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Homework Help: Got the answer, not convinced possible in reality?

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Inertial frame S' moves at a speed of 0.6c with respect to frame S. Further, x = x' = 0 at t = t' = 0. Two events are recorded. In frame S, event 1 occurs at the origin at t = 0 and event 2 occurs on the x axis at x = 3000m at t = 4*10-6 s. According to observer S', what is the time of (a) event 1 and (b) event 2? (c) Do the two observers see the two events in the same sequence or the reverse sequence?

    2. Relevant equations

    Standard Lorentz Transformation Equations.

    3. The attempt at a solution

    x1 = 0, x2 = 3000m, t1 = 0, t2 = 4*10-6 s

    x1' = 0, x2' = 2850m, t1' = 0, t2' = -2.5*10-6s

    Based on the question, x2' is some distance away and event 1 occurs at t = 0 in frame S'.

    1) It requires some time for light to reach event 2 in S'
    2) Event 1 ALREADY occurs at t = 0

    Then shouldn't event 1 occur first regardless of whatever speed you are travelling?

    How does the concept of 'negative' time come into play here? I'm finding this very, very strange..
  2. jcsd
  3. Sep 25, 2012 #2
    If event 1 had caused - or had been able to cause - event 2 in S, then both observers would agree that e1 were the first and e2 the second. However, e2 is so far away from e1 and the time between them is so short that no signal could possibly have reached e2 from e1 (or vice versa in in S'). The events are truly independent and separated by a space-like interval, so may occur in different orders depending on the frame of reference.

    Note also, that if the observers are at the origins of S and S', the observer in S' will still see e2 after e1. This is because e2, even if occurring earlier than e1 in S', is away from the observer, and the signal from that event still needs to reach the origin of S'; and that will take more time than is left before e1. The observer in S' will infer, based on his knowledge of the distance to e2, that it must have occurred 2.5 microseconds before e1. If the observers are halfway between e1 and e2 in their respective frames, then they will see the events exactly how they will think they happened, i.e., e1 before e2 in S, and e2 before e1 in S'.
  4. Sep 25, 2012 #3

    Doc Al

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    Staff: Mentor

    What light?
    Already according to whom?

    If you're thinking that somehow light from event 1 reaches event 2, ask yourself if that's possible. Can the two events be causally connected?
    If I call 'right now' on my watch to be t = 0, then 5 minutes ago is t = - 5min. Nothing all that strange.

    (Realize that built into the Lorentz transformations is the interesting fact that simultaneity is relative. That's pretty strange.)
  5. Sep 25, 2012 #4

    This is exactly what I mean. How is it that the observer sees e1 first, followed by e2 and still say that e2 occurred before e1? As e2 is farther away, the signal from e2 arrives later than the signal from e1.
  6. Sep 25, 2012 #5
    So what? We just keep track of two different notions: when the light from the events reach the observer and when the events must've happened according to his clock. The observer can measure distance and figure out when the events must've happened.
  7. Sep 25, 2012 #6

    Doc Al

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    Staff: Mentor

    Looks like you've answered your own question. If an observer sees two flashes of light he cannot decide which flash occurred first unless he knows how far away they occurred.

    (The Lorentz transformations deal with measurements after already accounting for light travel time.)
  8. Sep 25, 2012 #7


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    Science Advisor
    Gold Member

    This is counter intuitive so confusion is always the first step toward understanding. Start with the (intuitive) non-relativistic case. You see two events, flash here and now and another flash distant and later. But you measure the speed of light from the distant flash and triangulate the distance and figure that it happened at the same time as the first.

    I moving relative to you in a non-relativistic universe would see the same sort of thing, (as we pass at t=0) I would see the first flash then measuring a different speed of light (your measurement plus our relative motion) and a different duration to observing the 2nd flash, and I'd triangulate its position to get agreement with you on the simultaneity of events. Critical to my agreeing with you as I move is my seeing the light signal traveling at a different speed relative to me than you see relative to you. We see light traveling (in this non-relativistic universe) in the same way as baseball's travel.

    Understand the non-relativistic situation then ask the question, what does insisting that the speed of light is the same for all observers change?
  9. Sep 25, 2012 #8
    As I remarked, what the observers see is determined not only by the "sequence" of events, but also by their positions. Forget two observers, forget moving frames. If an event occurs in a mile from you, and another one in two, simultaneously, you will see the closest event before the other one. Relativity does not change this. But it does change what you call simultaneous.
  10. Sep 26, 2012 #9
    I think that explains it. Even if event 1 and 2 happens together, given that event 2 is farther away, to the observer event 1 happens first (since light travels at a finite speed).

    But i'm still perplexed by the concept of 'negative' time...Unless in the question where it states that event 1 occurs at t = 0, but it says NOTHING about event 2.

    The question says event 1 happens at t = 0. Does this set the restriction that event 2 has not happened yet?? (Meaning, even before reading the question event 2 might have already happened)
  11. Sep 26, 2012 #10
    Just to check:

    Suppose an event P happens at (x1,t1) according to frame S. Does this mean that the time that the event actually happened should be t1 - tL where tL = x1/c ?
  12. Sep 26, 2012 #11
    Once again, the interval between e1 and e2 is space-like. That means no signal can pass between them. So, depending on the frame, they can be simultaneous, or either one can follow the other. This will never lead to any paradoxical causality effects.

    Negative time has no particular meaning. Zero time is an arbitrary reference, so anything that happened before that arbitrary reference is negative.

    The time an event happens in a frame is not the same as the time an observer is reached by a signal sent by the event. The former time is the same everywhere in the frame, the latter, as already discussed, depends on the position of the observer in the frame. The coordinate time is the "frame time", not the "observer time". For an observer stationary in the frame, the difference between the two is the distance between the event and the observer (in the frame) divided by c. The coordinate time is an abstraction, not physically measurable, and is deduced from the relation with the "observer time". Yet this is the time used in Lorentz's transformations; one reason for that is there are infinitely many observers in any given frame, so specifying time with respect to a particular observer introduces an unnecessary complication.

    In general relativity, however, that becomes necessary, too, because "observer time" is no longer just a simple function of the observer's position.
  13. Sep 27, 2012 #12
    Ok, that cleared things up! Thanks voko.
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