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Special relativity/velocity transformations

  1. Feb 16, 2009 #1
    1. A cosmic-ray proton streaks through the lab with velocity 0.85c at an angle of 50o with the +x direction (in the xy plane of the lab). Compute the magnitude and direction of the proton’s velocity when viewed from frame S’ moving with β=0.72 in the +x direction.

    2. Ux'= Ux-v/1-vUx/c^2

    3. My attempt at the solution was to simply in an example of 1-D motion. From considering that a proton is moving at 0.85c in the positive x-direction (ignoring the angle), this can be compared when viewed from the S' frame, moving relative to S with 0.72c in the +x direction. The observer in S' measures the proton to be approaching in the +x-direction at 0.85c. Using the first equation U1x' = 0.85c - 0.72c/1-(0.85c)(0.72c)/c^2 = .34c. Introducing a y component makes the problem much more challenging and I'm not quite sure what to do. Do I work out the x and y directions independently? In which case, would the velocity for the y-direction be 0.85c*sin(50 degrees) and the x direction be 0.85c*cos(50 degrees). How would this appear to an observer in S'? That is, how would I calculate the direction as observed in S' and the magnitude?

    Thank you.
  2. jcsd
  3. Feb 16, 2009 #2
    It's not very clear to me what you are asking. The proton has a certain velocity relative to the lab which is given. You can work out the x and y components of that velocity. The other frame's velocity relative to the lab is given. This has only an x component. You know how to calculate relative velocities Ux and Uy in that case. If you know that, can you work out the magnitudes and directions?
  4. Feb 17, 2009 #3
    Sorry for the ambiguity, the question just asks to compute the magnitude and direction of the proton as observed in S'.

    I worked more on the problem and completed the x and y components.

    Using the velocity transformations, the x-component was 0.85c*cos(50), doing the calculation out with the above equation for the x-direction I got -0.287c.

    The Y-component was 0.85c*sin(50). Doing this calculation out with the above equation for the y-direction, I got 0.569c.

    To find the magnitude I used P-theorem. c^2=a^2+b^2. In this example, c^2 =0.569c^2+(-0.287)^2. c=0.637c.

    To find the direction I used theta = tan^-1 y'/x'. y' = 0.569c x' = -0.287c. Theta = 63.2 degrees.

    Is this correct? Thank you!
  5. Jan 6, 2010 #4
    Seems correct.
    But I'd like to calculate it again to make sure :)
  6. Jan 6, 2010 #5
    Seems correct.
    But I'd like to calculate it again to make sure :)
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