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Lorentz velocity transformation problem

  1. Aug 9, 2012 #1
    An observer on Earth observes two spacecraft moving
    in the same direction toward the Earth. Spacecraft A appears
    to have a speed of 0.50c, and spacecraft B appears
    to have a speed of 0.80c. What is the speed of spacecraft
    A measured by an observer in spacecraft B?


    So if S is the reference frame of the Earth, S' is the reference frame of plane B, v is the velocity with which S' approaches S (according to S), u is the velocity with which plane A approaches Earth (according to Earth) and u' is the velocity with which plane A approaches plane B (according to B), then:

    v = -0.8c
    u = -0.5c

    u' = (u-v) / 1-(uv/c2)) = 0.5c

    I am interpreting this as meaning that according to plane B, plane A is actually moving away from B at 0.5c. My book, however, has the answer as -0.5c, implying the opposite, i think?...is the book wrong or are my signs wrong? It makes sense to me making v and u negative because the formula for the velocity transformation was derived assuming that S' and the event (plane A here) are moving away from S, not towards...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 9, 2012 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi Aziza! :smile:

    I think you're over-thinking this …
    … if we ignore relativity, and choose 50mph and 80mph instead,

    then obviously B is going faster than A, so A's speed measured by B will be negative :wink:

    (of course, this is a non-standard usage of "speed" :redface: … the standard usage is as the magnitude of velocity, so that it's always non-negative)
     
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