Consider the standard arrangement in special relativity. Let S' move in the +ve 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?

  • $\begingroup$ 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. $\endgroup$ – AWanderingMind Jun 8 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 at 11:40
  • $\begingroup$ @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 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 at 11:43
  • $\begingroup$ @jacob1729 i think this is very basic for me $\endgroup$ – aman_cc Jun 8 at 11:48

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.

  • $\begingroup$ 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 at 13:29

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.

  • $\begingroup$ 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 at 12:08
  • $\begingroup$ 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 at 12:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.