= Solution
Inside the horizon, $g^{ab}\nabla_ar\nabla_br=f<0$, so $\nabla r$ is timelike. The chosen black-hole time orientation declares $-\nabla r$ future directed; therefore every future-directed timelike vector $V$ has $V(r)<0$. Thus $r$ decreases strictly along every future-directed timelike curve.
Write $F=2M/r-1=-f>0$. Along such a curve,
$$
d\tau^2=\frac{dr^2}{F}-Fdt^2-r^2d\Omega^2
\leq\frac{dr^2}{F}.
$$
It cannot remain at any $r>0$ indefinitely, because $r$ is a time function and the displayed bound gives finite remaining proper time. From a starting radius $r_p<2M$,
$$
\tau\leq\int_0^{r_p}\frac{dr}{\sqrt{2M/r-1}}
<\int_0^{2M}\frac{dr}{\sqrt{2M/r-1}}.
$$
Putting $r=2M\sin^2\chi$ evaluates the last integral as $\pi M$. Hence every such curve reaches the $r=0$ curvature singularity with
$$
\boxed{\tau<\pi M}.
$$
Solved by gpt-5.6-sol high.
Back to article page
