What Lies Beyond a Black Hole: Event Horizons, Spaghettification, and Evaporation

0. Key points

• A black hole is a region whose escape velocity exceeds the speed of light, and its boundary is the event horizon. For a body of mass $M$ the boundary is the sphere of radius $r_s = 2GM/c^2$.
• The size of the horizon is proportional to the mass. Compress the Sun into a black hole and the radius is 3 km; the Earth gives 8.9 mm; a 60 kg person gives one ten-billionth of a proton.
• The event horizon is not a wall. For a sufficiently large black hole, someone falling in crosses it without feeling a thing. What kills you is not the strength of gravity but the difference in gravity — the tidal force.
• Black holes are not perfectly black. Hawking radiation gives them a temperature $T \propto 1/M$, and they evaporate in a time proportional to $M^3$. For a solar mass that is $10^{67}$ years, which is effectively forever.
• Wormholes really do exist as solutions of the Einstein equations. But one built out of ordinary matter pinches off faster than light can cross it.

1. Motivation — the “star from which light cannot escape” was an eighteenth-century idea

Black holes are usually thought of as a twentieth-century product of Einstein’s general relativity. Yet the germ of the idea was already there in the age of Newtonian mechanics.

In 1783 the English clergyman and natural philosopher John Michell wrote the following in the Philosophical Transactions of the Royal Society. If a star is heavy enough and small enough, light emitted from its surface will be pulled back and returned. Such a “dark star” would be permanently invisible to us. Thirteen years later Laplace, in France, performed the same calculation independently and reached the same conclusion in his book.

Their argument is exactly the argument about firing a cannonball straight up. Fire it slowly and it falls back. Fire it fast enough and it leaves for good. The speed at the boundary between the two is called the escape velocity; for the Earth it is 11.2 km per second. So, they asked, if a star were heavy and small enough that its escape velocity exceeded the speed of light $c$, could light escape at all? Naive, but a sharp question.

Seen from today, though, the reasoning has a hole in it. In Newtonian mechanics light is a particle whose speed can change, so it would rise while slowing down, turn around somewhere, and fall back. In other words, light leaving the surface of Michell’s dark star should linger for a while in the space above the star, and you would see it if you got close enough. A real black hole does not behave like that. Under the relativistic premise that the speed of light is $c$ for every observer, light cannot slow down at all — and yet it still cannot get out. That situation is far stranger.

The precise description came in 1916, from Karl Schwarzschild. A year after Einstein published general relativity, Schwarzschild — then serving on the Eastern Front of the First World War — solved exactly for the shape of spacetime around a spherically symmetric body. He died of illness a few months later, but the radius $r_s = 2GM/c^2$ that appeared in his solution still bears his name. The label “black hole” only became standard half a century after that, in the 1960s.

What is striking is that the Newtonian calculation of Michell and Laplace gives numerically the right answer. In this article we first follow that calculation, then ask why it is right and how much of the agreement is a coincidence. From there we take up three questions: what happens if you fall in, whether black holes last forever, and whether you can come out the other side.

2. Preliminaries — escape velocity

Let us first fix the piece of high-school physics we start from.

Definition 2.1（Escape velocity）

Let there be a spherically symmetric body of mass $M$ and radius $R$. The escape velocity $v_{\text{esc}}$ is the smallest launch speed for which an object of mass $m$, fired straight up from the surface, reaches arbitrarily large distances without being pulled back. Air resistance and the gravity of every other body are neglected.

Proposition 2.2（Formula for the escape velocity）

In the setting of Definition 2.1,

$$v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$$

holds, where $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ is the gravitational constant. This value does not depend on the mass $m$ of the launched object.

Proof（Proposition 2.2）

The gravitational potential energy of an object of mass $m$ at distance $r$ from the centre is $U(r) = -GMm/r$. The sign is negative because we take infinity as the reference point.

Mechanical energy is conserved between the moment of launch ($r = R$, speed $v$) and the moment the object is infinitely far away ($r \to \infty$, speed $v_\infty$), so

$$\frac{1}{2}mv^2 - \frac{GMm}{R} = \frac{1}{2}mv_\infty^2 - 0 .$$

The right-hand side is non-negative since $v_\infty^2 \ge 0$. Hence the condition for the object to reach infinity is

$$\frac{1}{2}mv^2 - \frac{GMm}{R} \ge 0 , \qquad\text{that is}\qquad v^2 \ge \frac{2GM}{R} .$$

Equality — “arriving at infinity with exactly zero speed” — gives the minimum, so $v_{\text{esc}} = \sqrt{2GM/R}$. The common factor $m$ cancelled on both sides, so the value is the same whether we launch a feather or an elephant.

∎

Example 2.3（Escape velocities of the Earth and the Sun）

For the Earth, $M = 5.97 \times 10^{24}\ \mathrm{kg}$ and $R = 6.37 \times 10^{6}\ \mathrm{m}$. Substituting into Proposition 2.2,

$$v_{\text{esc}} = \sqrt{\frac{2 \times 6.674\times10^{-11} \times 5.97\times10^{24}}{6.37\times10^{6}}} = \sqrt{1.25\times10^{8}} = 1.12\times10^{4}\ \mathrm{m/s}$$

that is, 11.2 km per second. This is the “second cosmic velocity” familiar from rocket launches.

For the Sun, $M = 1.99 \times 10^{30}\ \mathrm{kg}$ and $R = 6.96 \times 10^{8}\ \mathrm{m}$, so

$$v_{\text{esc}} = \sqrt{\frac{2 \times 6.674\times10^{-11} \times 1.99\times10^{30}}{6.96\times10^{8}}} = \sqrt{3.81\times10^{11}} = 6.18\times10^{5}\ \mathrm{m/s}$$

or 618 km per second. That is about $0.2\ \%$ of the speed of light, so light leaving the Sun’s surface escapes with room to spare.

