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SR Velocity Transformation

  1. May 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Two rockets A and B are moving away from the earth in opposite directions at 0.85c and -0.75c respectively.
    How fast does A measure B to be travelling?

    Now I have worked out v = -0.85-0.75/(1- -0.85*-0.75) = -0.997. This is correct.

    Now I would like to work it out backwards to check my understanding so:
    According to A:
    B travels at 0.997c
    The earth frame travels at 0.75c

    Q - How fast does the earth measure B to be travelling?

    2. Relevant equations
    w=u-v/(1-uv)

    3. The attempt at a solution

    I expected the answer to be 0.85c but:
    v=0.997-0.75/(1- 0.997*0.75) == 0.979 and not 0.85.

    Thanks for any help
     
  2. jcsd
  3. May 29, 2016 #2

    PeroK

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    Are you sure?

    Technically you're missing some brackets in that equation.
     
  4. May 29, 2016 #3
    As far are I know everything is correct 'mathematically'. You are right about the brackets but I have used the equation correctly in both cases as such.
     
  5. May 29, 2016 #4

    stevendaryl

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    Your problem is just keeping your variables straight. In my opinion, it's always good to make up variables for all the values in a problem, and do as much of the derivation as possible symbolically, and only substitute numbers into the equations at the last step.

    So, let's make up some variable names:
    [itex]v_{AE} = [/itex] velocity of [itex]A[/itex] as measured in the Earth's rest frame
    [itex]v_{BE} = [/itex] velocity of [itex]B[/itex] as measured in Earth's rest frame
    [itex]v_{BA} = [/itex] velocity of [itex]B[/itex] as measured in A's rest frame

    The velocity addition formula tells you that:

    [itex]v_{BE} = \dfrac{v_{BA} + v_{AE}}{1 + \frac{v_{BA} v_{AE}}{c^2}}[/itex]

    You have: [itex]v_{BA} = -0.997c[/itex], [itex]v_{AE} = 0.85c[/itex].

    [edit]: [itex]v_{BA}[/itex] is negative, so it should be [itex]-0.997c[/itex].
    [second edit]: As PeroK points out, it should actually be [itex]-0.977c[/itex]
     
  6. May 29, 2016 #5

    PeroK

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    I think it should be 0.977 not 0.997. That's the problem.
     
  7. May 29, 2016 #6
    haha I think you're right - typo in the text book!
     
  8. May 29, 2016 #7
    yup f*s sry for bothering!
     
  9. May 29, 2016 #8

    PeroK

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    I prefer:

    ##v' = \frac{u+v}{1+uv}##

    ##v'' = \frac{v'-u}{1-v'u} = \frac{v(1-u^2)}{1-u^2} = v##

    Then you can relax!
     
  10. May 29, 2016 #9

    stevendaryl

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    But my point is that for each of the variables [itex]u, v, v', v''[/itex] you need to get clear in your mind: Whose velocity is it, and relative to which frame? And you also have to keep in mind that velocity has a direction, as well as a magnitude. (In these 1-D problems, "direction" means "sign")
     
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