71
=−−+−
−−−++ ∫ −
tKetBtKetKztKtBz
etBtAcxdteetBtAxczyExctB
tttB
tt
tAtt
tAtAtty (2.2.33)
Differentiating equation (2.2.33) with respect to ,x we get
.0)]()([)]()([2 )(2
)()(11 =−−−+ ∫ − tA
tttAtA
tt etBtAcdteetBtAcc Solving this equation
and remember that in this case ,0)()( ≠− tBtA tt we get ,01 =c which on back
substitution gives us .02 =c Substituting the above information in equation (2.2.33)
and differentiating with respect to ,y we get
),()(),(0),( 111077 zKzyKzyEzyEyy +=⇒= where )(10 zK and )(11 zK are
functions of integration. Substituting the above value in (2.2.33) and differentiating
with respect to z twice, we get .,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzK zz
Substituting back this value in equation (2.2.33) and differentiating with respect to ,z
we get .0)()()( 557 =+− tKtKtBc tt Solving this equation we get
.,)( 9)(
9)()(
75 ℜ∈+−= ∫ − cecdteectK tBtBtB Now substituting all the above
34
information in equation (2.2.33) we get ⇒=−− 0)()()( 7)(27)(28 tKetKetBc t
tBtBt
.,)( 10)(
10)()(
87 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations
(2.2.32) we get
),(
,
,
),(
6)(9
)()(7
)(5
)()(3
3
)(10
)()(8
)(9
)()(7
2
)(6
)()(4
)(5
)()(3
1
118743
0
tKeycdteeycexcdteexcX
ecdteecezcdteezcX
ecdteecezcdteezcX
zKycyzcxcxzcX
tBtBtBtAtAtA
tBtBtBtBtBtB
tAtAtAtAtAtA
++−+−=
++−=
++−=
++++=
∫∫∫∫∫∫
−−
−−−−−−
−−−−−−
(2.2.34)
where .,,,,,,, 109876543 ℜ∈cccccccc Now considering equation (2.2.11) and using
equation (2.2.34) we get
.0)()()(
)()()(226)(
9)()(
7
)(5
)()(3
1173
=−−+
−+++
∫∫
−
−
tKetBycdteetByc
etAcxdteetAxczKycxc
ttB
ttBtB
t
tAt
tAtAtz
(2.2.35)
Differentiating equation (2.2.35) with respect to ,x we get
.0)()(2 )(5
)()(33 =−+ ∫ − tA
ttAtA
t etAcdteetAcc Solving this equation and remember that
in this case ,0)( ≠tAt we get ,03 =c which on back substitution gives us .05 =c
Substituting the above information in equation (2.2.35) and differentiating with
respect to ,y we get .0)()(2 )(9
)()(77 =−+ ∫ − tB
ttBtB
t etBcdteetBcc Solving this
equation and remember that in this case ,0)( ≠tBt we get ,07 =c which on back
substitution gives us .09 =c Now substituting all the above values in equation
(2.2.35) and differentiating with respect to ,z we get
.,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzK zz Substituting the above value in
equation (2.2.35) we get ⇒=− 0)(611 tKc t .,)( 131311
6 ℜ∈+= cctctK Refreshing the
system of equations (2.2.34) we get
35
,,
,,)(
13113)(
10)()(
82
)(6
)()(4
1121184
0
tCtBtBtB
tAtAtA
ectcXecdteecX
ecdteecXczcycxcX−−−−
−−−
+=+=
+=+++=
∫∫
(2.2.36)
where .,,,,,, 13121110864 ℜ∈ccccccc The line element for Bianchi type I space-times is
given as
.22)(22)(222 dzdyedxedtds tBtA +++−= (2.2.37)
The above space-time admits seven linearly independent teleparallel Killing vector
fields which are ,t∂∂ ,)(
xe tA
∂∂− ,)(
ye tB
∂∂− ,
z∂∂ ,)(1
xtG
tx
∂∂
+∂∂
ytG
ty
∂∂
+∂∂ )(2
and ,z
tt
z∂∂
+∂∂ where ∫ −−= dteetG tAtA )()(1 )(
and .)( )()(2 ∫ −−= dteetG tBtB Killing
vector fields in general relativity are ,x∂∂
y∂∂ and .
z∂∂ It is evident that only one
teleparallel Killing vector field is the same and other are different from Killing vector
fields in general relativity. Also teleparallel Killing vector fields are more in number
to the Killing vector fields in general relativity. Cases (II)(b) and (II)(c) can be solved
exactly the same as in the above case.
Case (III)(a):
In this case we have )(),(),( tCCtBBtAA === and ).()( tCtB = Now substituting
equation (2.2.12) in equation (2.2.6), we get
.0),,(),,( 2)(23)(2 =+ zytPezxtPe ytA
xtB (2.2.38)
Differentiating equation (2.2.38) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (2.2.38) we get
⇒−= − ),(),,( 1)(2)(22 ztEezytP tAtBy ),,(),(),,( 31)(2)(22 ztEztEeyzytP tAtB +−= − where
36
),(3 ztE is a function of integration. Now refreshing the system of equations (2.2.12)
we get
).,,(),,(),(),,(),(),,,(
43212
31)(2)(2110
yxtPXztEztxEXztEztEeyXzyxPX tAtB
=+=
+−== −
(2.2.39)
Considering equation (2.2.7) and using equation (2.2.39) we get
.0)],(),([),,( 31)(2)(2)(24)(2 =+−+ − ztEztEeyeyxtPe zztAtBtA
xtB (2.2.40)
Differentiating equation (2.2.40) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (2.2.40) we get
.0),(),(),( 3)(21)(24)(2 =+− ztEeztEeyytEe ztA
ztBtB Differentiating this equation with
respect to y twice we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting
back the above value in equation (2.2.40) and solving, we get
)()(),( 311 tKtKzztE += and ),()(),( 42)(2)(23 tKtKezztE tAtB +−= − where ),(1 tK
),(2 tK )(3 tK and )(4 tK are functions of integration. Substituting back the above
information in equation (2.2.39) we get
).,()()(),,()()(
),()()()(),,,(
5213
2312
42)(2)(23)(2)(21)(2)(21
10
ytEtxKtKyxXztEtxKtKzxX
tKtKeztKeytKezyXzyxPX
tAtBtAtBtAtB
++=
++=
+−−−=
=−−−
(2.2.41)
Considering equation (2.2.8) and using equation (2.2.41) we get
.0),(),()(2 251 =++ ztEytEtxK zy (2.2.42)
Differentiating the above equation with respect to x we get .0)(1 =tK Substituting
the above value in (2.2.42) and differentiating the resulting equation with respect to
,y we get ),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are
functions of integration. Substituting the above values back in (2.2.42) we get
37
),()(),( 752 tKtKzztE +−= where )(7 tK is a function of integration. Refreshing
the system of equations (2.2.41) we get
).()()(),()()(),()()(
),,,(
65237532
42)(2)(23)(2)(21
10
tKtKytxKXtKtKztxKXtKtKeztKeyX
zyxPXtAtBtAtB
++=+−=
+−−=
=−− (2.2.43)
Considering equation (2.2.9) and using equation (2.2.43) we get
.0)()(),,()()(
)()]()(2[)()()]()(2[4)(214)(22)(2
2)(23)(23)(2
=−+−+
−++−
tKetAzyxPtKetKez
tKetAtBztKeytKetAtBytA
txttA
ttB
tBttt
tBtBtt (2.2.44)
Differentiating (2.2.44) with respect to ,x we get
),,(),(),,(0),,( 7611 zyEzyxEzyxPzyxPxx +=⇒= where ),(6 zyE and ),(7 zyE
are functions of integration. Substituting the above value back in (2.2.44) and
differentiating twice with respect to ,y we get
),()(),(0),( 9866 zKzKyzyEzyEyy +=⇒= where )(8 zK and )(9 zK are functions
of integration. Substituting back the above value in (2.2.44) and differentiating with
respect to ,y we get .0)()()()]()(2[ 83)(23)(2 =++− zKtKetKetAtB ttBtB
tt
Differentiating this equation with respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzK z
Substituting back this value in the above equation and solving we get
.,)( 2)(2)(
2)()(2)(
13 ℜ∈+−= −−− ∫ cecdteectK tBtAtAtBtA Now substituting all the above
information in (2.2.44) and solve after differentiating twice with respect to ,z we get
.,,)(0)( 434399 ℜ∈+=⇒= ccczczKzK zz Substituting back in equation (2.2.44) and
solve after differentiating with respect to ,z we get
.,)( 5)(2)(
5)()(2)(
32 ℜ∈+−= −−− ∫ cecdteectK tBtAtAtBtA Now substituting all the above
information in equation (2.2.44) and solving we get
.,)( 6)(
6)()(
44 ℜ∈+= −−− ∫ cecdteectK tAtAtA Refreshing the system of equation
(2.2.43) with the help of above information, we get
38
),()(
),()(
,
),,(
65)(2)(5
)()(2)(3
3
75)(2)(2
)()(2)(1
2
)(6
)()(4
)(5
)()(3
)(2
)()(1
1
7431
0
tKtKyexcdteexcX
tKtzKexcdteexcX
ecdteec
ezcdteezceycdteeycX
zyExcxzcxycX
tBtAtAtBtA
tBtAtAtBtA
tAtAtA
tAtAtAtAtAtA
+++−=
+−+−=
++
−+−=
+++=
−−−
−−−
−−−
−−−−−−
∫∫
∫∫∫
(2.2.45)
where .,,,,, 654321 ℜ∈cccccc Now considering equation (2.2.10) and using equation
(2.2.45) we get
.0)()()()()()(
)]()([)]()([),(27)(25)(27)(25)(2
)(2
)()(1
71
=−+−+
−−−++ ∫ −
tKetBtKetzBtKetKez
etBtAcxdteetBtAxczyExctB
ttB
tttB
ttB
tAtt
tAtAtty (2.2.46)
Differentiating equation (2.2.46) with respect to ,x we get
.0)]()([)]()([2 )(2
)()(11 =−−−+ ∫ − tA
tttAtA
tt etBtAcdteetBtAcc Solving this equation
and remember that in this case ,0)()( ≠− tBtA tt we get ,01 =c which on back
substitution gives us .02 =c Substituting the above information in equation (2.2.46)
and differentiating with respect to ,y we get
),()(),(0),( 111077 zKzyKzyEzyEyy +=⇒= where )(10 zK and )(11 zK are
functions of integration. Substituting the above value in (2.2.46) and differentiating
with respect to z twice, we get .,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzK zz
Substituting back this value in equation (2.2.46) and differentiating with respect to ,z
we get .0)()()( 5)(25)(27 =+− tKetKtBec t
tBt
tB Solving this equation we get
.,)( 9)(
9)()(
75 ℜ∈+−= −−− ∫ cecdteectK tBtBtB Now substituting all the above
information in equation (2.2.46) we get ⇒=−− 0)()()( 7)(27)(28 tKetKetBc t
tBtBt
.,)( 10)(
10)()(
87 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations
(2.2.45) we get
39
),(
,
,
),(
6)(9
)()(7
)(2)(5
)()(2)(3
3
)(10
)()(8
)(9
)()(7
2
)(6
)()(4
)(5
)()(3
1
118743
0
tKeyc
dteeycexcdteexcX
ecdteecezcdteezcX
ecdteecezcdteezcX
zKycyzcxcxzcX
tB
tBtBtBtAtAtBtA
tBtBtBtBtBtB
tAtAtAtAtAtA
++
−+−=
++−=
++−=
++++=
−
−−−−−
−−−−−−
−−−−−−
∫∫∫∫∫∫
(2.2.47)
where .,,,,,,, 109876543 ℜ∈cccccccc Now considering equation (2.2.11) and using
equation (2.2.47) we get
.0)()()()]()([
)]()([)(226)(26)(2)(
5
)()(3
1173
=−−−−
−+++ ∫ −
tKtBetKeetBtAcx
dteetBtAxczKycxc
ttB
ttBtA
tt
tAtAttz (2.2.48)
Differentiating equation (2.2.48) with respect to ,x we get
.0)()([)()([2 )(5
)()(33 =−−−+ ∫ − tA
tttAtA
tt etBtAcdteetBtAcc Solving this equation
and remember that in this case ,0)()( ≠− tBtA tt we get ,03 =c which on back
substitution gives us .05 =c Substituting the above information in equation (2.2.48)
and differentiating with respect to ,y we get .07 =c Now substituting all the above
values in equation (2.2.48) and differentiating with respect to ,z we get
.,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzK zz Substituting the above value in
equation (2.2.48) we get ⇒=−− 0)()()( 6)(26)(211 tKtBetKec t
tBt
tB
.,)( 13)(
13)()(
116 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations
(2.2.47) we get
,
,
,,
)(13
)(9
)()(11
3
)(10
)(9
)()(8
2
)(6
)()(4
1121184
0
tBtBtBtB
tBtBtBtB
tAtAtA
eceycdteecX
ecezcdteecX
ecdteecXczcycxcX
−−−−
−−−−
−−−
++=
+−=
+=+++=
∫∫
∫ (2.2.49)
where .,,,,,, 13121110864 ℜ∈ccccccc The line element for Bianchi type I space-times is
given as
40
).( 22)(22)(222 dzdyedxedtds tBtA +++−= (2.2.50)
The above space-time admits eight linearly independent teleparallel Killing vector
fields which are ,t∂∂ ,)(
xe tA
∂∂− ,)(
ye tB
∂∂− ,)(
ze tB
∂∂− ,)(1
xtG
tx
∂∂
+∂∂
,)(2
ytG
ty
∂∂
+∂∂
ztG
tz
∂∂
+∂∂ )(2 and ),()(
yz
zye tB
∂∂
−∂∂− where
∫ −−= dteetG tAtA )()(1 )(
and .)( )()(2 ∫ −−= dteetG tBtB Killing vector fields in general
relativity are ,x∂∂ ,
y∂∂
z∂∂ and .
yz
zy
∂∂
−∂∂ Teleparallel Killing vector fields are
different and more in number to the Killing vector fields in general relativity. Cases
(III)(b) and (III)(c) can be solved exactly the same as in the above case.
Case (IV):
In this case we have )(),(),( tCCtBBtAA === and ).()()( tCtBtA == Now
substituting equation (2.2.12) in equation (2.2.6), we get
.0),,(),,( 23 =+ zytPzxtP yx (2.2.51)
Differentiating equation (2.2.51) with respect to ,y we get
),(),(),,(0),,( 2122 ztEztyEzytPzytPyy +=⇒= ),(1 ztE and ),(2 ztE are functions
of integration. Substituting back this value in equation (2.2.51) we get
),,(),(),,(),(),,( 31313 ztEztxEzxtPztEzxtPx +−=⇒−= where ),(3 ztE is a
function of integration. Now refreshing the system of equations (2.2.12) we get
).,,(),,(),(),,(),(),,,(
43312
21110
yxtPXztEztxEXztEztEyXzyxPX
=+−=
+== (2.2.52)
Considering equation (2.2.7) and using equation (2.2.52) we get
.0)],(),(),,( 214 =++ ztEztEyyxtP zzx (2.2.53)
41
Differentiating equation (2.2.53) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (2.2.53) we get
.0),(),(),( 214 =++ ztEztEyytE zz Differentiating this equation with respect to y
twice we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting back the above
value in equation (2.2.53) and solving, we get )()(),( 311 tKtKzztE +−= and
),()(),( 422 tKtKzztE +−= where ),(1 tK ),(2 tK )(3 tK and )(4 tK are functions of
integration. Substituting back the above information in equation (2.2.53) we get
).,()()(),,()()(),()()()(),,,(
52133312
4231110
ytEtxKtKyxXztEtxKtKzxXtKtKztKytKzyXzyxPX
++=+−=
+−+−== (2.2.54)
Considering equation (2.2.8) and using equation (2.2.54) we get
.0),(),()(2 351 =++ ztEytEtxK zy (2.2.55)
Differentiating the above equation with respect to x we get .0)(1 =tK Substituting
the above value in (2.2.55) and differentiating the resulting equation with respect to
,y we get ),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are
functions of integration. Substituting the above values back in (2.2.55) we get
),()(),( 753 tKtKzztE +−= where )(7 tK is a function of integration. Refreshing the
system of equations (2.2.55) we get
).()()(),()()(),()()(),,,(
65237532
423110
tKtKytxKXtKtKztxKXtKtKztKyXzyxPX
++=+−−=
+−== (2.2.56)
Considering equation (2.2.9) and using equation (2.2.56) we get
.0)()(),,()()(
)()()()()(4)(214)(22)(2
2)(23)(23)(2
=−+−+
+−−
tKetAzyxPtKetKez
tKetAztKeytKetAytA
txttA
ttA
tAtt
tAtAt (2.2.57)
Differentiating (2.2.57) with respect to ,x we get
),,(),(),,(0),,( 7611 zyEzyxEzyxPzyxPxx +=⇒= where ),(6 zyE and ),(7 zyE
42
are functions of integration. Substituting the above value back in (2.2.57) and
differentiating twice with respect to ,y we get
),()(),(0),( 9866 zKzKyzyEzyEyy +=⇒= where )(8 zK and )(9 zK are functions
of integration. Substituting back the above value in (2.2.57) and differentiating with
respect to ,y we get .0)()()()( 83)(23)(2 =+−− zKtKetKetA ttAtA
t Differentiating this
equation with respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzK z Substituting back
this value in the above equation and solving we get
.,)( 2)(
2)()(
13 ℜ∈+= −−− ∫ cecdteectK tAtAtA Now substituting all the above
information in (2.2.57) and solve after differentiating twice with respect to ,z we get
.,,)(0)( 434399 ℜ∈+=⇒= ccczczKzK zz Substituting back in equation (2.2.57) and
solve after differentiating with respect to ,z we get
.,)( 5)(
5)()(
32 ℜ∈+−= −−− ∫ cecdteectK tAtAtA Now substituting all the above
information in equation (2.2.57) and solving we get
.,)( 6)(
6)()(
44 ℜ∈+= −−− ∫ cecdteectK tAtAtA Refreshing the system of equation
(2.2.56) with the help of above information, we get
),()(
),()(
,
),,(
65)(5
)()(3
3
75)(2
)()(1
2
)(6
)()(4
)(5
)()(3
)(2
)()(1
1
7431
0
tKtKyexcdteexcX
tKtzKexcdteexcX
ecdteec
ezcdteezceycdteeycX
zyExcxzcxycX
tAtAtA
tAtAtA
tAtAtA
tAtAtAtAtAtA
+++−=
+−−−=
++
−++−=
+++=
−−−
−−−
−−−
−−−−−−
∫∫
∫∫∫
(2.2.58)
where .,,,,, 654321 ℜ∈cccccc Now considering equation (2.2.10) and using equation
(2.2.58) we get
.0)()()()()()(),(2 7)(25)(27)(25)(271 =−+−++ tKetAtKetzAtKetKezzyExc tA
ttA
tttA
ttA
y (2.2.59)
Differentiating equation (2.2.59) with respect to ,x we get .01 =c Substituting the
above information in equation (2.2.59) and differentiating with respect to ,y we get
43
),()(),(0),( 111077 zKzyKzyEzyEyy +=⇒= where )(10 zK and )(11 zK are
functions of integration. Substituting the above value in (2.2.59) and differentiating
with respect to z twice, we get .,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzK zz
Substituting back this value in equation (2.2.59) and differentiating with respect to ,z
we get .0)()()( 5)(25)(27 =++ tKetKtAec t
tAt
tB Solving this equation we get
.,)( 9)(
9)()(
75 ℜ∈+−= −−− ∫ cecdteectK tAtAtA Now substituting all the above
information in equation (2.2.59) we get ⇒=−− 0)()()( 7)(27)(28 tKetKetAc t
tAtAt
.,)( 10)(
10)()(
87 ℜ∈+= −−− ∫ cecdteectK tAtAtA Refreshing the system of equations
(2.2.58) we get
),(
,
,
),(
6)(9
)()(7
)(5
)()(3
3
)(10
)()(8
)(9
)()(7
)(2
2
)(6
)()(4
)(5
)()(3
)(2
1
118743
0
tKeycdteeycexcdteexcX
ecdteecezcdteezcexcX
ecdteecezcdteezceycX
zKycyzcxcxzcX
tAtAtAtAtAtA
tAtAtAtAtAtAtA
tAtAtAtAtAtAtA
++−+−=
++−+−=
++−+=
++++=
−−−−−−
−−−−−−−
−−−−−−−
∫∫∫∫∫∫
(2.2.60)
where .,,,,,,, 109876543 ℜ∈cccccccc Now considering equation (2.2.11) and using
equation (2.2.60) we get
.0)()()()(22 6)(26)(21173 =−−++ tKtAetKezKycxc t
tAt
tAz (2.2.61)
Differentiating equation (2.2.61) with respect to x and y respectively, we get
.073 == cc Substituting the above information in equation (2.2.61) and
differentiating with respect to ,z we get
.,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzK zz Substituting the above value in
equation (2.2.61) we get ⇒=−− 0)()()( 6)(26)(211 tKtAetKec t
tAt
tA
.,)( 13)(
13)()(
116 ℜ∈+= −−− ∫ cecdteectK tAtAtA Refreshing the system of equations
(2.2.60) we get
44
,
,
,
,
)(13
)(9
)(5
)()(11
3
)(10
)(9
)(2
)()(8
2
)(6
)(5
)(2
)()(4
1
1211840
tAtAtAtAtA
tAtAtAtAtA
tAtAtAtAtA
eceycexcdteecX
ecezcexcdteecX
ecezceycdteecX
czcycxcX
−−−−−
−−−−−
−−−−−
+++=
++−−=
++−+=
+++=
∫∫∫
(2.2.62)
where .,,,,,,,,, 13121110986542 ℜ∈cccccccccc The line element for Bianchi type I
space-times is given as
).( 222)(222 dzdydxedtds tA +++−= (2.2.63)
The above space-time admits ten linearly independent teleparallel Killing vector
fields which are ,t∂∂ ,)(
xe tA
∂∂− ,)(
ye tA
∂∂− ,)(
ze tA
∂∂− ,)(1
xtG
tx
∂∂
+∂∂
,)(1
ytG
ty
∂∂
+∂∂ ,)(1
ztG
tz
∂∂
+∂∂ ),()(
xz
zxe tA
∂∂
−∂∂− ),()(
yx
xye tA
∂∂
−∂∂− and
),()(
yz
zye tA
∂∂
−∂∂− where .)( )()(1 ∫ −−= dteetG tAtA
Killing vector fields in general
relativity are ,x∂∂ ,
y∂∂ ,
z∂∂ ),(
yx
xy
∂∂
−∂∂ )(
zy
yz
∂∂
−∂∂ and ).(
zx
xz
∂∂
−∂∂ It is
clear that teleparallel Killing vector fields are different and more in number to the
Killing vector fields in general relativity.
Case (V)(a):
In this case we have )(),(,tan tCCtBBtconsA === and ).()( tCtB = Now
substituting equation (2.2.12) in equation (2.2.6), we get
.0),,(),,( 23)(2 =+ zytPzxtPe yxtB (2.2.64)
Differentiating equation (2.2.64) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (2.2.64) we get
⇒−= ),(),,( 1)(22 ztEezytP tBy ),,(),(),,( 31)(22 ztEztEeyzytP tB +−= where
45
),(3 ztE is a function of integration. Now refreshing the system of equations (2.2.12)
we get
).,,(),,(),(),,(),(),,,(
43212
31)(2110
yxtPXztEztxEXztEztEeyXzyxPX tB
=+=
+−== (2.2.65)
Considering equation (2.2.7) and using equation (2.2.65) we get
.0),(),(),,( 31)(24)(2 =+− ztEztEeyyxtPe zztB
xtB (2.2.66)
Differentiating equation (2.2.66) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (2.2.66) we get
.0),(),(),( 31)(24)(2 =+− ztEztEeyytEe zztBtB Differentiating this equation with
respect to y twice we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting
back the above value in equation (2.2.66) and solving, we get
)()(),( 311 tKtKzztE += and ),()(),( 42)(23 tKtKezztE tB +−= where ),(1 tK
),(2 tK )(3 tK and )(4 tK are functions of integration. Substituting back the above
information in equation (2.2.65) we get
).,()()(),,()()(
),()()()(),,,(
5213
2312
42)(23)(21)(21
10
ytEtxKtKyxXztEtxKtKzxX
tKtKeztKeytKezyXzyxPX
tBtBtB
++=
++=
+−−−=
=
(2.2.67)
Considering equation (2.2.8) and using equation (2.2.67) we get
.0),(),()(2 251 =++ ztEytEtxK zy (2.2.68)
Differentiating the above equation with respect to x we get .0)(1 =tK Substituting
the above value in (2.2.68) and differentiating the resulting equation with respect to
,y we get ),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are
functions of integration. Substituting the above values back in (2.2.68) we get
46
),()(),( 752 tKtKzztE +−= where )(7 tK is a function of integration. Refreshing
the system of equations (2.2.67) we get
).()()(),()()(),()()(),,,(
65237532
42)(23)(2110
tKtKytxKXtKtKztxKXtKtKeztKeyXzyxPX tBtB
++=+−=
+−−== (2.2.69)
Considering equation (2.2.9) and using equation (2.2.69) we get
.0),,()()(
)()(2)()()(2142)(2
2)(23)(23)(2
=+−+
++
zyxPtKtKez
tKetBztKeytKetBy
xtttB
tBtt
tBtBt (2.2.70)
Differentiating (2.2.70) with respect to ,x we get
),,(),(),,(0),,( 7611 zyEzyxEzyxPzyxPxx +=⇒= where ),(6 zyE and ),(7 zyE
are functions of integration. Substituting the above value back in (2.2.70) and
differentiating twice with respect to ,y we get
),()(),(0),( 9866 zKzKyzyEzyEyy +=⇒= where )(8 zK and )(9 zK are functions
of integration. Substituting back the above value in (2.2.70) and differentiating with
respect to ,y we get .0)()()()(2 83)(23)(2 =++ zKtKetKetB ttBtB
t Differentiating this
equation with respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzK z Substituting back
this value in the above equation and solving we get
.,)( 2)(2
2)(2
13 ℜ∈+−= −− cecetctK tBtB Now substituting all the above information
in (2.2.70) and solve after differentiating twice with respect to ,z we get
.,,)(0)( 434399 ℜ∈+=⇒= ccczczKzK zz Substituting back in equation (2.2.70) and
solve after differentiating with respect to ,z we get
.,)( 5)(2
5)(2
32 ℜ∈+−= −− cecetctK tBtB Now substituting all the above information
in equation (2.2.70) and solving we get .,)( 6644 ℜ∈+= cctctK Refreshing the
system of equation (2.2.69) with the help of above information, we get
47
),()(
),()(
,),,(
65)(25
)(23
3
75)(22
)(21
2
64532117
4310
tKtKyexcetxcX
tKtzKexctexcX
ctczctzcyctycXzyExcxzcxycX
tBtB
tBtB
+++−=
+−+−=
++−+−=+++=
−−
−− (2.2.71)
where .,,,,, 654321 ℜ∈cccccc Now considering equation (2.2.10) and using equation
(2.2.71) we get
.0)()()()()()(
)()(),(27)(25)(27)(25)(2
217
1
=−+−+
+−+
tKetBtKetzBtKetKez
tBcxtBtxczyExctB
ttB
tttB
ttB
tty (2.2.72)
Differentiating equation (2.2.72) with respect to ,x we get
.0)()(2 211 =+− tBctBtcc tt Solving this equation and remember that in this case
,0)( ≠tBt we get ,01 =c which on back substitution gives us .02 =c Substituting the
above information in equation (2.2.72) and differentiating with respect to ,y we get
),()(),(0),( 111077 zKzyKzyEzyEyy +=⇒= where )(10 zK and )(11 zK are
functions of integration. Substituting the above value in (2.2.72) and differentiating
with respect to z twice, we get .,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzK zz
Substituting back this value in equation (2.2.72) and differentiating with respect to ,z
we get .0)()()( 5)(25)(27 =+− tKetKtBec t
tBt
tB Solving this equation we get
.,)( 9)(
9)()(
75 ℜ∈+−= −−− ∫ cecdteectK tBtBtB Now substituting all the above
information in equation (2.2.72) we get ⇒=−− 0)()()( 7)(27)(28 tKetKetBc t
tBtBt
.,)( 10)(
10)()(
87 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations
(2.2.71) we get
),(
,
,
),(
6)(9
)()(7
)(25
)(23
3
)(10
)()(8
)(9
)()(7
2
64531
118743
0
tKeycdteeycexcectxX
ecdteecezcdteezcX
ctczcctzX
zKycyzcxcxzcX
tBtBtBtBtB
tBtBtBtBtBtB
++−+−=
++−=
++−=
++++=
−−−−−
−−−−−−
∫∫∫
(2.2.73)
48
where .,,,,,,, 109876543 ℜ∈cccccccc Now considering equation (2.2.11) and using
equation (2.2.73) we get
.0)()()()()()(22 6)(26)(253
1173 =−−+−++ tKtBetKetBtcxtBtxczKycxc t
tBt
tBttz (2.2.74)
Differentiating equation (2.2.74) with respect to ,x we get
.0)()(2 533 =+− tBctBtcc tt Solving this equation and remember that in this case
,0)( ≠tBt we get ,03 =c which on back substitution gives us .05 =c Substituting the
above information in equation (2.2.74) and differentiating with respect to ,y we get
.07 =c Now substituting all the above values in equation (2.2.74) and differentiating
with respect to ,z we get .,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzK zz Substituting
the above values in equation (2.2.74) we get ⇒=−− 0)()()( 6)(26)(211 tKtBetKec t
tBt
tB
.,)( 13)(
13)()(
116 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations
(2.2.73) we get
,
,
,,
)(13
)(9
)()(11
3
)(10
)(9
)()(8
2
641
1211840
tBtBtBtB
tBtBtBtB
eceycdteecX
ecezcdteecX
cctXczcycxcX
−−−−
−−−−
++=
+−=
+=+++=
∫∫ (2.2.75)
where .,,,,,,, 131211109864 ℜ∈cccccccc The line element for Bianchi type I space-
times is given as
).( 22)(2222 dzdyedxdtds tB +++−= (2.2.76)
The above space-time admits eight linearly independent teleparallel Killing vector
fields which are ,t∂∂ ,
x∂∂ ,)(
ye tB
∂∂− ,)(
ze tB
∂∂− ,
xt
tx
∂∂
+∂∂ ,)(1
ytG
ty
∂∂
+∂∂
ztG
tz
∂∂
+∂∂ )(1 and ),()(
yz
zye tB
∂∂
−∂∂− where
.)( )()(1 ∫ −−= dteetG tBtB Killing
vector fields in general relativity are ,x∂∂ ,
y∂∂
z∂∂ and .
yz
zy
∂∂
−∂∂ Only one
49
teleparallel Killing vector field is the same as in general relativity and other
teleparallel Killing vector fields are different and more in number to the Killing
vector fields in general relativity. Cases (V)(b) and (V)(c) can be solved exactly the
same as in the above case.
Case (VI)(a):
In this case we have )(tAA = and .tan tconsCB == Substituting equation (2.2.12)
in equation (2.2.6), we get
.0),,(),,( 2)(23 =+ zytPezxtP ytA
x (2.2.77)
Differentiating equation (2.2.77) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (2.2.77) we get
⇒−= − ),(),,( 1)(22 ztEezytP tAy ),,(),(),,( 31)(22 ztEztEeyzytP tA +−= − where
),(3 ztE is a function of integration. Now refreshing the system of equations (2.2.12)
we get
).,,(),,(),(),,(),(),,,(
43212
31)(2110
yxtPXztEztxEXztEztEeyXzyxPX tA
=+=
+−== −
(2.2.78)
Considering equation (2.2.7) and using equation (2.2.78) we get
.0)],(),(),,( 3)(214 =+− ztEeztEyyxtP ztA
zx (2.2.79)
Differentiating equation (2.2.79) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (2.2.79) we get
.0),(),(),( 3)(214 =+− ztEeztEyytE ztA
z Differentiating this equation with respect to
y twice we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting back the
above value in equation (2.2.79) and solving, we get )()(),( 311 tKtKzztE += and
50
),()(),( 42)(23 tKtKezztE tA +−= − where ),(1 tK ),(2 tK )(3 tK and )(4 tK are
functions of integration. Substituting back the above information in equation (2.2.78)
we get
).,()()(),,()()(),()()()(
),,,(
52132312
42)(2)3)(21)(2)1
10
ytEtxKtKyxXztEtxKtKzxXtKtKeztKeytKezyX
zyxPXtAtAtA
++=++=
+−−−=
=−−− (2.2.80)
Considering equation (2.2.8) and using equation (2.2.80) we get
.0),(),()(2 251 =++ ztEytEtxK zy (2.2.81)
Differentiating the above equation with respect to x we get .0)(1 =tK Substituting
the above value in (2.2.81) and differentiating the resulting equation with respect to
,y we get ),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are
functions of integration. Substituting the above values back in (2.2.81) we get
),()(),( 752 tKtKzztE +−= where )(7 tK is a function of integration. Refreshing
the system of equations (2.2.80) we get
).()()(),()()(),()()(
),,,(
65237532
42)(23)(21
10
tKtKytxKXtKtKztxKXtKtKeztKeyX
zyxPXtAtA
++=+−=
+−−=
=−− (2.2.82)
Considering equation (2.2.9) and using equation (2.2.82) we get
.0)()(),,()()(
)()()()()(4)(214)(22
233
=−+−+
−+−
tKetAzyxPtKetKz
tKtAztKytKtAytA
txttA
t
ttt (2.2.83)
Differentiating (2.2.83) with respect to ,x we get
),,(),(),,(0),,( 7611 zyEzyxEzyxPzyxPxx +=⇒= where ),(6 zyE and ),(7 zyE
are functions of integration. Substituting the above value back in (2.2.83) and
differentiating twice with respect to ,y we get
),()(),(0),( 9866 zKzKyzyEzyEyy +=⇒= where )(8 zK and )(9 zK are functions
51
of integration. Substituting back the above value in (2.2.83) and differentiating with
respect to ,y we get .0)()()()( 833 =++− zKtKtKtA tt Differentiating this equation
with respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzK z Substituting back this value
in the above equation and solving we get .,)( 2)(
2)()(
13 ℜ∈+−= ∫ − cecdteectK tAtAtA
Now substituting all the above information in (2.2.83) and solve after differentiating
twice with respect to ,z we get .,,)(0)( 434399 ℜ∈+=⇒= ccczczKzK zz
Substituting back in equation (2.2.83) and solve after differentiating with respect to
,z we get .,)( 5)(
5)()(
32 ℜ∈+−= ∫ − cecdteectK tAtAtA Now substituting all the above
information in equation (2.2.83) and solving we get
.,)( 6)(
6)()(
44 ℜ∈+= −−− ∫ cecdteectK tAtAtA Refreshing the system of equation
(2.2.82) with the help of above information, we get
),()(
),()(
,
),,(
65)(5
)()(3
3
75)(2
)()(1
2
)(6
)()(4
)(5
)()(3
)(2
)()(1
1
7431
0
tKtKyexcdteexcX
tKtzKexcdteexcX
ecdteec
ezcdteezceycdteeycX
zyExcxzcxycX
tAtAtA
tAtAtA
tAtAtA
tAtAtAtAtAtA
+++−=
+−+−=
++
−+−=
+++=
∫∫∫
∫∫
−
−
−−−
−−−−−−
(2.2.84)
where .,,,,, 654321 ℜ∈cccccc Now considering equation (2.2.10) and using equation
(2.2.84) we get
.0)()()()(),(2 75)(2
)()(1
71 =−+−++ ∫ − tKtKzetAcxdteetAxczyExc tt
tAt
tAtAty (2.2.85)
Differentiating equation (2.2.85) with respect to ,x we get
.0)()(2 )(2
)()(11 =−+ ∫ − tA
ttAtA
t etAcdteetAcc Solving this equation and remember that
in this case ,0)( ≠tAt we get ,01 =c which on back substitution gives us .02 =c
Substituting the above information in equation (2.2.85) and differentiating with
respect to ,y we get ),()(),(0),( 111077 zKzyKzyEzyEyy +=⇒= where )(10 zK
52
and )(11 zK are functions of integration. Substituting the above value in (2.2.85) and
differentiating with respect to z twice, we get
.,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzK zz Substituting back this value in equation
(2.2.85) and differentiating with respect to ,z we get ⇒=+ 0)(57 tKc t
.,)( 9975 ℜ∈+−= cctctK Now substituting all the above information in equation
(2.2.85) we get ⇒=− 0)(78 tKc t .,)( 10108
7 ℜ∈+= cctctK Refreshing the system of
equations (2.2.84) we get
),(
,
,
),(
697
)(5
)()(3
3
108972
)(6
)()(4
)(5
)()(3
1
118743
0
tKyctycexcdteexcX
ctczctzcX
ecdteecezcdteezcX
zKycyzcxcxzcX
tAtAtA
tAtAtAtAtAtA
++−+−=
++−=
++−=
++++=
∫
∫∫
−
−−−−−−
(2.2.86)
where .,,,,,,, 109876543 ℜ∈cccccccc Now considering equation (2.2.11) and using
equation (2.2.86) we get
.0)()()()(22 6)(5
)()(3
1173 =−−+++ ∫ − tKetAcxdteetAxczKycxc t
tAt
tAtAtz (2.2.87)
Differentiating equation (2.2.87) with respect to ,x we get
.0)()(2 )(5
)()(33 =−+ ∫ − tA
ttAtA
t etAcdteetAcc Solving this equation and remember that
in this case ,0)( ≠tAt we get ,03 =c which on back substitution gives us .05 =c
Substituting the above information in equation (2.2.87) and differentiating with
respect to ,y we get .07 =c Now substituting all the above values in equation
(2.2.87) and differentiating with respect to ,z we get
.,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzK zz Substituting the above values in
equation (2.2.87) we get ⇒=− 0)(611 tKc t .,)( 131311
6 ℜ∈+= cctctK Refreshing the
system of equations (2.2.86) we get
53
,,
,,
139113
10982
)(6
)()(4
1121184
0
cyctcXczctcX
ecdteecXczcycxcX tAtAtA
++=+−=
+=+++= −−− ∫ (2.2.88)
where .,,,,,,, 131211109864 ℜ∈cccccccc The line element for Bianchi type I space-
times is given as
).( 222)(222 dzdydxedtds tA +++−= (2.2.89)
The above space-time admits eight linearly independent teleparallel Killing vector
fields which are ,t∂∂ ,)(
xe tA
∂∂− ,
y∂∂ ,
z∂∂ ,)(1
xtG
tx
∂∂
+∂∂ ,
yt
ty
∂∂
+∂∂
zt
tz
∂∂
+∂∂
and ),(y
zz
y∂∂
−∂∂ where .)( )()(1 ∫ −−= dteetG tAtA
Killing vector fields in general
relativity are ,x∂∂ ,
y∂∂
z∂∂ and .
yz
zy
∂∂
−∂∂ It is evident that three teleparallel
Killing vector fields are the same as Killing vector fields in general relativity and all
other teleparallel Killing vector fields are different. Teleparallel Killing vector fields
are more in number to the Killing vector fields in general relativity. Cases (VI)(b) and
(VI)(c) can be solved exactly the same as in the above case.
2.3. Teleparallel Killing Vector Fields in Bianchi
Type II Space-Times
Consider Bianchi type II space-times in usual coordinates ),,,( zyxt (labeled by
),,,,( 3210 xxxx respectively) with the line element [76, 78]
,)(2)]()([)()( 22222222222 dydztxBdztCtBxdytBdxtAdtds −++++−= (2.3.1)
where ,A B and C are no where zero functions of t only. The tetrad components
and its inverse components by using the relation (1.3.4) is obtained as [66]
54
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
−=
)(000)()(00
00)(00001
tCtxBtB
tAS a
µ .
)()(000)(0000)(00001
11
1
1
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=
−−
−
−
tCtxCtB
tASa
µ (2.3.2)
The non vanishing torsion components are obtained by (1.3.11) as
,011
AAT
•
= ,022
BBT
•
= ,033
CCT
•
= ,032
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
••
BB
CCxT ,113
2 −=T (2.3.3)
where ‘dot’ denotes differentiation with respect to .t A vector field X is said to be
teleparallel Killing vector field if it satisfies equation (1.3.17). One can write (1.3.17)
explicitly using (2.3.1) and (2.3.3) as
,0,0 1,1
0,0 == XX (2.3.4)
,02,3
2,2 =− XxX (2.3.5)
,011,
00,
12 =−− • XAAXXA (2.3.6)
,0322,
00,
320,
22 =−+−− •• XBBxXBBXXxBXB (2.3.7)
,0322,
121,
321,
22 =−+− XBXAXBxXB (2.3.8)
( ),0332
20,
32223,
00,
22
=−−
++−+••
•
XCCXBBxXBxBXCBxXXBx
(2.3.9)
( ) ,0321,
32221,
223,
12 =+++− XBxXCBxXxBXA (2.3.10)
( ) ,0,332
2,3222
2,22
3,22 =−++− XBxXCBxXxBXB (2.3.11)
( ) ,0,322
3,3222 =−+ XBxXCBx (2.3.12)
Solving equation (2.3.4), (2.3.5) and (2.3.11) we get
55
),,,(),,(
),,,(),,(),,(
),,,(),,,(
432
3
3432
2
2110
zxtPzxtPCByX
zxtPzxtxPzxtPCByxX
zytPXzyxPX
z
z
+⎟⎠⎞
⎜⎝⎛−=
++⎟⎠⎞
⎜⎝⎛−=
==
(2.3.13)
where ),,,(1 zyxP ),,,(2 zytP ),,,(3 zxtP and ),,,(4 zxtP are functions of integration
which are to be determined. Considering equation (2.3.12) and using equation
(2.3.13) we get
.0),,()(),,()(),,()( 324232 =−+− zxtPtBxzxtPtCzxtPtyB zzzz (2.3.14)
Differentiating (2.3.14) with respect to ,y we get
),,(),(),,(0),,( 2133 xtKxtKzzxtPzxtPzz +=⇒= where ),(1 xtK and ),(2 xtK are
functions of integration. Substituting back this value in equation (2.3.14) and
differentiating the resulting equation with respect to ,z we get
),,(),(),,(0),,( 4344 xtKxtKzzxtPzxtPzz +=⇒= where ),(3 xtK and ),(4 xtK are
functions of integration. Substituting back the above values we get
).,()()(),( 1
23 xtK
tCtBxxtK ⎟⎟⎠
⎞⎜⎜⎝
⎛= Therefore ).,(),(
)()(),,( 41
24 xtKxtK
tCtBzxzxtP +⎟⎟⎠
⎞⎜⎜⎝
⎛=
Refreshing the system of equations (2.3.13) we have
).,(),(][
),,(),(),(][
),,,(),,,(
4122
3
24122
22
2110
xtKxtKCBy
CBzxX
xtKxtxKxtKzCByx
CBzxX
zytPXzyxPX
+⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
+++⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
==
(2.3.15)
Considering equation (2.3.8) and using equation (2.3.15) we get
.0),,()(),()(),()( 222212 =++ zytPtAxtKtBxtKtzB yxx (2.3.16)
56
Differentiating (2.3.16) first with respect to z and then with respect to ,x we get
),()(),(0),( 2111 tDtDxxtKxtK xx +=⇒= where )(1 tD and )(2 tD are functions of
integration. Substituting back the above value in (2.3.16) then solve after
differentiating with respect to ,z we get
),,(),()()()(),,( 651
22 ytKztKtD
tAtByzzytP ++⎟⎟⎠
⎞⎜⎜⎝
⎛−= where ),(5 ztK and ),(6 ztK are
functions of integration. Substituting back these values in (2.3.16) and solving we get
)()(),( 432 tDtDxxtK += and ),()()()(),( 53
26 tDtD
tAtByytK +⎟⎟⎠
⎞⎜⎜⎝
⎛−= where ),(3 tD
)(4 tD and )(5 tD are functions of integration. Substituting all these information in
(2.3.15) we get
).,())()(]([
),()(),())()((][
),()()()(),()(
)()(
),,,(
42122
3
4342122
22
532
512
1
10
xtKtDtxDCBy
CBzxX
tDtxDxtxKtDtxDzCByx
CBzxX
tDtDtAtByztKtD
tAtBzyX
zyxPX
++⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
+++++⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
+⎟⎟⎠
⎞⎜⎜⎝
⎛−+⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
=
(2.3.17)
Considering equation (2.3.10) and using equation (2.3.17) we get
.0),()(),()()()()()()()]()(2[
4252
3222122
=++
−++−
xtKtCztKtAtDtxBtDtBztDtzBxtyB
xz
(2.3.18)
Differentiating (2.3.18) with respect to ,z we get
.0),()()()()()( 522212 =++ ztKtAtDtBtDtBx zz (2.3.19)
Differentiating the above equation with respect to ,x we get .0)(1 =tD Substituting
back in (2.3.19) we get ⇒=+ 0),()()()( 5222 ztKtAtDtB zz
57
),()()()()(
21),( 762
225 tDtzDtD
tAtBzztK ++⎟⎟⎠
⎞⎜⎜⎝
⎛−= where )(6 tD and )(7 tD are
functions of integration. Now substituting back the above value in (2.3.18) then
differentiating the resulting equation with respect to x and solving we get
),()()()()(
21),( 983
224 tDtxDtD
tCtBxxtK ++⎟⎟⎠
⎞⎜⎜⎝
⎛= )(
)()()( 8
26 tD
tAtCtD ⎟⎟⎠
⎞⎜⎜⎝
⎛−= and
),()()()()(
)()(
21),( 78
22
225 tDtD
tAtCztD
tAtBzztK +⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−= where )(8 tD and )(9 tD
are functions of integration. Refreshing the system of equations (2.3.17) we get
).()()()()(
21)(][
),()()]()(
)()()(
21[)(][
),()()()()(
)()()(
)()(
21
),,,(
9832
2222
3
4398
32
2222
22
1032
82
22
21
10
tDtxDtDtCtBxtD
CBy
CBzxX
tDtxDtDtxD
tDtCtBxxtDz
CByx
CBzxX
tDtDtAtBytD
tAtCztD
tAtBzX
zyxPX
++⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
++++
⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
+⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−+⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
=
(2.3.20)
Considering equaiton (2.3.7) and using equation (2.3.20) we get
.0)()()()()()(
)()()(),,()()()()()()(43
21423222
=++
+−++
tDtBtBtDtBtxB
tDtBtzBzyxPtDtBtDtBxtDtzB
tt
tyttt (2.3.21)
Differentiating (2.3.21) with respect to ,z we get
.0)()()(),,()()( 2122 =+− tDtBtBzyxPtDtB tzyt In order to solve this equation we
consider ),,(),(),,(0),,( 8711 yxKzxKzyxPzyxP zy +=⇒= where ),(7 zxK and
),(8 yxK are functions of integration. Substituting back this value in the above
equation we get .,)(
1)(0)()()()( 11222 ℜ∈=⇒=+ cc
tBtDtDtBtDtB tt Substituting
back the above values in (2.3.21) and differentiating with respect to ,x we get
58
.0)()()(),,()()( 3832 =+− tDtBtBzyxKtDtB tyxt Solving this equation by considering
0),,(8 =zyxK yx we get ),()(),(0),,( 121188 yDxDyxKzyxK yx +=⇒= where
)(11 xD and )(12 yD are functions of integration and .,)(
1)( 223 ℜ∈= cc
tBtD Now
substituting back all the above values in (2.3.21) and solving we get ,)(
1)( 34 c
tBtD =
and .,,)(0)( 153151212 ℜ∈=⇒= cccyDyD Refreshing the system of equations
(2.3.20) by using the above information, we get
).()()(
)(21]
)()(
)()([
,)(
1)()(
])(
)(21[]
)()(
)()([
),()(
)()()()(
)()(
21
,)(),(
9822
2122
3
3982
222
12222
1022
82
1221
151170
tDtxDctC
tBxctC
tBytC
tBzxX
ctB
txDtDx
cxtC
tBxxcztC
tByxtC
tBzxX
tDctAtBytD
tAtCzc
tAtBzX
cxDzxKX
++⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+++
+⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
++=
(2.3.22)
Now differentiating (2.3.6) with respect to y and using equation (2.3.22) we get
( ) .0)()()()(2 =− tBtAtAtBc tt We will discuss here three different possibilities to
solve this equation:
(I) ( ) ,0)()()()( ≠− tBtAtAtB tt ,02 =c (II) ( ) ,0)()()()( =− tBtAtAtB tt ,02 ≠c
(III) ( ) ,0)()()()( =− tBtAtAtB tt .02 =c
We will discuss each case in turn. In case (III) we will obtain the same teleparallel
Killing vector fields as in case (II) with the reduction of one teleparallel Killing
vector fileds generated from .2c Hence we will not discuss the case (III).
59
Case (I):
In this case we have ( ) 0)()()()( ≠− tBtAtAtB tt and .02 =c Substituting 02 =c in
the system of equations (2.3.22) we get
).()(])(
)()(
)([
,)(
1)()(])(
)()(
)([
),()()()(
)()(
21
,)(),(
98122
3
3982
12222
1082
1221
151170
tDtxDctC
tBytC
tBzxX
ctB
txDtDxcztC
tByxtC
tBzxX
tDtDtAtCzc
tAtBzX
cxDzxKX
++⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
++++⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
++=
(2.3.23)
Differentiating equation (2.3.6) with respect to z and using (2.3.23), we get
.0),()()(
)()()()()()(2)()()()(
782
82
1
=−−
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−⎟
⎠⎞
⎜⎝⎛ −
−
zxKtDtC
tDA
tAtCtCtCtAA
tAtBtBtAcz
zxt
tttt
(2.3.24)
We solve this equation by letting ),()(),(0),( 141377 zDxDzxKzxK zx +=⇒= where
)(13 xD and )(14 zD are functions of integration. Substituting this value back in
(2.3.24) and differentiating with respect to ,z we get .01 =c We are also left with
equation which we will solve later on.
.0)()()()()()()()(2 828
2
=+⎟⎟⎠
⎞⎜⎜⎝
⎛ −tDtCtD
AtAtCtCtCtA
ttt (2.3.25)
Refreshing the system of equations with the help of above information, we get
),()(,)(
1)()(
),()()()(,)()(
9833
9822
1082
115
14150
tDtxDXctB
txDtDxX
tDtDtAtCzXczDxDX
+=++=
+⎟⎟⎠
⎞⎜⎜⎝
⎛−=++=
(2.3.26)
60
where ).()()( 111315 xDxDxD += Considering equation (2.3.9) and using equation
(2.3.26) we get
.0)()()()()()()()()()()( 98928214 =−−−− tDtCtCtDtCtxCtDtCtDtxCzD ttttz (2.3.27)
Differentiating this equation with respect to ,x we get ⇒=+ 0)()()()( 88 tDtCtDtC tt
.,)(
1)( 448 ℜ∈= cc
tCtD Now using this value in (2.3.27) and taking
ℜ∈=⇒= 551414 ,)(0)( cczDzDz we get .,
)(1)( 66
9 ℜ∈= cctC
tD Now substituting
back these values in (2.3.25) we get ( ) .0)()()()(4 =− tAtCtCtAc tt In order to solve
this equation we consider the following three cases:
(a) ( ) ,0)()()()( ≠− tAtCtCtA tt ,04 =c (b) ( ) ,0)()()()( =− tAtCtCtA tt ,04 ≠c
(c) ( ) ,0)()()()( =− tAtCtCtA tt .04 =c
We will discuss each case in turn. In case (I)(c) we will obtain the same teleparallel
Killing vector fields as in case (I)(b) with the reduction of one teleparallel Killing
vector fileds generated from .4c Hence we will not discuss the case (I)(c).
Case (I)(a):
In this case we have ( ) ,0)()()()( ≠− tBtAtAtB tt ( ) 0)()()()( ≠− tAtCtCtA tt and
.04 =c Substituting 04 =c with all the above information in system of equations
(2.3.26) we get
,)(
1,)(
1)(
1),(,)( 63
362101
16150 c
tCXc
tBcx
tCXtDXcxDX =+==+= (2.3.28)
where .,, 1551663 ℜ∈+= ccccc Considering equation (2.3.9) and using equation
(2.3.28) we get 715 )( cxD = and .,,
)(1)( 878
10 ℜ∈= ccctA
tD The line element for
61
Bianchi type II space-times is given in equation (2.3.1) and the solution of equations
from (2.2.4) to (2.3.12) is [66]
,)(
1,)(
1)(
1,)(
1, 63
362
81
90 c
tCXc
tBcx
tCXc
tAXcX =+=== (2.3.29)
where .,,, 1679863 ℜ∈+= cccccc The above space-times (2.3.1) admits four linearly
independent teleparallel Killing vector fields which are ,t∂∂ ,
)(1
xtA ∂∂
ytB ∂∂
)(1 and
).()(
1zy
xtC ∂
∂+
∂∂ Killing vector fields in general relativity are ,
y∂∂ ,
z∂∂
).(y
zx ∂
∂+
∂∂ One can easily see that the teleparallel Killing vector fields are different
from Killing vector fields in general relativity.
Case (I)(b):
In this case we have ( ) ,0)()()()( ≠− tBtAtAtB tt ( ) 0)()()()( =− tAtCtCtA tt and
.04 ≠c Equation ( ) ).()(0)()()()( tCtAtAtCtCtA tt =⇒=− System of equations
(2.3.26) become
,)(
1)(
1,)(
1)(
1)(
1
),()(
1,)(
643
36422
104
116
150
ctC
ctC
xXctB
cxtC
cxtC
X
tDctC
zXcxDX
+=++=
+−=+= (2.3.30)
where .,,, 16643 ℜ∈cccc Considering equation (2.3.9) and using (2.3.30) we get
715 )( cxD = and .,,
)(1)( 878
10 ℜ∈= ccctA
tD The line element for Bianchi type II
space-times takes the form
,)(2)]()([)()( 22222222222 dydztxBdztAtBxdytBdxtAdtds −++++−= (2.3.31)
Solution of equations from (2.2.4) to (2.3.12) is given as [66]
62
,)(
1)(
1,)(
1)(
1)(
1
,)(
1)(
1,
643
36422
841
170
ctA
ctA
xXctB
cxtA
cxtA
X
ctA
ctA
zXcX
+=++=
+−== (2.3.32)
where .,,,, 167178643 ℜ∈+= ccccccc The above space-time (2.3.31) admits five
linearly independent teleparallel Killing vector fields which can be written as ,t∂∂
,)(
1xtA ∂∂ ,
)(1
ytB ∂∂ )(
)(1
zyx
tA ∂∂
+∂∂ and ).(
)(1 2
zx
yx
xz
tA ∂∂
+∂∂
+∂∂
− Killing
vector fields in general relativity are ,y∂∂ ,
z∂∂ )(
yz
x ∂∂
+∂∂ and
).)22
(22
zx
yxz
xz
∂∂
+∂∂
++∂∂ One can easily compare that teleparallel Killing vector
fields are different from Killing vector fields in general relativity.
Case (II):
In this case we have ( ) 0)()()()( =− tBtAtAtB tt and .02 ≠c Equation
( ) ).()(0)()()()( tBtAtBtAtAtB tt =⇒=− The system of equations (2.3.22) can be
written as
).()()(
)(21]
)()(
)()([
,)(
1)()(
)(1]
)()(
21[]
)(1
)()(
)()([
),()(
1)()()(
)(1
21
,)(),(
9822
2122
3
3982
22
22
12222
102
82
121
151170
tDtxDctCtBxc
tCtBy
tCtBzxX
ctB
txDtDx
ctB
xtCtBxxc
tBz
tCtByx
tCtBzxX
tDctB
ytDtAtCzc
tBzX
cxDzxKX
++⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+++
+⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+−⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
++=
(2.3.33)
63
Differentiating (2.3.6) with respect to z then using (2.3.33) and remember that in this
case ),()( tBtA = we get ),()(),( 13127 zDxDzxK += where )(12 xD and )(13 zD are
functions of integration and
.0)()()()()()()()(2 828
2
=−⎟⎟⎠
⎞⎜⎜⎝
⎛ −− tDtCtD
AtAtCtCtCtA
ttt (2.3.34)
We will solve this equation later on. Refreshing the system of equations with the
above information, we get
).()()(
)(21]
)()(
)()([
,)(
1)()(
)(1]
)()(
21[]
)(1
)()(
)()([
),()(
1)()()(
)(1
21
,)()(
9822
2122
3
3982
22
22
12222
102
82
121
1513140
tDtxDctCtBxc
tCtBy
tCtBzxX
ctB
txDtDx
ctB
xtCtBxxc
tBz
tCtByx
tCtBzxX
tDctB
ytDtAtCzc
tBzX
czDxDX
++⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+++
+⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+−⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
++=
(2.3.35)
where ).()()( 121114 xDxDxD += Considering equation (2.3.9) and using equation
(2.3.35) then after differentiating with respect to y we get
( ) .0)()()()(1 =− tCtBtBtCc tt Here we have to discuss three possible cases:
(d) ( ) ,0)()()()( ≠− tBtCtCtB tt ,01 =c (e) ( ) ,0)()()()( =− tBtCtCtB tt ,01 ≠c
(f) ( ) ,0)()()()( =− tBtCtCtB tt .01 =c
We will discuss each case in turn. In case (II)(f) we will obtain the same teleparallel
Killing vector fields as in case (II)(e) with the reduction of one teleparallel Killing
vector fileds generated from .1c Hence we will not discuss the case (II)(f).
64
Case (II)(d):
In this case we have ),()( tBtA = ( ) 0)()()()( ≠− tCtBtBtC tt and .01 =c Substituting
all the above information in (2.3.9) and differentiating twice with respect to ,x we
reach to the equation ( ) .0)()()()(2 =− tCtBtBtCc tt In this case neither 02 =c nor
( ) .0)()()()( =− tCtBtBtC tt Hence this case leads to a contradiction and is not
possible.
Case (II)(e):
In this case we have ),()( tBtA = ( ) 0)()()()( =− tCtBtBtC tt and .01 ≠c Equation
( ) ).()(0)()()()( tCtBtCtBtBtC tt =⇒=− Therefore we have ).()()( tCtBtA ==
Now substituting all the above information in (2.3.9) and differentiating the resulting
equation first with respect to z then with respect to x we get ,)( 1613 czD =
ℜ∈= 1716178 ,,
)(1)( ccctB
tD and .,)(
1)( 18189 ℜ∈= cc
tBtD Refreshing the system of
equations (2.3.35) we get
.)(
1)(
1)(
121
)(1)(
,)(
1)(
1)(
1)(
1]21[
)(1][
),()(
1)(
1)(
121
,)(
181722
13
318172
23
122
102171
21
19140
ctB
ctB
xctB
xctB
yzxX
ctB
ctB
xctB
x
ctB
xxctB
zyxzxX
tDctB
yctB
zctB
zX
cxDX
+++−=
+++
+++−=
+−−⎟⎟⎠
⎞⎜⎜⎝
⎛−=
+=
(2.3.36)
Solve after considering equation (2.3.6) and using equation (2.3.36) we get
2014 )( cxD = and .,,
)(1)( 212021
10 ℜ∈= ccctB
tD The line element is this case
becomes
65
].2)1([)( 2222222 dzxdydzxdydxtAdtds −++++−= (2.3.37)
Refreshing the system of equations we get [66]
.)(
1)(
1)(
121
)(1)(
,)(
1)(
1)(
1)(
1)21(
)(1)(
,)(
1)(
1)(
1)(
121,
181722
13
318172
23
122
21217121
220
ctB
ctB
xctB
xctB
yzxX
ctB
ctB
xctB
x
ctB
xxctB
zyxzxX
ctB
ctB
yctB
zctB
zXcX
+++−=
+++
+++−=
+−−⎟⎟⎠
⎞⎜⎜⎝
⎛−==
(2.3.38)
where .,,,,,, 201922211817321 ℜ∈+= ccccccccc Here the above space-time (2.3.37)
admits seven linearly independent teleparallel Killing vector fields which can be
written as ,t∂∂ ,
)(1
xtA ∂∂ ),(
)(1
zyx
tA ∂∂
+∂∂ ),(
)(1 2
zx
yx
xz
tA ∂∂
+∂∂
+∂∂
−
),2
)2
(()(
1 23
zx
yxx
xy
tA ∂∂
+∂∂
++∂∂
− ytA ∂∂
)(1 and
].)()(2
[)(
1 22
zyxz
yzxyzx
xz
tA ∂∂
−+∂∂
+−+∂∂
− Killing vector fields in general
relativity are ,y∂∂ ,
z∂∂ )(
yz
x ∂∂
+∂∂ and ).)
22(
22
zx
yxz
xz
∂∂
+∂∂
++∂∂ One can
easily see that teleparallel Killing vector fields are different from general relativity in
this case and there are three more teleparallel Killing vector fields in teleparallel
theory of gravitation.
In the following we will consider now the case when .0\, ℜ∈=== λλCBA
The line element is this case after a suitable rescaling, becomes
.2)1( 222222 dzxdydzxdydxdtds −++++−= (2.3.39)
In this case the only non vanishing torsion components are .1 313
132 TT −=−= Our
system of equations (2.3.13) for the space-time (2.3.39) becomes
66
).,,(),,(
),,,(),,(),,(
),,,(),,,(
433
3432
2110
zxtPzxtPyX
zxtPzxtxPzxtPyxX
zytPXzyxPX
z
z
+−=
++−=
==
(2.3.40)
Substituting (2.3.40) in (2.3.12) and differentiating with respect to ,y we get
),,(),(),,(0),,( 2133 xtKxtKzzxtPzxtPzz +=⇒= where ),(1 xtK and ),(2 xtK are
functions of integration. Substituting the above value back in (2.3.12) we get
),,(),(),,(0),(),,( 41414 xtKxtKzxzxtPxtKxzxtPz +=⇒=− where ),(4 xtK is a
function of integration. Refreshing the system of equations (2.3.40) we get
),,(),(][),,(),(),(][
),,,(),,,(
413
24122
2110
xtKxtKyzxXxtKxtxKxtKzyxzxX
zytPXzyxPX
+−=
+++−=
==
(2.3.41)
Considering equation (2.3.8) and using equation (2.3.41) we get
.0),,(),(),( 221 =++ zytPxtKxtzK yxx Differentiating this equation first with respect to
z and then with respect to ,x we get ),()(),(0),( 2111 tDtDxxtKxtK xx +=⇒=
where )(1 tD and )(2 tD are functions of integration. Substituting back we get
),,(),()(),,(0),,()( 651221 ytKztKtDyzzytPzytPtD zy ++−=⇒=+ where
),(5 ztK and ),(6 ytK are functions of integration. Once again substituting back
these values in the above equation we get .0),(),( 62 =+ ytKxtK yx Differentiating this
equation with respect to ,x we get ),()(),(0),( 4322 tDtDxxtKxtK xx +=⇒= where
)(3 tD and )(4 tD are functions of integration. Substituting back and solving we get
),()(),( 536 tDtDyytK +−= where )(5 tD is a functions of integration. Substituting
all these information in (2.3.41) we get
67
).,())()(]([),()(),())()((][
),()(),()(),,,(
4213
4342122
53511
10
xtKtDtxDyzxXtDtxDxtxKtDtxDzyxzxX
tDtyDztKtDzyXzyxPX
++−=
+++++−=
+−+−=
=
(2.3.42)
Considering equation (2.3.10) and using equation (2.3.42) we get
0),(),()()()(]2[ 45321 =++−++− xtKztKtxDtDztDzxy xz (2.3.43)
Differentiating (2.3.43) with respect to ,y we get .0)(1 =tD Now differentiating
(2.3.43) with respect to ,z we get ⇒=+ 0),()( 52 ztKtD zz
),()()(21),( 76225 tDtzDtDzztK ++−= where )(6 tD and )(7 tD are functions of
integration. Now substituting back this value in (2.3.43) then differentiating with
respect to x and solving, we get ),()()(21),( 98324 tDtxDtDxxtK ++= where
)(8 tD and )(9 tD are functions of integration and
).()()(21),()()( 7822586 tDtzDtDzztKtDtD +−−=⇒−= Refreshing the system of
equations (2.3.42) we get
).()()(21)(][
),()()]()()(21[)(][
),()()()(21),,,(
983223
439832222
103822110
tDtxDtDxtDyzxX
tDtxDtDtxDtDxxtDzyxzxX
tDtyDtzDtDzXzyxPX
+++−=
++++++−=
+−−+−==
(2.3.44)
Considering equation (2.3.7) and using equation (2.3.7) we get
.0),,()()()( 1432 =−++ zyxPtDtDxtzD yttt Differentiating the above equation with
respect to z and ,t we get .,,)(0)( 212122 ℜ∈+=⇒= ccctctDtDtt Substituting
back this value in the above equation and differentiating with respect to ,z we get
),,(),(),,(),,( 871
11
1 zxKyxKzcyzyxPczyxP zy ++=⇒= where ),(7 yxK and
68
),(8 zxK are functions of integration. Substituting back these values in the above
equation and solving we get ,,,)( 43433 ℜ∈+= ccctctD
),()(),( 12113
7 yDxDcxyyxK ++= where )(11 xD and )(12 yD are functions of
integration, ℜ∈+= 65654 ,,)( ccctctD and ,,)( 775
12 ℜ∈+−= ccycyD Refreshing
the system of equations (2.3.44) we get
).()()(21)]([
,)()()](21[)(][
),()()()(21
,),()(
9843
221
3
65982
432
2122
1043
821
21
78
511
310
tDtxDctcxctcyzxX
ctctxDtDxctcxxxctczyxzxX
tDctcytzDctczX
czxKycxDcyxczyX
+++++−=
+++++++++−=
++−−++−=
++−++=
(2.3.45)
Considering equation (2.3.6) and using equation (2.3.45) then differentiating the
resulting equation with respect to y we get .03 =c Substituting back the above value
and differentiating the resulting equation first with respect to x then with respect to
z twice we get ),()()(),( 1312118 zDzxDxDzxK ++−= where )(13 zD is a function
of integration and .,,21)( 98981
212 ℜ∈++−= cccczczzD Substituting back the
above values we get ,)( 1088 ctczD += ℜ∈+= 1110119
10 ,,)( ccctctD and
.2)( 813 czzD = Refreshing the system of equations (2.3.45) we get
),()(21)]([
,)()(]21[)(][
,)()(21
,221
91084
221
3
659
1082
42
2122
119108212
41
789812
510
tDctcxcxctcyzxX
ctctxDctcxcxxxctczyxzxX
ctcctczctczycX
czcxcxzccxzcyczyX
+++++−=
+++++++++−=
+++−+−−=
++++−−=
(2.3.46)
where .,,,,,,,,, 111098765421 ℜ∈cccccccccc Considering equation (2.3.9) and using
equaiton (2.3.46) then differentiating the resulting equation with respect to ,y z and
69
x we get ,021 == cc 85 cc = and .,21)( 15158
29 ℜ∈+= cccttD Now considering
equations (2.3.6) and (2.3.7) and using (2.3.46) we get .085 == cc Refreshing the
system (2.3.46) teleparallel Killing vector fields are obtained as [66]
.21
,]21[
,,
1510423
615102
422
1191041
790
cxccxX
cxccxcxxxX
ctczcycXcxcX
++=
++++=
++−−=+=
(2.3.47)
where .,,,,,, 1511109764 ℜ∈ccccccc Here the above space-time (2.3.39) admits seven
linearly independent teleparallel Killing vector fields which can be written as ,t∂∂
,x∂∂ ),(
zyx
∂∂
+∂∂ ),( 2
zx
yx
xz
∂∂
+∂∂
+∂∂
− ),2
)2
((23
zx
yxx
xy
∂∂
+∂∂
++∂∂
− y∂∂
and ].)()(2
[ 22
zyxz
yzxyzx
xz
∂∂
−+∂∂
+−+∂∂
− Killing vector fields in general
relativity are ,y∂∂ ,
z∂∂ )(
yz
x ∂∂
+∂∂ and ).)
