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Relativity of magnetic and electric fields Masatsugu Sei Suzuki
Department of Physics, SUNY at Bimghamton (Date: January 13, 2012)
________________________________________________________________________ 1. Lorentz transformation 1.1 Derivation of Lorentz transformation
We consider a Galilean transformation given by
tt
vtxx
vtxx
'
''
'
vuu
vdt
dx
dt
dtv
dt
dx
dt
dx
''''
'
We know that the velocity of light remains unchanged under a transformation (so-called the Lorentz transformation) satisfying the principle of relativity. This implies that the Lorentz transformation is not the same as the Galilean transformation. Here we assume that
)''(
)('
vtxx
vtxx
from the symmetry of transformation What is the value of ?
2
(i) The light is emitted at
0'
0'
xx
tt
(initially). The speed of light (in vacuum) is the same in all internal reference frames; it always has the value c.
ct
x
t
x
'
'
((Mathematica))
Derivation of Lorentz transformation eq1=x== (x'+v t');eq2=x'== (x- v t) x t vx eq3=Solve[{eq1,eq2},{x',t'}]//Simplify//Flatten
t
x 1
v t, x t v x
eq4
x't'
xt.eq3 Simplify
vt v x 2
x t v2 x2
x
t eq5=eq4/.{xc t}//Simplify
vcv 2
cc2 v2 c
eq6=Solve[eq5,]
c c2 v2
, c
c2 v2
T'=t'/.eq3/.eq6[[2]]//Simplify
c2 t v xc
c2 v2 x'/.eq3 (-t v+x) X'=x'/.eq3/.eq6[[2]]//Simplify
ct vx c2 v2
_____________________________________________________________ Then we have
)('
)()('
xc
tt
ctxvtxx
where
3
21
1
,
with
c
v
Note that is expanded as
)(8
3
21 6
42
O
in the limit of →0. For convenience, we introduce
ictx 4 or
tc
xi 4
Then we have
)('
)('
414
411
xxix
xixx
or
4
3
2
1
4
3
2
1
00
0100
0010
00
'
'
'
'
x
x
x
x
i
i
x
x
x
x
or
axx '
)''(
)''(
414
411
xxix
xixx
4
or in the matrix form,
'
'
'
'
00
0100
0010
00
4
3
2
1
4
3
2
1
x
x
x
x
i
i
x
x
x
x
'1xax
Note that Taa 1
')()( 1 xaaax T
1.2 Lorentz contraction
Imagine a stick moving to the right at the velocity v. Its rest length (that is, its length measured in S’) is '1x .
We measure the distance of the stick under the condition that 04 x . Since
1411 )(' xxixx or
02
12
11 1'1'1
xxxx
The length of the stick measure in S (x1) is shorter than that observed in S’ (x1, proper length) 1.3 Time dilation
We are watching one moving clock moving to the right at the velocity v.
)''( 414 xxix with 0'1 x . Then we have
'' 444 xxx . or
0022 1
1'
1
1tttt
5
The time in S (t) is longer than that observed in S’ (t0, proper time). The moving clocks run slow 1.4 Proper time
22' dxdxdxdxdxaadx
We define the proper time as
23
22
21
2223
22
21
222 )'()'()'()'()()()()( dxdxdxdtcdxdxdxdtcds
]1[)(]})()()[(1
1{)( 2222322212
222 dtcdt
dx
dt
dx
dt
dx
cdtcds
or
21 dtc
dsd
where is a proper time. 1.5 Notation of four vector
Four vector notation
0
3
2
1
4
3
2
1
ib
b
b
b
b
b
b
b
b
b ( = 1, 2, 3, and 4)
where
b1, b2, b3: real b4 = ib0 purely imaginary
((Note))
We use the Einstein convention, in which repeated indices are summed.
i, j, k (= 1 – 3) , , (= 1, - 4)
The co-ordinate vector
6
0
3
2
1
4
3
2
1
ix
x
x
x
x
x
x
x
x
b ( = 1, 2, 3, and 4)
Under a Lorentz transformation, we have
xax ' instead of
xax '
where
aa
aaaaaaaa T )()( 11
Note that
Taa 1 (transpose matrix) (1)
xxxxxxaaxaxaxx ''
(2)
xax '
xxxaaxa '
or
' xax or ' xax A four vector, by definition, transforms in the same way as x under the Lorentz
transformation.
7
xa
xx
x
x
''
where
' xax
The scalar product cb is defined by
cbcb
It is invariant under the Lorentz transformation
bababaaababacb ''
1.6 Four dimensional Laplacian operator
2
2
22
24
2
23
2
22
2
21
2 1
tcxxxxxx
is invariant under the Lorentz transformation: Lorentz scalar
xxxxxxxx
aax
ax
axx
''
1.7 A tensor of second rank
A tensor of second rank, t , transforms as
taat '
1.8 A tensor of third rank
A tensor of third rank, t , transforms as
taaat '
((Note)) We make no distinction between a covariant and contravariant vector. We do not define the metric tensor g .
2. Velocity, acceleration, and force 2.1 Lorentz velocity transformation
8
')( 1 xax
So we have
)()( 1 Taaa
)('
'
'
)('
1
33
22
11
xc
tt
xx
xx
vtxx
)''(
'
'
)''(
1
33
22
11
xc
tt
xx
xx
vtxx
Suppose that an object has velocity components as measured in S’ and S.
1
333
1
222
1
111
1
1
'
''
1
1
'
''
1'
''
uc
u
dt
dxu
uc
u
dt
dxu
uc
vu
dt
dxu
'1
'1
'1
'1
'1
'
'
1
333
1
222
1
111
uc
u
dt
dxu
uc
u
dt
dxu
uc
vu
dt
dxu
The Lorentz transformation of a velocity less than c never leads to a velocity greater than c. The relations reduce to the Galilean transformation for v<<c.
Suppose that the particle is s photon, and u1 = c in the frame S. Then we have
c
c
vvc
cc
vc
uc
vuu
111'
1
11
((Example))
cv
cu
10
910
9'1
cc
cc
uc
vuu
181
180
100
811
10
9
10
9
'1
'
1
11
whereas the Galilean transformation would have given
ccccvuu 5
9
10
9
10
9'11
9
2.2 Lorentz acceleration transformation Similarly we have the acceleration components as measured in S’ and S.
31
31
23
1
32
1
3
1
2
1
333
31
21
22
1
22
1
2
1
2
1
222
31
13
31
12
1
1
11
111
)'1(
'1
)'1(
'1
'1
'
')'1(
11
'1
'
'
'1
)'1(
'1
)'1(
'1
'1
'
')'1(
11
'1
'
'
'1
)'1(
'1
)'1(
')1(1
'1
'
')'1(
11
'1
'
'
'
uc
caa
uc
a
uc
u
dt
d
uc
uc
u
dt
d
dt
dt
dt
dua
uc
caa
uc
a
uc
u
dt
d
uc
uc
u
dt
d
dt
dt
dt
dua
uc
a
uc
a
uc
vu
dt
d
uc
uc
vu
dt
d
dt
dt
dt
dua
where
)'1(
11'
1uc
dt
dt
The acceleration is a quantity of limited and questionable value in special relativity. Not only is it not an invariant, but the expressions for it are in general cumbersome, and moreover its different components transform in different ways. 2.3 Force F under the Lorentz transformation
Lorentz transformation
)('
'
'
)('
12tx
c
vt
zz
yy
vtxx
or
10
)''(
'
'
)''(
12tx
c
vt
zz
yy
vtxx
12
1
12
1
12
1
12
11
11)1(
)(
)(
)(
'
'
uc
vdt
dE
cF
dt
dx
c
vdt
dE
cdt
dp
dt
dx
c
vdt
dE
cdt
dp
dtdxc
v
dEc
dp
dt
dp
Since uF dt
dE
12
11
1
1
)(
'
''
uc
vc
F
dt
dpF
Fu
Similarly
)1()1()('
''
12
2
12
2
12
222
uc
vF
dt
dx
c
vdt
dp
dtdxc
vdp
dt
dpF
)1()1()('
''
12
3
12
3
12
333
uc
vF
dt
dx
c
vdt
dp
dtdxc
vdp
dt
dpF
We consider one special case when the particle is instantaneously at rest in S. So that u = 0.
