I'm just starting to learn about special relativity, and I'm a little bit confused about something. Take the example of an observer in $S$ on the ground observing a train move at constant velocity $v$ relative to $S$, an observer in $S'$ is on the train, and this observer in $S'$ flashes a light that reflects from the ceiling and returns to him in a time, $t'$ he measures.

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I know that in general, if two frames $S$ and $S'$ are in relative uniform motion with respect to each other, and an observer in $S$ can see the clock of an observer $S'$ and the observer in $S'$ can see the clock of the observer $S$, then the observer in $S$ will see the clock of the observer in $S'$ run slower than his own, and vice versa. But I also know that $S'$ time is proper time, and so $t'\leq t$. This seems very strange to me. Observer $S$ sees his clock running faster, so intuitively, observer $S$ expects to measure less time for the event, but the time he measures for the light to return is no less than the time that observer $S'$ measures.

Am I understanding this correctly?

  • $\begingroup$ You just managed to confuse me about special relativity... can you draw this scenario, please? I think you may actually discover the source of your problem, when you do. $\endgroup$
    – CuriousOne
    Sep 11, 2014 at 7:31
  • $\begingroup$ "but the time he measures for the light to return is no greater than the time that observer S measures." By no greater do you mean equal to or less than? $\endgroup$
    – BMS
    Sep 11, 2014 at 7:52
  • $\begingroup$ @BMS Yes, I mean that the time observed by the observer in the train is $\leq$ than the time observed by the observer on the ground $\endgroup$
    – user153582
    Sep 11, 2014 at 7:56

2 Answers 2


I agree with your statements up through the claim that $t'<t$. That's all fine.

Here I think is the issue you're running into:

The quantity $t'$ in the relationship above represents the time interval as measured in frame $S'$. It does not represent the number of ticks by the moving clock as measured in frame $S$. That's a subtle but important distinction, read it again.

I think this latter (incorrect) reasoning is leading to your apparent contradiction.

  • $\begingroup$ Oh jeez, I think I asked my question the wrong way: $S$ also sees his clock as ticking faster, but $S$ observes a larger time than does $S'$. I think that's what I meant to ask :) is there still no contradiction? $\endgroup$
    – user153582
    Sep 11, 2014 at 7:58
  • $\begingroup$ That makes sense. I'll have to think about it more, but I have a feeling that this is indeed the source of my confusion. Thank you very much for taking the time to explain this :) $\endgroup$
    – user153582
    Sep 11, 2014 at 8:38

It seems you feel that the example should be symmetrical because from each point of view (S or S'), time is running slower in the other. However, the experience is not symmetrical because as you said : S' is proper time and S is not.

If you make an other "experiment" of which S is proper time, then S will measure t2 and S' will measure t2' and you will have t2


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