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Does simulataneity contradicts the constancy of speed of light

  1. Sep 8, 2011 #1
    This is my first post here,

    Consider the typical simultaneity scenario:

    An observer `O' right in the middle of two light signal sources `A' & `B', all three has their clocks synchronized to each other. Now, A & B both emit a light signal simultaneously and then again one second later. After one second from receiving the first signal the observer starts moving(does not matter how fast) towards one of the sources. Now there are two contradicting statements as
    1. According to simultaneity problem the second signal would be non-simultaneous w.r.t the moving observer O.
    2. And as the speed of light signal is same in every inertial frame, this means w.r.t the moving observer O, speeds of both the second signals should be same. As the moment the signals were emitted the distance between the observer and the sources was same. This means same speed and same distance should result in same time to arrive, therefore the second event should also be simultaneous.
  2. jcsd
  3. Sep 8, 2011 #2


    Staff: Mentor

    Have you tried to draw a spacetime diagram of your scenario? If you do, I think it may help you to see the error in your reasoning. However, I'll try to explain the error verbally.

    The fact that the distance between O and B (or O and A) is the same at time t = 1 as at time t = 0, does *not* imply that the signals emitted at time t = 1 should take the same time to arrive at O as the signals emitted at time t = 0. This is because O starts moving towards B and away from A at time t = 1, so he is moving towards the signal coming from B and away from the signal coming from A. So the B signal has to cover less distance to reach O (because O covers some of that distance himself, since he is moving) and the A signal has to cover more distance (because it has to cover the extra distance that O moved towards B). The key point is that it's not the instantaneous distance at the time of emission that matters, but the distance that the light actually has to cover, which must take into account the movement of O between the times of emission and reception.

    All the above analysis is done in the original frame (the one in which A, B, and O all start out at rest, and in which A and B remain at rest). However, once done, it makes clear that the B signal and the A signal (meaning the signals emitted at time t = 1 in the original frame) arrive at O at *different events*, which is the key point. That is, the B signal crosses O's worldline at a different point than the A signal does. Drawing a spacetime diagram makes this easy to see. This difference in arrival events is *invariant*; it must hold in any frame, because actual physical events, like the meeting of two worldlines (a light signal and an observer), are frame-independent.

    But if the two signals cross O's worldline at different points once O starts moving, then the signal emission events cannot be simultaneous to O, by definition (that is, they will not be simultaneous in O's moving frame). According to O, once O starts moving, B's signal is emitted *before* A's signal. This is what your statement #1 asserts, and it is correct.

    So to summarize: in the original frame, the difference in arrival times of the signals at O (B first, then A) is due to the difference in distance traveled; the signals are both emitted at the same time and travel at the same speed, but B's travels a shorter distance than A's does because O is moving toward B and away from A. But in the moving frame (O's moving frame), the difference in arrival times is due to B's signal being emitted before A's; both signals travel the same distance at the same speed, but B's is emitted before A's is, so it arrives first.

    So there is no contradiction; instead, your statement #2 is simply not correct, and when it is corrected it agrees with #1, that the events are not simultaneous to O when O is moving. Again, drawing a spacetime diagram (or better yet, two diagrams, one from the viewpoint of the original frame and one from O's moving frame) makes it easy to see how everything fits together.
  4. Sep 8, 2011 #3
    Thanks for the reply,

    This is exactly what other people i talked with said, that in the frame of reference of the observer O the emitting of second signals are themselves non-simultaneous

    But, what about the synchronization of clocks, or better yet what if the observer starts moving at time t=1.5 instead of 1. This means when the second signals were emitted at t=1 they were perfectly simultaneous even in observer's frame of reference.
  5. Sep 8, 2011 #4


    Staff: Mentor

    If the observer O is at rest with respect to A and B at time t = 0, but starts moving with respect to A and B at time t = 1 (both times with respect to A and B's mutual rest frame), then "the observer's frame of reference" does not name a single inertial frame. The A and B signals emitted at time t = 1 in their mutual rest frame are simultaneous in that frame (which is O's frame *before* he starts moving), but are *not* simultaneous in O's moving frame (after he starts moving).

    In other words, you can change whether a pair of events "look simultaneous" to you by changing your state of motion. That seems counterintuitive, I know, but it's a necessary consequence of simultaneity being frame-dependent. When you change your state of motion, you change which inertial frame is "at rest" for you, so you change which events look simultaneous to you.

    This is another reason why I like spacetime diagrams: they help you to focus on what does *not* change when you change your state of motion, which is where the real physics is. No actual physical result depends on whether or not two events look simultaneous to you.
  6. Sep 8, 2011 #5
    My thanks to you for giving me the time.

    And I guess, I understand it now

    Thanks again,
  7. Sep 8, 2011 #6


    Staff: Mentor

    You don't sound quite sure. :wink: Feel free to ask more questions if you need to.
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