1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativity, trains, and running passenger

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data

    A train of proper length L moves at speed v1 with respect to the ground. A passenger runs from the back of the train to the front at speed v2 with respect to the train. How much time does this take as viewed by some one on the ground?

    a) Find the relative speed of the passenger and the train (as viewed by someone on the ground), and then find the time it takes for the passenger to erase the initial "head start" that the front of the train had.

    b) Find the time elapsed on the passenger's clock (by working in whatever frame you want), and then use time dilation to get the time elapsed on a ground clock.

    2. Relevant equations


    Vector addition:
    u=(v1+v2)/(1+v1*v2/c^2)

    time dilation:
    t=gamma*t'

    length contraction:
    x=x'/gamma

    gamma=1/sqrt(1-v^2/c^2)

    3. The attempt at a solution

    a)

    For an observer on the ground the passenger is moving at speed, u=(v1+v2)/(1+v1*v2/c^2) and the train is moving at speed v1. The length of the train is L/sqrt(1-(v1/c)^2).

    The time is just the length divided by the difference in the velocities:

    t1=L'/(v1-u)=[L/sqrt(1-(v1/c)^2)]/[v1-(v1+v2)/(1+v1*v2/c^2)]


    b)

    According to the passenger, the length of the train is L/sqrt(1-(v2/c)^2) and he is run at velocity v2. So he thinks he reachs the end of the train in time (L/sqrt(1-(v2/c)^2))/v2.

    According to someone standing still on the train, the time is
    [(L/sqrt(1-(v2/c)^2))/v2]*(sqrt(1-(v2/c)^2)).

    So according to someone on the ground t2=L/v2*(sqrt(1-(v1/c)^2))

    ...

    My problem is that t1=/=t2 but it should. I am pretty sure that I did part a) correct. I am not sure if I am transforming the times correctly in part b). I am probably doing something really stupid but I just do not understand enough to know what is going on.
     
    Last edited: Mar 9, 2009
  2. jcsd
  3. Mar 10, 2009 #2
    In the second part, be sure that the time dilation takes place between the frame of the runner and the frame of a person standing on the ground. The relative velocity between those two frames isn't v1 or v2, so you'll have to use velocity addition.
     
  4. Mar 10, 2009 #3
    So are you saying that the time in the passengers frame is L/sqrt(1-(v2/c)^2)/v2 but to transform it to the grounds frame I should do something like
    L/sqrt(1-(v2/c)^2)/v2*sqrt(1-((v1+v2)/(1+v1*v2/c^2))^2/(c)^2)?

    In other words, my gamma is 1/sqrt(1-(u/c)^2)?
     
  5. Mar 10, 2009 #4
    Really, I think the best thing to do is to use the full Lorentz Transformation. The second part of the problem is tricky because the relative positions of the frames change as a function of time. Here's a hint:

    Assume that all frames start at the same position at time t = 0 when the runner starts. In the frame of the train, the runner reaches position L at time L/v2. You can then Lorentz transform this space-time vector backward to the frame of a person standing on the ground to determine the time passed for them.

    The tricky part is the fact that changing between frames, or Lorentz transforming, mixes time and space. So, if you are comparing frame A to frame B, the time passed in time coordinate of frame A depends on both the time and space coordinate of frame B.
     
  6. Mar 10, 2009 #5
    Well, once I get over all of the algerbra mistakes, my answer for part a and b are the same. I think that I am kind of using the Lorentz transforms in a way. One glaringly wrong statement that I made earlier is that the speed of the passenger in the passenger's frame is v2 when it should be zero.

    Anyway, thanks for the help.
     
  7. Mar 10, 2009 #6
    To be specific, here's what I had in mind when thinking about the Lorentz transformation:

    Consider frame S' which is attached to the back of the train. Frame S is attached to the ground, and frame S' moves with velocity v1 with respect to frame S. The two origins for these frames coincide at the origin at time t = 0.

    In frame S', the initial space-time position of the person is (t',x') = (0,0). In the frame of S', the length of the train is L and it therefore takes L/v2 for the person to reach the end of the train. The final space-time position of the person is (t',x') = (L/v2,L).

    Then, we use the lorentz transformation going from frame S' back to frame S:

    t = [itex]$\gamma (t' + v1 x') = \gamma (L/v2 + v1 L)$[/itex], which we can then simplify to what you got for answer 1 (c = 1).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Relativity, trains, and running passenger
  1. Subterranean trains (Replies: 4)

Loading...