Let $A$ be the person who is standing on the Earth (ignoring gravitational field, etc.). Let $B$ be the person who is moving to the Proxima Centaura and back. By the time dilation equation we get: $$t_A=\gamma \cdot t_B$$ because the time is running slower for $B$ than for $A$.

But the motion of $B$ back and forth is equivalent to the motion of $A$ forth and back, isn't it? Thus, by the same equation: $$t_B=\gamma \cdot t_A$$ and clearly, $\gamma$ must be 1 for both equations to be true, so $v=0$ and so they are both stationary, which is a paradox. Where is an error in my process of thinking?

  • $\begingroup$ We have many existing questions on this topic. B is changing reference frames, A isn't. $\endgroup$
    – PM 2Ring
    Jun 21, 2021 at 11:54
  • $\begingroup$ @PM2Ring What about the reverse? (A is changing reference frames, B isn't) By Einstein's postulates, all inertial systems are equivalent. $\endgroup$
    – User123
    Jun 21, 2021 at 11:55
  • 1
    $\begingroup$ @User123 Changing (inertial) reference frames means that the observer is not inertial. $\endgroup$
    – J. Murray
    Jun 21, 2021 at 12:14
  • $\begingroup$ This is called the twin paradox, it's very old. If you include the extra time dilation from acceleration, everything works out. $\endgroup$ Jun 21, 2021 at 12:22
  • 3
    $\begingroup$ Please read some of the many many questions already on this topic. If your question is indeed different then you need to refer to the previous questions and explain what exactly is different about your question. $\endgroup$
    – Dale
    Jun 21, 2021 at 12:28

2 Answers 2


A hint that the evolution of the physical state of $A$ and $B$ are not indiscernable (not interchangeable, not symmetrical):

Let’s make the experiment a bit more specific and let $A$ and $B$ be two identical rockets $A$ and $B$, each fitted with an accelerometer and some recording device.

$A$ stays on planet $Pa$ and $B$ goes to planet $Pb$ and back. Both will give their data to an observer located in a inertial frame of reference $C$ at the end of the experiment.

Le’t make the thought experiment even a little more specific for the sake of clarity/simplicity: lets assume that $B$‘s journey consists of 4 simple phases:

  1. Acceleration from $Pa$ to halfway between $Pa$ and $Pb$.
  2. Deceleration from halfway between $Pa$ and $Pb$ and $Pb$.
  3. And 4. : ditto in reverse.

When $C$ analyses the data once $B$ is back is, he sees.

  • $A$’s accelerometer recorded nothing.
  • $B$’s accelerometer recorded 4 constant acceleration phases of magnitude, say $G_0$ (or even just a single acceleration phase for the duration of the entire trip, if the device records only the module of the acceleration).

Without knowing any additional details, merely from the inspection of the data from the recording devices from $A$ and $B$, $C$ can’t tell if $B$ has taken a trip around the galaxy or if its ship has been dipped in a gravitational well (spent some time on a planet heavier than $Pa$), but he will be able to tell that $B$‘s clock will be lagging behind $A$‘s clock at the end of the experiment, regardless of the cause (travel or gravity).

  • $\begingroup$ Thank you! The system of observer B isn't inertial because it was undergoing acceleration, that's what I missed. Also, congratulations with your newly 1000+ reputation. $\endgroup$
    – User123
    Jun 21, 2021 at 13:19
  • $\begingroup$ Thank you very much! $\endgroup$ Jun 21, 2021 at 14:01
  • $\begingroup$ @User123 Acceleration is a red herring. The asymmetry is due to one observer staying in one inertial frame but the other observer occupying 2 (or more) inertial frames. Acceleration is merely the mechanism by which the change of frames is achieved. $\endgroup$
    – PM 2Ring
    Jun 23, 2021 at 3:28
  • $\begingroup$ @PM2Ring Yes, because of the acceleration. They are basically equivalent formulations. $\endgroup$
    – User123
    Jun 23, 2021 at 6:39

Following is the extract from Wikipedia page on Time Dilation

Given a certain frame of reference, and the "stationary" observer described earlier, if a second observer accompanied the "moving" clock, each of the observers would perceive the other's clock as ticking at a slower rate than their own local clock, due to them both perceiving the other to be the one that is in motion relative to their own stationary frame of reference.

Common sense would dictate that, if the passage of time has slowed for a moving object, said object would observe the external world's time to be correspondingly sped up. Counterintuitively, special relativity predicts the opposite. When two observers are in motion relative to each other, each will measure the other's clock slowing down, in concordance with them being in motion relative to the observer's frame of reference.

While this seems self-contradictory, a similar oddity occurs in everyday life. If two persons A and B observe each other from a distance, B will appear small to A, but at the same time A will appear small to B. Being familiar with the effects of perspective, there is no contradiction or paradox in this situation.

In our context although you have stated the equation and the correct idea, you made a huge mistake of mixing up two frames and creating a round trip. A and B represents two entirely different frames related only by a Lorentz transformation. So what you should have written is $$t_B=\gamma \cdot t_A$$ and $${t'_A}=\gamma \cdot t'_B$$

Alternate way to look at is to realize that the initial point(Earth) and the final point(Proxima Centaura) are at two different lengths for A and B because of lenght contraction.

The above comment is just for a single sided trip. And a very important part of round trip is the fact that its not symmetric. You may look up Twin Paradox.

  • $\begingroup$ Thank you as well, but @SergeHulne stated it a bit more clearly, so I accepted his answer. Analogy with perspective is really good, have you come up with it? $\endgroup$
    – User123
    Jun 21, 2021 at 13:20

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