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Relativistic addition of orthogonal velocities

  1. Jul 18, 2011 #1
    1. The problem statement, all variables and given/known data
    In a given inertial frame, two particles are shout out simultaneously from a given point, with equal speeds v, in orthogonal directions. What is the speed of each particle relative to the other?

    Answer: [itex]v{(2-\frac{v^{2}}{c^{2}})}^{1/2}[/itex]

    2. Relevant equations
    Velocity Transformation
    [itex]v_{total}=\sqrt{v_{x}^{2}+v_{y}^{2}}[/itex]

    3. The attempt at a solution
    Let the given inertial frame be denoted by S, and let the particles be emitted in the +x and +y directions. Let S' be the rest frame of the particle emitted in the +x direction. These two frames will coincide at t=0 (time of emission), hence the Lorentz transforms are applicable.

    In S', the x component of the velocity of the particle going in the +y direction would -v, as that particle has no x component of velocity. The y component would be given by the velocity transformation:

    [itex]v'_{y}=\large \frac{(v) \sqrt{1-\frac{v^{2}} {c^{2}}} } {1- \frac{(v)(v)}{c^{2}}} = \frac{v}{\sqrt{1-\frac{v^{2}} {c^{2}}}}[/itex]

    The relative velocity would then be-

    [itex]v_{rel} = \sqrt{v'_{x}^{2}+v'_{y}^{2}} =\large \sqrt{v^{2}+\frac{v^{2}} {1- \frac{ v^{2} }{ c^{2} } } }[/itex]

    Which would yield

    [itex]v_{rel}= v{(2-\frac{v^{2}}{c^{2}})}^{1/2} \large \frac{1}{\sqrt{1- \frac{ v^{2} }{ c^{2} }}}[/itex]

    So I have an extra factor in my answer. What am I doing wrong?
     
  2. jcsd
  3. Jul 18, 2011 #2

    Dick

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    I think the (v)(v)/c^2 in your formula for v'_y should be the dot product of the two original velocities. That's zero since they are orthogonal vectors.
     
  4. Jul 19, 2011 #3
    Is that so? How is the expression with the dot product derived then? I just assumed that the velocity in frame S' is given by [itex]\frac{y'}{t'}[/itex]. And from the Lorentz transforms:

    [itex]y'=y[/itex]
    [itex]t'=\gamma (t - \frac {vx}{c^{2}}) [/itex]

    where v is the velocity of frame S' with respect to S. Where does the dot product come in?
     
  5. Jul 19, 2011 #4

    vela

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    The dot product arises because only the component of velocity in the direction of the boost is contracted. I'm sure you could find a general derivation somewhere.

    What I would do, though, is transform the coordinates of the second particle using the Lorentz transformations and express x' and y' as a function of t' (not t). From that expression, you can see what the components of the velocity are to an observer at rest in the moving frame.
     
  6. Jul 19, 2011 #5
    Ok, I got the answer following vela's suggestion. In the frame S, the coordinates of the particle going up in the +y direction are given by:

    [itex]x=0[/itex]
    [itex]y=vt[/itex]

    Transforming these to the frame S' gives the coordinates:

    [itex]x'= - \gamma vt[/itex]
    [itex]y'=vt[/itex]

    The inverse Lorentz transform from S' to S gives:

    [itex]t= \gamma (t'+\frac{vx}{c^{2}})[/itex]

    Which on substituting x yields:

    [itex]t=t' \sqrt{1-\frac{v^{2}} {c^{2}} }[/itex]

    Substituting this in the original equations for x' and y', and differentiating with respect to t' yields the velocity components of the particle moving in the +y direction in S':

    [itex]v'_{x}=-v[/itex] (as expected)
    [itex]v'_{y}=v\sqrt{1-\frac{v^{2}} {c^{2}} }[/itex]

    And these give the required result. I assume you suggested to put x' and y' in terms of t' so that the reading would be at the same time in S', right?

    Also, I don't see why the dot product is required in the y-velocity transformation. I assumed that the "v" there was the velocity in the y direction, not any arbitrary direction....none of my books shed light on this.
     
  7. Jul 19, 2011 #6

    vela

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    Not really. I suggested it because [itex]v'_x = dx'/dt'[/itex], not [itex]dx'/dt[/itex].
     
  8. Jul 20, 2011 #7
    Ok - Thanks for the help!
     
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