33
22
11
'
'
'
FF
FF
FF
11
The component of F parallel to the motion of S’ is unchanged, whereas the components perpendicular are divided by . 3. Charge and current density 3.1 Charge density
We measure the distance of the cylinder under the condition that 04 x . Since
1411 )(' xxixx or
'1'1
12
11 xxx
we have
'1'1 2 LLL
but with the same area A (since dimension transverse to the motion are unchangeable. If we call ' (= 0) the density of charges in the S’ frame in which charges momentarily at rest, the total charge Q is the same in any system,
LAALALQ ''''' 0
or
LL '0
or
00
' L
L
12
3.2 Current density J
The current density J is defined as
),(),( icicJ uJ
where u is the velocity of the particle in the S frame. Evidently the charge density and current density go together to make a 4 vector.
JaJ '
)('
)()('
1
1411
Jc
vJJiJJ
or
' JaJ
)''(
)''()''(
1
1411
Jc
vJJiJJ
Then we have
)''( 1 Jc
Note that
' when 0'1 J . 3.3 Invariance under the Lorentz transformation
We know that JJ is invariant under the Lorentz transformation
JJJJJJaaJaJaJJ ''
or
222222 '' ccJJ JJ
13
Suppose that J’ = 0 (or u’ = 0) in the S’ frame, where the point charge is at rest. vu J (the frame S’ moves at the velocity v relative to the frame S). Then we have
2
02222222 '0 cccv
or
02
2
1 c
v, or 0
2
2
0
1
c
v
4 Maxwell’s equation field fensor 4.1 Four vectors for the vector potential and scalar potential
))(
,(
)1
,(
),(
ict
ciA
icJ
A
J
The equation of continuity;
0))(1
(
tic
tciJ JJ
Maxwell’s equation;
D 0 B
t
B
E
t
D
JH
where
PED 0
)(0 MHB
14
AB
AE
t
4.2. Gauge transformation
)1
,( ciAA
AA' ,
t ' ,
AA'
((Note))
444 '
)(
1'
1
xAA
ictci
ci
Lorentz gauge:
01
)( 2
tcci
ictA
x
A
AA
4.3 Electromagnetic field tensor F
We define the field tensor as
x
A
x
AF
This tensor satisfies the Jacobi identity;
0
x
F
x
F
x
F
This equation holds automatically for the antisymmetric tensor The magnetic field;
15
32
1
1
212 B
x
A
x
AF
13
2
2
323 B
x
A
x
AF
21
3
3
131 B
x
A
x
AF
The electric field;
11
4
1
1
414 E
c
i
ic
E
x
A
x
AF
22
4
2
2
424 E
c
i
ic
E
x
A
x
AF
33
4
3
3
434 E
c
i
ic
E
x
A
x
AF
The field tensor is an anti-symmetric tensor of second rank and hence, has 6 independent components. Electromagnetic field tensor;
0
0
0
0
321
312
213
123
Ec
iE
c
iE
c
i
Ec
iBB
Ec
iBB
Ec
iBB
F
We show that
Faa
x
A
x
AF
'
'
'
''
xa
x
', and AaA '
16
x
Aaa
x
Aaa
x
Aa
x
Aa
x
A
x
AF
)()(''
'
'
'
''
or
Faa
x
A
x
AaaF
)('
4.4 Maxwell’s equation (1)
The Maxwell's equation is given by
Jx
F0
)()(
1)(
1010101
001
121
1
3
2
2
3
4
14
3
13
2
12
1
111
EJJt
E
t
E
ct
E
icc
i
x
B
x
B
x
F
x
F
x
F
x
F
x
F
B
B
)()(
1)(
2020202
002
222
2
3
1
1
3
4
24
3
23
2
22
1
212
EJJt
E
t
E
ct
E
icc
i
x
B
x
B
x
F
x
F
x
F
x
F
x
F
B
B
)()(
1)(
3030303
003
323
3
2
1
1
2
4
34
3
33
2
32
1
313
EJJt
E
t
E
ct
E
icc
i
x
B
x
B
x
F
x
F
x
F
x
F
x
F
B
B
17
0
20
0
403
3
2
2
1
1
4
44
3
43
2
42
1
414
c
icc
i
Jc
i
x
E
c
i
x
E
c
i
x
E
c
i
x
F
x
F
x
F
x
F
x
F
E
E
E
((Note))
),,(
ˆˆˆ
2
1
1
2
1
3
3
1
3
2
2
3
321
321
321
x
A
x
A
x
A
x
A
x
A
x
A
AAAxxx
xxx
AB
),,(4
3
34
2
24
1
1 x
Aic
xx
Aic
xx
Aic
xt
A
E
or
),,(4
3
3
4
4
2
2
4
4
1
1
4
x
A
x
A
x
A
x
A
x
A
x
Aic
E
where
4Ai
c .
4.5 Invariants of the field
FF is invariant under the Lorentz transformation
FFFFFFFFaaaaFF ''
iantinEEEc
BBBFF var)](1
[2 23
22
212
23
22
21
A further invariant is obtained by contraction of the field tensor with the “completely anti-symmetric unit tensor of fourth rank” defined by
=
0 if two indices are equal,
18
1 if )( is an even permutation of (1234), and -1 if )( is an odd permutation of (1234).
(Levi-Civita tensor)
One may be convinced easily that is a tensor of rank 4 because
'''''''' ' aaaa
Now we consider
BE c
iFFFFFF
8...2413132434121234
So the scalar product BE is Lorentz invariant, 4.6 Equation of continuity
FF
22
FFFFF
0)(2
1)(
2
1
Fxx
Fxx
FFxxx
F
x
Since
Jx
F0
we have
0
Jx
4.7 Maxwell's equation using dual tensor
Using the electromagnetic tensor
19
0
0
0
0
)(
321
312
213
123
Ec
iE
c
iE
c
i
Ec
iBB
Ec
iBB
Ec
iBB
F
the dual tensor G is defined as
FG2
1
or
0
0
0
0
0
0
0
0
)(
321
312
213
123
123123
121442
311434
234234
BBB
BEc
iE
c
i
BEc
iE
c
i
BEc
iE
c
i
FFF
FFF
FFF
FFF
G
Note that
GF2
1 .
Using the Jacobi identity
0
x
F
x
F
x
F
we get
20
0)(
3
x
F
x
F
x
F
x
F
x
F
x
F
x
G
since
.
Then we have the Maxwell's equation,
0
x
G.
(a)
0123
4
14
3
13
2
12
1
111
tic
B
z
E
c
i
y
E
c
i
x
G
x
G
x
G
x
G
x
G
or
11)( Bt
E
(b)
0213
4
24
3
23
2
22
1
212
tic
B
z
E
c
i
x
E
c
i
x
G
x
G
x
G
x
G
x
G
or
22)( Bt
E
(c)
0312
4
34
3
33
2
32
1
313
tic
B
y
E
c
i
x
E
c
i
x
G
x
G
x
G
x
G
x
G
or
33)( Bt
E
21
(d)
0321
4
44
3
43
2
42
1
414
z
B
y
B
x
B
x
G
x
G
x
G
x
G
x
G
or
0 B Note
BE c
iEBEBEB
c
iFG
4)(
4332211
4.8 Summary
The Maxwell's equation can be expressed by
Jx
F0
,
and
0
x
G
using the tensors F and G. 5. Vector potential under the Lorentz transformation
)'1
,'('
)1
,(
ciA
ciA
A
A
AaA '
')( 1
AaA
22
2
14
33
22
2
11
1
)('
'
'
1'
Ac
c
iA
AA
AA
c
cAA
2
1
33
22
2
11
1'
'
'
1'
Ac
AA
AA
c
cAA
and
2
14
33
22
2
11
1
)''(
'
'
1
''
Ac
c
iA
AA
AA
c
cAA
2
1
33
22
2
11
1
)''(
'
'
1
''
Ac
AA
AA
c
cAA
6. E and B under the Lorentz transformation 6.1 Transformation
FaaF '
FFFaaaaFaaaaFaa '
or
')()(')()(' 11 FaaFaaFaaF TT
____________________________________________________________
)('
)('
'
233
322
11
BcEE
BcEE
EE
33
22
11
)('
)('
'
BvE
BvE
E
E
EE
)('
)('
'
233
322
11
Ec
BB
Ec
BB
BB
33
222
11
)1
('
)1
('
'
EvB
EvB
cB
cB
BB
23
____________________________________________________________
)''(
)''(
'
233
322
11
BcEE
BcEE
EE
33
22
11
)''(
)''(
'
BvE
BvE
E
E
EE
)''(
)''(
'
233
322
11
Ec
BB
Ec
BB
BB
323
222
11
)'1
'(
)'1
'(
'
EvB
EvB
cB
cB
BB
6.2 Choice of the frame S’ which has pure electric or pure magnetic fields
From the Sec.3.5, we find that
(1) 22
2 1EB
c = invariant under the Lorentz transformation
(2) BE = invariant under the Lorentz transformation
Here we assume that BE =0 and 01 2
22 EB
c
Then one can find a frame S’ in which (E’ = 0 and B’ ≠ 0) [pure magnetic field], or (B’ = 0 and E’ ≠ 0) [pure electric field]. The proof is given in the following. (a) Pure magnetic field (E’ = 0)
We assume that E’ = 0. From the Lorentz transformation, we have
0)('
0)('
0'
233
322
11
BcEE
BcEE
EE
or
223
332
1 0
vBBcE
vBBcE
E
The condition BE =0 is satisfied since
03232332211 BvBBvBBEBEBEBE
The condition 0'11 2
22'2
22 EBEB
cc can be rewritten as
0'1 22
22 BEB
c
This implies that one can find the frame where 0'2 B and E’ = 0.
