So my textbook says the following - roughly translated - in the context of Special Relativity:

"Assume we have two observers, A and B, moving relative to each other. Observer A measures a velocity v for observer B. Because of the symmetry of the situation, observer B measures the same velocity for observer A. If this isn't clear, note that A and B can be replaced by one another. If A and B measured a different velocity, then either of the two would be in a 'special reference frame'. This is in contradiction with the postulates of special relativity."

Alright, so I get this intuitively, of course. However, I can't follow their line of reasoning. How would we know that the laws of physics are different for A and B, if they don't measure the same velocity? Velocity is already relative, so how do we know for sure there would have to be a different set of laws for, say, A to get this result?

I'm familiar with the following argument: We've already deduced that a light clock yields different time intervals between its ticks for different observers. Assume there'd be another 'invariant' clock that would tick the same for any observer (a supposed 'universal time' clock). Then any observer would be able to measure their absolute speed, by comparing the time interval measured by the universal clock and a licht clock (that are assumed to not be movingg relative to one another). This is a contradiction, for velocity can't be absolute by our postulates.

Is it perhaps possible to give the same kind of argument with the symmetry problem? Would it somehow be possible for either of the observers to deduce an 'absolute velocity', if they don't measure the same velocity (which would yield the desired contradiction).

I'm hoping someone could help me out with either of the arguments!

  • $\begingroup$ Well, intuitively I'd say, their frame of reference is different, so in theory it's possible to measure a different velocity, because velocity is relative. But I get what you're saying with the law/rule thing. If they measured a different velocity, we would have to be able to deduce that from some law of rule, but they are the same, so it's impossible. I'm still not getting it entirely, but this is helpful. $\endgroup$
    – Sha Vuklia
    Jan 4 '17 at 18:48
  • $\begingroup$ I think the argument as it stands is untenable. With the same reasoning one also concludes that the velocities have the same sign. And this is not true. $\endgroup$ Jan 4 '17 at 18:49
  • $\begingroup$ @AlbertAspect Question though. Why would we need such a law (which we can't think of, obviously) for the two observers to measure a different velocity? Why does the fact that we don't have a law for it a good argument that it can't be true? Or is this a case of; as long as it's not proven wrong, we can assume it's true? $\endgroup$
    – Sha Vuklia
    Jan 4 '17 at 19:00
  • $\begingroup$ @ValterMoretti Good point. You could even be using different units. $\endgroup$
    – user126422
    Jan 4 '17 at 19:01
  • 1
    $\begingroup$ @ShaVuklia In Galilean transformations, you need the invariance of time differences as well as the invariance of distances to make that argument. $\endgroup$
    – Ali
    Jan 4 '17 at 19:43

Alright, I got a more intuitive/physical answer (instead of just plugging in values in Galileo/Lorentz transformations!).

We are going to assume two things: object A and object B are moving relative to each other (acceleration=0). Now, we can fix object A and make A face in the direction of B. Assume A tells us that B moves with speed v. Now, we maintain a completely identical situation if we fix B instead of A. In that case we have an object (B, in this case) that is facing another object that moves relative to it [in the same way as above]. This is literally the same situation as above, so if A gives us a value v for B, then B must give us the same value for A.

In short: Situation I: An object (A) is moving relative to another (B). Situation II: An object (B) is moving relative to another (A), in the same way. Conclusion: Those situations are identical, so any results concocted in situation I must also apply to situation II.

I know this is an overly cumbersome explanation, but this is the only way I believe I truly get it. :)

  • $\begingroup$ So for instance can you also conclude that if A finds B red coloured also B finds A red coloured? It does not work... $\endgroup$ Jan 5 '17 at 8:23
  • 1
    $\begingroup$ Moreover the discussed issue has not to do with inertial reference frames, Galileian invariance and all that which concern dynamics not kinematics. Indeed, the fact that the absolute value of the relative velocity is equal is valid also when A is inertial and B is not. More generally it holds regardless the relative acceleration. A better starting point in classical physics, in my view, is to notice or to assume as a basic experimental fact that the distance between A and B is symmetric. If there is a shared notion of time, all that leads to the wanted statement about relative velocities. $\endgroup$ Jan 5 '17 at 8:45
  • $\begingroup$ @ValterMoretti In response to your first comment; no, I don't think that's how one could go about it. You cannot deduce the colour of an object directly from its relative motion, so you would need more information, which is not given in this case. It is however possible - in theory - to express their relative motion in terms of their relative velocity. So, if B is able to measure A's velocity, he will give us the same value as A does, because of the symmetry of the situation. $\endgroup$
    – Sha Vuklia
    Jan 5 '17 at 11:02
  • $\begingroup$ Your claim is untenable, sorry, why colours do not enter the game? What you are saying is completely arbitrary, you are assuming lots of physical suppositions about kinematical relations. This is bad physics. Please be more careful this is not the right way to learn physics. $\endgroup$ Jan 5 '17 at 11:05
  • $\begingroup$ @ValterMoretti In response to your second comment: of course the discussion has to do with inertial reference frames, because, as I've already stated clearly, I'm using a symmetry argument. So we cannot have an inertial and a non-inertial reference frame... $\endgroup$
    – Sha Vuklia
    Jan 5 '17 at 11:05

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