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Consider the standard arrangement in special relativity. Let $S'$ move in the +ve $x$-axis with a velocity $v$ with respect to $S$. This implies that $S$ 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?

@RogerJBlarlow below provided an answer with basically says $f(f(v))=v$ hence $f(v)= -v$. There are many other solutions to this so why do we only choose where $f(v)=-v$

This question has been puzzling me for a while now. Any help is much appreciated. Thanks

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  • $\begingroup$ This is the principle of relativity. All inertial frames are equal. Observers from whatever inertial frame can consider themselves being stationary. So, if $S'$ moves on the $x$ axis with $v$ relative to $S$, and both are inertial frames, then saying that $S$ moves on the $x$ axis with $-v$ relative to $S'$ is perfectly valid. $\endgroup$ Jun 8 '19 at 11:37
  • $\begingroup$ well this argument is based on additive rule of velocities which itself is questioned in the theory. so for e.g. why -V and not some other value? $\endgroup$
    – aman_cc
    Jun 8 '19 at 11:40
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    $\begingroup$ @AWanderingMind I'm not sure the principle of relativity can be easily leveraged to say the speeds need to be exactly opposite each other. What would go wrong if Alice is moving at velocity $v$ in Bob's frame and Bob was moving at velocity $-v/2$ in Alice's frame? $\endgroup$
    – jacob1729
    Jun 8 '19 at 11:42
  • $\begingroup$ Is there any particular reason you care if it's a postulate or a theorem? Usually the choice of which statements to make postulates is subjective. The way I think about SR I wouldn't explicitly postulate that, so it would be a theorem, but I'm sure somebody has an axiomatization of SR that has that as an explicit assumption. $\endgroup$
    – jacob1729
    Jun 8 '19 at 11:43
  • $\begingroup$ @jacob1729 i think this is very basic for me $\endgroup$
    – aman_cc
    Jun 8 '19 at 11:48
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There is already a good answer by @Dale. Though if you want to do little math, you can do it this way. Write the Lorentz transformation $$x'=\gamma \ (x-vt)$$ $$t'=\gamma \left(t-\frac{vx}{c^2}\right)$$ Use these two find $(x,t)$ as function of $(x',t')$ which would lead to


$$x=\gamma \ (x'+vt')$$ $$t=\gamma \left( t'+\frac{vx'}{c^2}\right)$$ That's what you are asking for.


You can go even further through the basics remains the same. We know that co-ordinate transformations are just rotations in Hyperbolic space. Thus $S$ to $S'$ is a rotation through an angle, say $\beta$, then from $S'$ to $S$ is a rotation through an angle $-\beta$. If you recall $$\begin{pmatrix} x \\ ct \end{pmatrix}=\begin{pmatrix} \cosh\beta & \sinh\beta \\ \sinh\beta & \cosh\beta \end{pmatrix} \begin{pmatrix} x' \\ ct' \end{pmatrix}$$


Edit :

$$x'=\gamma \ (x-vt)\Rightarrow \frac{x'}{\gamma}+vt=x$$ Use this in $$t'=\gamma \left(t-\frac{vx}{c^2}\right)$$ $$\Rightarrow t'=\gamma \left( t-\frac{v}{c^2}\left(\frac{x'}{\gamma}+vt\right)\right) $$ Solve this for $t$ and you would find $$t=\gamma \left( t'+\frac{vx'}{c^2}\right)$$ In a similar manner, you can find the other equation for $x$ eliminating $t$.

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  • $\begingroup$ Hi,IMO, In writing the second set of equations you've implicitly assumed that the velocity in Frame $S'$ is $-v$ $\endgroup$
    – Kashmiri
    Feb 28 at 14:08
  • $\begingroup$ @YasirSadiq See the edit that what I mean. $\endgroup$ Feb 28 at 15:53
  • $\begingroup$ Yes got it. Thank you :) $\endgroup$
    – Kashmiri
    Feb 28 at 16:23
  • $\begingroup$ so can you derive Lorentz Transformation (which is your starting point) with assuming the statement about velocities? $\endgroup$
    – aman_cc
    Mar 4 at 15:33
  • $\begingroup$ Yes, You can. As the Lorenz transformation takes postulates of relativity as input. and that is granted. $\endgroup$ Mar 4 at 15:45
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It's a theorem which comes from the basic symmetry between S and S'.

Let $v$ be the relative velocity between O and O', as determined in S.

