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Lorentz Transformation

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Two identical spaceship A and B, each 200m long in its rest frame, pass one another travelling in opposite directions. According to a passenger in spaceship B, the relative velocity of the two ships is 0.58c. What is the length of each spaceship in a reference frame in which both are of equal length?


    2. Relevant equations

    I'm kinda new to the interface, can't seem to type out the formulas but i used:

    Lorentz Factor = 1 / *square root* [1 - beta2]



    3. The attempt at a solution

    I found the length of space ship A in view of the passenger in B, which is 163m.
    That would, by right be the length of B in view of A, right? What perplexes me the most is the las part of the question; Length of each spaceship in a reference frame in which both are of equal length. Would that be the frame of Earth?

    Please advise =S
     
  2. jcsd
  3. Sep 24, 2011 #2

    vela

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    Yes, that's correct, but it doesn't help you solve the problem.
    Considering Earth isn't mentioned in the problem at all, I'd say no. :wink:

    There are three reference frames in this problem: the rest frame of A, the rest frame of B, and the frame where A and B have equal lengths. You want to find the velocities of A and B with respect to this third reference frame. Once you know those, you can use the length contraction formula to calculate the length of the ships in that reference frame.

    Think about the symmetry in this problem. Can you infer anything about the velocities of A and B seen by an observer in the third reference frame?

    Hint: You'll also want to use the velocity-addition formula.
     
    Last edited: Sep 24, 2011
  4. Sep 24, 2011 #3
    haha ahh, I see. thought as much for the first part.

    So if I look in the perspective of the 3rd frame where both ships are of equal length,
    would the relative speed of each spaceship be 0.29c moving towards each other? (0.58c / 2)

    So relative to the 3rd observer the length of each spaceship should be


    200 / ( 1 / *squareroot*(1 - 0.292) = 191 m


    thanks for taking the time to help =)

    sorry if i'm a little slow at getting it though haha
     
  5. Sep 25, 2011 #4

    vela

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    No, while that may sound reasonable, it's not correct. You can see this if you try using the velocity-addition formula to calculate the speed of A relative to B assuming both are moving with speed 0.29c relative to the third frame.
     
    Last edited: Sep 25, 2011
  6. Sep 25, 2011 #5
    Ok, I think I have it. I can't just assume that they can divided by two. So using some algebra,

    if the observer sees the two ships as equal in length, therefore to the observer, they are travelling at an equal speed, V

    using Velocity Addition formula:

    0.58c = -2v/ *squareroot*(1 - [(-v)(v)/c2])

    Solving for V, I got 0.32c

    Then using the length contraction formula, I get a length of 189.5m
     
  7. Sep 25, 2011 #6

    vela

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    Good work!
     
  8. Sep 25, 2011 #7
    Awesome! Thank you Dr. Vela =)
     
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