# Rephrasing the train and platform thought experiment

1. Dec 16, 2013

### ScottTracy

Hi all,

I'm sure that things similar to this have been covered before. Also, I'm sure that a Minkowski diagram would probably clear things up no end but, anyway, please bear with me.

Let's assume that we have our familiar railway carriage packed with observers who are satisfied that they have synchronised their watches and that they are all separated by the same distance as measured by a rigid metal rod. The origin of their co-ordinate system is the centre of the train which can be called O'. Can we say that the ends of the train are at x' = -10 and x = 10? Finally, we place two opposing flashlights at O' in line with the x' axis.

On the platform we have a load of observers with watches and they spread out from the origin O making up the platform frame of reference.

O' is approaching O at some velocity v which has only an x component.

The usual thought experiment is that the flashlights are fired as O' passes O. Let's pretend that this is achieved by some electrical contact on the track that closes a circuit or something. The observers on the train get together and conclude that the flashes reached the ends of the train at the same time. The observers on the platform get together and conclude that the flash of light going in the -x direction reached the back of the train before the forward (+ve x) travelling light pulse reached the front.

This shows nicely that two events (the light pulses reaching the ends of the train) may be simultaneous when viewed from one reference frame but not from another.

Now I think it can be shown that there is a way that both observers could agree that two pulses of light reach the two ends of the carriage at the same time. If we work backwards from the assumption, then I think that the platform observers would have to see the forward-facing flashlight fire before the backwards facing one. This would compensate for the fact that the front of the carriage is moving away from the wavefront, as the platform observers see things.

In practice, maybe this could be achieved by having two separate contacts on the train and two on the track, (i.e. one each on the train and the track for each flashlight). If we placed the contact for the forward flashlight more to the left (-ve x direction) on the track than the other, then this would trigger the two lights at seperate times as seen from the platform.

However, the train observers see the light reach the ends of the carriage at the same time and they can easily confirm that the flashlights are in the centre of the train. Given that they measure the speed of light to be constant, wouldn't the train passenger at O' (in the centre of the carriage) confirm to the others that the flashlights fired at the same time ("simultaneously")?

So far, I hope that the example is, if a little strained, still acceptable. The disagreement about the simultaneity of the forward flash and the backward flash are as predicted by the relativity of simultaneity.

What I'm wondering is: If the train passenger at O' sees the flashes simultaneously then they must see themselves passing the two contacts on the track simultaneously (for that is how the lights are triggered). These contacts have been separated by the platform observers by some distance Δx in order to trigger the lights at separate times. Presumably with (Δt = v * Δx). How is it that the observers see Δx' to be zero?

My intuition (I know, a dangerous thing) tells me that Lorentz-Fitzgerald length contraction is probably not the compensating factor. Instead, I have probably just made a logical error.

Anyway, if all of the above is correct, could someone demonstrate the calculation whereby the track contact separation distance Δx is calculated as a function of velocity v so that all observers agree that the light pulses reached the ends of the train simultaneously?

Thanks in advance for the patient explanation and the pretty diagrams

Last edited: Dec 16, 2013
2. Dec 16, 2013

### K^2

I think you are assuming that signal from triggers to flashlights travels either instantly or in fixed time. Keep in mind that in order for two events to be casually related, they must be time-like, or at least, null-separated. In other words, you must allow for the signal to propagate from triggers to flashlights. And it must be a real signal propagation. Not just a fixed delay. For simplicity, lets say that this signal also propagates at the speed of light. The two triggers are separated by equal distance on the train, and so trivially, they are tripped at the same time and both flashlights go on at the same time.

Observers on the platform also see both flashlights go on at the same time, and they see two triggers tripped at different times. However, they still observe signal from each trigger travel at speed of light. That's speed of light relative to them, the observers. And that means, time to reach flashlights will be different from two triggers, according to platform observers. If you do the math, you'll find that this difference compensates exactly for difference between timing on each trigger.

Finally, one note on coordinate systems. Yes, you can always find a coordinate system that has its origin fixed at the platform, and yet describes any preserving simultaneity of events on the train. These are coordinate systems, and how you draw your time and position grids is entirely up to you. And that might be the source of your intuition on this. However, the trade off is going to be that this new system isn't inertial. Two inertial coordinate systems that preserve order of all events can only differ by rotation. This is important. This is what Special Relativity is built on. Inertial systems are nice and easy to work with. But there are ways of working with accelerated frames of reference. That's what General Relativity is all about. And there, you can bend coordinate system any way you like. You'll just have to account for fictitious forces that arise from it.

3. Dec 16, 2013

### ScottTracy

Thanks for the reply K^2. Can I ask what you mean by "in order for two events to be casually related, they must be time-like, or at least, null-separated". It's this kind of general statement that I wish I could understand better without having to resort to time-slice by time-slice examples as I did above.