3. The event horizon and the Schwarzschild radius

Proposition 3.1（Schwarzschild radius）

For a spherically symmetric body of mass $M$, the radius at which the escape velocity is exactly equal to the speed of light $c = 2.998\times10^{8}\ \mathrm{m/s}$ is

$$r_s = \frac{2GM}{c^2} .$$

This is called the Schwarzschild radius. When the actual radius of the body is smaller than $r_s$, the body is a black hole.

Proof（Proposition 3.1）

Set $v_{\text{esc}} = c$ in $v_{\text{esc}} = \sqrt{2GM/R}$ from Proposition 2.2.

$$c = \sqrt{\frac{2GM}{R}} \ \Longrightarrow\ c^2 = \frac{2GM}{R} \ \Longrightarrow\ R = \frac{2GM}{c^2} .$$

We write this $R$ as $r_s$. Since $v_{\text{esc}} = \sqrt{2GM/R}$ grows as $R$ shrinks, a body whose radius drops below $r_s$ has a surface escape velocity exceeding $c$.

∎

Remark 3.2

All we used was eighteenth-century Newtonian mechanics. Nevertheless, the horizon radius that Schwarzschild derived from general relativity is exactly $2GM/c^2$, coefficient included.

This is a happy coincidence, not a proof. In the Newtonian computation, light is a particle that rises while slowing down and eventually turns around, so there ought to be a region just outside the horizon where light hovers. In relativity, light always travels at $c$ and still cannot make outward progress. The formulas agree; their meanings do not.

It is worth knowing $r_s$ by heart, since it speeds up estimates. For a solar mass $M_\odot = 1.99\times10^{30}\ \mathrm{kg}$ we get $r_s = 2.95\ \mathrm{km}$, so roughly

$$r_s \approx 3.0\ \mathrm{km} \times \frac{M}{M_\odot} ,$$

and most objects can then be handled in your head.

Definition 3.3（Event horizon）

The event horizon of a black hole is the boundary surface such that information about events occurring inside it never again reaches the outside. For a spherically symmetric black hole with no rotation and no charge, it is the sphere of radius $r_s = 2GM/c^2$ about the centre.

The horizon is not a surface made of matter; it is the place where the character of spacetime changes. There is no wall, no membrane, no signpost there.

The metaphor of a “horizon” is well chosen. Whatever happens beyond the Earth’s horizon is invisible to you; yet as you walk, the horizon retreats, and you never collide with the horizon itself. A black hole’s horizon is the same: at the instant you cross it there is no marker telling you “that was it”. The one difference is that once across, you can never come back.

As one approaches a black hole, the directions in which light can travel tip inward. At the event horizon, even light emitted outward barely manages to stay where it is.

Example 3.4（Schwarzschild radii of the Sun, the Earth, and you）

We put each mass into Proposition 3.1.

Object	Mass $M$	$r_s = 2GM/c^2$	Actual radius
Sun	$1.99\times10^{30}\ \mathrm{kg}$	$2.95\ \mathrm{km}$	$6.96\times10^{5}\ \mathrm{km}$
Earth	$5.97\times10^{24}\ \mathrm{kg}$	$8.9\ \mathrm{mm}$	$6.37\times10^{3}\ \mathrm{km}$
A 60 kg person	$60\ \mathrm{kg}$	$8.9\times10^{-26}\ \mathrm{m}$	about $0.2\ \mathrm{m}$
Sagittarius A*	$4.3\times10^{6}\,M_\odot$	$1.3\times10^{7}\ \mathrm{km}$	—
M87*	$6.5\times10^{9}\,M_\odot$	$1.9\times10^{10}\ \mathrm{km}$	—

Let us verify the row for a person.

$$r_s = \frac{2 \times 6.674\times10^{-11} \times 60}{(2.998\times10^{8})^2} = \frac{8.01\times10^{-9}}{8.99\times10^{16}} = 8.9\times10^{-26}\ \mathrm{m} .$$

A proton is about $1.7\times10^{-15}\ \mathrm{m}$ across, so this $r_s$ is smaller than a ten-billionth of a proton. Turning you into a black hole would require squeezing you that thoroughly. It is not an everyday concern.

Supermassive black holes at galactic centres are in another regime entirely. The value $r_s = 1.9\times10^{10}\ \mathrm{km}$ for M87* is $130$ astronomical units, more than four times the orbital radius of Neptune ($30$ astronomical units). The whole solar system would fit inside the event horizon.

Let us now clear up a point that many people get wrong. A black hole is not “infinitely dense”.

Proposition 3.5（Mean density of a black hole）

For a black hole of mass $M$, define the volume inside the event horizon as $V = \frac{4}{3}\pi r_s^3$ and the mean density as $\rho = M/V$. Then

$$\rho = \frac{3c^6}{32\pi G^3 M^2} .$$

That is, the mean density is inversely proportional to the square of the mass: heavier black holes are more tenuous.

Proof（Proposition 3.5）

By Proposition 3.1 we have $r_s = 2GM/c^2$, so

$$r_s^3 = \frac{8G^3M^3}{c^6}, \qquad V = \frac{4}{3}\pi r_s^3 = \frac{32\pi G^3 M^3}{3c^6} .$$

Therefore

$$\rho = \frac{M}{V} = M \cdot \frac{3c^6}{32\pi G^3 M^3} = \frac{3c^6}{32\pi G^3 M^2} .$$

With $M^2$ in the denominator, increasing $M$ tenfold divides $\rho$ by a hundred. The reason is simple: since $r_s$ is proportional to $M$, the volume grows like $M^3$, while the mass grows only like $M$, so the ratio is diluted by a factor $M^2$.

