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Consider the standard arrangement in special relativity. Let $S'$ move in the positive $x$-axis with a velocity $V$ with respect to $S$

Question: $S$ then moves with a velocity $-V$ with respect to S'. Is this an assumption or a theorem of special relativity?

If theorem - How can it be derived? If assumption - Why is it justified - what happens if you do not make this assumption?

I have asked this question before on this forum - and the answer I have got is broadly - symmetry of $S$ and $S'$

But I would like to have a deeper / more rigorous understanding of this.

So for example - Which law of physics will break if velocity of $S$ relative to $S'$ is $(-V+1)$?

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  • $\begingroup$ Which law of physics will break if velocity of S relative to S' is (-V+1)? +1 what? $\endgroup$
    – G. Smith
    Commented May 16, 2020 at 5:37
  • $\begingroup$ @G.Smith - That was a generic statement. It could be anything - so for e.g what argument do we have that relative to S', velocity of S can't be = -V/2 or - (V +10) etc. Why should it be -V? $\endgroup$
    – aman_cc
    Commented May 16, 2020 at 7:15
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    $\begingroup$ Does this answer your question? Special Relativity - Reference Frames $S$ and $S'$ $\endgroup$
    – Paul T.
    Commented May 16, 2020 at 13:48
  • $\begingroup$ @PaulT. - Actually no. I only asked that Q sometime back. I haven't fully understood it yet. Thanks $\endgroup$
    – aman_cc
    Commented May 16, 2020 at 14:19

2 Answers 2

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$A. From Einstein's derivation of the Lorentz Transformation

Einstein proved first that the transformation between the inertial frames $\rm S_1,\rm S_2$ of the Figure(1) must be, except an undetermined factor $\phi\left(\boldsymbol{\upsilon}\right)$ \begin{align} x_2 &\boldsymbol{=}\phi\left(\boldsymbol{\upsilon}\right)\gamma_\upsilon \left(x_1\boldsymbol{-}\upsilon t_1\right) \tag{01a}\label{01a}\\ y_2 & \boldsymbol{=}\phi\left(\boldsymbol{\upsilon}\right)y_1 \tag{01b}\label{01b}\\ z_2 & \boldsymbol{=}\phi\left(\boldsymbol{\upsilon}\right)z_1 \tag{01c}\label{01c}\\ t_2 &\boldsymbol{=}\phi\left(\boldsymbol{\upsilon}\right)\gamma_\upsilon \left(t_1\boldsymbol{-}\dfrac{\upsilon}{c^2}x_1\right) \tag{01d}\label{01d}\\ \gamma_\upsilon & \boldsymbol{=}\left(1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-}\frac12} \tag{01e}\label{01e} \end{align} Then using the Figure he proved that $\phi\left(\boldsymbol{\upsilon}\right)\boldsymbol{=+}1$ based on the assumption that $\,\rm S_1\,$ is moving with respect to $\,\rm S_2\,$ with velocity $\,\boldsymbol{-\upsilon}$. If this assumption is false then the Lorentz transformation is false, the in-variance of the speed $\,c\,$ of light in empty space in all inertial frames is not valid, so the co-variance of the Maxwell equations is not valid.

