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Ok, so I know there have been some variations of this passed around and answered already, but I still can't quite understand how this works, so I want to clarify some particular points in this.

Lets start with Einstein's relativity of simultaneity thought experiment. Specifically the variant where a person stands in the middle of a train car with a lightbulb and another one is stationary on the ground next to the train as it passes by. When the moving observer lights the lightbulb he will wee the light reach both ends of the rail car at the same time. Meanwhile the stationary observer will see the light reach the rear end of the train car first.

I get the experiment itself. However I can't quite understand how it works if we start trying to measure the difference in relativity. Say we put a light detector and an attosecond-accurate clock in each end of the train car and sync the clocks before the experiment (when both observers start in the same "stationary" frame of reference and can agree that the clocks a synced). When the light detector registers a photon, the clocks save the current timestamp and send it to a central computer. When the central computer receives both timestamps it compares them and outputs "EQUAL" or "DIFFERENT" on a screen.

In the original thought experiment we talk of the observers "observing" all the events take place, so, using the same terminology, the observer in the train car will, obviously see the light hit the detectors at the same time. The detectors will both be showing the same time (both will be showing t0 for example) when the light reaches both of them. He will, therefore, see the detectors record identical timestamps at that moment and send them to the central computer.

The stationary observer will observe the light hit the detectors at different moments in time. As both clocks are in the same frame of reference there is no time dilation between them, and the stationary observer will see them showing the same time at any specific point in time (even if they are slightly diverged from a similar clock he might have). Therefore he will see them both showing, say, t1 when light hits the first one, and both showing t2 when light hits the second one, which is fundamentally different from the first case. He will see them record different timestamps and send them to the central computer.

So now we have one person who saw the computer receive 2 identical timestamps, and another who saw it receive 2 different timestamps. So they will observe the computer perform the calculations and output different answers.

I understand there must be an error in such logic, but I can't understand where and why. Theoretically both observers could see the whole process happening up to and including the displaying of the result on the screen. At the same time, if the train stops and both observers walk up to the screen you would expect them to agree as to what is output on it.

Basically the specific questions I want to understand are:

  1. If you are the stationary observer, then you should theoretically be able to observe the whole process and see "DIFFERENT" on the screen. Or if not, then how could you see anything else?
  2. If you are the moving observer, then you should theoretically be able to observe the whole process and see "EQUAL" on the screen. Or if not, then how could you see anything else?

If anyone can explain this I would be extremely grateful, because this is frying my brain.

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The clocks are synched in the platform frame, so can't be synched in the train frame.

The story in the platform frame: The light beam takes longer to get to one of the synched clocks than to the other. Therefore it hits one clock when it says 1PM and the other when it says 2PM.

The story in the train frame: The light beam hits both clocks at the same time. However, one clock runs an hour behind the other, so the readings on the clocks when it hits are 1PM and 2PM.

(Or, if the clocks are synched in the train frame --- and therefore not in the platform frame --- you can tell essentially the same story in reverse.)

Edited to add in response to the OP's comment:

Why the clocks can't be synchronized in both frames: Because two points determine a line. If there is a frame in which both both clocks read 12:00 at the same time, then the line connecting those two events is a line of simultaneity for that frame. With just one spatial dimension, this is enough to uniquely determine the frame.

How the clocks get out of synch: You haven't given us enough information to answer this question. It depends on how the train starts moving.

Scenario A: In the platform frame, all parts of the train suddenly start moving rightward at the same time. Then in the (final) train frame, the train is initially moving left (so its synchronized clocks were both running slow) but then it stops. Moreover, the left side of the train stops before the right side does. Therefore there's a period when the right clock is running slow and the left clock is running normally. Therefore they get out of synch. (And incidentally, during the time when the right side is moving but the left side isn't, the train stretches.)

Scenario B: In the (final) train frame, all parts of the train suddenly start moving rightward at the same time. Then in the platform frame, the train is initially still (so its synchronized clocks run normally) but eventually starts moving. Moreover, the left side of the train starts moving before the right side does. Therefore there's a period in which the left clock runs slow while the right clock runs normally. Therefore they get out of synch. (And incidentally, during the period when the left side is moving but the right side isn't, the train shrinks.)

