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Special Relativity question (from non-physicist)

  1. May 6, 2008 #1

    uby

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    I took a course in special relativity as an undergrad, but am now almost 6 years removed from it and have not retained much of the info.

    Let me pose a scenario I came up with that is a modification of a problem discussed in that class:

    Suppose there is a train moving at some fraction of the speed of light relative to an observer outside the train. There is a light source at either end of the train set to emit a single photon on separate timers. They are set to emit photons at the same clock time as perceived by an observer in the train.

    The observer waits in the middle of the train. He's a sad guy, so he's decided to potentially give his life to science in a grandiose experiment in collusion with the savvy non-exploding stationary observer. (He's constructed a device that will detonate a bomb if it detects the photons arriving at the same time.)

    I have two questions:
    1) Do the photons arrive at the same time as perceived by the observer in the middle of the moving train? How about the stationary observer? (in other words, does the bomb detonate?)
    2) Related to question #1: would the stationary observer who sees the train explode/not explode scratch his head at the outcome of their little experiment? [ie - would he observe a seemingly contradictory event prior to the train exploding/not exploding]

    If the answer to #2 is yes ... then I am deeply troubled.
     
  2. jcsd
  3. May 7, 2008 #2

    jtbell

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    Staff: Mentor

    How about if you tell us what you think, and your reasoning? Even if you can't give a definite answer yourself, you can at least describe the conflicting "arguments" in your mind for each outcome. That will make it easier for us to zero in on whatever conceptual problems you might be having with this problem.

    This is pretty vague. It depends on what the stationary observer knows in advance about what is going to happen, and how much he knows about relativity. What sort of "seemingly contradictory event" do you think might occur here?
     
  4. May 7, 2008 #3

    uby

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    what i think happens:

    the observer in the train sees that the timers on the lights are synchronized. they are equidistant from the center of the train. the photons are emitted at the same time and arrive at the center of the train at the same time. train goes boom.

    the stationary observer sees that the timers on the lights are not synchronized. he sees the timer on the front of the ship is slower than the one on the rear of the ship. the lights are equidistant from the center of the train, although the distance is different to him than to the observer in the train due to length contraction. the photons are emitted at different times, travel an equal distance, and do not arrive at the center of the train at the same time. but, he still sees the train explode.

    simultaneity is relative. ok, that's great and all. but the stationary observer still saw the train explode. metaphorically, his models of the universe are based on observed events which are not the true event. since the train exploded, i consider that to be a true event and that the photons truly must have arrived at the same time. if the stationary observer knows this ahead of time, he can deduce relativity and correct for its effects in his observations. if he does not, as would be the case when observing most astronomical events, then he cannot hope to come up with a model consistent with the true events.

    i explained the last paragraph poorly. i'm in a rush right now, and i will return later this evening hopefully after gathering more thoughts on the problem.
     
  5. May 7, 2008 #4

    paw

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    I'll take a shot at this one. The observer on the train sees both photons arrive at the same time and gets blown to bits.

    The observer on the track also sees the train explode but will disagree with our deceased train observer as to the simultaneity of the lights. He will see the front light flash first since the light has less distance to travel than the back light.
     
  6. May 7, 2008 #5

    jtbell

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    Correct.

    I'd prefer to say that the timer on the front lags behind the one on the rear, but they run at the same rate which is slower than an identical timer on the ground. We have to be careful about using the words "fast" and "slow" because physicists usually associate them with a rate or speed, whereas in everyday life people commonly refer to a badly-set clock as "fast" or "slow" even though it runs at the same rate as other clocks.

    No, they do not travel equal distances. While the photons are in flight, the center of the train is moving forwards, towards the backwards-traveling photon from the front end, and away from the forwards-traveling photon from the back end. Therefore the photon emitted from the back end must travel a greater distance than the one emitted from the front end, which exactly compensates for the fact that it was emitted first.

    Therefore the two photons arrive at the center of the train simultaneously, according to both the observer on the ground and the observer on the train.
     
  7. May 7, 2008 #6
    Hi jtbell. When I first read the problem, I thought that since the front of the train moves at the same speed as the rear of the train that the timers would 1) run at the same rate AND 2) be in synch.

    Is the reason that they're out of synch because the front of the train starts moving (a very small fraction of a second) before the rear starts moving?

    Thanks - btw, I've appreciated your responses to a lot of the questions on here - I find them very informative.
     
  8. May 7, 2008 #7

    jtbell

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    No, they're out of synch because of the relativity of simultaneity. Consider two clocks that are stationary in one inertial reference frame, separated by distance [itex]\Delta x[/itex], and synchronized. In another inertial reference frame moving with speed v relative to the first one, along the line connecting the two clocks, those clocks are out of synchronization by [itex]\Delta t = v \Delta x / c^2[/itex]. That is, in the second reference frame, at any instant the readings on the two moving clocks differ by that amount. The clock that is "in the lead" in terms of their motion is "behind" the other clock in terms of their readings.

    The formula above can be derived from the Lorentz transformation, similarly to the way the familiar formulas for length contraction and time dilation can be derived.
     
    Last edited: May 7, 2008
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