Naked Science Forum
On the Lighter Side => New Theories => Topic started by: Hal on 12/03/2021 12:04:35
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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’’ = γu 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’’ = γw γv 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 ?
Note that we will have three gamma factors: γu , γv , γw
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I got help from elsewhere. I think it is because I wrongly used linear addition of velocities.