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## 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 $|\mathbf{u}'| > c$.

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:
$$u_x' = \frac{u_x-v}{1-\frac{vu_x}{c^2}} \;\;\; \mathrm{and} \;\;\; u_y' = \frac{u_y}{1-\frac{vu_x}{c^2}}\sqrt{1-\frac{v^2}{c^2}} \;\;\; \mathrm{and} \;\;\; u_z' = \frac{u_z}{1-\frac{vu_x}{c^2}}\sqrt{1-\frac{v^2}{c^2}}$$
These three equalities should allow for the computation of $|\mathbf{u}'|$ 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:
$$|\mathbf{u}'|^2 = \left( \frac{u_x - v}{1 - \frac{vu_x}{c^2}} \right)^2 + \left( \frac{u_y}{1-\frac{v u_x}{c^2}} \sqrt{1-\frac{v^2}{c^2}} \right)^2+ \left( \frac{u_z}{1 -\frac{vu_x}{c^2}} \sqrt{1-\frac{v^2}{c^2}} \right)^2 = \left( \frac{1}{1 - \frac{vu_x}{c^2}} \right)^2 \left[ (u_x-v)^2 + u_y^2\left(1 - \frac{v^2}{c^2}\right) + u_z^2\left(1 - \frac{v^2}{c^2}\right) \right]$$
Does any one have some suggestion how to simplify that any further in a useful direction?

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