Falling into a black hole

Outside any spherically symmetric body of mass $M$ — a star, or a black hole — spacetime has the metric found by Karl Schwarzschild:

Here $\theta$ (from 0 at the north pole to $\pi$ at the south pole) and $\phi$ (from 0 to $2\pi$, like longitude) are ordinary polar angles. Some features:

• No mass, no gravity. Set $M = 0$ and this is the Minkowski metric of special relativity in spherical coordinates.
• Only the mass matters. As in Newtonian gravity, the field outside a spherical body depends on its total mass, not on how the mass is spread out.
• A special radius. The factor $1 - 2GM/(rc^2)$ vanishes at the Schwarzschild radius

where the $dt^2$ coefficient becomes 0 and the $dr^2$ coefficient infinite.

A metric component that is zero or infinite is called a singularity, but this one is removable: it is caused only by Schwarzschild's (sensible, simple) choice of coordinates. Other coordinates, best known today the Kruskal coordinates, show nothing singular there. Still, $r_s$ is special: it is the horizon. Anything that falls inside can never get out again (stated, not shown, in the lecture).

For the Sun, $r_s \approx 2.95$ km. But at 2.95 km from the Sun's centre you are deep inside the Sun, where the Schwarzschild metric does not apply, so the Sun has no horizon. Only a body squeezed inside its own $r_s$ has one. Answering questions, Guth adds that a very dense spot inside a star could in principle form a horizon (it would not be stable: the material would fall in). A student points out that our Galaxy has a supermassive black hole at its centre, with matter outside it that keeps falling in, and Guth agrees.

Release an object at rest at radius $r_0$ and let it fall straight in. By symmetry it cannot drift in $\theta$ or $\phi$, so only $r$ and $t$ change. Write $h(r) = 1 - r_s/r = 1 - 2GM/(rc^2)$. Dividing the metric by $d\tau^2$ (with $d\theta = d\phi = 0$) gives one equation relating the two rates:

So once $dr/d\tau$ is known, $dt/d\tau$ follows. For $dr/d\tau$, take the $\mu = r$ component of the geodesic equation. The metric is diagonal, so only $\nu = r$ survives on the left, and on the right only $g_{rr}$ and $g_{tt}$ depend on $r$:

with $g_{rr} = h^{-1}$ and $g_{tt} = -h\,c^2$. Expanding the left side, solving for $d^2r/d\tau^2$ and replacing $dt/d\tau$ by the previous equation is tedious, but a string of cancellations leaves something remarkably simple:

This is Newton's law of gravity, letter for letter. But the symbols mean different things.

• $r$ is not the distance from the centre. For a black hole that distance isn't even defined: getting to $r = 0$ means passing the singularity. What $r$ does measure is circumference: the angular part of the metric is $r^2(d\theta^2 + \sin^2\theta\,d\phi^2)$, as in flat space, so a circle at fixed $r$ is $2\pi r$ around. That is why $r$ is called the circumferential radius.
• $\tau$ is not Newton's universal time. It is the proper time along the geodesic: the time on the falling object's own watch.

Because the form is Newtonian, the 8.01 trick works. The equation says $\frac{d}{d\tau}\left[\frac12\left(\frac{dr}{d\tau}\right)^2 - \frac{GM}{r}\right] = 0$ (conservation of energy), and at the release point the bracket is $-GM/r_0$. Hence

Separating variables and integrating gives the proper time to fall from $r_0$ to $r$ (the inverse, $r(\tau)$, has no closed form):

It is 0 at $r = r_0$ and grows as the object falls. Nothing special happens at $r_s$, and even at $r = 0$ the answer is finite: by its own watch, the object falls through the horizon and reaches the centre in a finite time. There, $r = 0$ is a true singularity. Approaching it, the tidal forces — the difference in pull between the front and back of the object — grow without limit and stretch it out along its motion: it is "spaghettified".

Now describe the same fall in the Schwarzschild coordinate time $t$, the external time variable of the metric (far from the hole it runs at the rate of a distant observer's clock). By the chain rule, $dr/dt = (dr/d\tau)/(dt/d\tau)$, with $dt/d\tau$ from the metric equation above:

Near the horizon $h^{-1} = r/(r - r_s) \approx r_s/(r - r_s)$ blows up, and the $h^{-2}$ term dominates the denominator. Then $dr/d\tau$ cancels, leaving (Guth gives $dr/dt$ only in magnitude; it is negative, since the object falls in)

(The approximation only holds near $r_s$, so the lower limit of the integral is left open: only the behaviour as $r \to r_s$ matters.) As $r$ approaches $r_s$, the logarithm goes to $-\infty$ and $t$ to $+\infty$. Seen from outside, the object never reaches the horizon; it creeps ever closer.

How can the fall be over in 5 ms for the object but never finish for the outside observer? In general relativity time is measured locally: you measure your time, I measure mine, and they need not agree — here, by an infinite amount.

According to classical general relativity, nothing special happens to the object as it crosses the horizon. In 2013 Guth called this "controversial": attempts to include quantum theory had suggested that something dramatic — a firewall — might happen at the horizon. That question was, in his words, not settled. Any such effect would be quantum mechanical, like Hawking's 1974 result that black holes evaporate, which Guth notes would vanish as $\hbar \to 0$. In classical general relativity the black hole described here is stable, and the infalling object passes the horizon without incident.

• The Schwarzschild metric $ds^2 = -h\,c^2dt^2 + h^{-1}dr^2 + r^2(d\theta^2 + \sin^2\theta\,d\phi^2)$, $h = 1 - 2GM/(rc^2)$, describes spacetime outside any spherical mass.
• At $r_s = 2GM/c^2$ (2.95 km for the Sun) the metric has a removable coordinate singularity; for a black hole it is the horizon. $r = 0$ is a true singularity.
• Radial free fall gives $d^2r/d\tau^2 = -GM/r^2$: Newton's form, but with $\tau$ the faller's proper time and $r$ the circumferential radius.
• By its own watch the object reaches the centre in finite time, $\frac{\pi}{2}\sqrt{r_0^3/(2GM)}$; in the outside time $t \approx -(r_s/c)\ln(r - r_s)$ it never reaches the horizon.