 Solution

Take spacetime coordinates $X=(ct,x,y,z)$. Differentiating the initial segment gives the four-velocityThe identity ${\cosh }^{2}r-{\sinh }^{2}r=1$ gives $g(U,U)=c^2$, and $dt/ds=\cosh \omega s>0$. The proper time element therefore satisfies $d{\tau }^2=g(dX,dX)/c^2=d{s}^2$, with the future orientation selecting $d\tau =ds$. Thus $s$ is proper time along this path.