I'm getting interested in special relativity and almost all people and explanations about the subject I found on the internet seem to think that when A and B would be in relative motion, they would both observe the other's clock tick slower than their own. However what I found about this is following, in § 4. (Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks) of Einstein's On the electrodynamics of moving bodies (1905):

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity $v$ along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by $\frac{1}{2}tv^2/c^2$ (up to magnitudes of fourth and higher order), $t$ being the time occupied in the journey from A to B.

This sounds like complete opposite to what everyone seems to think.

Now this time discrepancy doesn't have anything to do with Doppler effects or isn't just some observers illusion. This is supposed to be physical fact, after stopping the traveler has physically aged less than the rest of the world.

The question: Why should the stationary clock seem slower to the traveller?

The quote i provided explicitly says that his own clock will have experienced less time. Please note that A and B can be arbitrarily close and there can be an infinite amount of such points for which the quote applies. Also the train need not stop to witness this effect

Therefore it seems to me certain that the traveller would observe the stationary clocks move faster.

  • $\begingroup$ Comments are for clarifying and improving the post, not for extended discussion; this conversation has been moved to chat. $\endgroup$ – rob Aug 6 '19 at 5:57
  • $\begingroup$ I was writing a rather lengthy answer to the edited question when it was closed. Since I put a good bit of work into this, I took a screenshot of the preview which you can find here $\endgroup$ – Alfred Centauri Aug 6 '19 at 15:51
  • $\begingroup$ I've voted to reopen this question since the edited question clearly isn't a duplicate of "How can time dilation be symmetric?" In this thought experiment, clock A is not inertial - clock A changes reference frames from the unprimed rest frame of clock B to the primed frame with relative speed $v$. That clock B is ahead of clock A when they meet is rooted in this change of frames. It should be pointed out that, once clock A has accelerated to the primed frame, clock B is ahead in this frame but the difference is decreasing. That is, clock B runs slower than clock A in the primed frame. $\endgroup$ – Alfred Centauri Aug 6 '19 at 22:33
  • $\begingroup$ So many downwotes for very reasonable question! @AlfredCentauri, (or anyone else) what do you think about Figure 2 (b) and Eq. 3 in this article en.wikipedia.org/wiki/Relativistic_Doppler_effect? The observer moves in the frame of the clock- source. Does this clock - source is running slower or faster according to him? $\endgroup$ – Albert Aug 16 '19 at 18:18
  • $\begingroup$ The question before edit was never duplicate of "How can time dilation be symmetric?" either. You can find it in edit history.. $\endgroup$ – Katz389 Aug 17 '19 at 19:33

Both observers really do see the other set of clocks as ticking slow. But the thought experiment from Einstein doesn't contradict this.

The key is that according to the person on the train the set of clocks along the track are not synchronized. They are only synchronized according to an observer standing on the ground, not for anyone in relative motion to the ground.

An example: If the person on the train looks at any individual clock they will see the clock ticking slowly as compared to their watch. Say there are two clocks anchored to the track A and B. At the moment the train passes clock A the rider's watch reads 0 sec, the time on clock A reads 0 sec and the time on clock B reads 10 sec (according to the rider; A and B are not synchronized in the rider's frame). Say $\gamma=2$ so, according to the train rider, clocks A and B tick at half the rate of their watch. The observer might take (according to their own watch) 2 seconds to reach clock B. At the moment they pass clock B it will read 11 sec (it is ticking half as fast as the watch). But the watch reads 2 sec. So even though the clocks A and B are ticking slowly clock B has "run ahead" of the watch when the train reaches B.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Aug 6 '19 at 13:27

So the observer on the train ($\gamma=10$) has a clock. He looks out the window and sees a single clock (A). Let's say they are momentarily synchronized, so they both read 12:00. The instantaneous speed of the second hand on the A clock is moving more slowly than his clock (by a factor of 10), but for a brief moment, the 2 agree: it is Twelve Noon.

Ten minutes pass on the train. If the conductor sticks his head out the window, he looks back and sees the A-clock reading something like 12:00 and a few seconds. This doesn't mean much, because he knows he is looking back into the past (the A-clock is about 9'57" times $c$ far away). Anyway, when discussing distant clocks in Gedanken Experiments, light delays are never considered. "seeing" means "what he knows is true", not what delayed light looks like.

What he knows is true is that 10 minutes have passed and that the A clock is running slow by a factor of 10: The A clock reads 12:01 right now.

He looks out the window and sees the B-clock. The A and B clocks are synchronized in the track frame. The B-clock reads 1:40. We know that because in the track-frame, 100 minutes have passed since the A clock read 12:00, and the A and B clocks are synchronized in the track frame, it must show 100 minutes later. Note that he observes that the B-clock is also ticking slowly by a factor of 10. He also observes that the B clock leads the A clock by 99 minutes.

