# Time dilation and simultaneity: easier to see with a train

1. Feb 1, 2011

### Erland

Many people have difficulties to understand how time dilation can be consistent with he special relativity principle: "According to the observer on the Earth, the observer at the space ship is aging slower, but according to the observer at the space ship, the observer on the Earth is aging slower. Who is really aging slower?"

Those asking this type of questions have not really grasped the relatvitivity of simultaneity. I think it is better to explain this using a thought experiment with a train than with a space ship, because a space ship has a quite small extension and it is difficult to intuitively associate it with an entire coordinate system. A train, on the other hand, is perceived as "long", with many cars, and easier to associate with a coordinate system. The disadvantage is that such a thought experiment is even more unrealistic than a space ship thought experiment; we need an absurdely fast and long train and an absurdely long platform. Still, I think the benefits outweigh the disadvantages.

So, here comes the thought experiment (where we use the Lorentz transformation to go from one sytsem to the other one):

Consider a train running east, passing, without stopping, a platform extending in the east-west direction. The velocity of the train, relative to the platform, is 0.87c (= sqrt(3)/2 c, making the factor sqrt(1-v^2/c^2) = 1/2). There are two clocks at the west and east ends of the platform, respectively. They are synchronized, all observers at the platform agree that these two clocks show the same time. Also, there are two clocks in the train, one in the front, in the locomotive, and one in the back, in the last car. They are synchronized too, all observers on the train agree that these two clocks show the same time.
The train driver is, of course, located in the locomotive all the time. The station master is located at the west end of the platform and remains there during all the time when train passes.
Assume that at the event when the locomotive arrives at the west end of the platform, the clock at the west end of the platform and the clock in the locomotive both show t=0 (I prefer not specifying time units, it just becomes ridicolously unrealistic otherwise).
Also, assume that at the event when the locomotive arrives to the east end of the platform, the clock at the east end shows t=2. Then, the clock in the locomotive shows t=1.
Finally, assume that at the event when the back of the train arrives to the west end of the platform, the clock at the west end of the platform also shows t=2.
According to an observer at the platform, these two events are simultaneous (t=2 for both). The station master would therefore say that the train driver has aged 1 unit while himself has aged 2 units, so the train driver has aged slower than himself.

But, by the special principle of relativity, the train driver can, with same right, say that the station master ages slower than himself.
So, how can that be? Didn't we just show that the train driver is really aging slower than the station master?

Well, no. What we showed was that the train driver is aging slower than the station master according to the station master (and other observers on the platform). The solution lies the fact that simultaneity is not absolute. The two events when the locomotive arrives to the east end and the back of the train arrives to the west end of the platform are simultaneous (t=2), but only according to an abserver on the platform. Recall that at the first of these events, the clock in the locomotive shows t=1. But what does the clock in the last car of the train show at the second of these events? Answer: t=4! An observer on the train would therefore say that the two events are far from simultaneous. The latter event would, according to an abserver of the train, be simultaneous to an event when t=4 in the locomotive, and then the locomotive is far east of the entire platform. The train driver then would say that the station master is aging slower than himself (2 units for the station master and 4 units for the train driver).

Tom summarize:

Event 1: Locomotive arrives to east end. Clock at east end shows 2, clock in locomotive shows 1

Event 2: Last car arrives to west end. Clock at west end shows 2, clock in last car shows 4.

According to an observer at the platform, the two evenst are simultaneous, and the train driver has been aging slower than the station master.

According to an observer on the train, the two events are not simultaneos, and also, the station master is aging slower than the train driver.

They both say that the other one is aging slower, and they are equally right, and that is no contradiction.

2. Dec 3, 2017

### JulianM

"all observers on the train agree that these two clocks show the same time"

Surely this would only be true if they are standing at precisely the same place on the train. Light travels about 1 ft in a nanosecond. Let's assume a train of 1000 ft long. An observer standing at either the front or back of the train will perceive a difference of a microsecond between the 2 clocks. The only person who could see the clocks as simultaneous would be so one located in the middle of the train and that person would see the 2 clocks as 1/2 microsecond behind his own clock.