Transforming from S to S' gives some $v'=f(v)$

Transforming again from S' to S gives $v''=f(v')$, with the same $f$ because S from S' is the same as S' from S.

But we are now back where we started, so $v''=v$

Hence $f(f(v))=v$ for all $v$. This (given that $v$ has dimensions) means $f(v)=v$ and $f(v)=-v$ are the only possibilities. the former applies to the speed, the latter to the velocity.

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  • $\begingroup$ can i generalize this then to any observable property of S' from S and vice a versa. And hence claim they will be same. Question them becomes what are the observable properties of a ref frame. $\endgroup$
    – aman_cc
    Jun 8 '19 at 13:29
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    $\begingroup$ How do you know $v'=f(v)$ can be written as only function of speed? $\endgroup$
    – Kashmiri
    Feb 26 at 13:28
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    $\begingroup$ Assuming space is isotropic. And there is a symmetry between S and S'. It can depend on nothing else. $\endgroup$ Feb 26 at 17:15
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This follows directly from the principle of relativity. The principle of relativity states that there is no way physically to distinguish between different reference frames. If the velocities were different then you could distinguish frames on the basis of their relative speeds. In principle there would in principle be a reference frame where the relative speed was minimal or maximal, and this frame would be a unique “privileged” frame.

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  • $\begingroup$ im not sure i agree. relativity allows you to have different uniform velocities. maybe im missing the argument here. for e.g. there can be a ref frame whose orgin coincides with the event. that might make it special but not special in the sense that laws of physics will be different $\endgroup$
    – aman_cc
    Jun 8 '19 at 12:08
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    $\begingroup$ The analogy with the location of the origin is flawed. Relative velocities are directly measurable, coordinates are not (displacements are). So the correct analogy would be with the relative displacement of the origins, which is indeed the same as required by the principle of relativity. $\endgroup$
    – Dale
    Jun 8 '19 at 12:27
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    $\begingroup$ The principle of relativity states that the laws of physics are the same in every inertial frame, having different velocities doesn't break that principle. I am not able to understand. Could you please elaborate on it please. $\endgroup$
    – Kashmiri
    Feb 28 at 16:58
  • $\begingroup$ Having different velocities does break that principle. How can you know which measures a lower velocity if the laws are the same? If the laws of physics are the same in both frames then the transformation law to the other frame is also the same in both frames. That precludes one frame transforming to a higher velocity and the other transforming to a lower velocity. $\endgroup$
    – Dale
    Feb 28 at 17:37
  • $\begingroup$ I have plenty of sympathy with this answer but I'm struggling to square it with what you have written in other answers about the so-called "one way speed of light" (which I think is just an example of a coordinate speed). $\endgroup$ Feb 28 at 20:02
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Ever since 1910 the Russian physicist Vladimir Ignatowsky noticed that some assumptions about the nature of inertial frames (linearity, homogeneity, isotropy, reciprocity etc.) lead uniquely to either the Lorentz transformations of special relativity or to Galileo’s transformations of classical Newtonian mechanics.

The so-called reciprocity principle (i.e. the object of your question) is an axiom for the inertial reference frames: it can't be proved, but it's implicitly assumed in the definition of such reference systems.

On Journal of High Energy Physics volume 2012, Article number: 119 (2012) or also in the pre-print you can find an in-depth discussion of your question.

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  • $\begingroup$ I wasn't able to find official sources referring to the Reciprocity Principle you mentioned. Can you provide more sources please? $\endgroup$
    – Noumeno
    Mar 3 at 17:40
  • $\begingroup$ The Reciprocity Principle is mentioned in the paper arxiv.1112.1466 section 7.2 (and in its first references). $\endgroup$
    – Pangloss
    Mar 3 at 20:53
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It is a postulate in my opinion.

Suppose there is an atomic clock in the Mars robot programmed to send signals every $\Delta t$. The radial velocity $v$ between earth and mars changes with time. The time between signals received at earth is:$$\Delta t_e = \Delta t + \frac{r+v\Delta t}{c} - \frac{r}{c} = \left(1 + \frac{v}{c}\right)\Delta t$$

Now suppose that another atomic clock at earth is sending signals at the same rate $\Delta t$. And the result of $\Delta t_m$ (that is informed to us) shows a persistent difference compared to $\Delta_e$.

I believe that the research, if any experimental errors were discarded, would target first the possibility that the average $c$ between the planets is not the same for both directions, due to some gravitational effects. Rather than assuming that $v$ is not symmetric.

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