A commuter is riding a train that is moving at a speed of 0.75c. When the train passes a station platform, a clerk at the station picks up a magazine and sets it down. If the commuter observes that the clerk held the magazine for 8.0 s, how long does the clerk think that he held it? Ans: 12.1 s.

Since all motions are relative, and if we put ourselves on the reference frame of a commuter, which to him, he is the one who's stationary and that everything outside the train, including the clerk, the station platform, and the surrounding, are the ones that are moving at a rate of 0.75c, then we would identify 8.0 s to be the reverse and that the time the clerk thinks he held it is 5.3 s.

Since I cannot comment to the answer of this question: Symmetric Time Dilation in Uniform Relative Motion, I would have commented the same question: but if all motions are relative, and that there's no way of telling which one is moving or accelerating, how can we tell which frame experiences the time dilation?

Please enlighten my ignorance of the matter. Thanks.

  • $\begingroup$ I don't understand why we must write down a truck of paragraphs to explain time dilation, a so so so simple consequence of the Lorentz Transformation and here : $$ \Delta \tau =\Delta t_{clerk} = \dfrac{1}{\gamma} \Delta t_{commuter} = \sqrt{1-\left(\dfrac34\right)^{2}}\cdot8\,\rm{sec}= 5,2915\,\rm{sec} \tag{01} $$ See my answer therein :How do I know which observer is running the time faster or slower? $\endgroup$ – Frobenius Sep 29 '17 at 19:09

So, we have a clock on the platform and a clock on the train. They are of the same model and perform well. You stay on a platform and can make sure the clock on the platform is ok using your heartbeat as reference.

Then you jump on a train and you make sure the clock on a train is working good again using your heartbeat as reference.

The "problem" is that in a frame of reference where the clock is moving the pace of the clock is different. The faster clock is moving in some frame of reference the slower it ticks in this frame of reference.

So, in the frame of reference of the guy on the platform the clock on the train ticks slower. In the frame of reference of the train - other way round.

Probably I should stop here. This is the answer. "which frame experiences the time dilation" or "which clock is slower" depends on frame of reference.

But it sounds crazy, isn't it? I am standing on a platform, now both clocks show 0:00. In 1 minute one of them (mine) shows exactly 1:00. The other one is behind and shows 0:45. Ok, stop the experiment now, stop both clocks! They will show 1:00 and 0:45 forever and it doesn't depend on frame of reference! Even the guy on the train will agree, that your clock is ahead by 15 seconds! But from his point of view his clock should be ahead because at some moment they showed the same, he claims that his watch is ticking faster and and a little bit later we simultaneously stopped both watches. Paradox!

Actually no. In his frame of reference his clock is going faster, but the clocks were not stopped simultaneously. In 45 seconds his clock was stopped, the clock on the platform showed 0:34 (approximately) and continue clicking until they showed 1:00. The events "clock on the platform stopped" and "clock on the train stopped" happened in different points in space, so it's not possible to tell if they happened simultaneously or not - it depends on frame of reference.

You can try to think of some more complicated situations and convert "this sounds crazy" into some actual logical contradiction. But every time they will think of some explanation why it is not a paradox.

This is because there are no logical contradictions there, special relativity theory is quite simple. It just postulates rules how to calculate space-time coordinates of some event given space-time coordinates of this event in another frame of reference. The rules are quite simple, but some mathematics is necessary to realize why they are logically consistent.


Now back to the original question about commuter on a train.

We have 2 frames of references: first one is "Station", the other one is "Train".

We have 2 events: "magazine is taken" and "magazine was put back on a table".

Let's say coordinates of the first event in "Station" system of reference are: $$x_1=0, t_1=0$$. Coordinates of the second event would be: $$x_2=0, t_2=?$$ ($x$ coordinate is the same, because the clerk was staying still in "Station" frame of reference.)