All clocks are set at the same time and travelling at the same speed but they do not agree that the clocks show the same time.

3. Dec 3, 2017

### Ibix

Observers are assumed to be smart enough to subtract out the travel time of light. The thing is that different frames will not agree on the results of that process (as Erland's post shows), so they may come to different conclusions about whether or not two clocks are synchronised.

4. Dec 7, 2017

### JulianM

Just trying to understand this complex subject. Does the observer also do the same with the moving object? that is to say does he recognize that he is seeing the far end of the rod at a different time from when he sees the near end and make an adjustment for that? I am not seeing that in the math. Any help?

5. Dec 7, 2017

### Ibix

There doesn't seem to be any mention of a rod anywhere in this page before your post, nor is there much maths, so I can't comment specifically on what you are talking about. You'll have to provide a reference or a detailed description if you want specific help.

As a general comment, typically light travel time is subtracted out of experiments because that way you don't need to specify where the observer is standing. Indeed, it may not be possible to see all of an experiment from one place. Instead, we use a frame of reference where all observers using that frame agree on a notion of "at rest" and a notion of what "at the same time" means, as well as things like the position of the origin of coordinates and their orientation. So the Lorentz transforms relate the time and place at which things happened according to two different frames, not the time at which they were observed by two particular observers.

6. Dec 7, 2017

### JulianM

Sorry, this example is about a train so I guess I really meant train, but was just thinking of something moving through space so rod or train.

My initial question was about how the clocks are observed. Now I am trying to work through the rest.

The posting talks about events taking place at both the front and the end of the train and compares train clocks to station clocks without talking about the relationship between platform length, train length, etc. and without reference to what an observer sees so I am trying to draw a diagram for what Erland is explaining.

7. Dec 7, 2017

### Janus

Staff Emeritus
Consider the following. Observer by the tracks as train goes by, when the ends of the trains align with the red dots, light flashes are emitted. The observer half way between the dots sees the flashes simultaneously. However, anyone situated anywhere else along the tracks would not. All of them would, however, agree that the flashes originated simultaneously. This is true for any light leaving the ends of the train also. It doesn't matter whether we consider the light flashes as being emitted by the red dots or by the ends of the train.

8. Dec 7, 2017

### Ibix

Erland carefully sets things up so that clocks that are compared are compared as they pass each other - so he talks about what the train rear clock reads when it passes the west platform clock, and so on. So the details don't actually matter. Whoever is looking and wherever they are they will agree what any two clocks said when they passed each other.

If you want to draw a diagram, I recommend looking up Minkowski diagrams. They are basically displacement-time graphs, and are probably the best form of diagram for understanding special relativity.

9. Dec 7, 2017

### PeroK

Two suggestions. If you are trying to learn about SR, then textbooks have much more time, space and diagrams to go through the subject step-by-step than a single PF post. That's why, in general, we cannot teach things here, but rather supplement proper study materials with answers to specific questions or problems.

My other suggestion is to use "reference frames" rather than "observers". The way I would describe a reference frame is a large number of observers - for 1D problems they are all in a line and for 3D problems there is a whole lattice of them. These observers are all at rest with respect to each other, they all know the coordinates of where they are standing and they all have clocks that are synchronised.

When a series of events takes place, there is always a local observer in this reference frame ready to note down the time of each event. Afterwards, they can all get together and compare notes (or send their data to a single master observer who puts together everything that happened, where and when in that reference frame).

In the case of a moving train, therefore, you have the "platform" reference frame: observers all along the platform. If the train has a clock in it, each observer can note down the time on their own clock when the train clock passes and what time the train clock read when it passed. And, if something happens on the train, then there is always a local observer in the platform reference frame who can record the time and place that event took place (in the platform reference frame).

Trying to think in terms of a single observer can lead to confusion about when and where things happened in their reference frame.

