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Relativity of Simultaneity - Question about the Classic Train Example

  1. Jan 6, 2010 #1
    This is my first time posting so it's nice to meet everyone!

    I'm not trained in physics, but lately I've been very interested in and reading a lot about both Relativity and Quantum Mechanics. With regard to relativity, I found the topic of relativity of simultaneity very interesting. The example often given (which I'm sure everyone here knows) of a train passing an observer on a platform, and light from a source inside reaching the train car's ends at different times from the two reference frames, was very helpful in my understanding.

    So to help myself better understand relativity of simultaneity, I drew out a couple diagrams of the train example. In general, everything made sense, except for one situation that I'm wondering if someone here could tell me the answer to. I'm sure to someone with a firm grasp of special relativity, the answer to this will be obvious.

    I'll describe the situation first (which is a variation of the train passing a stationary observer example), and then ask the question.

    Imagine, just as in the aforementioned example, you are standing in a field with train tracks in the distance. Over these train tracks passes a locomotive that moves very fast (at whatever speed is used in the other examples so this phenomenon is apparent), and pulls behind it a glass train car you can see inside.

    In the exact middle of the train car is a light source that sends a beam of light towards the front of the train car, and a beam of light towards the back of the train car, at the exact same time (to a passenger on the train of course). Again, exactly like in the train example given by Einstein and others.

    Here's what I was thinking that I don't quite know the answer to. Let's say one places on the front and back walls of the train photon detectors. These photon detectors are connected to an apparatus on top of the train that consists of the following: timers that measure in nanoseconds, a jack-in-the-box, and a digital billboard. The way the setup would work is that the timers would start the moment the light from the source on the train was emitted (so always at the same time, since it's the same source), and stop when their respective photon detector detected the light. There is a timer for each photon sensor, so the beams' travel times are measured independently. Additionally, as the timers runs, their elapsed times are displayed independently on the billboard (and you, as the outside observer, can see this).

    The jack-in-the-box does one thing: if the final times from the two timers are the same, it will pop out of its box. If not, it stays inside. Obviously this whole apparatus is part of the train's reference frame.

    So here's my question. In the usual examples, it is said that to an observer on the train, the light would reach the walls, and thus the photon sensors at the same time. This would be reflected in the timers (they would both display, say, 8 nanoseconds), and the jack-in-the-box would pop out.

    How would you, the outside observer, see things? It is always said that the observer will see that the light hits the back wall of the train car first. How would the position of the two light beams relate to the number of nanoseconds on the timers that the observer can see? They would obviously appear to be running slower, but would the timer timing the front seem to run slower than the one timing the back (they're both in the same reference frame though, so this I don't get)? And would the jack-in-the-box pop out when the back photon hit for the outside observer, or would it wait for the front photon to hit? I assume that the times would still show up on the billboard the same and the jack-in-the-box would pop out, but I can't grasp how the whole thing would appear to the outside observer.

    I apologize in advance if there is something super basic that I have overlooked or if this question has been answered multiple times in different texts.

    Thanks much in advance!
     
  2. jcsd
  3. Jan 7, 2010 #2

    A.T.

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    The problem here is that the timers cannot be at same place with the source and the detector.

    - If you place the timers far apart at the detectors, the observers will disagree if they were started simultaneously with the emission of the signal.

    - If you place both timers at the source you cannot stop them simultaneously with the detections. And the observers will disagree on the delays between detection and timer stop.

    The above possibilities "cancel out" the effect of relative simultaneity, so both observes agree if the jack comes out of the box. But they disagree how it happened that the timers show the same time at the end.
     