∎

Example 3.6（M87* is lighter than air）

Let us put numbers into Proposition 3.5. First for a solar mass, where $r_s = 2953\ \mathrm{m}$:

$$V = \frac{4}{3}\pi (2953)^3 = 1.08\times10^{11}\ \mathrm{m^3}, \qquad \rho = \frac{1.99\times10^{30}}{1.08\times10^{11}} = 1.8\times10^{19}\ \mathrm{kg/m^3} .$$

Nuclear density is $2.3\times10^{17}\ \mathrm{kg/m^3}$, so this is eighty times denser. That really is an outrageous density.

But by Proposition 3.5, multiplying the mass by $N$ relative to $M_\odot$ divides the density by $N^2$. For M87*, $N = 6.5\times10^{9}$, so

$$\rho = \frac{1.8\times10^{19}}{(6.5\times10^{9})^2} = \frac{1.8\times10^{19}}{4.2\times10^{19}} = 0.44\ \mathrm{kg/m^3} .$$

Air at ground level has density $1.2\ \mathrm{kg/m^3}$. The mean density of M87* is about a third that of air. If you could float near the horizon of M87* without a spacesuit, the space around you would be an unremarkable near-vacuum. Thinking of a black hole as a crushed lump of matter is wrong. A black hole is not matter at all; it is the property of spacetime that nothing gets out.

Tip

The relation $\rho \propto 1/M^2$ has one further implication. If you compute the mass of a black hole whose density equals the mean density of the universe (about $10^{-26}\ \mathrm{kg/m^3}$, of the order of the critical density（Definition 5.5）［Dark Matter and Dark Energy］), you get roughly the mass of the observable universe. This is where the suggestion that “the whole universe might be inside a giant black hole” comes from, but an expanding universe and a black hole have different spacetime geometries, so this single coincidence settles nothing. For the size and age of the universe, see The Edge and Age of the Universe, in particular the definition of the particle horizon（Definition 4.1）［The Edge and the Age of the Universe］.

4. What happens if you fall in — tidal forces and spaghettification

Now to the most popular question of all. What happens if you fall into a black hole?

The danger is not the strength of gravity itself. While you are in free fall through strong gravity, you are weightless. By the same logic that lets astronauts float on the International Space Station, you, the air around you, and the shoes on your feet are all falling identically, so there is no sensation of being crushed.

The trouble is that your feet and your head do not fall in quite the same way. Your feet are slightly closer to the centre, so they are pulled slightly harder. That difference stretches you lengthwise. Since it is the same mechanism that produces the ocean tides, it is called the tidal force.

Proposition 4.1（Tidal acceleration）

Suppose a person of height $h$ is in free fall, feet pointed towards the centre, at distance $r$ from the centre of a body of mass $M$. When $h \ll r$, the difference $\Delta g$ between the gravitational acceleration at the feet and at the head is

$$\Delta g \simeq \frac{2GMh}{r^3} .$$

In particular, at the event horizon ($r = r_s = 2GM/c^2$),

$$\Delta g \simeq \frac{h\,c^6}{4G^2M^2} ,$$

which is inversely proportional to the square of the mass.

Proof（Proposition 4.1）

The gravitational acceleration at distance $r$ from the centre is $g(r) = GM/r^2$. The feet are at $r$ and the head at $r + h$, so

$$\Delta g = g(r) - g(r+h) = \frac{GM}{r^2} - \frac{GM}{(r+h)^2} .$$

Factor $r$ out of the second term. Setting $x = h/r$,

$$\frac{GM}{(r+h)^2} = \frac{GM}{r^2}\cdot\frac{1}{(1+x)^2} .$$

Here $x$ is small (a person’s height is far shorter than the fall distance), so we approximate by dropping terms of order $x^2$ and higher. First, $(1+x)^2 = 1 + 2x + x^2 \simeq 1 + 2x$. Next, $(1+2x)(1-2x) = 1 - 4x^2 \simeq 1$, so $\dfrac{1}{1+2x} \simeq 1 - 2x$. Together,

$$\frac{1}{(1+x)^2} \simeq 1 - 2x = 1 - \frac{2h}{r} .$$

Therefore

$$\Delta g \simeq \frac{GM}{r^2} - \frac{GM}{r^2}\left(1 - \frac{2h}{r}\right) = \frac{GM}{r^2}\cdot\frac{2h}{r} = \frac{2GMh}{r^3} .$$

For the second statement we simply substitute $r = r_s = 2GM/c^2$. By Proposition 3.1, $r_s^3 = 8G^3M^3/c^6$, so

$$\Delta g \simeq \frac{2GMh}{r_s^3} = 2GMh \cdot \frac{c^6}{8G^3M^3} = \frac{h\,c^6}{4G^2M^2} .$$

Again $M^2$ appears in the denominator — the same structure as in Proposition 3.5. The larger $M$ is, the further away the horizon lies, and the gentler the gradient of gravity there.

∎

Example 4.2（When does spaghettification begin?）

Using Proposition 4.1, we compute the tidal acceleration felt by a person of height $h = 1.8\ \mathrm{m}$ at the instant of crossing the event horizon.

A solar-mass black hole ($M = 1.99\times10^{30}\ \mathrm{kg}$, $r_s = 3.0\ \mathrm{km}$):

$$\Delta g \simeq \frac{1.8 \times (2.998\times10^{8})^6}{4 \times (6.674\times10^{-11})^2 \times (1.99\times10^{30})^2} = \frac{1.31\times10^{51}}{7.05\times10^{40}} = 1.9\times10^{10}\ \mathrm{m/s^2} .$$

Dividing by $g = 9.8\ \mathrm{m/s^2}$ gives about $2\times10^{9}\,g$ — a difference of two billion $g$ between head and feet. The human body tolerates at most a few tens of $g$, so you would be drawn out long and thin far before reaching the horizon. Physicists call this, with a straight face, spaghettification.

