Relativist Addition of Velocities

[jgens]

Homework Statement

Fix an inertial reference frame and consider a particle moving with velocity \mathbf{u} in this frame. Let \mathbf{u'} denote the velocity of the particle as measured in an inertial frame moving at velocity \mathbf{v} with respect to the original frame. Show the following:

• If |\mathbf{u}| < c, then |\mathbf{u}'| < c.
• If |\mathbf{u}| = c, then |\mathbf{u}'| = c.
• If |\mathbf{u}| > c, then |\mathbf{u}'| > c.

Homework Equations

N/A

The Attempt at a Solution

By choosing coordinates appropriately, it suffices to assume that \mathbf{v} is directed along the x-axis of the original frame, so write \mathbf{v} = v. Now write \mathbf{u} = (u_x,u_y,u_z) and \mathbf{u}' = (u_x',u_y',u_z') and recall that the relativistic addition of velocities implies:
<br /> u_x&#039; = \frac{u_x-v}{1-\frac{vu_x}{c^2}} \;\;\; \mathrm{and} \;\;\; u_y&#039; = \frac{u_y}{1-\frac{vu_x}{c^2}}\sqrt{1-\frac{v^2}{c^2}} \;\;\; \mathrm{and} \;\;\; u_z&#039; = \frac{u_z}{1-\frac{vu_x}{c^2}}\sqrt{1-\frac{v^2}{c^2}}<br />
These three equalities should allow for the computation of |\mathbf{u}&#039;| in terms of the known speeds |\mathbf{u}| and |\mathbf{v}|, but for whatever reason, I am having difficulty simplifying the expression below to a useful point:
<br /> |\mathbf{u}&#039;|^2 = \left( \frac{u_x - v}{1 - \frac{vu_x}{c^2}} \right)^2 + \left( \frac{u_y}{1-\frac{v u_x}{c^2}} \sqrt{1-\frac{v^2}{c^2}} \right)^2+ \left( \frac{u_z}{1 -\frac{vu_x}{c^2}} \sqrt{1-\frac{v^2}{c^2}} \right)^2 = \left( \frac{1}{1 - \frac{vu_x}{c^2}} \right)^2 \left[ (u_x-v)^2 + u_y^2\left(1 - \frac{v^2}{c^2}\right) + u_z^2\left(1 - \frac{v^2}{c^2}\right) \right]<br />
Does anyone have some suggestion how to simplify that any further in a useful direction?

 

[tiny-tim]

hi jgens!

try proving it for the easy |u| = c case first