Mining a spinning hole

Leans on Feeding a black hole, What is a black hole?.

What does it mean for a black hole to be a flywheel?

A flywheel is a heavy wheel that keeps spinning. Spin it up and you have stored energy in it. Brake it against a generator and the energy comes back out as electricity, and the wheel slows down. Its rotation provides an energy reserve.

A black hole has no wheel in it. It has no parts at all. Even so, a hole that formed from a spinning star keeps the spin, and the spin is real. It drags space around with it, so close to the hole that nothing can hold still against the swirl, not even a beam of light. And a spinning thing carries energy.

Vesper turns at 0.7 of the fastest rate a hole can have. At that spin, about 7.4 % of Vesper's total mass-energy is available as spin energy. Vesper weighs 12 Suns, so the store is the mass equivalent of 0.89 Suns of energy. Turn it all into radiation and you have 1.6 × 10⁴⁷ joules. The Sun, at its steady output, would need 1.3 × 10¹³ years to put out that much.

Run the Vigil on it at the plan's goal of 1 × 10²⁰ watts, far beyond the petawatt of First Light in 2540. At that rate, the store lasts 5 × 10¹⁹ years, which is 3.7 × 10⁹ times the age of the universe. The spin reserve is therefore far larger than the energy the Vigil would use while Ember still shines.

Extraction has two constraints.

The first is that you can never take all of it. Only the spin comes out. The hole is left with a mass it can never give up, the irreducible mass, and it is 11.11 Suns for Vesper. Removing the rotation leaves a non-spinning black hole.

The second is the extraction mechanism. You need a way for the energy to leave. There are two main ideas, and both work by letting something fall into the hole in a way that slows the spin. One uses a particle that breaks in two. The other uses a magnetic field that acts as a brake and a generator at once. Neither has been caught in the act on a particular hole. Physicists in the 2020s take both seriously, and the second is their leading explanation for the most powerful jets in the sky.

Guess first Take every bit of spin energy out of Vesper. Is the surface of its horizon then larger, smaller or the same?

The same, if the job is done perfectly, and larger if it is not. This is the rule that decides everything on this page: the horizon's area can grow but never shrink. Spin energy is what you can take out while the area stays put. Take any more and the area would have to fall, which is not allowed.

How much of a hole's mass-energy is stored in its spin?

A hole's spin is one number, a, from 0 (not turning) to 1 (turning as fast as a hole can). With the mass, it fixes the spin energy:

spin energy = (1 − M_irr ÷ M) × M c²

Here M c² is the hole's whole mass counted as energy (A gram is a fortune) and M_irr is the irreducible mass, the part that can never be taken out. Their ratio depends on spin alone:

M_irr ÷ M = √((1 + √(1 − a²)) ÷ 2)

At a = 0 the ratio is 1 and there is nothing to take. At a = 1 it is 1 ÷ √2, and 29.3 % of the total mass-energy is spin energy. In between:

Spin a	Spin energy as a share of total mass-energy	Mass equivalent in Suns
0	0 %	0 Suns
0.5	3.41 %	0.41 Suns
0.7 (Vesper)	7.42 %	0.89 Suns
0.9	15.3 %	1.8 Suns
0.998 (the most a feeding disc can reach)	27.1 %	3.25 Suns
1	29.3 %	3.51 Suns

The share climbs slowly, then fast. Slow Vesper from 0.7 to 0.5 and less than half of its 7.42 % is left. The efficiency ladder of Feeding a black hole has the same shape: a hole that turns fast is a much better furnace, and a much better battery.

What does Vesper's spin look like up close?

Without spin, a hole of Vesper's mass would have its horizon 35.4 km from the centre. The spin draws it in to 30.4 km, Vesper's horizon. The gap is the ergosphere, where space is dragged round so fast that nothing can stay still. Its outer edge at the equator stays at 35.4 km whatever the spin, so the shell there is about 5 km thick; at the poles it closes onto the horizon.

