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Length contractionLength measured differs from frame to frame – another consequence of relativistic effect Gedankan experiment again!

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Two observers: O on Earth, O’ traveling to and fro from Earth and alpha centauri with speed uTotal distance between Earth - alpha centauri – Earth, according to O (Earth observer), = L0

O sees O’ return to Earth after t0

Observer O’ in a spaceship is heading AC with speed u and returns to Earth after t’ according to his clock

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Use some simple logics…

In O: 2L0 = ut0

In O’: 2L0’ = ut0’

Due to time dilation effect, t0’ is shorter than t0 , i.e. t0 > t0’

t0 is related to t0’ via a time dilation effect, t0’ = t0 / , hence

L0’ / L0 = t0’ /t0 = 1 / , or

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L0’ = L0 / L0 is defined as the proper length = length of object measured in the frame in which the object (in this case, the distance btw Earth and AC) is at rest

L0’ is the length measured in the O’ frame, which is moving wrp to the object

The length of a moving objected is measured to be shorter than the proper length – length contraction

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If an observer at rest wrp to an object measures its length to be L0 , an observer moving with a relative speed u wrp to the object will find the object to be shorter than its rest length by a foctor 1 /

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A stick moves to the right with a speed u. (a) The stick as viewed by a frame attached to it (b) The stick as seen by an observer in a frame at rest relative to the stick. The length measured in the rest frame is shorter than the proper length by a factor 1/

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Length contraction only happens along the direction of motion

In 3-D, the length contraction effect is a shortening of length plus a rotation

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An observer on Earth sees a spaceship at an altitude of 435 moving downward toward the Earth with a speed of 0.97c. What is the altitude of the spaceship as measured by an observer in the spaceship?

Solution One can consider the altitude seen by thestationary (Earth) observer as the proper length (say, L'). The observer in the spaceship should sees a contracted length, L, as compared to the proper length. Hence the moving observer in the ship finds the altitude to be L = L' / = 435 m x [1- (0.97)2]-1/2 = 106 m

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Relativistic kinematics

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Lorentz Transformation

All inertial frames are equivalentAll physical processes analysed in one frame can also be analysed in other inertial frame and yield consistent resultsA transformation law is required to related the space and time coordinates from one frame to another

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O' frame uses {x',y',z‘;t‘} to denote the coordinates of an event, whereas O frame uses {x,y,z;t}

How to related {x',y',z',t‘} to {x,y,z;t}

In Newtonian mechanics, we use Galilean transformation

But GT must not be valid when u c because it is not consistent with the constancy of the light speed postulate

The relativistic version of the transformation law is given by Lorentz transformation

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Derivation of Lorentz transformation

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Consider a rocket moving with a speed u (O' frame) along the xx' direction wrp to the stationary O frameA light pulse is emitted at the instant t' = t =0when the two origins of the two reference frames coincidethe light signal travels as a spherical wave at a constant speed c in both framesAfter some times t, the origin of the wave centered at O has a radius r = ct, wherer 2 = x2 + y2 + z2

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From the view point of O', after some times t'the origin of the wave, centered at O' has a radius r' = ct‘ , (r’ )'2 = (x)‘ 2 + (y)2 + (z)2

y'=y, z' = z (because the motion of O' is along the xx‘) axis – no change for y,z coordinatesThe transformation from x to x’ (and vice versa) must be linear, i.e. x’ xFrom the view point of O: x = ct corresponds to x ’ = 0, so we assume the form x’ = k(x - ct ); k some proportional constant;Likewise, from the view point of O’,x’ = -ct’ corresponds to x = 0, so we set x = k(x’ + ct’ )

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With r = ct , r’ = ct ’, x = k(x’ + ct’ ), x’ = k(x - ct),

we solve for {x',t'} in terms of {x,t } to obtain

)(

1

'2

utx

cu

utxx

xcut

cu

xcutt )/(

1

/' 2

2

2

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the constant k is identified as the Lorentz factor, Note that, now, the length and time interval measured become dependent of the state of motion (in terms of ) – in contrast to Newton’s viewpointLorentz transformation reduces to Galilean transformation when u << c (show this yourself)

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To recap

the LT given in the previous analysis

)(

1

'2

utx

cu

utxx

xcut

cu

xcutt )/(

1

/' 2

2

2

Which relates x’ , t’ to x, t’ , for which O’ is moving at velocity +u wrp to O

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O’O

I see O’ moving with a velocity +u

+u

I measures the coordinates of M as {x,t} Object

M

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From the view point of O’

Equivalently, we could also perform the analysis from the view point of O’ that O is moving in the –u direction.