M87* ($M = 6.5\times10^{9}\,M_\odot$): since $\Delta g \propto 1/M^2$ in Proposition 4.1, we divide the value above by $(6.5\times10^{9})^2 = 4.2\times10^{19}$:

$$\Delta g \simeq \frac{1.9\times10^{10}}{4.2\times10^{19}} = 4.5\times10^{-10}\ \mathrm{m/s^2} .$$

That is $10^{-11}\,g$. There is no way to feel it. The event horizon of M87* can be crossed with no warning at all, in complete calm.

Where is the dividing line? Solving for the mass at which $\Delta g$ equals $100\,g \simeq 10^{3}\ \mathrm{m/s^2}$,

$$M^2 = \frac{1.8 \times (2.998\times10^{8})^6}{4 \times (6.674\times10^{-11})^2 \times 10^{3}} = 7.3\times10^{67}, \qquad M = 8.6\times10^{33}\ \mathrm{kg} \simeq 4\times10^{3}\,M_\odot .$$

So for a black hole heavier than a few thousand solar masses, your body survives the moment of horizon crossing. Of course, as you then approach the centre, $\Delta g \propto 1/r^3$ makes the tidal force climb steeply — you have only been granted a reprieve.

Remark 4.3

How much time is left after crossing the horizon? According to general relativity, inside the event horizon there is an upper bound, no matter how you move, on the proper time (the time on your own wristwatch) before you reach the central singularity. The maximum is

$$\tau_{\max} = \frac{\pi GM}{c^3} .$$

Since $GM_\odot/c^3 = 4.93\times10^{-6}\ \mathrm{s}$, we get $\tau_{\max} = 1.55\times10^{-5}\ \mathrm{s} \times (M/M_\odot)$. In numbers:

Black hole	Mass	$\tau_{\max}$
Stellar mass	$1\,M_\odot$	$1.5\times10^{-5}$ s
Sagittarius A*	$4.3\times10^{6}\,M_\odot$	about 67 s
M87*	$6.5\times10^{9}\,M_\odot$	about 28 hours

At M87* you would have more than a full day after crossing the horizon. Struggling is counterproductive: the harder you fire your engines to hold your ground, the shorter your proper time becomes. Falling freely, doing nothing, is the way to live longest.

Remark 4.4

Someone watching from outside sees an entirely different scene. Light emitted by the infalling person is redshifted more and more strongly as the horizon is approached, and the intervals between arrivals stretch out. The frequency received by a distant observer is $\sqrt{1 - r_s/r}$ times the original, which tends to $0$ as $r \to r_s$.

It is often said that from outside, the falling observer appears frozen forever at the horizon, but that is not accurate. The light fades away exponentially fast while growing dim and red. The time constant of the fading is of order $r_s/c$, which for a solar mass is on the order of $10^{-5}$ seconds. So in reality nothing “freezes”; it simply blinks out. The phrase is a leftover from the era when Soviet physicists called these objects “frozen stars”.

About the very same event, the person falling in says “I passed through with nothing happening”, and the outside observer says “they vanished before reaching the horizon”. Neither is lying. Time simply runs differently for the two of them.

5. Black holes evaporate — Hawking radiation

So far a black hole has been perfectly black: it emits nothing and only swallows. In the 1970s, however, a serious theoretical difficulty was found in exactly that picture.

The second law of thermodynamics says that entropy (disorder) never decreases. So what happens if you throw a hot cup of coffee into a black hole? The coffee’s entropy disappears from the universe. The entropy of the universe outside the black hole has decreased, and the second law is violated.

In 1972 Jacob Bekenstein, then a graduate student, argued that the only way to avoid this contradiction is for the black hole itself to carry entropy — and moreover that this entropy is proportional to the area of the event horizon. But if it has entropy it must have a temperature, and anything with a temperature must radiate heat. A supposedly black object would glow. Many researchers initially thought this was absurd.

In 1974 Stephen Hawking redid the calculation and concluded that black holes really do glow. In quantum mechanics the vacuum is not empty: pairs of particles and antiparticles are constantly appearing and vanishing. When this happens just outside the event horizon, one member of a pair can fall in while the other escapes. From outside it looks as though the black hole emitted a particle, and the black hole’s mass decreases by the corresponding amount.

Caution

This picture of “one half of a created pair escaping” is the metaphor Hawking himself used for general audiences, but the actual calculation is subtler. What matters is the property of quantum field theory that particle number depends on the observer’s state of motion; the radiation does not originate at a single point on the horizon but seeps out of a broad region surrounding it. The metaphor is a convenient way in, but it is best not to demand more of it than that.

Definition 5.1（Hawking temperature）

A Schwarzschild black hole of mass $M$ emits thermal radiation with exactly the spectrum of a black body（Definition 2.3）［Was There Really a Big Bang? Expansion, Background Radiation, and the Ratio of the Elements］ at absolute temperature

$$T_H = \frac{\hbar c^3}{8\pi G M k_B} ,$$

where $\hbar = 1.055\times10^{-34}\ \mathrm{J\,s}$ is the reduced Planck constant and $k_B = 1.381\times10^{-23}\ \mathrm{J/K}$ is the Boltzmann constant. This radiation is called Hawking radiation.

Substituting numbers gives $T_H = 1.23\times10^{23}\ \mathrm{K} / (M/\mathrm{kg})$: the temperature is inversely proportional to the mass.

