 Solution

Write the four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$ and four-acceleration $A^{\mu }=d{U}^{\mu }/d\tau$. Since $U_{\mu }{A}^{\mu }=0$, spatial orthogonality $\mathbf{U}\cdot \mathbf{A}=0$ implies $U^0{A}^0=0$. Thus $A^0=cd\gamma /d\tau =0$ and the speed is constant. If “acceleration” is read as the laboratory three-acceleration instead, $\gamma \mathbf{v}\cdot d\mathbf{v}/dt=0$ immediately gives the same conclusion.

For four-acceleration in uniform circular motion, the speed is $r|\omega |<c$, so $\gamma =(1-r^2{\omega }^2/c^2{)}^{-1/2}$ is constant. Differentiating the four-velocity using $d/d\tau =\gamma d/dt$ givesThe proper acceleration is $\sqrt{-A_{\mu }{A}^{\mu }}={\gamma }^{2}r{\omega }^2$. The vanishing time component reflects constant energy; the spatial components describe the changing momentum direction.