= Solution
For a radial timelike <geodesic>, the static Killing energy is $E=f\dot t$ and normalization gives $\dot r^2=E^2-f$. Choose $E=\sqrt2$ and start at $r(0)=0$ with outward velocity one. An explicit trajectory is
$$
\boxed{r(\tau)=a\sinh(\tau/a).}
$$
It solves $\dot r^2=1+r^2/a^2$ and reaches $r=a$ at $\boxed{\tau_H=a\operatorname{arsinh}(1)<\infty}$. Differentiating the energy equation gives $\ddot r=r/a^2$, the radial geodesic equation, so this is an actual freely falling trajectory.
The regular coordinate satisfies $\dot u=(E-\dot r)/f=1/(E+\dot r)$, which is finite and positive at the crossing. Together with $u(\tau)=u(0)+\int_0^\tau [\sqrt2+\cosh(s/a)]^{-1}ds$, the trajectory extends through the horizon in the regular chart. Divergence of the static time there does not imply infinite <proper time>.
Back to article page