Special theory of relativity states that given coordinates $(t, x)$ of some event in "Station" frame of reference, we can calculate coordinates $(t', x')$ of this event in "Train" frame of reference as follows: $$x' = \gamma * (x-v*t)$$ $$t' = \gamma * (t - v*x/c^2)$$, where $$\gamma=\sqrt{(1-v^2/c^2)}$$ Let's use this formula. $$\gamma = \sqrt{1-0.75^2} \approx 1.51$$ $$x_1'= \gamma * (0-v*0) =0$$ $$t_1' = \gamma * (0 - v*0/c^2) = 0$$. So far nothing special: it only means that the zero point of both frames of reference coincide. Now let's calculate when the second event happened in the "Train" frame of reference: $$t_2' = \gamma * (t_2 - v*x_2/c^2)$$.

Pay attention to this formula. There could be lot of different events happening at different places at the moment $t_2$ of the "Station" frame of reference. But in the "Train" frame of reference all these events happened not simultaneously. $t'$ depends not only from $t$, but also from $x$! Given only time difference (in one frame of reference) between two events we can't tell the time difference between these two events in some other frame of reference. It can be longer or shorter - it depends on coordinates of these events, so one should be careful when talking about "time dilation".

We know coordinates, so go on: $$t_2' = \gamma * (t_2 - v*0/c^2) = \gamma * t_2 = 8 sec.$$ ($t_2'$ is time span between our two events and is given: 8 sec). So the time span between the two events in "Station" frame of reference is: $$t_2 = t_2' / \gamma \approx 5.34 sec.$$

I encourage you to go on and calculate $x_2'$. You will know coordinates of both events in "Train" frame of reference. Knowing these space-time coordinates calculate the space-time coordinates of the events in the frame of reference which moves with velocity $-v$. Train is staying still in this frame of reference, time difference between the two events is 5.34 sec, you shift to a moving frame of reference, but because the space coordinates of these two events are different, the time difference in the moving frame of reference would become not shorter, but longer. You should get 8 sec, because effectively you returned to the "Station" frame of reference.

  • $\begingroup$ Great explanation! $\endgroup$ – Luke Sep 27 '17 at 15:26
  • $\begingroup$ Why I couldn't get the point. I imagine myself as the commuter who thinks he is stationary, and for me it is the clerk who is traveling at 0.75c, and if I am aware of SR, I would think that the clerk's time ticks slower, since I am think that he is the one who is moving. Thus, my measurement of his time would be 5.3s. $\endgroup$ – Juan Dela Cruz Sep 28 '17 at 13:21
  • $\begingroup$ @JuanDelaCruz Hmm, I should have read your question more carefully. Looks like the answer 5.3 is the correct one. Commuter sees a station moving along very fast, clock on the station ticks slower, so when his clock shows 8.00 sec., the clock on the station must show 5.3 sec. Not 12.1 sec. Could you please check if you write down the problem correctly? Could you give the exact words as in textbook? $\endgroup$ – lesnik Sep 28 '17 at 13:58
  • $\begingroup$ But the question again is who has the proper time and who experiences time dilation since both think they are stationary. I couldn't find it in a book. I just found it in a workbook of my former teacher in physics. $\endgroup$ – Juan Dela Cruz Sep 29 '17 at 1:22
  • $\begingroup$ @JuanDelaCruz You ask a good valid question. I'll update my answer and will try to explain it. But most probably I'll be able to do it tomorrow only. Sorry for delay. $\endgroup$ – lesnik Sep 29 '17 at 6:35

User lesnik has addressed your specific thought experiment. However, your wording seems to suggest you may have a more general misconception,

Since all motions are relative

This is not a postulate of special relativity. In special relativity, inertial motion is relative--i.e., any two observers who are travelling at a constant velocity with respect to each other are said to be inertial observers. That is to say, there's no physical experiment you could do in an inertial frame to decide whether you were sitting still or moving at a constant velocity.

There is, however, an experiment you could do to decide that something is accelerating--in particular, you would notice that in an accelerating frame any massive object would experience a measurable force proportional to its mass times the acceleration.

The invariance of inertial frames is, by the way, NOT an original insight of "special" relativity, but is a fact that was known to Galileo. Einstein's contribution to relativity theory was to postulate that there is a particular speed that all inertial observers would agree upon which is the speed of light (or, more precisely, the speed of any massless object/particle). Phrased differently, Einstein postulates the speed of light is the same in all inertial frames of reference.


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