recapitulate...prof. shiva prasad, department of physics, iit bombay 3. symmetry arguments can be...

3Prof. Shiva Prasad, Department of Physics, IIT Bombay

Recapitulate• Postulates of Special theory of

Relativity.• Discussed Galilean Transformation.• Discussed the concept of time in an

example and shown that its equality in different frames assures that simultaneous events in a frame are also simultaneous in another frame.


Postulates of Special Theory of Relativity

1. Laws of Physics are same in all inertial frames of references. No preferred inertial frame exists.

2. The speed of light ‘c’ is same in all inertial frames.


Direct Transformation Galilean Transformation

′ = − = ==

, ' , ''

x x vt y y z zt t

Inverse Transformation

′ ′ ′ ′= + = =′=

, , x x vt y y z zt t


Velocity Transformation

, x x y y z zu u v u u u u′ ′ ′= − = =

Inverse Velocity Transformation

′ ′ ′= + = =, x x y y z zu u v u u u u


Galilean Transformation and

second Postulate• We now show that Galilean

transformation is not consistent with the second postulate.

• We shall also show that simultaneity of events is also relative under second postulate.


Classical Treatment of a pulse of light

Example 1

xO O’

z

x’

y y’

z’

S S’c


An observer in frame S sees a pulse of light emitted from origin at t=0, which is moving with a speed of ‘c’ in x-y plane making an angle of

with x-axis. Find the position

of the pulse in S at t=2x10-6 s, assuming it to be highly localized.

1 3tan4

−


Assume another observer in S’ which is moving relative to S with a speed of 0.6 c. Assume that the two frames obey the conditions of Galilean transformation described earlier. Find the speed of the light pulse and its co-ordinates at t=2x10-6 s in S’, under Galilean transformation.


The components of the velocity of the pulse in S frame are given as follows:

4c 0.8 c 53c 0.6 c 5

0

x

y

z

u

u

u

= × =

= × =

=


The co-ordinate of the pulse in S frame at t=2x10-6 s are given as follows:

6

6

0.8c 2 10 480 m

0.6 2 10 360 m

0

x

y

x u t

y u t c

z

−

−

= × = × × =

= × = × × =

=

Here we have taken the speed of light ‘c’ to be 3x108 m/s.


The co -ordinates of the pulse in S’ at a time t=2x10-6 s as given by Galilean transformation as follows.

6480 0.6c 2 10 120 m360 m

z =z=0

x x vty y

−′ = − = − × × =′ = =′


The resultant of speed of the light pulse in S’ frame can be obtained as follows

86

86

120 0.6 10 m/s2 10

360 1.8 10 m/s2 100

x

y

z

u

u

u

−

−

′ = = ××

′ = = ××

′ =


2 2 2 81.9 10 m/sx y zu u u u′ ′ ′ ′= + + ≈ ×

The speed of the pulse in S’ is given as follows.


• We thus see that speed of light is different in S’, violating second postulate.

• If we have to find a transformation in which the speed of light is maintained as ‘c’ even in S’, then it may have to change x’ and probably also t’ from what is given in Galilean transformation.


An observer is exactly half way in a running compartment of length L’. He shines light instead of throwing balls at t’=0, which travels both in the front and the back direction.

Event 1: Light reaches the front wall

Event 2: Light reaches the back wall

Example 2


v

S

S’

cc


According to S’

Both events are simultaneous, i.e., they occur at the same time which is given as follows:

2Ltc′

′ =


According to S

The two events again would have turned out to be simultaneous, if we had used the classical velocity transformation formula as in the case of balls.


However, under the second postulate, the speed of light is still ‘c’ in both the directions. But it has to travel a larger distance to reach the front wall than the back wall.

Hence Event 2 occurs before Event 1.


Time• Time is often related to the

simultaneity of two events.• We have just seen that

simultaneity also depends on frame under the second postulate.

• Probably we have a reason to believe that

t t′ ≠


Lorentz Transformation• We start afresh collecting all the

clues that we have.

