# Schwarzschild Black Hole

## Schwarzschild coordinates

The spacetime metric is

$ds^2 = - \left(1 - \frac{2M}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{2M}{r}\right)} + r^2 d\Omega^2$

where $d\Omega^2 \equiv d\theta^2 + \sin^2\theta\, d\phi^2$ is the metric for the unit two-sphere. The horizon is $r = 2M$ and the singularity is $r=0$. The lapse function, shift vector, and extrinsic curvature defined by the $t = {\rm const}$ slices and time flow vector field $\partial_t$:

$\begin{array}{rcl} \alpha & = & \sqrt{1 - \frac{2M}{r} } \\ \beta^a & = & 0 \\ K_{ab} & = & 0 \end{array}$

## Geodesic Motion

The equations of geodesic motion for a test mass in Schwarzschild coordinates with respect to time according to the test mass $\tau$ yield

$\frac{dt}{d\tau} =\frac{\gamma}{\left(1-\frac{2M}{r}\right)}$

where $\gamma$ is a constant of the motion called the energy parameter, the energy per mass for the test mass.

$\frac{d\phi}{d\tau} =\frac{l_z}{r^2 sin^2 \theta}$

where $l_z$ is the conserved angular momentum per mass for the test mass.

$\frac{d}{d\tau}\left(r^2 \frac{d\theta}{d\tau}\right) = r^2 sin\theta cos\theta \left(\frac{d\phi}{d\tau}\right)^2$

and finally

$\frac{\gamma^2 - 1}{2} =\frac{1}{2}\left(\frac{dr}{d\tau}\right)^2 -\frac{M}{r}+\frac{1}{2}\left(\frac{l_z ^2}{r^2 sin^2 \theta}+r^2 \left(\frac{d\theta}{d\tau}\right)^2 \right)\left(1-\frac{2M}{r}\right)$

The final equation of motion looks much like a Newtonian gravitation conservation of energy equation with the exception of the $\left(1-\frac{2M}{r}\right)$ term multiplying the angular part and the time derivatives being with respect to the test mass time. This factor multiplying the angular part perturbs the motion of nearly elliptical orbits so that they process. Orienting the coordinates so that the motion of an orbit is equatorial and defining $u=\frac{1}{r}$ the equations of motion yield

$\frac{d^2 u}{d\phi ^2}-3Mu^2 +u-\frac{M}{l_z ^2}=0$

which in a weak field the solution can be approximated by

$u=\frac{M}{l_z ^2}\left(1+ecos\left(\left(1-3\frac{M^2}{l_z ^2}\right)\phi \right) \right)$

where $e$ is the eccentricity So perihelion occurs at

$\left(1-3\frac{M^2}{l_z ^2}\right)\phi _p = 2n\pi$

which for the weak field can be approximated by

$\phi _p = 2n\pi +\frac{6n\pi M^2}{l_z ^2}$

and given an orbital period of T this implies that after a time t the orbit will have processed by an amount given by

$\Delta \phi _p =\frac{6\pi M^2}{l_z ^2}\frac{t}{T}=\frac{6\pi M}{a\left(1-e^2 \right)}\frac{t}{T}$

where $a$ is the semi-major axis. This effect was first observed in nature for Mercury which processes 575" per 100 earth years, 534" of which are accounted for by the gravitational effects from other planets.

## Geodesics For Light

Writing the geodesic motion in terms of Schwarzschild time t instead of time for the test mass $\tau$ and taking the limit as $\gamma$ and $l_z$ go to $\infty$ yields the motion for a massless test particle such as a photon. Orienting the coordinates so that the orbital plane of the photon is equatorial and defining $u=\frac{1}{r}$ results in

$\frac{d^2 u}{d\phi ^2}-3Mu^2 +u=0$

and

$\frac{dr}{dt}=\pm \left(1-\frac{2M}{r}\right)\sqrt{1-\left(\frac{r_m}{r}\right)^2 \frac{\left(1-\frac{2M}{r}\right)}{\left(1-\frac{2M}{r_m}\right)}}$

where $r_m$ is the distance of closest approach for a deflected photon. For small deflection of light the first of these two yields a deflection angle of

$\Delta \phi = \frac{4M}{r_m}$

And for circular orbit of a photon yields

$r_p = 3M$

which is a location referred to as the photon sphere. Integrating the second with appropriate weak field approximation and writting the result in terms of a lab's time instead of remote observer Schwarzschild time for a photon following geodesics between earth and another planet at superior conjunction yields the round trip Shapiro delay equation of

$\Delta t_l =4M\left(-\frac{1+\delta _1}{2}+\frac{1+\delta _2}{2}ln\left(\frac{4ab}{r_m ^2}\right) \right)$

where $a$ and $b$ are the orbital distance from sun of the planets and the $\delta$s are curve fit parameters allowing for perturbances such as the gravitational time dilation from the earth's mass itself.

