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Time Dilation Paradox (Not resolved)

  1. Sep 25, 2012 #1

    bgq

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    Hi,

    I am new to this forum, and I wish I get some help here.

    I really can't find any solution to this paradox; I am going to deny all the special theory of relativity because of this. The paradox is very simple, here it is:

    Consider two persons sitting on two adjacent trains at rest. Each person has a timer, and both of them are set to zero. The timers are designed in a way so that when the timer reaches 2 minutes, the other train will explode (and of course damaging the timer inside)! The two trains start moving in opposite directions with same constant speed; the timers start exactly as the trains starts to move.

    Now here is the paradox: For each person frame of reference, he is at rest, and the other is in motion, then the other timer will run slower, so he will be very happy that his timer will reach 2 minutes first ... destroying the other train. The paradoxical question which of the trains will be destroyed?! Although the whole situation is completely symmetrical, yet each person will assure that his timer will reaches 2 minutes first destroying the other train.

    Thanks in advance for any ideas.
     
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  3. Sep 25, 2012 #2

    mfb

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    How does train 1 decide "the other timer reached 2 minutes"? It depends on the position and velocity of the timer and train. Remember: There is no universal simultaneity in special relativity.
    If you add specific position data, I can show you how that works. Both trains will explode at the same time, if you fix some method to determine "the time the other timer shows".

    Relativity is self-consistent, if a paradox does not work out this just means you did something wrong.
     
  4. Sep 25, 2012 #3

    bgq

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    Thank you for your reply.

    The train may know that the timer reaches 2 minutes by electromagnetic signal for example. My point is whatever the mechanism of communication, the situation is completely symmetrical.

    Anyway, each person can predict that his timer will reach 2 minutes before the other although the symmetry of the situation assures that both will be at the same instant (lets say with respect to a third observer which is in complete symmetry with respect to the two trains).
     
    Last edited: Sep 25, 2012
  5. Sep 25, 2012 #4

    ghwellsjr

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    The problem is your statement that "the other train will explode (and of course damaging the timer inside)!", as if this can happen in an instant of time. Assuming that you have explosives along the length of each train that can be set off by the other train at any point and then the explosion propagates along the rest of the train at some finite speed, then both trains will explode. It's perfectly symmetrical.
     
  6. Sep 25, 2012 #5

    PAllen

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    Such a signal travels at a finite speed (<= c). In that case, both trains explode, each some time after their own timer passes 2 minutes. Each sees their own timer reaching two minutes, knowing the other train will explode. Each explodes before seeing the other explode.

    Adding a little to this: each train sees their own clock reach 2 minutes first, knowing the other guy is doomed. However, some time later, each sees the other clock approaching 2 minutes, and knows they are doomed. If signals travel at light speed, each blows up the moment they see the other clock reach 2 minutes (and theirs reads later). If signals travel by current over long wires, each blows up some time after the other clock reaches 2 minutes, and theirs is well beyond two minutes.
     
    Last edited: Sep 25, 2012
  7. Sep 25, 2012 #6
    The situation is not symmetrical. It's another instance of the 'twin' scenario.
    Each observes the other clock to run slower, but only a reunion will allow a comparison for accumulated time. There are three possible explosive outcomes, 1, 2, or both.
     
  8. Sep 25, 2012 #7

    ghwellsjr

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    I thought the whole idea here was that the timer (or the observer) on each train causes the other train to explode (or initiates the explosion) and that when the timer (and the observer) explodes, it prevents the other train from exploding. Wouldn't each observer see immediately the other train begin to explode at his 2-minute mark, not some time later? What do you mean by "Each explodes before seeing the other explode"? I guess you must mean each observer explodes before seeing the other observer explode, correct?
     
  9. Sep 25, 2012 #8

    ghwellsjr

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    The situation is symmetrical. Why do you think it is not? Where is the asymmetry? This has nothing to do with the 'twin' scenario.

    It also has nothing to do with each observer observing the other ones clock. Each observer only watches his own clock and when it reaches 2 minutes, it causes the other train to explode. Since the clocks are separated in space when they both cause the other train to explode, both trains will explode. There is only one possible outcome.
     
  10. Sep 25, 2012 #9
    I don't think it is the same as with Twin Paradox. Here we have straight velocities in opposite directions starting at the same time, the situation is completely symmetrical. Insert an embankment at the origin (where the trains start moving). From this frame, you would observe both trains explode at the same.