  4. Jan 7, 2010 #3
    The timers at the back and front are running slow according to the outside observer but they are also running at the same rate as each other according to the outside observer. The difference is that the outside observer sees a permanent offset between the front and back clocks with the back clock permanently ahead of the front clock by a fixed amount of nanoseconds. *
    You are forgetting that it takes time for the information that the clocks have stopped to be transmitted back to the digital readouts and jack-in-the-box at the centre. Let us say that the electrical signals are transmitted at the speed of light. This is the fastest rate at which the electrical signals could be transmitted and in practice the signal transmission rate would probably be slower. When the light signals start their journey the outside observer might see +5 nanoseconds on the back clock and -5 nanoseconds on the front clock because to the outside observer the clocks do not look synchronised. The outside observer sees a light signal hit the back clock first. Let us say it takes 3 nanoseconds of train time for the light to arrive at the back then the total time indicated on the back clock is 3+5 = 8 nanoseconds. The digital displays at the centre of the train do not yet know the light signal has arrived at the back because of the time delay in sending the electrical signal back to the middle. The light signal going forward takes longer to arrive at the front because the front is going away from the light signal. Let us say that it takes 13 nanoseconds of train time for the light signal to arrive at the front. The time indicated by the arrival of the forward going light signal is 13-5 = 8 nanoseconds because of the original offset. Now the information of the arrival of the front signal is transmitted back to the centre which is not yet aware that the light signal has arrived. The electrical signal from the front now only takes 3 nanosends train time to return to the centre while the electrical signal from the back takes 13 nanoseconds train time to return to the centre so they both have a round trip time of 16 nanoseconds train time and return at to centre simultaneously and both indicate 8 nanosends on the digital displays. The outside observer also sees both central digital displays light up and read 8 nanoseconds at the same time (although this is some time later than he saw the light signals arrive at either the back or front of the train) and he also sees the jack-in-the-box pop out.

    * If you want to work out the exact time offset on the front and back clocks the formula is t(offset) = -L*v/c^2 where L is half the length of the train as measured by someone onboard the train. For an offset of +/- 5 nanoseconds as I have used, the speed of the train would have to be 5/8 = 0.625c.
     
  5. Jan 7, 2010 #4

    jtbell

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    Also, if those signals don't travel at c, then you have to use the "velocity addition" equation to calculate how fast those signals travel in the ground observer's reference frame. For example, the signals might be baseballs shot from guns located at the clocks, with a fixed speed of (say) 50 m/sec in the train's reference frame. If the balls hit the box simultaneously, Jack pops up.

    An observer on the train and an observer on the ground disagree on whether the balls are shot simultaneously, and on whether they're traveling at the same speed or not, and on how far they have to travel. Nevertheless, both observers in the end agree on whether Jack pops up or not.
     
  6. Jan 7, 2010 #5
    Thanks for the replies everyone!

    I was always thinking in my head that the timer would be at the source, and somehow never thought that the information that the photon hit the detector would have to travel back (in the same apparent fashion as the opposite beam is traveling) to the source.

    Clearly, the information transfer will cancel out the part I was confused about.

    Just out of curiosity, is there a way that quantum entanglement could be used in this scenario (though I imagine the apparatus would be much, much more fanciful) to transfer the information that the photon hit the detector back to the source? Could anyone recommend a good paper or book that describes how the worlds of special relativity and quantum entanglement would interact?

    And thanks for the equations. I'll have to try these out!

    Thanks again!
     
  7. Jan 7, 2010 #6
    I don't believe this is right. Maybe I'm wrong, but I believe the observer on the train would see the light hit the front first, as length is contracted in the direction of velocity.
     
  8. Jan 7, 2010 #7

    jtbell

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    In the reference frame of the observer on the train, the train is not length contracted.
     
  9. Jan 7, 2010 #8
    Hmm, then what about the train example where...

    An observer outside the train is perfectly in line with a passenger in the exact middle of the train when two lightning flashes occur at the front and back of the train. The observer sees the light flashes at the same time, as he is equi-distance from the source of the flashes. And the observer in the train sees the front flash first. And both conclude different scenarios of simultaneity.

    However, I posed the argument that in the passenger's reference frame, being inertial, and following the postulate that c is the same in all inertial reference frames, the flashes must reach the passenger at the same time.

    Someone once told me that the train is contracted in the direction of velocity.
     
  10. Feb 24, 2010 #9
    It is given that the flashes occur simultaneously in the ground time frame. In the train time frame, not nercessarily so. Time changs, not light pseed, so time has to change to make the distance covered the same. The train observer sees the front flash occur first in the ground time frame and this is an essential fact and is constant throughout all time frames. What the on board train observer sees is the same in all time frames - front flash first. What changes between time frames is the distance the flash must travel to reach to the onboard observer (depending on the speed of the train.) When the train is considered the frame of refernmce, the two distances traveled by the flash in order to hit the observer are the same. Therefore the onset of the flashes must be different to maintain the relativity of the events (front flash first to on board observer.)

