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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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Qmechanic
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aman_cc
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  • 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. – AWanderingMind Jun 08 '19 at 11:37
  • 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? – aman_cc Jun 08 '19 at 11:40
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    @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? – jacob1729 Jun 08 '19 at 11:42
  • 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. – jacob1729 Jun 08 '19 at 11:43
  • @jacob1729 i think this is very basic for me – aman_cc Jun 08 '19 at 11:48
  • The relativity principle states that there exists an infinite continuous class of reference frames in space-time, in which the laws of physics take on the same form. This doesn't mean that physical quantities have the same value in all these frames. This principle equally applies in General Relativity where the relative speed is not invariant. If you observe me falling to a black hole horizon, you see me moving very slowly while I see you moving away near the speed of light. The invariance of the relative speed is a direct mathematical consequence of the isotropy of space (e.g. a flat space). – safesphere Feb 26 '21 at 23:08

5 Answers5

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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.

RogerJBarlow
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  • 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. – aman_cc Jun 08 '19 at 13:29
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    How do you know $v'=f(v)$ can be written as only function of speed? – Kashmiri Feb 26 '21 at 13:28
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    Assuming space is isotropic. And there is a symmetry between S and S'. It can depend on nothing else. – RogerJBarlow Feb 26 '21 at 17:15
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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 $t$.

Young Kindaichi
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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.

Urb
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Pangloss
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  • I wasn't able to find official sources referring to the Reciprocity Principle you mentioned. Can you provide more sources please? – Noumeno Mar 03 '21 at 17:40
  • The Reciprocity Principle is mentioned in the paper arxiv.1112.1466 section 7.2 (and in its first references). – Pangloss Mar 03 '21 at 20:53
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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.

Dale
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  • 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 – aman_cc Jun 08 '19 at 12:08
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    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. – Dale Jun 08 '19 at 12:27
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    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. – Kashmiri Feb 28 '21 at 16:58
  • 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. – Dale Feb 28 '21 at 17:37
  • 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). – Andrew Steane Feb 28 '21 at 20:02
  • @Andrew Steane there is no conflict. The principle of relativity here is in reference to inertial frames. Synchronization conventions in which the one-way speed of light is anisotropic are not inertial. – Dale Feb 28 '21 at 20:23
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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.

Claudio Saspinski
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