22(
22
zx
yxz
xz
∂∂
+∂∂
++∂∂ One can
easily see that teleparallel Killing vector fields are different and more in number from
Killing vector fields in general relativity.
2.4. Teleparallel Killing Vector Fields in Bianchi
Types VIII and IX Space-Times
Consider Bianchi types VIII and IX space-times in usual coordinates ),,,( zyxt
(labeled by ),,,,( 3210 xxxx respectively) with the line element [38, 78]
70
,)()(2)()()()()()( 2222222
dzdxygtAdzyGtBygtAdytBdxtAdtds
+++++−=
(2.4.1)
where A and B are no where zero functions of t only. The above space-time (2.4.1)
is called Bianchi type VIII if yyg cosh)( = and yyG sinh)( = while (2.4.1) becomes
Bianchi type IX when yyg cos)( = and .sin)( yyG = The above space-times admit
four linearly independent Killing vector fields in general relativity which are ,x∂∂
,z∂∂
zz
yGyg
yz
xz
yG ∂∂
−∂∂
+∂∂ sin
)()(cossin
)(1
and
.cos)()(sincos
)(1
zz
yGyg
yz
xz
yG ∂∂
+∂∂
+∂∂
− The tetrad components and its inverse can be
obtained by using the relation (1.3.4) as [67]
,
sin)(0000)(00
cos)(0)(00001
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=
ytBtB
ytAtAS a
µ
⎟⎟⎟⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜⎜⎜⎜
⎝
⎛−
=
ytB
tB
ytBy
tASa
sin)(1000
0)(
100
sin)(cos0
)(10
0001
µ (2.4.2)
Using equation (1.3.5), the corresponding non-vanishing Weitzenböck connections
are obtained as [67]
,2
101
AA•
=Γ ,2
202
BB•
=Γ ,2
303
BB•
=Γ ,)()(
323
yGyg
=Γ ,)(21
301
BB
AAyg
••
−=Γ
,)(
132
1
yG−=Γ where dot denotes the derivative with respect to .t The non vanishing
torsion components by using equation (1.3.11) are ,2
011
AAT
•
=
,)(21
031
BB
AAygT
••
−= ,2
022
BBT
•
= ,2
033
BBT
•
= ,)(
123
1
yGT −= .
)()(
233
yGygT =
71
A vector field X is said to be teleparallel Killing vector field if it satisfies equation
(1.3.17). One can write (1.3.17) explicitly using (2.4.1) and the above torsion
components as:
,0,,0, 22
00 == XX (2.4.3)
,0,)(, 13
11 =+ XygX (2.4.4)
,0,2,2 22
00
2 =+− • XBXXB (2.4.5)
,0)()(,, 32,
32
11
2 =−++ XyGAXygAXAXB (2.4.6)
,0)(21)(
21, 3
0,31
1,0
01 =+++− •• XygAXygAXAXXA (2.4.7)
,0)()(21
)(21)()(,)(
32
13,
00,
3220
1
=++
+−++
••
•
XyGBygA
XygAXXyBGyAgXygA (2.4.8)
,0)()()(,)( 3,3
3,1
1,322
11 =++++ XygAAXXyBGyAgXygA (2.4.9)
,0)()()()()(,)( 33,
22,
3222
1 =−−+++ XyGygBABXXyBGyAgXygA (2.4.10)
.0)()(,)( 3,322
31 =++ XyBGyAgXygA (2.4.11)
Integrating equation (2.4.3) we get
,),,(),,,( 2210 zxtEXzyxEX == (2.4.12)
where ),,(1 zyxE and ),,(2 zxtE are functions of integration which are to be
determined. Considering equation (2.4.9) and using equation (2.4.4) we get
.0)()( 3,3
3,1
1,32 =++ XygXAXyGB (2.4.13)
Now differentiating equation (2.4.4) with respect to ,z we get
0,)(, 133
131 =+ XygX (2.4.14)
72
Differentiating equation (2.4.13) with respect to x and using equation (2.4.14) we get
.011,3 =X Now integrating this equation with respect to ,x we get
),,,(),,( 433 zytEzytExX += (2.4.15)
where ),,(3 zytE and ),,(4 zytE are functions of integration. Substituting back
equation (2.4.15) into equation (2.4.4), we get
),,,(),,()( 531 zytEzytEyxgX +−= (2.4.16)
where ),,(5 zytE is a functions of integration which is to be determined. We have the
following system of equations:
).,,(),,(,),,(),,,(),,()(),,,(
43322
53110
zytEzytExXzxtEXzytEzytEyxgXzyxEX
+==
+−== (2.4.17)
Considering equation (2.4.5) and using equation (2.4.3) then differentiating the
resulting equation with respect to y we get
),,(),(),,(0),,( 2111 zxFzxFyzyxEzyxEyy +=⇒= where ),(1 zxF and ),(2 zxF
are functions of integration. Refreshing the system of equations (2.4.17) we get
).,,(),,(,),,(),,,(),,()().,,(),(
43322
531210
zytEzytExXzxtEXzytEzytEyxgXzxFzxFyX
+==
+−=+= (2.4.18)
Considering equation (2.4.6) and using equation (2.4.18) then differentiating with
respect to ,x we get ),,(),(),,(0),,( 4322 ztFztxFzxtEzxtExx +=⇒= where
),(3 ztF and ),(4 ztF are functions of integration. Refreshing (2.4.18) we get
).,,(),,(),,(),(),,,(),,()().,,(),(
433432
531210
zytEzytExXztFztxFXzytEzytEyxgXzxFzxFyX
+=+=
+−=+= (2.4.19)
Considering equation (2.4.7)and using equation (2.4.19) then differentiating the
resulting equation with respect to ,x we get .0),(),( 21 =+ zxFzxyF xxxx Differentiating
the above equation with respect to ,y we get )()(),(0),( 2111 zKzxKzxFzxFxx +=⇒=
73
and ),()(),(0),( 4322 zKzxKzxFzxFxx +=⇒= where ),(1 zK ),(2 zK )(3 zK and
)(4 zK are functions of integration. Substituting these values in (2.4.19) we get
( )
).,,(),,(),,(),(
),,,(),,()(),()()()(
433
432
531
43210
zytEzytExXztFztxFX
zytEzytEyxgXzKzxKzKzxKyX
+=
+=
+−=
+++=
(2.4.20)
Considering equation (2.4.6) and using equation (2.4.20) then differentiating the
resulting equation with respect to y we get
( ) ),,(),(),,()(),,(0),,()(),,( 654522,
45 ztFztyFzytEygzytEzytEygzytE ++−=⇒=+
where ),(5 ztF and ),(6 ztF are functions of integration. Equation (2.4.20) becomes
( )
).,,(),,(),,(),(
),,(),(),,()(),,()(),()()()(
433
432
65431
43210
zytEzytExXztFztxFX
ztFztyFzytEygzytEyxgXzKzxKzKzxKyX
+=
+=
++−−=
+++=
(2.4.21)
Considering equation (2.4.11) and using (2.4.21) then differentiating with respect to
,x we get ),,(),,(0),,( 733 ytFzytEzytEz =⇒= where ),(7 ytF is a function of
integration. Now using these information in (2.4.8) and differentiate first with respect
to x then with respect to ,z we get .0)()( 31 =+ zKzyK zzzz Solving this equation by
differentiating with respect to ,y we get .,,)(0)( 212111 ℜ∈+=⇒= ccczczKzK zz
Substituting back we get .,,)(0)( 434333 ℜ∈+=⇒= ccczczKzK zz The system of
equations (2.4.21) become
( )
),,,(),,(),,(),(),,(),()],,(),([)(
),()()()(
473432
65471
443
221
0
zytEzytFxXztFztxFXztFztyFzytEytxFygX
zKczcxzKczcxyX
+=+=
++−−=
+++++=
(2.4.22)
where .,,, 4321 ℜ∈cccc Considering equation (2.4.10) and using (2.4.22) then
differentiating the resulting equation first with respect to x then with respect to ,z we
74
get ),()(),(0),( 2133 tRtzRztFztFzz +=⇒= where )(1 tR and )(2 tR are functions of
integration. Also considering equation (2.4.10) and using equation (2.4.22) then
differentiating the resulting equation first with respect to x then with respect to ,y
we get
.0),())()((),()()(3),()( 722772 =−++ ytFyGygytFyGygytFyG yyy (2.4.23)
We will make use of this equation later on. Refreshing the system (2.4.22) we get
( )
).,,(),,(),,()]()([),,(),()],,(),([)(
),()()()(
4734212
65471
443
221
0
zytEzytFxXztFtRtzRxXztFztyFzytEytxFygX
zKczcxzKczcxyX
+=++=
++−−=
+++++=
(2.4.24)
Considering equation (2.4.9) and using equation (2.4.24), we get
.0),(),(),()( 6572 =++ ztAFztFyAytFyGB zz (2.4.25)
Differentiating (2.4.25) with respect to ,z we get .0),(),( 65 =+ ztFztyF zzzz Solving
this equation by differentiating with respect to ,y we get
),()(),(0),( 4355 tRtzRztFztFzz +=⇒= where )(3 tR and )(4 tR are functions of
integration. Substituting back the above value we get ⇒= 0),(6 ztFzz
),()(),( 656 tRtzRztF += where )(5 tR and )(6 tR are functions of integration. Using
the above values in (2.4.25) and differentiating twice with respect to ,y we get
.0))()()(,(2),()()(4),()( 227772 =−++ yGygytFytFyGygytFyG yyy (2.4.26)
Now subtracting equation (2.4.23) from (2.4.26) we get
.0))()()(,(),()()( 2277 =−+ yGygytFytFyGyg y (2.4.27)
Suppose ,0),(7 ≠ytF then ),()()(
1),()()(
)()(),(),( 77
22
7
7
tRyGyg
ytFyGyg
ygyGytFytFy =⇒
−=
where )(7 tR is the function of integration. Now substituting back all the above
information in (2.4.9) and differentiating the resulting equation with respect to ,y we
75
get 0)(7 =tR which implies ,0),(7 =ytF a contradiction to our assumption. Hence
we must take .0),(7 =ytF From equation (2.4.25) we see that ⇒= 0)(3 tR
),(),( 45 tRztF = ).(),(0)( 665 tRztFtR =⇒= Substituting back the above values in
(2.4.10) and differentiating with respect to ,x we get .0)(1 =tR Also substituting
back the above values in (2.4.5) and differentiating first with respectg to x then with
respcect to ,z we get .01 =c Refreshing (2.4.24) with all the abvoe information, we
get
( )
).,,(),,()(),()(),,()(
),()()(
43422
6441
443
22
0
zytEXztFtxRXtRtyRzytEygX
zKczcxzKxcyX
=+=
++−=
++++=
(2.4.28)
Considering equation (2.4.11) and using (2.4.28) then solving we get
⇒= 0),,(4 zytEz ),,(),,( 84 ytFzytE = where ),(8 ytF is a function of integration.
Refreshing the system of equations (2.4.28) with the help of above information, we
get
( )
).,(),,()(),()(),()(
),()()(
83422
6481
443
22
0
ytFXztFtxRXtRtyRytFygX
zKczcxzKxcyX
=+=
++−=
++++=
(2.4.29)
Differentiating equation (2.4.10) first with respect to x then with respect to z and
using (2.4.29) we get ),()(),(0),( 9844 tRtzRztFztFzz +=⇒= where )(8 tR and
)(9 tR are functions of integration. The system of equations (2.4.29) become
( )
),,(),()()(),()(),()(
),()()(
8310822
6481
443
22
0
ytFXtRtzRtxRXtRtyRytFygX
zKczcxzKxcyX
=++=
++−=
++++=
(2.4.30)
where ).()()( 9610 tRtRtR += Now considering equation (2.4.7) and using equation
(2.4.30) then differentiating the resulting equation with respect to z we get .03 =c
Also considering equation (2.4.8) and using equation (2.4.30) then differentiating the
76
resulting equation with respect to z we get .0)()( 42 =+ zKzyK zzzz Solving this
equation by differentiating with respect to ,y we get ⇒= 0)(2 zK zz
.,,)( 65652 ℜ∈+= ccczczK Substituting back we get ,)(0)( 87
44 czczKzKzz +=⇒=
., 87 ℜ∈cc Refreshing equation (2.4.30) we get
( )
),,(),()()(),()(),()(
,
8310822
6481874652
0
ytFXtRtzRtxRXtRtyRytFygX
czcxcczcxcyX
=++=
++−=
+++++=
(2.4.31)
where .,,,,, 876542 ℜ∈cccccc Considering equation (2.4.6) and using equation
(2.4.31) we get
.0)()( 42 =+ tARtBR (2.4.32)
In order to sovle this equation we will consider the following possibilities:
(I) ,BA ≠ constant≠A and constant.≠B
(II) ,BA ≠ constant≠A and .0\, ℜ∈= ηηB
(III) ,BA ≠ 0\, ℜ∈= λλA and constant.≠B
(IV) ,BA = constant≠A and constant.≠B
(V) ,BA ≠ 0\, ℜ∈= λλA and .0\, ℜ∈= ηηB
(VI) .0\, ℜ∈== λλBA
We will discuss each case in turn.
Case I:
In this case we have ,BA ≠ constant≠A and constant.≠B From equation (2.4.32)
we get .0)()( 42 == tRtR Substituting the above values in (2.4.5) and differentiating
with respect to ,x we get .02 =c Now substituting 02 =c in (2.4.5) and
differentiating with respect to ,z we get ⇒=−+ 02)()(2 588 ctRBtBR tt 05 =c and
77
.,)(
1)( 998 ℜ∈= cc
tBtR Substituting all the above information back in (2.4.5) we
get ⇒=−+ 02)()(2 61010 ctRBtBR tt 06 =c and .,
)(1)( 1010
10 ℜ∈= cctB
tR Also
using the above information in (2.4.7) we get ⇒=−+ 0)(21)( 4
66 ctRAtAR tt 04 =c
and .,)(
1)( 11116 ℜ∈= cc
tAtR Refreshing the system of equations (2.4.31) we get
),,(,)(
1)(
1
,)(
1),()(,
83109
2
1181
870
ytFXctB
ctB
zX
ctA
ytFygXczcX
=+=
+−=+=
(2.4.33)
Considering equation (2.4.8) and using equation (2.4.33) we get
⇒=−⎥⎦⎤
⎢⎣⎡ + 0)(),(
21),( 7
288 cyGytFBytBF tt 07 =c and 0),(21),( 88 =+ ytFBytBF tt
.)(),( 21
18 −=⇒ ByDytF Considering equation (2.4.10) and using equation (2.4.33)
with all the above information we get
.)(
1)()(
)(1ln)(
)()()()()( 9
1911
yGyGyg
yGcyD
yGc
yDygyDyG y ⎥⎦
⎤⎢⎣
⎡+=⇒−=+ The
solution of equations (2.4.3) to (2.4.11) is given as [67]
,)(
1)()(
)(1ln
)(,
)(1
)(1
,)(
1)()(
)()(
)(1ln
)(
,
93109
2
1191
80
yGyGyg
yGtBc
XctB
ctB
zX
ctAyG
ygyGyg
yGtBc
X
cX
⎥⎦
⎤⎢⎣
⎡+=+=
+⎥⎦
⎤⎢⎣
⎡+−=
=
(2.4.34)
where .,,, 111098 ℜ∈cccc In this case the space-time (2.4.1) admits four linearly
independent teleparallel Killing vector fields, which can be written as ,t∂∂ ,1
xA ∂∂
78
yB ∂∂1 and ].
)(1
)()(
)(1ln
)()(
)()(
)(1ln[1
zyGyGyg
yGyz
xyGyg
yGyg
yGB ∂∂
++∂∂
+∂∂
+−
Killing vector fields in general relativity are ,x∂∂ ,
z∂∂
zz
yGyg
yz
xz
yG ∂∂
−∂∂
+∂∂ sin
)()(cossin
)(1 and .cos
)()(sincos
)(1
zz
yGyg
yz
xz
yG ∂∂
+∂∂
+∂∂
− On
comparison to the Killing vector fields in general relativity we see that all the
teleparallel Killing vector fields are different from Killing vector fields in general
relativity.
Case II:
In this case we have ,BA ≠ constant≠A and .0\, ℜ∈= ηηB From the above
equation (2.4.32) we get .0)()( 42 == tRtR Substituting these values in (2.4.5) and
differentiating with respect to ,x we get .02 =c Now substituting 02 =c in (2.4.5)
and differentiating with respect to ,z we get 05 =c and .,)( 998 ℜ∈= cctR
Substituting all the above information back in (2.4.5) we get 06 =c and
.,)( 101010 ℜ∈= cctR Also using the above information in (2.4.7) we get 04 =c and
.,)(
1)( 11116 ℜ∈= cc
tAtR Refreshing the system of equations (2.4.31) we get
),,(,
,)(
1),()(,
83109
2
1181
870
ytFXczcX
ctA
ytFygXczcX
=+=
+−=+= (2.4.35)
Considering equation (2.4.8) and using equation (2.4.35) we get 07 =c and
).((),(0),( 2188 yDytDytFytFtt +=⇒= Substituting the above information and the
system of equations (2.4.35) in (2.4.8) we get 0)(1 =yD and .07 =c Now using the
above values in (2.4.10) we get
79
.)(
1)()(
)(1ln)(
)()()()()( 9
2922
yGyGyg
yGcyD
yGc
yDygyDyG y ⎥⎦
⎤⎢⎣
⎡+=⇒−=+ The
solution of equations (2.4.3) to (2.4.11) is given as
,)(
1)()(
)(1ln,
,)(
1)()(
)()(
)(1ln,
93
1092
1191
80
yGyGyg
yGcXczcX
ctAyG
ygyGyg
yGcXcX
⎥⎦
⎤⎢⎣
⎡+=+=
+⎥⎦
⎤⎢⎣
⎡+−==
(2.4.36)
where .,,, 111098 ℜ∈cccc In this case the line element for Bianchi type VIII and IX
space-time becomes
,)()(2)()()()( 2222222
dzdxygtAdzyGygtAdydxtAdtds
+++++−= ηη
(2.4.37)
The above space-time (2.4.37) admits four linearly independent teleparallel Killing
vector fields, which can be written as ,t∂∂ ,1
xA ∂∂
y∂∂ and
.)(
1)()(
)(1ln
)()(
)()(
)(1ln
zyGyGyg
yGyz
xyGyg
yGyg
yG ∂∂
++∂∂
+∂∂
+− Killing vector
fields in general relativity are ,x∂∂ ,
z∂∂
zz
yGyg
yz
xz
yG ∂∂
−∂∂
+∂∂ sin
)()(cossin
)(1 and
.cos)()(sincos
)(1
zz
yGyg
yz
xz
yG ∂∂
+∂∂
+∂∂
− On comparison to the Killing vector fields in
general relativity we see that all the teleparallel Killing vector fields are different
from Killing vector fields in general relativity.
Case III:
In this case we have ,BA ≠ 0\, ℜ∈= λλA and constant.≠B In this case the
line element for Bianchi type VIII and IX space-time becomes
,)(2)()()()( 2222222 dzdxygdzyGtBygdytBdxdtds λλλ +++++−= (2.4.38)
80
This case can be solved similarly as the two cases above and the result is given as
[67]
,)(
1)()(
)(1ln
)(,
)(1
)(1
,)()(
)()(
)(1ln
)(
,
93109
2
1191
80
yGyGyg
yGtBc
XctB
ctB
zX
cyGyg
yGyg
yGtBc
X
cX
⎥⎦
⎤⎢⎣
⎡+=+=
+⎥⎦
⎤⎢⎣
⎡+−=
=
(2.4.39)
where .,,, 111098 ℜ∈cccc In this case the above space-time (2.4.38) admits four
linearly independent teleparallel Killing vector fields, which can be written as ,t∂∂
,x∂∂
yB ∂∂1 and ].
)(1
)()(
)(1ln
)()(
)()(
)(1ln[1
zyGyGyg
yGyz
xyGyg
yGyg
yGB ∂∂
++∂∂
+∂∂
+−
Killing vector fields in general relativity are ,x∂∂ ,
z∂∂
zz
yGyg
yz
xz
yG ∂∂
−∂∂
+∂∂ sin
)()(cossin
)(1 and .cos
)()(sincos
)(1
zz
yGyg
yz
xz
yG ∂∂
+∂∂
+∂∂
− On
comparison to the Killing vector fields in general relativity we see that in this case
only one teleparallel Killing vector field is the same as in general relativity and all
others are different from Killing vector fields in general relativity.
Case IV:
In this case we have ,BA = constant≠A and constant.≠B From the above
equation (2.4.32) we get ).()( 42 tRtR −= Using these information in (2.4.7) and solve
after differentiating with respect to ,y we get 02 =c and ℜ∈= 994 ,
)(1)( cc
tAtR
which in turn implies .)(
1)( 92 c
tAtR −= Now substituting these information in
81
(2.4.7) and solving we get 04 =c and .,)(
1)( 10106 ℜ∈= cc
tAtR Our system of
equations (2.4.31) becomes
( )
),,(),()()(
1
,)(
1)(
1),()(
,
831089
2
10981
87650
ytFXtRtzRctA
xX
ctA
ctA
yytFygX
czcczcyX
=++−=
++−=
+++=
(2.4.40)
Considering equation (2.4.5) and using equation (2.4.40) then solve after
differentiating with respect to ,z we get 05 =c and .,)(
1)( 11118 ℜ∈= cc
tAtR Now
using this value in (2.4.5) we get 06 =c and .,)(
1)( 121210 ℜ∈= cc
tAtR Now using
all the above information in (2.4.8) and solving we get 07 =c and ,)()(),(
18
tAyDytF =
where )(1 yD is a function of integration. Now using the above information in
(2.4.10) we get [ ] ⇒−−=)(
1)()()(sin 1192,
1
yGc
yGygcyyD
.)(
1)()(
)(1ln)(ln)( 119
1
yGyGyg
yGcyGcyD ⎥
⎦
⎤⎢⎣
⎡++−= In this case the line element for
Bianchi type VIII and IX space-time becomes
,)()(2)]()([)(])[( 2222222 dzdxygtAdzyGygtAdydxtAdtds +++++−= (2.4.41)
Teleparallel Killing vector fields for the above space-time is given as [67]
82
,)(
1)()(
)(1ln)(ln1
,
,)()(
)()(
)(1ln)(ln1
,
1193
121192
109119
1
80
yGyGyg
yGcyGc
AX
Ac
Azc
Axc
X
Ac
Acy
yGyg
yGyg
yGcyGc
AX
cX
++=
++−=
+−++−−=
=
(2.4.42)
where .,,,, 12111098 ℜ∈ccccc The above space-time (2.4.41) admits five linearly
independent teleparallel Killing vector fields, which can be written as ,t∂∂ ,1
xA ∂∂
,1yA ∂∂ ]
)(1
)()(
)(1ln
)()(
)()(
)(1ln[1
zyGyGyg
yGyz
xyGyg
yGyg
yGA ∂∂
++∂∂
+∂∂
+−
and ].)(ln1)(ln1[z
yGAyA
xxA
yyGA ∂
∂+
∂∂
−∂∂
− Killing vector fields in
general relativity are ,x∂∂ ,
z∂∂
zz
yGyg
yz
xz
yG ∂∂
−∂∂
+∂∂ sin
)()(cossin
)(1 and
.cos)()(sincos
)(1
zz
yGyg
yz
xz
yG ∂∂
+∂∂
+∂∂
− On comparison to the Killing vector fields in
general relativity we see that all the teleparallel Killing vector fields are different
from Killing vector fields in general relativity. It is important to note that in this case
the space-time admits one more teleparallel Killing vector field than Killing vector
fields in general relativity.
Case V:
In this case we have ,BA ≠ 0\, ℜ∈= λλA and .0\, ℜ∈= ηηB From the
above equation (2.4.32) we get ).()( 42 tRtRηλ
−= Using these information in (2.4.7)
and solve after differentiating with respect to ,y we get 02 =c and
ℜ∈= 994 ,)( cctR which in turn implies .)( 9
2 ctRηλ
−= Now substituting these
83
information in (2.4.7) and solving we get 04 =c and .,)( 10106 ℜ∈= cctR Our system
of equations (2.4.31) becomes
( )
),,(),()(
,),()(,
831089
2
10981
87650
ytFXtRtzRxcX
cycytFygXczcczcyX
=++−=
++−=+++=
ηλ (2.4.43)
Considering equation (2.4.5) and using eqution (2.4.43) then solve after
differentiating with respect to ,z we get 05 =c and .,)( 11118 ℜ∈= cctR Now using
this value in (2.4.5) and solving we get 06 =c and .,)( 121210 ℜ∈= cctR Now using
all the above information in (2.4.8) and solving we get 07 =c and ),(),( 18 yDytF =
where )(1 yD is a function of integration. Now using the above information in
equation (2.4.10) we get [ ] ⇒−−=)(
1)()()(sin 1192,
1
yGc
yGygcyyD
ηλ
.)(
1)()(
)(1ln)(ln)( 119
1
yGyGyg
yGcyGcyD ⎥
⎦
⎤⎢⎣
⎡++−=
ηλ In this case the line element
for Bianchi type VIII and IX space-time becomes
,)(2)()( 2222222 dzdxygdzyGygdydxdtds ληληλ +++++−= (2.4.44)
Teleparallel Killing vector fields for the above space-time are given as [67]
.)(
1)()(
)(1ln)(ln
,
,)()(
)()(
)(1ln)(ln,
1193
121192
1091191
80
yGyGyg
yGcyGcX
czccxX
ccyyGyg
yGyg
yGcyGcXcX
++−=
++−=
++++−−==
ηλ
ηλ
ηλ
(2.4.45)
where .,,,, 12111098 ℜ∈ccccc The above space-time (2.4.44) admits five linearly
independent teleparallel Killing vector fields, which can be written as ,t∂∂ ,
x∂∂
84
,y∂∂ ]
)(1
)()(
)(1ln
)()(
)()(
)(1ln[
zyGyGyg
yGyz
xyGyg
yGyg
yG ∂∂
++∂∂
+∂∂
+− and
].))(
1()(ln)(ln[zyG
yGy
xx
yyG∂∂
−+∂∂
−∂∂
+ηλ
ηλ
ηλ Killing vector fields in
general relativity are ,x∂∂ ,
z∂∂
zz
yGyg
yz
xz
yG ∂∂
−∂∂
+∂∂ sin
)()(cossin
)(1 and
.cos)()(sincos
)(1
zz
yGyg
yz
xz
yG ∂∂
+∂∂
+∂∂
− On comparison to the Killing vector fields in
general relativity we see that only one teleparallel Killing vector field is the same and
all others are different from Killing vector fields in general relativity. It is important
to note that in this case the space-time admits one more teleparallel Killing vector
field.
Case VI:
In this case we have .0\, ℜ∈== λλBA The space-time in this case becomes
,)(2)()()( 2222222 dzdxygdzyGygdydxdtds λλλ +++++−= (2.4.46)
This case can be solved exactly the same as case (IV). Teleparallel Killing vector
fields in this case are given as
.)(
1)()(
)(1ln)(ln
,
,)()(
)()(
)(1ln)(ln,
1193
121192
1091191
80
yGyGyg
yGcyGcX
czccxX
ccyyGyg
yGyg
yGcyGcXcX
++−=
++−=
++++−−==
(2.4.47)
where .,,,, 12111098 ℜ∈ccccc The above space-time (2.4.46) admits five linearly
independent teleparallel Killing vector fields, which can be written as ,t∂∂ ,
x∂∂
,y∂∂ ]
)(1
)()(
)(1ln
)()(
)()(
)(1ln[
zyGyGyg
yGyz
xyGyg
yGyg
yG ∂∂
++∂∂
+∂∂
+− and
85
].))(
1()(ln)([lnzyG
yGy
xx
yyG∂∂
−+∂∂
−∂∂
+ Killing vector fields in general
relativity are ,x∂∂ ,
z∂∂
zz
yGyg
yz
xz
yG ∂∂
−∂∂
+∂∂ sin
)()(cossin
)(1 and
.cos)()(sincos
)(1
zz
yGyg
yz
xz
yG ∂∂
+∂∂
+∂∂
− On comparison to the Killing vector fields in
general relativity we see that only one teleparallel Killing vector field is the same and
all others are different from Killing vector fields in general relativity. It is important
to note that in this case the space-time admits one more teleparallel Killing vector
field.
2.5. Summary of the Chapter
In this chapter we explored Killing vector fields for Bianchi types I, II, VIII and IX
space-times in the context of teleparallel theory of gravitation using direct integration
technique. Following results are obtained from the above study:
(1) In Bianchi type I: Different possibilities for the existence of teleparallel Killing
vector fields have been found by using direct integration technique. It turns out that
the above space-times admit 7, 8 or 10 teleparallel Killing vector fields. When the
space-time becomes Minkowski then all the torsion components become zero and the
teleparallel Lie derivative for the metric gives the same equations as in general
relativity, hence the Killing vector fields are same in both the theories. The results for
above space time when it admits seven independent teleparallel Killing vector fields
are given in equations (2.2.24) and (2.2.36). When the above space time admits eight
independent teleparallel Killing vector fields, results are given in equations (2.2.49),
(2.2.75) and (2.2.88). When the above space time admits ten independent teleparallel
Killing vector fields, result is given in equation (2.2.62). It is important to note that
even the space time possesses ten linearly independent teleparallel Killing vector
fields, it does not become Minkowski and has torsion.
86
(2) In Bianchi type II: Teleparallel Killing vector fields have been explored by direct
integration technique. The above space-time admits teleparallel Killing vector fields
in different cases. It turns out that the above space-times admit 4, 5 or 7 teleparallel
Killing vector fields. It is important to note that when all the metric functions are
constant the torsion does not become zero and hence the teleparallel Killing vector
fields are different than Minkowski space-times in general relativity. When the above
space time admits four independent teleparallel Killing vector fields, result is given in
equation (2.3.29). The result for five independent teleparallel Killing vector fields is
given in equation (2.3.32) and the results for seven independent teleparallel Killing
vector fields are given in equations (2.3.38) and (2.3.47).
(3) In Bianchi type VIII and IX: Teleparallel Killing vector fields have been shown
by using direct integration technique. It turns out that the above space-times admit 4
or 5 teleparallel Killing vector fields. It is important to note that when all the metric
functions are constant the torsion does not become zero and hence the teleparallel
Killing vector fields are different than Killing vector fields in general relativity due to
non vanishing of torsion components. A brief comparison of Killing vector fields in
both the theories for Bianchi types VIII and IX space-times are also given and it
comes out that Killing vector fields in both the theories are different. The result for
four independent teleparallel Killing vector fields are given in equations (2.4.34),
(2.4.36) and (2.4.39). Also the results for five independent teleparallel Killing vector
fields are given in equations (2.4.42), (2.4.45) and (2.4.47).
87
Chapter 3
Teleparallel Killing Vector Fields in
Kantowski-Sachs, Bianchi Type III, Static
Cylindrically Symmetric and Spatially
Homogeneous Rotating Space-Times
3.1. Introduction
This chapter is devoted to investigate teleparallel Killing vector fields in Kantowski-
Sachs, Bianchi type III, static cylindrically symmetric and spatially homogenous
rotating space-times by using direct integration technique. This chapter is organized
as follows: In section (3.2) Killing vector fields of Kantowski-Sachs and Bianchi type
III space-times are investigated. In the next section (3.3) Killing vector fields in
cylindrically symmetric static space-times in the context of teleparallel theory have
been explored. In section (3.4) Killing vector fields of spatially homogeneous rotating
space-times are explored in the context of teleparallel theory. In the same section we
have also discussed some special cases of spatially homogeneous rotating space-
times. Last section (3.5) of the chapter is devoted to a detailed summary of the work.
88
3.2. Teleparallel Killing Vector Fields in
Kantowski-Sachs and Bianchi Type III
Space-Times
Consider Kantowski-Sachs and Bianchi type III space-times in usual coordinates
),,,( φθrt (labeled by ),,,,( 3210 xxxx respectively) with the line element [77]
),)(()()( 22222222 φθθ dfdtBdrtAdtds +++−= (3.2.1)
where A and B are no where zero functions of t only. For θθ sinh)( =f the space-
time (3.2.1) represents Bianchi type III space-time and for θθ sin)( =f it becomes
Kantowski-Sachs space-times. The above space-times (3.2.1) admit at least four
independent Killing vector fields [79], which are
,r∂∂ ,
φ∂∂ ,sin
)()(cos
φφ
θθ
θφ
∂∂′
−∂∂
ff ,cos
)()(sin
φφ
θθ
θφ
∂∂′
+∂∂
ff (3.2.2)
where prime denotes the derivative with respect to .θ The tetrad components µaS
and its inverse µaS can be obtained by using the relation (1.3.4) as [68]
)),()(),(),(,1(diag θµ ftBtBtAS a = ).)()(
1,)(
1,)(
1,1(diagθ
µ
ftBtBtASa = (3.2.3)
The corresponding non-vanishing Weitzenböck connections for (3.2.3) are obtained
by using the relation (1.3.5) as
,101
AA•
=Γ ,202
BB•
=Γ ,303
BB •
=Γ (3.2.4)
where “dot” denotes the derivative with respect to .t The non vanishing torsion
components can be obtained by using (1.3.11) as [68]
89
,011
AAT
•
= ,022
BBT
•
= .033
BBT
•
= (3.2.5)
A vector field X is said to be teleparallel Killing vector field if it satisfies equation
(1.3.17). One can write (1.3.17) explicitly using (3.2.1) and (3.2.5) as
,0,0,0 2,2
1,1
0,0 === XXX (3.2.6)
,02,12
1,22 =+ XAXB (3.2.7)
,0)( 1,322
3,12 =+ XfBXA θ (3.2.8)
,0)( 2,32
3,2 =+ XfX θ (3.2.9)
,0)()(
3,32 =+
′XX
ffθθ (3.2.10)
,011,
00,
12 =+− • XAAXXA (3.2.11)
,022,
00,
22 =+− • XBBXXB (3.2.12)
.0)()( 323,
00,
322 =+− • XfBBXXfB θθ (3.2.13)
Equation (3.2.6) and (3.2.10) gives
),,,(),,()()(,),,(
),,,(),,,(
43332
2110
θφφθθφ
φθφθ
rtPdrtPffXrtPX
tPXrPX
+′
−==
==
∫ (3.2.14)
where ),,,(1 φθrP ),,,(2 φθtP ),,(3 φrtP and ),,(4 θrtP are functions of integration
which are to be determined. Considering equation (3.2.7) and using equation (3.2.14)
then differentiating the resulting equation with respect to ,θ we get
),,(),(),,(0),,( 2122 φφθφθφθθθ tKtKtPtP +=⇒= where ),(1 φtK and ),(2 φtK are
functions of integration. Also differentiating (3.2.7) with respect to ,r we get
),,(),(),,(0),,( 4333 φφφφ tKtKrrtPrtPrr +=⇒= where ),(3 φtK and ),(4 φtK are
90
functions of integration. Now substituting back the above information in (3.2.7) we
get
.0),(),( 3212 =+ φφ tKBtKA (3.2.15)
In order to solve this equation we impose the following conditions.
(I) ),(tAA = )(tBB = and ,BA ≠ (II) )(tAA = and 0),\R( ∈= ηηB
(III) )(tBB = and 0),\R( ∈= ηηA (IV) ),(tAA = )(tBB = and ,BA =
(V) β=A and η=B where 0).\R,( ∈ηβ
Case (I):
In this case we have ),(tAA = )(tBB = and .BA ≠ The space-time is given in
equation (3.2.1). Substituting the above information in equation (3.2.15) and solving
we get .0),(),( 31 == φφ tKtK The system of equations (3.2.14) becomes
).,,(),()()(,),(
),,(),,,(
44342
2110
θφφθθφ
φφθ
rtPdtKffXtKX
tKXrPX
+′
−==
==
∫ (3.2.16)
Considering equation (3.2.9) and using equation (3.2.16) then differentiating the
resulting equation with respect to ,φ we get ⇒=+ 0),(),( 44 φφφφ tKtK
,sin)(cos)(),( 214 φφφ tEtEtK += where )(1 tE and )(2 tE are functions of
integration. Now substituting back the above value in equation (3.2.9) we get
),,(),,(0),,( 544 rtKrtPrtP =⇒= θθθ where ),(5 rtK is a function of integration.
Substituting back all the above information in (3.2.16), we get
( ) ).,(cos)(sin)()()(
,sin)(cos)(),,(),,,(
5213
2122110
rtKtEtEffX
tEtEXtKXrPX
+−′
−=
+===
φφθθ
φφφφθ (3.2.17)
Considering equation (3.2.8) and using equation (3.2.17) then differentiating with
respect to ,θ we get ),(),(0),( 355 tErtKrtKr =⇒= where )(3 tE is a function of
91
integration. Substituting back the above value in (3.2.8) we get
),(),(0),( 422 tEtKtK =⇒= φφφ where )(4 tE is a function of integration.
Refreshing the system of equations (3.2.17), we get
( ) ).(cos)(sin)()()(
,sin)(cos)(),(),,,(
3213
2124110
tEtEtEffX
tEtEXtEXrPX
+−′
−=
+===
φφθθ
φφφθ (3.2.18)
Considering equation (3.2.11) and using equation (3.2.18) then solving, we get
),,(),,(0),,( 611 φθφθφθ KrPrPr =⇒= where ),(6 φθK is a function of integration
and .,1)( 114 ℜ∈= cc
AtE Our system of equations (3.2.18) become
( ) ).(cos)(sin)()()(
,sin)(cos)(,1),,(
3213
2121
160
tEtEtEffX
tEtEXcA
XKX
+−′
−=
+===
φφθθ
φφφθ (3.2.19)
Considering equation (3.2.12) and using equation (3.2.19) then solving, we get
),(),(0),( 566 φφθφθθ EKK =⇒= where )(5 φE is a function of integration.
Substituting back the above value and solving we get 21 1)( c
BtE = and ,1)( 3
2 cB
tE =
., 32 ℜ∈cc Refreshing the system of equations (3.2.19) we get
( )
( ) ).(sincos)()(1
,sincos1,1),(
323
3
322
1150
tEccff
BX
ccB
XcA
XEX
+−′
=
+===
φφθθ
φφφ (3.2.20)
Considering equation (3.2.13) and using equation (3.2.20) then solving we get
0)(5 =φE and .,1)( 443 ℜ∈= cc
BtE Refreshing equation (3.2.20) we get
92
( )
( ) .1sincos)()(1
,sincos1,1,0
5233
322
110
cB
ccff
BX
ccB
XcA
XX
+−′
=
+===
φφθθ
φφ (3.2.21)
In this case the above space-time (3.2.1) admits four linearly independent teleparallel
Killing vector fields which are ,)(
1rtA ∂∂ ,
)(1
φ∂∂
tB )sin
)()((cos
)(1
φφ
θθ
θφ
∂∂′
−∂∂
ff
tB
and ).cos)()((sin
)(1
φφ
θθ
θφ
∂∂′
+∂∂
ff
tB Killing vector fields in general relativity are
,r∂∂ ,
φ∂∂
φφ
θθ
θφ
∂∂′
−∂∂ sin
)()(cos
ff and .cos
)()(sin
φφ
θθ
θφ
∂∂′
+∂∂
ff Comparison
shows that Killing vector fields in both theories are same in number and teleparallel
Killing vector fields are multiple of the corresponding elements of the inverse tetrad
field.
Case (II):
In this case we have )(tAA = and 0).\( ℜ∈= ηηB The line element for Bianchi
type III and Kantowski-Sachs space-time in this case is given as
),)(()( 22222222 φθθη dfddrtAdtds +++−= (3.2.22)
Substituting the above information in equation (3.2.15) we get
).,(),( 32
21 φηφ tK
AtK −= Now substituting all the above information in equation
(3.2.14), we get
( ) ).,,(),(),()()(
,),(),(),,(),(),,,(
4433
432232
2110
θφφφθθ
φφφφηθφθ
rtPdtKtrKffX
tKtrKXtKtKA
XrPX
++′
−=
+=+−==
∫ (3.2.23)
Considering equation (3.2.9) and using equation (3.2.23) then differentiating the
resulting equation first with respect to r then with respect to ,φ we get
93
⇒=+ 0),(),( 33 φφφφ tKtK ,sin)(cos)(),( 213 φφφ tEtEtK += where )(1 tE and
)(2 tE are functions of integration. Substituting back the above value in (3.2.9) we get
.0),,()(),(),( 4244 =++ ∫ θθφφφ θφ rtPfdtKtK Differentiating this equation with
respect to ,r we get ),,(),(),,(0),,( 6544 rtKtKrtPrtPr +=⇒= θθθθ where
),(5 θtK and ),(6 rtK are functions of integration. Substituting back the above values
in (3.2.9) and differentiating with respect to ,φ we get
⇒=+ 0),(),( 44 φφφφ tKtK ,sin)(cos)(),( 434 φφφ tEtEtK += where )(3 tE and
)(4 tE are functions of integration. Substituting back the above values in (3.2.9) we
get ),(),(0),( 555 tEtKtK =⇒= θθθ where )(5 tE is a function of integration. Now
refreshing the system of equations (3.2.23) we get
( )( )
( ) ( )).,()(
]sin)(cos)(sin)(cos)([)()(
,sin)(cos)(sin)(cos)(
),,(sin)(cos)(),,,(
65
43213
43212
2212
2110
rtKtE
dtEtEtEtErffX
tEtEtEtErX
tKtEtEA
XrPX
++
+++′
−=
+++=
++−==
∫ φφφφφθθ
φφφφ
φφφηθφθ
(3.2.24)
Considering equation (3.2.8) and using equation (3.2.24) then differentiating the
resulting equation twice with respect to θ then with respect to φ and solving, we get
.0)()( 21 == tEtE Substituting back the above values in (3.2.8) and solving we get
)(),( 66 tErtK = and ),(),( 72 tEtK =φ where )(6 tE and )(7 tE are functions of
integration. Refreshing the system of equations (3.2.24) we get
( ) ),(sin)(cos)()()(
,sin)(cos)(),(),,,(
8433
4327110
tEdtEtEffX
tEtEXtEXrPX
++′
−=
+===
∫ φφφθθ
φφφθ (3.2.25)
where ).()()( 658 tEtEtE += Considering equation (3.2.12) and using equation
(3.2.25) then solving the resulting equation we get ,)(0)( 133 ctEtEt =⇒=
94
ℜ∈=⇒= 21244 ,,)(0)( ccctEtEt and ),,(),,(0),,( 811 φφθφθθ rKrPrP =⇒=
where ),(8 φrK is a function of integration. Our system of equations (3.2.25) become
( ) ).(cossin)()(
,sincos),(),,(
821
3
2127180
tEccffX
ccXtEXrKX
+−′
−=
+===
φφθθ
φφφ (3.2.26)
Considering equation (3.2.13) and using equation (3.2.26) then solving we get
ℜ∈=⇒= 3388 ,)(0)( cctEtEt and ),(),(0),( 988 rErKrK =⇒= φφφ where )(9 rE
is a function of integration. Now considering equation (3.2.11) and using equation
(3.2.26) then solving, we get 0)(9 =rE and .,1)( 447 ℜ∈= cc
AtE The solution of
equations (3.2.6) to (3.2.13) becomes [68]
( ) ,cossin)()(
,sincos,1,0
3213
212
510
cccffX
ccXcA
XX
+−′
−=
+===
φφθθ
φφ (3.2.27)
where .,,, 4321 ℜ∈cccc The above space-time (3.2.22) admits four linearly
independent teleparallel Killing vector fields which are ,)(
1rtA ∂∂ ,
φ∂∂
φφ
θθ
θφ
∂∂′
−∂∂ sin
)()(cos
ff and .cos
)()(sin
φφ
θθ
θφ
∂∂′
+∂∂
ff Killing vector fields in
general relativity are ,r∂∂ ,
φ∂∂
φφ
θθ
θφ
∂∂′
−∂∂ sin
)()(cos
ff and
.cos)()(sin
φφ
θθ
θφ
∂∂′
+∂∂
ff Comparison shows that only one teleparallel Killing
vector field is different from Killing vector fields in general relativity which is a
multiple of the corresponding element of the inverse tetrad field.
95
Case (III):
In this case we have )(tBB = and 0.\, ℜ∈= ηηA The line element for
Kantowski-Sachs and Bianchi type III space-time after a suitable rescaling of ,r is
given as
),)(()( 2222222 φθθ dfdtBdrdtds +++−= (3.2.28)
Substituting the above values in equation (3.2.15) we get ).,(),( 12
23 φηφ tK
BtK −=
Now substituting all the above information in (3.2.14) we get
).,,(),(),()()(
,),(),(),,(),(),,,(
4412
23
412
2221110
θφφφηθθ
φφηφφθφθ
rtPdtKtKBr
ffX
tKtKBrXtKtKXrPX
+⎟⎟⎠
⎞⎜⎜⎝
⎛+−
′−=
+−=+==
∫ (3.2.29)
Considering equation (3.2.9) and using equation (3.2.29) then differentiating the
resulting equation first with respect to φ then with respect to r and solving we get
⇒=+ 0),(),( 11 φφφφ tKtK ,sin)(cos)(),( 211 φφφ tEtEtK += where )(1 tE and
)(2 tE are functions of integration. Substituting back the above value in equation
(3.2.9) and differentiating with respect to ,φ we get ⇒=+ 0),(),( 44 φφφφ tKtK
φφφ sin)(cos)(),( 434 tEtEtK += where )(3 tE and )(4 tE are functions of
integration. Once again substituting back the above information in equation (3.2.9)
we get ),,(),,(0),,( 544 rtKrtPrtP =⇒= θθθ where ),(5 rtK is the function of
integration. Now refreshing the system of equations (3.2.29) we get
( )( )
( ) ( )).,(
]sin)(cos)(sin)(cos)([)()(
,sin)(cos)(sin)(cos)(
),,(sin)(cos)(),,,(
5
43212
23
43212
22
221110
rtK
dtEtEtEtEBr
ffX
tEtEtEtEBrX
tKtEtEXrPX
+
+++−′
−=
+++−=
++==
∫ φφφφφηθθ
φφφφη
φφφθφθ
(3.2.30)
96
Considering equation (3.2.8) and using equation (3.230) then differentiating the
resulting equation twice with respect to θ then with respect to φ and solving, we get
.0)()( 21 == tEtE Substituting back the above values in (3.2.8) we get
⇒= 0),(2 φφφ tK ),()(),( 652 tEtEtK += φφ where )(5 tE and )(6 tE are functions of
integration. Now substituting back all the above information in (3.2.8) and solving we
get 0)(5 =tE and ),(),(0),( 1055 tErtKrtKr =⇒= where )(10 tE is a function of
integration. Refreshing the system of equations (3.2.30) we get
( ) ).(sin)(cos)()()(
,sin)(cos)(),(),,,(
10433
4326110
tEdtEtEffX
tEtEXtEXrPX
++′
−=
+===
∫ φφφθθ
φφφθ (3.2.31)
Considering equation (3.2.11) and using equation (3.2.31) then solving we get
),,(),(),,(0),,( 7611 φθφθφθφθ KKrrPrPrr +=⇒= where ),(6 φθK and ),(7 φθK
are functions of integration. Substituting back the above value in (3.2.11) and solving
we get ),(),(0),( 766 φφθφθθ EKK =⇒= where )(7 φE is the function of integration.
Substituting all the above information in (3.2.11) we get 0)(7 =φE and
.,)( 336 ℜ∈= cctE Our system of equations (3.2.31) become
( ) ).(sin)(cos)()()(
,sin)(cos)(,),,(
10433
4323
170
tEdtEtEffX
tEtEXcXKX
++′
−=
+===
∫ φφφθθ
φφφθ (3.2.32)
Considering equation (3.2.12) and using equation (3.2.32) then solving we get
),(),(0),( 877 φφθφθθ EKK =⇒= where )(8 φE is the function of integration and
,1)( 43 c
BtE = .,,1)( 545
4 ℜ∈= cccB
tE Refreshing the system of equations we get
( )
( ) ),(cossin)()(1
,sincos1,),(
1054
3
542
3180
tEccff
BX
ccB
XcXEX
+−′
−=
+===
φφθθ
φφφ (3.2.33)
97
where .,, 543 ℜ∈ccc Now considering equation (3.2.13) and using equation (3.2.33)
then solving we get ,0)(8 =φE .,1)( 6610 ℜ∈= cc
BtE Finally, the solution of
equations (3.2.6) to (3.2.13) becomes
( )
( ) ,1cossin)()(1
,sincos1,,0
6543
542
310
cB
ccff
BX
ccB
XcXX
+−′
−=
+===
φφθθ
φφ (3.2.34)
where .,,, 6543 ℜ∈cccc The above space-time (3.2.28) admits four linearly
independent teleparallel Killing vector fields which are ,r∂∂ ,
)(1
φ∂∂
tB
)sin)()((cos
)(1
φφ
θθ
θφ
∂∂′
−∂∂
ff
tB and ).cos
)()((sin
)(1
φφ
θθ
θφ
∂∂′
+∂∂
ff
tB In this case
Killing vector fields in general relativity are ,r∂∂ ,
φ∂∂
φφ
θθ
θφ
∂∂′
−∂∂ sin
)()(cos
ff and
.cos)()(sin
φφ
θθ
θφ
∂∂′
+∂∂
ff Teleparallel Killing vector field are same in number and
one is exactly the same as in general relativity. Other three teleparallel Killing vector
fields are different from Killing vector fields in general relativity.
Case (IV):
In this case we have ),(tAA = )(tBB = and .BA = The line element for Kantowski-
Sachs and Bianchi type III space-time in this case is given as
).)(()( 2222222 φθθ dfddrtAdtds +++−= (3.2.35)
Substituting back the above information in equation (3.2.15) we get
).,(),( 13 φφ tKtK −= Now substituting all the above information in (3.2.14) we get
( ) ).,,(),(),()()(
,),(),(),,(),(),,,(
4413
41221110
θφφφθθ
φφφφθφθ
rtPdtKtrKffX
tKtrKXtKtKXrPX
++−′
−=
+−=+==
∫ (3.2.36)
98
Considering equation (3.2.9) and using equation (3.2.36) then differentiating the
resulting equation first with respect to φ then with respect to r and solving we get
⇒=+ 0),(),( 11 φφφφ tKtK ,sin)(cos)(),( 211 φφφ tEtEtK += where )(1 tE and
)(2 tE are functions of integration. Substituting back the above value in equation
(3.2.9) and differentiating with respect to ,φ we get ⇒=+ 0),(),( 44 φφφφ tKtK
φφφ sin)(cos)(),( 434 tEtEtK += where )(3 tE and )(4 tE are functions of
integration. Once again substituting back the above information in equation (3.2.9)
we get ),,(),,(0),,( 544 rtKrtPrtP =⇒= θθθ where ),(5 rtK is the function of
integration. Refreshing the system of equations (3.2.36) we get
( )( )
( ) ( )).,(
]sin)(cos)(sin)(cos)([)()(
,sin)(cos)(sin)(cos)(),,(sin)(cos)(),,,(
5
43213
43212
221110
rtK
dtEtEtEtErffX
tEtEtEtErXtKtEtEXrPX
+
+++−′
−=
+++−=
++==
∫ φφφφφθθ
φφφφ
φφφθφθ
(3.2.37)
Now considering equation (3.2.8) and using equation (3.2.37) then differentiating the
resulting equation twice with respect to θ then with respect to φ and solving, we get
.0)()( 21 == tEtE Substituting back the above values in (3.2.8) and differentiating
with respect to ,φ we get ⇒= 0),(2 φφφ tK ),()(),( 652 tEtEtK += φφ where )(5 tE
and )(6 tE are functions of integration. Now substituting back all the above
information in (3.2.8) and solving we get ,0)(5 =tE
),(),(0),( 1055 tErtKrtKr =⇒= where )(10 tE is a function of integration.
Refreshing the system of equations (3.2.37) we get
( ) ).(sin)(cos)()()(
,sin)(cos)(),(),,,(
10433
4326110
tEdtEtEffX
tEtEXtEXrPX
++′
−=
+===
∫ φφφθθ
φφφθ (3.2.38)
99
Considering equation (3.2.11) and using equation (3.2.38) then solving, we get
),,(),(),,(0),,( 7611 φθφθφθφθ KKrrPrPrr +=⇒= where ),(6 φθK and ),(7 φθK
are functions of integration. Substituting back the above value in (3.2.11) and solving
we get ),(),(0),( 766 φφθφθθ EKK =⇒= where )(7 φE is the function of integration.
Substituting all the above information in (3.2.11) we get 0)(7 =φE and
.,)(
1)( 336 ℜ∈= cc
tAtE Our system of equations (3.2.38) becomes
( ) ).(sin)(cos)()()(
,sin)(cos)(,)(
1),,(
10433
4323
170
tEdtEtEffX
tEtEXctA
XKX
++′
−=
+===
∫ φφφθθ
φφφθ (3.2.39)
Considering equation (3.2.12) and using equation (3.2.39) then solving we get
),(),(0),( 877 φφθφθθ EKK =⇒= where )(8 φE is the function of integration,
43 1)( c
AtE = and .,,1)( 545
4 ℜ∈= cccA
tE Refreshing the above system of equations
we get
( )
( ) ),(cossin)()(1
,sincos1,)(
1),(
1054
3
542
3180
tEccff
AX
ccA
XctA
XEX
+−′
−=
+===
φφθθ
φφφ (3.2.40)
where .,, 543 ℜ∈ccc Considering equation (3.2.13) and using equation (3.2.40) then
solving we get 0)(8 =φE and .,1)( 6610 ℜ∈= cc
AtE Finally, the solution of
equations (3.2.6) to (3.2.13) becomes [68]
( )
( ) ,)(
1cossin)()(
)(1
,sincos)(
1,)(
1,0
6543
542
310
ctA
ccff
tAX
cctA
XctA
XX
+−′
−=
+===
φφθθ
φφ (3.2.41)
100
where .,,, 6543 ℜ∈cccc The above space-time (3.2.35) admits four linearly
independent teleparallel Killing vector fields which are ,)(
1rtA ∂∂ ,
)(1
φ∂∂
tA
)sin)()((cos
)(1
φφ
θθ
θφ
∂∂′
−∂∂
ff
tA and ).cos
)()((sin
)(1
φφ
θθ
θφ
∂∂′
+∂∂
ff
tA Killing vector
fields in general relativity are ,r∂∂ ,
φ∂∂
φφ
θθ
θφ
∂∂′
−∂∂ sin
)()(cos
ff and
.cos)()(sin
φφ
θθ
θφ
∂∂′
+∂∂
ff Comparison shows that Killing vector fields in both
theories are same in number and the teleparallel Killing vector fields are multiple of
the corresponding elements of the inverse tetrad field.
Case (V):
In this case we have β=A 0\ℜ∈β and η=B 0.\ℜ∈η The space-time
(3.2.1) in this case after a suitable rescaling of ,r becomes
),)(( 2222222 φθθη dfddrdtds +++−= (3.2.42)
Substituting the above information in equation (3.2.15) we get
).,(),( 123 φηφ tKtK −= Substituting back all the above information in (3.2.14) we get
( ) ).,,(),(),()()(
,),(),(),,(),(),,,(
44123
412221110
θφφφηθθ
φφηφφθφθ
rtPdtKtKrffX
tKtKrXtKtKXrPX
++−′
−=
+−=+==
∫ (3.2.43)
Considering equation (3.2.9) and using equation (3.2.43) then differentiating the
resulting equation first with respect to φ then with respect to r and solving we get
⇒=+ 0),(),( 11 φφφφ tKtK ,sin)(cos)(),( 211 φφφ tEtEtK += where )(1 tE and
)(2 tE are functions of integration. Substituting back the above value in equation
(3.2.9) and differentiating with respect to ,φ we get ⇒=+ 0),(),( 44 φφφφ tKtK
φφφ sin)(cos)(),( 434 tEtEtK += where )(3 tE and )(4 tE are functions of
101
integration. Once again substituting back the above information in equation (3.2.9)
we get ),,(),,(0),,( 544 rtKrtPrtP =⇒= θθθ where ),(5 rtK is the function of
integration. Refreshing the system of equations (3.2.43) we get
( )( )
( ) ( )).,(
]sin)(cos)(sin)(cos)([)()(
,sin)(cos)(sin)(cos)(),,(sin)(cos)(),,,(
5
432123
432122
221110
rtK
dtEtEtEtErffX
tEtEtEtErXtKtEtEXrPX
+
+++−′
−=
+++−=
++==
∫ φφφφφηθθ
φφφφη
φφφθφθ
(3.2.44)
Now considering equation (3.2.8) and using equation (3.2.44) then differentiating the
resulting equation twice with respect to θ then with respect to φ and solving, we get
.0)()( 21 == tEtE Substituting back the above values in (3.2.8) and differentiating
with respect to ,φ we get ⇒= 0),(2 φφφ tK ),()(),( 652 tEtEtK += φφ where )(5 tE
and )(6 tE are functions of integration. Now substituting back all the above values in
(3.2.8) and solving, we get 0)(5 =tE and ),(),(0),( 1055 tErtKrtKr =⇒= where
)(10 tE is a function of integration. Refreshing the system of equations (3.2.44) we get
( ) ).(sin)(cos)()()(
,sin)(cos)(),(),,,(
10433
4326110
tEdtEtEffX
tEtEXtEXrPX
++′
−=
+===
∫ φφφθθ
φφφθ (3.2.45)
Considering equation (3.2.11) and using equation (3.2.45) then solving we get
),,(),(),,(0),,( 7611 φθφθφθφθ KKrrPrPrr +=⇒= where ),(6 φθK and ),(7 φθK
are functions of integration. Substituting back the above value in (3.2.11) we get
⇒=⇒=− 0)(0),()( 666 tEKtE ttt φθ .,,)( 43436 ℜ∈+= ccctctE Substituting back
this value we get .),( 36 cK =φθ Our system of equations (3.2.45) becomes
( ) ).(sin)(cos)()()(
,sin)(cos)(,),,(
10433
43243
173
0
tEdtEtEffX
tEtEXctcXKrcX
++′
−=
+=+=+=
∫ φφφθθ
φφφθ (3.2.46)
102
Considering equation (3.2.12) and using equation (3.2.46) then solving we get
),(),(0),( 877 φφθφθθ EKK =⇒= where )(8 φE is the function of integration,
53 )( ctE = and .,,)( 656
4 ℜ∈= ccctE Refreshing the above system of equations
(3.2.46) we get
( )
( ) ),(cossin)()(
,sincos,),(
1065
3
652
4318
30
tEccffX
ccXctcXErcX
+−′
−=
+=+=+=
φφθθ
φφφ (3.2.47)
where .,,, 6543 ℜ∈cccc Now considering equation (3.2.13) and using equation
(3.2.47) then solving we get 78 )( cE =φ and .,,)( 878
10 ℜ∈= ccctE Finally the
solution of equations (3.2.6) to (3.2.13) becomes [68]
( )
( ) ,cossin)()(
,sincos,,
8653
652
431
730
cccffX
ccXctcXcrcX
+−′
−=
+=+=+=
φφθθ
φφ (3.2.48)
where .,,,,, 876543 ℜ∈cccccc In this case the above space-time (3.2.42) admits six
linearly independent teleparallel Killing vector fields which are ,t∂∂ ,
r∂∂ ,
φ∂∂
,t
rr
t∂∂
+∂∂
φφ
θθ
θφ
∂∂′
−∂∂ sin
)()(cos
ff and
φφ
θθ
θφ
∂∂′
+∂∂ cos
)()(sin
ff which are
exactly the same as Killing vector fields in general relativity.
