Black holes

Black holes glow

Quantum fields near a black hole produce thermal radiation. Larger holes are colder and radiate less power.

Leans on What is a black hole?, Entropy is missing information.

How can something that swallows light glow?

A black hole is a region that light cannot leave, as What is a black hole? explains. Classical general relativity therefore predicts no radiation from the hole itself.

In 1974 Stephen Hawking found that quantum fields change this prediction. Quantum physics says that a horizon glows, like a very cold, very dim stove. It has a temperature, and it radiates heat at that temperature into the space around it.

A common analogy describes particle pairs near the horizon. Empty space is not really empty. Pairs of particles flicker into being and vanish again all the time, each pair so brief that nobody sees them. Now put a horizon in the way. Sometimes one of a pair falls in and the other flies off. The one that flies off is Hawking radiation, and the energy it carries comes out of the hole, so the hole loses mass. (This analogy is not the derivation. The calculation treats quantum fields in the spacetime of a forming black hole.)

The temperature decreases as the mass increases. The temperature of a hole is inversely proportional to its mass:

T = ħc³ / (8π G M k)

A hole with the mass of the Moon glows at 1.7 K. A hole ten times heavier than that is ten times colder.

Vesper has 12 times the mass of the Sun, and it glows at 4.3 billionths of a kelvin. The surroundings are much warmer. In 2540 the cosmic microwave background, the afterglow of the Big Bang, is at 2.7 K, which is 6.4 × 10⁸ times warmer than Vesper's glow. Heat flows from warm to cold, so Vesper takes in far more from the sky than it gives out. It gains mass from that net absorption.

Guess first Vesper glows at a few billionths of a kelvin, and the cosmic background is at a few kelvin. Which way does the heat flow?

From the sky into Vesper. The sky is the warmer body by a factor of 6 × 10⁸. A black hole only loses mass to its own glow once the surroundings are colder than the glow, and for Vesper that is a very long wait.

How long? Once the sky cools below the glow and Vesper starts to shrink, it takes something like 4 × 10⁷⁰ years to radiate itself away. The universe itself is 14 billion years old. Vesper's lifetime is 3 × 10⁶⁰ times that. This is far longer than the other timescales considered here.

Hawking's temperature also made it possible to calculate black-hole entropy and include black holes in the second law. They turned out to hold more entropy for their size than anything else can, and that story, from Entropy is missing information to the memory limits in How much can a region remember?, is why they matter to a civilization that thinks for a living.

One thing about the glow is not settled. If a hole glows away completely, does everything that fell in come back out, written into the glow, or is it gone for good? Physicists still argue about this, and the last fold below says where the argument stands. frontier

How hot, how bright, how long?

Three numbers come out of one idea: a horizon is a heat source with a temperature fixed by its mass.

temperature T = ħc³ / (8π G M k) power P = ħc⁶ / (15360 π G² M²) lifetime = 5120 π G² M³ / (ħc⁴)

The constants are Planck's constant ħ, the speed of light c, Newton's G and Boltzmann's k. The mass M is the only knob. Turn it up and the temperature and power fall, as 1/M and 1/M². The lifetime rises, as M³: a hole ten times heavier lives a thousand times longer.

The table and calculator use a simple model: a hole without spin whose horizon radiates as a perfect blackbody emitting only photons. Including greybody factors and the emitted particle species, as in the first fold, gives a mass-loss rate for Vesper about 1.8 times the simple rate, so it lives about half as long. The mass scaling is unchanged.

Vesper spins, at 0.70 of the fastest a hole can turn. The spin draws its horizon in and cools its glow to 0.83 of the value for a hole of its mass without spin, so the Vesper row gives its real radius and temperature. The lifetime there is still the simple count: a spinning hole sheds its spin long before it has lost most of its mass, so the mass sets how long it lasts, and the simple count is an upper end.

The hole	Mass	Horizon radius	Glows at	Takes this long to evaporate
One kilogram	1 kg	1 × 10⁻²⁷ m	1 × 10²³ K	8 × 10⁻¹⁷ s
Lasts as long as the universe is old (simplest count)	2 × 10¹¹ kg	3 × 10⁻¹⁶ m	7 × 10¹¹ K	14 billion years
The Moon's mass	7.3 × 10²² kg	0.11 mm	1.7 K	1 × 10⁴⁵ years
The Sun's mass	2.0 × 10³⁰ kg	3.0 km	62 nK	2 × 10⁶⁷ years
Gaia BH1, the nearest known hole in 2026	9.6 Suns	28 km	6.4 nK	2 × 10⁷⁰ years
Vesper, with its spin	12 Suns	30 km	4.3 nK	at most about 4 × 10⁷⁰ years

The second row is the one cosmologists care about: a hole formed in the early universe with that mass would be finishing its evaporation at an age equal to the universe's. A lighter one would already be gone. A heavier one still has time. Here the simplest count is off by more than for Vesper. A hole that small glows at hundreds of billions of kelvin, hot enough to make neutrinos and electrons as well as light, so it sheds mass faster. Counting all of them, the mass that just lasts the universe's age is about 5 × 10¹¹ kg, 2.9 times the table's figure (Page in 1976, refined by Carr and colleagues in 2010). Its horizon would be smaller than an atomic nucleus, and as of 2026 no one has found any trace of such a hole.

