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Lorentz Transformation

Lorentz tried to replace Galilean transformation to give results in agreement with Michelson and Morley experiment and second postulate of the special theory of relativity and developed his transformation equations in 1890.

 

I. Coordinate transformation

 

Here we consider two inertial systems S and S’. We assume S’ moves in such a way that x and x’ are one straight line and have same positive directions. Also 0 and 0′ coincides at t = 0 and S’ moves with a velocity ‘v’ with respect to S. For our convenience, we take the axis of two coordinate systems coincide at t = t’ = 0. (Refer to Figure 8.1)

 

Let a pulse of light be generated at t = 0 at the origin which grows in the space.

 

Now we consider the situation when the pulse reaches at point P. Let (x, y, z, t) and (x’, y’, z’, t’) are the position and time coordinates of P measured by observers at 0 and 0′ in frames S and S’ respectively. When the pulse is observed from S, we have the velocity of light as distance divided by the time taken; hence ,

 

C  = √(x 2 + y2 + z2 c / t

 

à                              x 2 + y’2 + z2 – c2t 2 = 0

 

Similarly, when the pulse in observed from S’;

(x’)2 + (y’)2 + (z’)2 – c2(t’)2 = 0

Whereas                      y=y’ ; z=z’

 

Using equation (3) into equations (1) and (2) we have,

x2 + y2 + z2- c2t2 = (x’)2 + (y’)2 + (z’)2- c2(t’)2

x2 + y2 + z2 _ c2t2 = (x’)2+ y2 + z2 – c2(t’)2

x2 – c2t2 = (x’)2 – c2(t’)2

Let the transformation between x and x’ is represented by

x’ = A.(x – v t) … (5)

 

where λ. is independent of x and t. Suppose the system S in moving relative to S’ along x-direction (because the motion is relative) then,

 

x =λ ‘ (x’ +  v t’)

 

where A.’ being independent of x’ and t’

From equations (5) and (6) we have,

x = λ.’ (A.(x –v t) + v t’)

 

v t’ = x / λ’ (x- v t)

 

=λ .[x / λ λ -  x + v t]

 

= λ.[ v t- x( 1-1/λ λ)].

 

t’ = λ[t – x/v ( 1 – 1/λ λ) ]

 

Now using equations (7) and (5) into equation (4) we have

 

X2 – c2 t2 =[λ ( x –v t )]2 –c2 [ λ { t – x/ v ( 1 – 1 /λ λ )}] 2

 

X2 –c2 t2 –λ 2  [ x2 -2xvt +v2 t2  ] +c2 λ2 [  t2 -2xt /v ( 1 -1 /λ λ ) + (x /v)2  +1- 1 /λ λ)2 =0

 

This is an identity and hence coefficients of various powers of x and t must vanish separately. So, equating the coefficient of t2 from both sides to zero, we have,

This is an identity and hence coefficients of various powers of x and t must vanish separately. So, equating the coefficient of t2 from both sides to zero, we have

 

-c2  -  λ2 u2 +  c2 λ2 =0

C22 (c2 –u2)

Λ2 = c2 /c2 –u2

Now equating the coefficient of ‘xt’ from both sides to zero, we have,

 

2uλ2  +c2 λ2 [-2/u (1- 1/λλ)] =0

2uλ2 = 2c2 λ2 ( 1- 1/λλ’)]

U2 λλ c2 [λλ -1]

Λλ (c2 –u2 ) =c2

 

Comparing equations (8) and (11) we have,

 

Λλ (c2 –u2) =c2 =λ2 [c2 –v2]

Λλ=λ2 à                                λ =λ

 

So using equation  (10) λ =  λ c /√c2 –v2 =1 / √ 1-u2 /c2

 

Substituting   equation (9) into  equation (7) we have ,

 

T =1/√1-u2 /c2 [t –x/u (1/λ2)]

= 1/ √ 1-u2 /c2 [ t-x/u (t –x/u ( 1 –c2 u2 /c2 )]

= 1/ √ 1-u2 /c2  [t –x/u u2 /c2]

T = t- ux /c2 /√ 1-u2 /c2

 

Substituting equation (9) into equation (5) we have,

 

X = x ut / √ 1-u2 /c2

Hence  ,

II. Velocity Transformation

 

Let the velocities of the particle in S and S’ frames are u(u x, u y, u z) and u ‘(u/, u y‘, u z) respectively

 

So  , ux  =dx /dt and ux =dx /dt

 

However from equation (13)

X = λ (x –ut)

 

Where λ =1/ √(1 –u2 /c2 )

 

On differentiating

 

Dx = λ (dx –udt ) ; dy = dy ; dz = dz

 

Also from equation (12)   t  = λ ( t- xu /c2)

 

On differentiating

Dt = λ (dt –v/c2 dx )

 

Hence ,                        dx /dt =λ (dx –udt)  / λ ( dt –u c2 dx  = dx –udt /dt –v /c2 dx

 

Dy /dt = dy /λ ( dt – u/c2 dx)

Dz dt = dz  /( dt – u/c2 dx)

 

Using equation (14) we have,

Similarly,                 u y = uy/λ[1- u/c 2 u x]

u z = uy/λ[1- u/c 2 u x]

 

Also from equation (17)

 

U x –v/c2 u x u x =u x –v

From equation (18)

 

Λu y= 1 –v/c2 ux ] =uy

Λu y = [1-v/c2 { ux +v /1+(v/c2 ) ux }] =λu y [ 1 –(v/c2 u x +v2 /c2 /1+v /c2 ux]

= λu y [ 1- v2 /c2 /1+v/c2 u x]

 

= u y = u y √ 1- v2 /c2 /1+v /c2 u x

Similarly                       = u y = u y √ 1- v2 /c2 /1+v /c2 u x

 

(a) For Non-relativistic Case

here, v << c and A = 1

 

so                                                         u x =u x +v1 u y =u y ; u z =u z

 

(b) When the motion is along x-axis only

Here                                                    u x =u x; u y =0 u z =0

It proves the constancy of the speed of light in any inertial reference frame.

 

(d) For v > c

 

Here the left hand side becomes negative and it has no physical significance.