What is a black hole?

What makes a black hole black?

Throw a ball straight up. If it leaves your hand slowly, gravity wins and it comes back. Throw it fast enough and it never returns. The speed where it just escapes is the escape speed, and it depends on how much mass is pulling and how close you start. From the surface of the Earth it is about 11 kilometres a second. Squeeze the same mass into a smaller ball and the escape speed at its surface climbs.

Now keep squeezing. At some size the escape speed reaches the speed of light, and light is as fast as anything gets. Past that point not even light gets out. In 1784 an English clergyman, John Michell, saw this coming: a star heavy enough and compact enough would keep its own light, and we could find it only by its pull on anything that circled it.

General relativity gives a different explanation of this boundary. Around that size sits a boundary called the horizon. It is not a surface. There is no skin, no wall, no glow, and a traveller who crosses it feels nothing special at that instant. Crossing the horizon changes which paths a traveller can take. Outside the horizon you can always choose to go outward, if you spend enough fuel. Inside a non-spinning hole, every possible future path leads toward a smaller radius.

Think of a river speeding up toward a waterfall. Far upstream you can paddle wherever you like. Closer in, you can still make headway upstream, but only if you work for it. Somewhere along the way the current reaches the fastest anyone can paddle. Past that line every stroke, however hard, still carries you downstream toward the falls. The horizon is that line, with light as the fastest paddler there is.

A black hole, then, is the region inside a horizon. The physical state at the centre is unknown. The equations predict unbounded density, which most physicists interpret as a limit of the theory. frontier

How big is Vesper's horizon?

Vesper holds the mass of 12 Suns, and it spins. Its horizon is 30.4 kilometres from the centre, 60.7 kilometres across. At 70 kilometres an hour you could drive across it in 52 minutes: 12 Suns of mass inside a ball you could cross before lunch. A hole of the same mass without spin would have a horizon 35.4 kilometres out; the spin draws the edge in, and the last fold works out why.

Do black holes suck things in?

No. Far from the horizon, Vesper's gravity is the gravity of any other object of 12 Suns. Probes and minds can orbit it for as long as they like, at the speeds they would have around a star of that mass.

Guess first Swap the Sun for a black hole with exactly the Sun's mass. What happens to the Earth's orbit?

Nothing at all. The Earth would keep going around exactly as before, because gravity at that distance depends only on the mass, not on what the mass is made of. It would get very dark and very cold, though.

The strongest relativistic effects occur close to the hole. Inside a few horizon radii the orbits change character, and at the horizon the options for escape run out. The pages on Why the hole looks like that, Clocks near the horizon and Ships die of tides, not of curvature follow what that closeness does to light, clocks and bodies.

How big, how dense, and what defines it?

The size of the horizon comes from one line:

horizon radius = 2 × G × M ÷ c²

Here G is the gravitational constant, M the hole's mass and c the speed of light. Everything but M is fixed, so the radius is simply proportional to the mass: 2.95 kilometres for every Sun of mass. This is the Schwarzschild radius, written r_s after Karl Schwarzschild, who found the solution in 1916. The whole Earth, squeezed until it became a hole, would have a horizon 8.9 millimetres in radius.

A hole with the mass of	Mass	Horizon radius, without spin	Average density
One Sun	1 Sun	2.95 km	1.8 × 10¹⁹ kg/m³
Gaia BH1, the nearest known hole in 2026	9.6 Suns	28.4 km	2.0 × 10¹⁷ kg/m³
Vesper	12 Suns	35.4 km	1.3 × 10¹⁷ kg/m³
Sagittarius A*, the Milky Way's centre	4 × 10⁶ Suns	1.2 × 10⁷ km	1.2 × 10⁶ kg/m³
M87*	6.5 × 10⁹ Suns	128 AU	0.44 kg/m³

Water is 1,000 kg/m³ and the air you breathe is about 1.2. "Average density" means the mass divided by the volume of a ball with the horizon's radius.

