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Special relativity (overtaking)

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data

    This is an example problem although I don't really understand it.

    Train A is moving at 4c/5 and train B moves at 3c/5 with respect to observer C.

    Let E1 be the event "front of A passing back of B" and E2 be the event "back of A passing the front of B".

    An traveller D (who is in train B) walks from the back of B to the front of B and coincides with both E1 and E2. What is the speed he walks at and how long does D measure the overtaking process to be?

    2. ???

    I just need a little hint here. This is actually an example problem from my book, so I do have the working and answer provided, but I don't actually understand what the significance of the observer D coinciding with both events is.

    The book states immediately that as a result of the condition here, and looking at the point of view of D, A moves at speed v wrt D and B moves at speed -v wrt D. Then it uses the velocity addition again to get [tex]\frac{2v}{1+v^2/c^2} = \frac{5c}{13}[/tex]. The book concludes that v = c/5, upon solving the equation given earlier.

    Once this is done, the rest is easy: using the velocity of either A or B relative to D, find the contracted length of each train and divide by c/5 to get the required answer, which is [tex]2\sqrt{6}L/c[/tex].

    (From an earlier and easier half of the question, velocity addition finds that A moves at 5c/13 with respect to B.)
     
    Last edited: Mar 10, 2009
  2. jcsd
  3. Mar 10, 2009 #2
    I run into another problem once I go further along this question.

    If I solve the equation given here, I actually get v = 2c/5 rather than v = c/5 quoted in the book.
     
  4. Mar 10, 2009 #3

    Doc Al

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    Staff: Mentor

    Look at things from D's perspective. D observes the length of train A pass him (front to back) in exactly the same amount of time that the length of train B passes him (back to front). That fact allows you to immediately conclude that the speed of A wrt D is the same as the speed of B wrt D (except for the sign, of course). (Assuming the trains have the same rest length L, of course.)

    OK. Given my comment above, does this now make some sense?

    Double check your solution. (Using that same equation, I got the book's answer.)
     
  5. Mar 11, 2009 #4
    Yes, I understand now, thanks Al!
     
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