    There is no paradox: pick any train as your rest frame. As the trains depart (without outside references, you would only see the other as moving) you have to consider the time light from the other train takes to reach you. Say a light beam comes out of the train at a time T and position X, that light will take some time to reach you, and during that time, the train keeps moving forward. When the light reaches you, the time you will see on your clock will be:

    T + the time it takes light to travel the distant between both trains

    This light will bring you information from the position X, where the train was located when the light departed from it. But when this light reaches you, the train will not be at the position X anymore, it wil be at:

    X + the distance the train traveled during the time light took to reach you

    That's where length contraction and time dilation comes from. You will see the train at a shorter distance than it will see itself, at a longer time than the train would see itself to be when he sent the light beam. The relativity transforms allow you to transform your local measurements into proper numbers.

    Thus, both trains would explode, when each clock ticks 2 minutes, the signals to detonate the bombs would be sent at exactly the same proper times, travel to the other during the same interval, and explode both trains at their proper times (which will be higher than 2).
     
  11. Sep 25, 2012 #10

    PAllen

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    I was replying to the OP's clarification that instantaneous communication is not implied:

    "The train may know that the timer reaches 2 minutes by electromagnetic signal for example. My point is whatever the mechanism of communication, the situation is completely symmetrical."

    This means, that for each train, the sequence of events is:

    1) I see my timer reach two minutes. I see the the other timer before two minutes. I now know the the other train will explode when it gets the signal from my timer reaching 2 minutes

    2) If the detonation signal is light speed, then each train sees the other timer reach two minutes (while theirs is beyond two minutes) at the moment off their own detonation.

    3) If they survive their own explosion, they see the other explosion some time after theirs.
     
    Last edited: Sep 25, 2012
  12. Sep 25, 2012 #11

    ghwellsjr

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    I guess when the OP talked about trains I assumed that they were very long and even after 2 minutes, they still hadn't passed each other and the signal to explode the other train didn't have very far to travel, just across to the next track. The way the rest of you seem to be interpreting this is that it would be the same as if we were talking about a single train car and a signal has to propagate over a long distance.

    But the answer remains the same either way. In the frame that the OP defined, once each clock reaches 2 minutes, it initiates a signal to explode the other train and both signals will be in transit simultaneously and will reach the other train car after the 2-minute mark on each train. You can transform the scenario into any other frame but it won't change the result.
     
    Last edited: Sep 25, 2012
  13. Sep 25, 2012 #12

    bgq

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    2) If the detonation signal is light speed, then each train sees the other timer reach two minutes (while theirs is beyond two minutes) at the moment off their own detonation.

    This is exactly my point: We have three possible outcomes, and each of them is obtained with respect to a certain inertial frame of reference. Each person will observe that his timer reaches 2 minutes before the other (thus the other explodes, but his own not), and for a third inertial frame (symmetrical to both), both timers reach 2 minutes at the same time, and so the both trains will explode.
     
  14. Sep 25, 2012 #13

    ghwellsjr

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    Different reference frames do not allow observers to see different things or for different things to actually happen. They only allow us (or the observers) to calculate different coordinates for events. What you are overlooking is the fact that both signals to detonate the other train are in transit at the same time and once the signals are sent, there is no way to prevent the explosion of the other train and so they both explode. Once you set up the problem in one frame, that is all you need to determine the outcome. Just for the fun of it, you can transform the same scenario into any other frame to see how that frame also predicts the same outcome but you have to do this correctly using the Lorentz Transformation process. You can't just redefine the scenario in a different reference frame and assume a different outcome.
     
  15. Sep 25, 2012 #14
    I have to disagree. You will explode before you see the other train exploding. When the signal of detonation reaches you, it will be reaching the other at the same time (but you won't be able to measure or observe this anyway). Both will explode, but by the time light from the other explosion reaches you, you're already dead (so you can't see the other explosion, unless you believe that your spirit lives on, but that's outside the scope of relativity).
     
  16. Sep 25, 2012 #15
    You don't know what their speed is before parting. Say it's u.
    A leaves B at speed u+v, B leaves A at u -v.
    The speed difference is (u+v)-(u-v) = 2v. This is true for any value of u
    that keeps the result <c. Time dilation before parting is a function of u/c.
    After parting what each observer measures is the difference in their absolute
    speeds, thus they cannot determine the time dilation of each, unless there is a
    clock comparison. The clock rate is symmetrical because it's doppler effects.
    Each sees the other clock running fast when converging, and slow when diverging.
    The speeds (u+v) and (u-v) cannot be equal, therefore we cannot expect equal
    time dilation for each. One will accumulate 2 minutes before the other.
     
  17. Sep 25, 2012 #16

    PAllen

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    Nonsense. The OP clearly stated:

    "
    Consider two persons sitting on two adjacent trains at rest.... The two trains start moving in opposite directions with same constant speed; the timers start exactly as the trains starts to move."

