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Homework Help: What time do simultaneous events in rest frame occur in moving frame?

  1. Apr 3, 2014 #1
    1. The problem statement, all variables and given/known data

    A train is moving by a stationary observer, Bob, at 0.8c. An observer standing still in the middle of the train, Alice, measures the distance to each end of train as 9*10^11 m. Two lightning bolts hit each end of the train simultaneously, as seen by the observer at the station, Bob. Set t = 0 when the lightning bolts hit the ends of the train and t' = 0 when Alice passes by Bob.

    a) How far away from the ends of the train does Bob think Alice is?
    b) How long will it take for Bob to see the lightning bolts?
    c) At what times, in Bob's reference frame, does Alice see the lightning bolts?

    2. Relevant equations

    3. The attempt at a solution
    I have done a) and b) but am confused on how to proceed with c).

    For a) I have that Bob observes Alice to be standing 5.4 * 10^11 m away from each end of the train, and therefore it will 5.4 * 10^11 / c = 1.8 * 10^3 seconds for the lightning to be seen by Bob.

    For c) I don't really know how to find the time. I've drawn a spacetime diagram, and I've determined that Alice sees the lightning bolt from the front before she sees the lightning bolt from the back, but I don't know how to express this quantitatively, or even how to apply which Lorentz transformation. I'm getting myself more and more confused, so if anyone could point me in the right direction that would be great.
    Last edited: Apr 4, 2014
  2. jcsd
  3. Apr 3, 2014 #2
    Your result for part (a) is not correct. Bob sees the train length-contracted, so, as reckoned from Bob's reference frame, Alice is located <4.5 x 1011 from each end of the train. Part (c) is a little confusing. I think they want you to take into account the fact that Alice is moving, so that the flash from one end reaches her first, and the flash from the other end reaches her later. So they are asking at what times would Bob calculate that the flashes reached her according to his clock.

  4. Apr 3, 2014 #3

    Simon Bridge

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    I think it wants the time on Bob's clock that the light from the bolt-strikes reach Alice.

    The time of the strike on Bobs clock is t=0
    The time the light from each strike reaches Bob is t=+1.8x10^3s to reach Bob.
    Alice is in the middle of the carriage, but while the light-pulse is in flight, Alice also moves.
    So the light reaches Alice, in Bob's frame, when Alice and the light are at the same position, in Bob's frame.

    Note: the math is easier if you put distances in light-seconds (the train is very big!)
    The length of the train in Alics frame is: LA=3000ls
    ... and the gamma factor is: γ=1/√(1-0.82)=10/6

    This means that the length of the train in Bob's frame is: LB=lA/γ=1800ls.
    i.e. it take light 1800s to go from one end of the train to the other, in Bob's frame.

    So far so good: it looks to me like this is as far as you got working in meters.

    So where is Bob in relation to the lightning strikes?
    t=0 for his is also t'=0 for Alice right?

    [edit]I'm with chestermillar
  5. Apr 4, 2014 #4
    Can you please expand on why part a) is incorrect? I tried to do length contraction, so that proper length is 9 * 10^11 m to one end, or 3000 light seconds. Gamma is 0.6, which works outs to 1800 Ls as mentioned, so the contracted length is 1800 Ls or 5.4 * 10^11 m. Why would it be less than this? Sorry if this is a dumb question.

    Bob, at the station, is equidistant to both ends of the train and Alice is passing by Bob right when the lightning strikes. And in regards to the above, I think that's exactly what they would like. They want to know what time on Bob's clock will Alice have seen the strike from the front, and the strike from the back, given that she's moving towards the direction of the light at the front and away from the light at the back.

    How should I go about doing this? I feel like I'm getting more tangled up the longer I look at it. Should I figure out at what time the flashes hit in Alice's frame, ie when in t' the flash at the front occurred and when in t' the flash at the back occurred?
    Last edited: Apr 4, 2014
  6. Apr 4, 2014 #5

    Simon Bridge

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    Is Bob 5.4*10^11m from the lightning strikes? Isn't that the entire (contracted) length of the train?

