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Lorentz Transformation of relative velocity

  1. Jan 9, 2015 #1
    1. The problem statement, all variables and given/known data
    If two particles have velocities u and v in frame S, find their relative speed in frame S'.


    2qbrhj9.png
    2. Relevant equations


    3. The attempt at a solution
    Isn't it strange that the relative speed doesn't depend on the velocity of the frame, ##\vec s##?

    Since the two particles have velocities ##\vec u## and ##\vec v## in some reference frame S, I am to find the relative velocity and speed in frame S'.

    Letting the relative velocity of the frame be ## \vec s##, the transformation of velocities are:

    [tex]
    \vec u' = \frac{1}{1 - \frac{\vec u \cdot \vec s}{c^2}}\left[ \frac{1}{\gamma_s} \vec u - \left( 1 - \frac{\vec u \cdot \vec s }{c^2}\frac{\gamma_s}{1 + \gamma_s} \right) \vec s \right] [/tex]

    [tex]
    \vec v' = \frac{1}{1 - \frac{\vec v \cdot \vec s}{c^2}}\left[ \frac{1}{\gamma_s} \vec v - \left( 1 - \frac{\vec v \cdot \vec s }{c^2}\frac{\gamma_s}{1 + \gamma_s} \right) \vec s \right] [/tex]

    Taking ##\vec u' - \vec v'## only gives me components in ##\vec u, \vec v, \vec s ##. How do I extract the magnitude?
     
  2. jcsd
  3. Jan 10, 2015 #2

    TSny

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    Note: the "speed of particle B relative to particle A" is the speed of particle B as measured in the reference frame moving with particle A.

    So, what should you choose for primed frame, S'? Hence, what should be the velocity ##\vec{s}##?
     
  4. Jan 10, 2015 #3
    Yes, I misread the question. ##\vec s## is simply ##\vec v##. Then all that remains is to find the magnitude. But it looks very messy due to the ##\gamma_v##

    [tex]
    \vec w = \frac{1}{1 - \frac{\vec u \cdot \vec v}{c^2}}\left[ \frac{1}{\gamma_v} \vec u - \left( 1 - \frac{\vec u \cdot \vec v }{c^2}\frac{\gamma_v}{1 + \gamma_v} \right) \vec v \right] [/tex]

    To find the magnitude I would take ##\sqrt{ \vec w \cdot \vec w}##
     
  5. Jan 10, 2015 #4

    TSny

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    OK, that looks good. Yes, it is messy to deal with all the gamma factors. It will work out.

    Another approach is to work out the x and y components of ##\vec{w}##, where the x direction is parallel to ##\vec{v}## and the y direction is perpendicular to ##\vec{v}## such that ##\vec{u}## is in the x-y plane.

    There are well-known formulas for calculating these components. See below. These can be derived from your general formula.
     

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    Last edited: Jan 10, 2015
  6. Jan 10, 2015 #5
    I derived the expression ##
    \vec w = \frac{1}{1 - \frac{\vec u \cdot \vec v}{c^2}}\left[ \frac{1}{\gamma_v} \vec u - \left( 1 - \frac{\vec u \cdot \vec v }{c^2}\frac{\gamma_v}{1 + \gamma_v} \right) \vec v \right] ## by using ## u_{||} = \frac{(\vec u \cdot \vec v) \vec v}{|\vec v|^2} ## and ## u_{\perp} = \vec u - \vec u_{||} ##.

    I thought it would be easier, because finding the magnitudes of the horizontal and vertical components would involve angles and such.
     
  7. Jan 10, 2015 #6

    TSny

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    If you want to work with the components, you can write ## u_{||} =u \cos \theta## and ## u_{\perp} =u \sin \theta##. Then the velocity-addition formulas for the two components will be fairly easy to work out.
     
  8. Jan 10, 2015 #7
    So do you reckon it's easier to simply take ##\left[ \frac{u cos \theta - v}{1 - \frac{uv cos\theta}{c^2}} \right]^2 + \left[ \frac{u sin\theta}{\gamma (1- \frac{uv cos \theta}{c^2})} \right]^2 ##?
     
  9. Jan 10, 2015 #8

    TSny

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    Yes. Note that in the denominators you have ##1 - \frac{uv cos\theta}{c^2}## which is part of the answer you want.
     
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