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OO’

I see O moving with a velocity -u

-u

From the view point of O’

We wish to express x, t in terms of x’, t’

I measures the coordinates of M as {x’,t’}

Object M

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Mathematically, this simply means making the permutation:

uuxx ,'

The two transformations above are equivalent; use which is appropriate in a given question

)''()(' utxxutxx

')/(')/(' 22 xcuttxcutt

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Length contraction can be recovered from the LTConsider the rest length of a ruler as measured in frame O’ is L’ = x’ = x’2 - x’1 (the proper length) measured at the same instant in that frame, hence t’2 = t’1What is the length of the rule as measured by O?The length in O, according the LT is

L’x’ = x’2 - x’1 = (x2 - x1) – u(t2 -t1)] The length of the ruler in O is simply the distance btw x2 and x1 measured at the same instant in that frame, hence t2 = t1, hence L’ = L

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How would you recover time dilation from the LT?

DIY as an exercise during Raya

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Lorentz velocity transformation

O’

M is moving with a velocity +ux from my point of view Object

M

I see M moving with a velocity +ux’

I see O’ moving with a velocity +u

+u

O

Object M

O

How to relate the velocity in the O’ (u’x) frame to that of the O frame (ux)?

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By definition, ux = dx/dt, u’x = dx’/dt’

The velocity in the O’ frame can be obtained by taking the differentials of the Lorentz transformation,

)('),('2dx

c

udtdtudtdxdx

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Combining

21cuuuu

x

x

)(

)(

'

'

2

'

dxcu

dt

udtdx

dt

dxux

dtdx

cu

dtdt

dtdtu

dtdx

dt

dt

2

where we have made used of the definition ux = dx/dt

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21'

''

cuuuu

dt

dxu

x

xx

Compare the Lorentz transformation of velocity

with that of Galilean transformation of velocity:

uuudt

dx

dt

utxd

dt

dxu xx

)(

'

''

GT reduces to LT in the limit u << c

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Please try to understand the definition of ux , u’x , u so that you wont get confused

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LT is consistent with the constancy of speed of light

in either O or O’ frame, the speed of light seen must be the same, cSay object M is moving with speed of light as seen by O, i.e. ux = c

According to LT, the speed of M as seen by O’ is

c

ucc

uc

cuuc

ccuuc

cuuuu

ux

xx

1111

'

22

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That is, in either frame, both observers agree that the speed of light they measure is the same, c = 3 x 108m/s

In contrast, according to GT, the speed of light seen by O’ would be ucuuu xx '

Which is inconsistent with constancy of speed of light postulate

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21'

cuuuu

ux

xx

To recap

the LT given in the previous analysis relates u’x to ux in which O’ is moving with +u wrp to O,

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From the view point of O’

Equivalently, we could also perform the analysis from the view point of O’ that O is moving in the –u direction.

We would be able to express ux in terms of u’x in the same spirit as we derive u’x in terms of ux

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O

Object M

I see M moving with a velocity +ux’

I see M moving with a velocity ux

O’

I see O moving with a velocity -u

-u

From the view point of O’

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Mathematically, this simply means making the permutation:

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'1

'

1'

cuuuu

u

cuuuu

ux

xx

x

xx

uuuuuu xxxx ,','

The two transformations above are equivalent; use which is appropriate in a given question

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Example

Rocket 1 is approaching rocket 2 on ahead-on collision course. Each is moving at velocity 4c/5 relative to an independent observer midway between the two. With what velocity does rocket 2 approaches rocket 1?

C.f. In GT, their relative speed would just be 4c/5 + 4c/5 = 1.6 c – which violates constancy of speed of light postulate. See how LT handle this situation:

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The observer in the middle is the stationary frame, O

Choose rocket 1 as the moving frame O‘Call the velocity of rocket 2 as seen from rocket 1 u’x. This is the quantity we are interested in

Frame O' is moving in the +ve direction as seen in O, so u = +4c/5The velocity of rocket 2 as seen from O is in the -ve direction, so ux = - 4c/5

Now, what is the velocity of rocket 2 as seen from frame O', u ’x = ? (intuitively, u ’x must be in the negative direction)

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Use the LT

c

c

cc

cc

cuuuu

ux

xx 41

40

54

54

1

54

54

1'

2

2

i.e. the velocity of rocket 2 as seen from rocket 1 (the moving frame, O’) is –40c/41, which means that O’ sees rocket 2 moving in the –ve direction (to the left in the picture), as expected.

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Read other examples from the texts and also the lecture notes

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Selamat Hari Raya Aidilfitri