24
((Note)) From the relation
223
332
1 0
vBBcE
vBBcE
E
we get
BvE (b) Pure electric field (B’ = 0)
Next we assume that B’ = 0. Then we have
0)('
0)('
0'
233
322
11
Ec
BB
Ec
BB
BB
, or
223
322
1 0
Ec
vB
Ec
vB
B
The condition BE =0 is satisfied since
0322322332211 EEc
vEE
c
vBEBEBEBE
The condition 0'11 2
22'2
22 EBEB
cc can be rewritten as
0'11 2
22
22 EEB
cc
This implies that one can find the frame where 0'2 E and B’ = 0. ((Note)) From the relation
223
322
1 0
Ec
vB
Ec
vB
B
we get
25
)(1
2EvB
c
7. Energy-momentum tensor and Maxwell’s stress 7.1 force density
We define the vector of the force density as f
fJF
Here we have
iii Ef )( BJ
where
321
321
ˆˆˆ
BBB
JJJ
zyx
BJ
)(
)(
)(
122133
311322
233211
BJBJEf
BJBJEf
BJBJEf
11
13223
41431321211111
)(
)(
E
icEc
iJBJB
JFJFJFJFJFf
JB
222 )( Ef JB
333 )( Ef JB
ci
c
i
JEc
iJE
c
iJE
c
iJFf
JEJE )(
33221144
7.2 Maxwell’s equation
The Maxwell’s equation is given by
26
Jx
F0
The current density:
),( icJ J
x
FFf
x
FFJFf
0
0
1
The left-hand side can be split into two terms,
Fx
FFFx
f
)(0
The second term:
Fx
FFx
FFx
FFx
FFx
F
2
1
2
1
2
1
2
1
or
)(4
1
2
1)(
2
1
FF
xF
xFF
xF
xFF
xF
Here we use the Jacobi identity;
0
Fx
Fx
Fx
(Jacobi identity)
Then we have
)(4
1
FF
xF
xF
The force density is rewritten as
x
TFFFF
xf
)4
1(
1
0
27
with the symmetric energy-momentum tensor (Maxwell’s stress tensor)
)4
1(
1
0
FFFFT
0)4
1(
1][
0
FFFFTTTr
7.3 Conservation law
JES
t
u
)1
(2
1 2
0
20 BE
u
)(00 TfS
t
where
)(1
0
BES
: pointing vector
SSG200
1
c : momentum of the field
)( BJEf
)1
(2
1)
1( 2
0
20
00 BE
ijjijiij BBEET
or
)1
(2
1)
1( 22
220 BE c
BBEEc
T ijjijiij
where 00
2 1
c
),()( icJ J
28
0
0
0
0
)(
321
312
213
123
Ec
iE
c
iE
c
i
Ec
iBB
Ec
iBB
Ec
iBB
F
FFFFT4
10
where
)](1
)[(2 23
22
212
23
22
21 EEE
cBBBFF
1313231
3232232
2121221
1
1
1
BBEEc
FF
BBEEc
FF
BBEEc
FF
30
122143
20
311342
10
323241
)(
)(
)(
Sc
iEBBE
c
iFF
Sc
iEBBE
c
iFF
Sc
iEBBE
c
iFF
)(1
)(1
)(1
)(1
23
22
21244
22
21
23233
21
23
22222
23
22
21211
EEEc
FF
BBEc
FF
BBEc
FF
BBEc
FF
The Maxwell’s stress tensor is given by
29
uEEEc
BBBT
cGiSc
iEBBE
c
iT
BBBEEEc
T
cGiSc
iEBBE
c
iT
BBEEc
T
BBBEEEc
T
cGiSc
iEBBE
c
iT
BBEEc
T
BBEEc
T
BBBEEEc
T
02
32
22
12
23
22
21440
3030
1221340
23
22
21
23
22
212330
2020
3113240
32322230
23
22
21
23
22
212220
1010
3232140
13132130
21212120
23
22
21
23
22
212110
)(2
1)(
2
1
)(
)(2
1)(
2
1
)(
1
)(2
1)(
2
1
)(
1
1
)(2
1)(
2
1
Explicitly, the elements of T are
uicGicGicG
icGTTT
icGTTT
icGTTT
T
321
3343331
2232221
1131211
)(
uTTT 332211
8. Lorentz force 8.1 Origin of the Lorentz force
Consider a particle of charge q moving with velocity v (along the x axis) with respect to the reference frame S in a region with electric and magnetic fields E and B. In the frame S, the Lorentz force on this charge is given by
))(),(,()( 23321 vBEqvBEqqEq BvEF
In the frame S’, the Lorentz force is given by
30
)',','('' 321 qEqEqEq EF
where q is a relativistic invariant and is at rest. The fields in S and S’ are related by
)('
)('
'
233
322
11
vBEE
vBEE
EE
Then we have
)(''
)(''
''
2333
3222
111
vBEqqEF
vBEqqEF
qEqEF
What is the relation between F and F’?
32333
23222
1111
)(''
)(''
''
FvBEqqEF
FvBEqqEF
FqEqEF
or
'
'
'
33
22
11
FF
FF
FF
8.2 force density and charge density
)( BJEf We choose the frame S’ in which the system with the charge density is at rest. We now calculate the force density vector
''' Ef when 0'J (the system is at rest). We note the Lorentz transformation of 4-dimensional vector, current density and charge density
31
),( icJ J
)''(
)''()''(
1
1411
Jc
vJJiJJ
Then we have
vvJ
'
'
1
The Lorentz transformation of E and B,
)('
)('
'
233
322
11
vBEE
vBEE
EE
Then we have
))('),(','(' 232
322
1 vBEvBEE f
or
))(),(,(' 23321 vBEvBEE f
In the frame S, the Lorentz force is given by
)(),(,()]([ 23321 vBEvBEE BvEf
Thus we have
33
22
11
'
'
'
ff
ff
ff
9. Lienard-Wiechert potential 9.1 Lienard-Wiechert potential
We now consider the Lienard-Wiechert potential
32
In the S’frame:
0'
'
1
4'
0
A
r
q
2
1
33
22
2
21
1
1
''
'
'
1
''
vA
AA
AA
c
vA
A
or
'
1
1
1
41
'
0
0
1
'
20
2
3
2
2
2
1
r
q
A
A
c
v
A
Then we get
23
22
21
20
2 '''
1
1
1
41
'
xxx
q
where
33
22
11
'
'
)('
xx
xx
vtxx
The scalar potential is given by
33
))(1()(
1
4)(
1
1
1
4 23
22
2210
23
22
21
220 xxvtx
q
xxvtx
q
or
*0
1
4 R
q
with
))(1()( 23
22
221
* xxvtxR
Similarly we have for the vector potential
)0,0,( 1AA with
*0
*0
22
2
1
1
4
1
41
'
R
qv
R
q
c
vc
v
A
The electric field E and the magnetic field B are given by
AEt
and AB
3*
2
0
)1(4 R
q RE
Ev
B 2c
where
),,( zyvtx R For a slow moving charge (v<<c), we can take for E the Coulomb field. Then w have
20
220
2 44 r
q
rc
q
c
rvrvE
vB
34
((Mathematica-10))
Lienard-Wiechert potential <<Calculus`VectorAnalysis` SetCoordinates[Cartesian[x,y,z]] Cartesian[x,y,z]
Rxvt212 y2 z2
t v x2 y2 z2 1 2
q4 0
1R
q
4 t vx2 y2 z2 1 2 0
A1
qv04
1R
q v0
4 t vx2 y2 z2 1 2
A={A1,0,0}
q v0
4 t vx2 y2 z2 1 2
, 0, 0
B1=Curl[A]//FullSimplify
0,q v z1 2 0
4 t v x2 y2 z2 1 232 ,
q v y12 0
4 t v x2 y2 z2 1 232 Electric field in the frame S
E1 Grad DA, t . 0 10c2 FullSimplify
qc2 v2 t v x4 c2 t v x2 y2 z2 1 2320
,
q y12
4 t v x2 y2 z2 1 2320,
q z12
4 t v x2 y2 z2 1 2320
V1={v,0,0} {v,0,0}
eq1
1c2
CrossV1, E1 Simplify
0,q v z1 2
4 c2 t v x2 y2 z2 1 2320,
q v y1 2
4 c2 t v x2 y2 z2 1 232 0
eq1 B1 .0
1c20
Simplify
{0,0,0}
