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Final Clarification

  1. Oct 5, 2009 #1
    Collisions in special relativity

    So we are studying special relativity physics at school right now and I have a question and was hoping you could answer it for me, this is not homework related.


    Lets assume that there is only movement along the x axis.

    A is positioned at x = 100 (arbitrary)

    At t = 0 for A (and B) B is moving towards A from x = 0, with constant velocity 0.999.

    Now lorentz tells me that A will observe that B spends lets say 5 seconds to reach him, while B will observe a much shorter time.

    1) Now how will these 2 ever meet??
    2) Will A say that he met B after 5 seconds and B will say a much shorter time? In that case, who is right? Is the whole point that both of them are right?
    3) Their meeting is inevitable so they must meet but a meeting requires time and space to be equal for both, how does this work out?
     
    Last edited: Oct 5, 2009
  2. jcsd
  3. Oct 5, 2009 #2

    Dale

    Staff: Mentor

    You are forgetting to account for the relativity of simultaneity.
    A and B are distant observers moving relative to each other so they will NOT agree on t = 0.
     
  4. Oct 5, 2009 #3
    So they must share the same starting position to be able to agree on t = 0? Why won't they agree? I am familiar with the famous train thought experiment but the same thing that confused me there leads me to these questions I posted here. In the train experiment both observers are right in their assumption of simultaneity? In one frame of reference the light must travel equal distances to the observer while in the other the observer is moving towards the light and since the speed of light is constant in all frames of reference the observer in the train naturally will see the lightning it is moving towards?

    So if we assume that t is NOT 0 for both, time is still experienced differently, how will they meet?
     
  5. Oct 5, 2009 #4

    Fredrik

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    In A's coordinates, B is at position x=0 at time t=0. This specifies the coordinates in A's rest frame of an event. It's OK to say that B assigns time coordinate t'=0 to that event. What you can't do is to assume that A and B would assign the same time coordinate to any event.

    How will they meet? How could they not? In A's coordinates (draw the t axis in the up direction, and the x axis to the right), A's motion is represented by a vertical line and B's motion is represented by a line with slope 1/0.999. Any two non-parallel lines in a plane must eventually meet.

    Both of them are just as right, or just as wrong if you prefer.

    The duration of their meeting isn't usually an issue, since we would normally treat them as point particles. If we treat them as extended objects, they can certainly disagree about the duration of their meeting.
     
  6. Oct 5, 2009 #5
    Wonderful, this make alot of sense. I just have another question.

    In the frame of A, if A observes the duration of the movement of B from x = 0 to x = 100 to be 5 seconds then B will in his own frame observe it to be 5*(sqrt(1-v**2))? But in Bs frame of reference he is at rest and A is moving towards him with v = 0.999, now B will observe the duration of the movement to be 5 seconds and will say that A will observe it to be 5*(sqrt(1-v**2))?? This seems like a paradox to me, what is the problem/solution?
     
  7. Oct 5, 2009 #6

    JesseM

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    Yes (in units where c=1; if not, you have to write it as [tex]5*\sqrt{1 - v^2/c^2}[/tex])
    Keep in mind that A will need to use two clocks to measure how long it takes for B to get from x=0 to x=100, the first clock at rest at x=0 and the second at rest at x=100; then A will note the time t1 on the clock at x=0 when B passes it, and the time t2 on the clock at x=100 when B passes it, and the time for B to make the journey in A's frame will then be (t2-t1). In order for this to work, of course the two clocks must be synchronized in A's frame. But because of the relativity of simultaneity, these two clocks are not synchronized in B's frame, instead the clock at x=100 is ahead of the clock at x=0 by a constant amount. So suppose that when B passes the clock at x=0, both his clock and A's clock at x=0 read a time of 0 seconds, and then when B passes the clock at x=100, his clock reads [tex]5*\sqrt{1 - v^2/c^2}[/tex] seconds while the clock at x=100 reads 5 seconds as he passes it. B's explanation for this is not that the clock at x=100 was ticking faster than his; rather, his explanation is that the clock at x=100 already had a "head start" at the time he was passing the clock at x=0, reading some time later than 0 seconds at that moment, so that even though it ticked slower than his own clock as it approached him, it was still ahead of his at the moment they passed. I gave a numerical example of how this all works out in post #65 of this thread if you're interested.
     
  8. Oct 5, 2009 #7

    Dale

    Staff: Mentor

    Pardon any sloppiness and please forgive my re-casting of the experiment in numbers that are easier to work out by hand. Also I will use units where c=1.

    Let's say that A and B are moving towards each other with a relative speed of 0.6 (Lorentz factor gamma = 1.25). Let's say that in A's reference frame at t=0 the distance between A and B is 6 and B's clock also reads 0. Because the distance is 6 and the speed is 0.6 they will meet in 10. A's clock is at rest so it will read 10 at the meeting, but B's clock is dilated by a factor of 1.25 so it will read 8.

    Now, in B's frame at t'=0, due to length contraction the distance is only 6/1.25=4.8. Due to the relativity of simultaneity A's clock reads 4.8*0.6*1.25=3.6. Because the distance is 4.8 and the speed is 0.6 they will meet at t'=8. B's clock is at rest so it will read 8, but due to time dilation A's clock will have only advanced by 8/1.25=6.4 so it will read 3.6+6.4=10 at the meeting.

    So both clocks see the other as running slow, yet due to the relativity of simultaneity they both agree what time each clock will show at the meeting.
     
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