Example 5.2（A solar-mass black hole is the coldest thing in the universe）

Put $M = 1.99\times10^{30}\ \mathrm{kg}$ into Definition 5.1:

$$T_H = \frac{1.23\times10^{23}}{1.99\times10^{30}} = 6.2\times10^{-8}\ \mathrm{K} .$$

Six hundredths of a millionth of a kelvin. Meanwhile the universe is filled with the cosmic microwave background at $2.725\ \mathrm{K}$ (see Example 4.1［Was There Really a Big Bang? Expansion, Background Radiation, and the Ratio of the Elements］ in Was There Really a Big Bang?). Taking the ratio,

$$\frac{2.725}{6.2\times10^{-8}} = 4.4\times10^{7} .$$

The cosmic background is 44 million times hotter. A stellar-mass black hole therefore absorbs far more energy from the background than it loses to radiation. In the present universe, a solar-mass black hole is not evaporating; it is growing fat.

Evaporation beats absorption when $T_H > 2.725\ \mathrm{K}$, that is, when

$$M < \frac{1.23\times10^{23}}{2.725} = 4.5\times10^{22}\ \mathrm{kg} .$$

The Moon has mass $7.35\times10^{22}\ \mathrm{kg}$, so a black hole somewhat lighter than the Moon is losing weight on net at this very moment. As the universe expands and the background cools further, this threshold mass will rise.

Proposition 5.3（Evaporation time is proportional to the cube of the mass）

For a black hole that absorbs nothing from its surroundings and loses mass only through Hawking radiation, the time $t_{\text{ev}}$ for its mass to go from $M_0$ to $0$ satisfies

$$t_{\text{ev}} \propto M_0^{\,3} .$$

Including the constant of proportionality, $t_{\text{ev}} = \dfrac{5120\,\pi G^2 M_0^{\,3}}{\hbar c^4}$.

Proof（Proposition 5.3）

Let us first estimate the radiated power. The black hole of Definition 5.1 is a black body, so by the Stefan–Boltzmann law the energy $P$ emitted per unit time is proportional to the product of the surface area $A$ and the fourth power of the temperature.

• The area is $A = 4\pi r_s^2$, and $r_s \propto M$ by Proposition 3.1, so $A \propto M^2$.
• The temperature satisfies $T_H \propto 1/M$ by Definition 5.1, so $T_H^4 \propto 1/M^4$.

Therefore

$$P \propto A\,T_H^4 \propto M^2 \cdot \frac{1}{M^4} = \frac{1}{M^2} .$$

The mass decreases by the amount of energy emitted ($E = Mc^2$). Writing the rate of mass loss per second as $\dfrac{dM}{dt}$, there is a positive constant $k$ with

$$\frac{dM}{dt} = -\frac{P}{c^2} = -\frac{k}{M^2} .$$

Multiplying both sides by $M^2$ gives $M^2 \dfrac{dM}{dt} = -k$. Now consider the rate of change of $M^3$; by the chain rule,

$$\frac{d(M^3)}{dt} = 3M^2 \frac{dM}{dt} = -3k .$$

The right-hand side is constant, so $M^3$ decreases linearly with time at a fixed rate. Taking $M = M_0$ at $t = 0$,

$$M(t)^3 = M_0^{\,3} - 3kt .$$

Evaporation is complete when this reaches $0$, so

$$t_{\text{ev}} = \frac{M_0^{\,3}}{3k} \propto M_0^{\,3} .$$

Writing the constant $k$ out explicitly in terms of the Stefan–Boltzmann constant and Definition 5.1 gives the formula in the statement.

∎

Example 5.4（The lifetime of a solar-mass black hole）

Substitute $M_0 = 1.99\times10^{30}\ \mathrm{kg}$ into the formula of Proposition 5.3:

$$t_{\text{ev}} = \frac{5120\pi \times (6.674\times10^{-11})^2 \times (1.99\times10^{30})^3}{1.055\times10^{-34} \times (2.998\times10^{8})^4} = \frac{5.64\times10^{74}}{0.852} = 6.6\times10^{74}\ \mathrm{s} .$$

In years that is $2.1\times10^{67}$. Compared with the age of the universe, $1.38\times10^{10}$ years (Example 3.7［The Edge and the Age of the Universe］), it is $10^{57}$ times longer — effectively eternal.

And then $t_{\text{ev}} \propto M_0^3$ from Proposition 5.3 takes over. M87* has $6.5\times10^{9}$ times the mass of the Sun, so its lifetime is the cube of that, $2.7\times10^{29}$ times longer: $6\times10^{96}$ years.

Light black holes, on the other hand, go quickly. A black hole of mass $1.7\times10^{11}\ \mathrm{kg}$ (about the mass of a $400\ \mathrm{m}$ cube of rock) has lifetime

$$t_{\text{ev}} = 2.1\times10^{67} \times \left(\frac{1.7\times10^{11}}{1.99\times10^{30}}\right)^3 = 2.1\times10^{67} \times 6.2\times10^{-58} = 1.3\times10^{10}\ \text{yr}$$

which is just about the age of the universe. If black holes of this mass had been made immediately after the Big Bang, they would be reaching their final evaporation at this very moment. Incidentally their $r_s$ is $2.5\times10^{-16}\ \mathrm{m}$, smaller than a proton: the mass of a mountain packed into a region smaller than a proton. Such primordial black holes are theoretically possible and are being searched for, but no definite detection has been made.

flowchart LR
A["Mass decreases"] --> B["Temperature rises"]
B --> C["Radiation strengthens"]
C --> A
C --> D["Explosive evaporation at the end"]

Evaporation accelerates. As the mass drops the temperature rises, and as the temperature rises the radiation strengthens — positive feedback.

Remark 5.5

Hawking radiation has an awkward side effect. Its spectrum is purely thermal, determined by the temperature alone, and carries no information whatsoever about what was swallowed. Then, after the black hole has evaporated completely, where did the contents of the book that fell in go? Quantum mechanics demands that information is never lost, so this is a clash between two theories. It is called the black hole information paradox.