The row for one kilogram shows how fast the end comes for a small hole. It would radiate 4 × 10³² W, and be gone in 8 × 10⁻¹⁷ s. That is the figure behind The ultimate computer is a black hole: a tiny black hole is as fast as a computer can be, and just as brief.

How much does Vesper radiate, in watts?

On the simple count Vesper radiates about 6 × 10⁻³¹ W, far less than it absorbs. The cosmic background arrives from every direction, and Vesper swallows every ray that comes within its 3 × 10¹⁰ m² of shadow (the capture area worked out in The black hole is not your freezer for a hole of Vesper's mass without spin). That adds up to 3 × 10⁵ W pouring in, 5 × 10³⁵ times what it gives back.

The balance tips only when the sky is cooler than Vesper's glow. The expansion of the universe stretches the background and cools it, and by the Vigil's own calendar that happens 351 billion years after 2540. From then on Vesper shrinks. The sky never gets colder than 2 × 10⁻³⁰ K, the floor set by the expansion itself, so the shrinking, once it starts, does not stop. Thinking in a cooling universe follows what that means for anyone still thinking.

Hawking temperature against mass, on log axes, without spin. A hole hotter than the sky around it loses more than it gains and shrinks; a colder one gains more than it loses and grows.

Open everything

Why does a horizon have a temperature?Surface gravity, an accelerating observer, and the evaporation equation

How steep is the pull at the horizon?

The first ingredient is the surface gravity $\kappa$ , the acceleration someone would need to hold still just outside the horizon, measured in the standard way for a far-off observer. For a hole of mass $M$ it comes out as

$\kappa = \frac{c^4}{4GM} = \frac{c^2}{2 r_s}$

where $r_s = 2GM/c^2$ is the horizon radius of a hole without spin. For a hole of Vesper's mass that is 1.3 × 10¹² m/s², or 1 × 10¹¹ times the pull at the surface of the Earth. Notice that it falls as $1/M$ : the bigger the hole, the gentler its horizon. It is the same factor the temperature carries.

Why $\kappa/2\pi$ ?

The temperature of the glow is the surface gravity, divided by $2\pi$ , and converted to kelvin with the constants:

$kT = \frac{\hbar \kappa}{2\pi c}$

Putting $\kappa = c^4/4GM$ in gives $T = \hbar c^3 / 8\pi G M k$ , the formula on this page. The $2\pi$ is the interesting part, and there are two ways to see it.

The Unruh picture. Someone accelerating through empty space at $a$ finds it warm. In 1976 William Unruh showed that an accelerating observer sees a thermal bath at

$kT_U = \frac{\hbar a}{2\pi c}$

The bath is faint for any ordinary acceleration: to feel even 1 K you would need 2 × 10²⁰ m/s². A hole's horizon is where the thing holding still is exactly this kind of accelerating observer. Close to the horizon, a hovering observer has acceleration $a \approx \kappa / \sqrt{1 - r_s/r}$ , which grows without limit as $r \to r_s$ . So the local bath there is hot, and it is

$T_\text{local} = \frac{\hbar \kappa}{2\pi c k} \cdot \frac{1}{\sqrt{1 - r_s/r}}$

Light climbing out of a gravitational field loses energy, and a temperature loses with it by the same factor, the Tolman factor $\sqrt{1 - r_s/r}$ . A far-off observer therefore measures $T_\infty = T_\text{local} \sqrt{1 - r_s/r} = \hbar \kappa / 2\pi c k$ , whatever the radius. That is the Hawking temperature. This heuristic reproduces the factors in the temperature formula. Hawking's calculation instead treats quantum fields during collapse.

The smooth-tip picture. Write the geometry near the horizon in terms of a distance $\rho$ from it, and let time run in the imaginary direction, the standard way to ask about thermal states. The metric becomes

$ds^2 = \rho^2 \kappa^2 \, d\tau^2 + d\rho^2$

which is a plane in polar coordinates, with the angle $\kappa\tau$ . It is smooth at $\rho = 0$ only if $\kappa \tau$ goes round once in $2\pi$ , so $\tau$ must repeat with period $2\pi/\kappa$ . A system in thermal equilibrium at temperature $T$ repeats in imaginary time with period $\hbar / kT$ . Equate the two and the temperature is $\hbar\kappa/2\pi k$ (with $c = 1$ ). The $2\pi$ is the angle of a full circle.

