Question

Using lorentz transformation obtain the velocity transformation. I am unable to calculate Ux', Uy', Uz'..... Where Ux'=dx'/dt'.... And x'=p(x-vt),,,p=(1-v^2/c^2)^-1/2

Answer 1

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 / tà                              x 2 + y’2 + z2 – c2t 2 = 0 
this is really hard question and lendhy i am solving this for proving that i know this concept 

(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

-c2  -  λ2 u2 +  c2 λ2 =0

C2 =λ2 (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

 dude it was written in my sheet so i just did it for u cause it developed my concept strong thanks for updating such questions