Strong gravity

Strong gravity spacetime is described by the Schwarzschild and Kerr metrics. We primarily use the Schwarzschild metric because it is simpler.

g_{00} = -\left(1 - \frac{2GM}{r}\right), \quad g_{11} = \left(1-\frac{2GM}{r}\right)^{-1}, \quad g_{22} = r^2, \quad g_{33} = r^2 \sin^2 \theta.

The Schwarzschild line element is therefore

\d s^2 = g_{\mu\nu} \dx^\mu \dx^\nu = -\left(1 - \frac{2GM}{r}\right) \d t^2 + \frac{\d r^2}{1-2GM/r} + r^2 (\d\theta^2 + \sin^2\theta\d\phi^2).

The mass $M$ exists only at the point $r=0$ . In all other spacetime the pressure and energy density is 0.

As $r \to \infty$ , the Schwarzschild metric approaches the Minkowski metric, and we reach special relativity. We will consider how events are measured by an observer in this limit to help with bookkeeping.

Consider an observer at rest at some radius $r$ . Let $\theta=\phi=0$ for simplicity. By equivalence, using the FFF near this observer, we can say that $\fv u \cdot \fv u = -1$ for this observer.

Let $\tau$ be the time measured by the observer at $r$ , and let $t$ be the time measured by the distant observer at $\infty$ . Using the metric, we write down

\begin{align*} \fv u \cdot \fv u &= -\left(1-\frac{2GM}{r}\right)\left(\frac{\dt}{\d\tau}\right)^2 = -1 \\ \frac{\dt}{\d\tau} &= \left(1-\frac{2GM}{r}\right)^{-1/2}. \end{align*}

Recall that in special relativity, $\dt/\d\tau = \gamma$ was used as a factor to relate measurements made by observers with some relative velocity. In the Schwarzschild spacetime we use $\dt/\d\tau$ similarly to relate measurements made by two observers, one of them distant.

This already gives some interesting results. A clock at smaller $r$ ticks more slowly than a clock at larger $r$ . This is similar to our results of gravitational redshift.

Something weird is that clocks appear to stop relative to a distant observer at $r = 2GM$ .

Light

Let us parametrize the path of light with a variable $\lambda$ such that $p = \frac{\dx}{\d\lambda}$ . We can do this by e.g. defining $\lambda$ to be the limit of $\d\tau/m$ as $m \to 0$ (formally we can define this as a sequence similarly to how we define the Dirac $\delta$ ).

Since $p$ still points along the light’s trajectory, $p \cdot p = 0$ .

\begin{align*} \fv p \cdot \fv p = g_{\alpha\beta} \frac{\dx^\alpha}{\d\lambda} \frac{\dx^\beta}{\d\lambda} = -\left(1-\frac{2GM}{r}\right) \left(\frac{\dt}{\d\lambda}\right)^2 + \left(1-\frac{2GM}{r}\right)^{-1} \left(\frac{\d r}{\d\lambda}\right)^2 = 0. \end{align*}

After some algebra we find

\begin{align*} \frac{\d r}{\dt} = \pm c \left(1-\frac{2GM}{r}\right). \end{align*}

To be clear $r$ is the radius coordinate, and $t$ is the time measured by a distant observer. In general relativity, $r$ is no longer the distance between two events occurring at the same time — we need to use the metric to find that. So this result doesn’t contradict the speed of light invariant.

It does mean that a distant observer will never observe light emitted from $r=2GM$ reach them.

Energy

Consider some light emitted from radius $R$ propagating outward. We first consider its Lagrangian

\begin{align*} L &= \frac12 \left[-(1-2GM/r) \dot t^2 + (1-2GM/r)^{-1} \dot r^2 + r^2 \dot\theta ^2 + r^2 \sin^2 \theta \dot\phi^2 \right], \\ 0 &= \frac{\partial L}{\partial x^\alpha} - \frac{\d}{\d\lambda} \frac{\partial L}{\partial \dot x^\alpha}; \\ & \text{where } \dot x^\alpha = \frac{\d x^\alpha}{\d\lambda}. \end{align*}

We see $\partial L/\partial t = 0$ , which tells us that $\partial L/\partial \dot t$ is conserved.

\begin{align*} \frac{\partial L}{\partial \dot t} = -\left(1 - \frac{2GM}{r}\right) \dot t = \text{constant}. \end{align*}

We note that from our definition of $\lambda$ , $\dot t = p^0$ and $-(1-2GM/r) = g_{00}$ . We can then rewrite as

g_{00} p^0 = p_0 = \text{constant}.