The horizon turns about 550 times a second, and the spin shrinks its surface by 14 %. How much can a region remember? counts a hole's memory by that surface: Vesper's 1.9 × 10⁷⁹ bits, against 2.2 × 10⁷⁹ for a hole of the same mass without spin.

How long would the reserve last?

Divide the store by the power drawn. At the plan's goal of 1 × 10²⁰ watts that gives the 5.0 × 10¹⁹ years above, and at the first petawatt 5.0 × 10²⁴ years. In the 30 billion years Ember shines, the Vigil at its goal would spend 6 × 10⁻¹⁰ of it, and the spin holds 2,700 times everything Ember will radiate in that time. For these timescales, the extraction rate is the practical constraint.

With JavaScript on, this is a calculator: slide the spin to see its energy, how long it runs the Vigil, and the horizon. For Vesper, 7.42 % is 1.6 × 10⁴⁷ joules, enough for 5.0 × 10¹⁹ years at 1 × 10²⁰ watts.

Which idea would the Vigil actually use?

A particle that breaks in two. In 1969 Roger Penrose saw that a particle falling into the ergosphere could split, one piece falling into the hole and the other flying out. If the piece that falls in has negative energy as the outside universe counts it, the piece that escapes carries more than the particle brought in. The surplus comes from the spin.

A particle with energy 10 enters the ergosphere and breaks in two. One piece falls in with energy minus 2, the other escapes with energy 12. The numbers are only to show the bookkeeping.

The particle-splitting process has severe practical constraints. One break-up gains at most about a fifth of the particle's energy, and only at the horizon of a hole spinning as fast as it can. Worse, the pieces must fly apart at more than half the speed of light, inside a shell a few kilometres thick.

A magnetic field. In 1977 Roger Blandford and Roman Znajek showed that a magnetic field threading a spinning hole turns it into a generator. The twisting of space winds up the field lines, and the wound field carries energy and angular momentum away along the axis as a jet. The magnetic field provides a continuous path for extracting rotational energy.

The power depends on the field. For a field of 1 × 10⁴ tesla at Vesper's horizon, an illustrative field strength design, the estimate is 1 × 10³⁰ watts, 1 × 10¹⁰ times the whole supply the plan aims for. To make exactly the goal's 1 × 10²⁰ watts the field needs to be only 0.084 tesla. The hard part is holding a field on the horizon and feeding it plasma so it does not slip away, and nobody has worked out how the Vigil would.

Open everything

Why is the limit M_irr, and where does the Penrose process get its energy?The area theorem, the mass formula and the ergosphere, with algebra

The horizon area can only grow

The rule behind this page is the area theorem. Stephen Hawking proved in 1971 that if general relativity holds, and matter never has negative energy density, then the total area of the horizons of black holes can never decrease, whatever happens to them. It looks like the second law of thermodynamics with area in place of entropy, and is related to black-hole entropy: How much can a region remember? ends up counting a hole's memory by its area.

For a hole of mass $M$ and spin $a$ (in units where $G = c = 1$ , so that lengths are measured in units of $GM/c^2$ ), the outer horizon sits at

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

and its area is

$A = 4\pi\left(r_+^2 + a^2 M^2\right) = 8\pi M^2\left(1 + \sqrt{1 - a^2}\right)$

where the second form follows by putting $a^2 = 1 - (1 - a^2)$ in the first. For a hole without spin this is $16\pi M^2$ , the surface of a sphere of radius $2M$ .