9.2 Distribution of the electric field
35
3*
2
0
)1(4 R
q pRE
where
))(1()( 2222* zyvtxR
)1,1,( 22* zyvtx R
),,( zyvtxp R
Rp is the relative coordinate of the field point and the charge point. The electric field is along the position vector Rp. Rp is a vector from the instantaneous location of the charge in S to the point where E is measured in S. ((Mathematica-11)) The electric field of a charge moving with the constant speed with v ( = v/c) on the unit circle of the real space
Lienard-Wiechert problem; field for a uniformly moving charge <<Graphics`PlotField`
E1X_, _:x12
x2 12y23
.x Cos, y Sin Simplify
E1Y_, _:y12
x2 12y23
.x Cos, y Sin Simplify
E1X[,]
1 2 CosCos2 1 2 Sin232
E1Y[,]
1 2 SinCos2 1 2 Sin232
s1[_]:=Table[{{E1X[,],E1Y[,]},{E1X[,],E1Y[,]}},{,0,2 ,/32}] s2[_]:=ListPlotVectorField[ Evaluate[s1[]]//N//Chop,ColorFunctionHue, AspectRatioAutomatic,ScaleFactor1,FrameTrue,PlotPoints20,AxesOrigin{0,0},DeFaultColorHue[0.6],DisplayFunctionIdentity]
General ::spell1 : Possible spelling error : new symbol
name "DeFaultColor " is similar to existing symbol "DefaultColor ". More… b = 0, 0.04, 0.08, 0.12, 0.16 b = 0.20, 0.24, 0.28, 0.32, 0.36
36
ps1=Evaluate[Table[s2[],{,0,0.36,0.04}]];Show[GraphicsArray[Partition[ps1,5]],DisplayFunction$DisplayFunction]
-1.5-1-0.500.511.5-2-1
012
-1.5-1-0.500.511.5-2-1
012
-1.5-1-0.500.511.5-2-1
012
-1.5-1-0.500.511.5-2-1
012
-1.5-1-0.500.511.5-2-1
012
-2-1 0 1 2-2-1
012
-2-1 0 1 2-2-1
012
-2-1 0 1 2-2-1
012
-2-1 0 1 2-2-1
012
-2-1 0 1 2-2-1
012
GraphicsArray b = 0.4, 0.42, 0.44, 0.46, 0.48 b = 0.50, 0.52, 0.54, 0.56, 0.58 ps2=Evaluate[Table[s2[],{,0.4,0.58,0.02}]];Show[GraphicsArray[Partition[ps2,5]],DisplayFunction$DisplayFunction]
-1-0.500.51-2
-1
0
1
2
-1-0.500.51-2
-1
0
1
2
-1-0.500.51-2
-1
0
12
-1-0.500.51-2
-1
01
2
-1-0.500.51-2
-1
01
2
-1.5-1-0.500.511.5-2-1
012
-1.5-1-0.500.511.5-2-1
012
-1.5-1-0.500.511.5-2-1
012
-1.5-1-0.500.511.5-2-1
012
-1.5-1-0.500.511.5-2-1
012
GraphicsArray b=0.6, 0.62, 0.64, 0.66, 0.68 b = 0.70, 0.72, 0.74, 0.76, 0.78 ps3=Evaluate[Table[s2[],{,0.6,0.78,0.02}]];Show[GraphicsArray[Partition[ps3,5]],DisplayFunction$DisplayFunction]
37
-0.75-0.5-0.2500.250.50.75
-2
-1
0
1
2
-0.75-0.5-0.2500.250.50.75
-2
-1
0
1
2
-0.75-0.5-0.2500.250.50.75
-2
-1
0
1
2
-0.75-0.5-0.2500.250.50.75
-2
-1
0
1
2
-0.75-0.5-0.2500.250.50.75
-2
-1
0
1
2
-1-0.500.51-2
-1
0
1
2
-1-0.500.51-2
-1
0
1
2
-1-0.500.5 1
-2
-1
0
1
2
-1-0.500.5 1
-2
-1
0
1
2
-1-0.500.51
-2
-1
0
1
2
GraphicsArray b = 0.80, 0.91, 0.82, 0.83, 0.84, b = 0.85, 0.86, 0.87, 0.88, 0.89 ps4=Evaluate[Table[s2[],{,0.8,0.89,0.01}]];Show[GraphicsArray[Partition[ps4,5]],DisplayFunction$DisplayFunction]
38
-0.6-0.4-0.200.20.40.6-3
-2
-1
0
1
2
3
-0.6-0.4-0.200.20.40.6-3
-2
-1
0
1
2
3
-0.6-0.4-0.200.20.40.6-3
-2
-1
0
1
2
3
-0.6-0.4-0.200.20.40.6-3
-2
-1
0
1
2
3
-0.6-0.4-0.200.20.40.6
-3
-2
-1
0
1
2
3
-0.75-0.5-0.2500.250.50.75
-2
-1
0
1
2
-0.6-0.4-0.200.20.40.6
-2
-1
0
1
2
-0.6-0.4-0.200.20.40.6
-2
-1
0
1
2
-0.6-0.4-0.200.20.40.6
-2
-1
0
1
2
-0.6-0.4-0.200.20.40.6
-2
-1
0
1
2
GraphicsArray b=0.9, 0.91, 0.92, 0.93, 0.94 b=0.95, 0.96, 0.97, 0.98, 0.99 ps5=Evaluate[Table[s2[],{,0.9,0.99,0.01}]];Show[GraphicsArray[Partition[ps5,5]],DisplayFunction$DisplayFunction]
39
-0.4-0.200.20.4
-4
-2
0
2
4
-0.4-0.200.20.4
-4
-2
0
2
4
-0.4-0.200.20.4
-4
-2
0
2
4
-0.4-0.200.20.4-6
-4
-2
0
2
4
6
-0.4-0.200.20.4
-7.5
-5
-2.5
0
2.5
5
7.5
-0.6-0.4-0.200.20.40.6
-3
-2
-1
0
1
2
3
-0.6-0.4-0.200.20.40.6
-3
-2
-1
0
1
2
3
-0.6-0.4-0.200.20.40.6
-3
-2
-1
0
1
2
3
-0.4-0.200.20.4
-3
-2
-1
0
1
2
3
-0.4-0.200.20.4-4
-2
0
2
4
GraphicsArray
10. Relativity of Electric field and magnetic field
40
E = 0 and B ≠ 0
We consider the charge q moving along the x axis in the presence of the magnetic field B (the frame S). In the frame S, there is only an external magnetic field B. Thus the magnetic force on the charge is given by
)( BvF q Suppose that there is no electric field (E = 0) in the frame S ( )0B . The E’ and B’ in the frame S’ are related to those in the frame S as
2233
3322
11
)('
)('
0'
vBBcEE
vBBcEE
EE
3233
2322
11
)('
)('
'
BEc
BB
BEc
BB
BB
or
')(' BvBvE (1) Then the force (electric force) on the charge q in the frame S’ is
)('' BvEF qq since the charge q is the same for any frame and the particle is at rest in the frame S’. There is no force due to B’ since the particle is at rest in the frame S’. 'F is the force of F’ in a direction perpendicular to the velocity v. Thus we have
'1
FF
41
11. Derivation of the Biot Savart law B’= 0 and E’≠ 0.
We consider that the magnetic field B’=0 in the frame S’. In the frame S’, there is only an external electric field E’ (the point charge is at rest). The E and B in the frame S are related to those in the frame S’ as
'
'
'
33
22
11
EE
EE
EE
''
'
0
2223
322
1
vEc
Ec
B
vEc
B
B
or
)(1
)'(22
EvEvB cc
, (2)
Using the result from the Lienard-Wiechert potential (<<1) (see Sec.8)
30
3*
2
0 4)1(
4 r
q
R
q rRE
30
30
22 44
1)(
1
r
q
r
q
cc
rvrvEvB
which is the application of the Biot-Savart law to a point charge. 12. Ampere’s law (Feynman 13-9)
We consider that the electrons located on the linear chain (the line density –0) moves at the velocity v. At the same time there are positive ions located on the same chain (the line density 0). We now consider the frame S’ which moves at the velocity v.
