The black hole is not your freezer

Leans on Cooling is the economy, Black holes glow.

Why not dump the heat into the coldest place there is?

Heat removal sets much of the Vigil's construction budget. Out here the only way to get rid of it is to glow, and a cold glow is faint, so a radiator working at 30 K has to be enormous: Cooling is the economy tells that story, and the figure is 43 Earths of radiating surface for the first petawatt.

Now look at Vesper. A black hole is not quite black. As Black holes glow explains, it glows very faintly, and the glow is cold: 4.3 billionths of a kelvin, far colder than the afterglow of the Big Bang that fills the sky at 2.7 K. Heat only ever flows toward the cold, and nothing within the Vigil's reach is colder than this. This makes it a candidate heat sink, provided it can absorb the required power.

The absorption rate is limited by the effective area that captures incoming radiation.

A cold reservoir connected through a small opening can receive only a limited flow of radiation. Lowering its temperature does not enlarge the opening.

Vesper's effective opening is its capture disc. Light that passes close enough is bent in and swallowed, and light that passes farther out is only bent a little and flies on, as the picture shows (the bending is the subject of Why the hole looks like that). The disc of rays that get caught is wider than the hole itself, but it is still small: a disc about 92 km in radius, against a swarm whose radiators are 8 × 10⁵ times bigger in area. This disc, and every limit this page builds on it, is worked out for a hole of Vesper's mass without spin.

Vesper's window. Rays of light that pass inside the dashed disc are swallowed. Rays outside it are bent a little and fly on. Whatever the Vigil points at Vesper, only what fits through this disc can go in.

Low-temperature radiation has a low power density. A surface at 30 K glows with only 0.046 watts on every square metre. Wrap Vesper in 30 K glow from every direction, so that every ray that could fall in is carrying it, and the hole takes at most 4.9 × 10⁹ watts. That is a few large power stations' worth, and it is 5 × 10⁻⁶ of the Vigil's petawatt.

Guess first Suppose you built a mirror far wider than Vesper's disc, to gather 30 K glow and funnel it toward the hole. How much more than that could the hole take?

No increase is possible once 30 K radiation fills every incoming direction. Passive optics conserve radiance: they can redirect light but cannot raise its brightness above the source. The capture area and source temperature then set the maximum absorbed power.

Vesper cannot absorb the Vigil's waste heat at the required rate. Accreting matter onto it can still supply the Vigil with power.

How much heat can a black hole take?

The ceiling comes from two quantities multiplied together: how big the hole's window is, and how much heat a bath of radiation at temperature T can push through each square metre of it.

The window first. Light coming in from far away is captured if it passes within 2.6 Schwarzschild radii of the centre, which is 3√3⁄2 r_s, with r_s = 35.4 km for a hole of Vesper's mass without spin. The capture area is then

capture area = π × (3√3 r_s / 2)² = 27π r_s² / 4 = 2.7 × 10¹⁰ m²

That is 27⁄4 times, nearly seven times, the horizon's own cross-section (the disc of radius r_s), because gravity bends light toward the hole.

Then the heat. A surface sitting in a bath of radiation at temperature T collects σT⁴ watts per square metre on each face, where σ is the Stefan-Boltzmann constant. A round target meets the bath from every direction, so for each square metre of its cross-section it collects four times that. The most Vesper can take is therefore

most heat in = 4 σ T⁴ × capture area

with nothing better possible: no mirror, lens or funnel can beat this (the fold explains why).

Where the heat comes from	Temperature	The most Vesper can take
The sky's afterglow	2.7 K	3.3 × 10⁵ W
The Vigil's radiators	30 K	4.9 × 10⁹ W
The Vigil's computers	300 K	4.9 × 10¹³ W
Hot enough to take the whole load	638 K	1 × 10¹⁵ W

The ceiling climbs as the fourth power of the temperature. Even at the computers' own 300 K the hole takes 4.9 × 10¹³ watts, still a small fraction of the load. To swallow the full petawatt the radiation must be at 638 K. That is hotter than the computers that made the heat, and no radiator can glow hotter than the thing it cools.

With JavaScript on, this is a calculator: choose how hot a radiator is and how big it is, with perfect optics aiming all of its glow at Vesper. The heat taken rises with the radiator's area until the area reaches four of the hole's capture discs, 1.1 × 10¹¹ m², and then it stops rising. The table above gives the ceiling for four temperatures: 4.9 × 10⁹ W at 30 K, 4.9 × 10¹³ W at 300 K, and the whole petawatt only at 638 K.

Can a heat pump or a bigger hole fix it?

A heat pump can lift the heat from 30 K to a hotter temperature, at the price of work. The page's heat-pump model estimates that an ideal one, lifting the Vigil's load until the hole can take all of it, has to raise it to 1,767 K, and then the work it spends is 58 times the heat it moved. The hole would be swallowing 6 × 10¹⁶ watts to get rid of one petawatt. A radiator at that temperature could also reject heat directly to space.

