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Distinguishing spacelike and timelike events

  1. Jun 15, 2013 #1
    I am having difficulties understanding some lines in the book.

    "An event can be later than another spacelike separated event in one inertial frame and earlier in another."

    So for example lets take the thought experiment where lighting strikes two sides of the train, and observers in different frames of reference do not agree which side of the train was stroked first.

    So events of the lighting strikes are spacelike separated?

    But it is also written: "Information can be received at P (point) only from events inside or on its past light cone, but not from events outside it."

    But spacelike separated events do not lie in the light cone. I find somehow contradictory.
     
  2. jcsd
  3. Jun 15, 2013 #2

    WannabeNewton

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    Hi amiras! Just because two events are simultaneous in one frame, doesn't mean they will be simultaneous in another as you noted. However this will not deter the potentially space-like nature of the separation of the events since the space-time interval is a Lorentz invariant.

    For example let's say we have an observer ##O## standing equidistant from two ends of a train of length ##2L## and that this train is moving with some velocity ##v## in the ##+x## direction relative to an observer ##O'## standing on the sidetracks. Now imagine that at the moment ##O## passes ##O'## (which is the event that both observers label as the origin of their coordinate systems), ##O## sends out a light pulse in both directions. In ##O##'s frame, the pulses coincide with the ends of the train at events ##P_1 = (t,-L)## and ##p_2 = (t,L)## so they are simultaneous in ##O##'s frame. The space-time interval between these two events in ##O##'s frame is given by ##\Delta s^{2} = (2L)^{2} > 0## i.e. it is space-like.

    Now let's boost to the frame of ##O'##. Now for the event corresponding to the light pulse hitting the back end of the train we have ##t'_1 = \gamma(t - \frac{vL}{c^{2}})## and for the front we have ##t'_2 = \gamma(t + \frac{vL}{c^{2}})##. In other words ##t'_2 = \gamma(t + \frac{vL}{c^{2}}), t'_1 = \gamma(t - \frac{vL}{c^{2}}), t'_1 - t'_2 = -\frac{2vL}{c^{2}}## meaning in the frame of ##O'## the light pulse hitting the back does so before the light pulse hitting the front. This is intuitively obvious because the constancy of the speed of light in all inertial frames will imply that ##O'## sees the back light pulse travel less to hit the end of the train since the end is "catching up" with this pulse whereas the front of the train is "moving away" from the forward light pulse. However as noted above, the space-time interval is a Lorentz invariant so boosting to another frame won't change the value of ##\Delta s^{2}## meaning we will still have ##\Delta s^{2} > 0## in the frame of ##O'## (you can verify this yourself if you wish). So events do not need to be simultaneous to be space-like separated is what they are saying.

    All the lightcone is saying is that you cannot send a signal from the event ##p_1 = (t,-L)## that will be received at event ##p_2 = (t,L)## (this would require the signal to travel instantaneously). In other words the light cone of ##p_1## does not contain ##p_2##. Even though the events don't look simultaneous in the frame of the observer standing on the side-tracks, they are still space-like separated as mentioned above and as such the same geometric meaning of the light cone carries over.
     
    Last edited: Jun 15, 2013
  4. Jun 15, 2013 #3
    Of course! Thank you for the answer, it is much clearer to me.

    EDIT: Sorry, I somehow managed to add word "not" into a sentence. Every time I read what you wrote, it gets more and more clear to me (maybe because you edit and add something new every time :))

    "light cone of p1 does not contain p2" - I somehow wrongly assumed that they are both contained, but that would not make sense. The way I understand it, is that future light cone contains the possible events including p1 and p2, but the timeline can only pass only through one of them (p1 or p2).
     
    Last edited: Jun 15, 2013
  5. Jun 15, 2013 #4

    WannabeNewton

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    It is still not clearer? What else needs clarifying, please feel free to ask my friend :)!
     
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