pigasuspig
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

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.

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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 $$u_x$$ and $$u_y$$ for particle A using trigonometry. $$u_x = u_a*cos(60)$$ $$u_y = u_a*sin(60)$$

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

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

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 $$U_a \cdot U_b$$ is an invariant that equals $$c^2 * \gamma (u)$$ where $$u$$ 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 $$U_a \cdot U_b$$ using the 60 degree angle you're given... then you can solve for $$\gamma (u)$$ and then solve for u.

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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.

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.

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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. I've been working through Rindler's Intro to Spec. Relativity, as we speak, so the formula is kind of fresh in my head.