42
((Formula))
' where for the frame where the particle moves at the velocity v along the x axis, and ' for the frame where the particle is at rest. We assume that (1) The line densities of electrons and positive ions are given by 0 and 0 in the
frame S. (2) The line densities of electrons and positive ions are given by and in the
frame S’
)()( 0 or 02
2
0 11
c
v for electrons
0 or 0
2
20
1
1
c
v
for ions
The net line charge density in the frame S’ is
43
02
2
02
2
2
202
2
0
2
2
1
11
1
1'
c
v
c
v
c
vc
v
c
v
______________________________________________________________________ ((Note))
This relation can be also derived from the Lorentz transformation of the 4-dimensional current density
),(),( icicJ vJ
JaJ '
)('
)()('
1
1411
Jc
vJJiJJ
' JaJ
)''(
)''()''(
1
1411
Jc
vJJiJJ
Here we define
''
A
A
A is the same for the S and S’, since the plane of A is perpendicular to v.
02
2
01 )(' c
vv
cJ
c
where vAJI )( 011 and = 0.
So the positive line density produces an electric field E’. We use the Gauss’s law.
44
The electric field E’ at the distance s from the axis of the cylinder,
)'(1
)2('0
hshE
where s is the radius of the Gaussian surface (cylinder). or
02
2
00 2
1
2
''
c
v
ssE
So there is an electrical force on q in S’;
02
2
02''
c
v
s
qqEF .
But if there is a force on the test charge q in S’, there must be one in S. In fact, one can calculate it by using the transformation rules for forces. Since q is at rest S’ and F is perpendicular to the x axis. Then we have
FF ' or '1
FF
Using this result we have
vs
vqvv
s
q
c
v
s
qFF
2
)()(
22'
1 000
002
2
0
where s
vB
2
)( 00 is a magnetic field due to the line current density 0v (Ampere’s
law). The force has a form as qvBF .
45
13. Derivation of the Ampere’s law from relativity
We analyze the fields and currents as viewed from two frames. S where the ions are at rest. S’ where the electrons are, on the average, at rest.
),( icJ J
'')( 1
JaJaJ
Multiplying the cross-sectional area (A) of the wires, we obtain the following transformation for currents and linear charge densities.
),(),( icicAAAJI IJ
' IaI
)''()''(
)''()''(
1414
1411
icIc
viIIiicI
vIIiII
or
)''(
)''(
2
Ic
v
vII
where A , the subscript 1 is neglected and the plus and minus subscript refer to the ions and the electrons, respectively. In S’ we know that 0'I since the electrons are at rest.
')''(2 I
c
v
In S the net charge per unit length must vanish.
'0 or
'
46
The fields in S’ due to ' are
0'
'2
''
0
B
eE rr
The fields in S due to are
0
2 0
B
eE rr
We now consider the field transformation for
0'
'2
''
0
B
eE rr
noting that 'ˆˆ rr (perpendicular to the x axis), we find that the fields in S are
'
')''(
0'
33
2322
11
EE
EBcEE
EE
')''(
')''(
0'
22323
32322
11
Ec
vcBE
cB
Ec
vEcB
cB
BB
or
47
)'(
,'2
2
0
EvB
eE
c
rr
Then the total fields in the frame S are
rc
v
rcc
r
rxr
r
)(
2
'
'2
')()'(
0)'(2
200
22
0
eeevEvBBBB
eEEE
Since )''( vII and 0'I , we have
' vI Using eee rx , we obtain
02
0
E
eB
r
I
We see that a magnetic field due to current flow is a relativistic effect. 14. Capacitance moving along the x axis with a uniform velocity 14.1 The capacitance moves along the x direction which is parallel to the electric
field of the capacitance.
In the frame S’ where the charges are at rest.
0'
0'
''
3
2
01
E
E
E
48
and
0'
0'
0'
3
2
1
B
B
B
0
0
'
3
2
11
E
E
EE
0
0
0
3
2
1
B
B
B
Thus we have
'11 EE where ' 14.2 The capacitance moves along the x direction which is perpendicular to the
electric field of the capacitance.
In the frame S’ where the charges are at rest.
03
2
1
''
0'
0'
E
E
E
0'
0'
0'
3
2
1
B
B
B
where
'
49
003323
322
11
'')''(
0)''(
0'
EEBcE
BcEE
EE
0)''(
'
1
')''(
0'
323
02
022
3322
11
BEc
B
c
v
c
v
c
EE
cBB
BB
15. Relativistic-covariant Lagrangian formalism 15.1 Lagrangian L (simple case) Proper time
22' dxdxdxdxdxaadx
We define the proper time as
23
22
21
2223
22
21
222 )'()'()'()'()()()()( dxdxdxdtcdxdxdxdtcds
)1()(]})()()[(1
1{)(2
222232221
2222
cdtc
dt
dx
dt
dx
dt
dx
cdtcds
u
or
2
21
cdt
c
dsd
u
where is a proper time and u is the velocity of the particle in the frame S.
The integral b
a
ds taken between a given pair of world points has its maximum value if it
is taken along the straight line joining two points.
b
a
b
a
t
t
t
t
b
a
Ldtc
dtcdsS2
2
1u
where
50
2
2
1c
cLu
Nonrelativistic case
ccc
cc
cL 22
22/1
2
2
2)
21()1( u
uu
In the classical mechanics,
22
m
c
or mc
Therefore the Lagrangian L is given by
2/12
22 )1(
cmcL
u
The momentum p is defined by
d
dt
dt
dm
d
dmm
c
mL rruu
u
u
up
)(
12
2
((Note)) This momentum coincides with the components of four-vector momentum p defined by
d
dxmp
15.2 Hamiltonian
The Hamiltonian H is defined by
E
c
mcmc
mcmmcmLH
2
2
22
22222
1
)()(
)(
)(
1)(
uu
u
uu
uuuup
or
51
2
2
2
1c
mcE
u
We have
222
2
2
222
222
2
2
22
2
2
1
)1(
1p
u
uu
u
cm
c
mc
cm
c
cm
c
E
15.3 Lagrangian form in the presence of an electromagnetic field
The action function for a charge in an electromagnetic field
)( dxqAmcdsSb
a
where the second term is invariant under the Lorentz transformation.
)1
,( ciA A , and ),,,( 321 icdtdxdxdxdx
Then we have
b
a
b
a
dtqc
mcdxqAmcdsS )](1[)(2
22 uA
u
The integrand in the Lagrangian function of a charge (q) in the electromagnetic field,
)(12
22 uA
uq
cmcL
Au
u
up q
c
mL
2
2
1
where
)1
,( ciA A
The Hamiltonian H is given by
52
)1(
12
22
2
2
2
qqc
mce
c
mLH
uAu
uAu
uup
or
)1(
12
22
2
2
2
qqc
mce
c
mLH
uAu
uAu
uup
q
c
mcH
2
2
2
1u
or
222
2
2
222
222
2
)(1
)1(Ap
u
uu
qcm
c
mc
cm
c
qH
15.4 Expression for the Lagrangian in terms of 4-dimensional velocity
Here we use d instead of dt in the expression of Lagrangian.
cdds
is a four-dimensional velocity defined by
))(,)(,)(,)(( 321 uuuu
icuuu
dt
dx
d
dt
d
dx
))((44332211 AuuAAAAA
since
)1
,( ciA A ,
d
dtic
dt
dx
d
dt 4
4
b
a
b
a
dqAmcdxqAmcdsS )()( 2
53
qAmcL 2
15.5 Lagrangian and Hamiltonian in terms of the field tensor F
)(2
)(2 23
22
212
23
22
21 EEE
cBBBFF
This is invariant under the Lorentz transformation. We may try the Lagrangian density
AJFFL
04
1
By regarding each component of A as an independent field, we find that
the Lagrange equation
])(
[
x
AL
xA
L
is equivalent to
Jx
F0
.
The Hamiltonian density Hem for the free Maxwell field can be evaluated as follows.
FFLem
04
1
)1
(2
1)( 2
22
0
44
0
4
4
4
EBcx
AF
FL
x
A
x
AL
H emem
em
or
EBE 02
0
20 2
1
2
1emH
54
drdddHem )2
1(
2
1)()
2
1(
2
1 2
0
200
2
0
20 BErErBEr
((Note))
0)()(])([)( aErErEErE dddd
where E vanishes sufficiently rapidly at infinity.
00
E (in this case).
