Einstein velocity transformations problem

  • #1
So I made this problem up to visualize the einstein velocity transformations between inertial frames.

Homework Statement



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?

Homework 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.

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?
 
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Answers and Replies

  • #2
Simon Bridge
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Is the bird really going toward the birdfeeder in both the frisbee's inertial frame and my inertial frame?
... to word "really" does not belong here.

You can check your setup by sketching the space-time diagrams for each observer.
 
  • #3
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You forgot about space contraction which changes the angle towards the location of the bird feeder.
 
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  • #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?
 
  • #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.
 
  • #6
Simon Bridge
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Well done - it's easier with the space-time diagrams though.
 
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  • #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?
 
  • #8
Simon Bridge
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