# The importance of velocity in simultaneity

1. Dec 31, 2014

### sjkelleyjr

Please take this post with a grain of salt as I am a computer scientist and not a mathematician or physicist. I started in on Einstein's Relativity: The Special and The General, and just finished the section on the train thought experiment involving simultaneity. I then started in on the Lorentz section following that, and I feel I understand the math involved fairly well. So I believe fully in what is being said in the thought experiment, however I feel Einstein had to kind of "dumb it down" for the general populace and I feel this is leading to my confusion. Here's why.

If we take a snapshot of the train at the moment the lightening strikes hit (i.e. the moment of simultaneity to the embankment observer) why does the velocity of the train have anything to do with this moment? In my mind I'm stopping everything at this exact moment in time (even though there isn't any one "exact" moment) and since the speed of light is constant, wouldn't the lightening bolts reach the midpoint of the train simultaneously regardless of the velocity of the train?

Writing it out now I feel like my problem is my attempt to "snapshot" the train. In reality time is still moving forward, and thus velocity does become a factor. Is this where my confusion lies?

Again, I understand the Lorentz transforms and how velocity effects these, and I must admit I feel safer in the realm of math than the thought experiment. As stated in this post "there is absolutely no ambiguity in the math, and it is quite clear what Einstein is saying." and I agree.

Thanks!

2. Dec 31, 2014

### ShayanJ

3. Dec 31, 2014

### Staff: Mentor

A snapshot is taken with a camera, and light has to travel from the events being captured in the snapshot to the camera before they can be recorded. So when you look at a snapshot, you are not seeing a picture of everything at the same time, you're seeing a picture in which the more distant stuff happened further back in time than the nearer stuff. Cameras moving relative to one another will capture different snapshots and different notions of what happened "at the same time" even if they are in the same place at the same time when they are triggered.

And if by "snapshot" you mean some magic process that captures everything at the same time without having to allow for light travel times... Well, how are you going to define "at the same time" in a way that works equally well when we assume the train is at rest and the platform is moving backwards and when we assume that the platform is at rest and the train is moving forward?

That is right.

4. Dec 31, 2014

### Joshua L

When I first started to learn Einstein's theory of relativity, I used a similar, yet different, thought experiment to understand the "issue" of simultaneity. In my opinion, it is easier to analyze.

Here are my two cents.

There is a train car moving with a constant velocity, $\vec v$ , on a straight railroad. There is an observer outside the train car standing right beside the track. There is another observer standing in the train car. Also, there is a strange device placed in the exact center the train car; it is a double-sided photon gun. When this photon gun activates, it will fire a photon towards the front end of the train car and it will fire another photon towards the opposite side of the train car.

Now, Maxwell's equation implies the speed of all electrodynamic radiation, or in our case, light, travels with a constant speed, $c$ (~3.00e8 m/sec).
But the question at the time was: relative to what?

Everything.

Light travels at the same speed for any inertial reference frame. This realization revealed humanity's misconception of the idea of simultaneity.

Now back to the thought experiment. The double photon gun is set to activate when the exact center of the train is aligned with the observer outside of the train car.
When that happens, the observer inside the train car would see the two photons traveling at the same speed, $c$, in different directions and hit the sides of the train car at the same time. To the observer inside the train car, the two photons would hit the walls simultaneously. This would be expected.

Now, lets imagine the event though the other observer's eyes. When the center of the train car is aligned with this observer, the double photon gun will activate. But the train car is traveling at velocity $\vec v$. What is the speed of each photon relative to this observer? The answer is $c$. The observer would see that the photons are traveling towards the walls of the train car with speed $c$. Since the observer also notices that the walls of the train car moving at velocity $\vec v$, he/she would find that the back wall of the train car would be struck by a photon before the front wall of the train car. To the observer outside the train car, this event would not be simultaneous.

This is one of my favorite thought experiments.

5. Dec 31, 2014

### Staff: Mentor

It's actually the exact same thought experiment, except with the light going in the other direction. The firing of your bidirectional light source is the common event analogous to the two flashes reaching the observer's eyes at the same time; and the arrival of the flashes at the two ends of the train is analogous to the two lightning strikes.
(This is a comment, not a criticism or a complaint.)

I presume that you've noticed that you could put the bidirectional light source on the platform and get the exact same result: signals reach the ends of the train simultaneously according to the train observer but not the platform observer?

6. Jan 1, 2015

### sjkelleyjr

But herein lies my intuitive confusion. It seems to me that if the photons fire at $c$ they too would hit each side of the train simultaneously with reference to the inside passenger, albeit farther down the tracks. I think I'm visualizing $\vec v$ as something roughly equal to a real train, in which the $\vec v$ is kind of inconsequential (my first error). I also feel also though $c$ is being added to $\vec v$ and thus traveling with the train (or carrying it down the railway with it, similar to a passenger), rather than being the limiting velocity (my second error).

