Transforming Relativistic Velocities

1. Nov 16, 2009

psholtz

1. The problem statement, all variables and given/known data
My question concerns what happens when one is transforming relativistic velocities in more than one dimension.

The Lorentz velocity addition formula is well known. Suppose there is an inertial reference frame F' moving at speed v in the +x direction with respect to a "stationary" lab frame F, and suppose that in F' the speed of a projectile, in the x-direction, is given by $$u_x'$$. Then the speed of this projectile, as measured in F, will simply be:

$$u_x = \frac{u_x' + v}{1 + \frac{u_x' v}{c^2}}$$

That's all well and good, but suppose that we don't launch the projectile in F' precisely parallel to the x-axis. Suppose rather, that we launch it at some angle to the x'-axis, say $$\theta'$$. The question is, what velocity is measured now in the frame F?

2. Relevant equations

$$u_y' = u' \sin \theta '$$

$$u_x' = u' \cos \theta '$$

3. The attempt at a solution
My "naive" attempt at a solution would be to take the "vertical" component, on the y-axis, and transform it "as is", and take the x-component and subject that to the Lorentz velocity addition law. In other words:

$$u_y = u_y' = u' \sin \theta '$$

$$u_x = \frac{u' \cos \theta' + v}{1 + \frac{v u' \cos \theta' }{c^2}}$$

The problem w/ this approach is in the limit as either u' or v approaches c.

Suppose, for the sake of illustration, that v approaches c. Then by the above formula, we would have:

$$u_x = c$$

as expected. But the problem is that there is still this $$u_y$$ component which is finite and non-zero, and when we add that in, by Pythagoras, we get an answer for the velocity in F which is in excess of c.

Apparently, I would think that as the $$u_x$$ component increases, then the $$u_y$$ component must somehow decrease... but I'm wondering how to specify this quantitatively..??

2. Nov 16, 2009