## Isotropic coordinates

Under the coordinate transformation

$r=\rho \left(1+\frac{M}{2\rho }\right)^2$

The line element can be expressed in isotropic coordinates as

$ds^2 = -\left( \frac{1 - \frac{M}{2\rho }}{1 + \frac{M}{2\rho }} \right)^2 dt^2 + \left( 1 + \frac{M}{2\rho } \right)^4 ( d\rho ^2 + \rho ^2 d\Omega^2 )$

The horizon is $\rho =M/2$. The lapse function, shift vector, and extrinsic curvature defined by the $t = {\rm const}$ slices and the time flow vector field $\partial_t$:

$\begin{array}{rcl} \alpha & = & \left( \frac{1 - \frac{M}{2\rho }}{1 + \frac{M}{2\rho}} \right)\\ \beta^a & = & 0 \\ K_{ab} & = & 0 \end{array}$

## Kruskal coordinates

Under the transformations

$u^2 - v^2 = \left( \frac{r}{2M} - 1 \right) e^{\frac{r}{2M}}$
$\frac{u}{v} = tanh\left(\frac{t}{4M}\right)$

The line element can be expressed as

$ds^2 = \frac{32M^3}{r} e^{-\frac{r}{2M}} (-dv^2 + du^2) + r^2 d\Omega^2$

The lapse function, shift vector, and extrinsic curvature defined by the $v = {\rm const}$ slices and the time flow vector field $\partial_v$:

$\begin{array}{rcl} \alpha & = & \sqrt{\frac{32M^3}{r} } e^{-\frac{r}{2M}} \\ \beta^u & = & \beta^\theta \ = \ \beta^\phi \ = \ 0 \\ K_{uu} & = & -2v(2M+r) \left( \frac{2M}{r}\right)^{5/2} e^{-\frac{3r}{4M}} \\ K_{\theta\theta} & = & v\sqrt{2Mr} e^{-\frac{r}{4M} } \\ K_{\phi\phi} & = & K_{\theta\theta} \sin^2\theta \\ K_{u\theta} & = & K_{u\phi} \ = \ K_{\theta\phi} \ = \ 0 \end{array}$

## Kerr-Schild coordinates

With the coordinate transformation

$t_{ks}=t + 2Mln|\frac{r}{2M}-1|+k$

the line element can be expressed

$ds^2 = -\left( 1 - \frac{2M}{r} \right) dt_{ks} ^2 + \left(\frac{4M}{r}\right)dt_{ks} \,dr + \left( 1 + \frac{2M}{r}\right) dr^2 + r^2 d\Omega^2$

The lapse function, shift vector, and extrinsic curvature defined by the $t_{ks} = {\rm const}$ slices and the time flow vector field $\partial_{t_{ks}}$:

$\begin{array}{rcl} \alpha & = & \frac{1}{\sqrt{1 + 2M/r} } \\ \beta^r & = & \frac{2M/r}{\left(1 + \frac{2M}{r}\right)} \\ \beta^\theta & = & \beta^\phi \ = \ 0 \\ K_{rr} & = & \frac{-2M(M+r)}{\sqrt{r^5(2M+r)} } \\ K_{\theta\theta} & = & 2M\sqrt{\frac{r}{(2M+r)} } \\ K_{\phi\phi} & = & K_{\theta\theta} \sin^2\theta \\ K_{r\theta} & = & K_{r\phi} \ = \ K_{\theta\phi} \ = \ 0 \end{array}$

Alternatively, the spacetime metric can be written as

$g_{\mu\nu} = \eta_{\mu\nu} + 2H\ell_\mu \ell_\nu$

where $\eta_{\mu\nu} = diag(-1,1,1,1)$ is the Minkowski metric, $\ell_\mu = \left(1,\frac{x}{r},\frac{y}{r},\frac{z}{r}\right)$ is a covector, and $H = M/r$. (The spatial coordinates are $x$, $y$, and $z$, and $r \equiv \sqrt{x^2+y^2+z^2}$ is the coordinate radius.) Also let $\ell^\mu \equiv \eta^{\mu\nu}\ell_\nu = \left(-1,\frac{x}{r},\frac{y}{r},\frac{z}{r}\right)$. Then the inverse metric is $g^{\mu\nu} = \eta^{\mu\nu} - 2H\ell^\mu \ell^\nu$. Note that the vector $\ell^\mu$ is null in both the Minkowski and physical metrics.

## The Vaidya Solution

The Vaidya solution is

$ds^2 = -\left( 1 - \frac{2M\left(r-t\right)}{r} \right) dt^2 - \left(\frac{4M\left(r-t\right)}{r}\right)dt\,dr + \left( 1 + \frac{2M\left(r-t\right)}{r}\right) dr^2 + r^2 d\Omega^2$

which is the exact solution to Einstein's field equations for an outflow of spherically symmetric electromagnetic radiation. Its time reversal would also be an exact solution, but would describe an influx of electromagnetic radiation:

$ds^2 = -\left( 1 - \frac{2M\left(r+t\right)}{r} \right) dt^2 + \left(\frac{4M\left(r+t\right)}{r}\right)dt\,dr + \left( 1 + \frac{2M\left(r+t\right)}{r}\right) dr^2 + r^2 d\Omega^2$

Notice in comparison with the previous section that the coordinates used to Express the Vaidya solution in the latter here are Kerr-Schild coordinates, and for the outflow case are the time reversal of Kerr-Schild coordinatess, so it is plain to see that for M being constant, the Vaidya solution is equivalent to the Schwarzschild black hole.

Using the latter case, describing an electromagnetic radiation inflow, a remote observer making use of the Kerr-Schild coordinates will reckon that it takes a finite time for a shell of radiation from somewhere outside to reach the horizon, but will never actually see the event where this occurs because the Kerr-Schild coordinate speed for radialy moving light for this case is

$\frac{dr}{dt}= - 1,+\frac{1-\frac{2M}{r}}{1+\frac{2M}{r}}$

Using these coordinates one reckons that the infall time is finite, but that it takes and infinite time for information about an event at the horizon to escape.