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Relativity Problem

  1. Aug 2, 2016 #1
    1. Two long trains pass each other head on with a relative speed of 0.97c. Bob, the driver of the first train notices that it takes 5.8 secs to pass the entire second train. How long do people on the second train say it takes for Bob to go by their train?

    2. t = t0/(1-v2/c2)1/2


    3. I understand how to plug in the numbers, but I am not sure if they are asking for t or t0. Thanks.
     
  2. jcsd
  3. Aug 2, 2016 #2
    Suppose Alice is in the second train. Alice sees Bob moving with a speed of ##0.97c##, in a direction that we will call the positive ##x## direction. We have two events, one in which the trains first meet, and one in which the trains first start separating. Bob measures the difference in time of the two events to be ##5.8s##, which you call ##t_0##. Then by the equation you wrote down (time dilation equation), the difference in time between the two events as measured by Alice will be ##t##.

    The justification for the time dilation equation is found in http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html
     
  4. Aug 2, 2016 #3
    Thank you for your help! :smile:
     
  5. Aug 2, 2016 #4
    After looking a bit more on the problem, I noticed that you can only use the the time dilation equation in the way I describe if the length of Bob's train is small enough, so that the two events happen approximately in the same position in space.

    More precisely we require that ##l_b<<c t_0##, where ##l_b## is the length of Bob's train according to Bob. This can be seen from Lorentz transformations. In general, we will require information about the length of one of the trains to compute ##t##, using a Lorentz transformation.

    Glad to help :biggrin:
     
  6. Aug 3, 2016 #5
    Since the reference frames for the trains are indistinguishable, does this mean the answer is 5.8?
     
  7. Aug 3, 2016 #6

    TSny

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    Treat Bob as a point object. Consider the two events:

    (1) Bob is coincident with the front of the second train
    (2) Bob is coincident with the rear of the second train.

    The sentence "Bob, the driver of the first train notices that it takes 5.8 secs to pass the entire second train" can be interpreted as saying that the time interval between the two events (1) and (2) is 5.8 s according to Bob's reference frame.

    The question "How long do people on the second train say it takes for Bob to go by their train?" can be interpreted as asking for the time interval between the same two events according to the reference frame of the second train.

    In this interpretation, we don't need to worry about the length of Bob's train.
     
  8. Aug 3, 2016 #7

    TSny

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    No. Have you covered the definitions of "proper time interval" and "nonproper time interval"?
     
  9. Aug 3, 2016 #8
    If any two events happen at the same position in a frame, the time between the two events (in that frame, called rest frame) is called the proper time. Proper time is the shortest time possible that can be measured between the two events. Time dilation quation allows you to compute the time difference in any frame, give you know the proper time.

    That is why, if one of the trains (which i chose to be Bob's in my previous explanation) is really short, the frame will be approximately the rest frame, and the time measured in that frame will be approximately the proper time.

    But in the general case, what yo can do is say that when the trains meet, Alice's and Bob's position in space coincide. The coordinate of such event is zero for both. Then when the trains separate, if ##l_1## is the length of the first train according to Bob, and ##l_2## is the length of the second train according to Alice, the coordinate of this event will be ##(t,l_2,0,0)## for alice and ##(t_0,-l_1,0,0)##. You may use Lorentz transformations to find a relationship between the quantities ##t,t_0,l_2,l_1##. The proper time will be the square root of ##c^2 t^2-(l_2)^2=c^2 (t_0)^2-(l_1)^2##, divided by ##c##.
     
    Last edited: Aug 3, 2016
  10. Aug 3, 2016 #9

    TSny

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    Actually, this would be correct if the question were a little different. Suppose the two trains have the same "rest length" and consider the following two events:

    (1) the front ends of the two trains are coincident
    (2) the rear ends of the two trains are coincident

    If Bob finds that the time interval between these two events is 5.8 s, then the time interval between these two events would also be 5.8 s according to observers on the second train. This would indeed be due to the equivalence of the two reference frames for these two events.

    But I don't think this is the interpretation that was meant. Note that the problem did not specify that the two trains have identical rest lengths.
     
  11. Aug 3, 2016 #10
    Ohh that makes sense - thank you! I'm only in physics 11 and we did not cover rest lengths, proper time interval etc
     
  12. Aug 3, 2016 #11

    TSny

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    You wrote the equation t = t0/(1-v2/c2)1/2. When this was covered in class or in your textbook, there should have been some discussion on the meaning of t0 versus t.
     
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