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

$\boldsymbol{\S}$ B. From the validity of the Lorentz Transformation

If we agree on the validity of the Lorentz Transformation between the inertial frames $\rm S_1,\rm S_2$ of the Figure, where $\rm S_2$ is moving along the common $x\boldsymbol{-}$axis with algebraic magnitude of the velocity $\upsilon \in [0,c)$ then \begin{equation} \mathbf R_2 \boldsymbol{=} \begin{bmatrix} x_2 \vphantom{\dfrac{a}{b}}\\ y_2 \vphantom{\dfrac{a}{b}}\\ z_2 \vphantom{\dfrac{a}{b}}\\ ct_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \gamma_\upsilon & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \boldsymbol{-} \gamma_\upsilon \dfrac{\upsilon}{c}\vphantom{\dfrac{a}{b}}\\ 0 & \hphantom{\gamma_\upsilon} 1 & \hphantom{\gamma_\upsilon} 0 & 0\vphantom{\dfrac{a}{b}}\\ 0 & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 1 & 0\vphantom{\dfrac{a}{b}}\\ \boldsymbol{-} \gamma_\upsilon \dfrac{\upsilon}{c} & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \gamma_\upsilon\vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} x_1 \vphantom{\dfrac{a}{b}}\\ y_1 \vphantom{\dfrac{a}{b}}\\ z_1 \vphantom{\dfrac{a}{b}}\\ ct_1 \vphantom{\dfrac{a}{b}} \end{bmatrix} \boldsymbol{=} \mathrm L_{12}\mathbf R_1 \tag{02}\label{02} \end{equation} Now, suppose that $\rm S_1$ is moving with respect to $\rm S_2$ along the common $x\boldsymbol{-}$axis with algebraic magnitude of the velocity not $\upsilon^\prime\boldsymbol{=-}\upsilon$ but of $\upsilon^\prime\boldsymbol{=-}\left(\upsilon\boldsymbol{+}\Delta \upsilon\right)$. Then \begin{equation} \mathbf R_1 \boldsymbol{=} \begin{bmatrix} x_1 \vphantom{\dfrac{\upsilon^\prime}{c}}\\ y_1 \vphantom{\dfrac{\upsilon^\prime}{c}}\\ z_1 \vphantom{\dfrac{\upsilon^\prime}{c}}\\ ct_1 \vphantom{\dfrac{\upsilon^\prime}{c}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \gamma_{\upsilon^\prime} & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \boldsymbol{-} \gamma_{\upsilon^\prime}\dfrac{\upsilon^\prime}{c}\vphantom{\dfrac{a}{b}}\\ 0 & \hphantom{\gamma_\upsilon} 1 & \hphantom{\gamma_\upsilon} 0 & 0\vphantom{\dfrac{\upsilon^\prime}{c}}\\ 0 & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 1 & 0\vphantom{\dfrac{\upsilon^\prime}{c}}\\ \boldsymbol{-} \gamma_{\upsilon^\prime}\dfrac{\upsilon^\prime}{c} & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \gamma_{\upsilon^\prime}\vphantom{\dfrac{\upsilon^\prime}{c}} \end{bmatrix} \begin{bmatrix} x_2 \vphantom{\dfrac{\upsilon^\prime}{c}}\\ y_2 \vphantom{\dfrac{\upsilon^\prime}{c}}\\ z_2 \vphantom{\dfrac{\upsilon^\prime}{c}}\\ ct_2 \vphantom{\dfrac{\upsilon^\prime}{c}} \end{bmatrix} \boldsymbol{=} \mathrm L_{21}\mathbf R_2 \tag{03}\label{03} \end{equation} From equations \eqref{02} and \eqref{03} \begin{equation} \mathbf R_2 \boldsymbol{=} \left(\mathrm L_{12}\mathrm L_{21}\right)\mathbf R_2 \tag{04}\label{04} \end{equation} This means that we must have \begin{equation} \mathrm L_{12}\mathrm L_{21} \boldsymbol{=}\mathrm I_{4}\boldsymbol{\equiv}\text{the identity $4\times 4$ matrix} \tag{05}\label{05} \end{equation} that is \begin{equation} \!\!\!\!\!\!\!\!\begin{bmatrix} \gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right) & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \boldsymbol{-} \gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(\dfrac{\upsilon }{c} \boldsymbol{+}\dfrac{\upsilon^\prime}{c}\right)\vphantom{\dfrac{a}{b}}\\ 0 & \hphantom{\gamma_\upsilon} 1 & \hphantom{\gamma_\upsilon} 0 & 0\vphantom{\dfrac{\upsilon^\prime}{c}}\\ 0 & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 1 & 0\vphantom{\dfrac{\upsilon^\prime}{c}}\\ \boldsymbol{-} \gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(\dfrac{\upsilon }{c} \boldsymbol{+}\dfrac{\upsilon^\prime}{c}\right) & \hphantom{\gamma_\upsilon} 0 & \hphantom{\gamma_\upsilon} 0 & \gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right)\vphantom{\dfrac{\upsilon^\prime}{c}} \end{bmatrix} \boldsymbol{=} \begin{bmatrix} 1 & 0 & 0 & 0 \vphantom{\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right)}\\ 0 & 1 & 0 & 0 \vphantom{\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right)}\\ 0 & 0 & 1 & 0 \vphantom{\dfrac{a}{b}}\\ 0 & 0 & 0 & 1 \vphantom{\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right)} \end{bmatrix} \tag{06}\label{06} \end{equation} Equation \eqref{06} is valid if \begin{equation} \gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(\dfrac{\upsilon }{c} \boldsymbol{+}\dfrac{\upsilon^\prime}{c}\right)\boldsymbol{=}0 \quad \textbf{and} \quad \gamma_{\upsilon}\gamma_{\upsilon^\prime}\left(1 \boldsymbol{+}\dfrac{\upsilon \upsilon^\prime}{c^2}\right)\boldsymbol{=}1 \tag{07}\label{07} \end{equation} The conditions \eqref{07} are satisfied if and only if \begin{equation} \boxed{\:\:\upsilon^\prime\boldsymbol{=-}\upsilon \vphantom{\dfrac{a}{b}}\:\:} \tag{08}\label{08} \end{equation}

$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

(1) This Figure is extracted from my answer here : Schutz's geometrical proof that spacetime interval is invariant.

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  • $\begingroup$ Thanks I'll read up you answer. I'm quite keen to understand why if that assumption doesn't hold leads to not a constant speed light. Maybe I do not fully understand the argument there. Thanks $\endgroup$
    – aman_cc
    Commented May 16, 2020 at 13:18
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    $\begingroup$ I know it's an answer from Frobenius when my phone lags while scrolling to the bottom ;) $\endgroup$ Commented May 16, 2020 at 17:48
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    $\begingroup$ @BioPhysicist : Oh ha ha, very funny, Aaron !!! Very good. $\endgroup$
    – Frobenius
    Commented May 17, 2020 at 15:28
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I think the transformation reasoning is quite persuasive.

If every time you move from one system to another, the relative velocity changes by some factor or constant, than by switching systems repeatedly you will get obscure results which would imply this transformation rule is not true.

Let me give an example :

Let's say that every time you switch between one inertial frame to another, V in S becomes (V + 1) in S'.

Then, switching from S' to S, (V + 1) becomes (V + 2), which is in contradiction to V that we started with. Moreover, if we keep switching systems repeatedly, the velocity will go to infinity which is obviously can not be true.

The same reasoning goes with transformations of the kind 2*V (goes to infinity) or V/2 (goes to zero).

The only way to avoid these results is to assume the speed doesn't change under transformations between systems.

The key argument in this reasoning is really the symmetry between the systems which dictates that the transformation rule would be identical.

This key argument is really an assumption.

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