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Perpendicular relativistic velocities

  1. Sep 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Imagine two motorcycle gang leaders racing at relativistic speeds along perpendicular paths from the local pool hall, as shown in Figure 1.21. How fast does pack leader Beta recede over Alpha’s right shoulder as seen by Alpha?

    Solution Figure 1.21 shows the situation as seen by a stationary police officer located in frame S, who observes the following:

    2. Relevant equations
    u'_x = (u_x - v)/(1 - v*u_x/c^2)

    u'_y = u_y/γ*(1 - v*u_x/c^2)

    3. The attempt at a solution
    I can follow the solution shown in the attachment.

    The problem is that I'm trying to verify the solution by attaching the s' frame of reference to the beta gang leader instead of the alpha gang leader. The relative speed of alpha to beta should be the same as the relative speed of beta to alpha, but that's not what im getting. Can you guys tell me what im doing wrong?
     

    Attached Files:

  2. jcsd
  3. Sep 4, 2016 #2

    vela

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    I suggest you put the solution away and try to reproduce it on your own. I expect you'll get the wrong answer because you'll make a similar mistake as you're doing now. Compare your solution to the book's and see what the mistake is.
     
  4. Sep 4, 2016 #3
    I've done that, and I've gone over it multiple times but still can't see what I'm doing wrong.
     
  5. Sep 4, 2016 #4
    The thing that gets me is that when s' is attached to the alpha leader, he measures a u'_x and u'_y relative to the beta but when I attach s' to the beta leader, I get that he measures a u'_x but no u'_y even though the beta leader should see the alpha receding in the positive y direction.
     
  6. Sep 4, 2016 #5

    vela

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    Then try typing up your solution here and explain your steps. I'm not trying to be difficult. It's just that you're making a really obvious mistake, and it's strange that you can't find it with a little thought.
     
  7. Sep 4, 2016 #6

    vela

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    Well, that's a good starting place. Why are you getting u'_y = 0?
     
  8. Sep 4, 2016 #7
    Ok so first I'll reproduce the result for the original case when s' is attached to the alpha, and s is the frame of reference of the officer who's at rest in s.

    u'_x will then be the velocity that alpha leader measures the beta leader to have in the x direction.
    that wil be equal to (u_x - v)/(1 - v*u_x/c^2) where u_x is the velocity in the x direction that the officer measures for the beta leader which is 0 because the officer only sees the beta traveling in the -y direction.
    v is the relative velocity between s and s' measured from s frame which is 0.75c.

    therefore u'_x = -0.75c

    u'_y will be the velocity in the y direction that the alpha measures for the beta.
    u'_y = u_y/γ*(1 - v*u_x/c^2) u_y is the velocity in the y direction that the officer measures for the beta which is -0.90c
    u_x and v are the same as before
    u'_y = -0.595c
    taking the sum of the squares and then square rooting gives 0.958c.
    so the relative of the alpha relative to the beta as measured by the alpha is 0.958c

    Now attaching s' to the beta instead.

    u'_x = (u_x - v)/(1 - v*u_x/c^2) now u_x is the velocity in the x direction that the officer measures for the alpha which is 0.75c.
    v is the relative velocity between the officer and the beta as measured by the officer which is -0.90c.

    plugging in those values i get u'_x = 0.985c

    u'_y = u_y/γ*(1 - v*u_x/c^2) u_y is the velocity in the y direction that the officer measures of the alpha which is 0 and gives me
    u'_y = 0.
     
  9. Sep 4, 2016 #8

    vela

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    When S' is Beta's rest frame, isn't it moving along the y-axis? Wouldn't the equations change?
     
  10. Sep 4, 2016 #9
    so if im understanding you correctly, you're saying that when I switch s' to beta's rest frame i'm essentially rotating my coordinate system 90 degrees clockwise to where the +x axis is now the +y axis and the -y axis is now the +x axis?
     
  11. Sep 4, 2016 #10

    vela

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    Yeah, pretty much. The expressions you used for ##u'_x## and ##u'_y## are for a boost along the x axis, but when you're looking from Beta's perspective, the boost needs to be along the y-axis. You can rotate the coordinate system to keep the formulas the same, or you could rewrite the formulas for a boost along the y-axis.
     
  12. Sep 4, 2016 #11
    hmm, i'm not familiar with this boost term. I see no mention of it in my lecture notes or book. From the context, it seems like you referring to applying the lorentz transformation to one dimensional cases?
     
  13. Sep 4, 2016 #12

    vela

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    Sorry for the unfamiliar terminology. A boost is a Lorentz transformation that mixes the time and spatial coordinates with no rotation in space involved. Lorentz transformations can also include plain old rotations in space.
     
  14. Sep 4, 2016 #13
    ah ok I see. Well thank you very much. I rotated my coordinate system for the beta frame and got the correct answer :)
     
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