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This is from A.P. French Special Relativity book, Chapter 3 (page 78)

Setup of the proof: $S$ and $S'$ be inertial reference frame. $S'$ move to the right with respect to $S$ at velocity $v$.

Let co-ordinates in $S$ be $(x,t)$ and co-ordinates in $S'$ be $(x',t')$

Equation (3-8) in the book, he writes that transformation will be of the form:

$x = ax' + bt'$ and by symmetry of the reference frames as implied by relativity principle, $x' = ax - bt$

My question:

How does symmetry of reference frame argument lead to the above conclusion? For e.g. why can't the second equation above be of the form $x' = -ax - bt$ or maybe $x' = -ax + bt$. These equations look as symmetric to me (mathematically!) as the one author uses. (I know I'm wrong but want to understand more clearly why am I wrong)

Thanks

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Without loss of generality, let's assume $v$ is positive. From the viewpoint of the S' frame, the S frame is moving to the left with velocity $v$. (Equivalently, S moves to the right with velocity $-v$). Now, if we make a video of this, and play the video backwards, it will look like S is moving to the right with velocity $v$.

Playing the video backwards is equivalent to replacing $t$ with $-t$. So the transformation from S coordinates to S' coordinates is the same as the transformation from S' coordinates to S coordinates, but with time reversed.

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  • $\begingroup$ thanks - i really liked this argument. but i guess it keeps coming back to a Q i have still not been able to reconcile with - "Equivalently, S moves to the right with velocity −v". Why? $\endgroup$
    – aman_cc
    May 24 '20 at 8:46
  • $\begingroup$ @aman_cc Maybe this will help. If you start at position O and face left and walk at 1 m/s for 3 seconds you end up 3 m to the left of O. And if you start at position O and face right but walk backwards at 1 m/s for 3 seconds you also end up 3 m to the left of O. That is, walking backwards at $v$ is the same as walking forwards at $-v$. And vice versa. $\endgroup$
    – PM 2Ring
    May 24 '20 at 8:55
  • $\begingroup$ yes of course as measured from a same reference frame S. But we now measure velocity of S from S' (and it is -v as you say in your answer - when every measurement of space time becomes reference frame dependent why should this velocity of S in the reference frame of S' be v) $\endgroup$
    – aman_cc
    May 24 '20 at 11:52
  • $\begingroup$ @aman_cc Ok. For the moment, let's assume that $v\ll c$, so we can ignore time dilation & length contraction, and see what happens in Galilean relativity. If I'm at rest in the S frame and you're at rest in the S' frame, we agree that the magnitude of our relative speed is $v$. But I say that you are moving with a speed of $v$ to the right in my frame, and you say that I'm moving to the left at a speed of $v$ in your frame. Do you agree? $\endgroup$
    – PM 2Ring
    May 24 '20 at 12:09
  • $\begingroup$ yes this is fine (and I can live with it). though I do think the Q we are trying to address is maybe a more basic thing (i think it is some symmetry argument ultimately that will break if we do not assume that each reference frame measures the speed of other as same but direction as opposite - what is that exact argument I don't know) - time dilation and length contractions are very natural consequences of lorentz transformations so in a way I think you are using consequences of the derivation to justify the derivation $\endgroup$
    – aman_cc
    May 24 '20 at 14:47
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$b$ has to have the dimension of a velocity to be consistent. When the reference frame $S$ looks at the reference frame $S^\prime$, he sees it has moving to the right with velocity $+v$. But if you swap the frames, in the $S^\prime$ frame, if you look at the $S$ frame, you'll see it moving with the same velocity as before but in the opposite direction, so $-v$. There's no reason to let even the $a$ change sign since swapping refrence frames just swaps the velocity of the other frame in that reference.

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