The time dilation experiment involves two frames in relative motion, let one at ground and other at train with velocity V. The light clock runs faster in rest frame, as seen by an observer A at rest in train ( just beside clock ) than that observed by an observer B in ground frame which observes moving clock. But here the discussion in both conditions are just about clocks everytime in Train's frame observed by both. So, how do I conclude that the time goes faster in ground frame while slower in moving frame,  that is the man at ground ages faster than the man in train. And, I know about the reciprocity and symmetricity for Twin's paradox. So, let's take ground to be at rest and train to be moving only.          One thing that I think that I can do is that the light clock, which is at rest in A's frame, can also be kept in ground frame and I also think that the time of rest-clock in ground frame will be same as that of rest-clock in train frame as the light covers same distance D with same speed of light, c.  Then, the time can be compared for the two light-clock set-ups after looking at both : one kept on the ground and the other at train, both observed from ground by observer B. And, phenomenon of Ages of observers A and B can be explained. But here is another problem that then the equation  

 ∆t'(observer:B) =   (gamma)*∆t(proper/observer:A) 

will not even be concluded because as I mentioned earlier that this one is for the same event placed at train's frame as seen by both observers A and B. Please tell me if my reasoning is wrong or right 😬 and help to sort out this issue i.e. Explain me the relativity of Time with this time dilation experiment.


2 Answers 2


You are right it is symmetric, So in the moving train you age the same as in the nit moving. It is just the other observer who thinks there is a difference. So if there were twins, they just age the same. Its only if one of them "returns" that you see an age difference.

  • $\begingroup$ the problem is not symmetric. the age difference comes from the fact that in order to return to their initial position, one of the twins has to change the direction of their velocity. change of velocity means that they are no longer in an inertial frame, and all observed age difference happens during this acceleration. the comment on the original post links to some very well written exposition on the subject $\endgroup$
    – paulina
    Commented Apr 9 at 19:35
  • $\begingroup$ Thats what I said in my answer, symmetry only if no return. $\endgroup$
    – trula
    Commented Apr 10 at 17:01

The point people often struggle to appreciate in SR is that time dilation is symmetric between two inertial reference frames. What that means is that if you are time dilated in my frame, then I am time dilated to exactly the same extent in yours.

To see this, imagine there is a light-clock with the people on the train and another light-clock with the people on the platform...

To the people on the train, the light in their clock travels straight up and down, but to the people on the platform it follows a longer diagonal path. Therefore it must take longer for the train light-clock to tick in the platform frame than it does in the train frame- ie it appears time dilated to the people on the platform.

However, to the people on the platform, the light in their clock travels straight up and down, but to the people on the train it follows a longer diagonal path. That means the platform clock takes longer to tick in the frame of the train than it does in the frame of the platform.

What I have described shows that the light-clock on the platform is time-dilated in the frame of the train, and the light-clock on the train is time-dilated in the frame of platform. The effect is entirely symmetrical.


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