How it shrinks

A body with temperature $T$ and surface area $A$ radiates $\sigma T^4 A$ , where $\sigma = \pi^2 k^4 / 60 \hbar^3 c^2$ is the Stefan-Boltzmann constant. The horizon has area $A = 4\pi r_s^2 = 16\pi G^2 M^2 / c^4$ . Multiply it out:

$P = \sigma T^4 A = \frac{\pi^2 k^4}{60 \hbar^3 c^2} \cdot \frac{\hbar^4 c^{12}}{(8\pi)^4 G^4 M^4 k^4} \cdot \frac{16\pi G^2 M^2}{c^4} = \frac{\hbar c^6}{15360 \pi G^2 M^2}$

The radiated energy comes out of the hole's mass, so $dM/dt = -P/c^2$ :

$M^2 \, dM = -\frac{\hbar c^4}{15360 \pi G^2} \, dt$

Integrate from $M$ down to zero and the time it takes is

$t_\text{life} = \frac{5120 \pi G^2 M^3}{\hbar c^4}$

which is 4 × 10⁷⁰ years at Vesper's mass, as in the table. The cube is what matters. It comes from two things: the power falls as $1/M^2$ , and the energy there is to radiate rises as $M$ .

The equation also shows that a hole loses mass faster as it gets lighter, because lighter means hotter. It speeds up all the way to the end, so the last stage is violent: the final kilogram goes in about 8 × 10⁻¹⁷ s. A black hole also has a negative heat capacity: losing energy makes it hotter, where an ordinary hot body cools. That is why a hole does not settle at the temperature of its surroundings, and why the sky's temperature decides whether it grows or shrinks.

What the simple count leaves out

The treatment above counts a hole as a perfect blackbody the size of its horizon, giving off photons and nothing else. A real hole differs in two ways. First, the curved space around it acts as a filter: long waves are partly bounced back in, while short waves see a target wider than the horizon, the same $27\pi r_s^2/4$ capture disc that The black hole is not your freezer works out. These corrections are called greybody factors. Second, a hole emits every kind of particle light enough to be made at its temperature. For Vesper that means photons and gravitons, and perhaps the lightest neutrino if it turns out to have no mass at all; every particle known to have mass is far too heavy to be made at a few billionths of a kelvin.

Don Page computed the true rates in 1976. A hole too cold to make anything with mass loses energy at the rate

$P_\text{Page} = \left(f_\gamma + f_g\right) \frac{ 2 \times 10^{-4}\,\hbar c^6}{G^2 M^2}$

where his total coefficient counts neutrinos too (taken as massless in 1976), and photons and gravitons carry the shares $f_\gamma$ = 17% and $f_g$ = 2% of it. Set that against $\hbar c^6/(15360\pi G^2 M^2)$ above: the fuller model gives a mass-loss rate 1.8 times the simple rate, and Vesper lasts about 2 × 10⁷⁰ years rather than 4 × 10⁷⁰. The $M^3$ is untouched, because the greybody factors depend only on the wavelength measured against the horizon, and so they scale away. A small hole is another matter. It is hot enough to make many more kinds of particle, each adding to the rate, which is why the universe-age mass in the numbers moves up by a factor of 2.9.

Why it counts as entropy

Hawking's temperature closes a loop that Jacob Bekenstein opened in 1973, when he argued from the second law that a hole's entropy must be proportional to the area of its horizon. The first law of thermodynamics, $c^2 \, dM = T \, dS$ , fixes the constant:

$dS = \frac{c^2 \, dM}{T} = \frac{8\pi G k M}{\hbar c} \, dM \quad\Rightarrow\quad S = \frac{4\pi G k M^2}{\hbar c} = \frac{k c^3 A}{4 G \hbar}$

A quarter of the horizon area, counted in Planck areas, gives the entropy in units of $k$ . The last form holds for any hole; the middle one is for a hole without spin, which at Vesper's mass would hold 2.2 × 10⁷⁹ bits.

What does the spin change?

A spinning hole of mass $M$ and spin $a$ (from 0 to 1) has its outer horizon at $r_+ = (GM/c^2)(1 + \sqrt{1-a^2})$ , with area and surface gravity

$A = 8\pi \left(\frac{GM}{c^2}\right)^{2}\left(1+\sqrt{1-a^{2}}\right), \qquad \kappa = \frac{c^4}{4GM}\cdot\frac{2\sqrt{1-a^{2}}}{1+\sqrt{1-a^{2}}}$

Both reduce to the formulas above at $a = 0$ . The same $kT = \hbar\kappa/2\pi c$ then gives

$T = \frac{\hbar c^{3}}{4\pi G M k}\cdot\frac{\sqrt{1-a^{2}}}{1+\sqrt{1-a^{2}}}$

At Vesper's spin of 0.70 the surface gravity is 1.1 × 10¹² m/s², the temperature 4.3 nK, 0.83 of the value without spin, and the area 0.86 of it, so Vesper holds 1.9 × 10⁷⁹ bits, the figure How much can a region remember? builds on. Mining a spinning hole works through where the area formula comes from.

