The ultimate computer is a black hole

Leans on How much can a region remember?, Black holes glow.

What is the best computer that physics allows?

The speed page asked how fast. A kilogram, spent entirely on thinking, can change state at most 5.4 × 10⁵⁰ times a second (How fast can anything think?). The memory page asked how much. A region of space can hold only so many bits, and the region that holds the most for its size is a black hole, whose memory is written on its horizon (How much can a region remember?).

Combining the speed and memory bounds gives a theoretical limiting case. Take that same kilogram and squeeze it, past every material there is, until nothing could be smaller or denser. It closes behind a horizon only 1.5 × 10⁻²⁷ metres in radius. Seth Lloyd, the physicist who made the ultimate laptop famous in 2000, described this limiting case in the same paper.

For the one-kilogram hole, its memory is 3.8 × 10¹⁶ bits, about 0.38 of one Vigil mind. Its speed is the full 5.4 × 10⁵⁰ steps a second the kilogram could ever manage. And it is so compact that a signal gets from one side to the other, half-way round the horizon, in 1.6 × 10⁻³⁵ seconds, while any one bit waits only 4.5 times as long between changes. Every part of it hears from every other part about as fast as the parts compute. No laptop is built like that, and the numbers below show why.

A hole this small is much hotter and shorter-lived than Vesper. The kilogram hole glows at 1.2 × 10²³ kelvin, pours out at least 9.3 × 10⁵ Suns' worth of light and particles, and is gone in less than 8.4 × 10⁻¹⁷ seconds. For all that speed, its memory is smaller than one mind. And the only thing that ever leaves it is the glow, which for more than half of the hole's short life carries almost nothing of what the hole has worked out. Its short lifetime and the distribution of information through its radiation make readout difficult.

Guess first A black hole twice as heavy as this one: how much longer does it live?

Eight times longer. Mass doubles, and lifetime goes as the cube of the mass. Memory goes up four times and speed two times, so the heavier hole is a bigger and slower-burning computer. The next section follows this thread.

No one has built such a device. The calculation does not provide a method for compressing a kilogram into a region far smaller than a proton. And whether a hole could be given a question and made to return the answer is, in 2026, an open problem in physics frontier. In 2540 the Vigil starts feeding Vesper for power and does not ask it to think: The black hole is not your freezer tells what happens when the Vigil tries to use the hole for anything else. The following calculation treats the hole as a physical bound.

How a black-hole computer changes with its mass, against a one-kilogram hole, on log axes. Memory grows as the square of the mass, lifetime as the cube, and the power the hole radiates falls as the square. The dashed line marks Vesper.

What would a one-kilogram black hole actually do?

Start from the formulas for a hole of mass M. All three come from earlier pages.

radius r = 2GM / c² memory N = 4πGM² / (ħ c ln 2) bits speed R = 2Mc² / (πħ) steps per second

Memory goes as the square of the mass, because it follows the horizon's area. Speed goes in step with the mass, because it follows the energy. Then add the glow. Its power falls as 1/M², and in the simplest count, which treats the horizon as a glowing surface that gives off light and nothing else, the lifetime comes out as

lifetime τ = 5120π G² M³ / (ħ c⁴)

so it goes as the cube. The fold below derives it. The three scalings together make bigger holes better computers by a wide margin:

The hole	Horizon radius (m)	Memory (bits)	Steps a second	Lifetime	Steps in its life
1 kg	1.5 × 10⁻²⁷	3.8 × 10¹⁶	5.4 × 10⁵⁰	8.4 × 10⁻¹⁷ s	3.4 × 10³⁴
1 tonne	1.5 × 10⁻²⁴	3.8 × 10²²	5.4 × 10⁵³	8.4 × 10⁻⁸ s	3.4 × 10⁴⁶
1.7 × 10¹¹ kg (lasts as long as the universe)	2.6 × 10⁻¹⁶	1.1 × 10³⁹	9.4 × 10⁶¹	the age of the universe	3.1 × 10⁷⁹
Vesper's mass (12 Suns), without spin	3.5 × 10⁴	2.2 × 10⁷⁹	1.3 × 10⁸²	3.6 × 10⁷⁰ years	1.1 × 10¹⁶⁰

The 1 kg hole lives only 8.4 × 10⁻¹⁷ seconds, but in that time it can take 3.4 × 10³⁴ steps, 8.9 × 10¹⁷ for every bit it holds. The steps available per stored bit grow with the mass.