Hawking himself initially took the side that information really is lost, and in 1997 he made a bet on it, conceding defeat in 2004. Recent quantum-gravity calculations (the reproduction of the so-called Page curve) support the side on which information comes back out, encoded in the radiation. Exactly how it comes out is not settled. This is the frontier of contemporary physics.

6. Wormholes — is there a way through?

Science fiction loves black holes because there might be an exit on the far side. That hope in fact has a respectable theoretical basis.

In 1935 Einstein and Nathan Rosen noticed that if the Schwarzschild solution is extended mathematically as far as it will go, a structure appears connecting two distant regions through a narrow “throat”. This is the Einstein–Rosen bridge, better known as a wormhole. As a solution of the equations, it certainly exists.

The question is whether you can get through it.

Theorem 6.1（The Schwarzschild wormhole is not traversable）

In the maximally extended spacetime of the vacuum Schwarzschild solution of the Einstein equations (the Kruskal extension), there is an Einstein–Rosen bridge joining two exterior regions. However, this throat forms, expands, contracts, and pinches off in time, and its lifetime is shorter than the time a light signal needs to cross it. Consequently no object — not even light — can travel from one exterior region through the bridge to the other.

Remark 6.2

The proof of Theorem 6.1 consists of drawing the entire spacetime in Kruskal coordinates and checking that light rays (straight lines at $45$ degrees) cannot connect the two exterior regions. It was given by Robert Fuller and John Wheeler in their 1962 paper “Causality and Multiply Connected Space-Time” (Physical Review 128, 919). The details are in Chapter 31 of Misner, Thorne and Wheeler’s Gravitation, listed in the references.

Intuitively: the throat collapses faster than light. The tunnel is there, but it closes before you can run through it.

Could one build a wormhole that does not close? In 1988, Michael Morris and Kip Thorne took the problem head on, prompted by a consultation about the premise of Carl Sagan’s novel Contact. Their conclusion was this.

To hold the throat open, matter with negative energy density is required. The reason can be seen from the behaviour of light. A light ray passing through the wormhole must converge at the entrance and diverge at the exit. But the gravity of ordinary matter can only act to make light rays converge. To turn convergence into divergence somewhere along the way, gravity must act repulsively — and that is what “negative energy density” means.

Does such matter exist? Not entirely absent. In quantum theory, bringing two metal plates extremely close makes the vacuum energy density between them negative (the Casimir effect, predicted in 1948 and measured in the 1990s). But the magnitude is unbelievably small. Supporting a wormhole large enough for a person to pass through is estimated to require an amount of negative energy comparable to the mass of the Earth or the Sun. Collecting that with the Casimir effect is like filling the ocean with a single glass of water.

Remark 6.3

Since 2017, Gao, Jafferis, Wall and others have proposed theoretical models of genuinely traversable wormholes built using quantum effects. But the transit time through such wormholes is always longer than the time to travel through the ordinary space outside. In other words, they are never shortcuts. Causality is preserved.

Even if wormholes exist, current theory rather strongly suggests that they will not serve as warp drives. Disappointing as a science-fiction premise, but healthy as physics.

White holes — objects into which nothing can enter and out of which things only emerge — also exist as solutions of the equations. But a white hole is a black hole with time reversed. For the same reason that we never see a broken cup reassemble itself, it requires initial conditions that violate the second law of thermodynamics, and it is not thought to form naturally.

7. Do they really exist — what the observations say

We have been discussing theory, but over the past decade the evidence that black holes exist has become conclusive.

First, gravitational waves. On 14 September 2015 the American LIGO detectors caught the ripples in spacetime produced when two objects of 36 and 29 solar masses merged into one of 62 solar masses. The missing 3 solar masses were released as gravitational waves in under $0.2$ seconds. Since objects of several tens of solar masses came within a few hundred kilometres of each other before merging, they cannot have been neutron stars or ordinary stars. The 2017 Nobel Prize in Physics was awarded for the work that made this observation possible.

Second, the orbits of stars at the galactic centre. The groups of Reinhard Genzel and Andrea Ghez tracked stars near the centre of the Milky Way for decades and showed that they orbit within a region much smaller than the solar system. Kepler’s laws (Keplerian rotation（Proposition 2.3）［Dark Matter and Dark Energy］) then give a central mass of $4.3$ million solar masses. The 2020 Nobel Prize in Physics was awarded for this work and for Roger Penrose’s singularity theorem.

Third, direct imaging. In April 2019 the Event Horizon Telescope (EHT) released an image of the “shadow” at the centre of the galaxy M87. In 2022 it also imaged Sagittarius A* at the centre of the Milky Way.

Example 7.1（Why is the shadow of M87* 42 microarcseconds across?）

According to general relativity, the radius of a black hole’s “shadow” is not $r_s$ itself. Light passing close by is bent and swallowed, so the shadow is somewhat larger:

$$b_{\text{crit}} = 3\sqrt{3}\,\frac{GM}{c^2} = \frac{3\sqrt{3}}{2}\,r_s \simeq 2.6\,r_s .$$

Using the value $r_s = 1.9\times10^{10}\ \mathrm{km}$ for M87* from Example 3.4, the diameter of the shadow is

$$D = 2 \times 2.6 \times 1.9\times10^{10} = 9.9\times10^{10}\ \mathrm{km} .$$

The distance to M87 is $16.8$ million parsecs, that is $5.2\times10^{20}\ \mathrm{km}$ (the parsec is fixed in Definition 3.1［How We Know the Distance to a Star］; for how such distances are measured, see Proposition 4.2［How We Know the Distance to a Star］ in How Do We Know the Distances to the Stars?). The subtended angle is

$$\theta = \frac{D}{d} = \frac{9.9\times10^{10}}{5.2\times10^{20}} = 1.9\times10^{-10}\ \mathrm{rad} .$$

Since $1$ radian $= 2.06\times10^{11}$ microarcseconds,

$$\theta = 1.9\times10^{-10} \times 2.06\times10^{11} = 40\ \text{microarcseconds} .$$

The ring diameter that EHT actually measured was $42 \pm 3$ microarcseconds. The value predicted from two independent observations — the mass and the distance — agrees with the value read off the image.

