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Let $f(r)=1-r_s/r$ and parametrize the timelike geodesic by proper time $\tau$. Time-translation and rotational symmetry conserve $E=f\dot{t}$ and $h=r^2\dot{\varphi }$ in the equatorial plane. The normalized four-velocity obeysSubstitution and multiplication by $f$ give ${\dot{r}}^2+f(1+h^2/r^2)=E^2$, orThe dot denotes a proper-time derivative, and $E,h$ are energy and angular momentum per unit rest mass.

A spacetime is geodesically complete if every maximal geodesic has an affine parameter ranging over all of $\mathbb{R}$. For timelike geodesics this is equivalent to unbounded proper time in both directions. An extendible geodesic segment can be prolonged in the same spacetime; an inextendible geodesic cannot. A finite coordinate endpoint need not imply finite affine parameter.

For the Kruskal spacetime, use $U,V$ withIn the right exterior $U<0<V$. A truncated ray $U=U_0<0$, $V_1<V<V_2$ is an extendible geodesic of radial null type: neither artificial endpoint is a spacetime boundary. A future ray $U=U_0>0$ in the black hole reaches $UV=1$, hence $r=0$, and is inextendible geodesic and future null-geodesically incomplete. Its Killing energy $E\ne 0$ gives $dr/d\lambda =-|E|$, so the Schwarzschild singularity occurs at finite affine parameter. The maximal continuation toward the past supplies the other half of this same null geodesic.

There is no inextendible, complete radial timelike geodesic in positive-mass Kruskal spacetime. This requested example is impossible as printed. For a radial timelike geodesic, the conserved Killing energy and normalization giveIf $E^2<1$, $r$ has at most one turning point, a maximum $r_{\max }=2M/(1-E^2)$; a maximal trajectory runs from the white hole Schwarzschild singularity to the Schwarzschild singularity. If $E^2\ge 1$, there is no finite turning point; one end can lie at infinity but the other reaches $r=0$. The exceptional $E=0$ trajectory through the bifurcation surface also reaches $r=0$ in both time directions. Near $r=0$,whose integral is finite. Constant-$r$ radial timelike curves are accelerated, not geodesics.

Two plausible repairs have different meanings. Removing “radial” permits a complete circular timelike geodesic at $r=R>6M$, with nonzero angular momentum and proper time ranging over $\mathbb{R}$. Replacing “timelike” by “null” permits a complete horizon null geodesic: $U=0$, $V\in \mathbb{R}$, with $V$ an affine parameter. The Penrose diagram shows both repairs explicitly, together with the two valid requested examples; the circular trajectory is only a radial projection and is labelled as nonradial.

The horizontal boundaries are the Schwarzschild singularities, diagonal dashed lines are the Killing horizons, and outer diagonal edges are null infinity.

For radial free fall, let $E=f{\dot{t}}_A>0$ be Alice's conserved Killing energy. The inward branch of the timelike geodesic has $\dot{r}=-\sqrt{E^2-f}$, henceWith $\delta =r-2M$, the near-horizon geodesic equations giveThe constants in the logarithms are understood to make their arguments dimensionless. Therefore $u_A=t_A-r_{\ast }=2t_A+O(1)$. For an outgoing signal reaching the fixed-radius Bob, $t_B=u_A+r_{\ast }(r_B)$, so $t_B=2t_A+O(1)$. The redshift consequently behaves asThis is the answer when $t$ is Bob's Schwarzschild time, as appropriate to the stated observation. It is the Schwarzschild surface gravity $a=\kappa =1/(4M)$. If $t$ instead means the emission Schwarzschild time $t_A$, the same redshift is proportional to $e^{t_A/(2M)}$ and gives $M=1/(2a)$. The two answers use different clocks, not different dynamics. If the measured exponential uses Bob's proper time ${\tau }_B$, its rate is $1/(4M\sqrt{f_B})$. In SI units the reception-time result is $M=c^3/(4Ga)$.

The reception-time exponent also holds for smooth radial infall crossing the future Schwarzschild event horizon with finite nonzero $dr/d\tau$: regular Ingoing Eddington-Finkelstein coordinates give finite $v$ there, $u=v-2r_{\ast }\sim -4M\log ({\tau }_H-\tau )$. Thus no special value of $E$ is needed.

Every maximal radial timelike geodesic in positive-mass Kruskal spacetime is incomplete in at least one direction. Its equations are ${\dot{r}}^2=E^2-1+2M/r$ and $\ddot{r}=-M/r^2$. A finite turning point is a maximum, and each trajectory reaches $r=0$ at one or both ends. The remaining proper time is finite because $d\tau \sim \sqrt{r/(2M)}dr$. Complete circular timelike geodesics exist, but they have nonzero angular momentum.

A timelike geodesic has timelike tangent and vanishing covariant acceleration in an affine parameter. Choosing proper time gives $g(\dot{\gamma },\dot{\gamma })=-1$.