3.3. Teleparallel Killing Vector Fields in Static
Cylindrically Symmetric Space-Times
Consider cylindrically symmetric static space-times in usual coordinates ),,,( zrt θ
(labeled by ),,,,( 3210 xxxx respectively) with the line element [36, 78]
,2)(2)(22)(2 dzededrdteds rCrBrA +++−= θ (3.3.1)
103
where ,A B and C are functions of r only. In general relativity the space-time
(3.3.1) admits minimum three Killing vector fields [36] which are .,,φθ ∂∂
∂∂
∂∂t
The
tetrad components and its inverse can be obtained by using the relation (1.3.3) as [69]
),,,1,(diag 2)(
2)(
2)( rCrBrA
a eeeS =µ ).,,1,(diag 2)(
2)(
2)( rCrBrA
a eeeS−−−
=µ (3.3.2)
Using equation (1.3.5), the corresponding non-vanishing Weitzenböck connections
are obtained as
,2
010 A′
=Γ ,2
212 B′
=Γ ,2
313 C ′
=Γ ,2221
B
e−=Γ (3.3.3)
where dash denotes the derivative with respect to .r The non vanishing torsion
components by using (1.3.11) are [69]
,2
100 AT
′= ,
213
3 CT′
= .2
122 BT
′= (3.3.4)
Now substituting (3.3.1) and (3.3.4) in (1.3.17) we get the teleparallel Killing
equations as follows:
,0,0,0,0 3,3
2,2
1,1
0,0 ==== XXXX (3.3.5)
,02,0)(
0,2)( =− XeXe rArB (3.3.6)
,03,0)(
0,3)( =− XeXe rArC (3.3.7)
,03,2)(
2,3)( =+ XeXe rBrC (3.3.8)
,02
0)(1,
0)(0,
1 =′
−− XeAXeX rArA (3.3.9)
,02
2)(2,
11,
2)( =′
++ XeBXXe rBrB (3.3.10)
,02
3)(3,
11,
3)( =′
++ XeCXXe rCrC (3.3.11)
104
Now integrating equations (3.3.5), we get
),,,(),,,(),,,(),,,(
4332
2110
θ
θθ
rtPXzrtPXztPXzrPX
==
== (3.3.12)
where ),,,(1 zrP θ ),,,(2 ztP θ ),,(3 zrtP and ),,(4 θrtP are functions of integration
which are to be determined. In order to find solution for equations (3.3.5) to (3.3.11)
we will consider each possible form of the metric for static cylindrically symmetric
space-times and then solve each possibility in turn. Following are the possible cases
for the metric where the above space-times admit teleparallel Killing vector fields:
(I) )(),(),( rCCrBBrAA === and .,, CBCABA ≠≠≠
(II)(a) ),(),( rCCrBB == and .tan tconsA =
(II)(b) ),(),( rCCrAA == and .tan tconsB =
(II)(c) ),(),( rBBrAA == and .tan tconsC =
(III)(a) )(),(),( rCCrBBrAA === and ).()( rCrB =
(III)(b) )(),(),( rCCrBBrAA === and ).()( rCrA =
(III)(c) )(),(),( rCCrBBrAA === and ).()( rBrA =
(IV) )(),(),( rCCrBBrAA === and ).()()( rCrBrA ==
(V)(a) )(),(,tan rCCrBBtconsA === and ).()( rCrB =
(V)(b) )(,tan),( rCCtconsBrAA === and ).()( rCrA =
(V)(c) tconsCrBBrAA tan),(),( === and ).()( rBrA =
(VI)(a) )(rAA = and .tan tconsCB ==
(VI)(b) )(rBB = and .tan tconsCA ==
(VI)(c) )(rCC = and .tan tconsBA ==
We will discuss each case in turn.
105
Case (I):
In this case we have ),(rAA = ),(rBB = ),(rCC = ,BA ≠ CA ≠ and .CB ≠ Now
substituting equation (3.3.12) in equation (3.3.6), we get
.0),,(),,( 1)(3)( =− zrPezrtPe rAt
rB θθ (3.3.13)
Differentiating equation (3.3.13) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (3.3.13) we get
⇒= − ),(),,( 11 zrEezrP ABθθ ),,(),(),,( 311 zrEzrEezrP AB += −θθ where ),(3 zrE is
a function of integration. Refreshing the system of equations (3.3.12) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θ
θθ
rtPXzrEzrEtXztPXzrEzrEeX AB
=+=
=+= −
(3.3.14)
Considering equation (3.3.7) and using equation (3.3.14) we get
.0)],(),([),,( 31)()()(4)( =+− − zrEzrEeertPe zzrArBrA
trC θθ (3.3.15)
Differentiating equation (3.3.15) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (3.3.15) we get
.0)],(),([),( 31)()()(4)( =+− − zrEzrEeerEe zzrArBrArC θθ Differentiating the above
equation with respect to θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ
where )(1 rK and )(2 rK are functions of integration. Substituting back the above
value in equation (3.3.15) and solving we get )()(),( 31)()(1 rKrKezzrE rBrC += − and
),()(),( 42)()(3 rKrKezzrE rArC += − where )(3 rK and )(4 rK are functions of
integration. Substituting all the above information in (3.3.14) we get
106
( )( )
( ) ).,()()(),,()()(),,,(
),()()()(
5213
231)()(221
42)()(3)()(1)()(0
θθ
θ
θ
rErKrKtXzrErKrKeztXztPX
rKrKezrKerKezXrBrC
rArCrArBrArC
++=
++==
+++=−
−−−
(3.3.16)
Considering equation (3.3.8) and using equation (3.3.16), we get
.0),(),()(2 2)(5)(1)( =++ zrEerEerKte zrBrCrC θθ (3.3.17)
Differentiating the above equation first with respect to t and then with respect to ,θ
we get 0)(1 =rK and ),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and
)(6 rK are functions of integration. Substituting back the above values in (3.3.17) we
get ),()(),()(),( 75)()(25)()(2 rKrKezzrErKezrE rBrCrBrCz +−=⇒−= −− where )(7 rK
is a function of integration. Substituting all the above information in (3.3.16) we get
( )).()()(),()()(
),,,(),()()(652375)()(32
2142)()(3)()(0
rKrKrtKXrKrKezrKtXztPXrKrKezrKeX
rBrC
rArCrArB
++=+−=
=++=−
−−
θ
θθ (3.3.18)
Considering equation (3.3.9) and using equation (3.3.18), we get
.0)()(21)()(
)(21)()(
21),,(
4)(4)(2)(
2)(3)(3)(2
=−−−
⎟⎠⎞
⎜⎝⎛ −−−⎟
⎠⎞
⎜⎝⎛ −−
rKerArKerKez
rKeACzrKerKeABztP
rArr
rAr
rC
rCrrr
rBrBrrt θθθ
(3.3.19)
Differentiating equation (3.3.19) with respect to t we get ⇒= 0),,(2 ztPtt θ
),,(),(),,( 762 zEzEtztP θθθ += where ),(6 zE θ and ),(7 zE θ are functions of
integration. Substituting back the above value in (3.3.19) and differentiating with
respect to ,θ we get
.0)()(21),( 3)(3)(6 =−⎟
⎠⎞
⎜⎝⎛ −− rKerKeABzE r
rBrBrrθθ (3.3.20)
Now differentiating the above equation with respect to ,θ we get
),()(),(0),( 9866 zKzKzEzE +=⇒= θθθθθ where )(8 zK and )(9 zK are functions
107
of integration. Substituting back the above value in (3.3.20) and differentiating with
respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzKz Now substituting this value in
equation (3.3.20) and solving we get,
.,)( 2
)(2
)(
22
)()(2
)(
13 ℜ∈+=
−−
−
∫ cecdreecrKrBrArArBrA
Substituting all the above
information in equation (3.3.19) and differentiating twice with respect to ,z we get
.,,)(0)( 434399 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in (3.3.19) and
solve after differentiating with respect to ,z we get
.,)( 5
)(2
)(
52
)()(2
)(
32 ℜ∈+=
−−
−
∫ cecdreecrKrCrArArCrA
Now substituting all the above
information in equation (3.3.19) and solving we get,
.,)( 62
)(
62
)(2
)(
44 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Refreshing the system of equations
(3.3.18) we get
).()(
),()(
),,(
,
65)(2
)(
52
)()(2
)(
33
75)()()(2
)(
22
)()(2
)(
12
7431
1
2)(
62
)(2
)(
42
)(
52
)(2
)(
32
)(
22
)(2
)(
10
rKrKetcdreectX
rKrKezectdreectX
zEtcctzctX
ecdreececzdreeczecdreecX
rCrArArCrA
rBrCrBrArArBrA
rArArArArArArArArA
+++=
+−+=
+++=
+++++=
−−
−
−−−
−
−−−−−−−−−
∫
∫
∫∫∫
θ
θθ
θθ
(3.3.21)
Considering equation (3.3.10) and using equation (3.3.21) we get
.0)()(21)()()()](
21)([
)]()([21)]()([
21),(2
7)(7)(5)(5)(
2)(
22
)(2
)(
17
1
=++−−−
−+−++ ∫−
rKerBrKerKezrKerBrCz
erBrActdreerBrActzEtc
rBrr
rBr
rCrCrr
rA
rr
rArA
rrθθ (3.3.22)
Differentiating equation (3.3.22) with respect to ,t we get
.0)]()([21)]()([
212 2
)(
22
)(2
)(
11 =−+−+ ∫− rA
rr
rArA
rr erBrAcdreerBrAcc Solving this
equation and remember that in this case ,0)()( ≠− rBrA rr we get ,01 =c which on
108
back substitution gives us .02 =c Substituting the above information in equation
(3.3.22) and differentiating with respect to ,θ we get
),()(),(0),( 111077 zKzKzEzE +=⇒= θθθθθ where )(10 zK and )(11 zK are functions
of integration. Now substituting the above information in equation (3.3.22) and
differentiating twice with respect to ,z we get
.,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzKzz Substituting back these information in
equation (3.3.22) and differentiating the resulting equation with respect to ,z we get
,)(0)()(21 )(
2)(
92
)()(2
)(
755)(5)(
7
rCrBrBrCrB
rrCrC
rr ecdreecrKrKerKeBCc−
−−
+=⇒=−⎟⎠⎞
⎜⎝⎛ −− ∫
.9 ℜ∈c Now substituting back all the above information in equation (3.3.22) we get
.,)(0)()()(21
102
)(
102
)(2
)(
877)(7)(
8 ℜ∈+−=⇒=++−−−
∫ cecdreecrKrKerKerBcrBrBrB
rrBrB
r
Refreshing the system of equations (3.3.21) we get
).(
,
),(
,
6)(2
)(
92
)()(2
)(
7
)(2
)(
52
)()(2
)(
33
2)(
102
)(2
)(
82
)(
92
)(2
)(
72
118743
1
2)(
62
)(2
)(
42
)(
52
)(2
)(
30
rKecdreecetcdreectX
ecdreececzdreeczX
zKczctcctzX
ecdreececzdreeczX
rCrBrBrCrBrCrArArCrA
rBrBrBrBrBrB
rArArArArArA
++++=
+−−−=
++++=
+++=
−−
−−−
−
−−−−−−
−−−−−−
∫∫
∫∫
∫∫
θθ
θθ
(3.3.23)
Considering equation (3.3.11) and using equation (3.3.23) we get
.0)()(21)()]()([
21
)]()([21)]()([
21
)]()([21)(22
6)(6)(2)(
9
2)(
2)(
72
)(
5
2)(
2)(
311
73
=++−+
−+−+
−+++
∫
∫−
−
rKerCrKeerCrBc
dreerCrBcerCrAct
dreerCrActzKctc
rCrr
rCrB
rr
rBrB
rr
rA
rr
rArA
rrz
θ
θ
θ
(3.3.24)
Differentiating equation (3.3.24) with respect to ,t we get
.0)]()([21)]()([
212 2
)(
52
)(2
)(
33 =−+−+ ∫− rA
rr
rArA
rr erCrAcdreerCrAcc Solving this
109
equation and remember that in this case ,0)()( ≠− rCrA rr we get ,03 =c which on
back substitution gives us .05 =c Substituting the above information in equation
(3.3.24) and differentiating with respect to ,θ we get
0)]()([21)]()([
212 2
)(
92
)(2
)(
77 =−+−+ ∫− rB
rr
rBrB
rr erCrBcdreerCrBcc Solving this
equation and remember that in this case ,0)()( ≠− rCrB rr we get ,07 =c which on
back substitution gives us .09 =c Now substituting the above information in equation
(3.3.24) and differentiating twice with respect to ,z we get
.,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzKzz Substituting back these information in
equation (3.3.24) and differentiating the resulting equation with respect to ,z we get
,)(0)()()(21 2
)(
132
)(2
)(
1166)(6)(
11
rCrCrC
rrCrC
r ecdreecrKrKerKerCc−−−
+−=⇒=++ ∫
.13 ℜ∈c Refreshing the system of equations (3.3.23) we get
,,
,,
2)(
132
)(2
)(
1132
)(
102
)(2
)(
82
12118412
)(
62
)(2
)(
40
rCrCrCrBrBrB
rArArA
ecdreecXecdreecX
czcctcXecdreecX−−−−−−
−−−
+−=+−=
+++=+=
∫∫
∫ θ
(3.3.25)
where .,,,,,, 13121110864 ℜ∈ccccccc The line element for static cylindrically
symmetric space-times is given in equation (3.3.1). The above space-time admits
seven linearly independent teleparallel Killing vector fields which are ,r∂∂ ,2
)(
te
rA
∂∂−
,2)(
θ∂∂− rB
e ,2)(
ze
rC
∂∂−
,)(1
trM
rt
∂∂
+∂∂
θθ
∂∂
+∂∂ )(2 rMr
and ,)(3
zrM
rz
∂∂
+∂∂
where ,)( 2)(
2)(
1 ∫−−
= dreerMrArA
dreerMrBrB
∫−−
−= 2)(
2)(
2 )( and
.)( 2)(
2)(
3 ∫−−
−= dreerMrCrC
Killing vector fields in general relativity are ,t∂∂
θ∂∂ and
110
.z∂∂ It is evident that teleparallel Killing vector fields are different and more in
number to the Killing vector fields in general relativity.
Case (II)(a):
In this case we have ,constant=A ),(rBB = )(rCC = and .CB ≠ Now
substituting equation (3.3.12) in equation (3.3.6), we get
.0),,(),,( 13)( =− zrPzrtPe trB θθ (3.3.26)
Differentiating equation (3.3.26) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (3.3.26) we get
⇒= ),(),,( 11 zrEezrP Bθθ ),,(),(),,( 311 zrEzrEezrP B +=θθ where ),(3 zrE is a
function of integration. Refreshing the system of equations (3.3.12) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θ
θθ
rtPXzrEzrEtXztPXzrEzrEeX B
=+=
=+= (3.3.27)
Considering equation (3.3.7) and using equation (3.3.27) we get
.0),(),(),,( 31)(4)( =−− zrEzrEertPe zzrB
trC θθ (3.3.28)
Differentiating equation (3.3.28) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (3.3.28) we get
.0),(),(),( 31)(4)( =−− zrEzrEerEe zzrBrC θθ Differentiating the above equation with
respect to θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ where )(1 rK
and )(2 rK are functions of integration. Substituting back the above value in equation
(3.3.28) and solving we get )()(),( 31)()(1 rKrKezzrE rBrC += − and
),()(),( 42)(3 rKrKezzrE rC += where )(3 rK and )(4 rK are functions of integration.
Substituting all the above information in (3.3.27) we get
111
( )( )
( ) ).,()()(),,()()(),,,(
),()()()(
5213
231)()(221
42)(3)(1)(0
θθ
θ
θ
rErKrKtXzrErKrKeztXztPX
rKrKezrKerKezXrBrC
rCrBrC
++=
++==
+++=− (3.3.29)
Considering equation (3.3.8) and using equation (3.3.29), we get
.0),(),()(2 2)(5)(1)( =++ zrEerEerKte zrBrCrC θθ (3.3.30)
Differentiating the above equation first with respect to t and then with respect to ,θ
we get 0)(1 =rK and ),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and
)(6 rK are functions of integration. Substituting back the above values in (3.3.30) we
get ),()(),()(),( 75)()(25)()(2 rKrKezzrErKezrE rBrCrBrCz +−=⇒−= −− where )(7 rK
is a function of integration. Substituting all the above information in (3.3.29) we get
).()()(),()()(),,,(),()()(
652375)()(32
2142)(3)(0
rKrKrtKXrKrKezrKtXztPXrKrKezrKeX
rBrC
rCrB
++=+−=
=++=− θ
θθ (3.3.31)
Considering equation (3.3.9) and using equation (3.3.31), we get
.0)()()()()()()(),,( 42)(2)(3)(3)(2 =−−−−− rKrKezrKerCzrKerKerBztP rrrCrC
rrrBrB
rt θθθ (3.3.32)
Differentiating equation (3.3.32) with respect to t we get ⇒= 0),,(2 ztPtt θ
),,(),(),,( 762 zEzEtztP θθθ += where ),(6 zE θ and ),(7 zE θ are functions of
integration. Substituting back the above value in (3.3.32) and differentiating with
respect to ,θ we get
.0)()()(),( 3)(3)(6 =−− rKerKerBzE rrBrB
rθθ (3.3.33)
Now differentiating the above equation with respect to ,θ we get
),()(),(0),( 9866 zKzKzEzE +=⇒= θθθθθ where )(8 zK and )(9 zK are functions
of integration. Substituting back the above value in (3.3.33) and differentiating with
respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzKz Now substituting this value in
equation (3.3.33) and solving we get, .,)( 2)(
2)(
13 ℜ∈+= −− cecrecrK rBrB
112
Substituting all the above information in equation (3.3.32) and differentiating twice
with respect to ,z we get .,,)(0)( 434399 ℜ∈+=⇒= ccczczKzKzz Substituting back
this value in (3.3.32) and solve after differentiating with respect to ,z we get
.,)( 5)(
5)(
32 ℜ∈+= −− cecrecrK rCrC Now substituting all the above information in
equation (3.3.32) and solving we get, .,)( 6644 ℜ∈+= ccrcrK Refreshing the
system of equations (3.3.31) we get
).()(),()(
),,(,65)(
5)(
3375)()()(
2)(
12
7431
1645321
0
rKrKetcectrXrKrKezecterctX
zEtcctzctXcrcczcrzcrcXrCrCrBrCrBrB +++=+−+=
+++=+++++=−−−−− θ
θθθθ(3.3.34)
Considering equation (3.3.10) and using equation (3.3.34) we get
.0)()(21)()(
)()](21)([)(
21)(
21),(2
7)(7)(5)(
5)(21
71
=++−
−−−−+
rKerBrKerKez
rKerBrCzrBctrBctrzEtc
rBrr
rBr
rC
rCrrrrθθ
(3.3.35)
Differentiating equation (3.3.35) with respect to ,t we get
.0)(21)(
212 211 =−− rBcrBcrc rr Solving this equation and remember that in this case
,0)( ≠rBr we get ,01 =c which on back substitution gives us .02 =c Substituting
the above information in equation (3.3.35) and differentiating with respect to ,θ we
get ),()(),(0),( 111077 zKzKzEzE +=⇒= θθθθθ where )(10 zK and )(11 zK are
functions of integration. Now substituting the above information in equation (3.3.35)
and differentiating twice with respect to ,z we get
.,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzKzz Substituting back these information in
equation (3.3.35) and differentiating the resulting equation with respect to ,z we get
,)(0)()(21 )(
2)(
92
)()(2
)(
755)(5)(
7
rCrBrBrCrB
rrCrC
rr ecdreecrKrKerKeBCc−
−−
+=⇒=−⎟⎠⎞
⎜⎝⎛ −− ∫
.9 ℜ∈c Now substituting back all the above information in equation (3.3.35) we get
113
.,)(0)()()(21
102
)(
102
)(2
)(
877)(7)(
8 ℜ∈+−=⇒=++−−−
∫ cecdreecrKrKerKerBcrBrBrB
rrBrB
r
Refreshing the system of equations (3.3.34) we get
).(
,
),(,
6)(2
)(
92
)()(2
)(
7)(
5)(
33
2)(
102
)(2
)(
82
)(
92
)(2
)(
72
118743
16453
0
rKecdreecetcerctX
ecdreececzdreeczX
zKczctcctzXcrcczcrzX
rCrBrBrCrBrCrC
rBrBrBrBrBrB
++++=
+−−−=
++++=+++=
−−
−−−
−−−−−−
∫
∫∫θθ
θθ
(3.3.36)
Considering equation (3.3.11) and using equation (3.3.36) we get
.0)()(21)()]()([
21
)]()([21)(
21)(
21)(22
6)(6)(2)(
9
2)(
2)(
75311
73
=++−+
−+−−++ ∫−
rKerCrKeerCrBc
dreerCrBcrCctrCctrzKctc
rCrr
rCrB
rr
rBrB
rrrrz
θ
θθ
(3.3.37)
Differentiating equation (3.3.37) with respect to ,t we get
.0)(21)(
212 533 =−− rCcrCrcc rr Solving this equation and remember that in this case
,0)( ≠rCr we get ,03 =c which on back substitution gives us .05 =c Substituting the
above information in equation (3.3.37) and differentiating with respect to ,θ we get
0)]()([21)]()([
212 2
)(
92
)(2
)(
77 =−+−+ ∫− rB
rr
rBrB
rr erCrBcdreerCrBcc Solving this
equation and remember that in this case ,0)()( ≠− rCrB rr we get ,07 =c which on
back substitution gives us .09 =c Now substituting the above information in equation
(3.3.37) and differentiating twice with respect to ,z we get
.,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzKzz Substituting back these information in
equation (3.3.37) and differentiating the resulting equation with respect to ,z we get
,)(0)()()(21 2
)(
132
)(2
)(
1166)(6)(
11
rCrCrC
rrCrC
r ecdreecrKrKerKerCc−−−
+−=⇒=++ ∫.13 ℜ∈c Refreshing the system of equations (3.3.36) we get
114
,,
,,
2)(
132
)(2
)(
1132
)(
102
)(2
)(
82
1211841
640
rCrCrCrBrBrB
ecdreecXecdreecX
czcctcXcrcX−−−−−−
+−=+−=
+++=+=
∫∫
θ
(3.3.38)
where .,,,,,, 13121110864 ℜ∈ccccccc The line element for static cylindrically
symmetric space-times is given as
,2)(2)(222 dzededrdtds rCrB +++−= θ (3.3.39)
The above space-time admits seven linearly independent teleparallel Killing vector
fields which are ,r∂∂ ,
t∂∂ ,2
)(
θ∂∂− rB
e ,2)(
ze
rC
∂∂−
,t
rr
t∂∂
+∂∂
θθ
∂∂
+∂∂ )(1 rMr
and
,)(2
zrM
rz
∂∂
+∂∂ where dreerM
rBrB
∫−−
−= 2)(
2)(
1 )( and .)( 2)(
2)(
2 ∫−−
−= dreerMrCrC
Killing vector fields in general relativity are ,t∂∂
θ∂∂ and .
z∂∂ It is evident that
teleparallel Killing vector fields are different and more in number to the Killing
vector fields in general relativity. Cases (II)(b) and (II)(c) can be solved exactly the
same as in the above case.
Case (III)(a):
In this case we have ),(rAA = ),(rBB = ),(rCC = ,BA ≠ CA ≠ and .CB = Now
substituting equation (3.3.12) in equation (3.3.6), we get
.0),,(),,( 1)(3)( =− zrPezrtPe rAt
rB θθ (3.3.40)
Differentiating equation (3.3.40) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (3.3.40) we get
⇒= − ),(),,( 11 zrEezrP ABθθ ),,(),(),,( 311 zrEzrEezrP AB += −θθ where ),(3 zrE is
a function of integration. Refreshing the system of equations (3.3.12) we get
115
).,,(),,(),(),,,(),,(),(
43212
21310
θ
θθ
rtPXzrEzrEtXztPXzrEzrEeX AB
=+=
=+= −
(3.3.41)
Considering equation (3.3.7) and using equation (3.3.41) we get
.0)],(),([),,( 31)()()(4)( =+− − zrEzrEeertPe zzrArBrA
trB θθ (3.3.42)
Differentiating equation (3.3.42) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (3.3.42) we get
.0)],(),([),( 31)()()(4)( =+− − zrEzrEeerEe zzrArBrArB θθ Differentiating the above
equation with respect to θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ
where )(1 rK and )(2 rK are functions of integration. Substituting back the above
value in equation (3.3.42) and solving we get )()(),( 311 rKrKzzrE += and
),()(),( 42)()(3 rKrKezzrE rArB += − where )(3 rK and )(4 rK are functions of
integration. Substituting all the above information in (3.3.41) we get
( )( )
( ) ).,()()(),,()()(),,,(
),()()()(
5213
231221
42)()(3)()(1)()(0
θθ
θ
θ
rErKrKtXzrErKrKztXztPX
rKrKezrKerKezX rArBrArBrArB
++=
++==
+++= −−−
(3.3.43)
Considering equation (3.3.8) and using equation (3.3.43), we get
.0),(),()(2 251 =++ zrErErtK zθθ Differentiating this equation first with respect to t
and then with respect to ,θ we get 0)(1 =rK and
),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and )(6 rK are functions
of integration. Substituting back the above values, we get
),()(),()(),( 75252 rKrzKzrErKzrEz +−=⇒−= where )(7 rK is a function of
integration. Substituting all the above information in (3.3.43) we get
( )).()()(),()()(
),,,(),()()(65237532
2142)()(3)()(0
rKrKrtKXrKrKzrKtXztPXrKrKezrKeX rArBrArB
++=+−=
=++= −−
θ
θθ (3.3.44)
116
Considering equation (3.3.9) and using equation (3.3.44), we get
.0)()(21)()(
)(21)()(
21),,(
4)(4)(2)(
2)(3)(3)(2
=−−−
⎟⎠⎞
⎜⎝⎛ −−−⎟
⎠⎞
⎜⎝⎛ −−
rKerArKerKez
rKeABzrKerKeABztP
rArr
rAr
rB
rBrrr
rBrBrrt θθθ
(3.3.45)
Differentiating equation (3.3.45) with respect to t we get ⇒= 0),,(2 ztPtt θ
),,(),(),,( 762 zEzEtztP θθθ += where ),(6 zE θ and ),(7 zE θ are functions of
integration. Substituting back the above value in (3.3.45) and differentiating with
respect to ,θ we get
.0)()(21),( 3)(3)(6 =−⎟
⎠⎞
⎜⎝⎛ −− rKerKeABzE r
rBrBrrθθ (3.3.46)
Now differentiating the above equation with respect to ,θ we get
),()(),(0),( 9866 zKzKzEzE +=⇒= θθθθθ where )(8 zK and )(9 zK are functions
of integration. Substituting back the above value in (3.3.46) and differentiating with
respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzKz Now substituting this value in
equation (3.3.45) and solving we get,
.,)( 2
)(2
)(
22
)()(2
)(
13 ℜ∈+=
−−
−
∫ cecdreecrKrBrArArBrA
Substituting all the above
information in equation (3.3.45) and differentiating twice with respect to ,z we get
.,,)(0)( 434399 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in (3.3.45) and
solve after differentiating with respect to ,z we get
.,)( 5
)(2
)(
52
)()(2
)(
32 ℜ∈+=
−−
−
∫ cecdreecrKrBrArArBrA
Now substituting all the above
information in equation (3.3.45) and solving we get,
.,)( 62
)(
62
)(2
)(
44 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Refreshing the system of equations
(3.3.44) we get
117
).()(
),()(
),,(
,
65)(2
)(
52
)()(2
)(
33
75)(2
)(
22
)()(2
)(
12
7431
1
2)(
62
)(2
)(
42
)(
52
)(2
)(
32
)(
22
)(2
)(
10
rKrKetcdreectX
rKrKzectdreectX
zEtcctzctX
ecdreececzdreeczecdreecX
rBrArArBrA
rBrArArBrA
rArArArArArArArArA
+++=
+−+=
+++=
+++++=
−−
−
−−
−
−−−−−−−−−
∫
∫
∫∫∫
θ
θθ
θθ
(3.3.47)
Considering equation (3.3.10) and using equation (3.3.47) we get
.0)()(21)()()()(
21
)]()([21)]()([
21),(2
7)(7)(5)(5)(
2)(
22
)(2
)(
17
1
=++−−
−+−++ ∫−
rKerBrKerKezrKerBz
erBrActdreerBrActzEtc
rBrr
rBr
rBrBr
rA
rr
rArA
rrθθ (3.3.48)
Differentiating equation (3.3.48) with respect to ,t we get
.0)]()([21)]()([
212 2
)(
22
)(2
)(
11 =−+−+ ∫− rA
rr
rArA
rr erBrAcdreerBrAcc Solving this
equation and remember that in this case ,0)()( ≠− rBrA rr we get ,01 =c which on
back substitution gives us .02 =c Substituting the above information in equation
(3.3.48) and differentiating with respect to ,θ we get
),()(),(0),( 111077 zKzKzEzE +=⇒= θθθθθ where )(10 zK and )(11 zK are functions
of integration. Now substituting the above information in equation (3.3.48) and
differentiating twice with respect to ,z we get
.,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzKzz Substituting back these information in
equation (3.3.48) and differentiating the resulting equation with respect to ,z we get
,)(0)()(21 2
)(
92
)(2
)(
755)(5)(
7
rBrBrB
rrBrB
r ecdreecrKrKerKeBc−−−
+=⇒=−− ∫ .9 ℜ∈c Now
substituting back all the above information in equation (3.3.48) we get
.,)(0)()()(21
102
)(
102
)(2
)(
877)(7)(
8 ℜ∈+−=⇒=++−−−
∫ cecdreecrKrKerKerBcrBrBrB
rrBrB
r
Refreshing the system of equations (3.3.47) we get
118
).(
,
),(
,
62)(
92
)(2
)(
7
)(2
)(
52
)()(2
)(
33
2)(
102
)(2
)(
82
)(
92
)(2
)(
72
118743
1
2)(
62
)(2
)(
42
)(
52
)(2
)(
30
rKecdreecetcdreectX
ecdreececzdreeczX
zKczctcctzX
ecdreececzdreeczX
rBrBrBrBrArArBrA
rBrBrBrBrBrB
rArArArArArA
++++=
+−−−=
++++=
+++=
−−−−
−−
−−−−−−
−−−−−−
∫∫
∫∫
∫∫
θθ
θθ
(3.3.49)
Considering equation (3.3.11) and using equation (3.3.49) we get
.0)()(21)()]()([
21
)]()([21)(22
6)(6)(2)(
5
2)(
2)(
311
73
=++−+
−+++ ∫−
rKerBrKeerBrAct
dreerBrActzKctc
rBrr
rBrA
rr
rArA
rrzθ
(3.3.50)
Differentiating equation (3.3.50) with respect to ,t we get
.0)]()([21)]()([
212 2
)(
52
)(2
)(
33 =−+−+ ∫− rA
rr
rArA
rr erBrAcdreerBrAcc Solving this
equation and remember that in this case ,0)()( ≠− rBrA rr we get ,03 =c which on
back substitution gives us .05 =c Substituting the above information in equation
(3.3.50) and differentiating with respect to ,θ we get .07 =c Now substituting the
above information in equation (3.3.50) and differentiating twice with respect to ,z we
get .,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzKzz Substituting back these information
in equation (3.3.50) and differentiating the resulting equation with respect to ,z we
get ,)(0)()()(21 2
)(
132
)(2
)(
1166)(6)(
11
rBrBrB
rrBrB
r ecdreecrKrKerKerBc−−−
+−=⇒=++ ∫
.13 ℜ∈c Refreshing the system of equations (3.3.49) we get
,
,
,,
2)(
132
)(
92
)(2
)(
113
2)(
102
)(
92
)(2
)(
82
12118412
)(
62
)(2
)(
40
rBrBrBrB
rBrBrBrB
rArArA
ececdreecX
ecezcdreecX
czcctcXecdreecX
−−−−
−−−−
−−−
++−=
+−−=
+++=+=
∫
∫
∫
θ
θ
(3.3.51)
119
where .,,,,,,, 131211109864 ℜ∈cccccccc The line element for static cylindrically
symmetric space-times is given as
),( 22)(22)(2 dzdedrdteds rBrA +++−= θ (3.3.52)
The above space-time admits eight linearly independent teleparallel Killing vector
fields which are ,r∂∂ ,2
)(
te
rA
∂∂−
,2)(
θ∂∂− rB
e ,2)(
ze
rB
∂∂−
),(2)(
θθ
∂∂
−∂∂−
zz
erB
,)(1
trM
rt
∂∂
+∂∂
θθ
∂∂
+∂∂ )(2 rMr
and ,)(2
zrM
rz
∂∂
+∂∂ where
∫−−
= dreerMrArA
2)(
2)(
1 )(
and .)( 2)(
2)(
2 dreerMrBrB
∫−−
−= Killing vector fields in
general relativity are ,t∂∂ ,
θ∂∂
z∂∂ and ).(
θθ
∂∂
−∂∂ zz
It is evident that teleparallel
Killing vector fields are different and more in number to the Killing vector fields in
general relativity. Cases (III)(b) and (III)(c) can be solved exactly the same as in the
above case.
Case (IV):
In this case we have )()()( rCrBrA == and .0)( ≠rAr Now substituting equation
(3.3.12) in equation (3.3.6), we get
.0),,(),,( 13 =− zrPzrtPt θθ (3.3.53)
Differentiating equation (3.3.53) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (3.3.53) we get
⇒= ),(),,( 11 zrEzrP θθ ),,(),(),,( 311 zrEzrEzrP +=θθ where ),(3 zrE is a function
of integration. Refreshing the system of equations (3.3.12) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θ
θθ
rtPXzrEzrEtXztPXzrEzrEX
=+=
=+= (3.3.54)
120
Considering equation (3.3.7) and using equation (3.3.54) we get
.0),(),(),,( 314 =−− zrEzrErtP zzt θθ (3.3.55)
Differentiating equation (3.3.55) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (3.3.55) we get
.0),(),(),( 314 =−− zrEzrErE zzθθ Differentiating the above equation with respect to
θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ where )(1 rK and
)(2 rK are functions of integration. Substituting back the above values in equation
(3.3.55) and solving we get )()(),( 311 rKrKzzrE += and
),()(),( 423 rKrKzzrE += where )(3 rK and )(4 rK are functions of integration.
Substituting all the above information in (3.3.54) we get
( )( ) ( ) ).,()()(),,()()(
),,,(),()()()(52132312
2142310
θθ
θθ
rErKrKtXzrErKrKztXztPXrKreKzrKrKzX++=++=
=+++= (3.3.56)
Considering equation (3.3.8) and using equation (3.3.56), we get
.0),(),()(2 251 =++ zrErErtK zθθ Differentiating this equation first with respect to t
and then with respect to ,θ we get 0)(1 =rK and
),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and )(6 rK are functions
of integration. Substituting back the above values, we get
),()(),()(),( 75252 rKrzKzrErKzrEz +−=⇒−= where )(7 rK is a function of
integration. Substituting all the above information in (3.3.56) we get
).()()(),()()(),,,(),()()(
65237532
214230
rKrKrtKXrKrKzrKtXztPXrKrKzrKX
++=+−=
=++=
θ
θθ (3.3.57)
Considering equation (3.3.9) and using equation (3.3.57), we get
121
.0)()(21)()(
)()(21)()()(
21),,(
4)(4)(2)(
2)(3)(3)(2
=−−−
−−−
rKerArKerKez
rKerAzrKerKerAztP
rArr
rAr
rA
rArr
rArArt θθθ
(3.3.58)
Differentiating equation (3.3.58) with respect to t we get ⇒= 0),,(2 ztPtt θ
),,(),(),,( 762 zEzEtztP θθθ += where ),(6 zE θ and ),(7 zE θ are functions of
integration. Substituting back the above value in (3.3.58) and differentiating with
respect to ,θ we get
.0)()()(21),( 3)(3)(6 =−− rKerKerAzE r
rArArθθ (3.3.59)
Now differentiating the above equation with respect to ,θ we get
),()(),(0),( 9866 zKzKzEzE +=⇒= θθθθθ where )(8 zK and )(9 zK are functions
of integration. Substituting back the above value in (3.3.59) and differentiating with
respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzKz Now substituting this value in
equation (3.3.59) and solving we get, .,)( 22
)(
22
)(2
)(
13 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Substituting all the above information in equation (3.3.58) and differentiating twice
with respect to ,z we get .,,)(0)( 434399 ℜ∈+=⇒= ccczczKzKzz Substituting back
this value in (3.3.58) and solve after differentiating with respect to ,z we get
.,)( 52
)(
52
)(2
)(
32 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Now substituting all the above
information in equation (3.3.58) and solving we get,
.,)( 62
)(
62
)(2
)(
44 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Refreshing the system of equations
(3.3.57) we get
122
).()(
),()(),,(
,
652)(
52
)(2
)(
33
752)(
22
)(2
)(
127
4311
2)(
62
)(2
)(
42
)(
52
)(2
)(
32
)(
22
)(2
)(
10
rKrKetcdreectX
rKrKzectdreectXzEtcctzctX
ecdreececzdreeczecdreecX
rArArA
rArArA
rArArArArArArArArA
+++=
+−+=+++=
+++++=
−−−
−−−
−−−−−−−−−
∫
∫
∫∫∫
θ
θθ
θθ
(3.3.60)
Considering equation (3.3.10) and using equation (3.3.60) we get
.0)()(21)()()()(
21),(2 7)(7)(5)(5)(7
1 =++−−+ rKerArKerKezrKerAzzEtc rArr
rAr
rArArθθ (3.3.61)
Differentiating equation (3.3.61) with respect to ,t we get .01 =c Substituting back
this information and differentiating with respect to ,θ we get
),()(),(0),( 111077 zKzKzEzE +=⇒= θθθθθ where )(10 zK and )(11 zK are functions
of integration. Now substituting the above information in equation (3.3.61) and
differentiating twice with respect to ,z we get
.,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzKzz Substituting back these information in
equation (3.3.61) and differentiating the resulting equation with respect to ,z we get
,)(0)()(21 2
)(
92
)(2
)(
755)(5)(
7
rArArA
rrArA
r ecdreecrKrKerKeAc−−−
+=⇒=−− ∫ .9 ℜ∈c Now
substituting back all the above information in equation (3.3.61) we get
.,)(0)()()(21
102
)(
102
)(2
)(
877)(7)(
8 ℜ∈+−=⇒=++−−−
∫ cecdreecrKrKerKerAcrArArA
rrArA
r
Refreshing the system of equations (3.3.60) we get
).(
,
),(
,
62)(
92
)(2
)(
72
)(
52
)(2
)(
33
2)(
102
)(2
)(
82
)(
92
)(2
)(
72
)(
22
118743
1
2)(
62
)(2
)(
42
)(
22
)(
52
)(2
)(
30
rKecdreecetcdreectX
ecdreececzdreeczectX
zKczctcctzX
ecdreecececzdreeczX
rArArArArArA
rArArArArArArA
rArArArArArArA
++++=
+−−−=
++++=
++++=
−−−−−−
−−−−−−−
−−−−−−−
∫∫
∫∫
∫∫
θθ
θθ
θ
(3.3.62)
Considering equation (3.3.11) and using equation (3.3.62) we get
123
.0)()(21)()(22 6)(6)(11
73 =++++ rKerArKezKctc rArr
rAzθ (3.3.63)
Differentiating equation (3.3.63) with respect to t and θ respectively, we get
.073 == cc Now substituting the above information in equation (3.3.63) and
differentiating twice with respect to ,z we get
.,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzKzz Substituting back these information in
equation (3.3.63) and differentiating the resulting equation with respect to ,z we get
0)()()(21 6)(6)(
11 =++ rKerKerAc rrArA
r
,)( 2)(
132
)(2
)(
116
rArArA
ecdreecrK−−−
+−=⇒ ∫ .13 ℜ∈c Refreshing the system of equations
(3.3.62) we get
.
,
,,
2)(
132
)(
92
)(
52
)(2
)(
113
2)(
102
)(
92
)(
22
)(2
)(
82
12118412
)(
62
)(2
)(
42
)(
22
)(
50
rArArArArA
rArArArArA
rArArArArA
ececetcdreecX
ececzectdreecX
czcctcXecdreecececzX
−−−−−
−−−−−
−−−−−
+++−=
+−+−=
+++=+++=
∫
∫
∫
θ
θθ
(3.3.64)
where .,,,,,,,,, 13121110986542 ℜ∈cccccccccc The line element for static cylindrically
symmetric space-times is given as
),( 222)(22 dzddtedrds rA ++−+= θ (3.3.65)
The above space-time admits ten linearly independent teleparallel Killing vector
fields which are ,r∂∂ ,2
)(
te
rA
∂∂−
,2)(
θ∂∂− rA
e ,2)(
ze
rA
∂∂−
),(2)(
θθ
∂∂
−∂∂−
zz
erA
),(2)(
θθ
∂∂
+∂∂−
tt
erA
),(2)(
zt
tze
rA
∂∂
+∂∂−
,)(t
rMr
t∂∂
+∂∂
θθ
∂∂
−∂∂ )(rMr
and
,)(z
rMr
z∂∂
−∂∂ where .)( 2
)(2
)(
∫−−
= dreerMrArA
Killing vector fields in general
124
relativity are ,t∂∂ ,
θ∂∂ ,
z∂∂ ),(
zt
tz
∂∂
+∂∂ )(
θθ
∂∂
+∂∂ tt
and ).(θ
θ∂∂
−∂∂ zz
It is
evident that teleparallel Killing vector fields are different and more in number to the
Killing vector fields in general relativity.
Case (V)(a):
In this case we have ,constant=A ),(rBB = )(rCC = and .CB = Now substituting
equation (3.3.12) in equation (3.3.6), we get
.0),,(),,( 13)( =− zrPzrtPe trB θθ (3.3.66)
Differentiating equation (3.3.66) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (3.3.66) we get
⇒= ),(),,( 11 zrEezrP Bθθ ),,(),(),,( 311 zrEzrEezrP B +=θθ where ),(3 zrE is a
function of integration. Refreshing the system of equations (3.3.12) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θ
θθ
rtPXzrEzrEtXztPXzrEzrEeX B
=+=
=+= (3.3.67)
Considering equation (3.3.7) and using equation (3.3.67) we get
.0),(),(),,( 31)(4)( =−− zrEzrEertPe zzrB
trB θθ (3.3.68)
Differentiating equation (3.3.68) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (3.3.68) we get
.0),(),(),( 31)(4)( =−− zrEzrEerEe zzrBrB θ Differentiating the above equation with
respect to θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ where )(1 rK
and )(2 rK are functions of integration. Substituting back the above value in equation
(3.3.68) and solving we get )()(),( 311 rKrKzzrE += and
125
),()(),( 42)(3 rKrKezzrE rB += where )(3 rK and )(4 rK are functions of integration.
Substituting all the above information in (3.3.67) we get
( )( )
( ) ).,()()(),,()()(),,,(
),()()()(
5213
231221
42)(3)(1)(0
θθ
θ
θ
rErKrKtXzrErKrKztXztPX
rKrKezrKerKezX rBrBrB
++=
++==
+++=
(3.3.69)
Considering equation (3.3.8) and using equation (3.3.69), we get
.0),(),()(2 251 =++ zrErErtK zθθ Differentiating this equation first with respect to t
and then with respect to ,θ we get 0)(1 =rK and
),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and )(6 rK are functions
of integration. Substituting back the above values, we get
),()(),()(),( 75252 rKrzKzrErKzrEz +−=⇒−= where )(7 rK is a function of
integration. Substituting all the above information in (3.3.69) we get
( )).()()(),()()(
),,,(),()()(65237532
2142)(3)(0
rKrKrtKXrKrKzrKtXztPXrKrKezrKeX rBrB
++=+−=
=++=
θ
θθ (3.3.70)
Considering equation (3.3.9) and using equation (3.3.70), we get
.0)()()()()()()(),,( 42)(2)(3)(3)(2 =−−−−− rKrKezrKerBzrKerKerBztP rrrBrB
rrrBrB
rt θθθ (3.3.71)
Differentiating equation (3.3.71) with respect to t we get ⇒= 0),,(2 ztPtt θ
),,(),(),,( 762 zEzEtztP θθθ += where ),(6 zE θ and ),(7 zE θ are functions of
integration. Substituting back the above value in (3.3.71) and differentiating with
respect to ,θ we get .0)()()(),( 3)(3)(6 =−− rKerKerBzE rrBrB
rθθ Now differentiating this
equation with respect to ,θ we get ),()(),(0),( 9866 zKzKzEzE +=⇒= θθθθθ
where )(8 zK and )(9 zK are functions of integration. Substituting back the above
value and differentiating with respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzKz
Now substituting back this value in the above equation and solving we get,
.,)( 2)(
2)(
13 ℜ∈+= −− cecercrK rBrB Substituting all the above information in
126
equation (3.3.71) and differentiating twice with respect to ,z we get
.,,)(0)( 434399 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in (3.3.71) and
solve after differentiating with respect to ,z we get
.,)( 5)(
5)(
32 ℜ∈+= −− cecercrK rBrB Now substituting all the above information in
equation (3.3.71) and solving we get, .,)( 6644 ℜ∈+= ccrcrK Refreshing the
system of equations (3.3.70) we get
).()(),()(
),,(,65)(
5)(
3375)(
2)(
12
7431
1645321
0
rKrKetcectrXrKrKzecterctX
zEtcctzctXcrcczcrzccrXrBrBrBrB +++=+−+=
+++=+++++=−−−− θ
θθθθ (3.3.72)
Considering equation (3.3.10) and using equation (3.3.72) we get
.0)()(21)(
)()()(21)(
21)(
21),(2
7)(7)(
5)(5)(21
71
=++
−−−−+
rKerBrKe
rKezrKerBzrBctrBrctzEtc
rBrr
rB
rrBrB
rrrθθ
(3.3.73)
Differentiating equation (3.3.73) with respect to ,t we get
.0)(21)(
212 211 =−− rBctrBrctc rr Solving this equation and remember that in this
case ,0)( ≠rBr we get ,01 =c which on back substitution gives us .02 =c
Substituting the above information in equation (3.3.73) and differentiating with
respect to ,θ we get ),()(),(0),( 111077 zKzKzEzE +=⇒= θθθθθ where )(10 zK and
)(11 zK are functions of integration. Now substituting the above information in
equation (3.3.73) and differentiating twice with respect to ,z we get
.,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzKzz Substituting back these information in
equation (3.3.73) and differentiating the resulting equation with respect to ,z we get
0)()(21 5)(5)(
7 =−− rKerKeBc rrBrB
r ,)( 2)(
92
)(2
)(
75
rBrBrB
ecdreecrK−−−
+=⇒ ∫ .9 ℜ∈c
Now substituting back all the above information in equation (3.3.73) we get
127
0)()()(21 7)(7)(
8 =++ rKerKerBc rrBrB
r
.,)( 102
)(
102
)(2
)(
87 ℜ∈+−=⇒
−−−
∫ cecdreecrKrBrBrB
Refreshing the system of equations
(3.3.72) we get
).(
,
),(,
62)(
92
)(2
)(
7)(
5)(
33
2)(
102
)(2
)(
82
)(
92
)(2
)(
72
118743
16453
0
rKecdreecetcerctX
ecdreececzdreeczX
zKczctcctzXccrczcrzX
rBrBrBrBrB
rBrBrBrBrBrB
++++=
+−−−=
++++=+++=
−−−−−
−−−−−−
∫
∫∫θθ
θθ
(3.3.74)
Considering equation (3.3.11) and using equation (3.3.74) we get
.0)()(21)()(
21)(
21)(22 6)(6)(
5311
73 =++−−++ rKerBrKerBctrBctrzKctc rBrr
rBrrzθ (3.3.75)
Differentiating equation (3.3.75) with respect to ,t we get
.0)(21)(
212 533 =−− rBcrBrcc rr Solving this equation and remember that in this case
,0)( ≠rBr we get .053 == cc Also differentiating equation (3.3.75) with respect to
,θ we get .07 =c Now substituting the above information in equation (3.3.75) and
differentiating twice with respect to ,z we get
.,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzKzz Substituting back these information in
equation (3.3.75) and differentiating the resulting equation with respect to ,z we get
,)(0)()()(21 2
)(
132
)(2
)(
1166)(6)(
11
rBrBrB
rrBrB
r ecdreecrKrKerKerBc−−−
+−=⇒=++ ∫ .13 ℜ∈c
Refreshing the system of equations (3.3.74) we get
,
,
,,
2)(
132
)(
92
)(2
)(
113
2)(
102
)(
92
)(2
)(
82
1211841
640
rBrBrBrB
rBrBrBrB
ececdreecX
ecezcdreecX
czcctcXcrcX
−−−−
−−−−
++−=
+−−=
+++=+=
∫
∫θ
θ
(3.3.76)
128
where .,,,,,,, 131211109864 ℜ∈cccccccc The line element for static cylindrically
symmetric space-times is given as
),( 22)(222 dzdedrdtds rB +++−= θ (3.3.77)
The above space-time admits eight linearly independent teleparallel Killing vector
fields which are ,r∂∂ ,
t∂∂ ,2
)(
θ∂∂− rB
e ,2)(
ze
rB
∂∂−
),(2)(
θθ
∂∂
−∂∂−
zz
erB
,t
rr
t∂∂
+∂∂
θθ
∂∂
−∂∂ )(rMr
and ,)(z
rMr
z∂∂
−∂∂ where .)( 2
)(2
)(
dreerMrBrB
∫−−
= Killing vector
fields in general relativity are ,t∂∂ ,
θ∂∂
z∂∂ and ).(
θθ
∂∂
−∂∂ zz
It is evident that
only one teleparallel Killing vector field is the same as Killing vector field in general
relativity and teleparallel Killing vector fields are more in number to the Killing
vector fields in general relativity. Cases (V)(b) and (V)(c) can be solved exactly the
same as in the above case.
Case (VI)(a):
In this case we have )(rAA = and .constant== CB Now substituting equation
(3.3.12) in equation (3.3.6), we get
.0),,(),,( 1)(3 =− zrPezrtP rAt θθ (3.3.78)
Differentiating equation (3.3.78) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (3.3.78) we get
⇒= − ),(),,( 11 zrEezrP Aθθ ),,(),(),,( 311 zrEzrEezrP A += −θθ where ),(3 zrE is a
function of integration. Refreshing the system of equations (3.3.12) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θ
θθ
rtPXzrEzrEtXztPXzrEzrEeX A
=+=
=+= −
(3.3.79)
Considering equation (3.3.7) and using equation (3.3.79) we get
129
.0),(),(),,( 3)(14 =−− zrEezrErtP zrA
zt θθ (3.3.80)
Differentiating equation (3.3.80) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (3.3.80) we get
.0),(),(),( 3)(14 =−− zrEezrErE zrA
zθθ Differentiating the above equation with
respect to θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ where )(1 rK
and )(2 rK are functions of integration. Substituting back the above value in equation
(3.3.80) and solving we get )()(),( 311 rKrKzzrE += and
),()(),( 42)(3 rKrKezzrE rA += − where )(3 rK and )(4 rK are functions of
integration. Substituting all the above information in (3.3.79) we get
( )( )
( ) ).,()()(),,()()(),,,(
),()()()(
5213
231221
42)(3)(1)(0
θθ
θ
θ
rErKrKtXzrErKrKztXztPX
rKrKezrKerKezX rArArA
++=
++==
+++= −−−
(3.3.81)
Considering equation (3.3.8) and using equation (3.3.81), we get
.0),(),()(2 251 =++ zrErErtK zθθ Differentiating this equation first with respect to t
and then with respect to ,θ we get 0)(1 =rK and
),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and )(6 rK are functions
of integration. Substituting back the above values, we get
),()(),()(),( 75252 rKrzKzrErKzrEz +−=⇒−= where )(7 rK is a function of
integration. Substituting all the above information in (3.3.81) we get
( )).()()(),()()(
),,,(),()()(65237532
2142)(3)(0
rKrKrtKXrKrKzrKtXztPXrKrKezrKeX rArA
++=+−=
=++= −−
θ
θθ (3.3.82)
Considering equation (3.3.9) and using equation (3.3.82), we get
130
.0)()(21)(
)()(21)()(
21),,(
4)(4)(
22332
=−−
−+−+
rKerArKe
rKzrKAzrKrKAztP
rArr
rA
rrrrt θθθ (3.3.83)
Differentiating equation (3.3.83) with respect to t we get ⇒= 0),,(2 ztPtt θ
),,(),(),,( 762 zEzEtztP θθθ += where ),(6 zE θ and ),(7 zE θ are functions of
integration. Substituting back the above value in (3.3.83) and differentiating with
respect to ,θ we get
.0)()(21),( 336 =−+ rKrKAzE rrθθ (3.3.84)
Now differentiating the above equation with respect to ,θ we get
),()(),(0),( 9866 zKzKzEzE +=⇒= θθθθθ where )(8 zK and )(9 zK are functions
of integration. Substituting back the above value in (3.3.84) and differentiating with
respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzKz Now substituting this value in
equation (3.3.84) and solving we get, .,)( 22
)(
22
)(2
)(
13 ℜ∈+= ∫
−
cecdreecrKrArArA
Substituting all the above information in equation (3.3.83) and differentiating twice
with respect to ,z we get .,,)(0)( 434399 ℜ∈+=⇒= ccczczKzKzz Substituting back
this value in (3.3.83) and solve after differentiating with respect to ,z we get
.,)( 52
)(
52
)(2
)(
32 ℜ∈+= ∫
−
cecdreecrKrArArA
Now substituting all the above
information in equation (3.3.83) and solving we get,
.,)( 62
)(
62
)(2
)(
44 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Refreshing the system of equations
(3.3.82) we get
131
).()(
),()(
),,(
,
652)(
52
)(2
)(
33
752)(
22
)(2
)(
12
7431
1
2)(
62
)(2
)(
42
)(
52
)(2
)(
32
)(
22
)(2
)(
10
rKrKetcdreectX
rKrKzectdreectX
zEtcctzctX
ecdreececzdreeczecdreecX
rArArA
rArArA
rArArArArArArArArA
+++=
+−+=
+++=
+++++=
∫
∫
∫∫∫
−
−
−−−−−−−−−
θ
θθ
θθ
(3.3.85)
Considering equation (3.3.10) and using equation (3.3.85) we get
.0)()()(21)(
21),(2 752
)(
22
)(2
)(
17
1 =+−+++ ∫−
rKrKzerActdreerActzEtc rr
rA
r
rArA
rθθ (3.3.86)
Differentiating equation (3.3.86) with respect to ,t we get
.0)(21)(
212 2
)(
22
)(2
)(
11 =++ ∫− rA
r
rArA
r erAcdreerAcc Solving this equation and
remember that in this case ,0)( ≠rAr we get .021 == cc Substituting the above
information in equation (3.3.86) and differentiating with respect to ,θ we get
),()(),(0),( 111077 zKzKzEzE +=⇒= θθθθθ where )(10 zK and )(11 zK are functions
of integration. Now substituting the above information in equation (3.3.86) and
differentiating twice with respect to ,z we get
.,,)(0)( 87871010 ℜ∈+=⇒= ccczczKzKzz Substituting back these information in
equation (3.3.86) and differentiating the resulting equation with respect to ,z we get
.,)(0)( 99755
7 ℜ∈+=⇒=− ccrcrKrKc r Now substituting back all the above
information in equation (3.3.86) we get
.,)(0)( 1010877
8 ℜ∈+−=⇒=+ ccrcrKrKc r Refreshing the system of equations
(3.3.85) we get
132
).(
,
),(
,
697
2)(
52
)(2
)(
33
108972
118743
1
2)(
62
)(2
)(
42
)(
52
)(2
)(
30
rKcrcetcdreectX
crcczrczX
zKczctcctzX
ecdreececzdreeczX
rArArA
rArArArArArA
++++=
+−−−=
++++=
+++=
∫
∫∫
−
−−−−−−
θθ
θθ
(3.3.87)
Considering equation (3.3.11) and using equation (3.3.87) we get
.0)()(21)(
21)(22 62
)(
52
)(2
)(
311
73 =+++++ ∫−
rKerActdreerActzKctc r
rA
r
rArA
rzθ (3.3.88)
Differentiating equation (3.3.88) with respect to ,t we get
.0)(21)(
212 2
)(
52
)(2
)(
33 =++ ∫− rA
r
rArA
r erAcdreerAcc Solving this equation and
remember that in this case ,0)( ≠rAr we get .053 == cc Substituting the above
information in equation (3.3.88) and differentiating with respect to ,θ we get .07 =c
Now substituting the above information in equation (3.3.88) and differentiating twice
with respect to ,z we get .,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzKzz Substituting
back these information in equation (3.3.88) and differentiating the resulting equation
with respect to ,z we get .,)(0)( 13131166
11 ℜ∈+−=⇒=+ ccrcrKrKc r Refreshing
the system of equations (3.3.87) we get
,,
,,
139113
10982
12118412
)(
62
)(2
)(
40
ccrcXczcrcX
czcctcXecdreecXrArArA
++−=+−−=
+++=+=−−−
∫θ
θ
(3.3.89)
where .,,,,,,, 131211109864 ℜ∈cccccccc The line element for static cylindrically
symmetric space-times is given as
),( 2222)(2 dzddrdteds rA +++−= θ (3.3.90)
133
The above space-time admits eight linearly independent teleparallel Killing vector
fields which are ,r∂∂ ,2
)(
te
rA
∂∂−
,θ∂∂ ,
z∂∂ ),(
θθ
∂∂
−∂∂ zz
,)(t
rMr
t∂∂
+∂∂
θθ
∂∂
−∂∂ rr
and ,z
rr
z∂∂
−∂∂ where .)( 2
)(2
)(
∫−−
= dreerMrArA
Killing vector fields
in general relativity are ,t∂∂ ,
θ∂∂
z∂∂ and ).(
θθ
∂∂
−∂∂ zz
It is obvious that in this
case three teleparallel Killing vector fields are the same as Killing vector fields in
general relativity. Other teleparallel Killing vector fields are different than Killing
vector fields in general relativity. The number of teleparallel Killing vector fields are
more than Killing vector fields in general relativity. Cases (VI)(b) and (VI)(c) can be
solved exactly the same as in the above case.
3.4. Teleparallel Killing Vector Fields in Spatially
Homogeneous Rotating Space-Times
Consider spatially homogeneous rotating space-times in usual coordinates ),,,( zrt φ
(labeled by ),,,,( 3210 xxxx respectively) with the line element [39]
,)(2)( 22222 φφ ddtrBdzdrAdrdtds −+++−= (3.4.1)
where A and B are no where zero functions of r only. The above space-times
(3.4.1) admit at least three linearly independent Killing vector fields in general
relativity which are ,t∂∂
φ∂∂ and .
z∂∂ The tetrad components and its inverse can be
obtained by using the relation (1.3.4) as [70]
134
,
10000)()(0000100)(01
2
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
+=
rBrA
rB
S aµ
⎟⎟⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜⎜⎜
⎝
⎛
+
+−
=
1000
0)()(
1000010
0)()(
)(01
2
2
rBrA
rBrArB
Saµ (3.4.2)
Using (3.4.2) in (1.3.5) the corresponding non-vanishing Weitzenböck connections
are obtained as
,)(2
22
012 BA
BAAB+
′+′−=Γ ,
)(22
22
12 BABBA
+′+′
=Γ (3.4.3)
where dash denotes the derivative with respect to .r The non vanishing torsion
components by using equation (1.3.11) are [70]
,)(2
2221
0
BABAABT
+′−′
= ,)(2
2212
2
BABBAT
+′+′
= (3.4.4)
A vector field X is said to be teleparallel Killing vector field if it satisfies equation
(1.3.17). One can write (1.3.17) explicitly using (3.4.1.) and (3.4.4) as follows
,0)( 0,2
0,0 =+ XrBX (3.4.5)
,0)()( 21,
21,
00,
1 =′−−− XrBXrBXX (3.4.6)
,0)()()( 2,2
2,0
0,2
0,0 =++− XrBXXrAXrB (3.4.7)
,0)( 3,2
3,0
0,3 =−− XrBXX (3.4.8)
,01,1 =X (3.4.9)
,02
)()()( 22,
11,
21,
0 =′
−−− XrAXXrAXrB (3.4.10)
,03,1
1,3 =+ XX (3.4.11)
,0)()( 2,0
2,2 =− XrBXrA (3.4.12)
135
,0)()( 3,2
3,0
2,3 =+− XrAXrBX (3.4.13)
,03,3 =X (3.4.14)
Integrating equations (3.4.9) and (3.4.14), we get
,),,(),,,( 2311 φφ rtEXztEX ==
where ),,(1 ztE φ and ),,(2 φrtE are functions of integrations which are to be
determined. Now differentiating equation (3.4.8) and (3.4.13) with respect to ,z we
get respectively
,0)( 33,2
33,0 =+ XrBX (3.4.15)
,0)()( 33,2
33,0 =− XrAXrB (3.4.16)
Multiplying (3.4.15) by A(r) and (3.4.16) by B(r) then adding and solving we get
),,,(),,( 430 φφ rtErtEzX += (3.4.17)
Substituting back (3.4.17) in equation (3.4.15) we get
),,,(),,( 652 φφ rtErtEzX += (3.4.18)
where ),,,(3 φrtE ),,,(4 φrtE ),,(5 φrtE and ),,(6 φrtE are functions of integration
which are to be determined. Our system of equations become
,),,(),,,(),,(),,,(),,,(),,(
23652
11430
φφφ
φφφ
rtEXrtErtEzXztEXrtErtEzX
=+=
=+= (3.4.19)
Now we have to solve the system (3.4.19) by using the remaining six equations.
Considering equation (3.4.11) and using equation (3.4.19) then differentiating with
respect to ,z we get ),,(),(),,(0),,( 2111 φφφφ tFtzFztEztEzz +=⇒= where
),(1 φtF and ),(2 φtF are functions of integration. Substituting back in (3.4.11) we
get ),,(),(),,(),(),,( 31212 φφφφφ tFtrFrtEtFrtEr +−=⇒−= where ),(3 φtF is a
function of integration. Refreshing the system of equations (3.4.19) we get
136
).,(),(),,,(),,(),,(),(),,,(),,(
313652
211430
φφφφ
φφφφ
tFtrFXrtErtEzXtFtzFXrtErtEzX
+−=+=
+=+= (3.4.20)
Considering equation (3.4.12) and using equation (3.4.20) then differentiating with
respect to z and ,φ we get
),,(),,()(),,()( 435 rtFrtErBrtErA =− φφ (3.4.21)
where ),(4 rtF is a function of integration. Substituting back (3.4.21) in (3.4.12) we
get
),,(),,()(),,()( 546 rtFrtErBrtErA =− φφ (3.4.22)
where ),(5 rtF is the function of integration. Now considering equation (3.4.13) and
using equations (3.4.20) and (3.4.22) then differentiating the resulting equation with
respect to ,r we get .0),(),( 41 =− rtFtF rφφ Differentiating this equation with respect
to ,φ we get ),()(),(0),( 2111 tRtRtFtF +=⇒= φφφφφ where )(1 tR and )(2 tR are
functions of integration. Substituting back in the above equation we get
⇒=− 0),()( 41 rtFtR r ),()(),( 314 tRtRrrtF += where )(3 tR is the function of
integration. Substituting back all the above information in (3.4.13) we get
⇒=+ 0)(),( 33 tRtF φφ ),()(),( 433 tRtRtF +−= φφ where )(4 tR is the function of
integration. Refreshing the system of equations (3.4.20) we get
( )( ) ).()()()(),,,(),,(
),,()()(),,,(),,(43213652
2211430
tRtRtRtRrXrtErtEzXtFtRtRzXrtErtEzX
+−+−=+=
++=+=
φφφφ
φφφφ (3.4.23)
Considering equation (3.4.5) and using equation (3.4.23) then differentiating with
respect to z we get ⇒=+ 0),,()(),,( 53 φφ rtErBrtE tt
),,(),,()(),,( 653 φφφ rFrtErBrtE +−= where ),(6 φrF is a function of integration.
Also considering (3.4.8) and using the above information and equation (3.4.23) then
differentiating with respect to φ we get .0),()()( 631 =−−− φφ rFtRtrR tt
137
Differentiating the above equation with respect to φ we get ⇒= 0),(6 φφφ rF
),()(),( 216 rGrGrF += φφ where )(1 rG and )(2 rG are functions of integration.
Substituting back this value in the above equation and solving we get
,)( 211 ctctR +−= 31
1 )( crcrG +−= and ,)( 433 ctctR +−= .,,, 4321 ℜ∈cccc Now
substituting back all the above information in (3.4.8) and solving we get
,)( 652 ctctR +−= 75
2 )( crcrG += and ,)( 874 ctctR += .,,, 8765 ℜ∈cccc Now
Substituting the above information in equation (3.4.5) we get
),,(),,()(),,( 764 φφφ rFrtErBrtE +−= where ),(7 φrF is the function of
integration. Refreshing the system of equations (3.4.23) by substituting all the above
information, we get
( )[ ]( )[ ]( )[ ] ( ) ,
),,,(),,(),,(
),,(),,()(),,()(
874365213
65226521
1
767531
50
ctcctcctcctcrX
rtErtEzXtFctcctczX
rFrtErBcrccrcrtErBzX
+++−−+−+−−=
+=++−+−=
+−++++−=
φφ
φφφφ
φφφφ
(3.4.24)
where .,,,,,,, 87654321 ℜ∈cccccccc Considering equation (3.4.12) and using equation
(3.4.24) then differentiating with respect to ,z we get
( )( )⇒++
= 3125 ),,( crc
BABrtE φφ ( ) ( ) ),,(),,( 8
3125 rtFcrc
BABrtE +++
= φφ where
),(8 rtF is a function of integration. Now substituting back the above value in
equation (3.4.12) we get ( ) ⇒+
= ),(),,( 72
6 φφ φφ rFBA
BrtE ( ) ),,(),(),,( 972
6 rtFrFBA
BrtE ++
= φφ
where ),(9 rtF is a function of integration. Refreshing the system of equations
(3.4.24) we get
138
( ) ( ) ( )( )[ ]
( ) ( ) ( )( )[ ] ( ) ,
),,(),(),(
),,(
),,(),(),(
874365213
972
8312
2
26521
1
97275
8312
0
ctcctcctcctcrX
rtFrFBA
BrtFcrcBA
BzX
tFctcctczX
rtBFrFBA
AcrcrtBFcrcBA
AzX
+++−−+−+−−=
++
+⎥⎦
⎤⎢⎣
⎡++
+=
++−+−=
−+
+⎥⎦
⎤⎢⎣
⎡++−+
+=
φφ
φφ
φφ
φφ
(3.4.25)
where .,,,,,,, 87654321 ℜ∈cccccccc Considering equation (3.4.7) and using equation
(3.4.25) then differentiating with respect to ,z we get ( )( )⇒++
= 3128 1),( crc
BArtFt
( )( ) ),(),( 3312
8 rGcrcBA
trtF +++
= where )(3 rG is a function of integration.