16. Relativistic form of Newton’s law 16.1 Relativistic force
We define the force F and the kinetic energy T as
uF
pF
Adt
dTdt
d
where u is the velocity of the particle.
2
2
2
2
2
2
2
1
1
1
1
1c
dt
dm
c
dt
dm
c
m
dt
d
dt
d
dt
dT
uu
u
uu
u
uu
puFu
or
2/3
2
22
22/1
2
2
11
1
cc
dt
d
m
c
dt
dm
dt
dTA
u
uu
uu
uu
or
2/3
2
2
2
2/3
2
22/3
2
2
2
2
2/3
2
2
2
2
1
2
1
]
11
1[
c
dt
dm
c
dt
dm
c
c
c
cdt
dm
dt
dTA
u
u
u
uu
u
u
u
uu
u ]
55
22/3
2
2
1
1
2u
ud
c
mdT
C
c
mcT
2
2
2
1u
where C is a constant of integration. Since the kinetic energy may be taken as zero for u = 0. 16.2 Relativistic energy
Then we have C = =mc2.
2
2
2
2
1
mc
c
mcT
u
It is convenient to introduce a quantity E defined by
2
2
2
22 )(
1
mc
c
mcmcTE u
u
E is the energy of a free particle 17. Four-dimensional momentum 17.1. Definition
)(1
2
2
u
u
cdt
ccdtds
)(u dt
c
dsd
where is a proper time.
2
2
1
1)(
c
uu
56
Here we define the four-dimensional momentum
),())(,)(,)(,)(()( 321 c
Eiimcumumum
dt
dxmx
d
dmp puuuuu
This is exactly the same as the expressions of p obtained from the Lagrangian.
2
2
22
1
)(
c
mcmcE
uu
17.2 Lorentz transformation
This momentum is clearly a four-vector since dx is a Lorentz four-vector and m and
d are Lorentz scalar. In fact, under the Lorentz transformation
pax
ds
dmax
d
dmp ''
22' pppppaap
So this is invariant under the Lorentz transformation.
),(c
Eip p
2
414
33
22
2
411
1'
'
'
1'
ppip
pp
pp
pipp
or
2
1
33
22
2
1
1
1'
'
'
1'
cpEE
pp
pp
Ec
pp
and
57
2
414
33
22
2
411
1
''
'
'
1
''
ppip
pp
pp
pipp
or
2
1
33
22
2
1
1
1
''
'
'
1
''
cpEE
pp
pp
Ec
pp
18. Four-dimensional velocity (or proper 4-velocity)
)(1
2
2
u
u
cdt
ccdtds
)(u dt
c
dsd
where is a proper time.
2
2
1
1)(
c
uu
Here we define the four-dimensional velocity by
))(,)(,)(,)(( 321 uuuu
icuuu
dt
dx
d
dt
d
dx
where ui is the 3-velocity
dt
dxu i
i (i = 1, 2, 3)
a'
4
3
2
1
4
3
2
1
00
0100
0010
00
'
'
'
'
i
i
or
' a
58
'
'
'
'
00
0100
0010
00
4
3
2
1
4
3
2
1
i
i
with
2
2
1
1
c
v
21. Force
)]'(''
'[
)'1(
1
'1
1
11 u dt
dmc
dt
dp
uc
dt
dp
dt
dt
dt
dp
)'1(
1
'
'
'
'
1
222
uc
dt
dp
dt
dp
dt
dt
dt
dp
)'1(
1
'
'
'
'
1
333
uc
dt
dp
dt
dp
dt
dt
dt
dp
When 0'1 u and 0'/'1 dtdu
)]'(''
'[
)'1(
1
'
' 1
1
11 u dt
dmc
dt
dp
uc
dt
dp
dt
dt
dt
dp
'
'1 22
dt
dp
dt
dp
'
'1 33
dt
dp
dt
dp
22. Minkowski force
We define the Minkowski force as
59
d
dpK
This is a 4-dimesional vector. The spatial components of K are related to the ordinary force by
Fu
p
u
ppK
22 1
1
1
1
dt
d
dt
d
d
dt
d
d
The 4th componenet
dt
dE
cdt
dp
d
dt
d
dpK
1444
23 Lorentz force in the relativistic mechanics 23.1
)]([ BvEp
F qdt
d
holds in an arbitrary frame S. This expression is the correct relativistic form for Newton’s second law. The momentum form is more fundamental.
The four-dimensional momentum is given by
2
2
1c
mv
vp
2
2
2
1c
mcEkin
v
222
2
2
222
222
2
2
22
2
2
1
)1(
1p
v
vv
v
cm
c
mc
cm
c
cm
c
E
or
2/1222 )( p cmcEkin
The final form of the equation of motion is given by
60
)]([ BvEp qdt
d (1)
2
2
1c
v
vp
,
)( EvvF qdt
dEkin
where
222
2
2
2
1
pv
cmc
c
mcEkin (2)
24. Cyclotron motion: a particle in a uniform magnetic field along the z axis.
We now consider the case of E = 0.
0kinEdt
d
Thus we have 2
2
2
1
1)(
mc
E
c
kin
v
v = constant
The momentum:
2c
Ekinvp
The equation of motion
)(2
Bvv qE
c
dt
d
kin
or
61
0
2
2
z
xkin
y
ykin
x
v
vE
qBcv
vE
qBcv
We use the complex plane for the solution.
)()(2
yxkin
yx ivvE
qBicivv
dt
d
or
)](exp[]exp[)()(2
00 tivE
qBticivvivv
kinyxyx
where
kinE
qBc2
i
yx veivv 00
Then we have
)sin(
)cos(
tvdt
dyv
tvdt
dxv
x
x
or
tconsvvv yx tan222
1
1
)cos(
)sin(
ytv
y
xtv
x
This equation describes a cyclotron motion (circular motion with radius R).
62
qB
p
qBc
vEvR kin
2
where w is the angular frequency,
kinE
qBc2
or
pq
BR1
The radius has a maximum when 2
1
c
v
In summary
)cos(
)sin(
220
20
220
20
tE
qBc
qB
ppy
tE
qBc
qB
ppx
kin
yx
kin
yx
25. The motion of the particle under an electric field ( E )
dt
dqqvqE
dt
dkin )( Ev
or
0)( qEdt
dkin
or
qEkin =constant
We now consider the capacitance consisting of two parallel planes. Suppose that the particle with charge q on the one plate moves to the other plate. The initial velocity is equal to zero. What is the velocity of the particle arriving at the other plate?
63
12
2 qmcqEkin
When 21 ,
22
2
2 )1(1
1
mc
q
c
v
or
2/1
22
])1(
11[
mc
qcv
26. Equation of motion under a constant electric field
We assume that E is along the y axis. The initial momentum p0 is in the (x, y) plane. The particle is at the origin at t = 0.
Ep qdt
d (1)
tqEpp 0
or
)0,,( 00 yx pqEtp p
2/1
022222
02
02422/1222 )]2()([)( qEtptEqcppccmcmcE yyxkin p
or
2/10
222220 )]2()[( qEtptEqcEE ykinkin
where 0kinE is the kinetic energy at the beginning of the motion (t = 0).
)( 20
20
2420yxkin ppccmE
22
2
2
1c
E
mc
Em
c
m kinkin vv
v
vp
64
Thus we have
)0,,()]2()[(
)0,,( 002/10
222220
2
00
22
yx
ykin
yxkinkin
pqEtpqEtptEqcE
cpqEtp
E
c
E
c
pv
or
0
)]2()[(
)(
)]2()[(
2/10
2222200
2
2/10
2222200
2
dt
dz
qEtptEqcE
qEtpc
dt
dy
qEtptEqcE
pc
dt
dx
ykin
y
ykin
x
Solving these differential equations (we use the Mathematica),
]2)([1 02
0222220
kinykin EEqtcptEqcEqE
y
or
])([0
20
2220 c
EpcmqEtp
qE
cy kin
xy
c
Ep
pcmqEtpqEtp
qE
cpx
kiny
xyyx0
0
20
222000
)(ln
0z
We now consider the special case when 00 yp .
