- #1

jgens

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## Homework Statement

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

- If [itex]|\mathbf{u}| < c[/itex], then [itex]|\mathbf{u}'| < c[/itex].
- If [itex]|\mathbf{u}| = c[/itex], then [itex]|\mathbf{u}'| = c[/itex].
- If [itex]|\mathbf{u}| > c[/itex], then [itex]|\mathbf{u}'| > c[/itex].

## Homework Equations

N/A

## The Attempt at a Solution

By choosing coordinates appropriately, it suffices to assume that [itex]\mathbf{v}[/itex] is directed along the [itex]x[/itex]-axis of the original frame, so write [itex]\mathbf{v} = v[/itex]. Now write [itex]\mathbf{u} = (u_x,u_y,u_z)[/itex] and [itex]\mathbf{u}' = (u_x',u_y',u_z')[/itex] and recall that the relativistic addition of velocities implies:

[tex]

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}}

[/tex]

These three equalities should allow for the computation of [itex]|\mathbf{u}'|[/itex] in terms of the known speeds [itex]|\mathbf{u}|[/itex] and [itex]|\mathbf{v}|[/itex], but for whatever reason, I am having difficulty simplifying the expression below to a useful point:

[tex]

|\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]

[/tex]

Does any one have some suggestion how to simplify that any further in a useful direction?