Yes this is what I meant. Whoa...that's and interesting way of looking at the embankment observer vs passenger perspective (and perhaps a more mature way).

But I guess what I'm saying is, if we assume this hadn't all been proven by the math and by Einstein (ie we don't assume there are different times with which to take the snapshot) if we had this ability to take a magic snapshot in both perspectives, wouldn't it show the same thing? lightning striking at each end of the train simultaneously to the outside observer, and lightening striking the front and the back of the train simultaneously to the insider observer. At the time of this magic snapshot, velocity of the train would be irrelevant would it not?

I suppose this point in time is kind of irrelevant though, because the thought experiment is observing $c$ throughout the experiment, and thus we need to "snapshot" even further in time after the initial strikes to understand fully the purpose of the thought experiment. I think all of these responses led me to a better understanding. Thanks to all who responded.

7. Jan 1, 2015

### Staff: Mentor

Looking at it that way (train at rest and platform moving backwards has to be equivalent to platform at rest and train moving forwards) is one of the essential breakthrough moments in understanding special relativity.

It's hard to rid ourselves of the habit of thinking that "at rest relative to the ground under my feet" means "not moving" - but you might consider that a hypothetical Martian, seated comfortably in his chair in front of a telescope on Mars and watching the Earth spinning on its axis and careening around the sun at several kilometers per second will consider the suggestion that either the platform or the train is "not moving" to be absurd.

8. Jan 1, 2015

### pervect

Staff Emeritus
That would be online at http://www.bartleby.com/173/9.html, correct?

Take a look at Figure 1 again, and read the text:

The added emphasis on the word embankment is mine, when we compare Einstein's words to yours, we see that you somehow switched from the embankment frame to the train frame.

Note that Einstein (and us in his footsteps) are doing the analysis in the embankment frame, and that because of this, the correct conclusions we can draw apply in the embankment frame.

When we compare your words to Einstein, we see a crucial difference, an intuitive leap perhaps.
The answer is no, the lightening bolts reach the midpoint of the embankment simultaneously.

Now, the next thing you/we need to ask is: When the light rays meet together at the midpoint of the embankment, where is the midpoint of the train? I'm torn between spelling it out and letting you think. As a compromise, I'll add the answer that I think you should be getting at the end of this post.

....

Maybe I should stop there, but I wanted to add something else.

An additional resource that might help (let me know if it does) "The challenge of changing deeply-held student beliefs about the relativity of simultaneity", by Scherr et al, available online at http://arxiv.org/ftp/physics/papers/0207/0207081.pdf

Of particular interest and relevance are the following sections:

Not quite stated in the above summary is the second observer Beth, at the midpoint of the train.

Note that this is similar to Einstein's original presentation, with a few elaborations, the addition of char marks and the specification of a particular observer Alan at the midpoint. Note that it is directly specified here that Alan receives the flashes at the same time (rather than relying on the reader to derive this from the definition of simultaneity).

The next section of interest suggests a further refinement, and the reason it was added.

-----

As promised, the answer to my question. The center of the train is not at the center of the embankment when the light flashes meet. The center of the train was at the center of the embankment when the light flashes were emitted (per figure 1), but because the light signals take time to meet, by the time they do meet, the center of the train has moved to the right of the center of the embankment.

9. Jan 1, 2015

### sjkelleyjr

Yes I understand all of this. I was absolutely making "intuitive leaps", perhaps because I was attempting to pretend as though none of this had already been proven thereby purposefully blinding myself logically.
My "snapshot" conception wasn't allowing for this time needed for the light signals to meet, and therein lay my misconception. You spelled it out elegantly with this bolded section.

I believe Nugatory also mentioned it in his/her first response, by asking whether these snapshots were camera snapshots or "magic" snapshots.

Thanks again for all the responses.

10. Mar 18, 2015

### Micheth

Hi.
What if you have two clocks, situated at the front and rear of the train (traveling at 99+c)
Lightning hits both of them simultaneously (as viewed by a person on the station platform).
The person on the train sees the front lightning hit much earlier than the rear lightning, but the lightning hitting each clock breaks it, leaving it displaying the respective times at which each was hit (which were exactly equal times, say at exactly 12:00:00: ... :00
The person on the train swears the first one hit first, but when the train comes to a stop and examines both broken clocks, the values are both exactly the same, no?

Reference frame-based simultaneity might say that the clocks would register different records at the time of their stopping, but what if the person on the platform took a (video/picture) of the clocks breaking and the times they displayed as viewed in his time frame? They would display the same times, and that can't be different from what they would find upon stopping the train and inspecting the stopped clocks times, right?

To me, this would seem to suggest that the non-simultaneity for the person on the train was merely an illusion and in fact the events occurred with an objective simultaneity...

11. Mar 18, 2015

### Ibix

To the person on the platform the two clocks were not correctly synchronised, so one would have read 12:00:00 and the other 12:00:01 (or whatever) when they were simultaneously struck. According to the person on the train, the clocks were correctly synchronised but were struck at 12:00:00 and 12:00:01 respectively.