Likewise, "Two inertial coordinate systems that preserve order of all events can only differ by rotation". What about ones that are just displaced relative to one another. Is that a trivial case? I presume that if my coordinates are just displaced "to the right" slightly then order is still preserved?

4. Dec 16, 2013

### K^2

Sorry, I was thinking of axis orientation. Rotation or translation it is, of course. Point is, if we established that all simultaneous events in one frame are simultaneous in another, then the direction of the time axes match. So one frame can't be boosted relative to the other.

Time-like separated means the two events are within each other's light cones. In simplest terms, light from one event has time to reach the other. And that means that the earlier event can be the cause of the later event. Null-separated means its exactly on the light cone. They are separated in space by exactly the right amount for light to reach from one to the other.

Because such events may be casually linked, their order cannot be switched due to a change of coordinate system. If I flip the switch, and the lights come on, switch is flipped before lights coming on in absolutely any inertial reference frame. (Faster-than-light travel notwithstanding.)

On the other hand, if two events are space-like separated, so that they are too far apart in space, but too close together in time for message from one event to reach the other, then one can never be the cause of the other. So there is no such limitation. And in fact, I can always choose a coordinate system in which order of these two events is reversed. I can also find a coordinate system in which two such events are simultaneous.

5. Dec 16, 2013

### Staff: Mentor

As I understand the setup, the distance between a flashlight and its trigger is purely vertical, i.e. perpendicular to the train's motion. In that case, you can idealize the setup to eliminate signal-propagation delays. Imagine putting the flashlights on the floor of the train, which is in turn only a very short distance above the track or ground. With an "ideal train" you can make this vertical distance as short as you like.

6. Dec 16, 2013

### K^2

In which case, triggering events are simultaneous in both frames and there is no problem.

7. Dec 16, 2013

### ScottTracy

That's right jtbell. As I was imagining it, the two trigger contacts were directly under the flashlights (ie an infinitesimal distance below) and are side by side along the z axis (parallel with the railway sleepers/axles).

The two triggering contacts on the track are displaced along the x axis so that they trigger the flashes at two non-simultaneous times (according to the platform).

8. Dec 16, 2013

### ScottTracy

Sorry for the terrible diagram

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9. Dec 17, 2013

### ghwellsjr

Hi, and welcome to PF.

Your last statement is excellent. You are pointing out that simultaneity is a reference frame issue and not an observer issue.

If by "both observers" you really mean "both reference frames", then I think you are mistaken.

So far, so good.

Not so good. Why did you backtrack from your previous excellent statement? Why do you think the light will reach the ends of the carriage at the same time in the train frame? If it's true for the platform frame, which you said it was, then it cannot be true for the train frame.

No. In order for two events at the same location in any frame to be "simultaneous", they have to be the same event. So if the two flashlights (which are at the same location) fired at the same time in the train frame, they also have to fire at the same time in the platform frame. But you already said that the two flashlights fire at different locations in the platform frame at different times, so they cannot be the same event.

Yes, pretty diagrams may help you understand what happens. First, is a spacetime diagram depicting your scenario in the rest frame of the platform. The train, traveling at 0.6c, is shown as the three thick diagonal lines and the platform observer at x-0 is shown in green. As you proposed, we work the problem backwards showing the two light flashes arriving simultaneously at t=0 at the two ends of the train. This determines when the light flashes occurred at the blue center of the train. The dots represent one-nanoseconds of Proper Time for each observer. The speed of light is one foot per nanosecond:

As you described, the forward flash is set to go off when the center of the train gets to the position of x=-12 and the backwards flash is set to go off at x=-3.

Now we use the Lorentz Transformation process (not our intuition) at a speed of 0.6c to see what the coordinates are for the rest frame of the train:

Now you can plainly see that the arrival of the two light flashes is not simultaneous in the train's frame. In fact, the forward flash is sent and arrives before the backwards flash is even sent.

Does this all make sense to you know?

Any questions?

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10. Dec 17, 2013

### ScottTracy

Yes, a slip of the tongue. I did mean both reference frames. I was actually picturing it as two actual observers on the train (one at each end) and two actual observers stood opposite them on the platform.

Since I was picturing the train-frame observers at the ends of the train lining up with the observers on the platform, I think that I was imagining them "sharing" clocks or at least being able to sneak a peak at each others. I think that this is a problem that a lot of beginners have; assuming that a shared x coordinate means that two observers can consult the same clock rather than using their own.

That's a very clear way of phrasing it. I guess I was trying to split one event up into two somehow.

Finally, it does make sense! Thanks for taking the time to explain it.

Can I ask what you used to draw the diagrams?

11. Dec 17, 2013

### ghwellsjr

12. Dec 17, 2013

### ScottTracy

A phrase that brings a tear to my eye.