    It really helps to put distances in light-seconds: seriously.
    You should be doing this for the space-time diagrams anyway.

    At t=0, what is the position coordinate of the flash-pulse from the back of the train in Bob's frame?
    At t=0, what is Alices position coordinate in Bob's frame?
    As t increases, what happens to the position coordinates?

    At what time is the position of the flash-pulse the same as the position of Alice?
  7. Apr 4, 2014 #6
    That is the contracted length for half of the train... I believe. Since Alice was originally 3000 ls from either end, she should now be 1800ls from either end, if I'm understanding this correctly.

    Thanks for the tip, I'm new at this.

    Right so at t = 0, the position of the back pulse is x = - 1800ls (again, if I'm correctly understanding a)), the front pulse is at x = +1800 ls. Alice is at x = 0. As t increases, the position coordinates shift to the right, so x' = gamma (x - vt), with gamma = 1/0.6 and v = 0.8c. Alice moves closer to the position where the first flash occurred, and farther away from where the second flash occured, so she should see the flash at the front first, right? As in, for both t and t', the event "Alice sees flash from front" will occur at a lower value than "Alice sees flash from back"?

    Edit: I'm trying to figure out at what time t' the flashes initiated.
    I'm using:
    t' = gamma (t - vx/c^2)
    So for the flash initiating at the back of the train, t = 0 and x = -1800ls.
    t' = -vx/0.6c^2
    t' = -0.8c (-1800ls)/0.6 c^2 = 2.4 * 10^3 seconds?
    Then for the flash from the front, it's -2.4 * 10^3 seconds?? How is this possible, I must be doing something wrong.
    Last edited: Apr 4, 2014
  8. Apr 4, 2014 #7
    Ok, maybe not. So for the flash at the front, I got t' = -2.4 *10 ^3 sec. I used simple velocity relations, and got that it would take 3000 seconds in Alice's frame for the light of the event to reach her, all good since it's 3000 light seconds away. So t' for when the light from the front reaches her is t' = 600 seconds.
    Using t = gamma(t' +vx'/c^2), and x' = 0, this means t when the light from the front reaches Alice is 1000 seconds.

    Then for the flash at the back, I got that it starts at t' = 2.4 * 10^3 sec. Again, it takes 3000 seconds to reach Alice, in Alice's frame. This puts the light from the back flash reaching her at t' = 5.4 * 10^3 seconds. Again, using t = gamma(t' +vx'/c^2), and x' = 0, this means t when the light from the front reaches Alice is 9000 seconds.

    Does this sound remotely plausible? Or as plausible as a it can, for a train moving at relativistic speeds that has a proper length of 6000 ls.
  9. Apr 4, 2014 #8

    Simon Bridge

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    But 1800ls is the entire length of the train!

    Why the gamma factor? You are computing the change in position in Bob's frame as seen by Bob.
    As long as you stay inside the same reference frame, normal physics applies.
  10. Apr 4, 2014 #9
    So if I fix the 1800 ls into 900 could I apply the same logic? Find out t' given t=0, x = +/- 900ls, then find out how long it would take light to travel to Alice from the ends using non relativistic relations, since I know the speed with be +/- c and the distance based on proper length. Then add this to the initial t' and convert back to t?
  11. Apr 4, 2014 #10

    Simon Bridge

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    You are almost there... no need to work with t' or convert anything, you don't need to know the proper length - you only need the length as measured by Bob.

    Working entirely in Bob's frame:
    At t=0, the pulse at the back to the train is at xp=-900ls, and Alice is at xa=0ls
    The pulse travels in the +x direction at speed c, and Alice travels in the +x direction at speed v.

    You can write an equation for xp(t) and xa(t).

    When xp=xa, Bob notices that Alice sees the pulse...

    Repeat for the front of the train.
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