Substituting back the above value in equation (3.4.7) we get
.0),(),()( 792 =−+ φφ rFrtFBA t Differentiating the above equation with respect to ,t
we get ⇒= 0),(9 rtFtt ),()(),( 549 rGrtGrtF += where )(4 rG and )(5 rG are
functions of integration. Substituting back this value in the above equation we get
( ) ),()(),( 6427 rGrGBArF ++= φφ where )(6 rG is a function of integration.
Refreshing the system of equation (3.4.25) we get
( ) ( ) ( )
( ) ( )( )[ ]
( ) ( ) ( )
( )( )[ ] ( ) .
),()()()()(
)(
),,(
,)()()()()(
)()()(
874365213
54462
3312312
2
26521
1
54462
753
3123120
ctcctcctcctcrX
rGrtGrGrBrGBA
B
rGcrcBA
tcrcBA
BzX
tFctcctczX
rGrGtBrGrArGBA
A
crcrGrBcrctBArBcrc
BAAzX
+++−−+−+−−=
++++
+
⎥⎦
⎤⎢⎣
⎡++
+++
+=
++−+−=
+−++
+
⎥⎦
⎤⎢⎣
⎡++−+
+−+
+=
φφ
φ
φ
φφ
φ
φ
(3.4.26)
Now considering equation (3.4.13) and using equation (3.4.26) then differentiating
with respect to ,t we get ⇒=+ 031 ccr .031 == cc Also considering equation
(3.4.6) and using equation (3.4.26) then differentiating the resulting equation with
139
respect to ,z we get .05 =c Substituting all the above information in system of
equations (3.4.26) we get
[ ] ( ) ( )[ ]
( )[ ] .
),()()()()()(
),,(
,)()()()()()()(
874623
54462
32
262
1
544627
30
ctccccrX
rGrtGrGrBrGBA
BrGzX
tFcczX
rGrGtBrGrArGBA
AcrGrBzX
++−+−=
++++
+=
++=
+−++
++−=
φφ
φ
φφ
φ
(3.4.27)
Considering equation (3.4.6) and using equation (3.4.27) then differentiating with
respect to ,t we get ⇒= 0),(2 φtFtt ),()(),( 212 φφφ KtKtF += where )(1 φK and
)(2 φK are functions of integration. Now substituting the above information and
equation (3.4.27) in equation (3.4.6) and differentiating the resulting equation with
respect to φ twice we get ⇒= 0)(1 φφφK .,,)( 1091091 ℜ∈+= ccccK φφ Also
substituting (3.4.27) and all the above information in (3.4.10) and differentiating with
respect to ,φ we get ⇒= 0)(2 φφφK ℜ∈+= 121112112 ,,)( ccccK φφ and
( ) .0)()(2)()()(4 =′−′ rBrArBrArG In order to solve this equation we need to discuss
the following three possibilities:
(I) ( ) ,0)()(2)()( ≠′−′ rBrArBrA ,0)(4 =rG
(II) ( ) ,0)()(2)()( =′−′ rBrArBrA .0)(4 ≠rG
(III) ( ) ,0)()(2)()( =′−′ rBrArBrA .0)(4 =rG
We will discuss each case in turn. It is important to note that in case (III) we will
obtain the same teleparallel Killing vector fields as in case (II) with the reduction of
one teleparallel Killing vector fields generated from .)(4 rG Hence we will not
discuss case (III).
140
Case (I):
In this case we have 0)(4 =rG and ( ) .0)()(2)()( ≠′−′ rBrArBrA Substituting the
above information in (3.4.6) and differentiating with respect to ,φ we get .09 =c
Refreshing the above system of equations (3.4.27) we get
[ ] ( )[ ] ( )[ ] .
),()()(,
),()()()(
874623
562
3212111062
1
5627
30
ctccccrX
rGrGBA
BrGzXcctccczX
rGBrGBA
AcrGrBzX
++−+−=
++
+=++++=
−+
++−=
φφ
φφ (3.4.28)
Considering equation (3.4.10) and using equation (3.4.28) then differentiating with
respect to z and solving we get 02 =c and .,)( 152
153 ℜ∈+
= cBA
crG Also
considering equation (3.4.6) and using equation (3.4.28) then differentiating with
respect to z and solving we get .,)( 1313106 ℜ∈+= cccrrG Substituting all the
above information in (3.4.13) and differentiating w.r.t ,r we get
( ) .0221
7152=′−
+
′+′ cBcBABBA (3.4.29)
Substituting the above equation (3.4.29) in the remaining equations and solving, after
some calculation we reach to the following three possibilities:
(i) ,0157 == cc ,0)( ≠′ rA ,0)( ≠′ rB
(ii) ,015 =c ,07 ≠c ,0)( ≠′ rA ,0)( =′ rB
(iii) ,0)( =′ rA .0)( ≠′ rB
We will discuss each possibility in turn.
141
Case (I)(i):
In this case we have ,0157 == cc 0)( ≠′ rA and .0)( ≠′ rB Substituting all the above
information in (3.4.28) we get
( )
( ) .),()(
,),()(
8635
131022
121110615
131020
ccrXrGcrcBA
BX
ccctczXrGBcrcBA
AX
+−=+++
=
+++=−++
= φ (3.4.30)
Considering equation (3.4.10) and using equation (3.4.30), the final solution of
equations (3.4.5) to (3.4.14) is given as
,,1,, 863
212
2126
122212
0 ccrXcBA
XcczXccBA
BX +−=+
=+=++
−= (3.4.31)
where .,,,, 22211286 ℜ∈ccccc The line element for spatially homogeneous rotating
space-time is given in equation (3.4.1). The above space-time (3.4.1) admits five
linearly independent teleparallel Killing vector fields which can be written as ,t∂∂
),(12 t
BBA ∂
∂−
∂∂
+ φ ,
r∂∂
z∂∂ and .
zr
rz
∂∂
−∂∂ Killing vector fields in general
relativity are ,t∂∂
φ∂∂ and .
z∂∂ On comparison to the Killing vector fields in general
relativity we see that only two teleparallel Killing vector field are same in both the
theories while others are totally different.
Case (I)(ii):
In this case we have ,015 =c ,07 ≠c ,0)( ≠′ rA .0)( =′ rB Substituting back the
above information in (3.4.10) we get 0131110 === ccc and .7
4
cc
B −= The line
element for space-time (3.4.1) in this case becomes
142
.2)(7
422222 φφ ddtcc
dzdrAdrdtds ++++−= (3.4.32)
Solution of equations (3.4.5) to (3.4.14) is given as [70]
,,1
,,
87463
2122
2
1261
2472122
230
ctcccrXccA
X
cczXcczccA
cX
++−−=+
=
+=+++
−=
φ
(3.4.33)
where ,0,0,,,,,,,,, 7424232221128764 ≠≠ℜ∈ ccccccccccc 7
423 c
cc
−= and
.)( 2
7
422 c
cc
−= The above space-time (3.4.32) admits five linearly independent
teleparallel Killing vector fields which can be written ,t∂∂ ),(1
2322 t
ccA ∂
∂−
∂∂
+ φ
,r∂∂
z∂∂ and .
zr
rz
∂∂
−∂∂ Killing vector fields in general relativity are ,
t∂∂
φ∂∂ and
.z∂∂ In this case also two teleparallel Killing vector field are same as in general
relativity.
Case (I)(iii):
In this case we have 0)( ≠′ rB and .0)( =′ rA Equation
.0\,0)( 2323 ℜ∈=⇒=′ ccArA Substituting back the above information in
(3.4.29) we get 0157 == cc which on back substitution gives .04 =c Now
substituting the above information in (3.4.10) we get .0131110 === ccc The line
element (3.4.1) in this case becomes
,)(22223
222 φφ ddtrBdzdcdrdtds −+++−= (3.4.34)
Solution of equations (3.4.5) to (3.4.14) is given as [70]
143
,,,, 863
223
242126
125242
23
0 ccrXBc
cXcczXcc
Bc
BX +−=+
=+=++
−= (3.4.35)
where ).0(,,,,, 232524231286 ≠ℜ∈ ccccccc The above space-time (3.4.34) admits five
linearly independent teleparallel Killing vector fields which can be written as ,t∂∂
),(12
23t
BBc ∂
∂−
∂∂
+ φ ,
r∂∂ ,
z∂∂ and .
zr
rz
∂∂
−∂∂ Killing vector fields in general
relativity are ,t∂∂
φ∂∂ and .
z∂∂ In this case two teleparallel Killing vector field are
the same as in general relativity.
Case (II):
In this case we have 0)(4 ≠rG and ( ) .0)()(2)()( =′−′ rBrArBrA Equation
( ) ⇒=′−′ 0)()(2)()( rBrArBrA .)]([)( 2rBrA = Substituting all the above
information in equation (3.4.6) and solving we get 09 =c and
.0\,)( 13134 ℜ∈= cA
crG Refreshing the system of equations (3.4.27) we get
[ ][ ]
[ ] .
),(1)(211)(
,
),(1)(21)()(
874623
5132
613
32
121110621
513
6137
30
cctcccrX
rGctB
rGB
cB
rGzX
cctccczX
rGBctB
rGccrGrBzX
++−+−=
++++=
++++=
−−+++−=
φφ
φ
φφ
φ
(3.4.36)
Considering equation (3.4.6) and using equation (3.4.36) we get ⇒= 106 )( crGr
.,)( 1414106 ℜ∈+= cccrrG Now substituting all the above information in (3.4.10)
and differentiating w.r.t ,z get 02 =c and .,)( 152153 ℜ∈= c
Bc
rG Our system of
equations (3.4.36) becomes
144
.
),(1)(211
,
),(1)(21
87463
51321410132
152
12111061
5131410137
150
cctcrcX
rGctB
crcB
cBB
czX
cctczcX
rGBctB
crcccBc
zX
++−−=
+++++=
+++=
−−+++⎥⎦⎤
⎢⎣⎡ +−
=
φ
φ
φ
φ
(3.4.37)
Considering equation (3.4.13) and using equation (3.4.37) we get .0)( 7 =′ crB We
will discuss here the following three different possibilities:
(a) ,0)( ≠′ rB ,07 =c (b) ,0)( =′ rB ,07 ≠c (c) ,0)( =′ rB .07 =c
We will discuss each case in turn. It is important to note that in case (II)(c) we will
obtain the same teleparallel Killing vector fields as in case (II)(b) with the reduction
of one teleparallel Killing vector fields generated from .7c Hence we will not discuss
case (II)(c).
Case (II) (a):
In this case we have ,2BA = 0)( ≠′ rB and .07 =c Substituting back the above
information in (3.4.13) we get .2 415 cc = Also substituting the above values in (3.4.6)
we get .010 =c Now substituting the above information in (3.4.10) and differentiating
with respect to φ we get .0)( 13 =′ crB In this case both 0)( ≠′ rB and .013 ≠c This
case leads to contradiction and hence is not possible.
Case (II) (b):
In this case we have ,2BA = 0)( =′ rB and .07 ≠c Equation ⇒=′ 0)(rB
0\b ℜ∈=B which in turn implies .0\, 2 ℜ∈== baaA The line element for
spatially homogeneous rotating space-time in this case becomes
,222222 φφ ddtbdzdadrdtds −+++−= (3.4.38)
It is important to note that in this case all the torsion components become zero. The
above space-times admit ten linearly independent teleparallel Killing vector fields. In
145
this case the teleparallel Killing vector fields are same as in general relativity and are
given below
,
,111,
,
32413
82212929252722
64571
102212929252720
ccrctcX
czcba
zcba
bcba
btcba
rcba
rcba
bX
czcctcX
czcba
bzcba
acba
atcba
brcba
brcba
aX
+++=
++
−+
++
++
−+
−+
=
+++=
++
++
++
++
−+
++
=
φ
φ
φ
φ
(3.4.39)
where .,,,,,,,,, 10987654321 ℜ∈cccccccccc These Killing vector fields can be written
as ,t∂∂ ,
r∂∂ ,
φ∂∂ ,
z∂∂ ),(2 φ∂
∂+
∂∂
++
∂∂ b
ta
bar
rt ),(2 φ
φ∂∂
−∂∂
++
∂∂
tb
bar
r
,)()(12 ι
φφ∂∂
−+∂∂
−+
tbt
btaba
),(2 φ∂∂
+∂∂
++
∂∂ b
ta
baz
zt
)(2 φφ
∂∂
−∂∂
++
∂∂
tb
baz
z and .
zr
rz
∂∂
−∂∂
Examples
In the following we will discuss teleparallel Killing vector fields of some well known
spatially homogeneous rotating space-times. We are listing the results and details are
omitted. These results are obtained by the same procedure as adopted above in section
(3.4). These examples are as follows:
(1). Reboucas space-time:
If we choose )2cosh31()( 2 rrA +−= and ,2cosh2)( rrB = the above space-time
(3.4.1) becomes Reboucas space-time and takes the form [70]
.2cosh4)2cosh31( 222222 φφ ddtrdzdrdrdtds −++−+−= (3.4.40)
For the above space-time (3.4.40) the teleparallel Killing vector fields are given as
,,)2(cos,,)2coth2( 423
12
321
510 ccrXcrechXcczXccrX +−==+=+−= (3.4.41)
146
where .,,,, 54321 ℜ∈ccccc
(2). Som-Raychaudhuri space-time:
If we choose )1()( 22 rrrA −= and ,)( 2rrB = the above space-time (3.4.1) becomes
Som-Raychaudhuri space-time and takes the form [70]
.2)1( 22222222 φφ ddtrdzdrrdrdtds −+−++−= (3.4.42)
For the above space-time (3.4.42) the teleparallel Killing vector fields are given as
,,1,, 423
12
321
510 ccrXc
rXcczXccrX +−==+=+−= (3.4.43)
where .,,,, 54321 ℜ∈ccccc
(3). Hoenselaers-Vishveshwara space-time:
If we choose )3)(cosh1(cosh21)( −−−= rrrA and ),1(cosh)( −= rrB the above
space-time (3.4.1) becomes Hoenselaers-Vishveshwara space-time and takes the form
.)1(cosh2)3)(cosh1(cosh21 22222 φφ ddtrdzdrrdrdtds −−+−−−+−= (3.4.44)
For the above space-time (3.4.44) the teleparallel Killing vector fields are given as
,,cos2,,2 423
12
321
510 ccrXcechrXcczXccX +−==+=+−= (3.4.45)
where .,,,, 54321 ℜ∈ccccc
(4). Gödel-Friedmann space-time:
Choosing )sinh1(sinh)( 22 rrrA −= and ,sinh2)( 2 rrB = the above space-time
(3.4.1) becomes Gödel-Friedmann space-time and takes the form
.sinh22)sinh1(sinh 22222222 φφ ddtrdzdrrdrdtds −+−++−= (3.4.46)
For the above space-time (3.4.46) the teleparallel Killing vector fields are given as
147
,,cosh
cos,,tanh2 423
12
321
510 ccrXc
rechrXcczXccrX +−==+=+−= (3.4.47)
where .,,,, 54321 ℜ∈ccccc
(5). Stationary Gödel space-time:
Choosing raerA 2
21)( −= and raerB =)( where .0\Ra∈ The above space-time
(3.4.1) becomes stationary Gödel space-time and takes the form
.221 222222 φφ ddtedzdedrdtds rara −+−+−= (3.4.48)
For the above space-time (3.4.48) the teleparallel Killing vector fields are given as
,,2,,2 423
12
321
510 ccrXceXcczXccX ar +−==+=+−= − (3.4.49)
where .,,,, 54321 ℜ∈ccccc
3.5. Summary of the Chapter
In this chapter we investigated teleparallel Killing vector fields for Kantowski-Sachs,
Bianchi type III, static cylindrically symmetric and spatially homogeneous rotating
space-times using direct integration technique. Following results are obtained from
the above study:
(1) In Kantowski-Sachs and Bianchi type III space-times: It turns out that the above
space-times admit only 4 or 6 teleparallel Killing vector fields. The space-time admits
six teleparallel Killing vector fields only when it’s metric function )(tA becomes
constant. The results for teleparallel Killing vector fields when it admits four
teleparallel Killing vector field are given in equations (3.2.21), (3.2.27), (3.2.34) and
(3.2.41). The result when the above space-time admits six teleparallel Killing vector
fields is given in equation (3.2.48).
148
(2) In static cylindrically symmetric space-times: Different possibilities for the
existence of teleparallel Killing vector fields have been found by using direct
integration technique. It turns out that the above space-times admit 7, 8 or 10
teleparallel Killing vector fields. The above space-time admits eight teleparallel
Killing vector fields when it becomes static plane symmetric space-time. The only
case when static cylindrically symmetric space-times admit ten teleparallel Killing
vector fields is the space like version of the Friedmann Robertson Walker 0=K
model. When the space-time becomes Minkowski then all the torsion components
become zero and the teleparallel Lie derivative for the metric gives the same
equations as in general relativity, hence the Killing vector fields are same in both the
theories. The results for above space time when it admits seven teleparallel Killing
vector fields are given in equations (3.3.25) and (3.3.38). When the above space time
admits eight teleparallel Killing vector fields, results are given in equations (3.3.51),
(3.3.76) and (3.3.89). When the above space time admits ten teleparallel Killing
vector fields, result is given in equation (3.3.64).
(3) In spatially homogeneous rotating space-times: Teleparallel Killing vector fields
have been explored in the above space-times and it turns out that these space-times
admit 5 or 10 teleparallel Killing vector fields. The results for five teleparallel Killing
vector fields are given in equations (3.4.31), (3.4.33) and (3.4.35). Also the result for
ten teleparallel Killing vector fields is given in equation (3.4.39). We also
investigated teleparallel Killing vector fields for some special classes of spatially
homogeneous rotating space-times and all these space-times possess five teleparallel
Killing vector fields which are given in equations (3.4.41), (3.4.43), (3.4.45), (3.4.47)
and (3.4.49).
149
Chapter 4
Teleparallel Proper Homothetic Vector
Fields in Bianchi Type I, Non Static Plane
Symmetric and Static Cylindrically
Symmetric Space-Times
4.1. Introduction
This chapter is devoted to investigate teleparallel proper homothetic vector fields in
Bianchi type I, non static plane symmetric and static cylindrically symmetric space-times
by using direct integration technique. We have discussed each possibility for the
existence of teleparallel proper homothetic vector fields in the above space-times. It turns
out that all the above space-times possess proper teleparallel homothetic vector fields for
special choice of the metric functions. This chapter is organized as follows: In section
(4.2) teleparallel proper homothetic vector fields of Bianchi type I space-times are
investigated. In the next section (4.3) teleparallel proper homothetic vector fields in non
static plane symmetric space-times in the context of teleparallel theory have been
explored. In section (4.4) teleparallel proper homothetic vector fields of static
cylindrically symmetric space-times are explored. Last section (4.5) of the chapter is
devoted to a detailed summary of the work.
150
4.2. Teleparallel Proper Homothetic Vector Fields In
Bianchi Type I Space-Times
The line element for Bianchi type I space-time in the usual coordinate system ),,,( zyxt
(labeled by ),,,,( 3210 xxxx respectively) is given by [75, 78]
.2)(22)(22)(222 dzedyedxedtds tCtBtA +++−= (4.2.1)
where BA, and C are functions of t only. The tetrad components and its inverse, non-
vanishing Weitzenböck connections and the non vanishing torsion components for (4.2.1)
are given in equations (4.2.2), (4.2.3) and (4.2.4). Now using (4.2.1) and (4.2.4) in
(1.3.17) we get the teleparallel homothetic equations as follows:
,3,3
2,2
1,1
0,0 α==== XXXX (4.4.2)
,01,2)(2
2,1)(2 =+ XeXe tBtA (4.2.3)
,01,3)(2
3,1)(2 =+ XeXe tCtA (4.2.4)
,02,3)(2
3,2)(2 =+ XeXe tCtB (4.2.5)
,01)(20,
1)(21,
0 =−− • XAeXeX tAtA (4.2.6)
,02)(20,
2)(22,
0 =−− • XBeXeX tBtB (4.2.7)
.03)(20,
3)(23,
0 =−− • XCeXeX tCtC (4.2.8)
Now integrating equations (4.4.2), we get
).,,(),,,(),,,(),,,(
4332
2110
yxtPzXzxtPyXzytPxXzyxPtX
+=+=
+=+=
αα
αα (4.2.9)
where ),,,(1 zyxP ),,,(2 zytP ),,(3 zxtP and ),,(4 yxtP are functions of integration
which are to be determined. In order to find solution for equations (4.4.2) to (4.2.8) we
will consider each possible form of the metric for Bianchi type I space-times and then
151
solve each possibility in turn. Following are the possible cases for the metric where the
above space-times admit teleparallel proper homothetic vector fields:
(I) )(),(),( tCCtBBtAA === and .,, CBCABA ≠≠≠
(II)(a) ),(),( tBBtAA == and .tan tconsC =
(II)(b) ),(),( tCCtAA == and .tan tconsB =
(II)(c) ),(),( tCCtBB == and .tan tconsA =
(III)(a) )(),(),( tCCtBBtAA === and ).()( tCtB =
(III)(b) )(),(),( tCCtBBtAA === and ).()( tCtA =
(III)(c) )(),(),( tCCtBBtAA === and ).()( tBtA =
(IV) )(),(),( tCCtBBtAA === and ).()()( tCtBtA ==
(V)(a) )(),(,tan tCCtBBtconsA === and ).()( tCtB =
(V)(b) )(,tan),( tCCtconsBtAA === and ).()( tCtA =
(V)(c) tconsCtBBtAA tan),(),( === and ).()( tBtA =
(VI)(a) )(tAA = and .tan tconsCB ==
(VI)(b) )(tBB = and .tan tconsCA ==
(VI)(c) )(tCC = and .tan tconsBA ==
We will discuss each possibility in turn.
Case (I):
In this case we have ),(tAA = ),(tBB = ),(tCC = ,BA ≠ CA ≠ and .CB ≠ Now
substituting equation (4.2.9) in equation (4.2.3), we get
.0),,(),,( 2)(23)(2 =+ zytPezxtPe ytA
xtB (4.2.10)
Differentiating equation (4.2.10) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (4.2.10) we get
152
⇒−= − ),(),,( 1)(2)(22 ztEezytP tAtBy ),,(),(),,( 31)(2)(22 ztEztEeyzytP tAtB +−= − where
),(3 ztE is a function of integration. Now refreshing the system of equations (4.2.9) we
get
).,,(),,(),(),,(),(),,,(
43212
31)(2)(2110
yxtPzXztEztxEyXztEztEeyxXzyxPtX tAtB
+=++=
+−=+= −
αα
αα (4.2.11)
Considering equation (4.2.4) and using equation (4.2.11) we get
.0)],(),([),,( 31)(2)(2)(24)(2 =+−+ − ztEztEeyeyxtPe zztAtBtA
xtC (4.2.12)
Differentiating equation (4.2.12) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (4.2.12) we get
.0),(),(),( 3)(21)(24)(2 =+− ztEeztEeyytEe ztA
ztBtC Differentiating this equation with
respect to y twice we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting back
the above value in equation (4.2.12) and solving, we get
)()(),( 31)(2)(21 tKtKezztE tBtC += − and ),()(),( 42)(2)(23 tKtKezztE tAtC +−= − where
),(1 tK ),(2 tK )(3 tK and )(4 tK are functions of integration. Substituting back the
above information in equation (4.2.11) we get
).,()()(),,()()(
),()()()(),,,(
5213
231)(2)(22
42)(2)(23)(2)(21)(2)(21
10
ytEtxKtKyxzXztEtxKtKezxyX
tKtKeztKeytKezyxXzyxPtX
tBtC
tAtCtAtBtAtC
+++=
+++=
+−−−=
+=
−
−−−
α
α
α
α
(4.2.13)
Considering equation (4.2.5) and using equation (4.2.13) we get
.0),(),()(2 2)(25)(21)(2 =++ ztEeytEetKxe ztB
ytCtC (4.2.14)
Differentiating the above equation with respect to x we get .0)(1 =tK Substituting the
above value in (4.2.14) and differentiating the resulting equation with respect to ,y we
get ),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are functions of
integration. Substituting the above values back in (4.2.14) we get
153
),()(),( 75)(2)(22 tKtKezztE tBtC +−= − where )(7 tK is a function of integration.
Refreshing the system of equations (4.2.13) we get
).()()(),()()(
),()()(),,,(
6523
75)(2)(232
42)(2)(23)(2)(21
10
tKtKytxKzXtKtKeztxKyX
tKtKeztKeyxXzyxPtX
tBtC
tAtCtAtB
+++=
+−+=
+−−=
+=
−
−−
α
α
α
α
(4.2.15)
Considering equation (4.2.6) and using equation (4.2.15) we get
.0)()()(),,()()(
)()]()(2[)()()]()(2[4)(2)(214)(22)(2
2)(23)(23)(2
=−−+−+
−++−
tKetAetxAzyxPtKetKez
tKetAtCztKeytKetAtBytA
ttA
txttA
ttC
tCttt
tBtBtt
α (4.2.16)
Differentiating (4.2.16) with respect to ,x we get .0)(),,( )(21 =− tAtxx etAzyxP α
Differentiating this equation with respect to ,x we get
),,(),(),(2
),,(0),,( 8762
11 zyEzyxEzyExzyxPzyxPxxx ++=⇒= where ),,(6 zyE
),(7 zyE and ),(8 zyE are functions of integration. Substituting back this value in the
above equation we get .0)(),( )(26 =− tAt etAzyE α Now differentiating this equation with
respect to ,y we get ),(),(0),( 866 zKzyEzyEy =⇒= where )(8 zK is a function of
integration. Now substituting back this value and differentiating with respect to ,z we get
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Substituting back this value we get
.,20)( 221)(2)(2
1 ℜ∈+=⇒=− cctceetAc tAtAt α
α Substituting the above values back in
(4.2.16) and differentiating twice with respect to ,y we get
),()(),(0),( 10977 zKzKyzyEzyEyy +=⇒= where )(9 zK and )(10 zK are functions of
integration. Substituting back the above value in (4.2.16) and differentiating with respect
to ,z we get .,)(0)( 3399 ℜ∈=⇒= cczKzKz Substituting back the above value in
(4.2.16) and differentiating with respect to ,y we get
.)()()]()(2[ 33)(23)(2 ctKetKetAtB t
tBtBtt −=+− solving this equation we get
.,)( 4)(2)(
4)()(2)(
33 ℜ∈+−= −−− ∫ cecdteectK tBtAtAtBtA Now substituting all the above
154
information in (4.2.16) and solve after differentiating twice with respect to ,z we get
.,,)(0)( 65651010 ℜ∈+=⇒= ccczczKzKzz Substituting back in equation (4.2.16) and solve
after differentiating with respect to ,z we get
.,)( 7)(2)(
7)()(2)(
52 ℜ∈+−= −−− ∫ cecdteectK tCtAtAtCtA Now substituting all the above
information in equation (4.2.16) and solving we get
.,)( 8)(
8)()(
64 ℜ∈+= −−− ∫ cecdteectK tAtAtA Refreshing the system of equation (4.2.15)
with the help of above information, we get
),()(
),()(
,
),,(2
65)(2)(7
)()(2)(5
3
75)(2)(2)(2)(4
)()(2)(3
2
)(8
)()(6
)(7
)()(5
)(4
)()(3
1
86531
20
tKtKyexcdteexczX
tKtKzeexcdteexcyX
ecdteec
ezcdteezceycdteeycxX
zyExcxzcxyccxtX
tCtAtAtCtA
tBtCtBtAtAtBtA
tAtAtA
tAtAtAtAtAtA
+++−=
+−+−=
++
−+−+=
+++++=
−−−
−−−−
−−−
−−−−−−
∫∫
∫∫∫
α
α
α
α
(4.2.17)
where .,,,,,, 8765431 ℜ∈ccccccc Now considering equation (4.2.10) and using equation
(4.2.17) we get
.0)()()()()()]()(2[
)()]()([)]()([),(27)(27)(25)(25)(2
)(2)(4
)()(3
83
=−−+−+
−−−−++ ∫ −
tKetBtKetKeztKetBtCz
etyBetBtAcxdteetBtAxczyExctB
tttB
ttCtC
tt
tBt
tAtt
tAtAtty α
(4.2.18)
Differentiating equation (4.2.18) with respect to ,x we get
.0)]()([)]()([2 )(4
)()(33 =−−−+ ∫ − tA
tttAtA
tt etBtAcdteetBtAcc Solving this equation and
remember that in this case ,0)()( ≠− tBtA tt we get .043 == cc Substituting the above
information in equation (4.2.18) and differentiating with respect to ,y we get
.0)(),( )(28 =− tBtyy etBzyE α Differentiating once again with respect to ,y we get
),()()(2
),(0),( 1312112
88 zKzyKzKyzyEzyEyyy ++=⇒= where ),(11 zK )(12 zK and
)(13 zK are functions of integration. Substituting back this value and differentiating the
resulting equation with respect to ,z we get .,)(0)( 991111 ℜ∈=⇒= cczKzKz Once
155
again substituting back we get ⇒=− 0)( )(29
tBt etBc α .,2
10109)(2 ℜ∈+= cctce tB
α Now
substituting the above information in equation (4.2.18) and differentiating with respect to
z twice, we get .,,)(0)( 121112111212 ℜ∈+=⇒= ccczczKzKzz Substituting back this value
in equation (4.2.18) and differentiating with respect to ,z we get
.0)()()]()(2[ 5)(25)(211 =+−+ tKetKetBtCc t
tCtCtt Solving this equation we get
.,)( 13)(2)(
13)()(2)(
115 ℜ∈+−= −−− ∫ cecdteectK tCtBtBtCtB Now substituting all the above
information in equation (4.2.18) we get ⇒=−− 0)()()( 7)(27)(212 tKetKetBc t
tBtBt
.,)( 14)(
14)()(
127 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations (4.2.17)
we get
),(
,
,
),(22
6)(2)(13
)()(2)(11
)(2)(7
)()(2)(5
3
)(14
)()(12
)(13
)()(11
2
)(8
)()(6
)(7
)()(5
1
1312119
2
651
20
tKeycdteeycexcdteexczX
ecdteecezcdteezcyX
ecdteecezcdteezcxX
zKycyzccyxcxzccxtX
tCtBtBtCtBtCtAtAtCtA
tBtBtBtBtBtB
tAtAtAtAtAtA
++−+−=
++−−=
++−+=
+++++++=
−−−−−−
−−−−−−
−−−−−−
∫∫∫∫∫∫
α
α
α
α
(4.2.19)
where .,,,,,,,,, 14131211987651 ℜ∈cccccccccc Now considering equation (4.2.11) and using
equation (4.2.19) we get
.0)()()(
)())()(())()((
))()(())()(()(22
6)(2)(26
)(2)(13
)()(11
)(7
)()(5
13115
=−−
−−−−+
−−−+++
∫∫
−
−
tKetCetK
etzCetCtBycdteetCtByc
etCtAcxdteetCtAxczKycxc
tCt
tCt
tCt
tBtt
tBtBtt
tAtt
tAtAttz
α (4.2.20)
Differentiating equation (4.2.20) with respect to ,x we get
.0)]()([)]()([2 )(7
)()(55 =−−−+ ∫ − tA
tttAtA
tt etCtAcdteetCtAcc Solving this equation and
remember that in this case ,0)()( ≠− tCtA tt we get .075 == cc Substituting the above
information in equation (4.2.20) and differentiating with respect to ,y we get
.0)]()([)]()([2 )(13
)()(1111 =−−−+ ∫ − tB
tttBtB
tt etCtBcdteetCtBcc Solving this equation and
remember that in this case ,0)()( ≠− tCtB tt we get .01311 == cc Now substituting all the
156
above information in equation (4.2.20) and differentiating with respect to ,z we get
.0)()( )(213 =− tCtzz etCzK α Again differentiating this equation with respect to z and
solving we get .,,,2
)(0)( 171615171615
21313 ℜ∈++=⇒= cccczcczzKzKzzz Substituting
back in the above equation and solving we get ⇒=− 0)( )(215
tCt etCc α
.,2181815
)(2 ℜ∈+= cctce tC
α Now substituting the above information in equation (4.2.20)
we get ⇒=−− 0)()()( 6)(26)(216 tKetKetCc t
tCtCt .,)( 19
)(19
)()(16
6 ℜ∈+= −−− ∫ cecdteectK tCtCtC
Refreshing the system of equations (4.2.19) we get
,
,,
,222
)(19
)()(16
3
)(14
)()(12
2)(8
)()(6
1
171612615
2
9
2
1
20
tCtCtC
tBtBtBtAtAtA
ecdteeczX
ecdteecyXecdteecxX
czcycxcczcycxtX
−−−
−−−−−−
++=
++=++=
+++++++=
∫∫∫
α
αα
α
(4.2.21)
where .,,,,,,,,, 1917161514129861 ℜ∈cccccccccc The line element for Bianchi type I space-
times in this case takes the form
,)2()2()2( 21815
2109
221
22 dzctcdyctcdxctcdtds ++++++−=ααα
(4.2.22)
where ).,,,0,0,0(,,,,, 159151911591181510921 ccccccccccccccc ≠≠≠≠≠≠ℜ∈ The above
space-time admits eight linearly independent teleparallel homothetic vector fields in
which seven are teleparallel Killing vector fields given as ,t∂∂ ,)(
xe tA
∂∂− ,)(
ye tB
∂∂−
,)(
ze tC
∂∂− ,)(1
xtG
tx
∂∂
+∂∂
ytG
ty
∂∂
+∂∂ )(2 and ,)(3
ztG
tz
∂∂
+∂∂ where
,)( )()(1 ∫ −−= dteetG tAtA
∫ −−= dteetG tBtB )()(2 )( and ∫ −−= dteetG tCtC )()(3 )( and one is
proper teleparallel homothetic vector field. Proper teleparallel homothetic vector field
after subtracting teleparallel Killing vector fields from (4.2.21) is given as
,,,,222
32115
2
9
2
1
20 zXyXxXczcycxtX αααα ===+++= (4.2.23)
157
Case (II)(a):
In this case we have tCtBBtAA tancos),(),( === and ).()( tBtA ≠ Now substituting
equation (4.2.9) in equation (4.2.3), we get
.0),,(),,( 2)(23)(2 =+ zytPezxtPe ytA
xtB (4.2.24)
Differentiating equation (4.2.24) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (4.2.24) we get
⇒−= − ),(),,( 1)(2)(22 ztEezytP tAtBy ),,(),(),,( 31)(2)(22 ztEztEeyzytP tAtB +−= − where
),(3 ztE is a function of integration. Now refreshing the system of equations (4.2.9) we
get
).,,(),,(),(),,(),(),,,(
43212
31)(2)(2110
yxtPzXztEztxEyXztEztEeyxXzyxPtX tAtB
+=++=
+−=+= −
αα
αα (4.2.25)
Considering equation (4.2.4) and using equation (4.2.25) we get
.0)],(),([),,( 31)(2)(2)(24 =+−+ − ztEztEeyeyxtP zztAtBtA
x (4.2.26)
Differentiating equation (4.2.26) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (4.2.26) we get
.0),(),(),( 3)(21)(24 =+− ztEeztEeyytE ztA
ztB Differentiating this equation with respect to
y twice we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting back the above
value in equation (4.2.26) and solving, we get )()(),( 31)(21 tKtKezztE tB += − and
),()(),( 42)(23 tKtKezztE tA +−= − where ),(1 tK ),(2 tK )(3 tK and )(4 tK are functions
of integration. Substituting back the above information in equation (4.2.25) we get
).,()()(),,()()(
),()()()(),,,(
5213
231)(22
42)(23)(2)(21)(21
10
ytEtxKtKyxzXztEtxKtKezxyX
tKtKeztKeytKezyxXzyxPtX
tB
tAtAtBtA
+++=
+++=
+−−−=
+=
−
−−−
α
α
α
α
(4.2.27)
158
Considering equation (4.2.5) and using equation (4.2.27) we get
.0),(),()(2 2)(251 =++ ztEeytEtxK ztB
y (4.2.28)
Differentiating the above equation with respect to x we get .0)(1 =tK Substituting the
above value in (4.2.28) and differentiating the resulting equation with respect to ,y we
get ),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are functions of
integration. Substituting the above values back in (4.2.28) we get
),()(),( 75)(22 tKtKezztE tB +−= − where )(7 tK is a function of integration. Refreshing
the system of equations (4.2.27) we get
).()()(),()()(
),()()(),,,(
6523
75)(232
42)(23)(2)(21
10
tKtKytxKzXtKtKeztxKyX
tKtKeztKeyxXzyxPtX
tB
tAtAtB
+++=
+−+=
+−−=
+=
−
−−
α
α
α
α
(4.2.29)
Considering equation (4.2.6) and using equation (4.2.29) we get
.0)()()(),,()(
)()()()()()]()(2[4)(2)(214)(2
223)(23)(2
=−−+−
+−+−
tKetAetxAzyxPtKe
tKztKtAztKeytKetAtBytA
ttA
txttA
ttttBtB
tt
α (4.2.30)
Differentiating (4.2.30) with respect to ,x we get .0)(),,( )(21 =− tAtxx etAzyxP α
Differentiating this equation with respect to ,x we get
),,(),(),(2
),,(0),,( 8762
11 zyEzyxEzyExzyxPzyxPxxx ++=⇒= where ),,(6 zyE
),(7 zyE and ),(8 zyE are functions of integration. Substituting back this value in the
above equation we get .0)(),( )(26 =− tAt etAzyE α Now differentiating this equation with
respect to ,y we get ),(),(0),( 866 zKzyEzyEy =⇒= where )(8 zK is a function of
integration. Now substituting back this value and differentiating with respect to ,z we get
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Substituting back this value we get
.,20)( 221)(2)(2
1 ℜ∈+=⇒=− cctceetAc tAtAt α
α Substituting the above values back in
(4.2.30) and differentiating twice with respect to ,y we get
159
),()(),(0),( 10977 zKzKyzyEzyEyy +=⇒= where )(9 zK and )(10 zK are functions of
integration. Substituting back the above value in (4.2.30) and differentiating with respect
to ,z we get .,)(0)( 3399 ℜ∈=⇒= cczKzKz Substituting back the above value in
(4.2.30) and differentiating with respect to ,y we get
.)()()]()(2[ 33)(23)(2 ctKetKetAtB t
tBtBtt −=+− solving this equation we get
.,)( 4)(2)(
4)()(2)(
33 ℜ∈+−= −−− ∫ cecdteectK tBtAtAtBtA Now substituting all the above
information in (4.2.30) and solve after differentiating twice with respect to ,z we get
.,,)(0)( 65651010 ℜ∈+=⇒= ccczczKzKzz Substituting back in equation (4.2.30) and solve
after differentiating with respect to ,z we get .,)( 7)(
7)()(
52 ℜ∈+−= ∫ − cecdteectK tAtAtA
Now substituting all the above information in equation (4.2.30) and solving we get
.,)( 8)(
8)()(
64 ℜ∈+= −−− ∫ cecdteectK tAtAtA Refreshing the system of equation (4.2.29)
with the help of above information, we get
),()(
),()(
,
),,(2
65)(7
)()(5
3
75)(2)(2)(4
)()(2)(3
2
)(8
)()(6
)(7
)()(5
)(4
)()(3
1
86531
20
tKtKyexcdteexczX
tKtKzeexcdteexcyX
ecdteecezcdteezceycdteeycxX
zyExcxzcxyccxtX
tAtAtA
tBtBtAtAtBtA
tAtAtAtAtAtAtAtAtA
+++−=
+−+−=
++−+−+=
+++++=
∫∫
∫∫∫
−
−−−−
−−−−−−−−−
α
α
α
α
(4.2.31)
where .,,,,,, 8765431 ℜ∈ccccccc Now considering equation (4.2.10) and using equation
(4.2.31) we get
.0)()()()()()(
)()]()([)]()([),(27)(27)(255
)(2)(4
)()(3
83
=−−+−
−−−−++ ∫ −
tKetBtKetKztKtBz
etyBetBtAcxdteetBtAxczyExctB
tttB
tt
tBt
tAtt
tAtAtty α
(4.2.32)
Differentiating equation (4.2.32) with respect to ,x we get
.0)]()([)]()([2 )(4
)()(33 =−−−+ ∫ − tA
tttAtA
tt etBtAcdteetBtAcc Solving this equation and
remember that in this case ,0)()( ≠− tBtA tt we get .043 == cc Substituting the above
information in equation (4.2.32) and differentiating with respect to ,y we get
.0)(),( )(28 =− tBtyy etBzyE α Differentiating once again with respect to ,y we get
160
),()()(2
),(0),( 1312112
88 zKzyKzKyzyEzyEyyy ++=⇒= where ),(11 zK )(12 zK and
)(13 zK are functions of integration. Substituting back this value and differentiating the
resulting equation with respect to ,z we get .,)(0)( 991111 ℜ∈=⇒= cczKzKz Once
again substituting back we get ⇒=− 0)( )(29
tBt etBc α .,2
10109)(2 ℜ∈+= cctce tB
α Now
substituting the above information in equation (4.2.32) and differentiating with respect to
z twice, we get .,,)(0)( 121112111212 ℜ∈+=⇒= ccczczKzKzz Substituting back this value
in equation (4.2.32) and differentiating with respect to ,z we get
.0)()()( 5511 =+− tKtKtBc tt Solving this equation we get
.,)( 13)(
13)()(
115 ℜ∈+−= ∫ − cecdteectK tBtBtB Now substituting all the above information
in equation (4.2.32) we get ⇒=−− 0)()()( 7)(27)(212 tKetKetBc t
tBtBt
.,)( 14)(
14)()(
127 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations (4.2.31)
we get
),(
,
,
),(22
6)(13
)()(11
)(7
)()(5
3
)(14
)()(12
)(13
)()(11
2
)(8
)()(6
)(7
)()(5
1
1312119
2
651
20
tKeycdteeycexcdteexczX
ecdteecezcdteezcyX
ecdteecezcdteezcxX
zKycyzccyxcxzccxtX
tBtBtBtAtAtA
tBtBtBtBtBtB
tAtAtAtAtAtA
++−+−=
++−−=
++−+=
+++++++=
∫∫∫∫∫∫
−−
−−−−−−
−−−−−−
α
α
α
α
(4.2.33)
where .,,,,,,,,, 14131211987651 ℜ∈cccccccccc Now considering equation (4.2.11) and using
equation (4.2.33) we get
.0)()()(
)()()(226)(
13)()(
11
)(7
)()(5
13115
=−−+
−+++
∫∫
−
−
tKetBycdteetByc
etAcxdteetAxczKycxc
ttB
ttBtB
t
tAt
tAtAtz
(4.2.34)
Differentiating equation (4.2.34) with respect to ,x we get
.0)()(2 )(7
)()(55 =−+ ∫ − tA
ttAtA
t etAcdteetAcc Solving this equation and remember that in
this case ,0)( ≠tAt we get .075 == cc Substituting the above information in equation
161
(4.2.34) and differentiating with respect to ,y we get
.0)()(2 )(13
)()(1111 =−+ ∫ − tB
ttBtB
t etBcdteetBcc Solving this equation and remember that in
this case ,0)( ≠tBt we get .01311 == cc Now substituting all the above information in
equation (4.2.34) and differentiating with respect to ,z we get ⇒= 0)(13 zKzz
Differentiating this equation with respect to z and solving we get
.,,)( 1615161513 ℜ∈+= ccczczK Now substituting the above information in equation
(4.2.34) we get ⇒=− 0)(615 tKc t .,)( 171715
6 ℜ∈+= cctctK Refreshing the system of
equations (4.2.33) we get
,,
,,22
17153)(
14)()(
122
)(8
)()(6
116151269
2
1
20
ctczXecdteecyX
ecdteecxXczcycxccycxtX
tBtBtB
tAtAtA
++=++=
++=++++++=
−−−
−−−
∫∫
αα
αα (4.2.35)
where .,,,,,,,, 17161514129861 ℜ∈ccccccccc The line element for Bianchi type I space-times
in this case takes the form
,)2()2( 22109
221
22 dzdyctcdxctcdtds +++++−=αα
(4.2.36)
where ).,0,0(,,, 919110921 cccccccc ≠≠≠ℜ∈ The above space-time admits eight linearly
independent teleparallel homothetic vector fields in which seven are teleparallel Killing
vector fields given as ,t∂∂ ,)(
xe tA
∂∂− ,)(
ye tB
∂∂− ,
z∂∂ ,)(1
xtG
tx
∂∂
+∂∂
ytG
ty
∂∂
+∂∂ )(2
and ,z
tt
z∂∂
+∂∂ where ∫ −−= dteetG tAtA )()(1 )(
and ∫ −−= dteetG tBtB )()(2 )( and one is
proper teleparallel homothetic vector field. Proper teleparallel homothetic vector field
after subtracting teleparallel Killing vector fields from (4.2.35) is given as
,,,,22
3219
2
1
20 zXyXxXcycxtX αααα ===++= (4.2.37)
Cases (II)(b) and (II)(c) can be solved exactly the same as in the above case.
162
Case (III)(a):
In this case we have ),(),(),( tCCtBBtAA === ),()( tBtA ≠ )()( tCtA ≠ and
).()( tCtB = Now substituting equation (4.2.9) in equation (4.2.3), we get
.0),,(),,( 2)(23)(2 =+ zytPezxtPe ytA
xtB (4.2.38)
Differentiating equation (4.2.38) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (4.2.38) we get
⇒−= − ),(),,( 1)(2)(22 ztEezytP tAtBy ),,(),(),,( 31)(2)(22 ztEztEeyzytP tAtB +−= − where
),(3 ztE is a function of integration. Now refreshing the system of equations (4.2.9) we
get
).,,(),,(),(),,(),(),,,(
43212
31)(2)(2110
yxtPzXztEztxEyXztEztEeyxXzyxPtX tAtB
+=++=
+−=+= −
αα
αα (4.2.39)
Considering equation (4.2.4) and using equation (4.2.39) we get
.0),(),(),,( 3)(21)(24)(2 =+− ztEeztEeyyxtPe ztA
ztB
xtB (4.2.40)
Differentiating equation (4.2.40) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (4.2.40) we get
.0),(),(),( 3)(21)(24)(2 =+− ztEeztEeyytEe ztA
ztBtB Differentiating this equation with
respect to y twice we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting back
the above value in equation (4.2.40) and solving, we get )()(),( 311 tKtKzztE += and
),()(),( 42)(2)(23 tKtKezztE tAtB +−= − where ),(1 tK ),(2 tK )(3 tK and )(4 tK are
functions of integration. Substituting back the above information in equation (4.2.39) we
get
163
).,()()(),,()()(
),()()()(),,,(
5213
2312
42)(2)(23)(2)(21)(2)(21
10
ytEtxKtKyxzXztEtxKtKzxyX
tKtKeztKeytKezyxXzyxPtX
tAtBtAtBtAtB
+++=
+++=
+−−−=
+=−−−
α
α
α
α
(4.2.41)
Considering equation (4.2.5) and using equation (4.2.41) we get
.0),(),()(2 251 =++ ztEytEtxK zy (4.2.42)
Differentiating the above equation with respect to x we get .0)(1 =tK Now
differentiating the above equation with respect to ,y we get
),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are functions of
integration. Substituting the above values back in (4.2.42) we get
),()(),( 752 tKtKzztE +−= where )(7 tK is a function of integration. Refreshing the
system of equations (4.2.41) we get
).()()(),()()(
),()()(),,,(
6523
7532
42)(2)(23)(2)(21
10
tKtKytxKzXtKtKztxKyX
tKtKeztKeyxXzyxPtX
tAtBtAtB
+++=
+−+=
+−−=
+=−−
α
α
α
α
(4.2.43)
Considering equation (4.2.6) and using equation (4.2.43) we get
.0)()()(),,()()(
)()]()(2[)()()]()(2[4)(2)(214)(22)(2
2)(23)(23)(2
=−−+−+
−++−
tKetAetxAzyxPtKetKez
tKetAtBztKeytKetAtBytA
ttA
txttA
ttB
tBttt
tBtBtt
α (4.2.44)
Differentiating (4.2.44) with respect to ,x we get .0)(),,( )(21 =− tAtxx etAzyxP α
Differentiating this equation again with respect to ,x we get
),,(),(),(2
),,(0),,( 8762
11 zyEzyxEzyExzyxPzyxPxxx ++=⇒= where ),,(6 zyE
),(7 zyE and ),(8 zyE are functions of integration. Substituting back this value in the
above equation we get .0)(),( )(26 =− tAt etAzyE α Now differentiating this equation with
respect to ,y we get ),(),(0),( 866 zKzyEzyEy =⇒= where )(8 zK is a function of
integration. Now substituting back this value and differentiating with respect to ,z we get
164
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Substituting back this value we get
.,20)( 221)(2)(2
1 ℜ∈+=⇒=− cctceetAc tAtAt α
α Substituting the above values back in
(4.2.44) and differentiating twice with respect to ,y we get
),()(),(0),( 10977 zKzKyzyEzyEyy +=⇒= where )(9 zK and )(10 zK are functions of
integration. Substituting back the above value in (4.2.44) and differentiating with respect
to ,z we get .,)(0)( 3399 ℜ∈=⇒= cczKzKz Substituting back the above value in
(4.2.44) and differentiating with respect to ,y we get
.)()()]()(2[ 33)(23)(2 ctKetKetAtB t
tBtBtt −=+− solving this equation we get
.,)( 4)(2)(
4)()(2)(
33 ℜ∈+−= −−− ∫ cecdteectK tBtAtAtBtA Now substituting all the above
information in (4.2.44) and solve after differentiating twice with respect to ,z we get
.,,)(0)( 65651010 ℜ∈+=⇒= ccczczKzKzz Substituting back in equation (4.2.44) and solve
after differentiating with respect to ,z we get
.,)( 7)(2)(
7)()(2)(
52 ℜ∈+−= −−− ∫ cecdteectK tBtAtAtBtA Now substituting all the above
information in equation (4.2.44) and solving we get
.,)( 8)(
8)()(
64 ℜ∈+= −−− ∫ cecdteectK tAtAtA Refreshing the system of equation (4.2.43)
with the help of above information, we get
),()(
),()(
,
),,(2
65)(2)(7
)()(2)(5
3
75)(2)(4
)()(2)(3
2
)(8
)()(6
)(7
)()(5
)(4
)()(3
1
86531
20
tKtKyexcdteexczX
tKtzKexcdteexcyX
ecdteecezcdteezceycdteeycxX
zyExcxzcxyccxtX
tBtAtAtBtA
tBtAtAtBtA
tAtAtAtAtAtAtAtAtA
+++−=
+−+−=
++−+−+=
+++++=
−−−
−−−
−−−−−−−−−
∫∫
∫∫∫
α
α
α
α
(4.2.45)
where .,,,,,, 8765431 ℜ∈ccccccc Now considering equation (4.2.10) and using equation
(4.2.45) we get
.0)()()()()()(
)()]()([)]()([),(27)(27)(25)(25)(2
)(2)(4
)()(3
83
=−−++
−−−−++ ∫ −
tKetBtKetKeztKetBz
etyBetBtAcxdteetBtAxczyExctB
tttB
ttBtB
t
tBt
tAtt
tAtAtty α
(4.2.46)
165
Differentiating equation (4.2.46) with respect to ,x we get
.0)]()([)]()([2 )(4
)()(33 =−−−+ ∫ − tA
tttAtA
tt etBtAcdteetBtAcc Solving this equation and
remember that in this case ,0)()( ≠− tBtA tt we get .043 == cc Substituting the above
information in equation (4.2.46) and differentiating with respect to ,y we get
.0)(),( )(28 =− tBtyy etBzyE α Differentiating once again with respect to ,y we get
),()()(2
),(0),( 1312112
88 zKzyKzKyzyEzyEyyy ++=⇒= where ),(11 zK )(12 zK and
)(13 zK are functions of integration. Substituting back this value and differentiating the
resulting equation with respect to ,z we get .,)(0)( 991111 ℜ∈=⇒= cczKzKz Once
again substituting back we get ⇒=− 0)( )(29
tBt etBc α .,2
10109)(2 ℜ∈+= cctce tB
α Now
substituting the above information in equation (4.2.46) and differentiating with respect to
z twice, we get .,,)(0)( 121112111212 ℜ∈+=⇒= ccczczKzKzz Substituting back this value
in equation (4.2.46) and differentiating with respect to ,z we get
.0)()()( 5)(25)(211 =++ tKetKetBc t
tBtBt Solving this equation we get
.,)( 13)(
13)()(
115 ℜ∈+−= −−− ∫ cecdteectK tBtBtB Now substituting all the above information
in equation (4.2.46) we get ⇒=−− 0)()()( 7)(27)(212 tKetKetBc t
tBtBt
.,)( 14)(
14)()(
127 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations (4.2.45)
we get
),(
,
,
),(22
6)(13
)()(11
)(2)(7
)()(2)(5
3
)(14
)()(12
)(13
)()(11
2
)(8
)()(6
)(7
)()(5
1
1312119
2
651
20
tKeycdteeycexcdteexczX
ecdteecezcdteezcyX
ecdteecezcdteezcxX
zKycyzccyxcxzccxtX
tBtBtBtBtAtAtBtA
tBtBtBtBtBtB
tAtAtAtAtAtA
++−+−=
++−−=
++−+=
+++++++=
−−−−−−
−−−−−−
−−−−−−
∫∫∫∫∫∫
α
α
α
α
(4.2.47)
where .,,,,,,,,, 14131211987651 ℜ∈cccccccccc Now considering equation (4.2.11) and using
equation (4.2.47) we get
166
.0)()()()(
)]()([)]()([)(226)(2)(26)(2
)(7
)()(5
13115
=−−−
−−−+++ ∫ −
tKetBetKetzB
etBtAcxdteetBtAxczKycxctB
ttB
ttB
t
tAtt
tAtAttz
α (4.2.48)
Differentiating equation (4.2.48) with respect to ,x we get
.0)]()([)]()([2 )(7
)()(55 =−−−+ ∫ − tA
tttAtA
tt etBtAcdteetBtAcc Solving this equation and
remember that in this case ,0)()( ≠− tBtA tt we get .075 == cc Substituting the above
information in equation (4.2.48) and differentiating with respect to ,y we get .011 =c
Now substituting all the above information in equation (4.2.48) and differentiating with
respect to ,z we get ⇒=⇒=− 913)(213 )(0)()( czKetBzK zz
tBtzz α
.,,2
)( 161516159
213 ℜ∈++= ccczcczzK Now substituting the above information in
equation (4.2.48) we get ⇒=−− 0)()()( 6)(26)(215 tKetKetBc t
tBtBt
.,)( 17)(
17)()(
156 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations (4.2.47)
we get
,
,,
,)22
(2
)(17
)(13
)()(16
3
)(14
)(13
)()(12
2)(8
)()(6
1
16151269
22
1
20
tBtBtBtB
tBtBtBtBtAtAtA
eceycdteeczX
ecezcdteecyXecdteecxX
czcycxcczycxtX
−−−−
−−−−−−−
+++=
+−+=++=
+++++++=
∫∫∫
α
αα
α
(4.2.49)
where .,,,,,,,, 17161514129861 ℜ∈ccccccccc The line element for Bianchi type I space-times
in this case takes the form
),()2()2( 22109
221
22 dzdyctcdxctcdtds +++++−=αα
(4.2.50)
where ).,0,0(,,, 919110921 cccccccc ≠≠≠ℜ∈ The above space-time admits nine linearly
independent teleparallel homothetic vector fields in which eight are teleparallel Killing
vector fields given as ,t∂∂ ,)(
xe tA
∂∂− ,)(
ye tB
∂∂− ,)(
ze tB
∂∂− ),()(
yz
zye tB
∂∂
−∂∂−
,)(1
xtG
tx
∂∂
+∂∂
ytG
ty
∂∂
+∂∂ )(2 and ,)(2
ztG
tz
∂∂
+∂∂ where ∫ −−= dteetG tAtA )()(1 )(
and
167
∫ −−= dteetG tBtB )()(2 )( and one is proper teleparallel homothetic vector field. Proper
teleparallel homothetic vector field after subtracting teleparallel Killing vector fields from
(4.2.49) is given as
,,,,)22
(2
3219
22
1
20 zXyXxXczycxtX αααα ===+++= (4.2.51)
Cases (III)(b) and (III)(c) can be solved exactly the same as the above case.