Spin matters much less for the lifetime. A spinning hole radiates particles that carry angular momentum away faster than energy, and Page showed in a second 1976 paper that it spins down to nearly no spin before most of its mass is gone. Between no spin and the fastest possible, the lifetime changes by a factor of only 2.0 to 2.7, shorter with spin. So the mass sets Vesper's lifetime, and the simple count's 3.6 × 10⁷⁰ years is an upper end.

Has anyone seen it, and where does the information go?Analogue experiments, the Page curve and islands

Seen in water and in cold atoms

As of 2026, no one has detected Hawking radiation from a real black hole. The reason is in the table: a hole of a few Suns glows at a few billionths of a kelvin and sits in a sky that is 6 × 10⁸ times warmer. Any signal would be buried.

What has been done is to build a horizon for something else. In 1981 Unruh pointed out that sound in a flowing fluid obeys the same equations as light near a black hole. Where the fluid speeds up past the speed of sound, sound cannot swim upstream, and that point is a horizon for sound. Waves in such a flow should be emitted from it, with a temperature set by how steeply the flow speeds up.

Three experiments tested this prediction. In 2011 Silke Weinfurtner and colleagues sent water waves up against an opposing current in a tank and saw the stimulated version of the effect, where a wave sent at the horizon comes out as the Hawking process says it should. In 2016 Jeff Steinhauer cooled a cloud of rubidium atoms to a Bose-Einstein condensate, let it flow past a sonic horizon, and saw the spontaneous version: pairs of sound quanta, one from each side of the horizon, correlated in the way the theory demands. In 2019 Juan Ramón Muñoz de Nova, Steinhauer and colleagues measured the spectrum of that radiation and found it thermal, with a temperature that matched the one predicted from the flow.

These experiments test wave and quantum-field behaviour at analogue horizons. They do not measure radiation from a gravitational horizon or the mass loss of an evaporating black hole.

What happens to what fell in?

If a hole evaporates completely, the radiation that comes out looks thermal, with nothing about what went in written on it. Burn a book in a fire and, in principle, you could recover the text from the ash and light. If you drop the book into a hole and wait for it to evaporate, Hawking argued in 1976 that you couldn't, and that quantum mechanics itself would have to give way. This is the information paradox, and it is the most famous open problem in the field.

The Page curve describes the information-preserving prediction for the radiation's entropy. Don Page showed in 1993 that, if the whole process is an ordinary quantum evolution that keeps information, the entropy of the radiation first rises, as Hawking's calculation says, and then must come back down to zero as the hole disappears. The turn, since called the Page time, comes when the hole has lost about two fifths of its entropy (Page 2013 puts it at about 40 %). For Vesper that lies more than halfway through its whole lifetime of 4 × 10⁷⁰ years.

In 2019 two groups showed how that downturn could appear in a calculation. frontier Geoff Penington, and independently Ahmed Almheiri, Netta Engelhardt, Donald Marolf and Henry Maxfield, computed the entropy of an evaporating hole with the quantum version of a rule from holography, which finds the surface that divides what the hole holds from what the radiation holds. At the Page time that surface jumps to just inside the horizon, the hole's entropy starts to fall as Page said it must, and part of the interior belongs, in the information sense, to the radiation outside. A few months later Almheiri, Raghu Mahajan, Juan Maldacena and Ying Zhao turned this into a rule for the radiation itself: to compute its entropy, count a region inside the hole, an island, as part of it. With the island, the radiation's entropy follows the Page curve. A review by Almheiri, Hartman, Maldacena, Shaghoulian and Tajdini in 2021 lays out the whole argument.

The calculations use simplified models of gravity, mostly in two dimensions or with a boundary that confines the system. They reproduce the Page curve without describing how information physically escapes from a real black hole. The island rule can also be derived from the gravitational path integral by including additional spacetime geometries called replica wormholes. A first-principles derivation for a hole like Vesper remains unavailable. The work is widely read as evidence that information is preserved. Whether it settles the paradox is still argued.

What Vesper can say about it

The two possible outcomes have different implications for information storage. If information is preserved, it could in principle be recovered from the radiation after the Page time, deep into a lifetime of 4 × 10⁷⁰ years. If information is destroyed, that recovery is impossible. The Vigil's plans do not depend on the answer in either case, because no one is waiting 3 × 10⁶⁰ universe-ages for a result. The distinction affects the relation between black-hole entropy and recoverable information.