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Relativity : Lorentz Transformation.

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data
    Star Trek Question: An enemy Klingon space ship moves away from Earth at a speed of 0.80c. The Starship Enterprise gives chase and pursues at a speed of 0.90c relative to the Earth. Observers on Earth see the Starship Enterprise overtaking the enemy ship at a relative speed of 0.1c. With what speed is the Starship Enterprise overtaking the Klingon spaceship, as observed by the crew of the Enterprise?


    2. Relevant equations
    [tex]V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}[/tex]

    [tex]U'_x = \frac{U_x - V}{1 - (U_x V)/c^2}[/tex]
    3. The attempt at a solution
    I used this formula:

    [tex]U'_x = \frac{U_x - V}{1 - (U_x V)/c^2}[/tex]

    Sub in the value where, [tex]U_x = 0.80c[/tex] [tex] V = 0.90c[/tex]

    Then i obtained the answer, [tex]U'_x = -0.357c[/tex]

    Is this correct?

    Also, can someone explain to me what each symbol of the 2 formula means?(the lecture wasn't very clear)

    [tex]V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}[/tex]

    [tex]U'_x = \frac{U_x - V}{1 - (U_x V)/c^2}[/tex]

    Any help will be greatly appreciated.
     
  2. jcsd
  3. Jan 27, 2007 #2
    Your answer corresponds with mine.

    And for the two equations:
    The second equation listed at the end, involving U':
    you have two inertial frames, lets say S, and S'. The S' frame is moving at a speed V(observers on spaceship), while frame S is motionless (observers on Earth).

    lets add your Klingon spaceship

    U is the velocity of the Klingon spaceship as observed in S.
    V is the speed of frame S' (i.e. the speed of the Starship Enterprise)
    U' is the speed of the Klingon Spaceship as observed in S'.

    and c is the speed of light.


    The first equation you have there,
    [tex]V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}[/tex]
    is the velocity addition formula. Lets say you measure the velocity ([tex]V_{b/c}[/tex]) of a particle in the moving S' frame, which is traveling at [tex]V_{a/b}[/tex]. Then from a "fixed" frame S, the velocity ([tex]V_{a/c}[/tex]) of the particle can be found using that equation.

    In star trek terms:
    Let S be the fixed, motionless frame of the observers on Earth, and S' be the frame of the people on board the Enterprise.
    The crew on board the Enterprise see the Klingon spaceship move with a velocity [tex]V_{b/c}[/tex] (which is the answer you found from the question). The Enterprise (frame S') itself is moving at a velocity [tex]V_{a/b}[/tex]. Then you can use these two velocities to find the velocity of the Klingon spaceship as observed from earth, [tex]V_{a/c}[/tex].


    I'm learning relativity right now too, in my mechanics course, so I hope this is helpful. I wish my professor made our homework questions this interesting too.
     
    Last edited: Jan 27, 2007
  4. Jan 28, 2007 #3
    Ah, i see, thanks for the help.
     
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