2
0222
0222)( xx pcmpcmqEt
c
qEy
65
20
22
20
2220
)(ln
x
xx
pcm
pcmqEtqEt
qE
cpx
or
20
22
20
222
0
)()exp(
x
x
x pcm
pcmqEtqEt
cp
qEx
20
22
20
222
20
222
20
22
0
)(
)()exp(
x
x
x
x
y pcm
pcmqEtqEt
pcmqEtqEt
pcm
cp
qEx
20
22
20
222
0
)()cosh(
x
x
x pcm
pcmqEt
cp
qEx
)cosh()(0
20
2220
22220
22
xxxx cp
qExpcmpcmqEtpcm
c
qEy
or
]1)[cosh(]1)[cosh(]1)[cosh(0
20
22
0
0
0
0
x
xx
kin
x
kin
cp
qExpcm
qE
c
cp
qEx
qE
E
cp
qEx
qE
Ey
Thus in a uniform electric field, a charge q moves along a catenary curve. ((Mathematica-13))
Zimmerman 2D motion of a relativistic particle in a uniform electric field
Y1 px0
m,
q E0tm
, 0
px0
m,
E0 q t
m, 0
eq12
1 2
c2
Y1.Y1
2
1 2c2
px02
m2
E02 q2t2
m2
eq2=Solve[eq1,]//Simplify
c2 px02 E02q2 t2
c2 m2 px02 E02 q2 t2
, c2 px02 E02 q2 t2
c2 m2 px02 E02 q2 t2
66
eq31
2
c2. eq22 Simplify
c2 m2
c2 m2 px02 E02q2 t2 V={x'[t],y'[t],z'[t]} xt, yt, zt eq4=eq3 Y1
px0 c2 m2
c2 m2px02E02 q2t2
m,
E0 q t c2 m2
c2 m2px02E02q2 t2
m, 0
eq5=Table[V[[i]]eq4[[i]],{i,1,3}]
xt
px0 c2 m2
c2 m2px02E02q2 t2
m,
yt
E0 q t c2 m2
c2 m2px02E02q2 t2
m, zt 0
eq6=DSolve[{eq5,{x[0]0,y[0]0,z[0]0}},{x[t],y[t],z[t]},t]//Simplify
xt 1
E0 m q
px0 c2m2
c2 m2 px02 c2m2 px02 Log2 c2m2 px02
px0 c2 m2
c2 m2 px02 E02q2 t2 c2 m2 px02 E02q2 t2
Log2E0 q t c2 m2 px02 E02q2 t2
,
yt c2 m
c2 m2
c2 m2px02 c2 m2c2 m2px02E02 q2t2
E0 c2 m2c2 m2px02 q c2 m2
c2 m2px02E02 q2t2
, zt 0
rule1={m1,q1,E00.1,px00.1,c1} {m1,q1,E00.1,px00.1,c1} x[t_]=x[t]/.eq6[[1,1]]/.rule1//Simplify
0.698122
10. 1.01 0.01 t2 1
101. 1. t2Log2
0.1 t
1.01 0.01 t2
y[t_]=y[t]/.eq6[[1,2]]/.rule1//Simplify
1. 10.0499 1101.1.t2
1101.1.t2
z[t_]=z[t]/.eq6[[1,3]]/.rule1//Simplify 0
67
pl1=Plot[{x[t],y[t]},{t,0,30},PlotStyleTable[Hue[0.5 i],{i,0,1}], PrologAbsoluteThickness[2], PlotPoints50,BackgroundGrayLevel[0.7]]
5 10 15 20 25 30
5
10
15
20
Graphics Nonrelativistic motion
xnont_ px0t
m. rule1
0.1 t
ynont_ q E0t2
2 m.rule1
0.05 t2
pl2=Plot[{xnon[t],ynon[t]},{t,0,30},PlotStyleTable[Hue[0.3 i],{i,1,2}], PrologAbsoluteThickness[1.6], PlotPoints50,BackgroundGrayLevel[0.7]]
5 10 15 20 25 30
10
20
30
40
Graphics Show[pl1,pl2]
5 10 15 20 25 30
5
10
15
20
25
30
Graphics
68
28. A particle in a uniform electric field and a magnetic field Let the electric field E be parallel to the y axis and the magnetic field B parallel to the
z axis. At t = 0 the particle is at the point (0,0,0) and has a momentum p0. Lorentz invariant:
)]([ BvEFp qdt
d (1)
According to the Lorentz invariance, we have
''
'1
'1 2
222
22
BEBE
EBEB
cc
Since 0BE , we have 0'' BE (1) We assume a frame that 0'B . In this case, we have
0'11 2
22
22 EEB
cc
or
22
2 1EB
c
or
23
1E
cB
This is a condition for E and B. Using the Lorentz transformation, we have
0)('
)('
0'
233
322
11
BcEE
BcEE
EE
0)('
0)('
0'
233
322
11
Ec
BB
Ec
BB
BB
69
We choose 0'3 B
0' 233 Ec
BB
or
cE
Bcv
2
32
In this case,
0'
0'
'1
)('
0'
3
1
2322
E
E
EE
EBcEE
B
The frame S’ move relative to the frame S with a velocity v along the x axis. We know the equation of motion for the particle in a uniform electric field 'E along the y axis.
c
Ep
pcmtqEptqEp
qE
cpx
kiny
xyyx
''
')'''('''ln
'
'' 0
0
20
222000
]'
')'''(['
'0
20
2220 c
EpcmtqEp
qE
cy kin
xy
2/12
02
0220 )''(' yxkin ppcmcE
with cE
Bcv
2
32
The Lorentz transformation between ),(0
00
c
Eip kinp and )
','('
0
00
c
Eip kinp is given
by
70
)('
'
'
)('
0100
3003
0202
00101
cpEE
pp
pp
Ec
pp
kinkin
kin
The required equations of motion for the particle in the frame S is obtained using the Lorentz transformation.
)''(
'
'
)''(
414
33
22
411
xxix
xx
xx
xixx
)''(
'
'
)''(
txc
t
zz
yy
vtxx
(2) We assume a frame S’ that 0'E . In this case, we have
0'1 22
22 BEB
c
or
22
2 1EB
c
or
23
1E
cB
This is a condition for E and B. Using the Lorentz transformation, we have
0)('
)('
0'
233
322
11
BcEE
BcEE
EE
)('
0)('
0'
233
322
11
Ec
BB
Ec
BB
BB
71
We choose 0'2 E
0)(' 322 BcEE
or
3
2
B
Ev
In this case,
0'
0'
1)('
0'
2
1
3333
B
B
BvBc
BB
E
The frame S’ move relative to the frame S with a velocity v (=E2/B3<c) along the x axis. We know the equation of motion for the particle in a uniform electric field 'B along the z’ axis.
)'''
'cos(
'
'''
)'''
'sin(
'
'''
220
20
220
20
tE
qBc
qB
ppy
tE
qBc
qB
ppx
kin
yx
kin
yx
with v =E2/B3<c.
The Lorentz transformation between ),(0
00
c
Eip kinp and )
','('
0
00
c
Eip kinp is given
by
)('
'
'
)('
0100
3003
0202
00101
cpEE
pp
pp
Ec
pp
kinkin
kin
72
The required equations of motion for the particle in the frame S is obtained using the Lorentz transformation.
)''(
'
'
)''(
414
33
22
411
xxix
xx
xx
xixx
)''(
'
'
)''(
txc
t
zz
yy
vtxx
19. Doppler shift and aberration 19.1.
),(c
ik
pk
where ck
txk rk =invariant under the Lorentz transformation.
or
tkykxxk sincos
This should be equal to
'''sin'''cos'''' tykxkxk
Note that
''
)''(
)''(
2
ck
ck
xc
vtt
vtxx
73
Substituting these parameters into the invariant form, we have
0)'sin'sin(')'cos'cos(')cos'(' kkykkc
kvxkvckckt
This should be satisfied for any t’, x’, and y’.
'sin'sin
'cos')(cos
')cos(
kk
kc
vk
ckvck
19.2. Doppler shift
Since '
2',
2
kk
'
2)cos1(
2
c
v
or
)cos1('
c
v
19.3 Derivation of the formula using Mathematica ((Mathematica-14))
eq1=(k (x Cos[]+y Sin[])- t -k1 (x1 Cos[1]+y1 Sin[1])+ 1 t1)/.{c k,1c k1}//Simplify -c k t+c k1 t1+k x Cos[]-k1 x1 Cos[1]+k y Sin[]-k1 y1 Sin[1]
rule1 x x1 vt1, t t1
vc2
x1, y y1
x t1 v x1 , t t1
v x1
c2 , y y1
eq2=eq1/.rule1//Simplify
c k1 t1 c kt1v x1
c2 kt1 v x1 Cos
k1 x1 Cos1 k y1 Sin k1 y1 Sin1 Collect[eq2,{x1,y1,t1}]
t1c k1 c k k v Cos x1 k v
c k Cos k1 Cos1 y1k Sin k1 Sin1
74
rule2 k
2
, k1
2
1
k
2
, k1
2
1
eq3=(c k1-c k +k v Cos[])/.rule2//Simplify
2 c 1 v 1 Cos 1
eq4
kvc
k Cos k1Cos1 . rule2 Simplify
2 v 1c 1 Cos c Cos1c 1
eq5=k Sin[]-k1 Sin[1]/.rule2//Simplify
2 Sin
2 Sin11
eq31=Solve[eq30,1]//Simplify
1
c
c v Cos
1_ 1. eq311 . 1
1 v2
c2
Simplify
c 1 v2
c2
c v Cos Longitudinal Doppler shift 1[0]
c 1 v2
c2
c v Series[1[0],{v,0,4}]
v
c v2
2 c2 v3
2 c3
3 v4
8 c4 Ov5
Transverse Doppler shift
1
2
1
v2
c2
Series1
2, v, 0, 4
v2
2 c2 v4
8 c4 Ov5
f1
1
. v c .