The table leaves spin out. Vesper spins, which draws its horizon in to 30.4 km and its memory down to 1.9 × 10⁷⁹ bits; spin barely changes the lifetime, which the mass sets. The lifetimes are the same simplest count that Black holes glow uses in its table. A real hole sheds mass faster, so every lifetime here, and every count of steps in a life, is an upper bound. For a cold hole like Vesper the careful count is about 1.8 times faster. A hole as hot as the kilogram one also makes every other kind of particle, and Lloyd's own estimate for it is a life several hundred times shorter.

With JavaScript on, this is a calculator: slide the mass of the hole and it gives the horizon radius, the memory, the speed, the temperature, the lifetime, the steps in a lifetime and the earliest moment the answer can come out. The table above gives the same numbers for four holes.

Why is a black hole one thinker and a laptop a crowd?

A signal cannot cross a black hole by going through it: whatever goes in stays in. The quickest way from one side to the other is half-way round the horizon, which takes 1.6 × 10⁻³⁵ seconds for the 1 kg hole. Now share the hole's speed out over its 3.8 × 10¹⁶ bits. Any single bit gets to change once every 7.1 × 10⁻³⁵ seconds, only 4.5 times longer. While one bit changes, news can go round the machine several times, so every bit hears from every other. The whole hole behaves like a single processor doing one thing at a time, very fast. Lloyd's word for it is serial.

A laptop is the opposite. Lloyd's ultimate laptop, a kilogram in a litre holding 2.1 × 10³¹ bits, is so roomy that each of its bits can change 8.5 × 10⁹ times while one message crosses it. It is a crowd of small thinkers that rarely hear from each other, like the swarm on the page about light delay. The hole is the far end of the same ladder, and the fold below shows that the ratio of 4.5 is the same for every black hole, whatever its mass.

When can the answer come out?

The only way out is the glow, and at first the glow is plain heat. Suppose, as most physicists expect, that the hole keeps everything it was given and hands it back in the end frontier. Even then you could collect every quantum that left it and learn almost nothing, until the glow holds as much entropy as the hole has left. That turning point is the Page time. Don Page's 2013 calculation, for a hole giving off light and gravitons, puts it a little past half-way: 54 percent of the way through the hole's life, when it has lost 40 percent of its entropy. For the 1 kg hole that is no later than 4.5 × 10⁻¹⁷ seconds in, with 2.3 × 10¹⁶ bits of entropy still inside. After it, the answer starts to come out. To read it you would have to catch quanta of a few joules each (the hole's own kT is 1.7 J), from all round a source at 1.2 × 10²³ kelvin, before it vanishes.

Open everything

Where do the lifetime, the seriality and the Page time come from?Stefan-Boltzmann, one integral, and a ratio that does not depend on size

The lifetime goes as the cube of the mass

A hole of mass $M$ has a horizon area $A = 16\pi G^2 M^2 / c^4$ and a Hawking temperature

$T = \frac{\hbar c^3}{8\pi G M k}$

Treat it as a black body radiating photons. The Stefan-Boltzmann law, with $\sigma = \pi^2 k^4 / 60\hbar^3 c^2$ , gives the power:

$P = \sigma A T^4 = \frac{\hbar c^6}{15360\,\pi G^2 M^2}$

The power falls as $1/M^2$ : the lighter the hole, the hotter and faster it radiates. The energy comes out of the mass, so $c^2\,dM/dt = -P$ , which rearranges to

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

Integrate from the starting mass $M_0$ down to zero:

$\frac{M_0^3}{3} = \frac{\hbar c^4}{15360\,\pi G^2}\,\tau \quad\Longrightarrow\quad \tau = \frac{5120\,\pi G^2 M_0^3}{\hbar c^4}$

That is the $M^3$ . It also gives the shape of the decline, $M(t) = M_0\,(1 - t/\tau)^{1/3}$ : the hole loses mass slowly at first and then in a rush. For a hole of 1 kg the formula returns 8.4 × 10⁻¹⁷ seconds.