Case (IV):
In this case we have )(),(),( tCCtBBtAA === and ).()()( tCtBtA == Now
substituting equation (4.2.12) in equation (4.2.6), we get
.0),,(),,( 23 =+ zytPzxtP yx (4.2.52)
Differentiating equation (4.2.52) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (4.2.52) we get
⇒−= ),(),,( 12 ztEzytPy ),,(),(),,( 312 ztEztEyzytP +−= where ),(3 ztE is a function
of integration. Now refreshing the system of equations (4.2.9) we get
).,,(),,(),(),,(),(),,,(
43212
31110
yxtPzXztEztxEyXztEztEyxXzyxPtX
+=++=
+−=+=
αα
αα (4.2.53)
Considering equation (4.2.4) and using equation (4.2.53) we get
.0),(),(),,( 314 =+− ztEztEyyxtP zzx (4.2.54)
Differentiating equation (4.2.54) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (4.2.54) we get
.0),(),(),( 314 =+− ztEztEyytE zz Differentiating this equation with respect to y twice
we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting back the above value in
equation (4.2.54) and solving, we get )()(),( 311 tKtKzztE += and
168
),()(),( 423 tKtzKztE +−= where ),(1 tK ),(2 tK )(3 tK and )(4 tK are functions of
integration. Substituting back the above information in equation (4.2.53) we get
).,()()(),,()()(
),()()()(),,,(
5213
2312
4231110
ytEtxKtKyxzXztEtxKtKzxyX
tKtzKtKytKzyxXzyxPtX
+++=
+++=
+−−−=+=
α
α
αα
(4.2.55)
Considering equation (4.2.5) and using equation (4.2.55) we get
.0),(),()(2 251 =++ ztEytEtxK zy (4.2.56)
Differentiating the above equation with respect to x we get .0)(1 =tK Now
differentiating the above equation with respect to ,y we get
),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are functions of
integration. Substituting the above values back in (4.2.56) we get
),()(),( 752 tKtKzztE +−= where )(7 tK is a function of integration. Refreshing the
system of equations (4.2.55) we get
).()()(),()()(),()()(),,,(
65237532
423110
tKtKytxKzXtKtKztxKyXtKtKztyKxXzyxPtX
+++=+−+=
+−−=+=
αα
αα (4.2.57)
Considering equation (4.2.6) and using equation (4.2.57) we get
.0)()()(),,()(
)()()()()()(4)(2)(214)(2
2)(22)(23)(23)(2
=−−+−
+++
tKetAetxAzyxPtKe
tKeztKetAztKeytKetaytA
ttA
txttA
ttAtA
tttAtA
t
α (4.2.58)
Differentiating (4.2.58) with respect to ,x we get .0)(),,( )(21 =− tAtxx etAzyxP α
Differentiating again this equation again with respect to ,x we get
),,(),(),(2
),,(0),,( 8762
11 zyEzyxEzyExzyxPzyxPxxx ++=⇒= where ),,(6 zyE
),(7 zyE and ),(8 zyE are functions of integration. Substituting back this value in the
above equation we get .0)(),( )(26 =− tAt etAzyE α Now differentiating this equation with
respect to ,y we get ),(),(0),( 866 zKzyEzyEy =⇒= where )(8 zK is a function of
integration. Now substituting back this value and differentiating with respect to ,z we get
169
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Substituting back this value we get
.,20)( 221)(2)(2
1 ℜ∈+=⇒=− cctceetAc tAtAt α
α Substituting the above values back in
(4.2.58) and differentiating twice with respect to ,y we get
),()(),(0),( 10977 zKzKyzyEzyEyy +=⇒= where )(9 zK and )(10 zK are functions of
integration. Substituting back the above value in (4.2.58) and differentiating with respect
to ,z we get .,)(0)( 3399 ℜ∈=⇒= cczKzKz Substituting back the above value in
(4.2.58) and differentiating with respect to ,y we get .)()()( 33)(23)(2 ctKetKetA t
tAtAt −=+
solving this equation we get .,)( 4)(
4)()(
33 ℜ∈+−= −−− ∫ cecdteectK tAtAtA Now
substituting all the above information in (4.2.58) and solve after differentiating twice with
respect to ,z we get .,,)(0)( 65651010 ℜ∈+=⇒= ccczczKzKzz Substituting back in
equation (4.2.58) and solve after differentiating with respect to ,z we get
.,)( 7)(
7)()(
52 ℜ∈+−= −−− ∫ cecdteectK tAtAtA Now substituting all the above information in
equation (4.2.58) and solving we get .,)( 8)(
8)()(
64 ℜ∈+= −−− ∫ cecdteectK tAtAtA
Refreshing the system of equation (4.2.57) with the help of above information, we get
),()(
),()(
,
),,(2
65)(7
)()(5
3
75)(4
)()(3
2
)(8
)()(6
)(7
)()(5
)(4
)()(3
1
86531
20
tKtKyexcdteexczX
tKtzKexcdteexcyX
ecdteecezcdteezceycdteeycxX
zyExcxzcxyccxtX
tAtAtA
tAtAtA
tAtAtAtAtAtAtAtAtA
+++−=
+−+−=
++−+−+=
+++++=
−−−
−−−
−−−−−−−−−
∫∫
∫∫∫
α
α
α
α
(4.2.59)
where .,,,,,, 8765431 ℜ∈ccccccc Now considering equation (4.2.10) and using equation
(4.2.59) we get
.0)()()(
)()()()(),(27)(27)(2
5)(25)(2)(283
=−−
++−+
tKetAtKe
tKeztKetzAetyAzyExctA
tttA
ttAtA
ttA
ty α (4.2.60)
Differentiating equation (4.2.60) with respect to ,x we get .03 =c Substituting the above
information in equation (4.2.60) and differentiating with respect to ,y we get
170
.0)(),( )(28 =− tBtyy etAzyE α Differentiating once again with respect to ,y we get
),()()(2
),(0),( 1312112
88 zKzyKzKyzyEzyEyyy ++=⇒= where ),(11 zK )(12 zK and
)(13 zK are functions of integration. Substituting back this value in the above equation,
we get .)(0)()( 111)(211 czKetAzK tA
t =⇒=−α Now substituting the above information in
equation (4.2.60) and differentiating with respect to z twice, we get
.,,)(0)( 111011101212 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in equation
(4.2.60) and differentiating with respect to ,z we get .0)()()( 5)(25)(210 =++ tKetKetAc t
tAtAt
Solving this equation we get .,)( 12)(
12)()(
105 ℜ∈+−= −−− ∫ cecdteectK tAtAtA Now
substituting all the above information in equation (4.2.60) we get
⇒=−− 0)()()( 7)(27)(211 tKetKetAc t
tAtAt .,)( 13
)(13
)()(11
7 ℜ∈+= −−− ∫ cecdteectK tAtAtA
Refreshing the system of equations (4.2.59) we get
),(
,
,
),()22
(
6)(12
)()(10
)(7
)()(5
3
)(13
)()(11
)(12
)()(10
)(4
2
)(8
)()(6
)(7
)()(5
)(4
1
131110651
220
tKeycdteeycexcdteexczX
ecdteecezcdteezcexcyX
ecdteecezcdteezceycxX
zKycyzcxcxzccyxtX
tAtAtAtAtAtA
tAtAtAtAtAtAtA
tAtAtAtAtAtAtA
++−+−=
++−++=
++−+−=
+++++++=
−−−−−−
−−−−−−−
−−−−−−−
∫∫∫∫∫∫
α
α
α
α
(4.2.61)
where .,,,,,,, 13121187651 ℜ∈cccccccc Now considering equation (4.2.11) and using
equation (4.2.61) we get
.0)()()()()(22 6)(2)(26)(213105 =−−−++ tKetAetKetzAzKycxc tA
ttA
ttA
tz α (4.2.62)
Differentiating equation (4.2.62) with respect to x and y respectively, we get 05 =c and
.010 =c Now differentiating equation (4.2.62) with respect to ,z we get
⇒=⇒=− 113)(213 )(0)()( czKetAzK zz
tAtzz α .,,
2)( 151415141
213 ℜ∈++= ccczcczzK Now
substituting the above information in equation (4.2.62) we get
⇒=−− 0)()()( 6)(26)(214 tKetKetAc t
tAtAt .,)( 16
)(16
)()(14
6 ℜ∈+= −−− ∫ cecdteectK tAtAtA
Refreshing the system of equations (4.2.61) we get
171
,
,
,
,)222
(
)(16
)(12
)(7
)()(14
3
)(13
)(12
)(4
)()(11
2
)(8
)(7
)(4
)()(6
1
15141161
2220
tAtAtAtAtA
tAtAtAtAtA
tAtAtAtAtA
eceycexcdteeczX
ecezcexcdteecyX
ecezceycdteecxX
czcycxcczyxtX
−−−−−
−−−−−
−−−−−
++++=
+−++=
+−−+=
+++++++=
∫∫∫
α
α
α
α
(4.2.63)
where .,,,,,,,,, 1615141312118761 ℜ∈cccccccccc The line element for Bianchi type I space-
times in this case takes the form
),)(2( 22221
22 dzdydxctcdtds ++++−=α
(4.2.64)
where ).0(, 121 ≠ℜ∈ ccc The above space-time admits eleven linearly independent
teleparallel homothetic vector fields in which ten are teleparallel Killing vector fields
given as ,t∂∂ ,)(
xe tA
∂∂− ,)(
ye tA
∂∂− ,)(
ze tA
∂∂− ),()(
yz
zye tA
∂∂
−∂∂− ),()(
xy
yxe tA
∂∂
−∂∂−
),()(
xz
zxe tA
∂∂
−∂∂− ,)(
xtG
tx
∂∂
+∂∂
ytG
ty
∂∂
+∂∂ )( and ,)(
ztG
tz
∂∂
+∂∂ where
∫ −−= dteetG tAtA )()()(
and one is proper teleparallel homothetic vector field. Proper
teleparallel homothetic vector field after subtracting teleparallel Killing vector fields from
(4.2.63) is given as
,,,,)222
( 3211
2220 zXyXxXczyxtX αααα ===+++= (4.2.65)
Case (V)(a):
In this case we have )(),(,constant tCCtBBA === and ).()( tCtB = Now substituting
equation (4.2.9) in equation (4.2.3), we get
.0),,(),,( 23)(2 =+ zytPzxtPe yxtB (4.2.66)
Differentiating equation (4.2.66) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (4.2.66) we get
172
⇒−= ),(),,( 1)(22 ztEezytP tBy ),,(),(),,( 31)(22 ztEztEeyzytP tB +−= where ),(3 ztE is a
function of integration. Now refreshing the system of equations (4.2.9) we get
).,,(),,(),(),,(),(),,,(
43212
31)(2110
yxtPzXztEztxEyXztEztEeyxXzyxPtX tB
+=++=
+−=+=
αα
αα (4.2.67)
Considering equation (4.2.4) and using equation (4.2.67) we get
.0),(),(),,( 31)(24)(2 =+− ztEztEeyyxtPe zztB
xtB (4.2.68)
Differentiating equation (4.2.68) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (4.2.68) we get
.0),(),(),( 31)(24)(2 =+− ztEztEeyytEe zztBtB Differentiating this equation with respect to
y twice we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting back the above
value in equation (4.2.68) and solving, we get )()(),( 311 tKtKzztE += and
),()(),( 42)(23 tKtKezztE tB +−= where ),(1 tK ),(2 tK )(3 tK and )(4 tK are functions of
integration. Substituting back the above information in equation (4.2.67) we get
).,()()(),,()()(
),()()()(),,,(
5213
2312
42)(23)(21)(21
10
ytEtxKtKyxzXztEtxKtKzxyX
tKtKeztKeytKezyxXzyxPtX
tBtBtB
+++=
+++=
+−−−=
+=
α
α
α
α
(4.2.69)
Considering equation (4.2.5) and using equation (4.2.69) we get
.0),(),()(2 251 =++ ztEytEtxK zy (4.2.70)
Differentiating the above equation with respect to x we get .0)(1 =tK Now
differentiating the above equation with respect to ,y we get
),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are functions of
integration. Substituting the above values back in (4.2.70) we get
),()(),( 752 tKtKzztE +−= where )(7 tK is a function of integration. Refreshing the
system of equations (4.2.69) we get
173
).()()(),()()(
),()()(),,,(
6523
7532
42)(23)(21
10
tKtKytxKzXtKtKztxKyX
tKtKeztKeyxXzyxPtX
tBtB
+++=
+−+=
+−−=
+=
α
α
α
α
(4.2.71)
Considering equation (4.2.6) and using equation (4.2.71) we get
.0),,()(
)()()(2)()()(214
2)(22)(23)(23)(2
=+−
+++
zyxPtK
tKeztKetBztKeytKetBy
xt
ttBtB
tttBtB
t (4.2.72)
Differentiating (4.2.72) with respect to ,x we get ⇒= 0),,(1 zyxPxx
),,(),(),,( 761 zyEzyxEzyxP += where ),(6 zyE and ),(7 zyE are functions of
integration. Substituting back this value in equation (4.2.72) and differentiating the
resulting equation with respect to ,y we get .0),()()()(2 63)(23)(2 =++ zyEtKetKetB yttBtB
t
Now differentiating this equation with respect to ,y we get
),()(),(0),( 9866 zKzyKzyEzyEyy +=⇒= where )(8 zK and )(9 zK are functions of
integration. Now substituting back this value and differentiating with respect to ,z we get
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Substituting back this value we get
⇒=++ 0)()()(2 13)(23)(2 ctKetKetB t
tBtBt .,)( 2
)(22
)(21
3 ℜ∈+−= −− cecetctK tBtB Substituting
the above values back in (4.2.72) and differentiating with respect to ,z we get
.0)()()(2)( 2)(22)(29 =++ tKetKetBzK ttBtB
tz Differentiating this equation with respect to ,z
we get ⇒= 0)(9 zKzz .,,)( 43439 ℜ∈+= ccczczK Substituting back this value in the
above equation we get ⇒=++ 0)()()(2 2)(22)(23 tKetKetBc t
tBtBt
.,)( 5)(2
5)(2
32 ℜ∈+−= −− cecetctK tBtB Now substituting all the above information in
equation (4.2.72) and solving we get .,)()( 6644
44 ℜ∈+=⇒= cctctKctKt Refreshing
the system of equation (4.2.71) with the help of above information, we get
),()(),()(
,),,(65)(2
5)(2
3375)(2
2)(2
12
64532117
4310
tKtKyexcextczXtKtzKexcextcyX
ctczcztcycytcxXzyExcxzcxyctXtBtBtBtB +++−=+−+−=
++−+−+=++++=−−−− αα
αα(4.2.73)
where .,,,, 65431 ℜ∈ccccc Now considering equation (4.2.10) and using equation (4.2.73)
we get
174
.0)()()()(
)()()()()(),(27)(27)(25)(2
5)(2)(221
71
=−−+
+−+−+
tKetBtKetKez
tKetBzetyBtBcxtBxtczyExctB
tttB
ttB
tBt
tBttty α
(4.2.74)
Differentiating equation (4.2.74) with respect to ,x we get .0)()(2 211 =+− tBctBtcc tt
Solving this equation and remember that in this case ,0)( ≠tBt we get .021 == cc
Substituting the above information in equation (4.2.74) and differentiating with respect to
,y we get .0)(),( )(27 =− tBtyy etBzyE α Differentiating once again with respect to ,y we
get ),()()(2
),(0),( 1211102
77 zKzyKzKyzyEzyEyyy ++=⇒= where ),(10 zK )(11 zK and
)(12 zK are functions of integration. Substituting back this value and differentiating the
resulting equation with respect to ,z we get .,)(0)( 771010 ℜ∈=⇒= cczKzKz Once
again substituting back we get ⇒=− 0)( )(27
tBt etBc α .,2
887)(2 ℜ∈+= cctce tB
α Now
substituting the above information in equation (4.2.74) and differentiating with respect to
z twice, we get .,,)(0)( 1091091111 ℜ∈+=⇒= ccczczKzKzz Substituting back this value
in equation (4.2.74) and differentiating with respect to ,z we get
.0)()()( 5)(25)(29 =++ tKetKetBc t
tBtBt Solving this equation we get
.,)( 11)(
11)()(
95 ℜ∈+−= −−− ∫ cecdteectK tBtBtB Now substituting all the above information
in equation (4.2.74) we get ⇒=−− 0)()()( 7)(27)(210 tKetKetBc t
tBtBt
.,)( 12)(
12)()(
107 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations (4.2.73)
we get
),(
,
,),(2
6)(11
)()(9
)(25
3
)(12
)()(10
)(11
)()(9
2
6453112
1097
2
430
tKeycdteeycexczX
ecdteecezcdteezcyX
ctczctzcxXzKycyzccyxcxzctX
tBtBtBtB
tBtBtBtBtBtB
++−+=
++−+=
++−+=++++++=
−−−−
−−−−−−
∫∫∫
α
α
αα
(4.2.75)
where .,,,,,,,, 121110976543 ℜ∈ccccccccc Now considering equation (4.2.11) and using
equation (4.2.75) we get
175
.0)(
)()()()()()(22)(2
36)(2)(26
512
93
=−
−−−+++tB
t
ttB
ttB
ttz
etzB
tBctxtKetBetKtBxczKycxc
α (4.2.76)
Differentiating equation (4.2.76) with respect to ,x we get .0)()(2 353 =−+ tBtctBcc tt
Solving this equation and remember that in this case ,0)( ≠tBt we get .053 == cc
Substituting the above information in equation (4.2.76) and differentiating with respect to
,y we get .09 =c Now substituting all the above information in equation (4.2.76) and
differentiating with respect to ,z we get ⇒=⇒=− 712)(212 )(0)()( czKetBzK zz
tBtzz α
.,,2
)( 141314137
212 ℜ∈++= ccczcczzK Now substituting the above information in
equation (4.2.76) we get ⇒=−− 0)()()( 6)(26)(213 tKetKetBc t
tBtBt
.,)( 15)(
15)()(
136 ℜ∈+= −−− ∫ cecdteectK tBtBtB Refreshing the system of equations (4.2.75)
we get
,
,
,,)22
(
)(15
)(11
)()(13
3
)(12
)(11
)()(10
2
641
14131047
220
tBtBtBtB
tBtBtBtB
eceycdteeczX
ecezcdteecyX
ctcxXczcycxcczytX
−−−−
−−−−
+++=
+−+=
++=++++++=
∫∫
α
α
αα
(4.2.77)
where .,,,,,,, 15141312111064 ℜ∈cccccccc The line element for Bianchi type I space-times in
this case takes the form
),()2( 2287
222 dzdyctcdxdtds ++++−=α
(4.2.78)
where ).0(, 787 ≠ℜ∈ ccc The above space-time admits nine linearly independent
teleparallel homothetic vector fields in which eight are teleparallel Killing vector fields
given as ,t∂∂ ,
x∂∂ ,)(
ye tB
∂∂− ,)(
ze tB
∂∂− ),()(
yz
zye tB
∂∂
−∂∂− ),(
xt
tx
∂∂
+∂∂
))((y
tGt
y∂∂
+∂∂ and ),)((
ztG
tz
∂∂
+∂∂ where ∫ −−= dteetG tBtB )()()( and one is proper
teleparallel homothetic vector field. Proper teleparallel homothetic vector field after
subtracting teleparallel Killing vector fields from (4.2.77) is given as
176
,,,,)22
( 3217
220 zXyXxXczytX αααα ===++= (4.2.79)
Cases (V)(b) and (V)(c) can be solved exactly the same as the above case.
Case (VI)(a):
In this case we have )(tAA = and constant.==CB Now substituting equation (4.2.9) in
equation (4.2.3), we get
.0),,(),,( 2)(23 =+ zytPezxtP ytA
x (4.2.80)
Differentiating equation (4.2.80) with respect to ,x we get
),,(),(),,(0),,( 2133 ztEztxEzxtPzxtPxx +=⇒= where ),(1 ztE and ),(2 ztE are
functions of integration. Substituting back this value in equation (4.2.80) we get
⇒−= − ),(),,( 1)(22 ztEezytP tAy ),,(),(),,( 31)(22 ztEztEeyzytP tA +−= − where ),(3 ztE is a
function of integration. Now refreshing the system of equations (4.2.9) we get
).,,(),,(),(),,(),(),,,(
43212
31)(2110
yxtPzXztEztxEyXztEztEeyxXzyxPtX tA
+=++=
+−=+= −
αα
αα (4.2.81)
Considering equation (4.2.4) and using equation (4.2.81) we get
.0),(),(),,( 3)(214 =+− ztEeztEyyxtP ztA
zx (4.2.82)
Differentiating equation (4.2.82) with respect to ,x we get
),,(),(),,(0),,( 5444 ytEytxEyxtPyxtPxx +=⇒= where ),(4 ytE and ),(5 ytE are
functions of integration. Substituting back the above value in (4.2.82) we get
.0),(),(),( 3)(214 =+− ztEeztyEytE ztA
z Differentiating this equation with respect to y
twice we get ).()(),(0),( 2144 tKtyKytEytEyy +=⇒= Substituting back the above
value in equation (4.2.82) and solving, we get )()(),( 311 tKtKzztE += and
),()(),( 42)(23 tKtKezztE tA +−= − where ),(1 tK ),(2 tK )(3 tK and )(4 tK are functions
of integration. Substituting back the above information in equation (4.2.81) we get
177
).,()()(),,()()(
),()()()(),,,(
5213
2312
42)(23)(21)(21
10
ytEtxKtKyxzXztEtxKtKzxyX
tKtKeztKeytKezyxXzyxPtX
tAtAtA
+++=
+++=
+−−−=
+=−−−
α
α
α
α
(4.2.83)
Considering equation (4.2.5) and using equation (4.2.83) we get
.0),(),()(2 251 =++ ztEytEtxK zy (4.2.84)
Differentiating the above equation with respect to x we get .0)(1 =tK Now
differentiating the above equation with respect to ,y we get
),()(),(0),( 6555 tKtKyytEytEyy +=⇒= where )(5 tK and )(6 tK are functions of
integration. Substituting the above values back in (4.2.84) we get
),()(),( 752 tKtKzztE +−= where )(7 tK is a function of integration. Refreshing the
system of equations (4.2.83) we get
).()()(),()()(),()()(),,,(
65237532
42)(23)(2110
tKtKytxKzXtKtKztxKyXtKtKeztKeyxXzyxPtX tAtA
+++=+−+=
+−−=+= −−
αα
αα (4.2.85)
Considering equation (4.2.6) and using equation (4.2.85) we get
.0)()()(
),,()()()()()()()(4)(2)(2
14)(22233
=−−
+−+−+−
tKetAetxA
zyxPtKetKztKtAztKytKtAytA
ttA
t
xttA
tttt
α (4.2.86)
Differentiating (4.2.86) with respect to ,x we get .0)(),,( )(21 =− tAtxx etAzyxP α
Differentiating this equation again with respect to ,x we get
),,(),(),(2
),,(0),,( 8762
11 zyEzyxEzyExzyxPzyxPxxx ++=⇒= where ),,(6 zyE
),(7 zyE and ),(8 zyE are functions of integration. Substituting back this value in the
above equation we get .0)(),( )(26 =− tAt etAzyE α Now differentiating this equation with
respect to ,y we get ),(),(0),( 866 zKzyEzyEy =⇒= where )(8 zK is a function of
integration. Now substituting back this value and differentiating with respect to ,z we get
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Substituting back this value we get
178
.,20)( 221)(2)(2
1 ℜ∈+=⇒=− cctceetAc tAtAt α
α Substituting the above values back in
(4.2.86) and differentiating twice with respect to ,y we get
),()(),(0),( 10977 zKzKyzyEzyEyy +=⇒= where )(9 zK and )(10 zK are functions of
integration. Substituting back the above value in (4.2.86) and differentiating with respect
to ,z we get .,)(0)( 3399 ℜ∈=⇒= cczKzKz Substituting back the above value in
(4.2.86) and differentiating with respect to ,y we get .)()()( 333 ctKtKtA tt −=+− solving
this equation we get .,)( 4)(
4)()(
33 ℜ∈+−= ∫ − cecdteectK tAtAtA Now substituting all the
above information in (4.2.86) and solve after differentiating twice with respect to ,z we
get .,,)(0)( 65651010 ℜ∈+=⇒= ccczczKzKzz Substituting back in equation (4.2.86) and
solve after differentiating with respect to ,z we get
.,)( 7)(
7)()(
52 ℜ∈+−= ∫ − cecdteectK tAtAtA Now substituting all the above information in
equation (4.2.86) and solving we get .,)( 8)(
8)()(
64 ℜ∈+= −−− ∫ cecdteectK tAtAtA
Refreshing the system of equation (4.2.85) with the help of above information, we get
),()(
),()(
,
),,(2
65)(7
)()(5
3
75)(4
)()(3
2
)(8
)()(6
)(7
)()(5
)(4
)()(3
1
86531
20
tKtKyexcdteexczX
tKtzKexcdteexcyX
ecdteecezcdteezceycdteeycxX
zyExcxzcxyccxtX
tAtAtA
tAtAtA
tAtAtAtAtAtAtAtAtA
+++−=
+−+−=
++−+−+=
+++++=
∫∫
∫∫∫
−
−
−−−−−−−−−
α
α
α
α
(4.2.87)
where .,,,,,, 8765431 ℜ∈ccccccc Now considering equation (4.2.10) and using equation
(4.2.87) we get
.0)()()()(),(2 75)(4
)()(3
83 =−+−++ ∫ − tKtKzetAcxdteetAxczyExc tt
tAt
tAtAty (4.2.88)
Differentiating equation (4.2.88) with respect to ,x we get
.0)()(2 )(4
)()(33 =−+ ∫ − tA
ttAtA
t etAcdteetAcc Solving this equation and remember that in
this case ,0)( ≠tAt we get .043 == cc Substituting the above information in equation
(4.2.88) and differentiating with respect to ,y we get ⇒= 0),(8 zyEyy
179
),()(),( 12118 zKzyKzyE += where )(11 zK and )(12 zK are functions of integration.
Substituting back this value and differentiating the resulting equation twice with respect
to ,z we get .,,)(0)( 1091091111 ℜ∈+=⇒= ccczczKzKzz Once again substituting back we
get ⇒=+ 0)(59 tKc t .,)( 11119
5 ℜ∈+= cctctK Now substituting all the above
information in equation (4.2.88) we get ⇒=− 0)(710 tKc t .,)( 121210
7 ℜ∈+= cctctK
Refreshing the system of equations (4.2.87) we get
),(
,
,
),(2
6119
)(7
)()(5
3
12101192
)(8
)()(6
)(7
)()(5
1
12109651
20
tKyctcyexcdteexczX
ctczcztcyX
ecdteecezcdteezcxX
zKycyzcxcxzccxtX
tAtAtA
tAtAtAtAtAtA
++++−=
++−−=
++−+=
++++++=
∫
∫∫
−
−−−−−−
α
α
α
α
(4.2.89)
where .,,,,,,,, 121110987651 ℜ∈ccccccccc Now considering equation (4.2.11) and using
equation (4.2.89) we get
.0)()()()(22 6)(7
)()(5
1295 =−−+++ ∫ − tKetAcxdteetAxczKycxc t
tAt
tAtAtz (4.2.90)
Differentiating equation (4.2.90) with respect to ,x we get
.0)()(2 )(7
)()(55 =−+ ∫ − tA
ttAtA
t etAcdteetAcc Solving this equation and remember that in
this case ,0)( ≠tAt we get .075 == cc Substituting the above information in equation
(4.2.90) and differentiating with respect to ,y we get .09 =c Now substituting all the
above information in equation (4.2.90) and differentiating with respect to ,z we get
.,,)(0)( 141314131212 ℜ∈+=⇒= ccczczKzK zz Now substituting the above
information in equation (4.2.90) we get ⇒=− 0)(613 tKc t .,)( 151513
6 ℜ∈+= cctctK
Refreshing the system of equations (4.2.91) we get
,,
,,2
1511133
1211102
)(8
)()(6
114131061
20
cyctczXczctcyX
ecdteecxXczcycxccxtX tAtAtA
+++=+−+=
++=+++++= −−− ∫αα
αα (4.2.91)
180
where .,,,,,,,, 151413121110861 ℜ∈ccccccccc The line element for Bianchi type I space-times
in this case takes the form
),()2( 22221
22 dzdydxctcdtds ++++−=α
(4.2.92)
where ).0(, 121 ≠ℜ∈ ccc The above space-time admits nine linearly independent
teleparallel homothetic vector fields in which eight are teleparallel Killing vector fields
given as ,t∂∂ ,)(
xe tA
∂∂− ,
y∂∂ ,
z∂∂ ),(
yz
zy
∂∂
−∂∂ ,
yt
ty
∂∂
+∂∂
zt
tz
∂∂
+∂∂ and
,)(x
tGt
x∂∂
+∂∂ where ∫ −−= dteetG tAtA )()()(
and one is proper teleparallel homothetic
vector field. Proper teleparallel homothetic vector field after subtracting teleparallel
Killing vector fields from (4.2.91) is given as
,,,,2
3211
20 zXyXxXcxtX αααα ===+= (4.2.93)
Cases (VI)(b) and (VI)(c) can be solved exactly the same as the above case.
4.3. Teleparallel Proper Homothetic Vector Fields in
Non Static Plane Symmetric Space-Times
Consider plane symmetric non static space-times in usual coordinates ),,,( zyxt (labeled
by ),,,,( 3210 xxxx respectively) with the line element
),( 22),(222),(22 dzdyedxdteds xtBxtA +++−= (4.3.1)
where A and B are functions of t and x only. The above space-time admits five
teleparallel Killing vector fields which are [74]
,),(
te xtA
∂∂− ,
x∂∂ ,),(
ye xtB
∂∂− ,),(
ze xtB
∂∂− ).(),(
zy
yze xtB
∂∂
−∂∂− (4.3.2)
181
The tetrad components µaS and the inverse tetrad components µ
aS for the space-time
(4.3.1) can be obtained by using equation (1.3.4) as
),,,1,(diag ),(),(),( xtBxtBxtAa eeeS =µ ).,,1,(diag ),(),(),( xtBxtBxtAa eeeS −−−=µ (4.3.3)
The corresponding non-vanishing Weitzenböck connections for (4.3.3) are obtained by
using equation (1.3.5) as
,000 •=Γ A ,01
0 A′=Γ ,202 •=Γ B ,30
3 •=Γ B ,212 B′=Γ ,31
3 B′=Γ (4.3.4)
where “dot” represents the derivative with respect to t and “dash” denotes the derivative
with respect to .x The non vanishing torsion components are obtained by using equation
(1.3.11) as
,100 AT ′= ,02
2 •= BT ,033 •= BT ,12
2 BT ′= ,133 BT ′= (4.3.5)
A vector field X is said to be teleparallel homothetic vector field if it satisfies equation
(1.3.17). Using equation (1.3.17) explicitly by using (4.3.1) and (4.3.5) we get the
following equations:
,,000 α=+• XXA (4.3.6)
,,,, 33
22
11 α=== XXX (4.3.7)
,0,, 23
32 =+XX (4.3.8)
,0,, 0),(21
0),(20
1 =′−− XeAXeX xtAxtA (4.3.9)
,0,, 2),(21
2),(22
1 =′++ XeBXeX xtBxtB (4.3.10)
,0,, 3),(21
3),(23
1 =′++ XeBXeX xtBxtB (4.3.11)
,0,, 2),(22
0),(20
2),(2 =+− • XeBXeXe xtBxtAxtB (4.3.12)
.0,, 3),(23
0),(20
3),(2 =+− • XeBXeXe xtBxtAxtB (4.3.13)
Solving equation (4.3.6) and (4.3.7) by simple integration, we get
),,,(,),,(
),,,(),,,(4332
211),(),(),(0
yxtPzXzxtPyX
zytPxXzyxPedteeX xtAxtAxtA
+=+=
+=+= −− ∫αα
αα (4.3.14)
182
where ),,,(1 zyxP ),,,(2 zytP ),,(3 zxtP and ),,(4 yxtP are functions of integration and
we need to determine these functions. In order to find solution for equations (4.3.6) to
(4.3.13) we will consider each possible form of the metric for plane symmetric non static
space-times and then solve each possibility in turn. Following are the possible cases for
the metric where the above space-times admit teleparallel proper homotheric vector
fields:
(I) ),(tAA = ),,( xtBB = (II) ),(tAA = ),(tBB = ,BA ≠
(III) ),(tAA = ),(tBB = ,BA = (IV) ),(tAA = constant,=B
(V) ),,( xtAA = ),(xBB = (VI) ),,( xtAA = constant,=B
(VII) ),(tAA = ),(xBB = (VIII) ),(xAA = ),(xBB = ,BA ≠
(IX) ),(xAA = ),(xBB = ,BA = (X) ),(xAA = ,constant=B
(XI) ,constant=A ),,( xtBB = (XII) ,constant=A ),(tBB =
(XIII) ,constant=A ),(xBB =
It is important to note that when constant=A and costant=B the space-time becomes
Minkowski and gives the same proper homothetic vector fields as in general relativity.
We will solve equations from (4.3.8) to (4.3.13) for the above lited cases.
Case I:
In this case we have )(tAA = and ).,( xtBB = Considering equation (4.3.8) and using
equation (4.3.14) then differentiating with respect to ,z we get
),,(),(),,(0),,( 2133 xtKxtzKzxtPzxtPzz +=⇒= where ),(1 xtK and ),(2 xtK are
functions of integration. Substituting back the above value in equation (4.3.8) we get
),,(),(),,(),(),,( 31414 xtKxtyKyxtPxtKyxtPy +−=⇒−= where ),(3 xtK is a function
of integration. Refreshing the system of equations (4.3.14) we get
).,(),(),,(),(
),,,(),,,(313212
211)()()(0
xtKxtyKzXxtKxtzKyX
zytPxXzyxPedteeX tAtAtA
+−=++=
+=+= −− ∫αα
αα (4.3.15)
183
Considering equation (4.3.10) and using equation (4.3.15) then differentiating with
respect to ,y we get ⇒=+ 0)(),,( 22 Bxyy exBzytP α ⇒= 0),,(2 zytPyyy
),,(),(),(2
),,( 6542
2 ztKztyKztKyzytP ++= where ),,(4 ztK ),(5 ztK and ),(6 ztK are
functions of integration. Substituting back this value in the above equation we get
⇒=+ 0)(),( 24 Bx exBztK α ⇒= 0),(4 ztK z ),(),( 14 tDztK = where )(1 tD is a
function of integration. Once again substituting back we get ⇒=+ 0)()( 21 Bx exBtD α
),()(2 212 tDtxDe B +−
=α
where )(2 tD is a function of integration. Substituting back the
above values in (4.3.10) and solving after differentiating with respect to ,z we get
),(),( 125 tDztK = )(),( 3),(1 tDextK xtB−= and ),(),( 4),(2 tDextK xtB−= where ),(3 tD
)(4 tD and )(12 tD are functions of integration. Refreshing equation (4.3.15) we get
).,()(),()(
),,()()(2
),,,(
33),(34),(3),(2
61212
11)()()(0
xtKtDeyzXtDetDezyX
ztKtyDtDyxXzyxPedteeX
xtBxtBxtB
tAtAtA
+−=++=
+++=+=
−−−
−− ∫αα
αα (4.3.16)
Considering equation (4.3.11) and using equation (4.3.16) we get
⇒=+ 0)(),( 26 Bxzz exBztK α ⇒= )(),( 16 tDztK zz ),()()(
2),( 651
26 tDtzDtDzztK ++= where
)(5 tD and )(6 tD are functions of integration. Substituting back the above values in
equation (4.3.11) and solving we get 0)(5 =tD and ),(),( 73 tDextK B−= where )(7 tD is
the function of integration. Substituting all the above information in (4.3.16) we get
).()(),()(
),()()()22
(
),,,(
7),(3),(3
4),(3),(2
612122
1
1)()()(0
tDetDyezXtDetDzeyX
tDtyDtDzyxX
zyxPedteeX
xtBxtB
xtBxtB
tAtAtA
−−
−−
−−
+−=
++=
++++=
+= ∫
α
α
α
α
(4.3.17)
Considering equation (4.3.9) and using equation (4.3.17) then differentiating with respect
to x we get ),,(),(),,(0),,( 8711 zyKzyxKzyxPzyxPxx +=⇒= where ),(7 zyK and
184
),(8 zyK are functions of integration. Substituting back the above value in (4.3.9) and
solve after differentiating with respect to y twice we get ℜ∈= 111 ,)( cctD and
),()(),( 987 zDzyDzyK += where )(8 zD and )(9 zD are functions of integration. Once
again substituting back the above information in (4.3.9) and solve after differentiating
with respect to y and z respectively, we get ,,)( 2212 ℜ∈= cctD ,0)(8 =zD 3
9 )( czD =
and .,,)( 11311)(
36 ℜ∈+= ∫ cccdtectD tA Refreshing the above system of equations (4.3.17)
we get
),()(),()(
,)22
(
,),(
7),(3),(3
4),(3),(2
11)(
321
221
)(3
8)()()(0
tDetDeyzXtDetDezyX
cdtecycczyxX
exczyKedteeX
xtBxtB
xtBxtB
tA
tAtAtAtA
−−
−−
−−−
+−=
++=
+++++=
++=
∫
∫
α
α
α
α
(4.3.18)
where ).0(,, 1321 ≠ℜ∈ cccc Considering equation (4.3.12) and using equation (4.3.18)
then differentiating with respect to ,y we get ⇒=− 0)(),( 28 Btyy
A exBzyKe α
⇒= 0),(8 zyK yyy ),()()(2
),( 1311102
8 zDzyDzDyzyK ++= where ),(10 zD )(11 zD and
)(13 zD are functions of integration. Substituting back this value in the above equation we
get ⇒=− 0)( 210 Bt
A eBzDe α .,)(0)( 441010 ℜ∈=⇒= cczDzDz Now
).(),(2)()(2 2),(2212 tDextBtDtxDe txtB
tB =⇒+
−=
α Substituting back this value in the
above equation we get ⇒=− 0)(2
2)(4 tDec t
tA α .,2
)( 55)(42 ℜ∈+= ∫ ccdte
ctD tA
α
Substituting back all the above values in (4.3.12) and solve after differentiating first with
respect to x then with respect to ,z we get ,)( 63 ctD = ℜ∈= 767
4 ,,)( ccctD and
.0)(11 =zD Substituting back all the above information in (4.3.18) we get
185
),(,
,)22
(
,))(2
(
7),(6
),(37
),(6
),(2
11)(
321
221
)(3
134
2)()()(0
tDeceyzXcecezyX
cdtecycczyxX
exczDcyedteeX
xtBxtBxtBxtB
tA
tAtAtAtA
−−−−
−−−
+−=++=
+++++=
+++=
∫
∫
αα
α
α
(4.3.19)
where ).0,0(,,,,,, 4111764321 ≠≠ℜ∈ ccccccccc Considering equation (4.3.13) and using
equation (4.3.19) then differentiating with respect to ,z we get ⇒=− 0)(134 zDc zz
.,,2
)( 1091094
213 ℜ∈++= ccczcczzD Substituting back the above value in (4.3.13) and
solving we get ℜ∈= 887 ,)( cctD and .092 == cc Substituting all the above
information in (4.3.19), we get
,,
,)22
(
,)22
(
8),(
6),(3
7),(
6),(2
11)(
31
221
)(310
)(4
22)()()(0
ceceyzXcecezyX
cdtecczyxX
excceczyedteeX
xtBxtBxtBxtB
tA
tAtAtAtAtA
−−−−
−−−−
+−=++=
++++=
++++=
∫
∫
αα
α
α
(4.3.20)
where ).0,0(,,,,,,, 411110876431 ≠≠ℜ∈ cccccccccc After a suitable rescaling of ,t the
metric for non static plane symmetric space-time becomes
),)(22( 2241222 dzdytcxcdxdtds ++−
++−=αα
(4.3.21)
where ).0,0(, 4141 ≠≠ℜ∈ cccc The solution of equations (4.3.6) to (4.3.13) becomes
,,
,)22
(,)22
(
8),(
6),(3
7),(
6),(2
131
221
1034
220
ceceyzXcecezyX
ctcczyxXcxcczytX
xtBxtBxtBxtB −−−− +−=++=
++++=++++=
αα
αα (4.3.22)
where ).0,0(,,,,,,, 411110876431 ≠≠ℜ∈ cccccccccc The above space-time (4.3.21)
admits seven linearly independent teleparallel homothetic vector fields in which six are
teleparallel Killing vector fields given as ,t∂∂ ,
x∂∂ ),(
xt
tx
∂∂
+∂∂ ,),(
ye xtB
∂∂−
,),(
ze xtB
∂∂− )(),(
zy
yze xtB
∂∂
−∂∂− and one is teleparallel proper homothetic vector field.
186
Teleparallel proper homothetic vector field after subtracting teleparallel Killing vector
fields from (4.3.22) is [74]
.,,)22
(,)22
( 321
221
4
220 zXyXczyxXczytX αααα ==++=++= (4.3.23)
Case (II):
In this case we have ,),(),( BAtBBtAA ≠== In this case the line element (4.3.1) after
a suitable rescaling of t becomes
),()2( 2232
222 dzdyctcdxdtds ++++−=α
(4.3.24)
where ).0(, 232 ≠ℜ∈ ccc Solution of equations (4.3.6) to (4.3.13) is given by
.,
,,)22
(
10)(
5)(3
6)(
5)(2
1111
912
220
ceceyzXcecezyX
cctxXccxczytX
tBtBtBtB −−−− +−=++=
++=++++=
αα
αα (4.3.25)
where .,,,,, 11109651 ℜ∈cccccc The above space-time (4.3.24) admits seven linearly
independent teleparallel homothetic vector fields in which six are teleparallel Killing
vector fields given as ,t∂∂ ,
x∂∂ ),(
xt
tx
∂∂
+∂∂ ,)(
ye tB
∂∂− ,)(
ze tB
∂∂−
)()(
zy
yze tB
∂∂
−∂∂− and one is teleparallel proper homothetic vector field. Teleparallel
proper homothetic vector field after subtracting teleparallel Killing vector fields from
equation (4.3.25) is
.,,,)22
( 3212
220 zXyXxXczytX αααα ===++= (4.3.26)
The results listed above can be obtained by the same process as in case (I).
Case (III):
In this case we have ),(tAA = )(tBB = and .BA = In this case the line element (4.3.1)
becomes
187
),()( 22223
222 dzyddtctcdxds ++−++=α
(4.3.27)
where ).0(, 232 ≠ℜ∈ ccc Solution of equations (4.3.6) to (4.3.13) is given by
.
,,
,)22
()(2
11)()(
96)(3
7)(
5)(
6)(2
11
)(10
)(9
)(5
)(2
22
32
2
20
ceetcceyzX
cectecezyXcxX
ecezcecyeczyctcc
X
tAtAtA
tAtAtA
tAtAtAtA
−−−
−−−
−−−−
++−=
+++=+=
++++++=
α
αα
αα
(4.3.28)
where .,,,,,,,, 11109765321 ℜ∈ccccccccc The above space-time (4.3.27) admits eight
linearly independent teleparallel homothetic vector fields in which seven are teleparallel
Killing vector fields given as ,)(
te tA
∂∂− ,
x∂∂ ),()(
yt
tye tA
∂∂
+∂∂− ),()(
zt
tze tA
∂∂
+∂∂−
,)(
ye tA
∂∂− ,)(
ze tA
∂∂− )()(
zy
yze tA
∂∂
−∂∂− and one is teleparallel proper homothetic
vector field. Teleparallel proper homothetic vector field after subtracting teleparallel
Killing vector fields from system of equations (4.3.28) is
.,,,)22
()(2
321)(2
22
32
2
20 zXyXxXeczyctc
cX tA ααα
αα
===+++= − (4.3.29)
The results listed above can be obtained by the same process as in case (I).
Case (IV):
In this case we have )(tAA = and .constant=B The line element (4.3.1) after a suitable
rescaling of ,t y and z becomes
,22222 dzyddxdtds +++−= (4.3.30)
Teleparallel homothetic vector fields in this case are
.,
,,
76513
43212
108521
98630
cctcxcyzXctccxczyX
cctczcyxXccxczcytX
++−−=++++=
+++−=++++=
αα
αα (4.3.31)
where .,,,,,,,,, 10987654321 ℜ∈cccccccccc The above space-time (4.3.30) admits eleven
linearly independent teleparallel homothetic vector fields in which ten are teleparallel
188
Killing vector fields given as ,t∂∂ ,
x∂∂ ),(
xt
tx
∂∂
+∂∂ ),(
xy
yx
∂∂
−∂∂ ),(
yt
ty
∂∂
+∂∂
),(z
tt
z∂∂
+∂∂ ),(
zx
xz
∂∂
−∂∂ ,
y∂∂ ,
z∂∂ )(
zy
yz
∂∂
−∂∂ and one is teleparallel proper
homothetic vector field. Teleparallel proper homothetic vector field after subtracting
Killing vector fields from equation (4.3.31) is
.,,, 3210 zXyXxXtX αααα ==== (4.3.32)
Case (V):
In this case we have ),( xtAA = and ).(xBB = Considering equation (4.3.8) and using
equation (4.3.14) then differentiating with respect to ,z we get
),,(),(),,(0),,( 2133 xtKxtzKzxtPzxtPzz +=⇒= where ),(1 xtK and ),(2 xtK are
functions of integration. Substituting back the above value in equation (4.3.8) we get
),,(),(),,(),(),,( 31414 xtKxtyKyxtPxtKyxtPy +−=⇒−= where ),(3 xtK is a function
of integration. Refreshing the system of equations (4.3.14) we get
).,(),(),,(),(
),,,(),,,(313212
211),(),(),(0
xtKxtyKzXxtKxtzKyX
zytPxXzyxPedteeX xtAxtAxtA
+−=++=
+=+= −− ∫αα
αα (4.3.33)
Considering equation (4.3.12) and using equation (4.3.33) then solve after differentiating
with respect to ,y we get ⇒= 0),,(1 zyxPyy ),,(),(),,( 541 zxKzxyKzyxP += where
),(4 zxK and ),(5 zxK are functions of integration. Substituting back the above value in
(4.3.12) and solve after differentiating with respect to ,z we get
),(),(0),( 144 xDzxKzxK z =⇒= where )(1 xD is a functions of integration. Substituting
back the above information in (4.3.12) and solving we get ),(),( 21 xDxtK = 0)(1 =xD
and ),(),( 32 xDxtK = where )(2 xD and )(3 xD are functions of integration. Refreshing
equation (4.3.33) we get
).,()(),()(
),,(),,(323322
215),(),(),(0
xtKxyDzXxDxzDyX
zytPxXzxKedteeX xtAxtAxtA
+−=++=
+=+= −− ∫αα
αα (4.3.34)
189
Considering equation (4.3.13) and using equation (4.3.34) then solving we get
)(),( 45 xDzxK = and ),(),( 53 xDxtK = where )(3 xD and )(5 xD are functions of
integration. Now substituting the above information and equation (4.3.34) in (4.3.10) and
solve after differentiating twice with respect to ,y we get
),,(),(),(2
),,( 8762
2 ztKztyKztKyzytP ++= where ),,(6 ztK ),(7 ztK and ),(8 ztK
are functions of integration. Substituting back the above value and solving we get
),(),( 66 tDztK = where )(6 tD is a function of integration. Now substituting the above
information in (4.3.10) and differentiating with respect to ,y then solving we get
ℜ∈= 116 ,)( cctD and ⇒=+ 0)( 2
1B
x exBc α ,22
12 cxce B +−
=α
).0(, 121 ≠ℜ∈ ccc
Now substituting all the above information in equation (4.3.34) we get
).(,
),,(2
),(
5)(3
3)(4
)(3
2
81
214),(),(),(0
xDeyczXecezcyX
ztKcyxXxDedteeX
xBxBxB
xtAxtAxtA
+−=++=
++=+=
−−−
−− ∫αα
αα (4.3.35)
Considering equation (4.3.11) and using equation (4.3.35) then solving after
differentiating with respect to ,z we get ),()(2
),( 21201
28 tDtzDczztK ++= where
)(20 tD and )(21 tD are functions of integration. Substituting back the above value in
equation (4.3.11) and solving, we get ℜ∈= −5
)(5
5 ,)( cecxD xB and .0)(20 =tD
Refreshing the system of equations (4.3.35) we get
.,
),()22
(),(
)(5
)(3
3)(4
)(3
2
211
2214),(),(),(0
xBxBxBxB
xtAxtAxtA
eceyczXecezcyX
tDczyxXxDedteeX
−−−−
−−
+−=++=
+++=+= ∫αα
αα (4.3.36)
Considering equation (4.3.9) and using equation (4.3.36) we get
.0)(),()( 4),(),(21 =−− ∫ xDedtextAetD xAxtA
xxtA
t α In order to solve this equation we take
844 )(0)( cxDxDx =⇒= and .,,,)()( 87676
216
21 ℜ∈+=⇒= cccctctDctDt
Substituting back these values in the above equation we get
190
⇒=− ∫ 0),( ),(),(6 dtextAec xtA
xxtAα .ln),( 6
tcx
xtAα
= Finally, the metric of non static
plane symmetric space-time in this case takes the form
( ).2 222
12262 dzdycxc
dxdtt
xcds +⎟
⎠⎞
⎜⎝⎛ +−
++⎟⎟⎠
⎞⎜⎜⎝
⎛−=
αα (4.3.37)
Teleparallel homothetic vector fields in this case are
.,
,)22
(,2
)(5
)(3
3)(4
)(3
2
71
22
61
8),(0
xBxBxBxB
xtA
eceyczXecezcyX
cczytcxXcetX
−−−−
−
+−=++=
++++=+=
αα
αα (4.3.38)
where ).0,0(,,,,,,, 6187654321 >≠ℜ∈ cccccccccc The above space-time (4.3.37) admits
six linearly independent teleparallel homothetic vector fields in which five are teleparallel
Killing vector fields given as ,),(
te xtA
∂∂− ,
x∂∂ ,)(
ye xB
∂∂− ,)(
ze xB
∂∂−
)()(
zy
yze xB
∂∂
−∂∂− and one is teleparallel proper homothetic vector field. Teleparallel
proper homothetic vector field after subtracting teleparallel Killing vector fields from
equation (4.3.38) is given as
.,,)22
(,2 321
22
610 zXyXczytcxXtX αααα ==+++== (4.3.39)
Case (VI):
In this case we have ),( xtAA = and constant.=B The metric for non static plane
symmetric space time after a suitable rescaling of y and z becomes
.222262 dzdydxdtt
xcds +++⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
α (4.3.40)
Solving equations (4.3.6) to (4.3.13) by the same procedure as explained above with the
help of equation (4.3.14), teleparallel homothetic vector fields become
.,
,,2
5413
3212
87421
6),(0
cxcyczXcxczcyX
ctczccyxXcetX xtA
+−−=+++=
+++−=+= −
αα
αα (4.3.41)
191
where .,,,,,,, 87654321 ℜ∈cccccccc The above space-time (4.3.40) admits eight linearly
independent teleparallel homothetic vector fields in which seven are teleparallel Killing
vector fields given as ,),(
te xtA
∂∂− ,
x∂∂ ),(
xy
yx
∂∂
−∂∂ ),(
zx
xz
∂∂
−∂∂ ,
y∂∂ ,
z∂∂
)(z
yy
z∂∂
−∂∂ and one is teleparallel proper homothetic vector field. Teleparallel proper
homothetic vector field after subtracting Killing vector fields from equation (4.3.41) is
given as
.,,,2 327
10 zXyXctxXtX αααα ==+== (4.3.42)
Case (VII):
In this case we have )(tAA = and ).(xBB = Considering equation (4.3.8) and using
equation (4.3.14) then solving after differentiating with respect to ,z we get
),(),(),,( 213 xtKxtzKzxtP += and ),,(),(),,( 314 xtKxtyKyxtP +−= where ),,(1 xtK
),(2 xtK and ),(3 xtK are functions of integration. Refreshing the system of equations
(4.3.14) we get
).,(),(),,(),(
),,,(),,,(313212
211)()()(0
xtKxtyKzXxtKxtzKyX
zytPxXzyxPedteeX tAtAtA
+−=++=
+=+= −− ∫αα
αα (4.3.43)
Considering equation (4.3.12) and using equation (4.3.43) then solving after
differentiating twice with respect to ,y we get ),(),(),,( 541 zxKzxyKzyxP += where
),(4 zxK and ),(5 zxK are functions of integration. Substituting back the above value in
(4.3.12) and solving after differentiating with respect to ,z we get ),(),( 14 xDzxK =
where )(1 xD is a functions of integration. Substituting back the above values in (4.3.12)
and solving we get ),(),( 21 xDxtK = 0)(1 =xD and ),(),( 32 xDxtK = where )(2 xD
and )(3 xD are functions of integration. Refreshing (4.3.43) we get
).,()(),()(
),,,(),,(323322
215)()()(0
xtKxyDzXxDxzDyX
zytPxXzxKedteeX tAtAtA
+−=++=
+=+= −− ∫αα
αα (4.3.44)
192
Considering (4.3.13) and using equation (4.3.44) then solving we get )(),( 45 xDzxK =
and ),(),( 53 xDxtK = where )(4 xD and )(5 xD are functions of integration. Now
substituting the above information and equation (4.3.44) in (4.3.10) and solve after
differentiating twice with respect to ,y we get
),,(),(),(2
),,( 8762
2 ztKztyKztKyzytP ++= where ),,(6 ztK ),(7 ztK and ),(8 ztK
are functions of integration. Substituting back and solving we get ),(),( 66 tDztK =
where )(6 tD is a function of integration. Now substituting the above information in
(4.3.10) and differentiating with respect to ,y then solving we get ℜ∈= 116 ,)( cctD
which on back substitution gives ⇒=+ 0)( 21
Bx exBc α
,22
12 cxce B +−
=α
).0(, 121 ≠ℜ∈ ccc Now substituting back the above values in
(4.3.10) and solve after differentiating with respect to ,z we get ),(),( 77 tDztK = where
)(7 tD is a function of integration and ,)( 32 BecxD −= ℜ∈= −
4343 ,,)( ccectD B and
.0)(7 =tD Refreshing (4.3.44) we get
).(,
),,(2
),(
5)(3
3)(4
)(3
2
81
214)()()(0
xDeyczXecezcyX
ztKcyxXxDedteeX
xBxBxB
tAtAtA
+−=++=
++=+=
−−−
−− ∫αα
αα (4.3.45)
Considering equation (4.3.11) and using equation (4.3.45) then solving after
differentiating with respect to ,z we get ),()(2
),( 21201
28 tDtzDczztK ++= where
)(20 tD and )(21 tD are functions of integration. Substituting back we get
ℜ∈= −55
5 ,)( cecxD B and .0)(20 =tD Refreshing the system of equations (4.3.45) we
get
.,
),()22
(),(
)(5
)(3
3)(4
)(3
2
211
2214)()()(0
xBxBxBxB
tAtAtA
eceyczXecezcyX
tDczyxXxDedteeX
−−−−
−−
+−=++=
+++=+= ∫αα
αα (4.3.46)
193
Considering equation (4.3.9) and using equation (4.3.46) then solving we get
874 )( cxcxD += and .,,,)( 8766
)(7
21 ℜ∈+= ∫ ccccdtectD tA Substituting the above
information in equation (4.3.46), we get
.,
,)22
(
),(
)(5
)(3
3)(4
)(3
2
6)(
71
221
87)()()(0
xBxBxBxB
tA
tAtAtA
eceyczXecezcyX
cdtecczyxX
cxcedteeX
−−−−
−−
+−=++=
++++=
++=
∫
∫
αα
α
α
(4.3.47)
The space-time (4.3.1) in this case after a suitable rescaling of ,t becomes
( ).2 2221
222 dzdycxcdxdtds +⎟⎠⎞
⎜⎝⎛ +−
++−=α
(4.3.48)
Teleparallel homothetic vector fields for the above space-time (4.3.48) becomes
.,
,)22
(,
)(5
)(3
3)(4
)(3
2
671
221
870
xBxBxBxB eceyczXecezcyX
ctcczyxXcxctX
−−−− +−=++=
++++=++=
αα
αα (4.3.49)
where ).0(,,,,,,, 187654321 ≠ℜ∈ ccccccccc The above space-time (4.3.48) admits seven
linearly independent teleparallel homothetic vector fields in which six are teleparallel
Killing vector fields given as ,t∂∂ ,
x∂∂ ,)(
ye xB
∂∂− ,)(
ze xB
∂∂− ),()(
zy
yze xB
∂∂
−∂∂−
xt
tx
∂∂
+∂∂ and one is teleparallel proper homothetic vector field. Teleparallel proper
homothetic vector field after subtracting teleparallel Killing vector fields from equation
(4.3.49) is [74]
.,,)22
(, 321
2210 zXyXczyxXtX αααα ==++== (4.3.50)
Case (VIII):
In this case we have ),(xAA = )(xBB = and .BA ≠ Considering eqution (4.3.8) and
using equation (4.3.14) then solve after differentiating with respect to ,z we get
),(),(),,( 213 xtKxtzKzxtP += and ),,(),(),,( 314 xtKxtyKyxtP +−= where ),,(1 xtK
194
),(2 xtK and ),(3 xtK are functions of integration. Refreshing the system of equations
(4.3.14) we get
).,(),(),,(),(),,,(),,,(
313212
211)(0
xtKxtyKzXxtKxtzKyXzytPxXzyxPetX xA
+−=++=
+=+= −
αα
αα (4.3.51)
Considering equation (4.3.9) and using equation (4.3.51) then solving after differentiating
twice with respect to ,t we get ),(),(),(2
),,( 6542
2 zyKzytKzyKtzytP ++= where
),,(4 zyK ),(5 zyK and ),(6 zyK are functions of integration. Substituting back the
above value in (4.3.9) and solving after differentiating with respect to ,y we get
),(),( 14 zDzyK = where )(1 zD is a functions of integration. Substituting back the above
values in (4.3.9) and solving we get 11 )( czD = and ,2
21)(2 cxce xA +=
α
).0(, 121 ≠ℜ∈ ccc Substituting all the above information in equation (4.3.9) and solving
we get 0),(5 =zyK and ),,(),,( 7)(1 zyKezyxP xA−= where ),(7 zyK is a function of
integration. Now substituting back the above information in (4.3.51), we get
).,(),(),,(),(
),,(2
),,(
313212
61
217)(0
xtKxtyKzXxtKxtzKyX
zyKctxXzyKetX xA
+−=++=
++=+= −
αα
αα (4.3.52)
Considering equation (4.3.10) and using equation (4.3.52) then solve after differentiating
with respect to ,y we get ),()()(2
),( 4322
6 zDzyDzDyzyK ++= where ),(2 zD )(3 zD
and )(4 zD are functions of integration. Substituting back the abvoe value in (4.3.10) and
solve after differentiating with respect to y and z we get .,)( 332 ℜ∈= cczD
Substituting back all the above information in equation (4.3.10) and solving we get
,243
)(2 cxce xB +−
=α
).0(, 343 ≠ℜ∈ ccc Now substituting all the above values in (4.3.9)
and solving we get ,0)(3 =zD )(),( 7)(1 tDextK xB−= and ),(),( 8)(2 tDextK xB−= where
)(7 tD and )(8 tD are functions of integration. Refreshing the system of equations
(4.3.52) we get
195
).,()(,)()(
),(22
),,(
3)(738)(72
43
2
1
217)(0
xtKetyDzXetDetzDyX
zDcyctxXzyKetX
xBBxB
xA
+−=++=
+++=+=
−−−
−
αα
αα(4.3.53)
Considering equation (4.3.11) and using (4.3.53) then solve after differentiating the
resulting equation with respect to z twice we get .,,2
)( 1555153
24 ℜ∈++= ccczcczzD
Substituting back the above value in (4.3.11) we get 015 =c and ),(),( 9)(3 tDextK xB−=
where )(9 tD is a function of integration. Substituting all the above information in
(4.3.53) we get
.)()(,)()(
,)22
(2
),,(
)(9)(738)(72
53
22
1
217)(0
xBxBBxB
xA
etDetyDzXetDetzDyX
cczyctxXzyKetX
−−−−
−
+−=++=
++++=+=
αα
αα (4.3.54)
Considering equation (4.3.12) and using equation (4.3.54) then solve after differentiating
twice with respect to ,y we get ),()()(2
),( 1211102
7 zDzyDzDyzyK ++= where ),(10 zD
)(11 zD and )(12 zD are functions of integration. Substituting back the above value and
solving we get .0)(10 =zD Now substituting the above information in (4.3.12) and
solving we get ,)( 87 ctD = ℜ∈= 989
8 ,,)( ccctD and .0)(11 =zD Also substituting the
above values and equation (4.3.54) in equation (4.3.13) we get 109 )( ctD = and
.,,)( 11101112 ℜ∈= ccczD Now refreshing the system of equations (4.3.54) the
teleparallel homothetic vector fields are given as [74]
.,
,)22
(2
,
)(10
)(8
39
)(8
2
53
22
1
21
11)(0
xBxBBxB
xA
ececyzXececzyX
cczyctxXcetX
−−−−
−
+−=++=
++++=+=
αα
αα (4.3.55)
where ).,0,0(,,,,, 313111109,8531 ccccccccccc ≠≠≠ℜ∈ The line element (4.3.1) in this
case takes the form
),()2
()2
( 224
3222
12 zdydcxc
dxdtcxc
ds ++−+++−=αα
(4.3.56)
196
where ).,0,0(,,, 31314321 cccccccc ≠≠≠ℜ∈ The above space-time (4.3.56) admits six
linearly independent teleparallel homothetic vector fields in which five are teleparallel
Killing vector fields given as ,)(
te xA
∂∂− ,
x∂∂ ,)(
ye xB
∂∂− ,)(
ze xB
∂∂−
)()(
zy
yze xB
∂∂
−∂∂− and one is teleparallel proper homothetic vector field. Teleparallel
proper homothetic vector field after subtracting teleparallel Killing vector fields from
equation (4.3.55) is
.,,)22
(2
, 323
22
1
210 zXyXczyctxXtX αααα ==+++== (4.3.57)
Case (IX):
In this case we have ),(xAA = )(xBB = and .BA = The line element (4.3.1) takes the
form
),)(2
( 2223
222 dzdydtcxc
dxds ++−+−+=α
(4.3.58)
where ).0(, 232 ≠ℜ∈ ccc Teleparallel homothetic vector fields are given as
.
,,)2t
22(
,
11)()(
96)(3
8)(
7)(
6)(2
52
2221
)(10
)(9
)(7
0
ceetcceyzX
cectecezyXcczyxX
ececzecytX
xAxAxA
xAxAxA
xAxAxA
−−−
−−−
−−−
++−=
+++=+−++=
+++=
α
αα
α
(4.3.59)
where ).0(,,,,,,, 21110987652 ≠ℜ∈ ccccccccc The above space-time (4.3.58) admits eight
linearly independent teleparallel homothetic vector fields in which seven are teleparallel
Killing vector fields given as ,)(
te xA
∂∂− ,
x∂∂ ),()(
yt
tye xA
∂∂
+∂∂− ),()(
zt
tze xA
∂∂
+∂∂−
,)(
ye xA
∂∂− ,)(
ze xA
∂∂− )()(
zy
yze xA
∂∂
−∂∂− and one is teleparallel proper homothetic
vector field. Teleparallel proper homothetic vector field after subtracting teleparallel
Killing vector fields from equation (4.3.59) is
197
.,,)2t
22(, 32
2
22210 zXyXczyxXtX αααα ==−++== (4.3.60)
The result listed above can be obtained by the same procedure as adopted above in case
(VIII).
Case (X):
In this case we have )(xAA = and .constant=B The line element (4.3.1) after a suitable
rescaling of y and z becomes
,)2
( 22222
12 dzyddxdtcxc
ds ++++−=α
(4.3.61)
where ).0(, 121 ≠ℜ∈ ccc Solution of equations (4.3.6) to (4.3.13) is given by [74]
.,
,2
,
9453
6352
10431
21
8)(0
ccxcyzXccxczyX
cczcyctxXcetX xA
+−−=+−+=
++++=+= −
αα
αα (4.3.62)
where ).0(,,,,,,, 1109865431 ≠ℜ∈ ccccccccc The above space-time (4.3.61) admits eight
linearly independent teleparallel homothetic vector fields in which seven are teleparallel
Killing vector fields given as ,)(
te xA
∂∂− ,
x∂∂ ),(
yx
xy
∂∂
−∂∂ ),(
zx
xz
∂∂
−∂∂ ,
y∂∂
,z∂∂ )(
zy
yz
∂∂
−∂∂ and one is teleparallel proper homothetic vector field. Teleparallel
proper homothetic vector field after subtracting teleparallel Killing vector fields from
equation (4.3.62) is
.,,2t, 32
1
210 zXyXcxXtX αααα ==+== (4.3.63)
The result listed above can be obtained by the same procedure as adopted above in case
(VIII).
Case (XI):
In this case we have constant=A and ).,( xtBB = The metric for non static plan
symmetric space-times in this case becomes
198
),)(22
( 2241222 dzdytc
xc
dxdtds ++−
++−=αα
(4.3.64)
where ).0,0(, 4141 ≠≠ℜ∈ cccc The solution of equations (4.3.6) to (4.3.13) becomes
[74]
,,
,)22
(,)22
(
8),(
6),(3
7),(
6),(2
1131
221
1034
220
ceceyzXcecezyX
ctcczyxXcxcczytX
xtBxtBxtBxtB −−−− +−=++=
++++=++++=
αα
αα (4.3.65)
where ).0,0(,,,,,,, 411110876431 ≠≠ℜ∈ cccccccccc The above space-time (4.3.64)
admits seven linearly independent teleparallel homothetic vector fields in which six are
teleparallel Killing vector fields given as ,t∂∂ ,
x∂∂ ),(
xt
tx
∂∂
+∂∂ ,),(
ye xtB
∂∂−
,),(
ze xtB
∂∂− )(),(
zy
yze xtB
∂∂
−∂∂− and one is teleparallel proper homothetic vector field.
Teleparallel proper homothetic vector field after subtracting teleparallel Killing vector
fields from (4.3.65) is
.,,)22
(,)22
( 321
221
4
220 zXyXczyxXczytX αααα ==++=++= (4.3.66)
The result listed above can be obtained by the same procedure as adopted above in case
(I).
Case (XII):
In this case we have constant=A and ).(tBB = The line element for the metric (4.3.1)
takes the form
),)(2
( 228
7222 dzdyctc
dxdtds ++−
++−=α
(4.3.67)
where ).0(, 787 ≠ℜ∈ ccc The solution of equations (4.3.6) to (4.3.13) becomes [74]
,,
,,)22
(
3)(
1)(3
2)(
1)(2
651
457
220
ceceyzXcecezyX
ctcxXcxcczytX
tBtBtBtB −−−− +−=++=
++=++++=
αα
αα (4.3.68)
199
where ).0(,,,,,, 77654321 ≠ℜ∈ cccccccc The above space-time (4.3.67) admits seven
linearly independent teleparallel homothetic vector fields in which six are teleparallel
Killing vector fields given as ,t∂∂ ,
x∂∂ ),(
xt
tx
∂∂
+∂∂ ,)(
ye tB
∂∂− ,)(
ze tB
∂∂−
)()(
zy
yze tB
∂∂
−∂∂− and one is teleparallel proper homothetic vector field. Teleparallel
proper homothetic vector field after subtracting Killing vector fields from (4.3.68) is
given as
.,,,)22
( 3217
220 zXyXxXczytX αααα ===++= (4.3.69)
The result listed above can be obtained by the same procedure as adopted above in case
(II).
Case (XIII):
In this case we have constant=A and ).(xBB = The line element for the metric (4.3.1)
takes the form
),)(2
( 224
3222 dzdycxc
dxdtds ++−
++−=α
(4.3.70)
where ).0(, 343 ≠ℜ∈ ccc Teleparallel homothetic vector fields in this case are given as
,,
,)22
(,
10)(
6)(3
7)(
6)(2
913
221
210
ceceyzXcecezyX
ctcczyxXcxctX
xBxBxBxB −−−− +−=++=
++++=++=
αα
αα (4.3.71)
where ).0(,,,,,, 310976321 ≠ℜ∈ cccccccc The above space-time (4.3.70) admits seven
linearly independent teleparallel homothetic vector fields in which six are teleparallel
Killing vector fields given as ,t∂∂ ,
x∂∂ ),(
xt
tx
∂∂
+∂∂ ,)(
ye xB
∂∂− ,)(
ze xB
∂∂−
)()(
zy
yze xB
∂∂
−∂∂− and one is teleparallel proper homothetic vector field. Teleparallel
200
proper homothetic vector field after subtracting Killing vector fields from (4.3.71) is
given as
.,,)22
(, 323
2210 zXyXczyxXtX αααα ==++== (4.3.72)
The result listed above can be obtained by the same procedure as adopted above in case
(VII).
4.4. Teleparallel Proper Homothetic Vector Fields in
Static Cylindrically Symmetric Space-Times
Consider cylindrically symmetric static space-times in usual coordinates ),,,( zrt θ
(labeled by ),,,,( 3210 xxxx respectively) with the line element [36, 78]
,2)(2)(22)(2 dzededrdteds rCrBrA +++−= θ (4.4.1)
where ,A B and C are functions of r only. The above space-time admits minimum
seven linearly independent teleparallel Killing vector fields which are ,r∂∂ ,2
)(
te
rA
∂∂−
,2)(
θ∂∂− rB
e ,2)(
ze
rC
∂∂−
,)(1
trM
rt
∂∂
+∂∂
θθ
∂∂
+∂∂ )(2 rMr
and ,)(3
zrM
rz
∂∂
+∂∂ where
,)( 2)(
2)(
1 ∫−−
= dreerMrArA
dreerMrBrB
∫−−
−= 2)(
2)(
2 )( and .)( 2)(
2)(
3 ∫−−
−= dreerMrCrC
The
tetrad components and its inverse, non-vanishing Weitzenböck connections and the non
vanishing torsion components for (4.4.1) are given in equations (3.3.2), (3.3.3) and
(3.3.4). Now using (4.4.1) and (3.3.4) in (1.3.17) we get the teleparallel homothetic
equations as follows:
,3,3
2,2
1,1
0,0 α==== XXXX (4.4.2)
,02,0)(
0,2)( =− XeXe rArB (4.4.3)
,03,0)(
0,3)( =− XeXe rArC (4.4.4)
201
,03,2)(
2,3)( =+ XeXe rBrC (4.4.5)
,02
0)(1,
0)(0,
1 =′
−− XeAXeX rArA (4.4.6)
,02
2)(2,
11,
2)( =′
++ XeBXXe rBrB (4.4.7)
,02
3)(3,
11,
3)( =′
++ XeCXXe rCrC (4.4.8)
Now integrating equations (4.4.2), we get
),,,(),,,(),,,(),,,(
4332
2110
θαθα
θαθα
rtPzXzrtPXztPrXzrPtX
+=+=
+=+= (4.4.9)
where ),,,(1 zrP θ ),,,(2 ztP θ ),,(3 zrtP and ),,(4 θrtP are functions of integration
which are to be determined. In order to find solution for equations (4.4.2) to (4.4.8) we
will consider each possible form of the metric for static cylindrically symmetric space-
times and then solve each possibility in turn. Following are the possible cases for the
metric where the above space-times admit teleparallel proper homothetic vector fields:
(I) )(),(),( rCCrBBrAA === and .,, CBCABA ≠≠≠
(II)(a) ),(),( rBBrAA == and .tan tconsC =
(II)(b) ),(),( rCCrAA == and .tan tconsB =
(II)(c) ),(),( rCCrBB == and .tan tconsA =
(III)(a) )(),(),( rCCrBBrAA === and ).()( rCrB =
(III)(b) )(),(),( rCCrBBrAA === and ).()( rCrA =
(III)(c) )(),(),( rCCrBBrAA === and ).()( rBrA =
(IV) )(),(),( rCCrBBrAA === and ).()()( rCrBrA ==
(V)(a) )(),(,tan rCCrBBtconsA === and ).()( rCrB =
(V)(b) )(,tan),( rCCtconsBrAA === and ).()( rCrA =
(V)(c) tconsCrBBrAA tan),(),( === and ).()( rBrA =
202
(VI)(a) )(rAA = and .tan tconsCB ==
(VI)(b) )(rBB = and .tan tconsCA ==
(VI)(c) )(rCC = and .tan tconsBA ==
We will discuss each possibility in turn.
