Relativistic addition question

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I understand that, given objects A and B moving away from a stationary point O and each other at relativistic speeds, you use the relativistic addition formula to find A's speed relative to B.

How would you find A's speed relative to B if A and B are moving away from O at a 60 degree angle from each other, and you know their speeds relative to O?

Thank you
 

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  • #2
JesseM
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Look at the velocity vector of B in A's frame and separate it into components--one component parallel to the direction that A is moving relative to O, and another component perpendicular to that direction. The magnitude of the perpendicular component will be the same in O's frame as it is in A's; and for the parallel component, just use the formula for addition of relativistic velocities to find its magnitude in O's frame.

edit: never mind, this answer is wrong, see learningphysics' answer below.
 
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  • #3
learningphysics
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There are two ways I can think of to find A's speed relative to B:

1) Rotate your coordinate system so that the x-axis is aligned along B's path...

then first calculate [tex]u_x[/tex] and [tex]u_y[/tex] for particle A using trigonometry. [tex]u_x = u_a*cos(60)[/tex] [tex]u_y = u_a*sin(60)[/tex]

Then use the velocity transformation equations for [tex]u_x[/tex] and [tex]u_y[/tex] to get [tex]u_x'[/tex] and [tex]u_y'[/tex] which are the components of A's velocity in B's rest frame.

[tex]u_x' = \frac{u_x - v}{1-u_x*v/c^2}[/tex] (the usual velocity addition equation)
[tex]u_y' = \frac{u_y}{\gamma*(1-u_x*v/c^2)}[/tex]

Here v is B's speed.

Then use the pythagorean theorem to get the magnitude of the speed.

2) If you've studied four tensors then there is a more elegant method. The product of the two four-velocities [tex]U_a \cdot U_b[/tex] is an invariant that equals [tex]c^2 * \gamma (u) [/tex] where [tex]u[/tex] is the relative speed (speed of B relative to A, or speed of A relative to B). I'm taking this argument is from "Introduction to Special Relativity" by Rindler p. 60. So if you calculate [tex]U_a \cdot U_b[/tex] using the 60 degree angle you're given... then you can solve for [tex]\gamma (u)[/tex] and then solve for u.
 
  • #4
learningphysics
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JesseM said:
Look at the velocity vector of B in A's frame and separate it into components--one component parallel to the direction that A is moving relative to O, and another component perpendicular to that direction. The magnitude of the perpendicular component will be the same in O's frame as it is in A's; and for the parallel component, just use the formula for addition of relativistic velocities to find its magnitude in O's frame.
JesseM, the perpendicular component isn't the same.
 
  • #5
JesseM
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learningphysics said:
JesseM, the perpendicular component isn't the same.
Ah, you're right, I didn't think that through very well--I was thinking that since the y and z coordinate are the same in both A and O's coordinate system that would mean the velocity is the same, but that doesn't take into account the different time-coordinates, not to mention the fact that something moving in the perpendicular direction in A's system will be moving diagonally in O's.
 
  • #6
learningphysics
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JesseM said:
Ah, you're right, I didn't think that through very well--I was thinking that since the y and z coordinate are the same in both A and O's coordinate system that would mean the velocity is the same, but that doesn't take into account the different time-coordinates, not to mention the fact that something moving in the perpendicular direction in A's system will be moving diagonally in O's.
It's cool. :wink: I've been working through Rindler's Intro to Spec. Relativity, as we speak, so the formula is kind of fresh in my head.
 

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