A real hole does two things this count leaves out. Its gravity filters what comes out, and it emits every kind of particle light enough to be made at its temperature. Page worked out both for the massless particles in 1976, and Black holes glow follows the details. Both raise the power, so the formula gives an upper bound on the life. A kilogram hole is hot enough to make every particle known, and Lloyd, counting tens to hundreds of kinds of them, gets about 1 × 10⁻¹⁹ seconds. The power of three survives once the hole is hot enough that the list of particles it can make stops growing, because then the prefactor is fixed.

Steps in a lifetime

The speed limit $R = 2Mc^2/\pi\hbar$ falls as the hole shrinks, so the total number of steps is the integral over the run:

$\int_0^\tau \frac{2 c^2}{\pi\hbar} M_0 \left(1 - \frac{t}{\tau}\right)^{1/3} dt = \frac{3}{4}\,\frac{2 M_0 c^2}{\pi\hbar}\,\tau$

Divide by the memory $N = 4\pi G M^2/(\hbar c \ln 2)$ and the steps for each bit come to

$\frac{\text{steps}}{N} = \frac{1920 \ln 2}{\pi}\,\frac{G M^2}{\hbar c}$

This grows as $M^2$ : a heavier hole not only has more bits, it can work on each of them for longer. For 1 kg it is 8.9 × 10¹⁷, an upper bound like the lifetime it comes from.

A ratio that does not depend on size

Lloyd's measure of how parallel a computer is compares two times. One is $t_\text{com}$ , the time a signal needs to get from one side to the other. For an ordinary computer of radius $R$ that is $2R/c$ . For a hole no path runs through the middle, and the shortest way between opposite points of the horizon is half-way round it, $t_\text{com} = \pi r/c$ . The other is $t_\text{flip}$ , the time between changes of any one bit: the time for one step, $\pi\hbar/2E$ , multiplied by the number of bits, since the steps are shared out among all of them. For a hole with $E = Mc^2$ and $r = 2GM/c^2$ :

$t_\text{flip} = N\,\frac{\pi\hbar}{2Mc^2} = \frac{2\pi^2 G M}{c^3 \ln 2}, \qquad t_\text{com} = \frac{\pi r}{c} = \frac{2\pi G M}{c^3}$

$\frac{t_\text{flip}}{t_\text{com}} = \frac{\pi}{\ln 2} \approx 4.5$

The mass cancelled. This is Lloyd's own result, which he writes the other way up as $t_\text{com}/t_\text{flip} = \ln 2/\pi$ : at the black-hole limit a bit flips in about the time a signal takes to go round the horizon, for every hole. Measured straight across, $2r/c$ , the ratio would be $\pi^2/2\ln 2 \approx$ 7.1. The path changes the constant, not the point.

The straight-across version carries over to anything else. A machine that fills the Bekenstein bound $N = 2\pi R E / \hbar c \ln 2$ , with all its energy inside radius $R$ , has $N\pi\hbar/2E = \pi^2 R/c\ln 2$ , and its ratio to $2R/c$ is again $\pi^2/2\ln 2$ . Anything holding fewer bits than the bound has a shorter $t_\text{flip}$ , so a message across it lasts more bit-changes, and it is more parallel. A black hole fills the bound exactly, which is why it sits at the serial end of the scale.

The Page time

Suppose the radiation carries off $\beta$ times the entropy the hole loses. As the hole goes from $S_0$ to $S$ , the radiation's thermal (coarse-grained) entropy grows to $\beta(S_0 - S)$ . If the whole process keeps information, the radiation's true (fine-grained) entropy measures its entanglement with what is left, so it can never exceed the hole's own $S$ . The two curves meet when

$\beta\,(S_0 - S) = S \quad\Longrightarrow\quad \frac{S}{S_0} = \frac{\beta}{1+\beta}$

With $S \propto M^2$ and $M = M_0(1 - t/\tau)^{1/3}$ , that moment comes at

$\frac{t}{\tau} = 1 - \left(\frac{\beta}{1+\beta}\right)^{3/2}$

The answer depends only on $\beta$ , not on the mass. If each quantum carried off exactly the entropy the hole loses, $\beta = 1$ , the turn would come when half the entropy has gone, at $t/\tau = 1 - 2^{-3/2} =$ 0.65: this is the half-way point of Hayden and Preskill's idealized model. A photon gas leaving at the hole's temperature carries $dS = \tfrac43\,dE/T$ , while by the first law the hole loses $dE/T$ , so $\beta = 4/3$ and $t/\tau =$ 0.57. Page's 1976 rates for photons and gravitons, filtered by the hole's gravity, give $\beta =$ 1.4847, and his 2013 calculation puts the turn at $t/\tau =$ 0.5381, with 60 percent of the entropy still in the hole. A hot hole making many kinds of particle has its own $\beta$ , and the number moves a little. After the turn the radiation's true entropy falls with the hole's, down to zero at the end. That rise and fall is the Page curve. Before the turn the radiation looks like heat, so reading the answer cannot begin earlier.