To get a feel for how small $40$ microarcseconds is: it is the angle subtended, seen from the Earth, by a doughnut $7.5\ \mathrm{cm}$ across placed on the Moon. Photographing it required operating eight radio telescopes around the world simultaneously and combining them into a single Earth-sized instrument.

Some things remain unknown: whether a singularity really exists at the centre (Penrose’s theorem guarantees it insofar as general relativity holds, but that theory itself breaks down near the singularity), how the information paradox is resolved, and whether primordial black holes were made in the earliest universe. The last point is also discussed in Dark Matter and Dark Energy, as a candidate for Definition 2.2［Dark Matter and Dark Energy］.

8. Exercises

Exercise 8.1Easy

To turn the Earth (mass $5.97\times10^{24}\ \mathrm{kg}$) into a black hole without changing its mass, below what radius, in millimetres, must it be squeezed? Use Proposition 3.1. Also compute the resulting mean density using the reasoning of Proposition 3.5 and compare it with nuclear density, $2.3\times10^{17}\ \mathrm{kg/m^3}$.

Solution

By Proposition 3.1,

$$r_s = \frac{2GM}{c^2} = \frac{2 \times 6.674\times10^{-11} \times 5.97\times10^{24}}{(2.998\times10^{8})^2} = \frac{7.97\times10^{14}}{8.99\times10^{16}} = 8.9\times10^{-3}\ \mathrm{m} .$$

That is $8.9\ \mathrm{mm}$, about the size of a marble.

The mean density is

$$V = \frac{4}{3}\pi (8.9\times10^{-3})^3 = 2.95\times10^{-6}\ \mathrm{m^3}, \qquad \rho = \frac{5.97\times10^{24}}{2.95\times10^{-6}} = 2.0\times10^{30}\ \mathrm{kg/m^3} .$$

This is $10^{13}$ times nuclear density. Just as Proposition 3.5 says with $\rho \propto 1/M^2$, lighter black holes are denser. For a solar mass the value was $1.8\times10^{19}\ \mathrm{kg/m^3}$ by Example 3.6, so as a check, the Earth’s density should be higher by the mass ratio squared, $(1.99\times10^{30}/5.97\times10^{24})^2 = 1.1\times10^{11}$.

Exercise 8.2Standard

Find, in units of the solar mass, the mass of a black hole whose mean density equals that of water, $1.0\times10^{3}\ \mathrm{kg/m^3}$. Does that mass lie between those of Sagittarius A* ($4.3\times10^{6}\,M_\odot$) and M87* ($6.5\times10^{9}\,M_\odot$)?

Solution

As computed in Example 3.6, a solar-mass black hole has mean density $\rho_\odot = 1.8\times10^{19}\ \mathrm{kg/m^3}$. By Proposition 3.5, $\rho \propto 1/M^2$, so for a mass $M = N M_\odot$,

$$\rho = \frac{1.8\times10^{19}}{N^2}\ \mathrm{kg/m^3} .$$

Setting this equal to $1.0\times10^{3}$,

$$N^2 = \frac{1.8\times10^{19}}{1.0\times10^{3}} = 1.8\times10^{16}, \qquad N = 1.3\times10^{8} .$$

About $1.3\times10^{8}$ solar masses. This is larger than the $4.3\times10^{6}$ of Sagittarius A* and smaller than the $6.5\times10^{9}$ of M87*, so it does lie between them. Indeed the mean density of Sagittarius A* is $1.8\times10^{19}/(4.3\times10^{6})^2 = 1.0\times10^{6}\ \mathrm{kg/m^3}$ (a thousand times that of water), while M87* is $0.44\ \mathrm{kg/m^3}$ by Example 3.6 (thinner than air).

Exercise 8.3Standard

For a probe that has crossed the event horizon of Sagittarius A* ($4.3\times10^{6}\,M_\odot$), compute in seconds the maximum remaining proper time $\tau_{\max} = \pi GM/c^3$. Also find the distance light travels in vacuum in that time, and compare it with the Schwarzschild radius of Sagittarius A*.

Solution

As in Remark 4.3, $GM_\odot/c^3 = 4.93\times10^{-6}\ \mathrm{s}$. Hence

$$\tau_{\max} = \pi \times 4.93\times10^{-6} \times 4.3\times10^{6} = 3.1416 \times 21.2 = 66.6\ \mathrm{s} .$$

About $67$ seconds — a little over a minute.

The distance light travels in that time is

$$c\,\tau_{\max} = 2.998\times10^{8} \times 66.6 = 2.0\times10^{10}\ \mathrm{m} = 2.0\times10^{7}\ \mathrm{km} .$$

The Schwarzschild radius of Sagittarius A* is $1.3\times10^{7}\ \mathrm{km}$ by Example 3.4, so the ratio is about $1.6$. Since $\tau_{\max} = \pi GM/c^3 = (\pi/2)(r_s/c)$, the ratio is necessarily $\pi/2 = 1.57$. A useful way to remember it: crossing the inside of the horizon takes about the time light needs to travel one horizon radius.