We consider the energy measured by a static observer at $r=R$ and one at $r=\infty$ . Recall that the energy of an object with 4-momentum $p$ , as measured by an observer whose 4-velocity is $v$ , is $-p\cdot v$ .

Here we compare the energy measured by a static observer at $r \to \infty$ and at $r = R$ . Recall from above that $\d t/\d\tau = 1/\sqrt{1-2GM/r}$ for an observer at rest at radius $r$ .

\begin{align*} \frac{E_\infty}{E_R} &= \frac{-\fv p \cdot \fv u |_{r\to\infty}}{-\fv p \cdot \fv u|_{r=R}} \\ &= \frac{p_t u^t_\infty}{p_t u^t_R} = \frac{u^t_\infty}{u^t_R} \tag{since observers at rest} \\ \text{Recall that}~~~~~~~u^t &= \frac{\dt}{\d\tau}, \\ \text{and}~\,u^t_{r\to\infty} &= 1 ~~~~~~~\text{ by definition},\\ \frac{E_\infty}{E_r} &= \sqrt{1-\frac{2GM}{r} }. \end{align*}

Falling in

Consider some poor soul falling from rest at $r=R$ toward $r=0$ . The Lagrangian describing their motion is

\begin{align*} L = \frac12 g_{\alpha\beta} u^\alpha u^\beta = \frac12\left[ -(1-2GM/r) \dot t^2 + \frac{\dot r^2}{1-2GM/r} \right]. \end{align*}

Since $\partial L/\partial t = 0$ , we know $\partial L/\partial\dot t$ is a constant (this follows from the Euler equation). We call this quantity $\hat E$ , the energy per mass, because of its asymptotic behavior.

\begin{align*} \hat E = - \frac{\partial L}{\partial \dot t} = \left(1-\frac{2GM}{r}\right) \dot t = \text{constant}. \end{align*}

Since this is a constant, we can evaluate it at a particularly convenient point, when our doomed protagonist is at rest at $r=R$ . We know $\dt/\d\tau$ for an observer at rest, which gives us

\begin{align*} \hat E = \sqrt{1 - \frac{2GM}{R}}. \end{align*}

To find $\dot r$ we will use the invariant $\fv u \cdot \fv u = -1$ .

\begin{align*} -1 &= -(1-2GM/r) \dot t^2 + \frac{\dot r^2}{1-2GM/r} \\ \dot r^2 &= \hat E^2 - (1-2GM/r) \\ \dot r &= - \sqrt{\frac{2GM}{r} - \frac{2GM}{R}}. \end{align*}

After some awful algebra, we find

\begin{align*} \tau(r) = \sqrt{\frac{1}{2GM}} \left[R^{3/2} \arctan \left(\sqrt{\frac{R-r}{r}}\right) + \sqrt{rR(R-r)} \right]. \end{align*}

As seen by a distant observer, we find

\begin{align*} \frac{\d r}{\d t} &= \frac{\d r}{\d \tau} \left(\frac{\dt}{\d\tau}\right)^{-1} = -\sqrt{2GM \left(\frac1r - \frac1R \right)} \frac{1}{\hat E} \left(1-\frac{2GM}{r}\right). \end{align*}

This integral is truly horrible, and I don’t even want to type the result. Parametrizing $r = 2GM + x$ and Taylor expanding to the first order, we find

\begin{align*} t(x) &= 2GM \ln \left[\frac{8GM(R-2GM)}{Rx} + \mathcal C_1 \right] + \mathcal C_2 \\ x(t) &= \frac{8GM(R-2GM)}{R} e^{-(t-\mathcal C_2)/2GM}. \end{align*}

According to the distant observer, our hero assymptotes to $2GM$ as $t \to \infty$ .

Although this appears inconsistent at first glance, we must recall that a distant observer cannot observe any light emitted from $r=2GM$ . The distant observer will never see their friend pass $r=2GM$ , and indeed will never observe their grim fate.