Why are big holes not dense?

The radius grows with M. The volume grows as M³. So the average density, mass over volume, falls as M ÷ M³:

average density ∝ 1 ÷ M²

Double the mass and the density drops to a quarter. Gaia BH1's hole, at 9.6 Suns, is 1.6 times denser than Vesper at 12. A hole like Vesper has an average density about half that of an atomic nucleus. Sagittarius A* is about 1,200 times denser than water, and M87* is thinner than air. Nothing is especially squeezed at a horizon either. What hurts a body near a hole is the difference in pull from head to foot, the tide: gentle at the horizon of a giant like M87*, deadly far outside one the size of Vesper. Ships die of tides, not of curvature takes that up.

What three numbers describe it?

Whatever fell in, whether stars, ships or minds, a settled black hole is described by just three numbers: its mass, its spin and its electric charge. The rest is forgotten, which is why physicists say black holes have "no hair". Real holes carry almost no charge, so two do the job. Vesper's are 12 Suns and a spin of 0.70 out of a maximum of 1; Mining a spinning hole shows what that spin is worth.

Where are the landmarks?

Everything near a hole is measured against r_s. For a hole of Vesper's mass without spin, there are three more circles worth knowing:

Photon sphere, 1.5 r_s or 53.2 km: light can follow an unstable circular orbit here.
Capture radius, 2.6 r_s or 92.1 km: a ray of light aimed to pass the centre closer than this is swallowed. It sets the size of the dark disc a distant camera sees.
Innermost stable circular orbit, 3 r_s or 106.3 km: inside it, nothing stays in orbit; it spirals in.

Vesper's spin moves the inner circles. Its horizon comes in to 30.4 km, and for orbits that go round the way the hole turns, the innermost stable orbit comes in to 60.1 km. Matter can get that much closer before it falls, which is why a spinning hole makes the better engine: Feeding a black hole runs the numbers.

For a hole of Vesper's mass without spin, light crosses the horizon radius in 118 millionths of a second. The toy below, which leaves out spin, lets you slide the mass and watch the horizon, the light-crossing time and the density change.

With JavaScript on, this is a calculator: pick a hole or slide the mass and it shows the horizon radius, the time light takes to cross it and the average density. The table above gives the same numbers for five holes.

Open everything

Why is the edge where it is, and why can nothing cross back?The Schwarzschild metric, light cones, effective potentials

From escape speed to general relativity

Set the kinetic energy of a thrown ball equal to the pull it must climb out of and let the speed be the speed of light:

$\tfrac{1}{2}v^{2} = \frac{GM}{r}\quad\text{with } v = c \;\;\Rightarrow\;\; r = \frac{2GM}{c^{2}}$

This reproduces Michell's radius and the general-relativistic result, although Newtonian mechanics does not describe light near a horizon. The relativistic derivation uses the Schwarzschild metric, the exact solution of Einstein's equations for empty space around a non-spinning mass. It says how far apart two nearby events are:

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

Here $t$ is the time of a clock far away, $r$ is the circumference of a circle divided by $2\pi$ , and $d\Omega$ is the angle around. Light travels along paths with $ds=0$ . Straight out or in, that gives

$\frac{dr}{dt} = \pm c\left(1-\frac{r_s}{r}\right)$

As $r$ approaches $r_s$ , the speed of light as seen from far away drops to zero. This is the root of the distant view that things freeze at the horizon, which Clocks near the horizon explores. This coordinate speed describes the distant observer's view. The traveller crosses the horizon in finite proper time.

Light cones tip over

Swap to coordinates that follow a falling ray of light. Define $v = ct + r + r_s\ln\left|r/r_s - 1\right|$ . The metric becomes

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

and nothing odd happens at $r=r_s$ . Eddington and Finkelstein found such coordinates, and Finkelstein's 1958 paper showed the horizon is a one-way membrane. In the time $\tilde t = (v - r)/c$ , in which ingoing light moves at 45 degrees, outgoing light moves at

$\frac{dr}{d\tilde t} = c\,\frac{1-r_s/r}{1+r_s/r}$

This is what the diagram shows: far away it is nearly $+c$ , at $r_s$ it is zero, and inside it is negative. Inside, both edges of the cone, and so every possible future, point inward.

