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Two lightnings that happen at the same time, a train and a passanger

  1. Oct 6, 2011 #1
    allright, I'm finishing up my final Gymnasium physics course and the weirdest chapters (QM and Einstein) are at the end.

    So here is a book-example I don't understand:

    Lightning strikes at point A and B at the same time (if an outside observer is watching). Between A and B is a train moving at a constant speed in the direction AB. An observer M is sitting right in the middle of the train, ie right in the middle of A and B.

    The observer measures that he light coming from B reaches him before the light from A.

    OK this is weird. I thought that that speed of light is always c to whoever the observer is. So since the light from A and B is approaching the dude at the same speed, shouldn't the light from A & B reach him at the same time?

    If the observer sees the light from B first, this would mean that the observer's relative speed to the light from B is > C.
     
  2. jcsd
  3. Oct 6, 2011 #2

    Doc Al

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    Note: Everyone (even the outside observer) agrees that light reaches M at different times.
    Only if the light starts out at the same time, which is the point. According to the train observers, the lightning strikes were not at the same time. Simultaneity is frame dependent.
     
  4. Oct 6, 2011 #3
    What? Really? I am so confused, why? Is this just some excuse used to claim that c is always constant?
     
  5. Oct 6, 2011 #4
    also can you help me with some math? how do I turn t02(1 - c2/v2) to the lorentz factor? I don't know how to convert (1 - c2/v2)-1 into (1 - v2/c2)
     
  6. Oct 6, 2011 #5

    Doc Al

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    The lightning strikes at the same time according to the outside observer on the tracks. But if we assume that light travels at the same speed in every frame, train observers are forced to concluded that the lightning struck B before it struck A.
     
  7. Oct 6, 2011 #6
    so this is just an excuse to say that c is always constant?

    well we know how that cern thing went...
     
  8. Oct 6, 2011 #7

    Doc Al

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    No, it's a consequence of c being invariant. (And overwhelmingly supported experimentally.)
    I assume you're joking.
     
  9. Oct 6, 2011 #8
    OK, can you help me with post #4?

    thanks for ur patience btw, I really appreciate the help
     
  10. Oct 6, 2011 #9

    Doc Al

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    Where is this coming from?

    In any case:

    (1 - c2/v2) = - (c2/v2)(1 - v2/c2)
     
  11. Oct 6, 2011 #10
    well the book says

    2) t2 = c2t02/(c2-v2)

    ==> 3) t = t0/sqrt(1 - v2/c2)

    I don't get how they went from 2 to 3
     
  12. Oct 6, 2011 #11

    Doc Al

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    t2 = c2t02/(c2-v2)

    t2 = t02/(1-v2/c2)

    Got it? (Divide top and bottom of the fraction by c2.)
     
  13. Oct 6, 2011 #12
    Oh oops thanks lol
     
  14. Oct 6, 2011 #13
    I got another problem, pls help: Inside a spaceship passing earth at 0.6c, time goes slower inside the ship from the POV of earth. But, from the spaceship's POV the earth is moving backwards at 0.6c so the earth is the one whose time is slowing down?

    The book claims this isn't a contradiction.. but how?
     
  15. Oct 6, 2011 #14

    Erland

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    I've said it before and I say it again: I don't like this example. I only get confused when I think of it. But there is a simpler example which demonstrates the relativity of simultaneity. I quote myself from another post:

    "... imagine a light being turned on in the middle of the train. The light reaches the front and the back of the train simultaneously, according to an observer on the train. But according to an observer on the embankment, the light will reach the back of the train before it reaches the front of the train, because according to such an observer, the light that reaches the back of the train will have travelled a shorter distance than the light that reaches the front of the train, since train is moving, according to this observer."
     
  16. Oct 6, 2011 #15

    Erland

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    I think it is easier to see this if we think of a train instead of a spaceship. Again, I refer to another post of my own, where I try to explain this:

    https://www.physicsforums.com/showthread.php?t=468826
     
  17. Oct 6, 2011 #16
    So if the earth gets accelerated so much that it catches up to the spaceship and moves parallel to it at the speed 0.6c, Earthlings who launch themselves in rockets and visit the spaceship find out that the spaceship guys are old?

    That.. can't be possible?
     
  18. Oct 7, 2011 #17

    Erland

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    As far as I undertand it, it will be as you write (provided that the rockets you mention have much smaller velocities than 0.6 c) because Earth is the body which accelerates in this case.

    And well, I guess it wouldn't be possible to accelerate the Earth to the speed 0.6 c.:smile:
     
  19. Oct 7, 2011 #18
    But how can simultaneity be frame dependent when events themselves are frame independent? In other words, since events occur independently of any coordinates or frame, i.e., since events occur in only one way, simultaneity cannot be frame dependent.

    I agree that light from events can reach observers in different frames differently, but what the heck has this got to do with the events themselves? I would say that it has everything to do with said observers physically separating during the observation, but nothing to do with how the events themselves actually occurred.

    What say you?
     
  20. Oct 7, 2011 #19

    Erland

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    In SR. time is just a coordinate. The phenomenon is analogous to the following: If we rotate a cartesian coordinate system in the plane, two points which had the same x-coordinate before the rotation will have unequal x-coordinates after the rotation.
     
  21. Oct 7, 2011 #20

    DaveC426913

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    Two events are separated by spacetime (otherwise there's only one event). How much time they are separated by is dependent on the observer's frame of reference.
     
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