Case (I):
In this case we have ),(rAA = ),(rBB = ),(rCC = ,BA ≠ CA ≠ and .CB ≠ Now
substituting equation (4.4.9) in equation (4.4.3), we get
.0),,(),,( 1)(3)( =− zrPezrtPe rAt
rB θθ (4.4.10)
Differentiating equation (4.4.10) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (4.4.10) we get
⇒= − ),(),,( 11 zrEezrP ABθθ ),,(),(),,( 311 zrEzrEezrP AB += −θθ where ),(3 zrE is a
function of integration. Refreshing the system of equations (4.4.9) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θαθα
θαθα
rtPzXzrEzrEtXztPrXzrEzrEetX AB
+=++=
+=++= −
(4.4.11)
Considering equation (4.4.4) and using equation (4.4.11) we get
.0)],(),([),,( 31)()()(4)( =+− − zrEzrEeertPe zzrArBrA
trC θθ (4.4.12)
Differentiating equation (4.4.12) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (4.4.12) we get
.0)],(),([),( 31)()()(4)( =+− − zrEzrEeerEe zzrArBrArC θθ Differentiating the above equation
with respect to θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ where
)(1 rK and )(2 rK are functions of integration. Substituting back the above value in
equation (4.4.12) and solving we get )()(),( 31)()(1 rKrKezzrE rBrC += − and
203
),()(),( 42)()(3 rKrKezzrE rArC += − where )(3 rK and )(4 rK are functions of
integration. Substituting all the above information in (4.4.11) we get
( )( )
( ) ).,()()(),,()()(),,,(
),()()()(
5213
231)()(221
42)()(3)()(1)()(0
θθα
θαθα
θα
rErKrKtzXzrErKrKeztXztPrX
rKrKezrKerKeztXrBrC
rArCrArBrArC
+++=
+++=+=
++++=−
−−−
(4.4.13)
Considering equation (4.4.5) and using equation (4.4.13), we get
.0),(),()(2 2)(5)(1)( =++ zrEerEerKte zrBrCrC θθ (4.4.14)
Differentiating the above equation first with respect to t and then with respect to ,θ we
get 0)(1 =rK and ),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and
)(6 rK are functions of integration. Substituting back the above values in (4.4.14) we get
),()(),()(),( 75)()(25)()(2 rKrKezzrErKezrE rBrCrBrCz +−=⇒−= −− where )(7 rK is a
function of integration. Substituting all the above information in (4.4.13) we get
( )
).()()(),()()(),,,(
),()()(
6523
75)()(3221
42)()(3)()(0
rKrKrtKzXrKrKezrKtXztPrX
rKrKezrKetXrBrC
rArCrArB
+++=
+−+=+=
+++=−
−−
θα
θαθα
θα
(4.4.15)
Considering equation (4.4.6) and using equation (4.4.15), we get
.0)()(21)(
21)()(
)(21)()(
21),,(
4)()(4)(2)(
2)(3)(3)(2
=−−−−
⎟⎠⎞
⎜⎝⎛ −−−⎟
⎠⎞
⎜⎝⎛ −−
rKerAertArKerKez
rKeACzrKerKeABztP
rAr
rArr
rAr
rC
rCrrr
rBrBrrt
α
θθθ (4.4.16)
Differentiating equation (4.4.16) with respect to t we get .0)(21),,( )(2 =− rA
rtt erAztP αθ
Differentiating this equation with respect to ,t we get ⇒= 0),,(2 ztPttt θ
),,(),(),(2
),,( 8762
2 zEztEzEtztP θθθθ ++= where ),,(6 zE θ ),(7 zE θ and ),(8 zE θ are
functions of integration. Substituting back the above value in (4.4.16) we get
.0)(21),( )(6 =− rA
r erAzE αθ Differentiating this equation with respect to ,θ we get
204
),(),(0),( 866 zKzEzE =⇒= θθθ where )(8 zK is a function of integration. Substituting back
in the above equation and differentiating with respect to ,z we get
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Once again substituting back and solving we get
.,20)(21
221)()(
1 ℜ∈+=⇒=− ccrceerAc rArAr α
α Substituting all the above information in
equation (4.4.16) and differentiating with respect to ,θ we get
.0)()(21),( 3)(3)(7 =−⎟
⎠⎞
⎜⎝⎛ −− rKerKeABzE r
rBrBrrθθ Differentiating this equation with respect to
,θ we get ⇒=0),(7 zE θθθ ),()(),( 1097 zKzKzE +=θθ where )(9 zK and )(10 zK are
functions of integration. Substituting back the above value and differentiating with
respect to ,z we get .,)(0)( 3399 ℜ∈=⇒= cczKzKz Now substituting back this value
and solving we get, .,)( 4
)(2
)(
42
)()(2
)(
33 ℜ∈+=
−−
−
∫ cecdreecrKrBrArArBrA
Substituting all the
above information in equation (4.4.16) and differentiating twice with respect to ,z we get
.,,)(0)( 65651010 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in (4.4.16) and
solve after differentiating with respect to ,z we get
.,)( 7
)(2
)(
72
)()(2
)(
52 ℜ∈+=
−−
−
∫ cecdreecrKrCrArArCrA
Now substituting all the above
information in equation (4.4.16) and solving we get,
.,)( 82
)(
82
)(2
)(
64 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Refreshing the system of equations (4.4.15)
we get
).()(
),()(
),,(2
,
65)(2
)(
72
)()(2
)(
53
75)()()(2
)(
42
)()(2
)(
32
86531
21
2)(
82
)(2
)(
62
)(
72
)(2
)(
52
)(
42
)(2
)(
30
rKrKetcdreectzX
rKrKezectdreectX
zEtcctzctctrX
ecdreececzdreeczecdreectX
rCrArArCrA
rBrCrBrArArBrA
rArArArArArArArArA
++++=
+−++=
+++++=
++++++=
−−
−
−−−
−
−−−−−−−−−
∫
∫
∫∫∫
θα
θα
θθα
θθα
(4.4.17)
Considering equation (4.4.7) and using equation (4.4.17) we get
205
.0)()(21)(
21
)()()()](21)([
)]()([21)]()([
21),(2
7)()(
7)(5)(5)(
2)(
42
)(2
)(
38
3
=++
+−−−
−+−++ ∫−
rKerBerB
rKerKezrKerBrCz
erBrActdreerBrActzEtc
rBr
rBr
rrB
rrCrC
rr
rA
rr
rArA
rr
θα
θθ
(4.4.18)
Differentiating equation (4.4.18) with respect to ,t we get
.0)]()([21)]()([
212 2
)(
42
)(2
)(
33 =−+−+ ∫− rA
rr
rArA
rr erBrAcdreerBrAcc Solving this
equation and remember that in this case ,0)()( ≠− rBrA rr we get .021 == cc
Substituting the above information in equation (4.4.18) and differentiating twice with
respect to ,θ we get ),()()(2
),(0),( 1312112
88 zKzKzKzEzE ++=⇒= θθθθθθθ where
),(11 zK )(12 zK and )(13 zK are functions of integration. Now substituting back this
information and differentiating with respect to ,z we get
.,)(0)( 991111 ℜ∈=⇒= cczKzKz Substituting back these information in equation
(4.4.18) and differentiating the resulting equation with respect to ,θ we get
.,20)(21
10109)()(
9 ℜ∈+−=⇒=+ ccrceerBc rBrBr α
α Substituting back the above information
in equation (4.4.18) and differentiating the resulting equation with respect to ,z we get
.0)()(21)( 5)(5)(12 =−⎟
⎠⎞
⎜⎝⎛ −− rKerKeBCzK r
rCrCrrz Differentiating this equation with respect to ,z
we get .,,)(0)( 121112111212 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in the above
equation and solving we get ,)()(
2)(
132
)()(2
)(
115 rCrBrBrCrB
ecdreecrK−
−−
+= ∫ .13 ℜ∈c Now
substituting back all the above information in equation (4.4.18) we get
.,)(0)()()(21
142
)(
142
)(2
)(
1277)(7)(
12 ℜ∈+−=⇒=++−−−
∫ cecdreecrKrKerKerBcrBrBrB
rrBrB
r
Refreshing the system of equations (4.4.17) we get
206
).(
,
),(22
,
6)(2
)(
132
)()(2
)(
11
)(2
)(
72
)()(2
)(
53
2)(
142
)(2
)(
122
)(
132
)(2
)(
112
131211659
2
1
21
2)(
82
)(2
)(
62
)(
72
)(2
)(
50
rKecdreecetcdreectzX
ecdreececzdreeczX
zKczctcctzcctrX
ecdreececzdreecztX
rCrBrBrCrBrCrArArCrA
rBrBrBrBrBrB
rArArArArArA
+++++=
+−−−=
+++++++=
++++=
−−
−−−
−
−−−−−−
−−−−−−
∫∫
∫∫
∫∫
θθα
θα
θθθα
α
(4.4.19)
Considering equation (4.4.8) and using equation (4.4.19) we get
.0)()(21)(
21)]()([
21
)()]()([21)]()([
21
)]()([21)(22
6)()(2)(
13
6)(2)(
2)(
112
)(
7
2)(
2)(
513
115
=++−+
+−+−+
−+++
∫
∫−
−
rKerCerzCerCrBc
rKedreerCrBcerCrAct
dreerCrActzKctc
rCr
rCr
rB
rr
rrC
rBrB
rr
rA
rr
rArA
rrz
αθ
θ
θ
(4.4.20)
Differentiating equation (4.4.20) with respect to ,t we get
.0)]()([21)]()([
212 2
)(
72
)(2
)(
55 =−+−+ ∫− rA
rr
rArA
rr erCrAcdreerCrAcc Solving this
equation and remember that in this case ,0)()( ≠− rCrA rr we get .075 == cc
Substituting the above information in equation (4.4.20) and differentiating with respect to
,θ we get .0)]()([21)]()([
212 2
)(
132
)(2
)(
1111 =−+−+ ∫− rB
rr
rBrB
rr erCrBcdreerCrBcc
Solving this equation and remember that in this case ,0)()( ≠− rCrB rr we get
.01311 == cc Now substituting the above information in equation (4.4.20) and
differentiating twice with respect to ,z we get
.,,,2
)(0)( 171615171615
21313 ℜ∈++=⇒= cccczcczzKzKzzz Substituting back these
information in equation (4.4.20) and differentiating the resulting equation with respect to
,z we get .,20)(21
181815)()(
15 ℜ∈+−=⇒=+ ccrceerCc rCrCr α
α Substituting back this value
in equation (4.4.20) and solving we get ⇒=++ 0)()()(21 6)(6)(
16 rKerKerCc rrCrC
r
207
,)( 2)(
192
)(2
)(
166
rCrCrC
ecdreecrK−−−
+−= ∫ .19 ℜ∈c Refreshing the system of equations (4.4.19)
we get
,,
,222
,
2)(
192
)(2
)(
1632
)(
142
)(2
)(
122
171612615
2
9
2
1
21
2)(
82
)(2
)(
60
rCrCrCrBrBrB
rArArA
ecdreeczXecdreecX
czcctcczcctrX
ecdreectX
−−−−−−
−−−
+−=+−=
+++++++=
++=
∫∫
∫
αθα
θθα
α
(4.4.21)
where .,,,,,,,,, 1917161514129861 ℜ∈cccccccccc The line element for static cylindrically
symmetric space-times in this case becomes
,)2()2()2( 21815
2109
2221
2 dzcrcdcrcdrdtcrcds +−++−+++−=α
θαα
(4.4.22)
where ).,,(,,,,, 15915191181510921 cccccccccccc ≠≠≠ℜ∈ The above space-time admits eight
linearly independent teleparallel homothetic vector fields in which seven are teleparallel
Killing vector fields given as ,r∂∂ ,2
)(
te
rA
∂∂−
,2)(
θ∂∂− rB
e ,2)(
ze
rC
∂∂−
,)(1
trM
rt
∂∂
+∂∂
θθ
∂∂
+∂∂ )(2 rMr
and ,)(3
zrM
rz
∂∂
+∂∂ where ,)( 2
)(2
)(1 ∫
−−
= dreerMrArA
dreerMrBrB
∫−−
−= 2)(
2)(
2 )( and .)( 2)(
2)(
3 ∫−−
−= dreerMrCrC
One teleparallel proper
homothetic vector field is given as
,,,222
, 3215
2
9
2
1
210 zXXczcctrXtX αθαθαα ==+++== (4.4.23)
Case (II)(a):
In this case we have ),(rAA = ),(rBB = constant=C and .BA ≠ Now substituting
equation (4.4.9) in equation (4.4.3), we get
.0),,(),,( 1)(3)( =− zrPezrtPe rAt
rB θθ (4.4.24)
208
Differentiating equation (4.4.24) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (4.4.24) we get
⇒= − ),(),,( 11 zrEezrP ABθθ ),,(),(),,( 311 zrEzrEezrP AB += −θθ where ),(3 zrE is a
function of integration. Refreshing the system of equations (4.4.9) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θαθα
θαθα
rtPzXzrEzrEtXztPrXzrEzrEetX AB
+=++=
+=++= −
(4.4.25)
Considering equation (4.4.4) and using equation (4.4.25) we get
.0)],(),([),,( 31)()()(4 =+− − zrEzrEeertP zzrArBrA
t θθ (4.4.26)
Differentiating equation (4.4.26) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (4.4.26) we get
.0)],(),([),( 31)()()(4 =+− − zrEzrEeerE zzrArBrA θθ Differentiating the above equation with
respect to θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ where )(1 rK and
)(2 rK are functions of integration. Substituting back the above value in equation (4.4.26)
and solving we get )()(),( 31)(1 rKrKezzrE rB += − and ),()(),( 42)(3 rKrKezzrE rA += −
where )(3 rK and )(4 rK are functions of integration. Substituting all the above
information in (4.4.25) we get
( )( )
( ) ).,()()(),,()()(),,,(
),()()()(
5213
231)(221
42)(3)()(1)(0
θθα
θαθα
θα
rErKrKtzXzrErKrKeztXztPrX
rKrKezrKerKeztXrB
rArArBrA
+++=
+++=+=
++++=−
−−−
(4.4.27)
Considering equation (4.4.5) and using equation (4.4.27), we get
.0),(),()(2 2)(51 =++ zrEerErtK zrBθθ (4.4.28)
Differentiating the above equation first with respect to t and then with respect to ,θ we
get 0)(1 =rK and ),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and
)(6 rK are functions of integration. Substituting back the above values in (4.4.28) we get
209
),()(),()(),( 75)(25)(2 rKrKezzrErKezrE rBrBz +−=⇒−= −− where )(7 rK is a function
of integration. Substituting all the above information in (4.4.27) we get
( )
).()()(),()()(),,,(
),()()(
6523
75)(3221
42)(3)()(0
rKrKrtKzXrKrKezrKtXztPrX
rKrKezrKetXrB
rArArB
+++=
+−+=+=
+++=−
−−
θα
θαθα
θα
(4.4.29)
Considering equation (4.4.6) and using equation (4.4.29), we get
.0)()(21)(
21)()(
)(21)()(
21),,(
4)()(4)(2
23)(3)(2
=−−−−
+−⎟⎠⎞
⎜⎝⎛ −−
rKerAertArKerKz
rKAzrKerKeABztP
rAr
rArr
rAr
rrrBrB
rrt
α
θθθ (4.4.30)
Differentiating equation (4.4.30) with respect to t we get .0)(21),,( )(2 =− rA
rtt erAztP αθ
Differentiating this equation with respect to ,t we get ⇒= 0),,(2 ztPttt θ
),,(),(),(2
),,( 8762
2 zEztEzEtztP θθθθ ++= where ),,(6 zE θ ),(7 zE θ and ),(8 zE θ are
functions of integration. Substituting back the above value in (4.4.30) we get
.0)(21),( )(6 =− rA
r erAzE αθ Differentiating this equation with respect to ,θ we get
),(),(0),( 866 zKzEzE =⇒= θθθ where )(8 zK is a function of integration. Substituting back
in the above equation and differentiating with respect to ,z we get
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Once again substituting back and solving we get
.,20)(21
221)()(
1 ℜ∈+=⇒=− ccrceerAc rArAr α
α Substituting all the above information in
equation (4.4.30) and differentiating with respect to ,θ we get
.0)()(21),( 3)(3)(7 =−⎟
⎠⎞
⎜⎝⎛ −− rKerKeABzE r
rBrBrrθθ Differentiating this equation with respect to
,θ we get ⇒=0),(7 zE θθθ ),()(),( 1097 zKzKzE +=θθ where )(9 zK and )(10 zK are
functions of integration. Substituting back the above value and differentiating with
respect to ,z we get .,)(0)( 3399 ℜ∈=⇒= cczKzKz Now substituting back this value
210
and solving we get, .,)( 4
)(2
)(
42
)()(2
)(
33 ℜ∈+=
−−
−
∫ cecdreecrKrBrArArBrA
Substituting all the
above information in equation (4.4.30) and differentiating twice with respect to ,z we get
.,,)(0)( 65651010 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in (4.4.30) and
solve after differentiating with respect to ,z we get
.,)( 72
)(
72
)(2
)(
52 ℜ∈+= ∫
−
cecdreecrKrArArA
Now substituting all the above information in
equation (4.4.30) and solving we get, .,)( 82
)(
82
)(2
)(
64 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Refreshing the system of equations (4.4.29) we get
).()(
),()(
),,(2
,
652)(
72
)(2
)(
53
75)()(2
)(
42
)()(2
)(
32
86531
21
2)(
82
)(2
)(
62
)(
72
)(2
)(
52
)(
42
)(2
)(
30
rKrKetcdreectzX
rKrKezectdreectX
zEtcctzctctrX
ecdreececzdreeczecdreectX
rArArA
rBrBrArArBrA
rArArArArArArArArA
++++=
+−++=
+++++=
++++++=
∫
∫
∫∫∫
−
−−−
−
−−−−−−−−−
θα
θα
θθα
θθα
(4.4.31)
Considering equation (4.4.7) and using equation (4.4.31) we get
.0)()(21)(
21)()()()(
21
)]()([21)]()([
21),(2
7)()(7)(55
2)(
42
)(2
)(
38
3
=+++−+
−+−++ ∫−
rKerBerBrKerKzrKrzB
erBrActdreerBrActzEtc
rBr
rBrr
rBrr
rA
rr
rArA
rr
θα
θθ (4.4.32)
Differentiating equation (4.4.32) with respect to ,t we get
.0)]()([21)]()([
212 2
)(
42
)(2
)(
33 =−+−+ ∫− rA
rr
rArA
rr erBrAcdreerBrAcc Solving this
equation and remember that in this case ,0)()( ≠− rBrA rr we get .021 == cc
Substituting the above information in equation (4.4.32) and differentiating twice with
respect to ,θ we get ),()()(2
),(0),( 1312112
88 zKzKzKzEzE ++=⇒= θθθθθθθ where
),(11 zK )(12 zK and )(13 zK are functions of integration. Now substituting back this
information and differentiating with respect to ,z we get
211
.,)(0)( 991111 ℜ∈=⇒= cczKzKz Substituting back these information in equation
(4.4.32) and differentiating the resulting equation with respect to ,θ we get
.,20)(21
10109)()(
9 ℜ∈+−=⇒=+ ccrceerBc rBrBr α
α Substituting back the above information
in equation (4.4.32) and differentiating the resulting equation with respect to ,z we get
.0)()(21)( 5512 =−+ rKrKBzK rrz Differentiating this equation with respect to ,z we get
.,,)(0)( 121112111212 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in the above
equation and solving we get ,)( 2)(
132
)(2
)(
115
rBrBrB
ecdreecrK += ∫−
.13 ℜ∈c Now substituting
back all the above information in equation (4.4.32) we get
.,)(0)()()(21
142
)(
142
)(2
)(
1277)(7)(
12 ℜ∈+−=⇒=++−−−
∫ cecdreecrKrKerKerBcrBrBrB
rrBrB
r
Refreshing the system of equations (4.4.31) we get
).(
,
),(22
,
62)(
132
)(2
)(
112
)(
72
)(2
)(
53
2)(
142
)(2
)(
122
)(
132
)(2
)(
112
131211659
2
1
21
2)(
82
)(2
)(
62
)(
72
)(2
)(
50
rKecdreecetcdreectzX
ecdreececzdreeczX
zKczctcctzcctrX
ecdreececzdreecztX
rBrBrBrArArA
rBrBrBrBrBrB
rArArArArArA
+++++=
+−−−=
+++++++=
++++=
∫∫
∫∫
∫∫
−−
−−−−−−
−−−−−−
θθα
θα
θθθα
α
(4.4.33)
Considering equation (4.4.8) and using equation (4.4.33) we get
.0)(21)()(
21
)(21)(
21)(22
2)(
1362
)(2
)(
11
2)(
72
)(2
)(
513
115
=+++
++++
∫
∫−
−
rB
rr
rBrB
r
rA
r
rArA
rz
erBcrKdreerBc
erActdreerActzKctc
θθ
θ
(4.4.34)
Differentiating equation (4.4.34) with respect to ,t we get
.0)(21)(
212 2
)(
72
)(2
)(
55 =++ ∫− rA
r
rArA
r erAcdreerAcc Solving this equation and remember
that in this case ,0)( ≠rAr we get .075 == cc Substituting the above information in
equation (4.4.34) and differentiating with respect to ,θ we get
212
.0)(21)(
212 2
)(
132
)(2
)(
1111 =++ ∫− rB
r
rBrB
r erBcdreerBcc Solving this equation and remember
that in this case ,0)( ≠rBr we get .01311 == cc Now substituting the above information
in equation (4.4.34) and differentiating twice with respect to ,z we get
.,,)(0)( 161516151313 ℜ∈+=⇒= ccczczKzKzz Substituting back and solving we get
.,)( 1717156 ℜ∈+−= ccrcrK Refreshing the system of equations (4.4.33) we get
,,
,22
,
171532
)(
142
)(2
)(
122
16151269
2
1
21
2)(
82
)(2
)(
60
crczXecdreecX
czcctccctrX
ecdreectX
rBrBrB
rArArA
+−=+−=
++++++=
++=
−−−
−−−
∫
∫
αθα
θθα
α
(4.4.35)
where .,,,,,,,, 17161514129861 ℜ∈ccccccccc The line element for static cylindrically
symmetric space-times in this case becomes
,)2()2( 22109
2221
2 dzdcrcdrdtcrcds ++−+++−= θαα
(4.4.36)
where ).(,,, 9110921 cccccc ≠ℜ∈ The above space-time admits eight linearly independent
teleparallel homothetic vector fields in which seven are teleparallel Killing vector fields
given as ,r∂∂ ,2
)(
te
rA
∂∂−
,2)(
θ∂∂− rB
e ,z∂∂ ,
zr
rz
∂∂
−∂∂
trM
rt
∂∂
+∂∂ )(1 and
θθ
∂∂
+∂∂ )(2 rMr
where ∫−−
= dreerMrArA
2)(
2)(
1 )(
and .)( 2)(
2)(
2 dreerMrBrB
∫−−
−= One
teleparallel proper homothetic vector field is given as
,,,22
, 329
2
1
210 zXXcctrXtX αθαθαα ==++== (4.4.37)
The cases (II)(b) and (II)(c) can be solved exactly the same as the above case.
Case (III)(a):
In this case we have ),(rAA = ),(rBB = ),(rCC = ,BA ≠ CA ≠ and .CB = Now
substituting equation (4.4.9) in equation (4.4.3), we get
213
.0),,(),,( 1)(3)( =− zrPezrtPe rAt
rB θθ (4.4.38)
Differentiating equation (4.4.38) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (4.4.38) we get
⇒= − ),(),,( 11 zrEezrP ABθθ ),,(),(),,( 311 zrEzrEezrP AB += −θθ where ),(3 zrE is a
function of integration. Refreshing the system of equations (4.4.9) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θαθα
θαθα
rtPzXzrEzrEtXztPrXzrEzrEetX AB
+=++=
+=++= −
(4.4.39)
Considering equation (4.4.4) and using equation (4.4.39) we get
.0)],(),([),,( 31)()()(4)( =+− − zrEzrEeertPe zzrArBrA
trB θθ (4.4.40)
Differentiating equation (4.4.40) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (4.4.40) we get
.0)],(),([),( 31)()()(4)( =+− − zrEzrEeerEe zzrArBrArB θθ Differentiating the above equation
with respect to θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ where
)(1 rK and )(2 rK are functions of integration. Substituting back the above value in
equation (4.4.40) and solving we get )()(),( 311 rKrKzzrE += and
),()(),( 42)()(3 rKrKezzrE rArB += − where )(3 rK and )(4 rK are functions of integration.
Substituting all the above information in (4.4.39) we get
( )( )
( ) ).,()()(),,()()(),,,(
),()()()(
5213
231221
42)()(3)()(1)()(0
θθα
θαθα
θα
rErKrKtzXzrErKrKztXztPrX
rKrKezrKerKeztX rArBrArBrArB
+++=
+++=+=
++++= −−−
(4.4.41)
Considering equation (4.4.5) and using equation (4.4.41), we get
.0),(),()(2 251 =++ zrErErtK zθθ (4.4.42)
Differentiating the above equation first with respect to t and then with respect to ,θ we
get 0)(1 =rK and ),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and
214
)(6 rK are functions of integration. Substituting back the above values in (4.4.42) we get
),()(),()(),( 75252 rKrKzzrErKzrEz +−=⇒−= where )(7 rK is a function of
integration. Substituting all the above information in (4.4.41) we get
( )
).()()(),()()(),,,(
),()()(
6523
753221
42)()(3)()(0
rKrKrtKzXrKrKzrKtXztPrX
rKrKezrKetX rArBrArB
+++=
+−+=+=
+++= −−
θα
θαθα
θα
(4.4.43)
Considering equation (4.4.6) and using equation (4.4.43), we get
.0)()(21)(
21)()(
)(21)()(
21),,(
4)()(4)(2)(
2)(3)(3)(2
=−−−−
⎟⎠⎞
⎜⎝⎛ −−−⎟
⎠⎞
⎜⎝⎛ −−
rKerAertArKerKez
rKeABzrKerKeABztP
rAr
rArr
rAr
rB
rBrrr
rBrBrrt
α
θθθ (4.4.44)
Differentiating equation (4.4.44) with respect to t we get .0)(21),,( )(2 =− rA
rtt erAztP αθ
Differentiating this equation with respect to ,t we get ⇒= 0),,(2 ztPttt θ
),,(),(),(2
),,( 8762
2 zEztEzEtztP θθθθ ++= where ),,(6 zE θ ),(7 zE θ and ),(8 zE θ are
functions of integration. Substituting back the above value in (4.4.44) we get
.0)(21),( )(6 =− rA
r erAzE αθ Differentiating this equation with respect to ,θ we get
),(),(0),( 866 zKzEzE =⇒= θθθ where )(8 zK is a function of integration. Substituting back
in the above equation and differentiating with respect to ,z we get
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Once again substituting back and solving we get
.,20)(21
221)()(
1 ℜ∈+=⇒=− ccrceerAc rArAr α
α Substituting all the above information in
equation (4.4.44) and differentiating with respect to ,θ we get
.0)()(21),( 3)(3)(7 =−⎟
⎠⎞
⎜⎝⎛ −− rKerKeABzE r
rBrBrrθθ Differentiating this equation with respect to
,θ we get ⇒=0),(7 zE θθθ ),()(),( 1097 zKzKzE +=θθ where )(9 zK and )(10 zK are
functions of integration. Substituting back the above value and differentiating with
215
respect to ,z we get .,)(0)( 3399 ℜ∈=⇒= cczKzKz Now substituting back this value
and solving we get, .,)( 4
)(2
)(
42
)()(2
)(
33 ℜ∈+=
−−
−
∫ cecdreecrKrBrArArBrA
Substituting all the
above information in equation (4.4.44) and differentiating twice with respect to ,z we get
.,,)(0)( 65651010 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in (4.4.44) and
solve after differentiating with respect to ,z we get
.,)( 7
)(2
)(
72
)()(2
)(
52 ℜ∈+=
−−
−
∫ cecdreecrKrBrArArBrA
Now substituting all the above
information in equation (4.4.44) and solving we get,
.,)( 82
)(
82
)(2
)(
64 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Refreshing the system of equations (4.4.43)
we get
).()(
),()(
),,(2
,
65)(2
)(
72
)()(2
)(
53
75)(2
)(
42
)()(
2)(
32
86531
21
2)(
82
)(2
)(
62
)(
72
)(2
)(
52
)(
42
)(2
)(
30
rKrKetcdreectzX
rKrKzectdreectX
zEtcctzctctrX
ecdreececzdreeczecdreectX
rBrArArBrA
rBrArA
rBrA
rArArArArArArArArA
++++=
+−++=
+++++=
++++++=
−−
−
−−
−
−−−−−−−−−
∫
∫
∫∫∫
θα
θα
θθα
θθα
(4.4.45)
Considering equation (4.4.7) and using equation (4.4.45) we get
.0)()(21)(
21)()()()(
21
)]()([21)]()([
21),(2
7)()(7)(5)(5)(
2)(
42
)(2
)(
38
3
=+++−−
−+−++ ∫−
rKerBerBrKerKezrKerBz
erBrActdreerBrActzEtc
rBr
rBrr
rBr
rBrBr
rA
rr
rArA
rr
θα
θθ (4.4.46)
Differentiating equation (4.4.46) with respect to ,t we get
.0)]()([21)]()([
212 2
)(
42
)(2
)(
33 =−+−+ ∫− rA
rr
rArA
rr erBrAcdreerBrAcc Solving this
equation and remember that in this case ,0)()( ≠− rBrA rr we get .033 == cc
Substituting the above information in equation (4.4.46) and differentiating twice with
respect to ,θ we get ),()()(2
),(0),( 1312112
88 zKzKzKzEzE ++=⇒= θθθθθθθ where
216
),(11 zK )(12 zK and )(13 zK are functions of integration. Now substituting back this
information and differentiating with respect to ,z we get
.,)(0)( 991111 ℜ∈=⇒= cczKzKz Substituting back these information in equation
(4.4.46) and differentiating the resulting equation with respect to ,θ we get
.,20)(21
10109)()(
9 ℜ∈+−=⇒=+ ccrceerBc rBrBr α
α Substituting back the above information
in equation (4.4.46) and differentiating the resulting equation with respect to ,z we get
.0)()(21)( 5)(5)(12 =−− rKerKeBzK r
rBrBrz Differentiating this equation with respect to ,z we
get .,,)(0)( 121112111212 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in the above
equation and solving we get ,)( 2)(
132
)(2
)(
115
rBrBrB
ecdreecrK−−−
+= ∫ .13 ℜ∈c Now
substituting back all the above information in equation (4.4.46) we get
.,)(0)()()(21
142
)(
142
)(2
)(
1277)(7)(
12 ℜ∈+−=⇒=++−−−
∫ cecdreecrKrKerKerBcrBrBrB
rrBrB
r
Refreshing the system of equations (4.4.45) we get
).(
,
),(22
,
62)(
132
)(2
)(
11
)(2
)(
72
)()(2
)(
53
2)(
142
)(2
)(
122
)(
132
)(2
)(
112
131211659
2
1
21
2)(
82
)(2
)(
62
)(
72
)(2
)(
50
rKecdreecetcdreectzX
ecdreececzdreeczX
zKczctcctzcctrX
ecdreececzdreecztX
rBrBrBrBrArArBrA
rBrBrBrBrBrB
rArArArArArA
+++++=
+−−−=
+++++++=
++++=
−−−−
−−
−−−−−−
−−−−−−
∫∫
∫∫
∫∫
θθα
θα
θθθα
α
(4.4.47)
Considering equation (4.4.8) and using equation (4.4.47) we get
.0)()(21)(
21)]()([
21
)()]()([21)(22
6)()(2)(
7
6)(2)(
2)(
513
115
=++−+
+−+++ ∫−
rKerBerzBerBrAct
rKedreerBrActzKctc
rBr
rBr
rA
rr
rrB
rArA
rrz
α
θ
(4.4.48)
Differentiating equation (4.4.48) with respect to ,t we get
.0)]()([21)]()([
212 2
)(
72
)(2
)(
55 =−+−+ ∫− rA
rr
rArA
rr erBrAcdreerBrAcc Solving this
217
equation and remember that in this case ,0)()( ≠− rBrA rr we get .075 == cc
Substituting the above information in equation (4.4.48) and differentiating with respect to
,θ we get .011 =c Now substituting the above information in equation (4.4.48) and
differentiating with respect to ,z we get ⇒−= 913 )( czKzz
.,,2
)( 161516159
213 ℜ∈++−= ccczcczzK Substituting back this value in equation (4.4.48) and
solving we get ⇒=++ 0)()()(21 6)(6)(
15 rKerKerBc rrBrB
r ,)( 2)(
172
)(2
)(
156
rBrBrB
ecdreecrK−−−
+−= ∫
.17 ℜ∈c Refreshing the system of equations (4.4.47) we get
,
,
,)22
(2
,
2)(
172
)(
132
)(2
)(
153
2)(
142
)(
132
)(2
)(
122
16151269
22
1
21
2)(
82
)(2
)(
60
rBrBrBrB
rBrBrBrB
rArArA
ececdreeczX
ececzdreecX
czcctcczctrX
ecdreectX
−−−−
−−−−
−−−
++−=
+−−=
++++−++=
++=
∫
∫
∫
θα
θα
θθα
α
(4.4.49)
where .,,,,,,,,, 1716151413129861 ℜ∈cccccccccc The line element for static cylindrically
symmetric space-times in this case becomes
),()2()2( 22109
2221
2 dzdcrcdrdtcrcds ++−+++−= θαα
(4.4.50)
where ).(,,, 9110921 cccccc ≠ℜ∈ The above space-time admits nine linearly independent
teleparallel homothetic vector fields in which eight are teleparallel Killing vector fields
given as ,r∂∂ ,2
)(
te
rA
∂∂−
,2)(
θ∂∂− rB
e ,2)(
ze
rB
∂∂−
),(2)(
θθ
∂∂
−∂∂−
zz
erB
,)(1
trM
rt
∂∂
+∂∂
θθ
∂∂
+∂∂ )(2 rMr
and ,)(2
zrM
rz
∂∂
+∂∂ where ∫
−−
= dreerMrArA
2)(
2)(
1 )(
and
.)( 2)(
2)(
2 dreerMrBrB
∫−−
−= One teleparallel proper homothetic vector field is given as
218
,,,)22
(2
, 329
22
1
210 zXXczctrXtX αθαθαα ==−++== (4.4.51)
The cases (III)(b) and (III)(c) can be solved exactly the same as the above case.
Case (IV):
In this case we have ),(rAA = ),(rBB = )(rCC = and .CBA == Now substituting
equation (4.4.9) in equation (4.4.3), we get
.0),,(),,( 13 =− zrPzrtPt θθ (4.4.52)
Differentiating equation (4.4.52) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (4.4.52) we get
⇒= − ),(),,( 11 zrEezrP ABθθ ),,(),(),,( 311 zrEzrEezrP AB += −θθ where ),(3 zrE is a
function of integration. Refreshing the system of equations (4.4.9) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θαθα
θαθα
rtPzXzrEzrEtXztPrXzrEzrEtX
+=++=
+=++= (4.4.53)
Considering equation (4.4.4) and using equation (4.4.53) we get
.0),(),(),,( 314 =−− zrEzrErtP zzt θθ (4.4.54)
Differentiating equation (4.4.54) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (4.4.54) we get
.0),(),(),( 314 =−− zrEzrErE zzθθ Differentiating the above equation with respect to θ
twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ where )(1 rK and )(2 rK are
functions of integration. Substituting back the above value in equation (4.4.54) and
solving we get )()(),( 311 rKrzKzrE += and ),()(),( 423 rKrKzzrE += where )(3 rK
and )(4 rK are functions of integration. Substituting all the above information in (4.4.53)
we get
219
( )( )
( ) ).,()()(),,()()(),,,(
),()()()(
5213
231221
42310
θθα
θαθα
θα
rErKrKtzXzrErKrKztXztPrX
rKrKzrKrKztX
+++=
+++=+=
++++=
(4.4.55)
Considering equation (4.4.5) and using equation (4.4.55), we get
.0),(),()(2 251 =++ zrErErtK zθθ (4.4.56)
Differentiating the above equation first with respect to t and then with respect to ,θ we
get 0)(1 =rK and ),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and
)(6 rK are functions of integration. Substituting back the above values in (4.4.56) we get
),()(),()(),( 75252 rKrKzzrErKzrEz +−=⇒−= where )(7 rK is a function of
integration. Substituting all the above information in (4.4.56) we get
).()()(),()()(),,,(),()()(
65237532
214230
rKrKrtKzXrKrKzrKtXztPrXrKrKzrKtX
+++=+−+=
+=+++=
θαθα
θαθα (4.4.57)
Considering equation (4.4.6) and using equation (4.4.57), we get
.0)()(21)(
21)(
)()()(21)()()(
21),,(
4)()(4)(
2)(2)(3)(3)(2
=−−−
−−−−
rKerAertArKe
rKezrKerzArKerKerAztP
rAr
rArr
rA
rrArA
rrrArA
rt
α
θθθ (4.4.58)
Differentiating equation (4.4.58) with respect to t we get .0)(21),,( )(2 =− rA
rtt erAztP αθ
Differentiating this equation with respect to ,t we get ⇒= 0),,(2 ztPttt θ
),,(),(),(2
),,( 8762
2 zEztEzEtztP θθθθ ++= where ),,(6 zE θ ),(7 zE θ and ),(8 zE θ are
functions of integration. Substituting back the above value in (4.4.58) we get
.0)(21),( )(6 =− rA
r erAzE αθ Differentiating this equation with respect to ,θ we get
),(),(0),( 866 zKzEzE =⇒= θθθ where )(8 zK is a function of integration. Substituting back
in the above equation and differentiating with respect to ,z we get
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Once again substituting back and solving we get
220
.,20)(21
221)()(
1 ℜ∈+=⇒=− ccrceerAc rArAr α
α Substituting all the above information in
equation (4.4.58) and differentiating with respect to ,θ we get
.0)()()(21),( 3)(3)(7 =−− rKerKerAzE r
rArArθθ Differentiating this equation with respect to ,θ
we get ⇒=0),(7 zE θθθ ),()(),( 1097 zKzKzE +=θθ where )(9 zK and )(10 zK are
functions of integration. Substituting back the above value and differentiating with
respect to ,z we get .,)(0)( 3399 ℜ∈=⇒= cczKzKz Now substituting back this value
and solving we get, .,)( 42
)(
42
)(2
)(
33 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Substituting all the
above information in equation (4.4.58) and differentiating twice with respect to ,z we get
.,,)(0)( 65651010 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in (4.4.58) and
solve after differentiating with respect to ,z we get
.,)( 72
)(
72
)(2
)(
52 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Now substituting all the above information
in equation (4.4.58) and solving we get, .,)( 82
)(
82
)(2
)(
64 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Refreshing the system of equations (4.4.57) we get
).()(
),()(
),,(2
,
652)(
72
)(2
)(
53
752)(
42
)(2
)(
32
86531
21
2)(
82
)(2
)(
62
)(
72
)(2
)(
52
)(
42
)(2
)(
30
rKrKetcdreectzX
rKrKzectdreectX
zEtcctzctctrX
ecdreececzdreeczecdreectX
rArArA
rArArA
rArArArArArArArArA
++++=
+−++=
+++++=
++++++=
−−−
−−−
−−−−−−−−−
∫
∫
∫∫∫
θα
θα
θθα
θθα
(4.4.59)
Considering equation (4.4.7) and using equation (4.4.59) we get
.0)()(21)(
21
)()()()(21),(2
7)()(
7)(5)(5)(83
=++
+−−+
rKerAerA
rKerKezrKerzAzEtc
rAr
rAr
rrA
rrArA
r
θα
θθ
(4.4.60)
221
Differentiating equation (4.4.60) with respect to ,t we get .03 =c Substituting the above
information in equation (4.4.60) and differentiating with respect to ,θ we get
⇒=+⇒=+ 0),(0)(21),( 1
8)(8 czEerAzE rAr θαθ θθθθ ),()(
2),( 1211
1
28 zKzKczE ++−= θθθ
where )(11 zK and )(12 zK are functions of integration. Substituting back the above
information in equation (4.4.60) and differentiating the resulting equation with respect to
,z we get .0)()()(21)( 5)(5)(11 =−− rKerKerAzK r
rArArz Differentiating this equation with
respect to ,z we get .,,)(0)( 1091091211 ℜ∈+=⇒= ccczczKzKzz Substituting back this value
in the above equation and solving we get ,)( 2)(
132
)(2
)(
95
rArArA
ecdreecrK−−−
+= ∫ .13 ℜ∈c
Now substituting back all the above information in equation (4.4.60) we get
.,)(0)()()(21
142
)(
142
)(2
)(
1077)(7)(
10 ℜ∈+−=⇒=++−−−
∫ cecdreecrKrKerKerAcrArArA
rrArA
r
Refreshing the system of equations (4.4.59) we get
).(
,
),()22
(
,
62)(
132
)(2
)(
92
)(
72
)(2
)(
53
2)(
142
)(2
)(
102
)(
132
)(2
)(
92
)(
42
12109651
221
2)(
82
)(2
)(
62
)(
72
)(2
)(
52
)(
40
rKecdreecetcdreectzX
ecdreececzdreeczectX
zKczctcctzctrX
ecdreececzdreeczectX
rArArArArArA
rArArArArArArA
rArArArArArArA
+++++=
+−−−+=
+++++−+=
+++++=
−−−−−−
−−−−−−−
−−−−−−−
∫∫
∫∫
∫∫
θθα
θα
θθθα
θα
(4.4.61)
Considering equation (4.4.8) and using equation (4.4.61) we get
.0)()(21)(
21)()(22 6)()(6)(12
95 =+++++ rKerAerzArKezKctc rAr
rArr
rAz αθ (4.4.62)
Differentiating equation (4.4.62) with respect to ,t we get .05 =c Substituting the above
information in equation (4.4.62) and differentiating with respect to ,θ we get .09 =c
Now substituting the above information in equation (4.4.62) and differentiating with
respect to ,z we get ⇒−=⇒=+ 112)(12 )(0)(
21)( czKerAzK zz
rArzz α
222
.,,2
)( 161516151
212 ℜ∈++−= ccczcczzK Substituting back this value in equation (4.4.62)
and solving we get ⇒=++ 0)()()(21 6)(6)(
15 rKerKerAc rrArA
r ,)( 2)(
172
)(2
)(
156
rArArA
ecdreecrK−−−
+−= ∫
.17 ℜ∈c Refreshing the system of equations (4.4.61) we get
.
,
,)222
(
,
2)(
172
)(
132
)(
72
)(2
)(
153
2)(
142
)(
132
)(
42
)(2
)(
102
16151061
2221
2)(
82
)(
72
)(
42
)(2
)(
60
rArArArArA
rArArArArA
rArArArArA
ececetcdreeczX
ececzectdreecX
czcctccztrX
ececzecdreectX
−−−−−
−−−−−
−−−−−
+++−=
+−+−=
++++−−+=
++++=
∫
∫
∫
θα
θα
θθα
θα
(4.4.63)
where .,,,,,,,,,, 17161514131087641 ℜ∈ccccccccccc The line element for static cylindrically
symmetric space-times in this case becomes
,)()2( 222221
2 drdzddtcrcds +++−+= θα (4.4.64)
where ).0(, 121 ≠ℜ∈ ccc The above space-time admits eleven linearly independent
teleparallel homothetic vector fields in which ten are teleparallel Killing vector fields
given as ,r∂∂ ,2
)(
te
rA
∂∂−
,2)(
θ∂∂− rA
e ,2)(
ze
rA
∂∂−
),(2)(
θθ
∂∂
+∂∂−
tt
erA
),(2)(
zt
tze
rA
∂∂
+∂∂−
),(2)(
θθ
∂∂
−∂∂−
zz
erA
,)(t
rMr
t∂∂
+∂∂
θθ
∂∂
−∂∂ )(rMr
and z
rMr
z∂∂
−∂∂ )( where
.)( 2)(
2)(
∫−−
= dreerMrArA
One teleparallel proper homothetic vector field is given as
,,,)222
(, 321
22210 zXXcztrXtX αθαθαα ==−−+== (4.4.65)
Case (V)(a):
In this case we have ,constant=A ),(rBB = )(rCC = and .CB = Now substituting
equation (4.4.9) in equation (4.4.3), we get
223
.0),,(),,( 13)( =− zrPzrtPe trB θθ (4.4.66)
Differentiating equation (4.4.66) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (4.4.66) we get
⇒= ),(),,( 11 zrEezrP Bθθ ),,(),(),,( 311 zrEzrEezrP B +=θθ where ),(3 zrE is a
function of integration. Refreshing the system of equations (4.4.9) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θαθα
θαθα
rtPzXzrEzrEtXztPrXzrEzrEetX B
+=++=
+=++= (4.4.67)
Considering equation (4.4.4) and using equation (4.4.67) we get
.0),(),(),,( 31)(4)( =−− zrEzrEertPe zzrB
trB θθ (4.4.68)
Differentiating equation (4.4.68) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (4.4.68) we get
.0),(),(),( 31)(4)( =+− zrEzrEerEe zzrBrB θθ Differentiating the above equation with
respect to θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ where )(1 rK and
)(2 rK are functions of integration. Substituting back the above value in equation (4.4.68)
and solving we get )()(),( 311 rKrKzzrE += and ),()(),( 42)(3 rKrKezzrE rB += where
)(3 rK and )(4 rK are functions of integration. Substituting all the above information in
(4.4.67) we get
( )( )
( ) ).,()()(),,()()(),,,(
),()()()(
5213
231221
42)(3)(1)(0
θθα
θαθα
θα
rErKrKtzXzrErKrKztXztPrX
rKrKezrKerKeztX rBrBrB
+++=
+++=+=
++++=
(4.4.69)
Considering equation (4.4.5) and using equation (4.4.69), we get
.0),(),()(2 251 =++ zrErErtK zθθ (4.4.70)
Differentiating the above equation first with respect to t and then with respect to ,θ we
get 0)(1 =rK and ),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and
224
)(6 rK are functions of integration. Substituting back the above values in (4.4.70) we get
),()(),()(),( 75252 rKrKzzrErKzrEz +−=⇒−= where )(7 rK is a function of
integration. Substituting all the above information in (4.4.69) we get
).()()(),()()(),,,(
),()()(
6523
753221
42)(3)(0
rKrKrtKzXrKrKzrKtXztPrX
rKrKezrKetX rBrB
+++=
+−+=+=
+++=
θα
θαθα
θα
(4.4.71)
Considering equation (4.4.6) and using equation (4.4.71), we get
.0)()(
)()()()()(),,(42)(
2)(3)(3)(2
=−−
−−−
rKrKez
rKerBzrKerKerBztP
rrrB
rBrr
rBrBrt θθθ
(4.4.72)
Differentiating equation (4.4.72) with respect to t we get ⇒= 0),,(2 ztPtt θ
),,(),(),,( 762 zEztEztP θθθ += where ),(6 zE θ and ),(7 zE θ are functions of
integration. Substituting back the above value in (4.4.72) and differentiating with respect
to ,θ we get .0)()()(),( 3)(3)(6 =−− rKerKerBzE rrBrB
rθθ Differentiating this equation with
respect to ,θ we get ),()(),(0),( 9866 zKzKzEzE +=⇒= θθθθθ where )(8 zK and )(9 zK are
function of integration. Substituting back in the above equation and differentiating with
respect to ,z we get .,)(0)( 1188 ℜ∈=⇒= cczKzKz Once again substituting back and
solving we get ⇒=−− 0)()()( 3)(3)(1 rKerKerBc r
rBrBr .,)( 2
)(2
)(1
3 ℜ∈+= −− cecercrK rBrB
Substituting all the above information in equation (4.4.72) and solve after differentiating
with respect to ,z we get .0)()()()( 2)(2)(9 =−− rKerKerBzK rrBrB
rz Now differentiating this
equation with respect to ,z we get .,,)(0)( 434399 ℜ∈+=⇒= ccczczKzKzz
⇒=−− 0)()()( 2)(2)(3 rKerKerBc r
rBrBr .,)( 5
)(5
)(3
2 ℜ∈+= −− cecercrK rBrB Now substituting all
the above information in equation (4.4.72) and solving we get, .,)( 6644 ℜ∈+= ccrcrK
Refreshing the system of equations (4.4.71) we get
).()(
),()(
),,(,
65)(5
)(3
3
75)(2
)(1
2
7431
1654321
0
rKrKetcerctzX
rKrKzecterctX
zEtctzcctrXcczcrcrzccrtX
rBrB
rBrB
++++=
+−++=
++++=++++++=
−−
−−
θα
θα
θθαθθα
(4.4.73)
Considering equation (4.4.7) and using equation (4.4.73) we get
225
.0)()(21)(
21)(
)()()(21)(
21)(
21),(2
7)()(7)(
5)(5)(21
71
=+++
−−−−+
rKerBerBrKe
rKezrKerBzrBctrBrctzEtc
rBr
rBrr
rB
rrBrB
rrr
θα
θθ
(4.4.74)
Differentiating equation (4.4.74) with respect to ,t we get
.0)(21)(
212 211 =−− rBcrBrcc rr Solving this equation and remember that in this case
,0)( ≠rBr we get .021 == cc Substituting the above information in equation (4.4.74) and
differentiating with respect to ,θ we get .0)(21),( )(7 =+ rB
r erBzE αθθθ Now
differentiating this equation with respect to ,θ we get ⇒= 0),(7 zE θθθθ
),()()(2
),( 1211102
7 zKzKzKzE ++= θθθ where ),(11 zK )(12 zK and )(12 zK are
functions of integration. Substituting back the above information in equation (4.4.74) and
differentiating the resulting equation with respect to ,z we get
.,)(0)( 771010 ℜ∈=⇒= cczKzKz Substituting back this value and solving we get
⇒=+ 0)(21 )(
7rB
r erBc α .,2887
2)(
ℜ∈+−= ccrcerB
α Substituting back this value in
equation (4.4.74) and differentiating the resulting equation twice with respect to ,z we
get .,,)(0)( 1091091111 ℜ∈+=⇒= ccczczKzKzz Substituting back this value and solving we
get .,)( 112
)(
112
)(2
)(
95 ℜ∈+=
−−−
∫ cecdreecrKrBrBrB
Now substituting all the above
information in equation (4.4.74) and solving we get
.,)( 122
)(
122
)(2
)(
107 ℜ∈+−=
−−−
∫ cecdreecrKrBrBrB
Refreshing the system of equations
(4.4.73) we get
).(
,
),(2
,
62)(
112
)(2
)(
9)(
5)(
33
2)(
122
)(2
)(
102
)(
112
)(2
)(
92
12109437
21
64530
rKecdreecetcerctzX
ecdreececzdreeczX
zKczctcctzcrXcrcczcrztX
rBrBrBrBrB
rBrBrBrBrBrB
+++++=
+−−−=
++++++=++++=
−−−−−
−−−−−−
∫
∫∫θθα
θα
θθθαα
(4.4.75)
226
Considering equation (4.4.8) and using equation (4.4.75) we get
.0)()(21)(
21
)(21)()(
21)(22
6)()(
56)(
312
93
=++
−+−++
rKerBerzB
rBctrKerBrctzKctc
rBr
rBr
rrrB
rz
α
θ
(4.4.76)
Differentiating equation (4.4.76) with respect to ,t we get
.0)(21)(
212 533 =−− rBcrBrcc rr Solving this equation and remember that in this case
,0)( ≠rBr we get .053 == cc Substituting the above information in equation (4.4.76) and
differentiating with respect to ,θ we get .09 =c Now substituting the above information
in equation (4.4.76) and differentiating with respect to ,z we get ⇒−= 712 )( czKzz
.,,2
)( 141314137
212 ℜ∈++−= ccczcczzK Substituting back this value in equation (4.4.76)
and solving we get ⇒=++ 0)()()(21 6)(6)(
13 rKerKerBc rrBrB
r ,)( 2)(
152
)(2
)(
136
rBrBrB
ecdreecrK−−−
+−= ∫
.15 ℜ∈c Refreshing the system of equations (4.4.75) we get
.
,
,)22
(,
2)(
152
)(
112
)(2
)(
133
2)(
122
)(
112
)(2
)(
102
14131047
221
640
rBrBrBrB
rBrBrBrB
ececdreeczX
ececzdreecX
czcctcczrXcrctX
−−−−
−−−−
++−=
+−−=
++++−+=++=
∫
∫θα
θα
θθαα
(4.4.77)
where .,,,,,,,, 151413121110764 ℜ∈ccccccccc The line element for static cylindrically
symmetric space-times in this case becomes
),()2( 2287
222 dzdcrcdrdtds ++−++−= θα
(4.4.78)
where ).0(, 787 ≠ℜ∈ ccc The above space-time admits nine linearly independent
teleparallel homothetic vector fields in which eight are teleparallel Killing vector fields
given as ,r∂∂ ,
t∂∂ ,2
)(
θ∂∂− rB
e ,2)(
ze
rB
∂∂−
),(2)(
θθ
∂∂
−∂∂−
zz
erB
,t
rr
t∂∂
+∂∂
227
θθ
∂∂
+∂∂ )(rMr
and ,)(z
rMr
z∂∂
+∂∂ where
.)( 2)(
2)(
dreerMrBrB
∫−−
−= One teleparallel
proper homothetic vector field is given as
,,,)22
(, 327
2210 zXXczrXtX αθαθαα ==−+== (4.4.79)
The cases (V)(b) and (V)(c) can be solved exactly the same as the above case.
Case (VI)(a):
In this case we have )(rAA = and .constant== CB Now substituting equation (4.4.9)
in equation (4.4.3), we get
.0),,(),,( 1)(3 =− zrPezrtP rAt θθ (4.4.80)
Differentiating equation (4.4.80) with respect to ,t we get
),,(),(),,(0),,( 2133 zrEzrEtzrtPzrtPtt +=⇒= where ),(1 zrE and ),(2 zrE are
functions of integration. Substituting back the above value in equation (4.4.80) we get
⇒= − ),(),,( 11 zrEezrP Aθθ ),,(),(),,( 311 zrEzrEezrP A += −θθ where ),(3 zrE is a
function of integration. Refreshing the system of equations (4.4.9) we get
).,,(),,(),(),,,(),,(),(
43212
21310
θαθα
θαθα
rtPzXzrEzrEtXztPrXzrEzrEetX A
+=++=
+=++= −
(4.4.81)
Considering equation (4.4.4) and using equation (4.4.81) we get
.0),(),(),,( 3)(14 =−− zrEezrErtP zrA
zt θθ (4.4.82)
Differentiating equation (4.4.82) with respect to ,t we get
),,(),(),,(0),,( 5444 θθθθ rErEtrtPrtPtt +=⇒= where ),(4 θrE and ),(5 θrE are
functions of integration. Substituting back the above value in (4.4.82) we get
.0),(),(),( 3)(14 =−− zrEezrErE zrA
zθθ Differentiating the above equation with respect to
θ twice, we get ),()(),(0),( 2144 rKrKrErE +=⇒= θθθθθ where )(1 rK and )(2 rK
are functions of integration. Substituting back the above value in equation (4.4.82) and
solving we get )()(),( 311 rKrKzzrE += and ),()(),( 42)(3 rKrKezzrE rA += − where
228
)(3 rK and )(4 rK are functions of integration. Substituting all the above information in
(4.4.81) we get
( )( )
( ) ).,()()(),,()()(),,,(),()()()(
5213
231221
42)(3)(1)(0
θθα
θαθα
θα
rErKrKtzXzrErKrKztXztPrXrKrKezrKerKeztX rArArA
+++=
+++=+=
++++= −−−
(4.4.83)
Considering equation (4.4.5) and using equation (4.4.83), we get
.0),(),()(2 251 =++ zrErErtK zθθ (4.4.84)
Differentiating the above equation first with respect to t and then with respect to ,θ we
get 0)(1 =rK and ),()(),(0),( 6555 rKrKrErE +=⇒= θθθθθ where )(5 rK and
)(6 rK are functions of integration. Substituting back the above values in (4.4.84) we get
),()(),()(),( 75252 rKrKzzrErKzrEz +−=⇒−= where )(7 rK is a function of
integration. Substituting all the above information in (4.4.83) we get
( )).()()(),()()(),,,(),()()(
65237532
2142)(3)(0
rKrKrtKzXrKrKzrKtXztPrXrKrKezrKetX rArA
+++=+−+=
+=+++= −−
θαθα
θαθα (4.4.85)
Considering equation (4.4.6) and using equation (4.4.85), we get
.0)()(21)(
21
)()()()(21)()()(
21),,(
4)()(
4)(22332
=−−
−−+−+
rKerAertA
rKerKzrKrzArKrKrAztP
rAr
rAr
rrA
rrrrt
α
θθθ (4.4.86)
Differentiating equation (4.4.86) with respect to t we get .0)(21),,( )(2 =− rA
rtt erAztP αθ
Differentiating this equation with respect to ,t we get ⇒= 0),,(2 ztPttt θ
),,(),(),(2
),,( 8762
2 zEztEzEtztP θθθθ ++= where ),,(6 zE θ ),(7 zE θ and ),(8 zE θ are
functions of integration. Substituting back the above value in (4.4.86) we get
.0)(21),( )(6 =− rA
r erAzE αθ Differentiating this equation with respect to ,θ we get
),(),(0),( 866 zKzEzE =⇒= θθθ where )(8 zK is a function of integration. Substituting back
229
in the above equation and differentiating with respect to ,z we get
.,)(0)( 1188 ℜ∈=⇒= cczKzKz Once again substituting back and solving we get
.,20)(21
221)()(
1 ℜ∈+=⇒=− ccrceerAc rArAr α
α Substituting all the above information in
equation (4.4.86) and differentiating with respect to ,θ we get
.0)()()(21),( 337 =−+ rKrKrAzE rrθθ Differentiating this equation with respect to ,θ we get
⇒=0),(7 zE θθθ ),()(),( 1097 zKzKzE +=θθ where )(9 zK and )(10 zK are functions of
integration. Substituting back the above value and differentiating with respect to ,z we
get .,)(0)( 3399 ℜ∈=⇒= cczKzKz Now substituting back this value and solving we get,
.,)( 42
)(
42
)(2
)(
33 ℜ∈+= ∫
−
cecdreecrKrArArA
Substituting all the above information in
equation (4.4.86) and differentiating twice with respect to ,z we get
.,,)(0)( 65651010 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in (4.4.86) and
solve after differentiating with respect to ,z we get
.,)( 72
)(
72
)(2
)(
52 ℜ∈+= ∫
−
cecdreecrKrArArA
Now substituting all the above information in
equation (4.4.86) and solving we get, .,)( 82
)(
82
)(2
)(
64 ℜ∈+=
−−−
∫ cecdreecrKrArArA
Refreshing the system of equations (4.4.85) we get
).()(
),()(
),,(2
,
652)(
72
)(2
)(
53
752)(
42
)(2
)(
32
86531
21
2)(
82
)(2
)(
62
)(
72
)(2
)(
52
)(
42
)(2
)(
30
rKrKetcdreectzX
rKrKzectdreectX
zEtcctzctctrX
ecdreececzdreeczecdreectX
rArArA
rArArA
rArArArArArArArArA
++++=
+−++=
+++++=
++++++=
∫
∫
∫∫∫
−
−
−−−−−−−−−
θα
θα
θθα
θθα
(4.4.87)
Considering equation (4.4.7) and using equation (4.4.87) we get
.0)()()(21)(
21),(2 752
)(
42
)(2
)(
38
3 =+−+++ ∫−
rKrKzerActdreerActzEtc rr
rA
r
rArA
rθθ (4.4.88)
230
Differentiating equation (4.4.88) with respect to ,t we get
.0)(21)(
212 2
)(
42
)(2
)(
33 =++ ∫− rA
r
rArA
r erAcdreerAcc Solving this equation and remember
that in this case ,0)( ≠rAr we get .043 == cc Substituting the above information in
equation (4.4.88) and differentiating with respect to ,θ we get
),()(),(0),( 121188 zKzKzEzE +=⇒= θθθθθ where )(11 zK and )(12 zK are functions of
integration. Substituting back the above information in equation (4.4.88) and
differentiating the resulting equation with respect to ,z we get .0)()( 511 =− rKzK rz
Differentiating this equation with respect to ,z we get
.,,)(0)( 121112111111 ℜ∈+=⇒= ccczczKzKzz Substituting back this value in the above equation
and solving we get ,)( 13115 crcrK += .13 ℜ∈c Now substituting back all the above
information in equation (4.4.88) we get .,)(0)( 14141277
12 ℜ∈+−=⇒=+ ccrcrKrKc r
Refreshing the system of equations (4.4.87) we get
).(
,
),(2
,
61311
2)(
72
)(2
)(
53
141213112
121211651
21
2)(
82
)(2
)(
62
)(
72
)(2
)(
50
rKcrcetcdreectzX
crczcrczX
zKczctcctzctrX
ecdreececzdreecztX
rArArA
rArArArArArA
+++++=
+−−−=
++++++=
++++=
∫
∫∫
−
−−−−−−
θθα
θα
θθα
α
(4.4.89)
Considering equation (4.4.8) and using equation (4.4.89) we get
.0)(21)()(
21)(22 2
)(
762
)(2
)(
512
115 =+++++ ∫− rA
rr
rArA
rz erActrKdreerActzKctc θ (4.4.90)
Differentiating equation (4.4.90) with respect to ,t we get
.0)(21)(
212 2
)(
72
)(2
)(
55 =++ ∫− rA
r
rArA
r erAcdreerAcc Solving this equation and remember
that in this case ,0)( ≠rAr we get .075 == cc Substituting the above information in
equation (4.4.90) and differentiating with respect to ,θ we get .011 =c Now substituting
the above information in equation (4.4.90) and differentiating with respect to ,z we get
231
⇒= 0)(12 zKzz .,,)( 1615161512 ℜ∈+= ccczczK Substituting back this value in equation
(4.4.90) and solving we get ⇒=+ 0)(615 rKc r ,)( 1715
6 crcrK +−= .17 ℜ∈c Refreshing the
system of equations (4.4.89) we get
.,
,2
,
1713153
1413122
16151261
212
)(
82
)(2
)(
60
ccrczXczcrcX
czcctcctrXecdreectXrArArA
++−=+−−=
+++++=++=−−−
∫θαθα
θαα
(4.4.91)
where .,,,,,,,, 171615141312861 ℜ∈ccccccccc The line element for static cylindrically
symmetric space-times in this case becomes
),()2( 222221
2 dzddrdtcrcds ++++−= θα
(4.4.92)
where ).0(, 121 ≠ℜ∈ ccc The above space-time admits nine linearly independent
teleparallel homothetic vector fields in which eight are teleparallel Killing vector fields
given as ,r∂∂ ,2
)(
te
rA
∂∂−
,θ∂∂ ,
z∂∂ ),(
θθ
∂∂
−∂∂ zz
,)(t
rMr
t∂∂
+∂∂ )(
θθ
∂∂
−∂∂ rr
and
),(z
rr
z∂∂
−∂∂ where .)( 2
)(2
)(
∫−−
= dreerMrArA
One teleparallel proper homothetic vector
field is given as
,,,2
, 321
210 zXXctrXtX αθααα ==+== (4.4.93)
The cases (VI)(b) and (VI)(c) can be solved exactly the same as the above case.
4.5. Summary of the Chapter
In this chapter we investigated teleparallel proper homothetic vector fields for Bianchi
type I, non static plane symmetric and static cylindrically symmetric space-times by
using direct integration technique. It follows that all of the above space-times admit
teleparallel proper homothetic vector fields for special classes of the metric. In all the
above space-times we have shown that the dimension of the teleparallel proper
232
homothetic vector field is one like in general relativity. Following results are obtained
from the above study:
(1) In Bianchi type I space-times: Teleparallel homothetic vector fields have been
explored and it is shown that the above space-times admit 8, 9 or 11 teleparallel
homothetic vector fields. In all the cases when the above space-times admit teleparallel
homothetic vector fields it admits only one proper teleparallel homothetic vector fields
for a special class of metric. When the space-time becomes Minkowski then all the
torsion components become zero and the teleparallel Lie derivative for the metric gives
the same equations as in general relativity, hence the homothetic vector fields are same in
both the theories.
