I am reading a lot about the theory of special relativity, but I have a very basic question about this theory I still don't understand. Consider a particle in two inertial reference frames $\Sigma$ and $\Sigma'$. The reference frame $\Sigma'$ is moving with uniform velocity $v$ relative to $\Sigma$. The particle is at rest in $\Sigma'$. Both reference frames have common axes $x$ and $x'$. When doing a certain calculation in both reference frames, which one of the obtained results is considered correct? Are they considered both correct or the one obtained by an observer in $\Sigma'$?


They are both correct. You can do physics in any inertial frame you want. (In fact, you can do it in non-inertial frames too, but it’s more complicated.) That’s basically what “relativity” means.

One thing worth mentioning is that not all physical quantities are relative, i.e., dependent on which reference frame you measure them in. There are lots of absolute quantities that are independent of reference frame. These are the Lorentz scalars. For example, the energy and the momentum of a particle are frame-dependent, but the invariant mass is frame-independent.

  • $\begingroup$ But we are obtaining two different results when applying Lorentz transformations! How come both are correct? $\endgroup$ – Naps Aug 1 '19 at 19:33
  • $\begingroup$ Because an observer at rest in one frame measures one result, and an observer in the other frame measures the other. Did you ever learn about Galilean transformations? $\endgroup$ – G. Smith Aug 1 '19 at 19:37
  • $\begingroup$ If I am standing on the street and you are driving by in a car, and we both observe a bicyclist, isn’t it obvious that the bicyclist has a different velocity relative to you than relative to me? $\endgroup$ – G. Smith Aug 1 '19 at 19:38
  • $\begingroup$ So we are both correct in these two cases. Thanks a lot! $\endgroup$ – Naps Aug 1 '19 at 19:40
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    $\begingroup$ “Proper” doesn’t mean “correct”; it means “self”. Time measured by your watch is your proper time. Time measured by a watch of someone else who is zooming by in a spaceship is their proper time but not yours. $\endgroup$ – G. Smith Aug 1 '19 at 20:31

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