According the train conductor, each equally spaced clock (A, B, C, D, ...) leads its alphabetical predecessor by 99 minutes. It is this position dependent bias that allows both reference frames to observer the others' clock moving more slowly than his own.

If we focus on just one clock, say the A-clock, we see that as the train moves down the track, the definition of "now" at the A-clock according to the train differs more and more than "now" according to the track clocks. Again, this is not what he sees looking at the clock's light, it is what he considers "now" there. Just as we can consider "now" at the Andromeda galaxy as being very different from the 2.2 million year old light we see tonight.

  • $\begingroup$ I dont know why you try to introduce doppler effect into this and obfuscate the issue. Always look only at the clock you pass by. Then the outside time will be visibly faster and every clock outside will move faster. Just imagine the clock could have been sun-clock and the traveller could have boarded at 12:00 and exited one minute later at 18:00. Then the sun (and the all clocks hes passing by) must be logically moving faster. Or do you perhaps suggest the sun would be stuck in the sky and then teleport to different location ? $\endgroup$ – Katz389 Aug 6 '19 at 8:15
  • $\begingroup$ @Katz389 What Doppler effect are you talking about? Where do you think my post is wrong? $\endgroup$ – JEB Aug 6 '19 at 22:06
  • $\begingroup$ @Katz389 now I get it: you appeal to "the outside time", well there is no one outside time: it is position dependent. You can call that "the clock reading out the window", and yes, it is increasing quickly according to the train, but it is not a time in the train frame. Moreover; each outside clock is moving slowly in the train frame. $\endgroup$ – JEB Aug 7 '19 at 1:19

First, your quote from Einstein's paper involves analysis from a single frame K. In frame K, clock A is in motion and clock B is not. This quote from Einstein does not consider the symmetric case of what an observer at rest with respect to clock A sees clock B doing, which is what the rest of your question is about. So this quote isn't relevant to the claim that time dilation isn't symmetric. Now, onto your ending questions.

1] Am I right in my assumption that the traveller inside the train observes the outside world go much faster, while the stationary observer outside the train would see the clock inside the train go much slower ?

No. The traveler on the train sees clocks that are at rest relative to the world as ticking slower. Similarly, an observer at rest relative to the world would see clocks on the train ticking slower. This is because there is no preferred reference frame. In either scenario, an observer sees a clock moving by them. Why should one view something different than the other?

2] Does this mean there must be an absolute speed (even if undetectable) for every object? For Example if an asteroid flies by then the universe must decide if the clock on asteroid will go faster or the clock on Earth will go faster, and so'remember' which gained more acceleration since the Big Bang.

Since the answer to the first question is no, this is invalid. SR assumes that the laws of physics are the same for all inertial observers and that the speed of light is constant independent of the relative motion between the observer and the source. The latter gives us time dilation, and the former tells us that this effect is the same for observers moving past each other. Given the experimental success of SR predictions as well as theories based on SR such as QED, I think the postulates hold true for our universe.

  • $\begingroup$ "Why should one view something different than the other?" Because the Einstein quote says so.. At the end of the journey, less time has passed for the moving clock. You realize that the same doesnt go juts for clock right? The clock could have been sun-clock and the traveller could have boarded at 12:00 and exited one minute later at 18:00. Now how in the world would anyone think that the sun would just stay at the same place for that minute? $\endgroup$ – Katz389 Aug 6 '19 at 8:09
  • $\begingroup$ @Katz389 Like I say in the answer, the Einstein quote only deals with a single frame. In frame K you will see that time has ticked slower for clock A than for clock B. Einstein's quote doesn't say anything about being in a different frame K' that moves with clock A. If you were to move with clock A so that you see clock B as being in motion you would see B ticking slower than A. I have not said anything about what type of clock it is. I agree with you that it doesn't matter what mechanism is used to keep track of the time. $\endgroup$ – Aaron Stevens Aug 6 '19 at 11:27
  • $\begingroup$ @Katz389 Related video $\endgroup$ – Aaron Stevens Aug 6 '19 at 11:37
  • $\begingroup$ I think the quote says a lot about frame K'. It explicitly says that the more time you spend in K' the less time you will experience compared to observer in K. But nothing at all prevents you to look how time runs in K while you are in K'. $\endgroup$ – Katz389 Aug 6 '19 at 11:55
  • $\begingroup$ Consider this: The railroad is in shape of O (like in particle accelerator) and there is a huge clock in the midle of the railroad. Under the clock there is a house being built. Now you go on the train and look at the center from a window. Now from the point of the house you drive around for a year, what exactly should prevent the photons from the construction of the house to reach you? Why you should not see the house built inside what you consider 1 minute on the train? Isnt this exactly what happens in particle accelerators? $\endgroup$ – Katz389 Aug 6 '19 at 11:59

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