The results for above space time when it admits eight teleparallel homothetic vector
fields, teleparallel proper homothetic vector fields are given in equations (4.2.23) and
(4.2.37). When the above space-time admits nine teleparallel homothetic vector fields,
teleparallel proper homothetic vector fields are given in equations (4.2.51), (4.2.79) and
(4.2.93). When the above space-time admits eleven teleparallel homothetic vector fields,
teleparallel proper homothetic vector field is given in equation (4.2.65).
(2) In non static plane symmetric space-times: Teleparallel homothetic vector fields have
been explored and it is shown that the above space-times admit 6, 7, 8 or 11 teleparallel
homothetic vector fields. The above space-time admits eleven teleparallel homothetic
vector fields only when the space-time becomes Minkowski and all the torsion
components become zero. In all the cases when the above space-times admit teleparallel
homothetic vector fields it admits only one proper teleparallel homothetic vector fields
for a special class of metric. When the above space-time admits six teleparallel
homothetic vector fields, results for teleparallel proper homothetic vector fields are given
in equations (4.3.39) and (4.3.57). When the above space-time admits seven teleparallel
homothetic vector fields, results for teleparallel proper homothetic vector fields are given
in equations (4.3.23), (4.3.26), (4.3.50), (4.3.66), (4.3.69) and (4.3.72). When the above
space-time admits eight teleparallel homothetic vector fields, results for teleparallel
proper homothetic vector fields are given in equations (4.3.29), (4.3.42), (4.3.60) and
233
(4.3.63). The result for teleparallel proper homothetic vector field when the space-time
admits eleven teleparallel homothetic vector fields is given in equation (4.3.32).
(3) In static cylindrically symmetric space-times: Teleparallel homothetic vector fields
have been explored and it turns out that the above space-times admit 8, 9 or 11
teleparallel homothetic vector fields. In all the cases when the above space-time admits
teleparallel homothetic vector fields, it possesses only one teleparallel proper homothetic
vector fields. The above space-times admit teleparallel homothetic vector fields for a
special class of the metric. When the space-time becomes Minkowski then all the torsion
components become zero and hence admit the same homothetic vector fields as in
general relativity. When the above space-time admits eight teleparallel homothetic vector
fields, results for teleparallel proper homothetic vector fields are given in equations
(4.4.23) and (4.4.37). When the above space-time admits nine teleparallel homothetic
vector fields, results for teleparallel proper homothetic vector fields are given in
equations (4.4.51), (4.4.79) and (4.4.93). When the above space-time admits eleven
teleparallel homothetic vector fields, teleparallel proper homothetic vector field is given
in equation (4.4.65).
234
Chapter 5
Proper Conformal Vector Fields in Non
Conformally Flat Non Static Cylindrically
Symmetric, Kantowski-Sachs and Bianchi
Type III Space-Times
5.1. Introduction
This chapter is devoted to find proper conformal vector fields in non conformally flat
non static cylindrically symmetric, Kantowski-Sachs and Bianchi type III space-times
using direct integration technique. All the above space-times possess proper
conformal vector fields for special class of space-times. It is important to note that
here the classification of non static cylindrically symmetric space-times will also
cover non static plane symmetric, static cylindrically symmetric and static plane
symmetric space-times [42]. This chapter is organized as follows. In section (5.2)
proper conformal vector fields of non conformally flat non static cylindrically
symmetric space-times are investigated. In the next section (5.3) proper conformal
vector fields in non conformally flat Kantowski-Sachs and Bianchi type III space-
times are investigated. In section (5.4) a detailed summary of the work is given.
235
5.2. Proper Conformal Vector Fields in Non
Conformally Flat Non Static Cylindrically
Symmetric Space-Times
The most general form of the metric for non static cylindrically symmetric space-time
is given by [78]
,2),(2),(22),(2 dzededrdteds rtCrtBrtA +++−= θ (5.2.1)
where ,A B and C are functions of rt and only. The above space-time admits two
independent Killing vector fields which are θ∂∂ and .
z∂∂ If in the above space-times
,CB = it becomes non static plane symmetric space-time and possess three Killing
vector fields which are ,θ∂∂
z∂∂ and ).(
zz
∂∂
−∂∂ θθ
A vector field X is said to be a
conformal vector field if it satisfies equation (1.2.4). One can write (1.2.4) explicitly
using (5.2.1)
,2 0,010 η=+′+• XXAXA (5.2.2)
,01,0),(
0,1 =− XeX rtA (5.2.3)
,02,0),(
0,2),( =− XeXe rtArtB (5.2.4)
,03,0),(
0,3),( =− XeXe rtArtC (5.2.5)
,2 1,1 η=X (5.2.6)
,02,1
1,2),( =+ XXe rtB (5.2.7)
,03,1
1,3),( =+ XXe rtC (5.2.8)
236
,2 2,210 η=+′+• XXBXB (5.2.9)
,03,2),(
2,3),( =+ XeXe rtBrtC (5.2.10)
,2 3,310 η=+′+• XXCXC (5.2.11)
where dot denotes the derivative with respect to t and dash denotes the derivative
with respect to .r Here it follows that [42] ).,( rtηη = Equations (5.2.3), (5.2.6),
(5.2.7) and (5.2.8) give
( )
),,,(),,(
),,,(),,(),,,(),(21
),,,(),,(),(21
4),(13
3),(1211
2),(1),(0
ztKdreztKX
ztKdreztKXztKdrrtX
ztKdreztKdrdrrteX
rtCz
rtB
rtAtt
rtA
θθ
θθθη
θθη
θ
+−=
+−=+=
++=
∫∫∫
∫∫ ∫
−
−
−−
(5.2.12)
where ),,(),,,(),,,( 321 ztKztKztK θθθ and ),,(4 ztK θ are functions of integration.
In order to determine ),,(),,,(),,,( 321 ztKztKztK θθθ and ),,(4 ztK θ we need to
integrate the remaining six equations. Considering equation (5.2.11) and using
equation (5.2.21) then solve after differentiating with respect to z we get
),,(),,(0),,( 111 θθθ tEztKztK z =⇒= where ),(1 θtE is a function of integration.
Substituting back the above value in (5.2.11) we get
0),,(2),,(),( 42 =+ ztKztKrtC zzzt θθ (5.2.13)
Differentiating the above equation with respect to ,r we get .0),,(),( 2 =ztKrtC zrt θ
In order to solve this equation we will discuss three possibilities:
(I) ,0),( ≠rtCrt ,0),,(2 =ztK z θ (II) ,0),( =rtCrt ,0),,(2 ≠ztK z θ
(III) ,0),( =rtCrt .0),,(2 =ztK z θ
We will discuss each possibility in turn.
237
Case I:
In this case we have 0),( ≠rtCrt and .0),,(2 =ztK z θ Equation ⇒= 0),,(2 ztK z θ
),,(),,( 22 θθ tEztK = where ),(2 θtE is a function of integration. Substituting back
the above value in (5.2.13) and solving we get ),,(),(),,( 434 θθθ tEtzEztK +=
where ),(3 θtE and ),(4 θtE are functions of integration. Substituting all the above
information in equation (5.2.12) we get
( )
),,(),(
,),,(),,(,),(),(21
,),(),(),(21
433
31211
210
θθ
θθθη
θθη
θ
tEtzEX
ztKdreztEXtEdrrtX
tEdretEdrdrrteX
B
Att
A
+=
+−=+=
++=
∫∫
∫ ∫∫−
−−
(5.2.14)
Now considering equation (5.2.9) and using equation (5.2.14) then solve after
differentiating with respect to ,θ we get ),(),(0),( 111 tFtEtE =⇒= θθθ where
)(1 tF is a function of integration. Substituting back the above value in (5.2.9) and
differentiating with respect to ,r we get .0),(),( 2 =θθ tErtBrt To solve this equation
we need to discuss three different possibilities:
(a) ,0),( ≠rtBrt ,0),(2 =θθ tE (b) ,0),( =rtBrt ,0),(2 ≠θθ tE
(c) ,0),( =rtBrt .0),(2 =θθ tE
We will discuss each case in turn.
Case I (a):
In this case we have 0),( ≠rtBrt and .0),(2 =θθ tE Equation
),(),(0),( 222 tFtEtE =⇒= θθθ where )(2 tF is a function of integration.
Substituting back the above value in (5.2.9) we get
),,(),(),,(0),,( 6533 ztEztEztKztK +=⇒= θθθθθ where ),(5 ztE and ),(6 ztE are
238
functions of integration. Substituting all the above information in the system of
equations (5.2.14) we get
( )
),,(),(
),,(),(,)(),(21
,)()(),(21
433
65211
210
θθ
θη
η
tEtzEX
ztEztEXtFdrrtX
tFdretFdrdrrteX Att
A
+=
+=+=
++=
∫
∫ ∫∫ −−
(5.2.15)
Considering equation (5.2.4) and using equation (5.2.15) then differentiating with
respect to ,θ we get ⇒= 0),(5 ztEt ),(),( 35 zFztE = where )(3 zF is a function of
integration. Substituting back the above value in (5.2.4) and solving we get
),(),(0),( 466 zFztEztEt =⇒= where )(4 zF is a function of integration.
Substituting the above information in (5.2.15) we get
( )
).,(),(),()(
,)(),(21
,)()(),(21
433432
11
210
θθθ
η
η
tEtzEXzFzFX
tFdrrtX
tFdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫ ∫∫ −−
(5.2.16)
Considering equation (5.2.5) and using equation (5.2.16) then solve after
differentiating with respect to ,z we get )(),( 53 θθ FtE = and ),(),( 64 θθ FtE =
where )(5 θF and )(6 θF are functions of integration. Substituting back all the above
information in (5.2.16) we get
( )
).()(),()(
),(),(21
,)()(),(21
653432
11
210
θθθ
η
η
FzFXzFzFX
tFdrrtX
tFdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫ ∫∫ −−
(5.2.17)
Now subtracting equation (5.2.9) from (5.2.11) we get
.0))()((2)()( 3510 =−+−+− zFFXBCXBC rrtt θ Now considering this equation
and using equation (5.2.17) then suppose ,0)( =− tt BC we get
239
( ).)()(2)(),(21)( 531 θη FzFtFdrrtBC rr −=⎥⎦
⎤⎢⎣⎡ +− ∫ Differentiating the above
equation with respect to ,t we get .0)(),(21)( 1 =⎥⎦
⎤⎢⎣⎡ +− ∫ tFdrrtBC ttrr η Solution of
this equation involves three different possibilities:
(i) ,0)( ≠− rr BC ,0)(),(21 1 =⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη
(ii) ,0)( =− rr BC ,0)(),(21 1 ≠⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη
(iii) ,0)( =− rr BC .0)(),(21 1 =⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη
We will discuss each case in turn.
Case I (a) (i):
In this case we have 0)( ≠− rr BC and .0)(),(21 1 =⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη Equation
⇒=⎥⎦⎤
⎢⎣⎡ +∫ 0)(),(21 1 tFdrrt ttη )(rηη = and .,)( 11
1 ℜ∈= cctF Refreshing the
system of equations (5.2.17) we get
).()(),()(
,)(21),(
653432
1120
θθθ
η
FzFXzFzFX
cdrrXtFX
+=+=
+== ∫ (5.2.18)
Considering equation (5.2.10) and using equation (5.2.18) then solve after
differentiating with respect to z and ,θ we get 3255 )(0)( ccFF +=⇒= θθθθθ and
⇒= 0)(3 zFzz ,)( 543 czczF += .,,, 5432 ℜ∈cccc Substituting back the above values
in (5.2.10) and differentiating with respect to ,z we get 02 =c and ⇒= 0)(4 zFzz
.,,)( 76764 ℜ∈+= ccczczF Now substituting back all the above values in (5.2.10)
240
we get 986 )( ccF += θθ and .0864 === ccc Substituting all the above information
in (5.2.18) we get
,,,)(21),( 93
375
21
120 czcXccXcdrrXtFX +=+=+== ∫ θη (5.2.19)
where .,,,, 97531 ℜ∈ccccc Considering equation (5.2.2) and using equation (5.2.19)
then differentiating with respect to t we get
.0)(2)(21),()(),()(),( 2
122 =+⎟
⎠⎞
⎜⎝⎛ +++ ∫ tFcdrrrtAtFrtAtFrtA ttrttttt η (5.2.20)
In order to solve (5.2.20) we discuss different possibilities. First, suppose 0),( =rtAt
then ⇒= 0)(2 tFtt .,,)( 111011102 ℜ∈+= ccctctF Substituting back all the above
information in (5.2.2) we get ⇒=+⎥⎦⎤
⎢⎣⎡ +∫ )(2)(21
101 rccdrrAr ηη
( ) .)(
21
2)(ln)(1
10
2
1 ∫∫
∫+
−+=cdrr
drccdrrrAη
η Now differentiating equation
(5.2.11) with respect to ,t we get
.0)(21),(),()(),( 1101110 =⎥⎦
⎤⎢⎣⎡ ++++ ∫ cdrrrtCrtCcctcrtC rtttt η In order to solve this
equation consider 0),( =rtCtt then from the above equation we get
⇒−=⎥⎦⎤
⎢⎣⎡ +∫ ),()(21),( 101 rtCccdrrrtC trt η ⇒∫=
− drrvctt etPrtC
)(10)(),(
),()(),( 1)(10 rKetPtrCdrrvc+∫=
− where )(1 rK is a function of integration and
.)(
21
1)(1∫ +
=cdrr
rVη
In this case ⇒= ),(),( rtBrtC tt ⇒∫=− drrvc
tt etPrtB)(10)(),(
),()(),( 2)(10 rKetPtrBdrrvc+∫=
− where )(2 rK is a function of integration. Now
differentiate the value of ),( trC with respect to r and substitute that value in (5.2.11)
241
then differentiating the resulting equation with respect to t and solving we get
.,,)( 13121312 ℜ∈+= ccctctP Substituting back the above value in equation (5.2.11)
and solving we get ⇒=+⎥⎦⎤
⎢⎣⎡ ++∫ ∫
−)(2)(
21)( 31
1)(14
10 rccdrrrKec rdrrVc
ηη
( ) ,)()(2)(ln)()(
143
2
11 10∫ ∫∫ ∫−−+=
−drerVcdrrVccdrrrK
drrVcη where
.01310121114 ≠−= ccccc Similarly substituting the above information in (5.2.9) and
solving we get ( ) .)()(2)(ln)()(
145
2
12 10∫ ∫∫ ∫−−+=
−drerVcdrrVccdrrrK
drrVcη Now
substituting the above information in equation (5.2.19) we get the conformal vector
fields [42]
933
752
11
11100 ,,)(
21, czcXccXcdrrXctcX +=+=+=+= ∫ θη (5.2.21)
The line element for non static cylindrically symmetric space-time in this case is
given as
,2),(2),(22)(2 dzededrdteds rtCrtBrA +++−= θ (5.2.22)
where
∫−= ,)(2)(ln)( 102 drrVcrUrA
∫∫ ∫−−+∫+=−−
,)()(2)(ln)(),()(
1452)(
13121010 drerVcdrrVcrUectcrtB
drrVcdrrVc
∫∫ ∫−−+∫+=−−
,)()(2)(ln)(),()(
1432)(
13121010 drerVcdrrVcrUectcrtC
drrVcdrrVc
,)(
1)(rU
rV = ,)(21)( 1∫ += cdrrrU η 53 cc ≠ and .01310121114 ≠−= ccccc Proper
conformal vector field after subtracting Killing vector fields is given as
( ).,),(, 351110 zccrUctcX θ+= (5.2.23)
We now consider another possibility for the solution of equation (5.2.20). Let in this
case ⇒= 0)(2 tFtt ,,,)( 76762 ℜ∈+= ccctctF 0),( =rtAtt and .0),( ≠rtArt
242
Substituting all the above information in equation (5.2.20) we get
⇒=⎟⎠⎞
⎜⎝⎛ ++ ∫ 0)(
21),(),( 16 cdrrrtAcrtA rtt η ),()(),( 1)(6 rEetgrtA
drrVc+∫=
− where
)(tg and )(1 rE are functions of integration and .)(
21
1)(1∫ +
=cdrr
rVη
Substituting
back the above information in (5.2.2) and solve after differentiating with respect to t
we get .,,)( 9898 ℜ∈+= ccctctg Substituting back the above value in (5.2.2) and
solving we get ( ) .)()(2)(ln)()(
106
2
11 6∫ ∫∫ ∫−−+=
−drerVcdrrVccdrrrE
drrVcη
Refreshing the value of ),( rtA we get
∫∫ ∫−−+∫+=−−
,)()(2)(ln)(),()(
1062)(
9866 drerVcdrrVcrUectcrtA
drrVcdrrVc where
,0968710 ≠−= ccccc )(
1)(rU
rV = and .)(21)( 1∫ += cdrrrU η Similarly solving
equations (5.2.9) and (5.2.11) we get
∫∫ ∫−−+∫+=−−
,)()(2)(ln)(),()(
1332)(
121166 drerVcdrrVcrUectcrtB
drrVcdrrVc
∫∫ ∫−−+∫+=−−
,)()(2)(ln)(),()(
1312)(
121166 drerVcdrrVcrUectcrtC
drrVcdrrVc
,012611713 ≠−= ccccc .,, 311636 cccccc ≠≠≠ The conformal vector field in this
case is given as
.,,)(21, 21
343
21
176
0 czcXccXcdrrXctcX +=+=+=+= ∫ θη (5.2.24)
The line element for non static cylindrically symmetric space-times becomes
,2),(2),(22),(2 dzededrdteds rtCrtBrtA +++−= θ (5.2.25)
where the metric functions are given above. Proper conformal vector field after
subtracting Killing vector fields from (5.2.24) is given as
( ).,),(, 1376 zccrUctcX θ+= (5.2.26)
243
Case I (a) (ii):
In this case we have 0)( =− rr BC and .0)(),(21 1 ≠⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη To solve
equations (5.2.2), (5.2.8) and (5.2.11) we must take .0)( =rη In this case no proper
conformal vector fields exists.
Case I (a) (iii):
In this case we have 0)( =− rr BC and .0)(),(21 1 =⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη Equation
⇒=− 0)( rr BC ),(),( rtBrtC rr = and equation ⇒=⎥⎦⎤
⎢⎣⎡ +∫ 0)(),(21 1 tFdrrt ttη
),(rηη = .,)( 111 ℜ∈= cctF The system of equations (5.2.17) becomes
).()(),()(
,)(21),(
653432
1120
θθθ
η
FzFXzFzFX
cdrrXtFX
+=+=
+== ∫ (5.2.27)
Now subtracting equation (5.2.11) from equation (5.2.9) and using the fact that in this
case ),(),( rtBrtC rr = and ),(),( rtBrtC tt = we get ⇒=− 0)()( 35 zFF θ
.,)()( 2235 ℜ∈== cczFF θ Substituting back the above values in equation (5.2.9)
and solving we get 546 )( ccF += θθ and .,,,)( 65464
4 ℜ∈+−= cccczczF
Substituting all the above information in equation (5.2.27), we get
5423
6422
1120
,
)(21),(
cczcXcczcX
cdrrXtFX
++=+−=
+== ∫θθ
η (5.2.28)
Considering equation (5.2.2) and using equation (5.2.28) then differentiating with
respect to ,t we get
.0)(2)(21)()( 2
122 =+⎥⎦
⎤⎢⎣⎡ +++ ∫ tFcdrrAtFAtFA ttrttttt η (5.2.29)
244
In order to solve this equation we consider different possibilities. First we take
⇒= 0),( rtAt ).(rAA = Substituting the above information in equation (5.2.29) we
get .,,)(0)( 878722 ℜ∈+=⇒= ccctctFtFtt Substituting back the above value in
(5.2.2) and solving we get ( ) ∫∫ −+= ,)(2)(ln)( 7
2
1 drrVccdrrrA η where
.)(
21
1)(1∫ +
=cdrr
rVη
Now substituting the above information (5.2.28) we get
5423
6422
11
870
,
)(21,
cczcXcczcX
cdrrXctcX
++=+−=
+=+= ∫θθ
η (5.2.30)
Considering equation (5.2.9) and using equation (5.2.30) then differentiating with
respect to ,t we get .0)(21)( 1787 =⎟
⎠⎞
⎜⎝⎛ ++++ ∫ cdrrBcBctcB rtttt η Substituting
0),( =rtBtt in the above equation we get ⇒∫=− drrvc
tt etgB)(7)(
),()(),( 1)(7 rKetgrtBdrrvc+∫=
− where )(1 rK is a function of integration. Now
differentiating the above value twice with respect to t and remember that 0),( =rtBtt
we get ⇒=∫=−
0)(),()(7 drrvc
tttt etgrtB .,,)( 109109 ℜ∈+= ccctctg Differentiating
the value of ),( rtB with respect to r and solving we get
( ) .)()(2)(ln)()(
112
2
11 7∫ ∫∫ ∫−−+=
−drerVcdrrVccdrrrK
drrVcη Finally the line
element for non static cylindrically symmetric space-times becomes
),( 22),(22)(2 dzdedrdteds rtBrA +++−= θ (5.2.31)
where
∫−= ,)(2)(ln)( 72 drrVcrUrA
∫∫ ∫−−+∫+=−−
,)()(2)(ln)(),()(
1122)(
10977 drerVcdrrVcrUectcrtB
drrVcdrrVc
245
,)(
1)(rU
rV = ,)(21)( 1∫ += cdrrrU η 72 cc ≠ and .7108911 ccccc −= Conformal
vector case in this case is given in equation (5.2.30). Proper conformal vector field
can be obtained by subtracting Killing vector fields from (5.2.30) as
( ).,),(, 2287 zccrUctcX θ+= (5.2.32)
Also for solving equation (5.2.29) we suppose that 0),( =rtAtt and ⇒= 0)(2 tFtt
.,,)( 87872 ℜ∈+= ccctctF Substituting back the above values in (5.2.29) and
solving we get ),()(),( 3)(7 rKetfrtAdrrvc+∫=
− where )(tf and )(3 rK are functions
of integration. Differentiating the above value twice with respect to ,t we get
⇒== 0)(),( tfrtA tttt .,,)( 13121312 ℜ∈+= ccctctf Similarly differentiating the
above value with respect to r and solving we get
∫∫ ∫−−+∫+=−−
,)()(2)(ln)(),()(
1472)(
131277 drerVcdrrVcrUectcrtA
drrVcdrrVc where
,)(
1)(rU
rV = ∫ += 1)(21)( cdrrrU η and .71381214 ccccc −= Solution of equations
(5.2.2) to (5.2.11) becomes [42]
.,
,)(21,
5423
6422
11
870
cczcXcczcX
cdrrXctcX
++=+−=
+=+= ∫θθ
η (5.2.33)
The line element for non static cylindrically symmetric space-time in this case
becomes non static plan symmetric space-times and is given as
),( 22),(22),(2 dzdedrdteds rtBrtA +++−= θ (5.2.34)
where
∫∫ ∫−−+∫+=−−
,)()(2)(ln)(),()(
1472)(
131277 drerVcdrrVcrUectcrtA
drrVcdrrVc
∫∫ ∫−−+∫+=−−
,)()(2)(ln)(),()(
1122)(
10977 drerVcdrrVcrUectcrtB
drrVcdrrVc
246
,)(
1)(rU
rV = ,)(21)( 1∫ += cdrrrU η 72 cc ≠ ,7108911 ccccc −= 71381214 ccccc −=
and .1411 cc ≠ Proper conformal vector fields for (5.2.34) can be obtained by
subtracting Killing vector fields form (5.2.33) as
( ).,),(, 2287 zccrUctcX θ+= (5.2.35)
Case I (b):
In this case we have 0),( =rtB tr and .0),(2 ≠θθ tE Substituting 0),( =rtB tr in
(5.2.9) with )(),( 11 tFtE =θ and solve after differentiating with respect to θ and ,z
we get ),,(),(),(),,( 8763 θθθ tEztEztEztK ++= where ),,(6 ztE ),(7 ztE and
),(8 θtE are functions of integration. Substituting back all the above information in
equation (5.2.14) we get
( )
).,(),(
),,(),(),(,)(),(21
,),()(),(21
433
876211
210
θθ
θθη
θη
tEtzEX
tEztEztEXtFdrrtX
tEdretFdrdrrteX Att
A
+=
++=+=
++=
∫
∫∫∫ −−
(5.2.36)
Considering equation (5.2.4) and using equation (5.2.36) then solve after
differentiating with respect to z and ,θ we get ),()(),( 326 zFtFztE += where
)(2 tF and )(3 zF are functions of integration. Substituting back the above value in
(5.2.4) and solving we get ),()(),( 547 zFtFztE += where )(4 tF and )(5 zF are
functions of integration. Substituting back all the above information in (5.2.4) and
solving after differentiating twice with respect to ,θ we get
)()()(),( 8768 θθθ FtFtFtE ++= and ),()()(21),( 1110922 tFtFtFtE ++= θθθ
where ),(6 tF ),(7 tF ),(8 θF ),(9 tF )(10 tF and )(11 tF are functions of integration.
Refreshing the system of equations (5.2.36) with the help of above information we
get
247
( )
( )).,(),(
),()()()()()()(
,)(),(21
,)()()(21)(),(
21
433
87654322
11
11109210
θθ
θθθ
η
θθη
tEtzEXFtFtFzFtFzFtFX
tFdrrtX
tFtFtFdretFdrdrrteX Att
A
+=
++++++=
+=
++++=
∫
∫∫∫ −−
(5.2.37)
Considering equation (5.2.10) and using equation (5.2.37) then solve after
differentiating twice with respect to z then with respect to ,θ we get
.,,,21)( 321321
23 ℜ∈++= cccczcczzF Substituting back the above value we get
.,,,21)( 654654
25 ℜ∈++= cccczcczzF Substituting back the above information in
(5.2.10) and solve after differentiating first with respect to z then with respect to ,θ
we get 01 =c and ),()()(21),( 14131223 tFtFtFtE ++= θθθ where ),(12 tF )(13 tF
and )(14 tF are functions of integration. Substituting back these values and solving
we get 04 =c and .0)()( 1312 == tFtF Substituting the above information in (5.2.10)
and solving we get ,052 == cc ),(),( 164 tFtE =θ ),(),( 143 tFtE =θ 65 )( czF = and
.)( 33 czF = Substituting all the above information in (5.2.37) we get
( )
( )).()(
),()()()()(
,)(),(21
,)()()(21)(),(
21
16143
8766
43
22
11
11109210
tFtzFX
FtFtFctFctFX
tFdrrtX
tFtFtFdretFdrdrrteX Att
A
+=
++++++=
+=
++++=
∫
∫∫∫ −−
θθθ
η
θθη
(5.2.38)
Considering equation (5.2.4) and using equation (5.2.38) then solve after
differentiating with respect to ,θ we get 0)(9 =tF and ).()( 62 tFtF tt −= Substituting
back the above values we get 0)(10 =tF and ).()( 74 tFtF tt −= Substituting all the
248
above values in the equation ⇒++= )()()(21),( 1110922 tFtFtFtE θθθ .0),(2 =θθ tE
Which is a contradiction to the fact that .0),(2 ≠θθ tE Hence this case is not possible.
Case I (c):
In this case we have 0),( =rtB tr and .0),(2 =θθ tE Equation
),(),(0),( 222 tFtEtE =⇒= θθθ where )(2 tF is a function of integration.
Substituting the above information in (5.2.9) and solve after differentiating with
respect to ,θ we get ),,(),(),,( 653 ztEztEztK += θθ where ),(5 ztE and ),(6 ztE
are functions of integration. Substituting all the above information in (5.2.14) we get
( )
).,(),(
),,(),(,)(),(21
,)()(),(21
433
65211
210
θθ
θη
η
tEtzEX
ztEztEXtFdrrtX
tFdretFdrdrrteX Att
A
+=
+=+=
++=
∫
∫∫∫ −−
(5.2.39)
Considering equation (5.2.4) and using equation (5.2.39) then solving we get
)(),( 35 zFztE = and ),(),( 46 zFztE = where )(3 zF and )(4 zF are functions of
integration. Also considering equation (5.2.5) and using equation (5.2.39) then
solving we get )(),( 53 θθ FtE = and ),(),( 64 θθ FtE = where )(5 θF and )(6 θF are
functions of integration. Substituting the above values in equation (5.2.39) we get
( )
).()(
),()(,)(),(21
,)()(),(21
653
43211
210
θθ
θη
η
FzFX
zFzFXtFdrrtX
tFdretFdrdrrteX Att
A
+=
+=+=
++=
∫
∫∫∫ −−
(5.2.40)
Considering equation (5.2.10) and using equation (5.2.40) then differentiating with
respect to ,r we get ( )( ) .0)()( 65 =+− θθ θθ FzFBC rr In this case we have
( ) ⇒≠− 0rr BC ( ) ⇒=+ 0)()( 65 θθ θθ FzF 15 )( cF =θ and .,,)( 212
6 ℜ∈= cccF θ
249
Substituting back the above information in (5.2.10) we get 33 )( czF = and
.,,)( 4344 ℜ∈= ccczF Refreshing the system of equations (5.2.40) we get
( ).,,)(),(
21
,)()(),(21
213
43211
210
czcXccXtFdrrtX
tFdretFdrdrrteX Att
A
+=+=+=
++=
∫
∫∫∫ −−
θη
η (5.2.41)
Now subtracting equation (5.2.9) from equation (5.2.11) and solving after
differentiating with respect to ,r we get .0),( =rtη Which means that in this case no
proper conformal vector field exist.
Case II:
In this case we have 0),,(2 ≠ztK z θ and we suppose .0),( =rtCt Substituting the
above value in equation (5.2.13) we get ⇒= 0),,(4 ztK zz θ
),,(),(),,( 324 θθθ tEtzEztK += where ),(2 θtE and ),(3 θtE are functions of
integration. Substituting the above information in equation (5.2.12) we get
( )
).,(),(),,,(),(
),,(),(21
),,,(),(),(21
3233),(12
11
2),(1),(0
θθθθ
θη
θθη
θ tEtzEXztKdretEX
tEdrrtX
ztKdretEdrdrrteX
rtB
rtAtt
rtA
+=+−=
+=
++=
∫∫
∫∫ ∫
−
−−
(5.2.42)
Considering equation (5.2.9) and using equation (5.2.42) then solve after
differentiating with respect to θ we get ),(),(0),( 111 tFtEtE =⇒= θθθ where )(1 tF
is a function of integration and .0),,(2),,(),( 32 =+ ztKztKrtBt θθ θθθ Differentiating
the above equation with respect to ,r we get .0),,(),( 2 =ztKrtBrt θθ We will discuss
here three different possibilities:
(a) ,0),( ≠rtBrt ,0),,(2 =ztK θθ (b) ,0),( =rtBrt ,0),,(2 ≠ztK θθ
(c) ,0),( =rtBrt .0),,(2 =ztK θθ
250
We will discuss each case in turn.
Case II (a):
In this case we have 0),( ≠rtBrt and .0),,(2 =ztK θθ Equation ⇒= 0),,(2 ztK θθ
),,(),,( 42 ztEztK =θ where ),(4 ztE is a function of integration. Substituting back
the above values in equation ,0),,(2),,(),( 32 =+ ztKztKrtBt θθ θθθ we get
⇒= 0),,(3 ztK θθθ ),,(),(),,( 653 ztEztEztK +=θθ where ),(5 ztE and ),(6 ztE are
functions of integration. Substituting all the above information in equation (5.2.42)
we get
( )
),,(),(),,(),(
,)(),(21
,),()(),(21
323652
11
410
θθθ
η
η
tEtzEXztEztEX
tFdrrtX
ztEdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫ ∫∫ −−
(5.2.43)
Considering equation (5.2.5) and using equation (5.2.43) then solve after
differentiating with respect to ,z we get )(),( 22 θθ FtE = and
),()(),( 434 tFtzFztE += where ),(2 θF )(3 tF and )(4 tF are functions of
integration. Substituting back the above values in (5.2.5) and solving, we get
0),(3 =θtEt and .0)(3 =tF Substituting back the above values in equation (5.2.5) we
get .0),,(2 =ztK z θ Which is a contradiction in this case. Hence this case is not
possible.
Case II (b):
In this case we have ,0),( =rtC tr ,0),,(2 ≠ztK z θ 0),( =rtBrt and .0),,(2 ≠ztK θθ
To solve the equation 0),,(2),,(),( 32 =+ ztKztKrtBt θθ θθθ we suppose 0),( =rtBt
which in turn implies ⇒= 0),,(3 ztK θθθ ),,(),(),,( 543 ztEztEztK += θθ where
251
),(4 ztE and ),(5 ztE are functions of integration. Refreshing the system of equations
(5.2.42) we get
( )
).,(),(),,(),(
,)(),(21
,),,()(),(21
323542
11
210
θθθ
η
θη
tEtzEXztEztEX
tFdrrtX
ztKdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫∫∫ −−
(5.2.44)
Considering equation (5.2.5) and using equation (5.2.44) then solve after
differentiating twice with respect to ,z we get
),,(),(),(21),,( 87622 θθθθ tEtEztEzztK ++= where ),,(6 θtE ),(7 θtE and
),(8 θtE are functions of integration. Substituting back the above value in (5.2.5) we
get 0),,(2 =ztK z θ which gives contradiction. Hence this case is also not possible.
Case II (c):
In this case we have ,0),( =rtC tr ,0),,(2 ≠ztK z θ 0),( =rtBrt and .0),,(2 =ztK θθ
Equation ⇒= 0),,(2 ztK θθ ),(),,( 42 ztEztK =θ where ),(4 ztE is a function of
integration. Substituting the above value in equation 0),,(2),,(),( 32 =+ ztKztKrtBt θθ θθθ
we get ),,(),(),,(0),,( 6533 ztEztEztKztK +=⇒= θθθθθ where ),(5 ztE and
),(6 ztE are functions of integration. Substituting all the above information in
equation (5.2.42) we get
( )
).,(),(),,(),(
,)(),(21
,),()(),(21
323652
11
410
θθθ
η
η
tEtzEXztEztEX
tFdrrtX
ztEdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫∫∫ −−
(5.2.45)
Considering equation (5.2.5) and using equation (5.2.45) then solve after
differentiating twice with respect to z we get
252
),()()(21),( 22212024 tFtFztFzztE ++= where ),(20 tF )(21 tF and )(22 tF are
functions of integration. Substituting back the above value in (5.2.5) and solving we
get ⇒== 0)()( 2120 tFtF ⇒= 0),(4 ztEz 0),,(2 =ztK z θ which gives contradiction.
Hence this case is also not possible.
Case III:
In this case we have 0),( =rtCrt and .0),,(2 =ztK z θ Equation ⇒= 0),,(2 ztK z θ
),,(),,( 22 θθ tEztK = where ),(2 θtE is a function of integration. Substituting back
the above value in (5.2.13) we get ),,(),(),,(0),,( 4344 θθθθ tEtzEztKztK zz +=⇒=
where ),(3 θtE and ),(4 θtE are functions of integration. Substituting all the above
information in (5.2.12) we get
( )
),,(),(,),,(),,(
,),(),(21
,),(),(),(21
433312
11
210
θθθθ
θη
θθη
θ tEtzEXztKdreztEX
tEdrrtX
tEdretEdrdrrteX
B
Att
A
+=+−=
+=
++=
∫∫
∫ ∫∫
−
−−
(5.2.46)
Considering equation (5.2.9) and using equation (5.2.46) then solve after
differentiating with respect to θ we get ⇒= 0),(1 θθ tE ),(),( 11 tFtE =θ where
)(1 tF is a function of integration. Substituting back the above value in equation
(5.2.9) and differentiating the resulting equation with respect to ,r we get
.0),(),( 2 =θθ tErtBrt To solve this equation we need to discuss three different
possibilities.
(a) ,0),( ≠rtBrt ,0),(2 =θθ tE (b) ,0),( =rtBrt ,0),(2 ≠θθ tE
(c) ,0),( =rtBrt .0),(2 =θθ tE
We will discuss each case in turn.
253
Case III (a):
In this case we have 0),( ≠rtBrt and .0),(2 =θθ tE Equation ⇒= 0),(2 θθ tE
),(),( 22 tFtE =θ where )(2 tF is a function of integration. Substituting back the
above value in (5.2.9) we get ),,(),(),,(0),,( 6533 ztEztEztKztK +=⇒= θθθθθ
where ),(5 ztE and ),(6 ztE are functions of integration. Substituting all the above
information in equation (5.2.46) we get
( )
),,(),(),,(),(
,)(),(21
,)()(),(21
433652
11
210
θθθ
η
η
tEtzEXztEztEX
tFdrrtX
tFdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫ ∫∫ −−
(5.2.47)
Considering equation (5.2.4) and using equation (5.2.47) then differentiating with
respect to ,θ we get ⇒= 0),(5 ztEt ),(),( 35 zFztE = where )(3 zF is a function of
integration. Substituting back the above value in (5.2.4) we get
),(),(0),( 466 zFztEztEt =⇒= where )(4 zF is a function of integration.
Substituting back all the above information in (5.2.47) we get
( )
).,(),(),()(
,)(),(21
,)()(),(21
433432
11
210
θθθ
η
η
tEtzEXzFzFX
tFdrrtX
tFdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫ ∫∫ −−
(5.2.48)
Considering equation (5.2.5) and using equation (5.2.48) then solve after
differentiating with respect to z we get ).(),(0),( 533 θθθ FtEtEt =⇒= Substituting
back the above value in (5.2.5) we get ),(),(0),( 644 θθθ FtEtEt =⇒= where
)(5 θF and )(6 θF are functions of integration. Substituting back all the above
information in equation (5.2.48) we get
254
( )
).()(),()(
),(),(21
,)()(),(21
653432
11
210
θθθ
η
η
FzFXzFzFX
tFdrrtX
tFdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫ ∫∫ −−
(5.2.49)
Considering equation (5.2.10) and using equation (5.2.49) we get
( ) ( ) .0)()()()( 4365 =+++− zFzFFFze zzBC θθθ θθ Differentiating the above equation
first with respect to r then with respect to ,t we get
( )( ) .0)()(),(),( 65 =+− θθ θθ FFzrtBrtC rtrt In this case we know that 0),( =rtCrt and
0),( ≠rtBrt which in turn gives ( ) ⇒=+ 0)()( 65 θθ θθ FFz 15 )( cF =θ and
.,,)( 2126 ℜ∈= cccF θ Substituting back these values in the above equation we get
33 )( czF = and .,,)( 434
4 ℜ∈= ccczF Refreshing our system of equations, we get
( ).,),(),(
21
,)()(),(21
213
43211
210
czcXccXtFdrrtX
tFdretFdrdrrteX Att
A
+=+=+=
++=
∫
∫ ∫∫ −−
θη
η (5.2.50)
Now subtracting equation (5.2.9) from equation (5.2.11) and solving after
differentiating with respect to ,r we get .0),( =rtη Which means that in this case no
proper conformal vector field exist.
Case III (b):
In this case we have 0),( =rtB tr and .0),(2 ≠θθ tE Substituting the above
information in equation (5.2.9) with 0),( =rtB t and solving we get
⇒= 0),,(3 ztK θθθ ),,(),(),,( 653 ztEztEztK += θθ where ),(5 ztE and ),(6 ztE are
functions of integration. Refreshing the system of equations (5.2.46) we get
255
( )
).,(),(),,(),(
,)(),(21
,),()(),(21
433652
11
210
θθθ
η
θη
tEtzEXztEztEX
tFdrrtX
tEdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫∫∫ −−
(5.2.51)
Considering equation (5.2.4) and using equation (5.2.51) then solve after
differentiating with respect to θ we get )(),( 25 zFztE = and
),()(),( 432 tFtFtE += θθ where ),(2 zF )(3 tF and )(4 tF are functions of
integration. Substituting back the above value in equation (5.2.4) and solve after
differentiating the resulting equation with respect to ,z we get
),()(),( 656 zFtFztE += where )(5 tF and )(6 zF are functions of integration.
Substituting back all the above information in (5.2.4) and solving we get
⇒= 0)(3 tF .0),(2 =θθ tE Which gives contradiction and hence this case is not
possible.
Case III (c):
In this case we have 0),( =rtB tr and .0),(2 =θθ tE Equation
),(),(0),( 222 tFtEtE =⇒= θθθ where )(2 tF is a function of integration.
Substituting the above information in (5.2.9) and solve after differentiating with
respect to ,θ we get ⇒= 0),,(3 ztK θθθ ),,(),(),,( 653 ztEztEztK += θθ where
),(5 ztE and ),(6 ztE are functions of integration. Substituting all the above
information in (5.2.46) we get
( )
).,(),(),,(),(
,)(),(21
,)()(),(21
433652
11
210
θθθ
η
η
tEtzEXztEztEX
tFdrrtX
tFdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫∫∫ −−
(5.2.52)
256
Considering equation (5.2.4) and using equation (5.2.52) then solve after
differentiating with respect to ,θ we get )(),(0),( 355 zFztEztEt =⇒= and
⇒= 0),(6 ztEt ),(),( 46 zFztE = where )(3 zF and )(4 zF are functions of
integration. Also considering equation (5.2.5) and using equation (5.2.52) then solve
after differentiating with respect to ,z we get )(),(0),( 533 θθθ FtEtEt =⇒= and
⇒= 0),(4 θtEt ),(),( 64 θθ FtE = where )(5 θF and )(6 θF are functions of
integration. Substituting all the above information in equation (5.2.52) we get
( )
).()(),()(
,)(),(21
,)()(),(21
653432
11
210
θθθ
η
η
FzFXzFzFX
tFdrrtX
tFdretFdrdrrteX Att
A
+=+=
+=
++=
∫
∫∫∫ −−
(5.2.53)
Considering equation (5.2.10) and using equation (5.2.53) then solve after
differentiating with respect to z and ,θ we get
ℜ∈+=⇒= 212155 ,,)(0)( ccccFF θθθθθ and .,,)(0)( 4343
33 ℜ∈+=⇒= ccczcFF θθθθ
Substituting back the above information in (5.2.10) and solving we get
ℜ∈+=⇒= 656544 ,,)(0)( ccczczFzFzz and ℜ∈+=⇒= 8787
66 ,,)(0)( ccccFF θθθθθ
which on back substitution give us .07531 ==== cccc Substituting all the above
information in equation (5.2.52) we get
( ).,,)(),(
21
,)()(),(21
823
64211
210
czcXccXtFdrrtX
tFdretFdrdrrteX Att
A
+=+=+=
++=
∫
∫∫∫ −−
θη
η (5.2.54)
Subtracting equation (5.2.9) from equation (5.2.11) we get
.0)(2)()( 4210 =−+−+− ccXBCXBC rrtt In order to solve this equation we
suppose ,0)( =− tt BC which in turn gives ( ).2)(),(21)( 24
1 cctFdrrtBC rr −=⎥⎦⎤
⎢⎣⎡ +− ∫η
Differentiating the above equation with respect to ,t we get
257
.0)(),(21)( 1 =⎥⎦
⎤⎢⎣⎡ +− ∫ tFdrrtBC ttrr η Solution of this equation involves three
different possibilities:
(l) ,0)( ≠− rr BC ,0)(),(21 1 =⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη
(m) ,0)( =− rr BC ,0)(),(21 1 ≠⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη
(n) ,0)( =− rr BC .0)(),(21 1 =⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη
We will discuss each case in turn.
Case III (c) (l):
In this case we have 0)( ≠− rr BC and .0)(),(21 1 =⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη Equation
⇒=⎥⎦⎤
⎢⎣⎡ +∫ 0)(),(21 1 tFdrrt ttη )(rηη = and .,)( 99
1 ℜ∈= cctF Refreshing the
system of equations (5.2.54) we get
.,,),(21),( 82
364
29
120 czcXccXcdrrtXtFX +=+=+== ∫ θη (5.2.55)
Considering equation (5.2.2) and using equation (5.2.55) then differentiating with
respect to t we get
.0)(2)(21),()(),()(),( 2
922 =+⎟
⎠⎞
⎜⎝⎛ +++ ∫ tFcdrrrtAtFrtAtFrtA ttrttttt η (5.2.56)
In order to solve (5.2.56) we suppose .0),( =rtAt Hence equation (5.2.56) gives
⇒= 0)(2 tFtt .,,)( 111011102 ℜ∈+= ccctctF Substituting back all the above
information in equation (5.2.2) we get ⇒=+⎥⎦⎤
⎢⎣⎡ +∫ )(2)(21
109 rccdrrAr ηη
258
( ) .)(
21
2)(ln)(1
10
2
9 ∫∫
∫+
−+=cdrr
drccdrrrAη
η Now differentiating equation (5.2.9)
with respect to t and using the fact that ,0),( =rtBrt we get
⇒=++ 0),()(),( 101110 rtBcctcrtB ttt ( ) ,)(),( 11110
1 −+= ctcrKrtBt where )(1 rK is
the function of integration. Differentiating the above equation with respect to ,r we
get .,)(0)(),( 121211 ℜ∈=⇒== ccrKrKrtB rrt Substituting this value in the above
equation and integrating with respect to ,t we get ( ) ),(ln),( 10
12
1110 rfctcrtB cc
++=
where )(rf is the function of integration. In this case 0)( =− tt BC therefore
( ) ),(ln),( 10
12
1110 rgctcrtC cc++= where )(rg is a function of integration. Substituting
back the above values in equation (5.2.9) then solving we get
( ) ,)()(ln)( 13
2
9 ∫∫ −+= drrVccdrrrf η ( ) ,)()(ln)( 14
2
9 ∫∫ −+= drrVccdrrrg η where
,)(
21
1)(9∫ +
=cdrr
rVη
41213 2ccc += and .2 12214 ccc += Finally, substituting all
the above information in equation (5.2.55) the conformal vector fields become [42]
.,,)(21, 82
364
29
11110
0 czcXccXcdrrXctcX +=+=+=+= ∫ θη (5.2.57)
The line element for non static cylindrically symmetric space-time in this case is
given as
,2),(2),(22)(2 dzededrdteds rtCrtBrA +++−= θ (5.2.58)
where
( ) ,)(
21
2)(ln)(1
10
2
9 ∫∫
∫+
−+=cdrr
drccdrrrAη
η
( ) ( ) ,)()(lnln),( 13
2
91110 10
12
∫∫ −+++= drrVccdrrctcrtB cc
η
259
( ) ( ) ,)()(lnln),( 14
2
91110 10
12
∫∫ −+++= drrVccdrrctcrtC cc
η ,)(
21
1)(9∫ +
=cdrr
rVη
41213 2ccc += and .2 12214 ccc += Subtracting Killing vector fields from (5.2.57) the
proper conformal vector fields for the above space-times (5.2.58) takes the form
( ).,),(, 241110 zccrUctcX θ+= (5.2.59)
Another possibility to solve the equation 0),()(),( 101110 =++ rtBcctcrtB ttt is to take
.0),( =rtBt Equation ).(0),( rBBrtBt =⇒= Substituting the above value in
equation (5.2.9) and solving we get ∫−= ,)(2)(ln)( 42 drrVcrUrB where
)(1)(rU
rV = and .)(21)( 9∫ += cdrrrU η Similarly considering equation (5.2.11) and
using equation (5.2.57) we get
).(2)(21),()(),( 291110 rccdrrrtCctcrtC rt ηη =+⎟
⎠⎞
⎜⎝⎛ +++ ∫ Solving this equation by
considering ,0),( =rtCt we get ∫−= ,)(2)(ln)( 22 drrVcrUrC where
)(1)(rU
rV =
and .)(21)( 9∫ += cdrrrU η In this case the space-time becomes static cylindrically
symmetric space-time and is given as
,2)(2)(22)(2 dzededrdteds rCrBrA +++−= θ (5.2.60)
where
∫−= ,)(2)(ln)( 102 drrVcrUrA ,)(2)(ln)( 4
2 ∫−= drrVcrUrB ,)(2)(ln)( 22 ∫−= drrVcrUrC
,)(
1)(rU
rV = ,)(21)( 9∫ += cdrrrU η ,010 ≠c ,02 ≠c ,04 ≠c ,410 cc ≠ 21 cc ≠ and
.42 cc ≠ Conformal vector fields for the above space-time are given in equation
(5.2.57). Proper conformal vector fields for the above space-times (5.2.60) takes the
form
260
( ).,),(, 241110 zccrUctcX θ+= (5.2.61)
Case III (c) (m):
In this case we have 0)( =− rr BC and .0)(),(21 1 ≠⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη To solve
equations (5.2.2), (5.2.8) and (5.2.11) we need to take .0)( =rη Which implies that in
this case no proper conformal vector fields exist.
Case III (c) (n):
In this case we have 0)( =− rr BC and .0)(),(21 1 =⎥⎦
⎤⎢⎣⎡ +∫ tFdrrt ttη Equation
⇒=− 0)( rr BC ),(),( rtBrtC rr = and equation ⇒=⎥⎦⎤
⎢⎣⎡ +∫ 0)(),(21 1 tFdrrt ttη
)(rηη = and .,)( 111 ℜ∈= cctF Substituting the above information in equation
(5.2.54) we get
).()(),()(
,)(21),(
653432
1120
θθθ
η
FzFXzFzFX
cdrrXtFX
+=+=
+== ∫ (5.2.62)
Now subtracting equation (5.2.11) from equation (5.2.9) and using the fact that in this
case ),(),( rtBrtC rr = and ),(),( rtBrtC tt = we get ⇒=− 0)()( 35 zFF θ
.,)()( 2235 ℜ∈== cczFF θ Substituting back the above values in equation (5.2.9)
and solving we get ℜ∈+=⇒= 545466 ,,)(0)( ccccFF θθθθθ and
.,,)(0)( 646444 ℜ∈+−=⇒= ccczczFzFzz Substituting all the above information in
equation (5.2.62) we get
5423
6422
1120
,
)(21),(
cczcXcczcX
cdrrXtFX
++=+−=
+== ∫θθ
η (5.2.63)
261
Considering equation (5.2.2) and using equation (5.2.63) then differentiating with
respect to ,t we get
.0)(2)(21)()( 2
122 =+⎥⎦
⎤⎢⎣⎡ +++ ∫ tFcdrrAtFAtFA ttrttttt η (5.2.64)
In order to solve this equation we suppose ⇒= 0),( rtAt ).(rAA = Substituting the
above value in equation (5.2.64) we get .,,)(0)( 878722 ℜ∈+=⇒= ccctctFtFtt
Substituting back the above value in equation (5.2.64) and solving we get
( ) ∫∫ −+= ,)(2)(ln)( 7
2
1 drrVccdrrrA η where .)(
21
1)(1∫ +
=cdrr
rVη
Now
substituting the above information in equation (5.2.63), we get
5423
6422
11
870
,
)(21,
cczcXcczcX
cdrrXctcX
++=+−=
+=+= ∫θθ
η (5.2.65)
Considering equation (5.2.9) and using equation (5.2.65) then differentiating with
respect to ,t we get ⇒=++ 0)( 787 cBctcB ttt ( ) ,)(),( 187
1 −+= ctcrKrtBt where
)(1 rK is a function of integration. Differentiating the above equation with respect to
,r we get ℜ∈=⇒== 9911 ,)(0)(),( ccrKrKrtB rrt therefore
( ) ),(ln),( 21110 7
9
rKctcrtB cc++= where )(2 rK is a function of integration. Substituting
back the above value in equation (5.2.9) and solving we get
( ) ∫∫ −+= ,)()(ln)( 10
2
12 drrVccdrrrK η where
∫ +=
1)(21
1)(cdrr
rVη
and
.2 2910 ccc += The line element for non static cylindrically symmetric space-time in
this case after a suitable rescaling of z is given as
),( 22),(22)(2 dzdedrdteds rtBrA +++−= θ (5.2.66)
where
262
∫−= ,)(2)(ln)( 72 drrVcrUrA ,)()(ln)ln(),( 10
287
7
9
∫−++= drrVcrUctcrtB cc
,)(
1)(rU
rV = ∫ += 1)(21)( cdrrrU η and .2 2910 ccc += In this case the proper
conformal vector fields can be obtained by subtracting Killing vector fields from
(5.2.65) as
( ).,),(, 2287 zccrUctcX θ+= (5.2.67)
Another possible solution for the equation 0)( 787 =++ cBctcB ttt is to take
0),( =rtBt which in turn gives ,)(2)(ln)( 22 ∫−= drrVcrUrB where
.)(
21
1)(1∫ +
=cdrr
rVη
Conformal vector fields are given as
5423
6422
11
870
,
)(21,
cczcXcczcX
cdrrXctcX
++=+−=
+=+= ∫θθ
η (5.2.68)
In this case the space-time becomes static plane symmetric space-time and is given as
),( 22)(22)(2 dzdedrdteds rBrA +++−= θ (5.2.69)
where
∫−= ,)(2)(ln)( 72 drrVcrUrA ,)(2)(ln)( 2
2 ∫−= drrVcrUrB ,)(
1)(rU
rV =
.)(21)( 1∫ += cdrrrU η Subtracting Killing vector fields from (5.2.68) we get the
proper conformal vector field as
( ).,),(, 227 zccrUtcX θ= (5.2.70)
263
5.3. Proper Conformal Vector Fields in Non
Conformally Flat Kantowski-Sachs and
Bianchi Type III Space-Times
Consider Kantowski-Sachs and Bianchi type III space-times in usual coordinates
),,,( φθrt (labeled by ),,,,( 3210 xxxx respectively) with the line element [77]
),)(()()( 22222222 φθθ dfdtBdrtAdtds +++−= (5.3.1)
where A and B are no where zero functions of t only. For θθ sinh)( =f the space-
time (5.3.1) represents Bianchi type III space-time and for θθ sin)( =f it becomes
Kantowski-Sachs space-times. The above space-times (5.3.1) admit at least four
independent Killing vector fields [79], which are
,r∂∂ ,
φ∂∂ ,sin
)()(cos
φφ
θθ
θφ
∂∂′
−∂∂
ff ,cos
)()(sin
φφ
θθ
θφ
∂∂′
+∂∂
ff (5.3.2)
where prime denotes the derivative with respect to .θ A vector field X is said to be a
conformal vector field if it satisfies equation (1.2.4). One can write (1.2.4) explicitly
using (5.3.1) as
,0,0 η=X (5.3.3)
,01,0
0,12 =− XXA (5.3.4)
,02,0
0,22 =− XXB (5.3.5)
,0)( 3,0
0,322 =− XXfB θ (5.3.6)
,1,10 η=+ XX
AAt (5.3.7)
,02,12
1,22 =+ XAXB (5.3.8)
264
,0)( 1,322
3,12 =+ XfBXA θ (5.3.9)
,2,20 η=+ XX
BBt (5.3.10)
,0)( 2,32
3,2 =+ XfX θ (5.3.11)
.)()(
3,320 η
θθ
=+′
+ XXffX
BBt (5.3.12)
Here it follows that [40] ).(tηη = Integrating (5.3.3), (5.3.4), (5.3.5) and (5.3.6) we
get
).,,(1),,()(
1
),,,(1),,(
),,,(1),,(),,,()(
42
12
3
32
12
22
1110
φθφθθ
φθφθ
φθφθφθη
φ
θ
rKdtB
rKf
X
rKdtB
rKX
rKdtA
rKXrKdttX r
+=
+=
+=+=
∫
∫
∫∫
(5.3.13)
where ),,,(1 φθrK ),,,(2 φθrK ),,(3 φθrK and ),,(4 φθrK are functions of
integration which are to be determined. Considering equation (5.3.11) and using
equation (5.3.13) then solve after differentiating with respect to t we get
),,(),()(),,( 211 θφφθφθ rFdrFfrK += ∫ where ),(1 φrF and ),(2 θrF are functions
of integration. Substituting the above value in equation (5.3.13) we get
( )( )
).,,(1),()(
1
),,,(1),(),()(
),,,(1),(),()(
),,(),()()(
42
13
32
212
22
211
210
φθφθ
φθθφφθ
φθθφφθ
θφφθη
θ
rKdtB
rFf
X
rKdtB
rFdrFfX
rKdtA
rFdrFfX
rFdrFfdttX
rr
+=
++′=
++=
++=
∫
∫∫
∫∫
∫∫
(5.3.14)
265
Considering equation (5.3.10) and using equation (5.3.14) then differentiating first
with respect to φ then with respect to ,t we get .0),(1 122
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
−φrF
BBBBB ttt In
order to solve this equation we will discuss three different possibilities:
(I) ,0122
2
≠⎟⎟⎠
⎞⎜⎜⎝
⎛+
−BB
BBB ttt ,0),(1 =φrF (II) ,0122
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
−BB
BBB ttt ,0),(1 ≠φrF
(III) ,0122
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
−BB
BBB ttt .0),(1 =φrF
We will discuss each possibility in turn.
Case I:
In this case we have 0122
2
≠⎟⎟⎠
⎞⎜⎜⎝
⎛+
−BB
BBB ttt and .0),(1 =φrF Substituting back the
above value in (5.3.10) and solve after differentiating with respect to ,φ we get
),,(),(),,(0),,( 4333 φθφθφθθφ rFrFrKrK +=⇒= where ),(3 θrF and ),(4 φrF are
functions of integration. Substituting all the above information in (5.3.14) we get
).,,(),,(),(1),(
),,,(1),(),,()(
43432
22
22
2120
φθφθθ
φθθθη
θ rKXrFrFdtB
rFX
rKdtA
rFXrFdttX r
=++=
+=+=
∫
∫∫ (5.3.15)
Considering equation (5.3.7) and using equation (5.3.15) then solve after
differentiating with respect to ,φ we get
),,(),(),,(0),,( 6522 φθθφθφθφ FrFrKrKr +=⇒= where ),(5 θrF and ),(6 φθF
are functions of integration. Considering equation (5.3.11) and using equation
(5.3.15) then solve after differentiating with respect to θ and we get
),,(),()()(),,( 874 φφ
θθφθ rFrF
ffrK +′
−= where ),(7 φrF and ),(8 φrF are functions
of integration. Now substituting all the above information in equation (5.3.15) we get
266
).,(),()()(),,(),(1),(
),,(),(1),(),,()(
873432
22
652
2120
φφθθφθθ
φθθθθη
θ rFrFffXrFrFdt
BrFX
FrFdtA
rFXrFdttX r
+′
−=++=
++=+=
∫
∫∫ (5.3.16)
Considering equation (5.3.10) and using equation (5.3.16) then differentiating first
with respect to r then twice with respect to ,t we get ,0),(22 =⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛•••
θrFBBB r
where dot denotes differentiation with respect to .t In this case ,02 ≠⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛•••
BBB
therefore ),(),(0),( 122 θθθ DrFrFr =⇒= where )(1 θD is the function of
integration. Substituting back the above value in (5.3.10) and solve after
differentiating with respect to ,r we get ),()(),( 323 θθ DrDrF += where )(2 rD
and )(3 θD are functions of integration. Substituting all the above information in
(5.3.16), we get
).,(),()()(
),,()()(1)(
),,(),(),()(
873
4322
12
65110
φφθθ
φθθ
φθθθη
θ
rFrFffX
rFDrDdtB
DX
FrFXDdttX
+′
−=
+++=
+=+=
∫
∫ (5.3.17)
Considering equation (5.3.7) and using equation (5.3.17) then solve after
differentiating with respect to r we get ),()(),( 545 θθθ DrDrF += where )(4 θD
and )(5 θD are functions of integration. Substituting all the above information in
equation (5.3.17) we get
267
).,(),()()(
),,()()(1)(
),,()()(),()(
873
4322
12
654110
φφθθ
φθθ
φθθθθη
θ
rFrFffX
rFDrDdtB
DX
FDrDXDdttX
+′
−=
+++=
++=+=
∫
∫ (5.3.18)
Now subtracting equation (5.3.10) from equation (5.3.12) and using equation (5.3.18)
in the resulting equation then solve after differentiating with respect to ,t we get
,)()( 211 cfcD +′= θθ ., 21 ℜ∈cc Refreshing the above equation (5.3.18) we get
).,(),()()(
),,()()(1)(
),,()()(,)()(
873
43221
2
654121
0
φφθθ
φθθ
φθθθθη
rFrFffX
rFDrDdtB
fcX
FDrDXcfcdttX
+′
−=
+++=
++=+′+=
∫
∫ (5.3.19)
Considering equation (5.3.8) and using equation (5.3.19) then differentiating the
resulting equation with respect to ,t we get ( ) .0),()()( 6542
2
=++⎟⎟⎠
⎞⎜⎜⎝
⎛•
φθθθ θθθ FDrDBA
The solution of this equation can be obtained by considering three different
possibilities:
(a) ,02
2
≠⎟⎟⎠
⎞⎜⎜⎝
⎛•
BA ( ) ,0),()()( 654 =++ φθθθ θθθ FDrD
(b) ,02
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛•
BA ( ) ,0),()()( 654 ≠++ φθθθ θθθ FDrD
(c) ,02
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛•
BA ( ) .0),()()( 654 =++ φθθθ θθθ FDrD
We will discuss each case in turn.
268
Case I (a):
In this case we have 02
2
≠⎟⎟⎠
⎞⎜⎜⎝
⎛•
BA and ( ) .0),()()( 654 =++ φθθθ θθθ FDrD Differentiating
the above equation with respect to ,r we get .,)(0)( 3344 ℜ∈=⇒= ccDD θθθ
Substituting back this value in the above equation we get
),()(),( 656 φθφθ DDF +−= where )(6 φD is a function of integration. Also
substituting all the above information in equation (5.3.11) we get
).,(),( 47 φφ φ rFrF −= Refreshing the system of equations (5.3.19) we get
).,(),()()(
),,()()(1)(
),(,)()(
843
43221
2
63
121
0
φφθθ
φθθ
φθη
φ rFrFffX
rFDrDdtB
fcX
DrcXcfcdttX
+′
=
+++=
+=+′+=
∫
∫ (5.3.20)
Considering equation (5.3.10) and using equation (5.3.20) then solve after
differentiating first with respect to θ then with respect to t twice we get
,)( 543 ccD += θθ ℜ∈54 ,cc and .01 =c Substituting all the above values in
(5.3.20) we get
).,(),()()(),,()(
),(,)(
843454
22
63
12
0
φφθθφθ
φη
φ rFrFffXrFccrDX
DrcXcdttX
+′
=+++=
+=+= ∫ (5.3.21)
Considering equation (5.3.8) and using equation (5.3.21) then solving we get
),()(),( 724 φφ DrDrF +−= where )(7 φD is the function of integration. Now
considering equation (5.3.12) and using equation (5.3.21) then solve after
differentiating the resulting equation first with respect to φ then with respect to ,θ
we get ,sincos)( 8767 φφφ cccD ++= .,, 876 ℜ∈ccc Substituting back the above
values in (5.3.8) and solving we get ),()(),( 988 rDrDrF += φφ where )(8 rD and
269
)(9 rD are functions of integration. Now substituting all the above information in
(5.3.21), we get
( ) ).()(sincos)()(
,sincos
),(,)(
9878
3
98742
63
12
0
rDrDccffX
ccccX
DrcXcdttX
++−′
=
+++=
+=+= ∫
φφφθθ
φφθ
φη
(5.3.22)
Considering equation (5.3.9) and using equation (5.3.22) then solve after
differentiating with respect to θ and φ we get ,)( 108 crD = 11
9 )( crD = and
,)( 126 cD =φ .,, 121110 ℜ∈ccc Substituting all the above information in equation
(5.3.22) we get
( ) .sincos)()(
,sincos
,,)(
1110783
98742
1231
20
ccccffX
ccccX
crcXcdttX
++−′
=
+++=
+=+= ∫
φφφθθ
φφθ
η
(5.3.23)
Subtracting equation (5.3.10) from equation (5.3.12) and using equation (5.3.23) in
the resulting equation then differentiating twice with respect to ,θ we get
.01094 === ccc Substituting all the above information in equation (5.3.23) we get
( ) .sincos)()(,sincos
,,)(
11783
872
1231
20
cccffXccX
crcXcdttX
+−′
=+=
+=+= ∫φφ
θθφφ
η (5.3.23)*
Considering equations (5.3.10) and (5.3.7) respectively and using equation (5.3.23)*
we get )()( tUtB = and ,)()( )(3 dttU
c
etUtA ∫=
−
where ( ).)()( 2cdtttU += ∫η Conformal
vector fields are given in equation (5.3.23)*. The line element for Kantowski-Sachs
and Bianchi type III space-times in this case is given in equation (5.3.1) with
270
dttU
c
etUtA ∫=
−)(
3
)()( and ),()( tUtB = where ( ).)()( 2cdtttU += ∫η Proper conformal
vector fields after subtracting Killing vector fields from (5.3.23)* is given as
( ).0,0,),( 3rctUX = (5.3.24)
Case I (b):
In this case we have ( ) 0),()()( 654 ≠++ φθθθ θθθ FDrD and .02
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛•
BA Equation
⇒=⎟⎟⎠
⎞⎜⎜⎝
⎛•
02
2
BA ),()( 22 tBtA α= where .0\ℜ∈α Substituting the above value in
(5.3.8) and solve after differentiating with respect to θ then with respect to ,r we get
434 )( ccD += θθ ., 43 ℜ∈cc Substituting back the above values in (5.3.8) and
solving we get )()()(),( 8766 θφφθφθ DDDF ++= and
),()()(),( 111094 rDDDrrF ++= φφφ where ),(6 φD ),(7 φD ),(8 θD ),(9 φD
)(10 φD and )(11 rD are all functions of integration. Refreshing the system of equations
(5.3.19) we get
( )
).,(),()()(
),()()()(1)(
),()()(
,)()(
873
10931221
2
761343
1
210
φφθθ
φφθθ
φφθθθ
θη
rFrFffX
DDrDrDdtB
fcX
DDDccrX
cfcdttX
+′
−=
++++=
++++=
+′+=
∫
∫
(5.3.25)
Considering equation (5.3.11) and using equation (5.3.25) then solving we get
),()(),( 1497 φφφ φ DDrrF +−= where )(14 φD is a function of integration. Also
substituting the above value and equation (5.3.25) in equation (5.3.10) and solve after
differentiating with respect to θ and then twice with respect to t we get .01 =c
271
Substituting back the above value we get .,,)( 65653 ℜ∈+= ccccD θθ Substituting all
the above information in equation (5.3.25) we get
( )
( ) ).,()()()()(
),()()(
),()()(
,)(
81493
10965
122
761343
1
20
φφφθθ
φφθ
φφθθθ
η
φ rFDDrffX
DDrccrDX
DDDccrX
cdttX
++−′
−=
++++=
++++=
+= ∫
(5.3.26)
Considering equation (5.3.11) and using equation (5.3.26) then solve after
differentiating with respect to ,φ we get .,)()( 771014 ℜ∈+−= ccDD φφ Also
considering equation (5.3.12) and using equation (5.3.26) then solve after
differentiating with respect to ,φ r and θ we get
.,,,sincos)( 109810989 ℜ∈++= ccccccD φφφ Substituting back the above value in
(5.3.12) and solving we get ( ) .,,sinhcosh)( 1211121110 ℜ∈−+= ccccD φφφ
Substituting all the above information in equation (5.3.26) we get
( )( )
( ) ( )[ ] ),,(sinhcoshsincos)()(
,sinhcosh)sincos()(
),()()(
,)(
81312910
3
111091285122
761343
1
20
φφφφφθθ
φφφφθ
φφθθθ
η
rFccccrffX
cccrcrccrDX
DDDccrX
cdttX
++−−−−′
−=
−++++++=
++++=
+= ∫
(5.3.27)
where .11713 ccc −= Considering equation (5.3.9) and using equation (5.3.27) then
solve after differentiating twice with respect to θ we get .0109 == cc Substituting
back the above value in (5.3.27) and solving the remaining equations we reach to the
contradiction that ( ) .0),()()( 654 =++ φθθθ θθθ FDrD Hence this case is not possible.