Where light from one event can go. Far out the cone is nearly upright. At the horizon its outward edge points straight up. Inside, both edges lean toward the centre.

Once $r<r_s$ , the radius $r$ acts like time: it can only decrease. Standing still there would mean moving faster than light.

The horizon is not a place where gravity is infinitely strong. A measure of curvature that does not depend on coordinates, the Kretschmann scalar, is

$K = \frac{48\,G^{2}M^{2}}{c^{4}r^{6}} = \frac{12\,r_s^{2}}{r^{6}}$

It is finite at $r_s$ and blows up only at $r=0$ . At the horizon it falls as $1/r_s^{4}$ , so as $1/M^{4}$ : smaller holes are rougher there, larger ones gentler, which is the subject of Ships die of tides, not of curvature.

Density and size, together

With $r_s = 2GM/c^2$ , the volume of a ball is $\tfrac{4}{3}\pi r_s^{3}$ , and the average density is

$\bar\rho = \frac{M}{\tfrac{4}{3}\pi r_s^{3}} = \frac{3c^{6}}{32\pi G^{3}M^{2}}$

The mass appears squared in the denominator. At Vesper's mass it comes to

$\bar\rho_{\text{Vesper}} = 1.3 \times 10^{17}\ \text{kg/m}^{3}$

Orbits and the photon sphere from effective potentials

A particle with mass moves as though in a one-dimensional potential. Work in units where $r$ is measured in $GM/c^{2}$ , and let $L$ be the angular momentum per unit mass. For a mass in orbit around a Schwarzschild hole,

$V(r) = -\frac{1}{r} + \frac{L^{2}}{2r^{2}} - \frac{L^{2}}{r^{3}}$

The first two terms are the Newtonian ones. The last term is the new thing from general relativity, and it is attractive and steepens fast at small $r$ . Circular orbits sit where $V'(r)=0$ :

$L^{2} = \frac{r^{2}}{r-3}$

so circular orbits exist only for $r>3$ , which is $1.5\,r_s$ . Whether the orbit is stable depends on $V''$ , and with that $L^{2}$ ,

$V'' = \frac{r-6}{r^{3}(r-3)}$

which is positive only for $r>6$ . The last stable circle is therefore at $6\,GM/c^{2} = 3\,r_s$ : for Vesper without spin, 106.3 km.

Light has no rest mass, so the $-1/r$ term drops out and only $\tfrac{L^{2}}{2r^{2}}(1-2/r)$ remains. That potential has a peak at $r=3$ (that is $1.5\,r_s$ , 53.2 km): the photon sphere. The peak makes the circular photon orbit unstable. Divide the light ray's equation through by $L^{2}$ and it reads

$\frac{1}{L^{2}}\left(\frac{dr}{d\lambda}\right)^{2} + \frac{1}{r^{2}}\left(1-\frac{2}{r}\right) = \frac{1}{b^{2}},$

where $b = L/E$ is the impact parameter, how far off-centre the ray is aimed. The hill in the middle term peaks at $1/27$ , at $r=3$ . A ray aimed close enough that $1/b^{2}$ tops the hill rolls over it and is captured; one aimed wider turns back out. The dividing line is the critical impact parameter

$b_{c} = \sqrt{27}\,\frac{GM}{c^{2}} = \frac{\sqrt{27}}{2}\,r_s,$

which for a non-spinning hole of Vesper's mass is 92.1 km: the shadow a distant observer sees is about 2.6 times the horizon radius, a result that Why the hole looks like that puts to work.