Exercise 8.4Hard

Suppose a black hole formed just after the Big Bang finishes evaporating exactly now, after the age of the universe $1.38\times10^{10}$ years. Find its initial mass. You may use the fact that a solar-mass black hole evaporates in $2.1\times10^{67}$ years (Example 5.4) together with the proportionality of Proposition 5.3. Also compute the Schwarzschild radius of that mass and compare it with the diameter of a proton, $1.7\times10^{-15}\ \mathrm{m}$.

Solution

By Proposition 5.3, $t_{\text{ev}} \propto M_0^3$, so taking the ratio against the solar-mass case,

$$\frac{1.38\times10^{10}}{2.1\times10^{67}} = \left(\frac{M_0}{1.99\times10^{30}}\right)^3 .$$

The left-hand side is $6.57\times10^{-58}$. We take the cube root. Rewriting the exponent as a multiple of 3, $6.57\times10^{-58} = 657\times10^{-60}$, gives

$$(657\times10^{-60})^{1/3} = 657^{1/3} \times 10^{-20} = 8.69\times10^{-20} .$$

Hence

$$M_0 = 1.99\times10^{30} \times 8.69\times10^{-20} = 1.7\times10^{11}\ \mathrm{kg} .$$

About $1.7\times10^{8}$ tonnes, the mass of a boulder some $400\ \mathrm{m}$ on a side.

The Schwarzschild radius, by Proposition 3.1, is

$$r_s = \frac{2 \times 6.674\times10^{-11} \times 1.7\times10^{11}}{8.99\times10^{16}} = \frac{22.7}{8.99\times10^{16}} = 2.5\times10^{-16}\ \mathrm{m} .$$

That is about $1/7$ of the proton diameter $1.7\times10^{-15}\ \mathrm{m}$. We get the strange object of a mountain’s worth of mass crammed into a region smaller than a proton.

Incidentally, computing this object’s temperature from Definition 5.1 gives $T_H = 1.23\times10^{23}/1.7\times10^{11} = 7\times10^{11}\ \mathrm{K}$, so the radiation is gamma rays. The basic idea behind searches of this kind is that the death of a primordial black hole should appear as a gamma-ray burst.

References

• Fumitaka Sato and Tetsuya Hara, Ippan Sotaiseiriron (General Relativity), Iwanami Shoten, 2000 (in Japanese) — the standard derivation of the Schwarzschild solution and the event horizon.
• Kip S. Thorne, Black Holes and Time Warps: Einstein’s Outrageous Legacy, W. W. Norton, 1994 — the history of black hole research and the chapter on wormholes. Written for a general audience, but substantial.
• S. W. Hawking, “Black hole explosions?”, Nature 248 (1974), 30–31. DOI: 10.1038/248030a0 — the original paper on Hawking radiation.
• M. S. Morris and K. S. Thorne, “Wormholes in spacetime and their use for interstellar travel: A tool for teaching general relativity”, American Journal of Physics 56 (1988), 395–412. DOI: 10.1119/1.15620 — traversable wormholes and the need for negative energy.
• C. W. Misner, K. S. Thorne, J. A. Wheeler, Gravitation, Princeton University Press, 2017 (original edition 1973) — Chapter 31 discusses the non-traversability of the Schwarzschild wormhole.
• Event Horizon Telescope Collaboration, “First M87 Event Horizon Telescope Results. I. The Shadow of the Supermassive Black Hole”, The Astrophysical Journal Letters 875 (2019), L1. DOI: 10.3847/2041-8213/ab0ec7 — the measured angular size of the shadow.

Appendix: A look at the Schwarzschild metric

Let us confirm, in formulas, where the discussion in the main text comes from. In general relativity the rule for measuring the “separation” between two points of spacetime — the metric — is gravity itself. Outside a spherically symmetric body of mass $M$ it takes the following form.

$$ds^2 = -\left(1 - \frac{r_s}{r}\right)c^2\,dt^2 + \frac{dr^2}{1 - r_s/r} + r^2\left(d\theta^2 + \sin^2\theta\, d\varphi^2\right), \qquad r_s = \frac{2GM}{c^2} .$$

Here $t$ is the clock of a distant observer, and $r$ is the coordinate fixed by the requirement that the sphere of that radius have area $4\pi r^2$.

Time dilation can be read off the first term. For a clock at rest at radius $r$ ($dr = d\theta = d\varphi = 0$) we have $ds^2 = -(1 - r_s/r)c^2 dt^2$, so the proper time $\tau$ ticked by that clock satisfies

$$d\tau = \sqrt{1 - \frac{r_s}{r}}\; dt .$$

Thus $d\tau$ is smaller than the distant $dt$: where gravity is strong, time runs slowly. The frequency of light emitted there drops by the same factor, and this is the redshift formula used in Remark 4.4. As $r \to r_s$ the factor goes to $0$, and to an outside observer time appears to stop.

At $r = r_s$ the denominator of the second term vanishes, but this is not spacetime breaking down. It is only a bad choice of coordinates; passing to Kruskal coordinates, for instance, makes $r = r_s$ a perfectly smooth, unremarkable surface. Indeed the tidal force there is finite, as Proposition 4.1 shows. What genuinely diverges is $r = 0$, and that divergence survives any change of coordinates.

What happens inside can be seen from the signs. For $r < r_s$ the quantity $1 - r_s/r$ is negative, so the coefficient of $dt^2$ becomes positive and the coefficient of $dr^2$ becomes negative. The roles of $t$ and $r$ are exchanged: $r$ becomes a time-like coordinate. In the same sense in which you cannot stop time, you cannot stop $r$ from decreasing. Moving towards the centre becomes the same thing as moving into the future. This is the most accurate way to state why nothing can leave the inside of the event horizon.

To the question “why can light not get out?”, the Newtonian answer was “because gravity is too strong and pulls it back”. The relativistic answer is entirely different. Nothing gets out because the outward direction no longer exists anywhere in the future.