12. Mar 18, 2015

### Micheth

I don't see why the platform viewer would not see the clocks synchronized? They're both moving at the same speed with respect to him, right?
And light from both of them take the same amount of time to reach him...

13. Mar 18, 2015

### Ibix

Observers in relative motion disagree on lengths, clock tick rates, and when clocks at different places where zeroed. All inertial observers will agree that the clocks tick at the same rate as each other, although they will not agree what rate that is. They will all agree that the clocks were never set to tht same time, except for the ones stationary with respect to the clocks.

This actually follows from the experiment you are proposing. If it weren't the case we'd end up with the result you thought of, with the implication that the platform is in an absolute rest frame. No evidence of any such rest frame has ever been found - every experiment has pointed to all inertial frames being equivalent.

Look up "relativity of simultaneity" to find out more detail.

14. Mar 18, 2015

### A.T.

It's not about what he actually sees with his eyes, but what happens in his frame when signal delays are accounted for. This might help:

Yes, they tick at the same rate in his frame, but with an offset.

That is not relevant. The key is that the signals that start the clocks, do not reach them simultaneously in his frame.

15. Mar 18, 2015

### Micheth

Hm? Why are they at an offset, when they get struck by the lightning at the same moment (in his frame), and are exactly the same distance away from him, moving at the same speed, at the moment they're struck?

16. Mar 18, 2015

### A.T.

Before you come to the lighting stopping the clocks, you need to start them first. How are the clocks synchronized initially?

17. Mar 18, 2015

### Micheth

I'm assuming they were synchronized initially when the train was at rest somewhere. Then they were accelerated together to their current uniform velocity of 99+c, so they must still be synchronized, right?
The moment they pass him, they are still synchronized, and at exact equal distances to him, and the lightning is in his own frame, so I would guess he would see both lightning bolts hit both clocks simultaneously and registering the same time when that happened.

18. Mar 18, 2015

### A.T.

They were accelerated together with the train, which contracts during the acceleration. So the clocks at its ends undergo different accelerations and don't stay synchronized.

19. Mar 18, 2015

### Staff: Mentor

Even if you had two guys on the ground at the exact locations where the lightning strikes hit recording the exact time that the strikes occurred, and you had two other guys on the train, also at the exact locations on the train where the lightning strikes hit recording the exact time that the two strikes occurred, if the guys on the ground report that the two strikes occurred simultaneously, the two guys on the train will report that the two strikes did not occur simultaneously. (That is, if all the clocks on the ground are synchronized with one another, while all the clocks on the train are independently synchronized with one another.)

Chet

Last edited: Mar 18, 2015
20. Mar 18, 2015

### Micheth

Why would the contraction of the train matter...?
Could we not dispense with the train altogether and just say the two clocks accelerate in the same direction (they have independent propulsion systems...)
They were at rest in the same frame, synchronized, and accelerated at the same time in the same direction, and register the same times at the instant the guy on the platform sees them at equal distances.

21. Mar 18, 2015

### bahamagreen

Now maybe I'm confused but I think your synch / non-sych polarity is backwards and your after 12 stopped clock times can't exist?

It looks to me like the data result is when the train and platform observers meet to examine the two clocks all together in the same place after the experiment and both stopped clocks read 12:00:00.000... , neither clock could have ever been showing a later time to either observer during the run of the experiment, only earlier times up to 12:00:00.000...

According to the person on the train, the strikes were staggered front first rear second, but at no point does he say he believes the clocks are synchronized. Clearly he must conclude they were not synchronized; the rear clock had to have been running behind the front clock in order to be struck after the first clock and have both clocks stopped reading 12.

According to the platform observer the clocks were synchronized and were both struck and stopped when reading 12.

If that is not correct, then I am confused.

But to Micheth's question... there is nothing "more real" about the clock synchrony. The Wiki link Shyan posted is the scenario where the clock synchrony is for the train observer rather than for the one on the platform. You can also take both versions and take the train to be at rest with respect to a moving platform... you can arrange the experiment to make clock synchrony for any third inertial observer and neither of the train or platform observers... nothing special about clock synchrony, it just makes the thought experiments clearer to arrange it for one of the observers.

22. Mar 18, 2015

### jbriggs444

This acceleration at the same time... is that at the same time according to the track frame or according to the train frame?

23. Mar 18, 2015

### Micheth

Both, since it is before they accelerate, so they are at rest with respect to the track & platform (though still being far, far away)

24. Mar 18, 2015

### A.T.

Because it's ends move at different speeds during the acceleration & contraction, and thus the clocks desynchronize.

Yes we can do this, but then the clocks will be desynchronized in their final rest frame.

25. Mar 18, 2015

### Micheth

Why is that?
I'm assuming they both accelerate and decelerate exactly in tandem.
(But of course in this example it doesn't matter what happens to them after they get hit by the lightning bolts since they're now broken and just continue to register the time they got hit).