What has been seen, and what is still argued?Kerr, the hair tests, the stellar and supermassive census

How the idea became real

For decades the horizon was treated as a mathematical oddity. In 1939 Robert Oppenheimer and Hartland Snyder showed that a heavy, burnt-out star collapses without limit and closes itself off from the outside. Finkelstein's 1958 coordinates showed the horizon was a one-way door and not a singular wall. Then in 1965 Roger Penrose proved that once a region traps light, a singularity must form, given reasonable conditions on matter, and the result does not depend on the collapse being perfectly round. The result extended the prediction beyond perfectly spherical collapse.

Spinning holes: the Kerr solution

Real stars spin, and Roy Kerr found the exact solution for a spinning hole in 1963. Its horizon moves inward as spin grows, to

$r_{+} = \frac{r_s}{2}\left(1+\sqrt{1-a^{2}}\right)$

with $a = Jc/GM^{2}$ the spin, from 0 to 1. At Vesper's spin of 0.70 that is 30.4 km, Vesper's horizon, against the 35.4 km of a hole of the same mass without spin. The Schwarzschild radius remains useful because outside the horizon there is a region, the ergosphere, where space itself is dragged around so fast that nothing can hold still against the distant stars, and at the equator its outer edge is at exactly $r_s$ , whatever the spin. For Vesper the ergosphere, at the equator, fills the shell between 30.4 and 35.4 km. That shell is where energy can be taken from the spin, covered in Mining a spinning hole. The innermost stable orbit also moves in, to 60.1 km for a prograde orbit at this spin (Bardeen, Press and Teukolsky, 1972).

Do all holes really have no hair?

Israel (1967), Carter (1971) and Robinson (1975) proved that a stationary hole in empty space is fully set by its mass and spin, plus charge in the general case. Observations began testing this prediction in the 2010s:

In 2019 the Event Horizon Telescope imaged the shadow of M87*, a hole of 6.5 × 10⁹ Suns, with a size matching what general relativity says for that mass. In 2022 it did the same for Sagittarius A*, about 4 × 10⁶ Suns.
After a merger the remnant rings, like a struck bell, at frequencies fixed only by its mass and spin. In 2019 Maximiliano Isi and colleagues reported the main tone plus one overtone in the ringdown of GW150914, agreeing with a hole of a single mass and spin to about 10 percent, and the overtone's own frequency to about 20 percent. Whether the overtone is really there is argued: in 2022 Roberto Cotesta and colleagues reanalysed the same data and found the evidence for it too sensitive to the chosen start time to trust. Stronger signals would help resolve that disagreement.

The census of stellar-mass holes

GW150914, detected in 2015, was the first direct sighting of two holes merging: about 36 and 29 Suns, leaving one of 62. In the decade that followed, LIGO, Virgo and KAGRA catalogued many more. A study of their third catalogue, published in 2023, found the black-hole pairs bunching at a chirp mass (the blend of the two masses that sets how fast the signal sweeps up in pitch) of about 8.3 Suns, and again near 28. Vesper, at 12 Suns, has the mass of a typical hole in those catalogues, but no detector of that kind could ever see it: its partner is a star 1.5 AU away, and A pair too wide to hear explains why that pair is silent.

There is also a puzzle at the bottom of the scale. Black holes weighed in X-ray binaries in the Milky Way cluster near 7.8 Suns, and none turned up between 2 and 5 Suns: the "mass gap" between the heaviest neutron stars and the lightest known holes (Özel and colleagues, 2010, from 16 holes). They checked whether the way targets were chosen could fake the gap and concluded it could not. In 2024 the LIGO-Virgo-KAGRA collaboration reported GW230529, a merger in which the heavier object weighs 2.5 to 4.5 Suns, inside the supposed gap, and is most probably a black hole. The population study of the third catalogue could neither confirm nor rule out a gap.

What nobody can see

Nothing outside a horizon can observe the singularity at the centre, and the equations that predict it stop being the right ones where quantum effects begin. Proposed replacements for the singularity remain untested. frontier The quantum side of the horizon is the subject of Black holes glow.