A bigger hole would do it, since the window grows as the square of the hole's size, and the size grows with the mass. The page's area-scaling model gives the requirement: to take the whole load at 30 K, the hole would need to be 452 times wider than Vesper, about 5,400 times the mass of the Sun. Vesper is a hole of 12 suns. The Vigil has no such hole.

What about the price of forgetting?

The cost of forgetting says forgetting a bit is cheaper the colder the place the heat goes, and Vesper's own glow is the coldest of all: 4.1 × 10⁻³² joules a bit. But a computer cold enough to exploit that would be glowing at Vesper's temperature, and the most it could ever push into the hole is 2 × 10⁻³⁰ watts. That pays for forgetting about 50 bits a second. The low erasure energy is accompanied by an extremely low heat-transfer rate.

Open everything

Where does the window come from, and why can nothing widen it?Light rays around a black hole, then the brightness theorem

The capture disc

Far from the hole, a ray of light travels in a straight line. Its impact parameter $b$ is how far that line would pass from the centre if gravity were switched off. In the Schwarzschild geometry, two quantities are conserved along a ray: its energy $E$ and its angular momentum $L$ , and $b = L/E$ . The radial motion of a ray then obeys

$\left(\frac{dr}{d\lambda}\right)^2 = E^2\left[\,1 - b^2\,V(r)\,\right], \qquad V(r) = \frac{1}{r^2}\left(1 - \frac{r_s}{r}\right)$

where $\lambda$ is a parameter that increases along the ray. A ray can only be at a radius where the bracket is not negative, that is where $1/b^2 \ge V(r)$ . The function $V$ is zero on the horizon, zero at infinity, and has one hump in between. Differentiating,

$V(r) = \frac{r - r_s}{r^3}, \qquad \frac{dV}{dr} = \frac{3 r_s - 2r}{r^4} = 0 \;\Rightarrow\; r = \frac{3}{2}\,r_s$

This is the photon sphere, where light can circle the hole. The height of the hump is

$V_{\max} = \frac{1 - 2/3}{(3 r_s/2)^2} = \frac{4}{27\,r_s^2}$

A ray that arrives from far away gets over the hump, and so falls in, exactly when $1/b^2 > V_{\max}$ , that is when

$b < b_c = \frac{3\sqrt{3}}{2}\,r_s$

Rays with a larger $b$ turn around at the hump and leave. Because the hole is round, every direction of approach looks the same, so the capture cross-section is the area of a disc of radius $b_c$ :

$\sigma_{\text{cap}} = \pi b_c^2 = \frac{27\pi}{4}\,r_s^2$

This is the number in the table: 2.7 × 10¹⁰ m². The bending that makes the disc wider than the horizon is the same bending that gives the shadow and photon ring in Why the hole looks like that.

Brightness cannot be concentrated

Radiance is the power per unit area per unit solid angle along a ray, $L$ in W m⁻² sr⁻¹. Two optical constraints set the bound.

First, in any lossless optical system, radiance is conserved along each ray. Lenses, mirrors, and the bending of light by gravity change where rays go but not how bright each ray is. The conserved thing for a whole beam is its étendue, the product of its cross-section and the solid angle it fills, which is the same before and after:

$G = A_{\text{in}}\,\Omega_{\text{in}} = A_{\text{out}}\,\Omega_{\text{out}}$

(for small angles in vacuum; the exact form uses $\sin^2\theta$ ). Squeezing a beam into a smaller area makes it spread over a wider range of angles, and that widening cannot go beyond the whole sky. This is why a magnifying glass in sunlight can make a spot hotter than the shade but never hotter than the Sun's surface.

Second, no source can be brighter than a blackbody at its own temperature: $L \le \sigma T^4/\pi$ . If optics could raise the radiance of a ray, we would reach a contradiction. Take a blackbody $A$ at temperature $T$ and another blackbody $B$ , also at $T$ , joined by some optics. Light paths are reversible, so each ray that carries light from $A$ to $B$ also carries light from $B$ to $A$ on the same path. If the optics made the $A \to B$ rays brighter than a blackbody at $T$ , $B$ would receive more from $A$ than it sent back, and heat would flow between two bodies at the same temperature, which the second law forbids. So no optics beats the blackbody radiance of the source.

Putting the two together

Follow every ray that reaches Vesper backwards. Each starts at a source with radiance no more than $L_{\max} = \sigma T^4/\pi$ , and the radiance cannot grow on the way. The rays that reach the hole fill an étendue of

$G_{\text{hole}} = \sigma_{\text{cap}} \times 4\pi$

because the hole presents the same disc to every direction of the sky. The power into the hole is at most radiance times étendue:

$P_{\max} = \frac{\sigma T^4}{\pi}\cdot 4\pi\,\sigma_{\text{cap}} = 4\,\sigma\,T^4\,\sigma_{\text{cap}}$

At $T = 30$ K this is 4.9 × 10⁹ W. Setting $P_{\max}$ equal to the Vigil's load and solving for the temperature gives 638 K.