Last edited: Dec 7, 2017
10. Dec 7, 2017

### Bartolomeo

Apparently Wolfgang Rindler, the author of the famous textbook “ Relativity: Special, General and Cosmological” has difficulties either.

Please have a look, what he teaches us in his textbook, page 79, first paragraph:

http://202.38.64.11/~jmy/documents/ebooks/Rindler 2006 Relativity (ISBN 44198567324).pdf

"A simple case in point is the frequency shift between a source at the center of a rapidly turning rotor, and an ‘observer’ (a piece of apparatus) at the rim, which moves, let us say, with linear velocity $v$. Setting $\alpha = 90$ and $\nu=\nu_0$ in (4.5), we find $\nu=\nu_0\gamma (v)$ for the observed frequency of a source of proper frequency $\nu_0$. This, of course, is entirely due to the time dilation of the moving observer."

On the same page, just below the formula (4.6)

"Time dilation is the only cause of the frequency shift whenever there is no radial motion between source and observer. This is the so-called Transverse Doppler Effect, and has long been considered as a possible basis for time dilation experiments."

Well, according to Rindler, "moving observer" actually dilates and measures, that clock in the center is ticking $\gamma$ times faster. So, Rindler claims, that according to observer at the space ship, the observer on the Earth is aging not slower, but faster, contrary to your assumption.

Mr. Rindler doesn’t even mention relativity of simultaneity here, but says that time dilation is due to “moving observer”. Hence, according to Mr. Rindler, since Kennedy – Thorndike interferometer was nothing different from that “ ‘observer’ (a piece of apparatus) at the rim” , the only explanation of null - result of this experiment is actual dilation of interferometer, fully in accordance with Lorentz Ether Theory, but Rindler probably thinks that it is "Special Relativity", because on the same page Rindler writes:

“Yet by a curious irony of fate, Ives and Stilwell were lone hold-outs for Lorentz’s ether theory and rejected special relativity”

Do not you think that Rindler himself is a “lone hold-out for Lorentz’s ether theory”?

11. Dec 7, 2017

### Ibix

Erland is talking about two inertial observers. Rindler is talking about one inertial and one non-inertial observer, which is a rather different situation.

12. Dec 7, 2017

### DrGreg

Rindler is talking about the special case "whenever there is no radial motion between source and observer". In this case the observer is moving round a circle and therefore not an inertial observer. This is a more advanced topic. Beginners in relativity really need to have a good understanding of the simpler case of inertial observers and inertial sources moving directly towards or away from each other before moving on to non-inertial observers.

13. Dec 7, 2017

### Bartolomeo

Erland does not make reservations whether the observer is inertial or not. Rindler doesn’t make any reservations also. He doesn’t tie effect of dilation with non – inertiality. At least I haven’t seen it. Rindler puts it straight – due to dilation of moving observer, or rotating absorber.

What will be different if instead of rotating there is inertial observer? Will he see redshift or dilation? Not at all.

Why we don’t make any reservations about Kennedy – Thorndike interferometer? What makes it different from rotating absorber in Mossbauer rotor experiments?

Last edited: Dec 7, 2017
14. Dec 7, 2017

### Ibix

Erland is clearly talking about inertial motion in one dimension. Rindler is able to make a non-arbitrary distinction between a moving and a non-moving observer precisely because he is discussing non-inertial motion of one of them.

If you wish to consider inertial motion with a transverse component, you will find that both observers measure light coming from the other having the same Doppler effect as a function of time as the other (they must do, otherwise there would be a violation of the principle of relativity). They just don't agree on time and place (so the relativity of simultaneity most definitely comes into it), which is how this is not contradictory.

15. Dec 7, 2017

### PeroK

I think Rindler would be horrified that you were quoting him in support of your ideas about SR.

16. Dec 7, 2017

### Staff: Mentor

You evidently aren't able to understand what you are reading.

This is not correct.

You have just earned yourself a thread ban. Please take some time to learn SR correctly.