180 FullSimplify
1 2
1 Cos 180
Plot[Evaluate[Table[f1,{,0,0.99,0.05}]],{,0,90},PlotStyleTable[Hue[0.05 i],{i,0,20}],PrologAbsoluteThickness[2],PlotRange{{0, 90},{0,7}},BackgroundGrayLevel[0.7],AxesLabel{"","'"}]
75
20 40 60 80
1
2
3
4
5
6
7'
Graphics
19.4. longitudinal Doppler shift 0 (red shift)
We suppose that a source is located at the origin of the reference frame S. An observer moves relative to S at velocity v. So that he is at rest in S’.
c
vc
v
c
v
1
1
)1('
If S’ moves toward S, rather than away from S, the signs in numerator and denominator of the radical would have been unterchanged. ((The red shift)) Wikipedia
The light from distant stars and more distant galaxies is not featureless, but has distinct spectral features characteristic of the atoms in the gases around the stars. When these spectra are examined, they are found to be shifted toward the red end of the spectrum. This shift is apparently a Doppler shift and indicates that essentially all of the galaxies are moving away from us. Using the results from the nearer ones, it becomes evident that the more distant galaxies are moving away from us faster. This is the kind of result one would expect for an expanding universe.
The building up of methods for measuring distance to stars and galaxies led Hubble to the fact that the red shift (recession speed) is proportional to distance. If this proportionality (called Hubble's Law) holds true, it can be used as a distance measuring tool itself.
The measured red shifts are usually stated in terms of a z parameter. The largest measured z values are associated with the quasars. ((Mathematica-16))
Red shift: l'/l vs b = v/c
eq11
1
1
1
76
Plot[eq1,{,0.4,0.99}, PlotStyle{Hue[0],Thickness[0.015]}, BackgroundGrayLevel[0.7],AxesLabel{"","'/"}]
0.4 0.5 0.6 0.7 0.8 0.9
2
4
6
8
10
12
14
'
Graphics z-parameter = (l'-l)/l=(l'/l)-1 z=eq1-1
1 1
1 Plot[z,{,0.4,0.99}, PlotStyle{Hue[0.5],Thickness[0.015]}, BackgroundGrayLevel[0.7],AxesLabel{"","z"}]
0.4 0.5 0.6 0.7 0.8 0.9
2
4
6
8
10
12
z
Graphics
19.5. Transverse Doppler shift (2
)
2
2
1'c
v
19.6. Aberration From these equations
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'cos')(cos
')cos1(
kc
vk
kc
vk
we have
cos1
cos'cos
c
vc
v
For low velocity we can neglect 22 / cv and higher-order terms. Setting '
sincos)cos( and
2sincos
cos1
cos
c
vc
v
Then we have
sin ((Mathematica-17))
Aberration eq1=Cos[+] Cos[+] Series[eq1,{,0,1}] Cos Sin O2
eq11=Cos[]-Sin[] Cos[]- Sin[]
eq2
Cos 1Cos
Cos1 Cos
Series[eq2,{,0,1}] Cos 1Cos2 O2
eq22 Cos 1Cos2 Cos 1Cos2 eq3=eq11eq22
Cos Sin Cos 1 Cos2 eq31=Solve[eq3,]//Simplify
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General ::spell1 : Possible spelling error : newsymbol name "eq31 " is similar to existing symbol "eq1 ". More…
{{ Sin[]}} REFERENCES 1. R.P. Feynman, R.B. Leighton, and M.L. Sands, The Feynman Lectures on
Physics: Commemorative Issue, Addison-Wesley Publishing Co. Inc., Reading, Massachussetts (1989).
2. C. Møller, Theory of Relativity 3. D.J. Griffiths, Introduction to Electrodynamics, Third Edition, Prentice Hall
(Upper Saddle River, New Jersey, 1999). 4. E.M. Purcell, Electricity and Magnetism, second edition, Berkley Physics Course-
Volume 2, McGraw-Hill Book Company (1985). 5. L.D. Landau and E.M. Lifshitz, The Classical Theory of Fields, Pergamon Press
1975). Appendix A-1 Capacitance The capacitor moves at the constant speed in the x direction. In the S’ system
0'
0'
'
1
3
0
02
E
E
E
From the Lorentz transformation, the electric field and the magnetic field in the S system are given by
0)''(
')''(
0'
233
0
02322
11
BcEE
EBcEE
EE
79
0
02233
322
11
')''(
0)''(
0'
cE
cE
cBB
Ec
BB
BB
Since E2 is expressed by
0
0
02
E
or
0
A-2 Faraday’s law 1. A conducting rod moving through a uniform magnetic field
We consider a metal rod (conductor) which moves at a constant velocity (v) in a direction perpendicular to its length. Pervading the space through which the rod moves there is a uniform magnetic field B (//z) constant in time. There is no electric field in the reference frame F.
The rod contains charged particles that will move if a force is applied to them. Any charged that is carried along with the rod, such as the particle of charge q moves through the magnetic field B and thus experience a force.
)( Bvf q The direction of the force is dependent on the sign of the charge q.
When the rod is moving at constant speed and things have settled to a steady state, the force f must be balanced, at every point inside the rod, by an equal and opposite force. This can only arise from an electric field in the rod. The electric field develops in the following way. The force f pushes negative charges toward one end of the rod, leaving the other end positively charges. This goes on until these separated charges themselves cause an electric field E such that, everywhere in the interior of the rod,
0 fEq , Then the motion of charge relative to the rod ceases. This charge distribution causes an electric field outside the rod, as well as inside. Inside the rod, there has developed an
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electric field BvE , exerting a force qE which just balances the force
)( Bvq . Let us observe the system from a frame F’ that moves with the rod. What is the
magnetic field B’ and the electric field E’? Note that there is no electric field (E = 0) in the frame F ( )0B . The E’ and B’ in the frame F’ are related to those in the frame F as
0)('
)('
0'
2233
3322
11
vBBcEE
vBBcEE
EE
,
3233
2322
11
)('
0)('
0'
BEc
BB
BEc
BB
BB
or
')(' BvBvE
where 1/1
122
cv for v<<c. The magnetic field B’ (= B) is almost equal to B.
The electric field E’ has only a component along the y’ axis (the same as y xis). The presence of the magnetic field B’ has no influence on the static charge distribution.
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2. A loop moving through a nonuniform magnetic field
F denotes the force which acts on a charge q that rides along with the loop. We evaluate the line integral of F, taken around the whole loop. On the two sides of the loop which lie parallel to the direction of motion, F is perpendicular to the path element ds. So there is no contribution to the line integral. Taking account of the contributions from the other two sides, each of length w, we have
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wBBqvdqdW banet )()( sBvsF
where
)(
)(
bb
aa
q
q
BvF
BvF
From the definition, we get
wBBqvdqqVW banet )( sE
or
wBBvdV ba )( sE
The electromotive force V is related in a very simple way to the rate of change of magnetic flux through the loop. The magnetic flux through a loop is the surface integral of B over a surface which has the loop for its boundary.
tvwBBtvwBtvwB baab )(
or
vwBB
ddt
d
dt
dddV
ba )(
)(
aBaEsE
We now consider the frame F’ attached to the loop.
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From the Lorentz transformation of E and B,
0)('
)('
0'
2233
3322
11
vBvBEE
vBvBEE
EE
,
3233
2322
11
)('
0)('
0'
BEc
BB
BEc
BB
BB
we have
aaa
aaa
BBB
vBvBE
'
',
bbb
bbb
BBB
vBvBE
'
'
For observers in the frame S’, Ea’ and Eb’ are genuine electric field. It is not an electrostatic field. The integral of E’ around the loop, which is the electromotive force V, is given by
)('' baba BBvwwvBwvBdV sE
which is the same as that obtained for the frame S. ________________________________________________________________________