• We do not derive but only give some arguments.

• Actual test lies in experimental verification.


1. Transformation should be linear.

1

2

3

4

xx xy xz xt

yx yy yz yt

zx zy zz zt

tx ty tz tt

x B x B y B z B t C

y B x B y B z B t C

z B x B y B z B t C

t B x B y B z B t C

′ = + + + +

′ = + + + +

′ = + + + +

′ = + + + +


Linearity is essential to maintain homogeneity of space. The length or rod should not depend on the origin chosen.

See: Introduction to Special Relativity by Robert Resnick, Wiley Eastern, 1988, for details


2. Special Choice of axes can make many coefficients zero.

O O’

z

x’

y y’

z’

S S’

Fixing the origins appropriately, we can get rid of constant terms.


Let us imagine that an event occurs at origin in S at t=0.This event would also appear to occur at the origin of S’ at t’=0. Substituting we get following.

1

2

3

4

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

xx xy xz xt

yx yy yz yt

zx zy zz zt

tx ty tz tt

B B B B C

B B B B C

B B B B C

B B B B C

= × + × + × + × +

= × + × + × + × +

= × + × + × + × +

= × + × + × + × +


This gives us

The transformation equations reduce to following.

1 2 3 4 0C C C C= = = =

1xx xy xz xtx B x B y B z B t C′ = + + + +

2yx yy yz yty B x B y B z B t C′ = + + + +

3zx zy zz ztz B x B y B z B t C′ = + + + +

4tx ty tz ttt B x B y B z B t C′ = + + + +


Fixing the planes appropriately, we can get rid of some other constants.

Let us imagine that an event occurs at an arbitrary time in x-y plane in Simplying z=0 for this event.

We can see that an observer in S’would also find it occurring in x’-y’ plane implying z’=0



2yx yy yz yty B x B y B z B t C′ = + + + +

30 0zx zy zz ztB x B y B B t C= + ×+ + +


This is possible only if Bzx ,Bzy and Bztare set to zero.


We can imagine similar event in x-zplane and get rid of three other constants finally getting following equations.


yxy B x′ = yy yzB y B z+ + ytB t+ 2C+

zxz B x′ = zyB y+ zz ztB z B t+ + 3C+



Now let us look at the y-z plane. At time t=0, if an event occurred in this plane, it would also appear to occur in y’-z’ plane to observer in S’. But at a later time the x-co-ordinate of this event would be shifted by v t. The transformation equation thus would appear as follows.


( )xx xyx B x vt B y′ = − + xzB z+ 1C+

yxy B x′ = yy yzB y B z+ + ytB t+ 2C+

zxz B x′ = zyB y+ zz ztB z B t+ + 3C+



Let us clean up the equations, which we could obtain without invoking any principle of relativity

( )xx

yy

zz

tx ty tz tt

x B x vty B y

z B zt B x B y B z B t

′ = −

′ =

′ =

′ = + + +


3. Symmetry arguments can be used to simplify the time transformation equation. Imagine two events occurring at the same time in S frame. Let the co-ordinates if the two events be (x0,y0,z0) and (x0,-y0,z0). If t’ depends on y, then time of these two events would appear to be different in different frame.


But what we call y-axis could have also been termed as –y. The choice of x-axis is unique as it is determined by the direction of relative velocity in our choice of axes, but not of y and z. Hence the transformation equations must appear as follows.


( )xx

yy

zz

tx ty

x B x vty B y

z B z

t B x B y

′ = −

′ =

′ =

′ = + tzB z+ ttB t+

Thus we are left with only with five constants.


( )xx

yy

zz

tx tt

x B x vty B y

z B zt B x B t

′ = −

′ =

′ =

′ = +

We note that Galilean Transformation is also a special case of these equations


• We discussed using examples that we have to attack the equation which makes time same in two frames.

• We then discussed the form of the transformation equations without invoking any postulates of relativity, only making time relative.

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recapitulate...prof. shiva prasad, department of physics, iit bombay 3. symmetry arguments can be...

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