    Then, in clarification, the OP asserted complete symmetry. The OP's reasoning that symmetry must mean either neither explodes or both explode is correct reasoning. The only flaw was believing the SR said anything different, and also that, for stated scenario, SR clearly says both trains explode.

    [EDIT: I missed that phyti has a much bigger problem than reading the OP. Velocity addition in SR is not u+v. Instead it is (u+v)/(1-uv). Further, if you analyze from a frame moving at u relative to the initial trains, it will have no effect on the result at all - that is a fundamental feature of SR.]
     
    Last edited: Sep 25, 2012
  18. Sep 25, 2012 #17

    Ibix

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    Shall we do some maths?

    Let's consider the rest frame of train A. In this frame, train A is stationary and train B is doing velocity v in the x direction. At time t=2, train A emits a signal at speed V (not necessarily the speed of light, but V>v or it'll never catch the other train), which catches up with train B and causes it to explode at time tB. That time is easily calculated by equating the distance travelled by the train (vtB) with that travelled by the signal in two minutes' less (V(tB-2)):
    [tex]
    t_B=\frac{2V}{V-v}
    [/tex]

    What time (according to train A) does train B emit its signal? Train B emits its signal when its clock reads two minutes. Per the time dilation formula, then, train A says that train B emits its signal at time T:
    [tex]
    T=\frac{2}{\sqrt{1-v^2/c^2}}
    [/tex]

    Do the trains explode?

    Train B does - since T>2 for any non-zero v, train A emits its signal first and train B is doomed (unless it can outrun the detonation signal).

    Train A only explodes if train B receives the signal from train A after it emitted its own signal - in other words, train A explodes if T<tB. That is:
    [tex]
    \begin{eqnarray}
    \frac{2}{\sqrt{1-v^2/c^2}}&<&\frac{2V}{V-v}\\
    1-\frac{v}{V}&<&\sqrt{1-v^2/c^2}
    \end{eqnarray}
    [/tex]
    Square both sides, rearrange, cancel the ones, and pull out a common factor of v to get:
    [tex]
    v\left(v\frac{c^2+V^2}{c^2V^2}-\frac{2}{V}\right)<0
    [/tex]
    The left hand side has two roots and is negative between, so train B gets to emit its signal as long as the velocity of train B lies between those roots:
    [tex]
    0<v<\frac{2V}{1+V^2/c^2}
    [/tex]
    Note that the upper bound is never less than V unless V>c, which isn't possible. That means that if the train is going fast enough to exceed the upper bound, it is also going fast enough to outrun the detonation signal. So train A also explodes unless it is fast enough to outrun the detonation signal.

    Conclusion: Both trains explode unless they are going fast enough to outrun the detonation signals. Observers may differ about which one emits a signal first, and which one explodes first, but the delay between the detonate signal being sent and being received always allows the later train time to send its detonation signal.
     
  19. Sep 25, 2012 #18
    But isn't the beams sent when each train's proper time equals 2? Picking either train frames as your rest frame will result in different local and relative times, that's correct, but locally both will send their beams at t=2 proper. If you pick a third frame at the origin, and assuming all signals travel at the same speed, and both trains at the same speed relative to the origin, every event on each train will be observed as simultaneous. That is sort of a preferred frame, in the sense that it is the only frame of the three that can observe simultaneity, and is at the midpoint of every other event, isn't it? Of course it is not a preferred frame in any other sense, since you can choose any and all calculations from it will be valid. I think einstein proposed this very scenario, with two mirrors at 45 degrees so you could see all events at the same time. In this sense, both trains are observed as late by the same amount, and both would explode at the same proper time from the origin. Is this correct?
     
  20. Sep 25, 2012 #19

    Nugatory

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    If I'm understanding you properly, the answer is "yes".

    You will also notice that Ibix carefully worded his conclusion as "Observers may differ"; he did that because it is possible to select a frame in which the two events are simultaneous.

    Of course that frame is not preferred or special in any way - I can find such a frame for any two space-like separated events, and this is no more interesting than the observation that I can draw a straight line between any two points in a Euclidean plane.
     
  21. Sep 26, 2012 #20

    Ibix

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    As altergnostic says, yes, there exists a frame in which the signals are sent simultaneously and received simultaneously. But as Nugatory says - not redundantly at all - that is only a special frame in that it happens to reflect the symmetry of the problem. It's often a good frame to work in because things are often simpler in this frame (altergnostic apparently correctly analysed the situation in this frame without any maths at all, for example), but that doesn't make it the 'right' frame. Merely the convenient frame.

    I don't think Nugatory's line analogy is quite right, however. A better one would be that I can always rotate my coordinate frame so that a straight line is parallel to the x axis. It's a useful frame, but it's not the One True Frame in any sense.
     
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