Could a black hole really be programmed, and could we read it?Lloyd's proposal, information loss, the Page curve, and what the 2019 calculations did

What Lloyd proposed, and what he said was unknown

In the 2000 paper that introduced the ultimate laptop, Lloyd added a box titled "Can a black hole compute?". Work in string theory suggested that a hole does not destroy the information about how it formed, but processes it and lets it out, altered, in the Hawking radiation. If that picture is right, he wrote, a hole could in principle be programmed: form one whose initial conditions encode the problem, let the dynamics at the horizon do the work, and read the answer from correlations in the radiation. A full account, he wrote, requires a theory of quantum gravity. His estimate for a kilogram hole was a life of about 1 × 10⁻¹⁹ seconds and about 1 × 10³² operations, and since a hole that hot probably makes even more kinds of particle than he counted, he called the operation count an upper bound. The simplest count on this page runs several hundred times longer than his, which is why its figures are upper bounds too. In 2004 Lloyd and Y. Jack Ng told the story for general readers in Scientific American: the hole as the most compressed computer there is, the open question of how its information gets out, and the universe itself as a computer.

Was the information destroyed?

If it is, the question has no answer: no programming can survive the hole. Stephen Hawking argued exactly that in 1976, that evaporation turns a pure state into a mixed one and predictability breaks down. In 2005 he changed his mind, arguing with Euclidean path integrals that the evolution is unitary and that information is preserved. The unitary side has support from string theory, which can count the microstates of some holes (How much can a region remember?), and from holography, and the 2019 calculations below assume it. The question is not closed. frontier

Getting the answer out

If the evaporation is unitary, the information is in the glow, but not in a form that is easy to take out. Don Page estimated in 1993 that it may come out so slowly at first, or be spread so thin, that no calculation built from small corrections would ever see it. Patrick Hayden and John Preskill gave the sharpest version in 2007. Assume the hole's inside is unitary and mixes quickly, and that the reader controls all of the radiation. Once the hole is past the half-way point, where half its initial entropy has been radiated away, anything thrown in comes back out in the radiation very quickly. Anything thrown in earlier stays hidden until the half-way point, and then emerges quickly. Their argument assumes the hole's dynamics acts like a random unitary on the information. Yasuhiro Sekino and Leonard Susskind conjectured in 2008 that black holes are the fastest scramblers in nature, mixing information in a time that grows only as the logarithm of the number of bits. For the 1 kg hole that is 3.7 × 10⁻³⁴ seconds, a fraction 4.5 × 10⁻¹⁸ of its life. Rapid scrambling does not remove the delay before the information can be recovered.

Calculating the Page curve

In 2019 two groups derived the Page curve from gravity in solvable models. Geoff Penington showed that, for a black hole in anti-de Sitter space evaporating through absorbing boundary conditions, the surface that computes the entropy (the quantum extremal surface) jumps to a new place at precisely the Page time, and that both the Page curve and the Hayden-Preskill rule follow from the geometry. Ahmed Almheiri, Netta Engelhardt, Donald Marolf and Henry Maxfield found a jump of the same kind in a two-dimensional model, with information thrown in leaving the black hole's reach after a scrambling time. The 2021 review by Almheiri, Hartman, Maldacena, Shaghoulian and Tajdini explains the method. These calculations use simplified geometries. Their applicability to a hole in flat space like Lloyd's remains disputed. frontier

Why the Vigil never uses one

Even if the physics worked, three practical problems would remain: squeezing a kilogram to a radius of 1.5 × 10⁻²⁷ metres, encoding a question in the collapse, and catching all the radiation. The simulation therefore treats Vesper as a power source rather than a programmable processor. Lloyd's proposed computer remains a frontier limiting case. Its memory and speed bounds come with unresolved requirements for formation, programming and readout.