Question: A particle with 4-momentum $P$ is detected by an observer whose four-velocity is $U$. Working in the rest frame of the observer, express $P\cdot U$ in terms of the mass of the particle and its speed $v$ in the observer's rest frame. Show that $\frac{v}{c} = \sqrt{1-\frac{(P\cdot P)c^2}{(P\cdot U)^2}}$.
I've managed to end up with the final equation but am unsure of my method and wanted some clarification on whether it is correct or not, and if there is a nicer solution. Anyways,
Solution: Let $S'$ be the stationary frame and $S$ be the moving one. We suppose that $S'$ and $S$ are related by a Lorentz boost in the $x$-axis direction. So if $U = \gamma(c,u_1 ,u_2, u_3)$ then we can find its 4-velocity in the stationary frame, $U'$, by $U' = \gamma \begin{pmatrix} \gamma & \frac{\gamma v}{c} & 0 & 0 \\ \frac{\gamma v}{c} & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} c \\ u_1 \\ u_2 \\ u_3 \end{pmatrix} = \begin{pmatrix} \gamma^2 c + \frac{\gamma^2 v u_1}{c} \\ \gamma^2 v + \gamma^2 u_1 \\ \gamma u_2 \\ \gamma u_3 \end{pmatrix}$.
But as we want the velocity component of $U'$ to be $\mathbf{0}$, we must have $u_1 = -v, u_2 = u_3 = 0$ So then $\gamma^2 c + \frac{\gamma^2 v u_1}{c} = \gamma^2 c - \frac{\gamma^2 v^2}{c} = \gamma^2 (\frac{c^2-v^2}{c}) = c \implies P\cdot U = c\cdot m\gamma c = m\gamma c^2 = \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$ Rearranging this for $\frac{v}{c}$ and using the fact that $P\cdot P = m^2 c^2$ got me the final result.