How I knew all this: I drew the spacetime diagram, which is the best way to solve almost any problem in relativity.

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  • $\begingroup$ Hmmm... But why though? If the train starts out as standing and the clocks are synced then they will be synced in bot frames, right (because its the same frame at the moment, as the train isn't moving)? So then the train accelerates. but why would that make them desync in either frame, if they are both accelerating identically? I mean yeah, they would desync from an identical stationary clock, but why from each other if they are moving in the exact same way at the same time? Or is it somehow just impossible to sync them for two different observers even if they are stationary? If so - why? $\endgroup$ – DFined Jul 12 '20 at 7:42
  • $\begingroup$ @DFined : I've added to the answer in response to your comment. $\endgroup$ – WillO Jul 12 '20 at 12:01
  • $\begingroup$ @PM2Ring : Thanks for this. I think the typos are fixed now, but I'll be glad to know if I missed any. $\endgroup$ – WillO Jul 12 '20 at 14:31
  • $\begingroup$ It looks ok to me now. I think the stuff about the different parts of the train not all starting to move simultaneously (in the train frame) makes it a bit more complicated, and potentially confusing. We still have the core issue of the relativity of simultaneity even if the train is rigid. OTOH, nothing can be perfectly rigid in SR... $\endgroup$ – PM 2Ring Jul 12 '20 at 14:57
  • $\begingroup$ Im afraid I still don't really get it. The rigidity of the train does seem a bit much to consider here. The core principle of relativity of simultaneity should still work with a perfectly rigid train as PM 2Ring said. Also the train could be accelerated from the front, back or middle and the result of the experiment would not change, but which parts accelerate first would, so I can't see how rigidity would play a role here. $\endgroup$ – DFined Jul 12 '20 at 15:44
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Is your paradox resolved by asking in which frame of reference the computer is located? For sure the events (light hitting ends of train) are simultaneous in the train frame but not in the platform frame. But you are adding another element - a computer that is somehow not in one frame or the other.

So let's say the computer is on the train, in the middle of the car, and at the moment the light signals hit the end of the car the sensors transmit signals to the computer wirelessly. They will reach the computer at the same time and the computer will read EQUAL. (If the computer is not in the middle of the car it would be programmed to take account of the travel time of the signals.)

If a similarly programmed computer is on the platform it must read DIFFERENT. That is, in the train frame the events are simultaneous (computer reads EQUAL) but in the platform frame the events are not (computer reads DIFFERENT). Each observer, if he knows special relativity, would be able to calculate what the other observer should measure and there would be no paradox.

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  • $\begingroup$ I was thinking a computer on the train but watched by both observers at the same time. So as to eliminate the propagation time of the signal I propose sending not just a signal but specifically a recorded timestamp. So as the computer is in the train frame and watched by two different observers just like the light and detectors I still don't get what it will compute $\endgroup$ – DFined Jul 12 '20 at 15:29
  • $\begingroup$ Whether it's sent to a computer, stored internally or printed out doesn't really matter - the timestamps of the train's clocks will show the same time, as they should. I think the problem arises when you say the platform observer "will see them both showing, say, t1...and both showing t2." In his frame, the events are not simultaneous. If he had clocks that recorded the events their times would be different. He can't change what the train's clocks read and, knowing the Lorentz Transformation, he would expect the train's clocks to have the same timestamp. $\endgroup$ – Not_Einstein Jul 12 '20 at 17:45
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The error in your logic is straightforward.

Let's say the detectors on the train each record an incoming photon at 4pm train time, flashing the time on a display when they do so. On the train, the clocks are in synch and the light arrives at each end at the same time.

An observer on the platform will see the light reach the rear of the train before it reaches the front. When the light reaches the rear, the detector there will be seen by the observer to flash 4pm. Later, when the light reaches the front, the detector there will be seen by the observer to flash 4pm. The observer will therefore conclude that the clocks on the train must be out of synch, since one flashed 4pm before the other did.

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