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While deriving the frame transformation equations, either the Galilean Transformation or Lorentz transformation. I have seen almost all authors mentioning/assuming that if an inertial frame $\textbf{S}'$ is moving with constant velocity $\textbf{V}$ with respect to another inertial frame $\textbf{S}$. Then the frame $\textbf{S}$ is moving with velocity exactly equal to $-\textbf{V}$ with respect to inertial reference frame $\textbf{S}'$.

What is the basis of this assumption. I mean is it an assumption/axiom or can it be proved? Note that since we haven't derived the frame transformation equations yet we can't use many tools that would have been otherwise available to us) few of the postulates that we assume prior to deriving the transformation equations are the existence of inertial reference frames (which are Euclidean, the clocks can be synchronized and all positions and time are equivalent in formulating physical laws), the postulates of relativity (frame moving with constant velocity with respect to inertial reference frame are also inertial, all inertial frames are equivalent in formulating physical laws and the law of inertia (Newton's first law)

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  • $\begingroup$ It is an extra assumption when defining an IRF. More precisely, you need to first spell out which working definition for an IRF you have. $\endgroup$
    – DanielC
    Commented Jul 15, 2021 at 13:46
  • $\begingroup$ Are you sure about this? I haven't seen any author explicitly stating it as an assumption. $\endgroup$ Commented Jul 15, 2021 at 13:49
  • $\begingroup$ After looking around in research papers, it actually is one of the assumptions for the derivation. Don't know why the texts I have used never mentioned it. For someone who might come here in the future here is a paper from AAPT(Paywall) aapt.scitation.org/doi/pdf/10.1119/1.1970244 and here is one I found on Researchgate(Free access) researchgate.net/publication/… $\endgroup$ Commented Jul 15, 2021 at 14:11
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    $\begingroup$ If you have an answer to your question, please post it as a answer rather than as a comment. $\endgroup$
    – Brick
    Commented Jul 20, 2021 at 18:52

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