= Solution
Take $\tau$ as particle <proper time>, dots as $d/d\tau$, and $f(r)=1-2\mu/r$ with $\mu=GM/c^2$. In the equatorial plane the <geodesic Lagrangian> is
$$
\mathcal L=\frac12\left(c^2f\dot t^2-f^{-1}\dot r^2-r^2\dot\phi^2\right).
$$
The <cyclic coordinates> give $d(c^2f\dot t)/d\tau=0$ and $d(-r^2\dot\phi)/d\tau=0$. Set the resulting constants to $c^2k$ and $-h$ respectively. The explicit second-order equations are
$$
\boxed{\ddot t+\frac{f'}f\dot r\dot t=0,\qquad \ddot\phi+\frac2r\dot r\dot\phi=0,}
$$
$$
\boxed{\ddot r-\frac{f'}{2f}\dot r^2+\frac{c^2ff'}2\dot t^2-rf\dot\phi^2=0,\qquad f'=\frac{2\mu}{r^2}.}
$$
The polar equation is satisfied by $\theta=\pi/2$, $\dot\theta=0$, consistently with <spherical symmetry>. The proper-time normalization $ds^2=c^2d\tau^2$ supplies the third first integral:
$$
\boxed{f\dot t=k,\qquad r^2\dot\phi=h,\qquad c^2f\dot t^2-f^{-1}\dot r^2-r^2\dot\phi^2=c^2.}
$$
The normalization is $2\mathcal L=c^2$; it is distinct from choosing an arbitrary affine tangent's norm.
Stationarity and axial symmetry make the covariant time and azimuthal momenta constant. For physical momentum and coordinate $t$ in seconds, $P_t=mc^2f\dot t=mc^2k$ and $P_\phi=-mr^2\dot\phi=-mh$. Thus \b[$mc^2k$ is the conserved <energy> normalized at infinity and $mh$ is the signed axial <angular momentum>]. This is <Killing energy with positive time signature>; a convention with negative timelike norm uses the opposite contraction sign. The supplied covariant-momentum form of the <geodesic equation> makes these cyclic constants immediate, but its parametrization must be understood: for $p^\mu=dx^\mu/d\lambda$ it is exactly as written. For physical $P^\mu=m\,dx^\mu/d\tau$, proper-time differentiation instead gives $dP_\mu/d\tau=(\partial_\mu g_{\nu\sigma})P^\nu P^\sigma/(2m)$; equivalently, the displayed mass-free form uses $\lambda=\tau/m$. This affine momentum normalization does not change either conservation law.
Multiply the normalization equation by $f$ and substitute the two cyclic integrals. Rearrangement gives the radial <energy> equation
$$
\boxed{\dot r^2+\frac{h^2}{r^2}\left(1-\frac{2\mu}r\right)-\frac{2c^2\mu}r=c^2(k^2-1).}
$$
For radial marginal escape $h=0$ and the future-directed constant is $k=1$, because the <speed> at infinity vanishes. Choose the outward root:
$$
\dot r=\sqrt{\frac{2GM}{r}},\qquad \frac23(r^{3/2}-r_0^{3/2})=\sqrt{2GM}\,\tau.
$$
Hence the <marginal escape geodesic in Schwarzschild spacetime> has
$$
\boxed{r(\tau)=\left(r_0^{3/2}+a\tau\right)^{2/3},\qquad a=\frac32\sqrt{2GM}=\frac{3c}{2}\sqrt{2\mu}.}
$$
The constant $a$ has units of <length> to the power $3/2$ divided by time. This radius is an areal coordinate, not accumulated local proper <distance>.
A <Schwarzschild static observer> at exterior radius $r$ uses local radial rulers $d\ell=dr/\sqrt f$ and a local clock $dT=\sqrt f\,dt$. Therefore the <local radial speed of a Schwarzschild geodesic> is
$$
v_{\rm loc}=\frac{d\ell}{dT}=\frac1f\frac{dr}{dt}=\frac{\dot r}{k}.
$$
For the marginal trajectory $k=1$ this gives
$$
\boxed{v_{\rm loc}=\sqrt{\frac{2GM}{r}}.}
$$
Its equality to $\dot r$ is particular to $k=1$; the local definition, rather than that accidental equality, is the required observer measurement. The Schwarzschild-coordinate <velocity> is instead $dr/dt=f\sqrt{2GM/r}$.
For nonzero asymptotic <speed> $0\le u<c$, the same <energy> constant is $k=\gamma_\infty=(1-u^2/c^2)^{-1/2}$. Combining the radial equation with the local-ruler relation gives <radial Schwarzschild escape with asymptotic speed>
$$
\boxed{v_{\rm loc}(r)=c\sqrt{1-\frac{f}{k^2}}=\sqrt{u^2+\frac{2GM}{r}\left(1-\frac{u^2}{c^2}\right)}.}
$$
The positive root describes the outward particle, at radii on its trajectory with $r\ge r_0$. Independently, a static observer has <four-velocity> $U=(1/\sqrt f,0,0,0)$ in these coordinates, giving relative <Lorentz factor> $g(U,u_{\rm particle})/c^2=\sqrt f\dot t=k/\sqrt f$ and the same <speed>. As $r\to\infty$ it tends to $u$; for $u=0$ it gives the preceding result. Its limiting value approaches $c$ as $r\downarrow2GM/c^2$, but that is a limit through exterior observers: \b[no static timelike observer exists on or inside the horizon].
\Image[/past-exam-of-the-relativity-course-of-the-university-of-cambridge/2016/paper-1-schwarzschild-speeds.png]
{title=Original comparison of local static-observer and Schwarzschild-coordinate radial speeds, for illustrative asymptotic speeds}
Back to article page
