 Solution

In the observer's rest frame, $U=(c,0)$ and $P=(E^{\prime }/c,{\mathbf{p}}^{\prime })$. The Lorentz invariance of the Minkowski inner product gives $P\cdot U=E^{\prime }$ andThe relation between four-momentum and velocity is ${\mathbf{p}}^{\prime }=E^{\prime }\mathbf{v}/c^2$, soTaking the nonnegative square root provesFor a future-directed massive particle and observer, $P\cdot U>0$ and the expression lies in $[0,1)$; the massless limit, with nonzero energy, gives $v=c$.