This also explains the picture of a wider mirror. A mirror of larger area gathers more power, but it delivers that power to the hole in a beam whose étendue is larger than the hole can accept; the surplus never enters. What gets in is set by the smaller of the two étendues, the beam's and the hole's.

The calculator uses exactly this. A flat radiator of area $A$ glowing from one face has étendue $\pi A$ and gives off $\sigma T^4 A$ . Perfect optics can deliver all of it while $\pi A \le 4\pi\,\sigma_{\text{cap}}$ , so

$P_{\text{in}} = \sigma T^4 \min\!\left(A,\; 4\,\sigma_{\text{cap}}\right)$

and a radiator bigger than four capture discs, 1.1 × 10¹¹ m², adds nothing.

The same limit, with a heat pump

An ideal pump takes heat $Q_c$ at $T_c$ and delivers $Q_h = Q_c\,T_h/T_c$ at $T_h$ , using work $Q_c(T_h/T_c - 1)$ . The hole must accept all of $Q_h$ , so $Q_h \le 4\sigma T_h^4\sigma_{\text{cap}}$ . Putting $Q_c$ equal to the load gives

$T_h^3 = \frac{Q_c}{4\,\sigma\,\sigma_{\text{cap}}\,T_c}$

which is where the 1,767 K in the numbers comes from.

Is the ceiling exact, and does a black-hole computer escape it?Where the geometry stops working, and the limits on concentrating light

Where the number holds tight

The capture area is the geometry of light rays, and it is accurate when the waves are much shorter than the hole; wave effects matter only once the wavelength comes within a few times r_s. A 30 K bath peaks at a wavelength of about 97 micrometres, and r_s at Vesper's mass is 4 × 10⁸ times that, so the 30 K ceiling sits deep in the regime of rays and is a tight upper bound. The exact absorption cross-section of a Schwarzschild hole, worked out wave by wave by Sanchez in 1978, behaves like a diffraction pattern: as the frequency rises it approaches this geometric value, oscillating around it.

Where it leaks

For waves longer than the hole, the geometry stops working and the greybody factors take over. These measure the fraction of a wave of each frequency and spin that actually gets through the barrier around the hole. Page used them in 1976 to work out how a hole emits neutrinos, photons and gravitons. For light, the barrier is high at long wavelengths, and a hole absorbs it poorly. So the ceiling at Vesper's own temperature, 2 × 10⁻³⁰ W in the numbers, is a bound and not a prediction: at 4.3 nK the glow peaks at about 676 km, 19 times r_s, and the true figure is lower still.

The limits of concentration

That optics cannot beat the radiance of the source is the founding result of nonimaging optics, the branch of optics built for collecting light rather than forming images. Winston's 1970 paper derived the limit from the conservation of phase space, as a sine inequality, and designed collectors that reach it; the textbook by Winston, Miñano and Benítez sets out the limits and the designs. The second law fixes how much light any concentrator can deliver, and the only way round it is to spend work, which is what the heat pump in the numbers does.

Matter does no better as a carrier. A hydrogen atom falling into Vesper is swallowed from a wider range of approach than light, since slow matter is pulled in. But each atom carries 4 × 10¹¹ times more energy in its rest mass than the thermal energy kT it would carry away at 30 K. Using matter to carry waste heat into Vesper would consume much more rest-mass energy than the thermal energy it removes. That matter is also the Vigil's fuel.

Black-hole computers

Lloyd's 2000 paper on the ultimate physical limits to computation takes the argument to its end. It bounds the speed and memory of a computer of one kilogram in one litre, then asks what happens when the same mass is squeezed into the smallest volume general relativity allows: a black hole. Such a computer would hold the most memory its mass permits. A black hole of one kilogram has a horizon of 1 × 10⁻²⁷ m and lasts 8 × 10⁻¹⁷ s before it evaporates, which leaves very little time to read out an answer.

Applying the absorption bound to an external computer gives the same heat-rejection limit. This is an inference from the model above; Lloyd's proposal concerns the hole's computation. Vesper's ceiling, 4.9 × 10⁹ W at 30 K, is a rate limit on any such scheme. A computer built into a hole's neighbourhood still has to cool the swarm around it by glowing.

Still argued in 2026

Whether information that falls into a black hole comes back out in the Hawking radiation is an open question, and whether the hole can serve as a memory at all depends on the answer frontier. The heat argument above does not depend on it: it needs only that the hole is a window of known size in a spacetime where light obeys the brightness theorem, and both of those are established.