Asymmetry and contradiction in Lorentz transformation and special relativity ?

[lidal]

 

 

Consider relatively moving inertial reference frames S, S’, S’’, with coordinates (x, y ), (x’, y’ ), ( x’’, y’’ ) , respectively. S’ and S’’ are moving with velocities v and u, respectively, relative to S, in the + x direction. At t = t’ = t”=0 , the origins of the reference frames coincide and the clocks in all frames are synchronized.

An event ( Event 1 ) occurs in S at: 

                               x = x₁   ,    t = t₁

To find the coordinates of Event 1 in S’ and in S’’, the traditional way is to use Lorentz transformation ( LT ) directly from S to S’, and directly from S to S’’.

My argument is that, if inertial frames have complete symmetry, we should get the same coordinates of the event in S’’ indirectly by using LT from S’ to S’’. This means that we first get the coordinates of the event in S’, by using LT from S to S’ . Then we use LT from S’ to S’’. We use the relative velocity w = u - v for the Lorentz transformation from S’ to S’’.

Traditional way

                  Use LT from S to S'

                  Use LT from S to S''

Alternative way required by symmetry

                  Use LT from S to S' ,  then from S'  to  S''  

                                       OR

                  Use LT from S to S'' ,  then from S''  to  S' 

 

 

Symmetry requires that the coordinates of the event in S’’ obtained using both approaches should agree. I find that this is not the case. The same argument applies to S’.

For example, using Lorentz transformation directly from S to S’’, the length of a rod in S’’ will be:

                                           L’’ = γ_u  L

However, using Lorentz transformation from S to S’, then from S’ to S’’ , the length of the same rod in S’’ will be:

                                  L’’ =  γ_w γ_v  L ( 1 + vw/c² )  

where L is the length of the rod in S.

This leads to a contradiction:

                                        L’’ ≠  L’’  

Is there any flaw in this argument ?

 

 

[studiot]

You haven't posted your working, but it would seem to me that you have not taken into account the different time coordiantes in the different frames.

If

[math]v = \frac{{dx}}{{dt}}\left( {S'} \right)\;and\;u = \frac{{dx}}{{dt}}\left( {S''} \right)[/math]

then what are u and v in terms of t' and t'' ?

[swansont]

46 minutes ago, lidal said:

S’ and S’’ are moving with velocities v and u, respectively, relative to S

...

We use the relative velocity w = u - v for the Lorentz transformation from S’ to S’’.

w, u and v are only valid for frame S. You have to use the relativistic velocity addition formula for other frames.

i.e. If you were in S' and S'' was moving at 0.6c in one direction relative to S', and S moving in the other at 0.6c, your formula would have S'' moving at 1.2c relative to S, because you naively used a linear addition of velocities

 

 

 

[lidal]

31 minutes ago, swansont said:

w, u and v are only valid for frame S. You have to use the relativistic velocity addition formula for other frames.

i.e. If you were in S' and S'' was moving at 0.6c in one direction relative to S', and S moving in the other at 0.6c, your formula would have S'' moving at 1.2c relative to S, because you naively used a linear addition of velocities

 

 

 

 

Yes, I think it is because I used linear addition of velocities. Thank you for the help.