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Einstein velocity transformations problem

  1. Jun 22, 2014 #1
    So I made this problem up to visualize the einstein velocity transformations between inertial frames.

    1. The problem statement, all variables and given/known data

    I throw a frisbee due north. It goes north at a constant velocity of .7c. At the same time I throw it, a bird flies in a straight line at a constant velocity of .5c at such an angle that its northward component is .3c and its eastward component is .4c, relative to the frisbee. It is going toward a birdfeeder located 1 light-minute north and .4 light-minutes east of where I stand. Is the bird really going toward the birdfeeder in both the frisbee's inertial frame and my inertial frame?

    2. Relevant equations

    vx = (vx' + β)/(1+vx'β)
    vy = (vy'(√1 - β2)/(1-vx'β)

    Apostrophied velocities are measured in the frisbee's frame, which moves at velocity "beta" relative to my frame.

    3. The attempt at a solution

    In the frisbee's frame, the birdfeeder is heading south at .7c. In one minute, it will be .7 light-minutes south of where it was before. The bird moves relative to the frisbee up .3 light-minutes and east .4 light-minutes, so it should meet the birdfeeder in one minute.

    In the home frame,

    The northward component of the bird is .3+.7 / 1+(.3)(.7) = .82645
    The eastward component of the bird is .4(sqrt(1-.49))/1+(.3)(.7) = .23608

    Since .82645/.23608 does not equal 1/.4, the bird is not heading toward the birdfeeder.

    I definitely did something wrong to get this contradiction. Would anyone like to try it?
     
    Last edited: Jun 22, 2014
  2. jcsd
  3. Jun 22, 2014 #2

    Simon Bridge

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    ... to word "really" does not belong here.

    You can check your setup by sketching the space-time diagrams for each observer.
     
  4. Jun 23, 2014 #3
    You forgot about space contraction which changes the angle towards the location of the bird feeder.
     
  5. Jun 25, 2014 #4
    Dauto,

    Thanks for the reply, but I'm not sure I get it. Wouldn't the Einstein velocity transformations already account for the space contraction between the points of view of me and the frisbee?
     
  6. Jun 25, 2014 #5
    Okay, I figured it out. From my point of view, the bird flew north at .82645c, but if I were to use a simple Lorentz contraction and multiply this velocity by sqrt(1-(.7)^2), you get a velocity of .5902c north, which, coupled with the .23608c component East, will get the bird to the birdfeeder.
     
  7. Jun 25, 2014 #6

    Simon Bridge

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    Well done - it's easier with the space-time diagrams though.
     
  8. Jun 26, 2014 #7
    I'm afraid I don't know space-time diagrams very well. Do you mean, a graph with perpendicular axes "time" and "distance" with a second "time" axis at slope 1/.7 and a second "distance" axis at slope .7, all from the origin?
     
  9. Jun 27, 2014 #8

    Simon Bridge

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