    Time can vary under relativity, not the relation of events (event A before event B, etc.) or the speed of light.
     
  11. Feb 24, 2010 #10

    JesseM

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    Well, the "when" in the phrase "when two lightning flashes occur" seems to mean "simultaneously with", but of course in relativity this depends on your choice of frame. In Einstein's version of the thought-experiment, he stipulated that the lightning strikes were simultaneous with each other, and simultaneous with the event of the passenger being in line with the outside observer, in the outside observer's rest frame.
    But as I explained in post #7 of the thread you started on this before, all frames must agree about local events:
    Since all frames agree on local events, it must still be true in the train-observer's frame that the light from the strikes reaches him at different times. Since light must travel at c in his frame and both strikes happened at equal distances from him, it must therefore be true that the strikes happened at different time-coordinates in his frame.

    But as I also pointed out in post #19 of that same thread:
     
  12. Feb 25, 2010 #11
    In section IX of "Relativity" things would have been a lot clearer had Einstein explained that at a given point the sequence of events remains the same in ALL frames of reference and that time and distance change to conform to the constancy of c.

    If we use the train as example of the "inertial" time frame or S, the given the flashes simultaneously at front and back and the equidistance to M', the observer, then, of course the time for light to reach M' from B' and A' (the front and back of the train, respectively) would be the same as danielletha4 points out. But, on the ground, which now is moving "backwards" at velocity v, the distances between the front strike to the observer (when he "sees" the flash) and the back strike to the observer when he sees the flash would be different. The observer would be closer to the front strike origin at the time he sees the flash than he would to the back strike origin. But, the observer sees the strikes together - that is preserved hence the times of origin of the flashes would be different in the two different frames of reference.

    The confusing part of Einstein's example is that the ground observer and the train observer are two different observers. Hence, the ground observer sees the strikes simultaneously, but the train observer sees the strikes - the front flash before the back. Now using the train observer as the inertial frame, this order or sequence of events is preserved. You can't go back to the ground observer and say "but he sees them together" because if you track him in the frame of reference, he will see the back strike first, then the front strike - just the opposite of the train observer.

    If Einstein had clarified this point this would not have been so tedious to understand.

    Distance and time can vary but c cannot

    The equation x^2 -ct^2 = (x')^2 - c(t'^2) derived from the Lorentz equations are the mathematical expressions of this which "tie together" the relationships between coordinates in space and time based on constant c and universal laws of physics in all time frames. Time and distance can both vary yet maintain the same original relationship in the original reference frame.

    To JesseM - It took a long time to figure this out but thanks for hammering this home again and again.
     
  13. Feb 25, 2010 #12

    JesseM

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    No problem, glad it's making sense to you now! Einstein's discussion goes pretty fast so it does take some time to see how all the pieces fit together...
     
  14. Feb 25, 2010 #13
    What blog do I go for this one?

    In the Einsteinian train example, it is given that the observer on the train sees the flash before the observer off the train does.

    If the the train is "1 c" in length (300,000 km) and is moving at 0.5 c to the right. And the flash occurs at t= 0, the train's relativistic speed is what? (It is NOT 0.5 c but is some what less - would that be by the factor [SQRT(1 + 0.5^2/1^2]^(-1)? Is the relativistic length likewise 1c *the same gamma factor?

    How would you calculate the meeting time and meeting point of the observer and the flash of light travelling back to meet the observer?

    If one uses "closure velocity" one would say Vc = (1 + 0.5)/[1 + ((0.5)(1)/1*1)] but that doesn't make sense or does it? I guess the closure velocity of ANY moving object on an advancing light wave is always c, no matter how fast the object is moving towards the light (surely can't be greater or less.) So I guess the light takes 0.5 sec to travel from the emanation point to the observer. But we say that the observer meets the light before the ground observer travelling at 0 does. Can someone show me the math on that as I can' get it to work out.

    Or, get me to another post where this is described as I can't find it.
     
    Last edited: Feb 25, 2010
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