272
Case I (c):
In this case we have ( ) 0),()()( 654 =++ φθθθ θθθ FDrD and .02
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛•
BA Equation
⇒=⎟⎟⎠
⎞⎜⎜⎝
⎛•
02
2
BA ),()( 22 tBtA α= where .0\ℜ∈α Differentiating
( ) 0),()()( 654 =++ φθθθ θθθ FDrD with respect to ,r we get
.,)(0)( 3344 ℜ∈=⇒= ccDD θθθ Substituting back this value in the above equation
we get ),()(),( 656 φθφθ DDF +−= where )(6 φD is a function of integration. Also
considering equation (5.3.11) and using equation (5.3.19) we get
).,(),( 47 φφ φ rFrF −= Now substituting all the above values in equation (5.3.19) we
get
).,(),()()(
),,()()(1)(
),(,)()(
843
43221
2
63
121
0
φφθθ
φθθ
φθη
φ rFrFffX
rFDrDdtB
fcX
DrcXcfcdttX
+′
=
+++=
+=+′+=
∫
∫ (5.3.28)
Considering equation (5.3.10) and using equation (5.3.28) then solve after
differentiating with respect to θ and with respect to t twice we obtain
,)( 543 ccD += θθ ℜ∈54 ,cc and .01 =c Substituting all the above information in
(5.3.28) we get
).,(),()()(),,()(
),(,)(
843454
22
63
12
0
φφθθφθ
φη
φ rFrFffXrFccrDX
DrcXcdttX
+′
=+++=
+=+= ∫ (5.3.29)
Considering equation (5.3.8) and using equation (5.3.29) then solving we get
),()(),( 724 φφ DrDrF +−= where )(7 φD is the function of integration. Now
considering equation (5.3.12) and using equation (5.3.29) then solving after
273
differentiating the resulting equation first with respect to φ then with respect to ,θ
we get ,sincos)( 8767 φφφ cccD ++= .,, 876 ℜ∈ccc Substituting back the above
value in equation (5.3.8) we get ),()(),( 988 rDrDrF += φφ where )(8 rD and
)(9 rD are functions of integration. Now substituting all the above values in (5.3.29),
we get
( ) ).()(sincos)()(
,sincos
),(,)(
9878
3
98742
63
12
0
rDrDccffX
ccccX
DrcXcdttX
++−′
=
+++=
+=+= ∫
φφφθθ
φφθ
φη
(5.3.30)
Considering equation (5.3.9) and using equation (5.3.30) then solve after
differentiating with respect to θ and φ we get ,)( 108 crD = 11
9 )( crD = and
,)( 126 cD =φ .,, 121110 ℜ∈ccc Substituting all the above information in (5.3.30) we
get
( ) .sincos)()(
,sincos
,,)(
1110783
98742
1231
20
ccccffX
ccccX
crcXcdttX
++−′
=
+++=
+=+= ∫
φφφθθ
φφθ
η
(5.3.31)
Subtracting equation (5.3.10) from (5.3.12) and substituting equation (5.3.31) in the
resulting equation then solve after differentiating twice with respect to ,θ we get
.01094 === ccc Refreshing the system of equations (5.3.31) we get the conformal
vector fields in this case as
( ) .sincos)()(,sincos
,,)(
11783
872
1231
20
cccffXccX
crcXcdttX
+−′
=+=
+=+= ∫φφ
θθφφ
η (5.3.31)*
274
Considering equations (5.3.10) and (5.3.7) respectively and using equation (5.3.31)*
we get )()( tUtB = and ,)()( )(3 dttU
c
etUtA ∫=
−
where ( ).)()( 2cdtttU += ∫η In this case
we have ).()( 22 tBtA α= Substituting back the values of A and B in the above
equation we conclude that 03 =c and .1=α The solution of equations (5.3.3) to
(5.3.12) is given as [40]
( ) .sincos)()(,sincos
,),(
11783
872
1210
cccffXccX
cXtUX
+−′
=+=
==
φφθθφφ
(5.3.32)
The line element for Kantowski-Sachs and Bianchi type III space-times in this case
becomes
),)()(( 2222222 φθθ dfddrtAdtds +++−= (5.3.33)
with )()( tUtA = where ( ).)()( 2cdtttU += ∫η Proper conformal vector fields after
subtracting Killing vector fields from (5.3.32) is given as
( ).0,0,0),(tUX = (5.3.34)
Case II:
In this case we have 02 =⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛•••
BBB and .0),(1 ≠φrF Considering equation
(5.3.10) and using equation (5.3.14) then solve after differentiating with respect to ,φ
we get ),,(),(),,(0),,( 4333 φθφθφθθφ rFrFrKrK +=⇒= where ),(3 θrF and
),(4 φrF are functions of integration. Refreshing the system of equations (5.3.14) we
get
275
( )( )
).,,(1),()(
1
),,(),(1),(),()(
),,,(1),(),()(
),,(),()()(
42
13
432
212
22
211
210
φθφθ
φθθφφθ
φθθφφθ
θφφθη
θ
rKdtB
rFf
X
rFrFdtB
rFdrFfX
rKdtA
rFdrFfX
rFdrFfdttX
rr
+=
+++′=
++=
++=
∫
∫∫
∫∫
∫∫
(5.3.35)
Considering equation (5.375) and using equation (5.3.35) then differentiating first
with respect to φ then with respect to t twice we get ⇒=⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛•••
02
AAA
,2∫=•
dtAA
A β where .0\ℜ∈β Substituting back the above value in equation
(5.3.7) and solving we get rDrDrF βφβφφ sin)(cos)(),( 211 += and
),,(),(),,( 652 θφθφθ rFFrK += where ),(1 φD ),(2 φD ),(5 φθF and ),(6 θrF are
functions of integration. Refreshing the system of equation (5.3.35) we get
[ ]( )[ ]( )
( )( )
( ) ).,,(1)(sin)(cos)(
1),,(),(
1),()(sin)(cos)(
),,(),(
1),()(sin)(cos)(
),,()(sin)(cos)()(
42
213
43
22212
65
22121
2210
φθφβφβθ
φθ
θφφβφφβθ
θφθ
θφφβφφββθ
θφφβφφβθη
θ
rKdtB
rDrDf
X
rFrF
dtB
rFdDrdDrfX
rFF
dtA
rFdDrdDrfX
rFdDrdDrfdttX
r
++=
++
++′=
++
+−=
+++=
∫
∫∫∫
∫∫∫
∫∫∫
(5.3.36)
Considering equation (5.3.8) and using equation (5.3.36) then solve after
differentiating first with respect to φ then with respect to ,t we get
,0)()( 21 == φφ DD which in turn implies .0),(1 =φrF Which gives a contradiction.
Hence this case is not possible.
276
Case III:
In this case we have 02 =⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛•••
BBB and .0),(1 =φrF Considering equation
(5.3.10) and using equation (5.3.14) then solve after differentiating with respect to ,φ
we get ),,(),(),,( 433 φθφθ rFrFrK += where ),(3 θrF and ),(4 φrF are functions
of integration. Refreshing the system of equations (5.3.14) we get
( )
( ) ).,,(),,(),(1),(
),,,(1),(
),,()(
43432
22
22
21
20
φθφθθ
φθθ
θη
θ rKXrFrFdtB
rFX
rKdtA
rFX
rFdttX
r
=++=
+=
+=
∫
∫
∫ (5.3.37)
Considering equation (5.3.7), (5.3.10), (5.3.11) and using equations (5.3.37), then
solving in a similar fashion, we get ),,(),(),,( 652 φθθφθ FrFrK +=
),,(),()()(
),,( 874 φφθθ
φθ rFrFff
rK +′
−= ,cos)(sin)()(),( 3212 θβθβθ rDrDrDrF ++=
)()(),( 543 rDrDrF += θθ and ),,(),( 47 φφ φ rFrF −= where ),,(5 θrF ),,(6 φθF
),,(7 φrF ),,(8 θrF ),(1 rD ),(2 rD ),(3 rD )(4 rD and )(5 rD are functions of
integration. Refreshing the system of equation (5.3.37) we get
( )
( )
).,(),()()(
),,()()(
1sin)(cos)(
),,(),(
1cos)(sin)()(
,cos)(sin)()()(
873
454
2322
65
23211
3210
φφθθ
φθ
θβθββ
φθ
θβθβ
θβθβη
rFrFffX
rFrDrD
dtB
rDrDX
rFrF
dtA
rDrDrDX
rDrDrDdttX
rrr
+′
−=
+++
−=
++
++=
+++=
∫
∫
∫
(5.3.38)
277
Subtracting equation (5.3.10) from equation (5.3.12) and substituting equation
(5.3.38) in the resulting equation then solve after differentiating with respect to t and
θ we get .cos)(sin)()(),( 8764 φφφ rDrDrDrF ++= Substituting back the above
value and solving we get ),()(),( 1098 rDrDrF += φφ where ),(6 rD ),(7 rD ),(8 rD
)(9 rD and )(10 rD are functions of integration. Substituting back all the above
information in equation [(5.3.12)-(5.3.10)] and solving we get
.0)()()()()( 119432 ===== rDrDrDrDrD Refreshing equation (5.3.38) we get
( )
( ) ).(sin)(cos)()()(
,cos)(sin)(
),,(),(1)(),()(
10873
872
652
1110
rDrDrDffX
rDrDX
rFrFdtA
rDXrDdttX r
+−′
−=
+=
++=+= ∫∫
φφθθ
φφ
φθη
(5.3.39)
Considering equation (5.3.8) and using equation (5.3.39) then solve after
differentiating the resulting equation with respect to ,φ we get ,)( 17 crD =
.,,)( 2128 ℜ∈= cccrD Substituting back the above values and solving we get
),(),( 125 rDrF =θ where )(12 rD is a function of integration. Considering equation
(5.3.10) and using equation (5.3.39) then solve after differentiating with respect to r
and ,t we get ℜ∈= 331 ,)( ccrD and ),()(),( 14136 φφ DrDrF += where )(13 rD and
)(14 φD are functions of integration. Substituting all the above information in
equation (5.3.39) we get
( ) ).(sincos)()(,cossin
),()(,)(
1021
321
2
141513
0
rDccffXccX
DrDXcdttX
+−′
−=+=
+=+= ∫φφ
θθφφ
φη (5.3.40)
Considering equation (5.3.7), (5.3.9) respectively and using equation (5.3.40) then
solving we get ,)( 410 crD = 5
14 )( cD =φ and .,,,,)( 76547615 ℜ∈+= cccccrcrD
Refreshing equation (5.3.40) we get
278
( ) ,sincos)()(,cossin
,,)(
4213
212
861
30
cccffXccX
crcXcdttX
+−′
−=+=
+=+= ∫φφ
θθφφ
η (5.3.41)
where ).(,,,,, 758864321 ccccccccc +=ℜ∈ Considering equation (5.3.7) and using
equation (5.3.41) then solving we get ( ) .)()( 3
6
)(3
dtcdtt
c
ecdtttA∫ ∫+=
+−
∫η
η Also
substituting the above information in equation (5.3.10) and solving we get
.)()( 2cdtttB += ∫η The solution of equations (5.3.3) to (5.3.12) is given as [40]
( ) ,sincos)()(,cossin
,),(
4213
212
8610
cccffXccX
crcXtUX
+−′
−=+=
+==
φφθθφφ
(5.3.42)
where ).(,,,,, 758864321 ccccccccc +=ℜ∈ ( ).)()( 2cdtttU += ∫η and the line
element for Kantowski-Sachs and Bianchi type III space-times is given as in (5.3.1)
with the metric functions given as above. Proper conformal vector fields after
subtracting Killing vector fields from (5.3.42) is given as
( ).0,0,),( 6rctUX = (5.3.43)
5.4. Summary of the Chapter In this chapter we explored proper conformal vector fields for non conformally flat
non static cylindrically symmetric, Kantowski-Sachs and Bianchi type III space-times
in general relativity theory. Following results are obtained from the above study:
(1) In non static cylindrically symmetric space-times: Different possibilities for the
existence of proper conformal vector fields have been found for non conformally flat
non static cylindrically symmetric space-times by using direct integration technique.
It has been shown that very special classes of these space-times admit proper
conformal vector fields. It is important to note that the classification of non static
cylindrically symmetric space-times also covered non static plane symmetric, static
279
cylindrically symmetric and static plane symmetric space-times. It turns out that the
above space-times admit 3, 4 or 5 conformal vector fields. In all the cases when the
above space-times admit conformal vector fields the number of proper conformal
vector field is one. When the above space-time admits three conformal vector fields,
results for proper conformal vector fields are given in equations (5.2.23),
(5.2.26),(5.2.59) and (5.2.61). When the above space-time admits four conformal
vector fields, results for proper conformal vector fields are given in equations
(5.2.32), (5.2.35) and (5.2.67). When the above space-time admits five conformal
vector fields, result for proper conformal vector fields is given in equation (5.2.70).
(2) In Kantowski-Sachs and Bianchi type III space-times: Proper conformal vector
fields are investigated in non conformally flat Kantowski-Sachs and Bianchi type III
space-times by using direct integration technique. It has been shown that special
classes of the above space-times admit proper conformal vector fields. It turns out
that the above space-times admit only five conformal vector fields. In all the cases
when the above space-times admit conformal vector fields the number of proper
conformal vector field is one. The results for proper conformal vector fields are given
in equations (5.3.24), (5.3.34) and (5.3.43).
280
Chapter 6
Self Similar Vector Fields in Kantowski-
Sachs, Bianchi Type III, Static Plane
Symmetric, Static Spherically Symmetric
and Static Cylindrically Symmetric Space-
Times
6.1. Introduction
This chapter is devoted to investigate self similar vector fields in Kantowski-Sachs,
Bianchi type III, static plane symmetric, static spherically symmetric and static
cylindrically symmetric space-times by using algebraic and direct integration method.
In this chapter we will discuss tilted and non tilted self similar vector fields admitted
by the above mentioned space-times. This chapter is organized as follows: In section
(6.2) self similar vector fields of Kantowski-Sachs and Bianchi type III space-times
are investigated. In the next section (6.3) self similar vector fields in static plane
symmetric space-times have been explored. In section (6.4) self similar vector fields
of static spherically symmetric space-times are explored. In section (6.5) self similar
vector fields of static cylindrically symmetric space-time are investigated. For the
above space-times we will study self similar vector fields of first, second, zeroth and
infinite kinds in tilted and non tilted cases. Last section (6.6) of the chapter is
dedicated to a detailed summary of the work.
281
6.2. Self Similar Vector Fields in Kantowski-Sachs
and Bianchi Type III Space-Times
Consider Kantowski-Sachs and Bianchi type III space-times in the spherical
coordinate system ),,,( φθrt (labeled by ),,,,( 3210 xxxx respectively) with the line
element [77]
),)(()()( 222222 φθθ dfdtBdrtAdtds +++−= (6.2.1)
where A and B are no where zero function of t only. For θθ sinh)( =f the space-
time (6.2.1) represents Bianchi type III space-time and for θθ sin)( =f it becomes
Kantowski-Sachs space-times. The above space-times (6.2.1) admit at least four
independent Killing vector fields [77, 79] given as
,r∂∂ ,
φ∂∂ ,sincos
φφ
θφ
∂∂′
−∂∂
ff ,cossin
φφ
θφ
∂∂′
+∂∂
ff (6.2.2)
where prime denotes the derivative with respect to .θ Here we shall take the four-
velocity vector as space like vector field u and define 1)( aa tAu δ= so that
.1=aauu The line element (6.2.1) becomes
).)()()(( 222222 φθθ dftBdtBdtduds ++−+= (6.2.3)
Clearly from equation (1.2.5) we get ⇒=αaa Xu ,
1 ,1 βα += uX where ., ℜ∈βα
Writing equation (1.2.6) explicitly using (6.2.3) we get
,0,0 δ=X (6.2.4)
,0,)(, 02
20 =− XtBX (6.2.5)
,0,)()(, 032
30 =− XftBX θ (6.2.6)
282
,2
2,20 δ=+
•
XXB
B (6.2.7)
,0)( 3,2
2,32 =+ XXf θ (6.2.8)
,)()(
23,
320 δθθ
=+′
+•
XXffX
BB (6.2.9)
where dot represents differentiation with respect to .t Solving equations (6.2.4),
(6.2.5) and (6.2.6) we get
,),(1),()(
1
,),(1),(),,(
312
3
21210
∫
∫
+=
+=+=
φθφθθ
φθφθφθδ
φ
θ
PdtB
Pf
X
PdtB
PXPtX (6.2.10)
where ),,(1 φθP ),(2 φθP and ),(3 φθP are functions of integration which are to be
determined. Considering equation (6.2.8) and using equation (6.2.10) we get
.0),(),()(1),(21),()()(2 23211 =+++
′− ∫∫ φθφθθφθφθ
θθ
φθθφφ PPfdtB
PdtB
Pff Differentiating
the above equation with respect to ,t and solving we get
),()()(),( 211 θφθφθ KKfP += where )(1 φK and )(2 θK are functions of
integration. Refreshing the system of equations (6.2.10) we get
.),(1)()(
1
,),(1))()()((
),()()(
313
2212
210
∫
∫
+=
++′=
++=
φθφθ
φθθφθ
θφθδ
φ
θ
PdtB
Kf
X
PdtB
KKfX
KKftX
(6.2.11)
Subtracting equation (6.2.7) from equation (6.2.9), we get
.0)()(
3,32
2,2 =+
′+− XX
ffXθθ Now considering this equation and using equation
(6.2.11) we get
283
.0)()()()()(
)(1)(
)()]([)()( 2211
21 =
′+−+
′+ θ
θθθφ
θφ
θθφθ θθθφφ K
ffKK
fK
ffKf Solving the
above equation we get φφφ cossin)( 211 ccK += and ,)( 3
2 cK =θ .,, 321 ℜ∈ccc
Substituting the above information in (6.2.11), we get
( )
.),(1)sincos()(
1
,),(1)cossin()(
,)cossin()(
321
3
221
2
3210
∫
∫
+−=
++′=
+++=
φθφφθ
φθφφθ
φφθδ
PdtB
ccf
X
PdtB
ccfX
cccftX
(6.2.12)
Considering equation (6.2.7) and using equation (6.2.12) then differentiating with
respect to φ we get .0)sincos(12 21 =−⎟
⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛∫
•
φφ ccdtBB
B Here we have to discuss
three different possibilities:
(I) 012
≠⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛∫
•
dtBB
B and ,0)sincos( 21 =− φφ cc
(II) 012
=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛∫
•
dtBB
B and ,0)sincos( 21 ≠− φφ cc
(III) 012
=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛∫
•
dtBB
B and .0)sincos( 21 =− φφ cc
Case I:
In this case we have 012
≠⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛∫
•
dtBB
B and .0)sincos( 21 =− φφ cc Equation
⇒=− 0)sincos( 21 φφ cc .021 == cc Refreshing the system of equations (6.2.12)
we get
).,(),,(, 33223
0 φθφθδ PXPXctX ==+= (6.2.13)
284
Considering equation (6.2.7) and using equation (6.2.13) we get
.),()(2
23 δφθδ θ =++
•
PctB
B Solving this equation we get ),(),( 32 φφθ KP = where
)(3 φK is the function of integration and .)()( 23cttB += δ Substituting all the above
information in equation (6.2.13) we get
),,(),(, 33323
0 φθφδ PXKXctX ==+= (6.2.14)
Considering equation (6.2.8) and using equation (6.2.13) we get
⇒=+ 0),()()( 323 φθθφ θφ PfK ),()()()(),( 433 φφ
θθφθ φ KK
ffP +′
= where )(4 φK is a
function of integration. Now considering equation (6.2.9) and using equation (6.2.14),
we get ⇒=++′
0)())()(()()( 433 φφφ
θθ
φφφ KKKff φφφ cossin)( 54
3 ccK += and
,)( 62 cK =φ .,, 654 ℜ∈ccc Substituting all the above information in equation (6.2.14)
we get
.)sincos()()(,cossin, 654
354
23
0 cccffXccXctX +−′
=+=+= φφθθφφδ (6.2.15)
The line element for Kantowski-Sachs and Bianchi type III space-times in this case is
given as
).)(()( 22223
222 φθθδ dfdctdudtds ++++−= (6.2.16)
Self similar vector fields in this case are given as
.)sincos()()(,cossin
,,
6543
542
13
0
cccffXccX
uXctX
+−′
=+=
+=+=
φφθθφφ
βαδ (6.2.17)
In the following we will discuss the self similar vector fields of different kinds.
(a) First consider that ,δα = then the line element for Kantowski-Sachs and Bianchi
type III space-times become
285
).)(()( 22223
222 φθθα dfdctdudtds ++++−= (6.2.18)
Proper self similar vector field after subtracting Killing vector fields from (6.2.17)
can be written as ).0,0,,( 3 uctX αα += In this case the proper self similar vector
field is tilted and of first kind.
(b) Now taking 0=α and ,0≠δ the line element for Kantowski-Sachs and Bianchi
type III space-times become
).)(()( 22223
222 φθθδ dfdctdudtds ++++−= (6.2.19)
Proper self similar vector field after subtracting Killing vector fields become
).0,0,0,( 3ctX += δ In this case the proper self similar vector field is of zeroth kind
orthogonal to the space like vector .au
(c) Now taking 0≠α 0≠δ and ,δα ≠ the line element for Kantowski-Sachs and
Bianchi type III space-times become
).)(()( 22223
222 φθθδ dfdctdudtds ++++−= (6.2.20)
Proper self similar vector field after subtracting Killing vector fields become
).0,0,,( 3 uctX αδ += In this case the proper self similar vector field is of second
kind tilted to the space like vector .au
(d) Taking 0≠α and ,0=δ the line element for Kantowski-Sachs and Bianchi type
III space-times become
),)(( 222222 φθθγ dfddrdtds +++−= (6.2.23)
where .0\ℜ∈γ Proper self similar vector field in this case is ).0,0,,0( uX α= In
this case the proper self similar vector field is of infinite kind parallel to the space like
vector .au
286
Case II:
In this case we have 012
=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛∫
•
dtBB
B and .0)sincos( 21 ≠− φφ cc Substituting
back the above information in (6.2.7) and solving we get ),(),( 32 φφθ KP =
,0321 ==== cccδ where )(3 φK is a function of integration. In this case 01 ≠c and
.02 ≠c Hence this case is not possible.
Case III:
In this case we have 012
=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛∫
•
dtBB
B and .0)sincos( 21 =− φφ cc Equation
⇒=− 0)sincos( 21 φφ cc .021 == cc It is important to note that in this case
equation 012
=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛∫
•
dtBB
B has no real solution. We therefore leave this case.
6.3. Self Similar Vector Fields in Static Plane
Symmetric Space-Times
Consider the static plane symmetric space-times in the usual coordinate system
),,,( zyxt (labeled by ),,,,( 3210 xxxx respectively) with the line element [78]
),( 22)(22)(2 dzdyedxdteds xBxA +++−= (6.3.1)
where A and B are arbitrary functions of x only. The above space-times (6.3.1)
admit at least four independent Killing vector fields [19] which are ,t∂∂ ,
y∂∂
z∂∂
and .z
yy
z∂∂
−∂∂ To find self similar vector fields for the above space-time we shall
287
take the four-velocity vector as time like vector and define 02a
A
a eu δ= where
.1−=aauu The line element (6.3.1) will now become [48]
).( 22)(222 dzdyedxduds xB +++−= (6.3.2)
Clearly from equation (1.2.5), we get
,0 βα += uX (6.3.3)
where ., ℜ∈βα Writing equation (1.2.6) explicitly using (6.3.2) we get
,1,1 δ=X (6.3.4)
,02,1
1,2)( =+ XXe xB (6.3.5)
,03,1
1,3)( =+ XXe xB (6.3.6)
,2,2)( 221 δ=+′ XXxB (6.3.7)
,03,2
2,3 =+ XX (6.3.8)
.2,2)( 331 δ=+′ XXxB (6.3.9)
where ‘dash’ represents differentiation with respect to .x Solving equations (6.3.4),
(6.3.5) and (6.3.6) we get
),,(),(
),,(),(),,(3)(13
2)(1211
zyKdxezyKX
zyKdxezyKXzyKxXxB
z
xBy
+−=
+−=+=
∫∫
−
−δ (6.3.10)
where ),(and),(),,( 321 zyKzyKzyK are functions of integration which are to be
determined. Considering equation (6.3.7) and using equation (6.3.10) then
differentiating the resulting equation with respect to y and ,x respectively we get
.0),(2),()( )(11 =−′′ − xByyyy ezyKzyKxB There exist the following two possibilities:
(I) ,0),(1 =zyK y (II) .0),(1 ≠zyK y
We will discuss each possibility in turn.
288
Case I:
In this case we have ⇒= 0),(1 zyK y ),(),( 11 zFzyK = where )(1 zF is a function of
integration. Substituting back the above value in equation (6.3.7), we get
.2),(2)()()( 21 δδ =+′+′ zyKzFxBxBx y Differentiating this equation with respect to
,y we get ⇒= 0),(2 zyK yy ),()(),( 322 zFzyFzyK += where )(2 zF and )(3 zF are
functions of integration. Substituting back this value in the above equation we get
.2)(2)()()( 21 δδ =+′+′ zFzFxBxBx Now differentiating this equation with respect
to z and ,x respectively we get .0)()( 1 =′′ zFxB z Here we have to discuss the
following three possibilities.
(a) 0)( ≠′′ xB and ,0)(1 =zFz (b) 0)( =′′ xB and ,0)(1 ≠zFz
(c) 0)( =′′ xB and .0)(1 =zFz
We will discuss each case in turn.
Case I(a):
In this case we have 0)( ≠′′ xB and .0)(1 =zFz Equation ⇒= 0)(1 zFz
.,)( 111 ℜ∈= ddzF Substituting back in equation (6.3.7) and differentiating with
respect to ,z we get .,)(0)( 2222 ℜ∈=⇒= ddzFzFz Substituting the above values
in equation (6.3.7) and solving we get ( ) .ln)()(2
1
2
δδ
δd
dxxB−
+= Substituting all the
above information in equation (6.3.10) we get
).,(),(, 3332
21
1 zyKXzFydXdxX =+=+= δ (6.3.11)
Considering equation (6.3.8) and using equation (6.3.11) we get
.0)(),( 33 =+ zFzyK zy Differentiating this equation with respect to ,y we get
),()(),(0),( 5433 zFzyFzyKzyK yy +=⇒= where )(4 zF and )(5 zF are functions
289
of integration. Equation ⇒=+ 0)(),( 33 zFzyK zy .0)()( 34 =+ zFzF z Substituting
back the above information in equation (6.3.11) we get
).()(),(, 53332
21
1 zFzyFXzFydXdxX z +−=+=+= δ (6.3.12)
Considering equation (6.3.9) and using equation (6.3.12), we get
.2))()((2))(( 531 δδ =+−++′ zFzyFdxxB zzz Differentiating this equation with respect
to ,y we get ⇒= 0)(3 zFzz .,,)( 43433 ℜ∈+= dddzdzF Substituting back in the
above equation and differentiating with respect to ,z we get
.,,)(0)( 656555 ℜ∈+=⇒= dddzdzFzFzz Substituting back the above values in
equation (6.3.9) we get .25 dd = Substituting all the above information in equation
(6.3.12) we get
.,, 6233
4322
11 dzdydXdzdydXdxX ++−=++=+= δ (6.3.13)
Finally the line element for static plane symmetric space-times becomes [48]
( ) ).( 22)(2
1222 2
dzdydxdxdudsd
++++−=−δ
δδ (6.3.14)
Self similar vector fields for the above space-times after subtracting Killing vector
fields are given as
.0,, 321
10 ==+== XXdxXuX δα (6.3.15)
In the following we will discuss the self similar vector fields of different kinds.
(i) First consider that ,δα = then the proper self similar vector fields for static plane
symmetric space-times can be written as
).0,0,,( 1dxuX += αα (6.3.16)
The line element in this case takes the form
( ) ).( 22)(2
1222 2
dzdydxdxdudsd
++++−=−δ
δδ (6.3.16)*
In this case the proper self similar vector field is tilted and of first kind.
290
(ii) Now taking 0=α and ,0≠δ then the proper self similar vector fields for static
plane symmetric space-times can be written as
).0,0,,0( 1dxX += δ (6.3.17)
The line element in this case takes the form
( ) ).( 22)(2
1222 2
dzdydxdxdudsd
++++−=−δ
δδ (6.3.17)*
In this case the proper self similar vector field is of zeroth kind orthogonal to the time
like vector .au
(ii) Now taking 0≠α 0≠δ and ,δα ≠ then the proper self similar vector fields for
static plane symmetric space-times can be written as
).0,0,,( 1dxuX += δα (6.3.18)
The line element in this case takes the form
( ) ).( 22)(2
1222 2
dzdydxdxdudsd
++++−=−δ
δδ (6.3.18)*
In this case the proper self similar vector field is of second kind tilted to the time like
vector .au
(iv) Taking 0≠α and ,0=δ then the proper self similar vector fields for static plane
symmetric space-times can be written as
)0,0,,( 1duX α= (6.3.19)
with line element
( ) ,22)(222 dzdyedxduds cax +++−= + (6.3.20)
where ,21
2
dd
a −= 01 ≠d and .,,, 21 ℜ∈cadd The proper self similar vector field is
tilted to the time like vector au and represents the self-similarity of infinite kind.
291
Case I(b):
In this case we have 0)(1 ≠zFz and .0)( =′′ xB Equation
.,)(0)( 11 ℜ∈=′⇒=′′ γγxBxB Substituting back in equation (6.3.7), we get
.2)(2)( 2111 δγγδ =++ zFzFx Differentiating this equation with respect to ,x we
get ⇒= 01γ .,)(0)( 22 ℜ∈=⇒=′ γγxBxB The line element for static plane
symmetric space-times after a suitable rescaling of y and z becomes
.22222 dzdydxduds +++−= (6.3.21)
The above space-time represents Minkowski space-time and we will leave this case.
Case I(c):
Ina this case we have 0)(1 =zFz and .0)( =′′ xB Equation
ℜ∈=⇒= 1111 ,)(0)( ddzFzFz and equation .,)(0)( 11 ℜ∈=′⇒=′′ γγxBxB
Substituting back the above values in equation (6.3.7) and differentiating with respect
to ,x we get ⇒= 01γ .,)(0)( 22 ℜ∈=⇒=′ γγxBxB Once again we reach to
Minkowski space-time.
Case II:
In this case we have .0),(1 ≠zyK y Therefore equation
⇒=−′′ − 0),(2),()( )(11 xByyyy ezyKzyKxB ⇒ℜ∈==′′ qq
zyKzyK
exBy
yyyxB ,),(),(
)(21
1
1)(
⇒=− 0),(),( 11 zyqKzyK yyyy ,cos)(sin)()(),( 3211 yqzFyqzFzFzyK ++=
where ),(1 zF )(2 zF and )(3 zF are functions of integration. Substituting back this
value in the above equation we get ⇒== 0)()( 32 zFzF .0),(1 =zyK y Which is a
contradiction. Hence this case is not possible.
292
6.4. Self Similar Vector Fields in Static Spherically
Symmetric Space-Times
Consider static spherically symmetric space-times in the usual coordinate system
),,,( φθrt (labeled by ),,,,( 3210 xxxx respectively) with the line element [16, 78]
),sin( 22222)(2)(2 φθθ ddrdredteds rBrA +++−= (6.4.1)
where A and B are functions of r only. The above space-times (6.4.1) admit at least
four independent Killing vector fields [16] which are
,t∂∂ ,
φ∂∂ ,sincotcos
φφθ
θφ
∂∂
−∂∂ .coscotsin
φφθ
θφ
∂∂
+∂∂ (6.4.2)
We are taking the four-velocity vector as time like vector field u and define
,02)(
a
rA
a eu δ= so that .1−=aauu The line element (6.4.1) after a suitable rescaling
becomes
).sin( 22222)(22 φθθ ddrdreduds rB +++−= (6.4.3)
Clearly from equation (1.2.5) we get ,0 βα += uX ., ℜ∈βα Writing (1.2.6)
explicitly using (6.4.3) we get
,2)( 1,11 δ=+′ XXrB (6.4.4)
,0,, 21)(
122 =+ XeXr rB (6.4.5)
,0,sin, 1322
31)( =+ XrXe rB θ (6.4.6)
,21
2,21 δrrXX =+ (6.4.7)
,0sin 3,2
2,32 =+ XXθ (6.4.8)
293
,21cot 3,
321 δθ rXrXrX =++ (6.4.9)
where ‘dash’ denotes differentiation with respect to .r Solving equations (6.4.4),
(6.4.5) and (6.4.6) we get
,),(),(sin
1
,),(),(
),,(2
32
21
23
22
212
12221
∫
∫
∫
+=
+−=
+=−−
φθφθθ
φθφθ
φθδ
φ
θ
PdrrePX
PdrrePX
PedreeX
B
B
BBB
(6.4.10)
where ),,(1 φθP ),(2 φθP and ),(3 φθP are functions of integration which are to be
determined. Considering equation (6.4.8) and using equation (6.4.10) then
differentiating with respect to ,r we get ⇒=− 0),(cot),( 11 φθθφθ φθφ PP
),()(sin),( 211 θφθφθ KKP += where )(1 φK and )(2 θK are functions of
integration. Refreshing the system of equations (6.4.10) we get
.),(),(sin
1
,),())()((cos
)),()((sin2
32
213
22
2212
212221
∫
∫
∫
+=
++−=
++=−−
φθφθθ
φθθφθ
θφθδ
φ
θ
PdrreKX
PdrreKKX
KKedreeX
B
B
BBB
(6.4.11)
Considering equation (6.4.7) and using equation (6.4.11) then differentiating with
respect to φ and ,r we get .0)( 2
221 =
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
′
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −
re
reK
BB
φφ In order to solve this equation
we need to discuss the following three possibilities:
294
(I) 0)(1 ≠φφK and ,02
22=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
′
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −
re
re
BB
(II) 0)(1 =φφK and ,02
22≠
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
′
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −
re
re
BB
(III) 0)(1 =φφK and .02
22=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
′
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −
re
re
BB
Case I:
In this case we have 0)(1 ≠φφK and .02
22=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
′
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −
re
re
BB
Equation
⇒=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
′
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −
02
22
re
re
BB
.0\,22
2)( ℜ∈
−= a
raae rB Substituting back the above
information in (6.4.7) and differentiating with respect to ,φ we get ⇒= 0),(2 φθθφP
),()(),( 432 φθφθ KKP += where )(3 θK and )(4 φK are functions of integration.
Now substituting the above values in (6.4.7) and differentiating with respect to θ and
,r we get ⇒=+ 0)()( 22 θθ θθθθ KK ,sincos)( 3212 θθθ cccK ++= .,, 321 ℜ∈ccc
Substituting back in equation (6.4.7) and solve after differentiating with respect to ,θ
we get ,)( 543 ccK += θθ ., 54 ℜ∈cc Finally, substituting all the above information
in equation (6.4.7) we get an equation
.2
1sin2 412
22122 δδ
=+−
+⎟⎠⎞
⎜⎝⎛− − cc
ara
rarra
r This equation will satisfy only if
.041 === ccδ Now substituting all the above information in (6.4.11), we get
295
( )
( )
).,(),(sin1
),(sincos)(cos1
,sincos)(sin
32
2213
4523
12
222
321
2
221
φθφθθ
φθθφθ
θθφθ
φ Pa
raKr
X
KcccKa
rar
X
ccKa
raX
+−
=
++−+−
=
++−
=
(6.4.12)
Considering equation (6.4.9) and using equation (6.4.12) and differentiating with
respect to ,r we get ⇒=+ 0)()( 11 φφφφ KK ℜ∈+= 76761 ,,sincos)( ccccK φφφ
and .03 =c Substituting back the above values in (6.4.9) and solving we get 05 =c
and ),()(cot),( 643 θφφθφθ KdKP +−= ∫ where )(6 θK is the function of
integration. Refreshing the system of equations (6.4.12) we get
( )
( )
),()(cot)sincos(sin1
),(sin)sincos(cos1
,cos)sincos(sin
64672
223
42762
222
2762
221
θφφθφφθ
φθφφθ
θφφθ
KdKcca
rar
X
Kccca
rar
X
ccca
raX
+−−−
=
+−+−
=
++−
=
∫
(6.4.13)
where .,, 762 ℜ∈ccc Considering equation (6.4.8) and using equation (6.4.13) then
differentiating with respect to ,φ we get ⇒=+ 0)()( 44 φφφφ KK
,sincos)( 984 φφφ ccK += ., 98 ℜ∈cc Substituting back we get
.,)(0)( 101066 ℜ∈=⇒= ccKK θθθ Refreshing the system of equations (6.4.13) we
get
296
( )
( )
,)cossin(cot)sincos(sin1
,sincossin)sincos(cos1
,cos)sincos(sin
1098672
223
982762
222
2762
221
ccccca
rar
X
ccccca
rar
X
ccca
raX
+−−−−
=
++−+−
=
++−
=
φφθφφθ
φφθφφθ
θφφθ
(6.4.14)
where .,,,,, 1098762 ℜ∈cccccc Finally the line element for static spherically
symmetric space-times in this case takes the form
),sin( 2222222
222 φθθ ddrdr
raaduds ++−
+−= (6.4.15)
and the self similar vector fields for the above space-time (6.4.15) is given as [49]
( )
( )
,)cossin(cot)sincos(sin1
,sincossin)sincos(cos1
,cos)sincos(sin
,
1098672
223
982762
222
2762
221
0
ccccca
rar
X
ccccca
rar
X
ccca
raX
uX
+−−−−
=
++−+−
=
++−
=
+=
φφθφφθ
φφθφφθ
θφφθ
βα
(6.4.16)
where .,,,,, 1098762 ℜ∈cccccc Since in this case 0=δ so the above space-time
(6.4.15) admits only proper self similar vector field of infinite kind given as
),0,0,0,( uX α= which is parallel to the time like vector .au
Case II:
In this case we have 0)(1 =φφK and .02
22≠
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
′
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −
re
re
BB
Equation ⇒= 0)(1 φφK
.,)( 111 ℜ∈= ccK φ Substituting back these information in (6.4.7) and differentiating
with respect to ,φ we get ⇒= 0),(2 φθθφP ),()(),( 432 φθφθ KKP += where )(3 θK
297
and )(4 φK are functions of integration. Now substituting the above values in (6.4.7)
we get an equation
.0)()()(sin12
32
2222
2
22
12
2=+−+
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛++
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛− ∫∫∫
−−−
θθθθδθθθ Kdr
reKKedr
reecdre
re
BB
BBB
B
In order to solve this equation we must take ⇒=∫−
122
drer
e BB
).0\( ℜ∈= ηηB
Substituting back this value in the above equation and solve after differentiating with
respect to r we get .,,sin)( 323212 ℜ∈++−= cccccK θθθ Also Substituting back
the above values in the above given equation we get
.,2
)( 4432
223 ℜ∈⎟⎟
⎠
⎞⎜⎜⎝
⎛++−=
−
cccceK θθθη
Substituting all these information in
equation (6.4.11) we get
( )
).,(),(2
,2
334432
22
2
22
3221
φθφθθ
θδ
ηη
η
PXKcccecr
eX
ccerX
=+⎟⎟⎠
⎞⎜⎜⎝
⎛++−=
++=
−
−
(6.4.17)
Considering equation (6.4.7) and using equation (6.4.17) we get .032 ==cc Also
considering equation (6.4.8) and using equation (6.4.17) with the above information
and solving we get ),()(cot),( 643 φφθφθ φ KKP += where )(6 φK is a function of
integration. Substituting all the above information in (6.4.17) we get
).()(cot),(,2
64344
221 φφθφδφ
η
KKXKceXrX +=+−==−
(6.4.18)
Considering equation (6.4.9) and using equation (6.4.18) then solve after
differentiating with respect to ,θ we get ,sincos)( 7654 φφφ cccK ++= 5
24 cec
η
=
298
and .,,,,)( 876586 ℜ∈= cccccK φ Now substituting all the above values in equation
(6.4.18) we get
( ) .sincoscot,sincos,2 867
376
21 cccXccXrX +−=+== φφθφφδ (6.4.19)
Thus the line element for static spherically symmetric space-time in this case
becomes
).sin( 2222222 φθθη ddrdreduds +++−= (6.4.20)
Self similar vector fields in this case becomes [49]
( ) .sincoscot,sincos
,2
,
8673
762
10
cccXccX
rXuX
+−=+=
=+=
φφθφφ
δβα (6.4.21)
In the following we will discuss the self similar vector fields of different kinds.
(i) First consider that ,δα = then the proper self similar vector fields for static
spherically symmetric space-times after subtracting Killing vector fields from
(6.2.21) can be written as
).0,0,2
,( ruX αα= (6.4.22)
In this case the proper self similar vector field is tilted and of first kind.
(ii) Now taking 0=α and ,0≠δ then the proper self similar vector fields for static
spherically symmetric space-times after subtracting Killing vector fields from
(6.2.21) can be written as
).0,0,2
,0( rX δ= (6.4.23)
In this case the proper self similar vector field is of zeroth kind orthogonal to the time
like vector .au
299
(iii) Now taking 0≠α 0≠δ and ,δα ≠ then the proper self similar vector fields for
static spherically symmetric space-times after subtracting Killing vector fields from
(6.2.21) can be written as
).0,0,2
,( ruX δα= (6.4.24)
In this case the proper self similar vector field is of second kind tilted to the time like
vector .au
(iv) Taking 0≠α and ,0=δ proper self similar vector fields for spherically
symmetric static space-times after subtracting Killing vector fields from (6.2.21) can
be written as
).0,0,0,( uX α= (6.4.25)
In this case the proper self similar vector field is of infinite kind parallel to the time
like vector .au
Case III:
In this case we have 0)(1 =φφK and .02
22=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
′
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −
re
re
BB
Equation
⇒=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+
′
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −
02
22
re
re
BB
0\,22
2)( ℜ∈
−= a
raae rB and equation ⇒= 0)(1 φφK
.,)( 111 ℜ∈= ccK φ Substituting back the above information in (6.4.7) and
differentiating with respect to ,φ we get ⇒= 0),(2 φθθφP ),()(),( 432 φθφθ KKP +=
where )(3 θK and )(4 φK are functions of integration. Now substituting the above
values in (6.4.7) and differentiating with respect to θ and ,r we get
⇒=+ 0)()( 22 θθ θθθθ KK ,sincos)( 4322 θθθ cccK ++= .,, 432 ℜ∈ccc Substituting
300
back in equation (6.4.7) and solve after differentiating with respect to ,θ we get
,)( 653 ccK += θθ ., 65 ℜ∈cc Finally, substituting all the above information in
equation (6.4.7) we get an equation .2
1sin2 522
22122 δδ
=+−
+⎟⎠⎞
⎜⎝⎛− − cc
ara
rarra
r
This equation will satisfy only if .052 === ccδ Now substituting all the above
information in (6.4.11), we get
( )
( )
).,(
,)(sincoscos1
,sincossin
33
64
3412
222
4312
221
φθ
φθθθ
θθθ
PX
cKccca
rar
X
ccca
raX
=
++−+−
=
++−
=
(6.4.26)
Considering equation (6.4.9) and using equation (6.4.26) and differentiating with
respect to ,φ we get ⇒−= )(cot),( 43 φθφθ φφφ KP
),()()(cot),( 6543 θθφφφθφθ KKdKP ++−= ∫ where )(5 θK and )(6 θK are
functions of integration. Substituting back the above values in equation (6.4.9) we get
⇒=++−
+ 0sin)(cos)( 562
22
41 θθθ rKrca
racc .0)(,0, 5641 ==−= θKccc
Refreshing the system of equations (6.4.26) we get
).()(cot
),(sin1,cos
643
432
222
32
221
θφφθ
φθθ
KdKX
Kca
rar
Xca
raX
+−=
+−
−=−
=
∫ (6.4.27)
Considering equation (6.4.8) and using equation (6.4.27) then differentiating with
respect to ,φ we get ⇒=+ 0)()( 44 φφφφ KK ,sincos)( 984 φφφ ccK += ., 98 ℜ∈cc
Substituting back we get .,)(0)( 101066 ℜ∈=⇒= ccKK θθθ Refreshing the system
of equations (6.4.27) we get
301
.)sincos(cot
,sincossin1,cos
10893
9832
222
32
221
cccX
ccca
rar
Xca
raX
+−=
++−
−=−
=
θθθ
φφθθ (6.4.28)
where .,,, 10983 ℜ∈cccc Also considering equation (6.4.4) and using equation
(6.4.28) we get .0)1( 23 =− ac If we take 03 ≠c then we must take .12 =a Finally,
the line element for static spherically symmetric space-times in this case takes the
form
),sin(1
1 222222
22 φθθ ddrdrr
duds ++−
+−= (6.4.29)
and the self similar vector fields for the above space-time (6.4.29) is given as [49]
.)sincos(cot,sincossin11,cos1,
10893
98322
3210
cccXcccrr
X
crXuX
+−=++−−=
−=+=
θθθφφθ
θβα (6.4.30)
where .,,, 10983 ℜ∈cccc Since in this case 0=δ so the above space-time (6.4.29)
admits only proper self similar vector field of infinite kind given as
),0,0,0,( uX α= which is parallel to the time like vector .au
6.5. Self Similar Vector Fields in Static
Cylindrically Symmetric Space-Times
Consider cylindrically symmetric static space-times in usual coordinates ),,,( zrt θ
(labeled by ),,,,( 3210 xxxx respectively) with the line element [36, 78]
,2)(2)(22)(2 dzededrdteds rCrBrA +++−= θ (6.5.1)
302
where ,A B and C are functions of r only. The space-time (6.5.1) admits minimum
three Killing vector fields [36] which are .,,φθ ∂∂
∂∂
∂∂t
It is important to note that the
above space-times (6.5.1) become static plane symmetric space-times when
).()( rCrB = We have already discussed self similar vector fields for static plane
symmetric space-times in section (6.3) so we will not discuss this case here in this
section. To find self similar vector fields for the above space-time we shall take the
four-velocity vector as time like vector field u and define 02a
A
a eu δ= so that
.1−=aauu The line element (6.5.1) will now become
,2)(2)(222 dzededrduds rCrB +++−= θ (6.5.2)
Clearly from equation (1.2.5), we get
,0 βα += uX (6.5.3)
where ., ℜ∈βα Writing equation (1.2.6) explicitly using (6.5.2) we get
,1,1 δ=X (6.5.4)
,02,1
1,2)( =+ XXe rB (6.5.5)
,03,1
1,3)( =+ XXe rC (6.5.6)
,2,2)( 221 δ=+• XXrB (6.5.7)
,03,2)(
2,3)( =+ XeXe rBrC (6.5.8)
,2,2)( 331 δ=+• XXrC (6.5.9)
where ‘dot’ represents differentiation with respect to .r Solving equations (6.5.4),
(6.5.5) and (6.5.6) we get
),,(),(
),,(),(),,(3)(13
2)(1211
zKdrezKX
zKdrezKXzKrXrC
z
rB
θθ
θθθδ θ
+−=
+−=+=
∫∫
−
−
(6.5.10)
303
where ),(and),(),,( 321 zKzKzK θθθ are functions of integration which are to be
determined. Considering equation (6.5.7) and using equation (6.5.10) then
differentiating the resulting equation with respect to θ and ,r respectively we get
.0),(2),()( )(11 =− −•• rBezKzKrB θθ θθθθ There exist the following two possibilities:
(I) ,0),(1 =zK θθ (II) .0),(1 ≠zK θθ
We will discuss each possibility in turn.
Case I:
In this case we have ⇒= 0),(1 zK θθ ),(),( 11 zFzK =θ where )(1 zF is a function of
integration. Substituting back the above value in equation (6.5.7), we get
.2),(2)()()( 21 δθδ θ =++ •• zKzFrBrrB Differentiating this equation with respect to
,θ we get ⇒= 0),(2 zK θθθ ),()(),( 322 zFzFzK += θθ where )(2 zF and )(3 zF
are functions of integration. Substituting back this value in the above equation we get
.2)(2)()()( 21 δδ =++ •• zFzFrBrrB Now differentiating this equation with respect
to z and ,r respectively we get .0)()( 1 =•• zFrB z Here we have to discuss the
following three possibilities.
(a) 0)( ≠•• rB and ,0)(1 =zFz (b) 0)( =•• rB and ,0)(1 ≠zFz
(c) 0)( =•• rB and .0)(1 =zFz
We will discuss each case in turn.
Case I(a):
In this case we have 0)( ≠•• rB and .0)(1 =zFz Equation ⇒= 0)(1 zFz
.,)( 111 ℜ∈= ddzF Substituting back in equation (6.5.7) and differentiating with
respect to ,z we get .,)(0)( 2222 ℜ∈=⇒= ddzFzFz Substituting the above values
in equation (6.5.7) and solving we get ( ) .ln)()(2
1
2
δδ
δd
drrB−
+= Substituting all the
above information in equation (6.5.10) we get
304
).,(),(, 3332
21
1 zKXzFdXdrX θθδ =+=+= (6.5.11)
Considering equation (6.5.8) and using equation (6.5.11) we get
.0)(),( 3)(3)( =+ zFezKe zrBrC θθ Differentiating this equation with respect to ,θ we get
),()(),(0),( 5433 zFzFzKzK +=⇒= θθθθθ where )(4 zF and )(5 zF are functions
of integration. Substituting back the above information in equation (6.5.11) we get
).()(),(, 54332
21
1 zFzFXzFdXdrX +=+=+= θθδ (6.5.12)
Considering equation (6.5.9) and using equation (6.5.12), we get
.2))()((2))(( 541 δθδ =+++• zFzFdrrC zz Differentiating this equation with respect
to ,θ we get ⇒= 0)(4 zFz .,)( 334 ℜ∈= ddzF Substituting back in the above
equation and differentiating with respect to ,z we get
.,,)(0)( 545455 ℜ∈+=⇒= dddzdzFzFzz Substituting back the above values in
equation ,2))()((2))(( 541 δθδ =+++• zFzFdrrC zz we get ⇒=++• δδ 22))(( 41 ddrrC
( ) .ln)()(2
1
4
δδ
δd
drrC−
+= Substituting all the above information in equation (6.5.12)
we get
.),(, 54333
22
11 dzddXzFdXdrX ++=+=+= θθδ (6.5.13)
Considering equation (6.5.8) and using equation (6.5.13) we get
.0)()()( 3
)(2
13
)(2
1
42
=+++−−
ddrzFdrd
z
dδ
δδ
δ
δδ Solving this equation we get 03 =d
and .,)(0)( 6633 ℜ∈=⇒= ddzFzFz Refreshing the system of equations (6.5.13)
we get
,,,, 543
622
110 dzdXddXdrXuX +=+=+=+= θδβα (6.5.14)
where ).0(,,,,,,, 65421 ≠ℜ∈ δδβαddddd Finally the line element for static
cylindrically symmetric space-times becomes [80]
.)()( 2)1(2
12)1(2
1222
42
dzdrddrdrdudsddδδ δθδ
−−+++++−= (6.5.15)
305
Self similar vector fields for the above space-times after subtracting Killing vector
fields are given as
.0,, 321
10 ==+== XXdrXuX δα (6.5.16)
In the following we will discuss the self similar vector fields of different kinds.
(i) First consider that ,0≠= δα then the proper self similar vector fields for static
cylindrically symmetric space-times can be written as
).0,0,,( 1druX += αα (6.5.17)
The line element in this case takes the form
.)()( 2)1(2
12)1(2
1222
42
dzdrddrdrdudsddαα αθα
−−+++++−= (6.5.18)
In this case the proper self similar vector field is tilted and of first kind.
(ii) Now taking 0=α and ,0≠δ then the proper self similar vector fields for static
cylindrically symmetric space-times can be written as
).0,0,,0( 1drX += δ (6.5.19)
The line element in this case takes the form
.)()( 2)1(2
12)1(2
1222
42
dzdrddrdrdudsddδδ δθδ
−−+++++−= (6.5.20)
In this case the proper self similar vector field is of zeroth kind orthogonal to the time
like vector .au
(ii) Now taking 0≠α 0≠δ and ,δα ≠ then the proper self similar vector fields for
static cylindrically symmetric space-times can be written as
).0,0,,( 1druX += δα (6.5.21)
The line element in this case takes the form [80]
.)()( 2)1(2
12)1(2
1222
42
dzdrddrdrdudsddδδ δθδ
−−+++++−= (6.5.22)
306
In this case the proper self similar vector field is of second kind tilted to the time like
vector .au
(iv) Taking 0≠α and ,0=δ then the proper self similar vector fields for static
cylindrically symmetric space-times can be written as
)0,0,,( 1duX α= (6.5.23)
with line element
,2)(2)(222 dzededrduds frcbar ++ +++−= θ (6.5.24)
where ,0,2 11
2 ≠−= ddda
1
42ddc −= and .,,,,,, 421 ℜ∈fcbaddd The proper self
similar vector field is tilted to the time like vector au and represents the self-
similarity of infinite kind.
Case I(b):
In this case we have 0)(1 ≠zFz and .0)( =•• rB Equation
.,,)(0)( 2121 ℜ∈+=⇒= ••• γγγγ rrBrB Substituting back in equation (6.5.7), we
get .2)(2)( 2111 δγγδ =++ zFzFr Differentiating this equation with respect to ,r
we get .01 =γ Substituting back in the above equation we get .)(2 δ=zF Refreshing
the system of equations (6.5.10), we get
).,()(),(),( 3)(133211 zKdrezFXzFXzFrX rCz θθδδ +−=+=+= ∫ − (6.5.25)
Considering equation (6.5.8) and using equation (6.5.25) then differentiating the
resulting equation with respect to ,θ we get ⇒= 0),(3 zK θθθ
),()(),( 543 zFzFzK += θθ where )(4 zF and )(5 zF are functions of integration.
Refreshing the system of equations (6.5.25) we get
).()()(
),(),(54)(13
3211
zFzFdrezFX
zFXzFrXrC
z ++−=
+=+=
∫ − θ
θδδ (6.5.26)
307
Now considering equation (6.5.9) and using equation (6.5.26) we get
.2))()()((2))()(( 54)(11 δθδ =++−++ ∫ −• zFzFdrezFzFrrC zzrC
zz Differentiating this
equation with respect to r and z respectively, we get
.0)(2)()( )(11 =− −•• rCzzzz ezFzFrC In this case we know that 0)(1 ≠zFz therefore we
have ⇒==•• qzFzFerC
z
zzzrC
)()()(
21
1
1)( ⇒=− 0)()( 11 zqFzF zzzz
.,,,cossin)( 3213211 ℜ∈++= dddzqdzqddzF Substituting back this value in
the above equation we get ⇒== 032 dd .0)(1 =zFz Which gives a contradiction.
Hence this case is not possible.
Case I(c):
In this case we have 0)(1 =zFz and .0)( =•• rB Equation
ℜ∈+=⇒= •••2121 ,,)(0)( γγγγ rrBrB and equation ⇒= 0)(1 zFz
.,)( 111 ℜ∈= ddzF Substituting back in equation (6.5.7), we get
.2)(2 2111 δγγδ =++ zFdr Differentiating this equation with respect to ,r we get
.01 =γ Substituting back in the above equation we get .)(2 δ=zF Refreshing the
system of equations (6.5.10), we get
).,(),(, 33321
1 zKXzFXdrX θθδδ =+=+= (6.5.27)
Considering equation (6.5.8) and using equation (6.5.27) then differentiating the
resulting equation with respect to ,θ we get ⇒= 0),(3 zK θθθ
),()(),( 543 zFzFzK += θθ where )(4 zF and )(5 zF are functions of integration.
Refreshing the system of equations (6.5.27) we get
).()(),(, 543321
1 zFzFXzFXdrX +=+=+= θθδδ (6.5.28)
Now considering equation (6.5.9) and using equation (6.5.28) we get
.2))()((2))(( 541 δθδ =+++• zFzFdrrC zz Differentiating this equation with respect
308
to θ we get .,)(0)( 3344 ℜ∈=⇒= ddzFzFz Substituting this value in the above
equation and differentiating the resulting equation with respect to ,z we get
.,,)(0)( 545455 ℜ∈+=⇒= dddzdzFzFzz Substituting back this value in (6.5.9)
we get ⇒=++• δδ 22))(( 41 ddrrC ( ) .ln)()(2
1
4
δδ
δd
drrC−
+= Substituting all the
above information in equation (6.5.28) we get
.),(, 543332
11 dzddXzFXdrX ++=+=+= θδθδ (6.5.29)
Considering equation (6.5.8) and using equation (6.5.29) we get
.0)( 3)(32 =+ dezFe rC
zγ Differentiating this equation with respect to ,r we get
.03 =d Substituting back we get .,)(0)( 6633 ℜ∈=⇒= ddzFzFz Finally, the
solution of equations (6.5.4) to (6.5.9) becomes
.,, 543
62
11 dzdXdXdrX +=+=+= δθδ (6.5.30)
The line element for static cylindrically symmetric space-times in this case after a
suitable rescaling of θ takes the form [80]
,)( 2)1(2
12222
4
dzdrddrdudsdδδθ
−++++−= (6.5.31)
where ).0(,,,, 5461 ≠ℜ∈ δδ dddd Proper self similar vector fields for the above
space-times after subtracting Killing vector fields are given as
.0,, 321
10 ==+== XXdrXuX δα (6.5.32)
In the following we will discuss the self similar vector fields of different kinds.
(i) First consider that ,0≠= δα then the proper self similar vector fields for static
cylindrically symmetric space-times can be written as
).0,0,,( 1druX += αα (6.5.33)
The line element in this case takes the form
309
.)( 2)1(2
12222
4
dzdrddrdudsdααθ
−++++−= (6.5.34)
In this case the proper self similar vector field is tilted and of first kind.
(ii) Now taking 0=α and ,0≠δ then the proper self similar vector fields for static
cylindrically symmetric space-times can be written as
).0,0,,0( 1drX += δ (6.5.35)
The line element in this case takes the form
.)( 2)1(2
12222
4
dzdrddrdudsdδδθ
−++++−= (6.5.36)
In this case the proper self similar vector field is of zeroth kind orthogonal to the time
like vector .au
(ii) Now taking 0≠α 0≠δ and ,δα ≠ then the proper self similar vector fields for
static cylindrically symmetric space-times can be written as
).0,0,,( 1druX += δα (6.5.37)
The line element in this case takes the form
.)( 2)1(2
12222
4
dzdrddrdudsdδδθ
−++++−= (6.5.38)
In this case the proper self similar vector field is of second kind tilted to the time like
vector .au
(iv) Taking 0≠α and ,0=δ then the proper self similar vector fields for static
cylindrically symmetric space-times can be written as
)0,0,,( 1duX α= (6.5.39)
with line element
,2)(2222 dzeddrduds frc ++++−= θ (6.5.40)
310
where ,21
4
ddc −= ,01 ≠d and .,,, 41 ℜ∈fcdd The proper self similar vector field is
tilted to the time like vector au and represents the self-similarity of infinite kind.
Case II:
In this case we have .0),(1 ≠zK θθ Equation ⇒=− −•• 0),(2),()( )(11 rBezKzKrB θθ θθθθ
⇒−==•• qzKzK
erB rB
),(),(
)(21
1
1)(
θθ
θ
θθθ ⇒=+ 0),(),( 11 zqKzK θθ θθθθ
.cos)(sin)()(),( 3211 θθθ qzFqzFzFzK ++= Substituting back this value in the
above equation we get .0)()( 32 == zFzF Therefore, ⇒= )(),( 11 zFzK θ
.0),(1 =zK θθ Which is a contradiction. Hence this case is not possible.
6.6. Summary of the Chapter
In this chapter we explored self similar vector fields for Kantowski-Sachs, Bianchi
type III, static plane symmetric, static spherically symmetric and static cylindrically
symmetric space-times by using algebraic and direct integration techniques.
Following results are obtained from the above study:
(1) In Kantowski-Sachs and Bianchi type III space-times: Different possibilities for
the existence of proper self similar vector fields have been found for the above space-
times by using direct integration technique. It turns out that the above space-times
admit tilted and non tilted proper self similar vector fields. It turns out that the above
space-times admit tilted proper self similar vector field of first kind and second kind.
The above space-time also admit non tilted orthogonal proper self similar vector
fields of zeroth kind and non tilted parallel proper self similar vector fields of infinite
kind. This space-times admit the above kinds of self similar vector fields for a special
choice of the metric functions.
311
(2) In static plane symmetric space-times: Proper self similar vector fields in both
tilted and non tilted cases have been explored for the above space-times by using
direct integration technique. It turns out that the above space-times admit tilted and
non tilted proper self similar vector fields. It turns out that the above space-times
admit tilted proper self similar vector field of first kind, second kind and infinite kind.
The above space-time also admits non tilted orthogonal proper self similar vector
fields of zeroth kind. This space-time admits the above kinds of self similar vector
fields for a special choice of the metric functions.
(3) In static spherically symmetric space-times: Proper self similar vector fields of
tilted and non tilted kinds have been found for the above space-times by using direct
integration technique. It turns out that the above space-times admit tilted and non
tilted proper self similar vector fields. The above space-times admit tilted proper self
similar vector field of first kind and second kind. The above space-time also admit
non tilted orthogonal proper self similar vector fields of zeroth kind and non tilted
parallel proper self similar vector fields of infinite kind. This space-time admits the
above kinds of self similar vector fields for a special choice of the metric functions.
(4) In static cylindrically symmetric space-times: Proper self similar vector fields of
tilted and non tilted kinds have been found for static cylindrically symmetric space-
times. The above space-time admits tilted proper self similar vector fields of first,
second and infinite kind. This space-time also admits non tilted zeroth kind proper
self similar vector field orthogonal to the time like vector field .au The above space-
time admits the above kinds of self similar vector fields for a special choice of metric
functions.
312
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