Defining the irreducible mass

If a hole's area can only grow, then the quantity $\sqrt{A / 16\pi}$ can only grow. Call it the irreducible mass:

$M_\mathrm{irr} = \sqrt{\frac{A}{16\pi}} = M\sqrt{\frac{1 + \sqrt{1 - a^2}}{2}}$

For a hole without spin $M_\mathrm{irr} = M$ . That is why spin is the part you can take: it is the difference between $M$ and $M_\mathrm{irr}$ . Demetrios Christodoulou, then a student of John Wheeler's, showed in 1970 that this is the right way to cut the mass. He proved that a transformation that keeps $M_\mathrm{irr}$ fixed can be undone and one that raises it cannot, so the energy you can take out is the energy above the $M_\mathrm{irr}$ floor, no more. The mass splits into two parts that depend on area and spin:

$M^2 = M_\mathrm{irr}^2 + \frac{J^2}{4 M_\mathrm{irr}^2}$

where $J = aM^2$ is the spin angular momentum. To check, put in $M_\mathrm{irr}^2 = M^2(1 + s)/2$ with $s = \sqrt{1 - a^2}$ :

$\frac{J^2}{4 M_\mathrm{irr}^2} = \frac{a^2 M^4}{2M^2(1 + s)} = \frac{M^2(1 - s^2)}{2(1 + s)} = \frac{M^2(1 - s)}{2}$

and the sum is $M^2(1 + s)/2 + M^2(1 - s)/2 = M^2$ . The first term is fixed by the irreducible mass. The second is the spin, and it can be taken out one dose at a time. At $a = 1$ , $s = 0$ , and $M_\mathrm{irr} = M/\sqrt{2}$ , so the share that comes out is $1 - 1/\sqrt{2} = 29.29\ \%$ .

Where the first law comes in

When a hole absorbs a small amount of energy $\delta E$ and angular momentum $\delta L$ , its mass and spin change by $\delta M = \delta E$ and $\delta J = \delta L$ . James Bardeen, Brandon Carter and Hawking showed in 1973 that these changes obey a law like the first law of thermodynamics:

$\delta M = \Omega_H\,\delta J + \frac{\kappa}{8\pi}\,\delta A$

Here $\Omega_H = a / (2 M (1 + s)) = a/(2r_+)$ is the angular velocity of the horizon, which for Vesper is 3,454 radians per second, and $\kappa$ is the horizon's surface gravity, which is positive. The area theorem says $\delta A \ge 0$ . So $\delta M \ge \Omega_H\,\delta J$ . If the hole absorbs something with $\delta J < 0$ , its mass can fall by at most $\Omega_H |\delta J|$ . That is a ceiling on how much energy you can take per unit of angular momentum. It is reached when $\delta A = 0$ , the reversible case.

The ergosphere and negative energy

Why can the piece that falls in have negative energy? Energy as measured far away is the particle's momentum projected along the direction of time at infinity. Where space is dragged around fast enough, that direction stops being a direction of time at all: in the ergosphere it points through space. A particle there can have a negative value of that projection, which is what negative energy at infinity means. A particle that stays outside the ergosphere never can.

Split a particle of energy $E_0$ into two, with $E_0 = E_1 + E_2$ by conservation. If $E_1 < 0$ , then $E_2 = E_0 - E_1 > E_0$ . The escaping piece has more than the particle that went in. The piece with $E_1 < 0$ falls through the horizon and lowers the hole's mass by $|E_1|$ . Its angular momentum $L_1$ is negative too: a piece can only cross the horizon if $E_1 \ge \Omega_H L_1$ , so $L_1 \le E_1/\Omega_H < 0$ . The hole loses both mass and spin, exactly as the first law above requires.

Why a fifth at most? The best single break-up happens right at the horizon, with the particle turning into two photons. The escaping photon then carries $\tfrac12\left(1 + \sqrt{2M/r_+}\right)$ of the energy sent in. For a hole at the maximal spin, $r_+ = M$ , and the gain is $(\sqrt{2} - 1)/2 = 20.7\ \%$ (the review by Brito, Cardoso and Pani works it through). Real break-ups do far worse, and Robert Wald showed why in 1974. A fragment's energy per unit of its mass can differ from the parent's only by an amount set by the speed at which the pieces separate. For matter on the orbits gas actually follows, one piece can only reach negative energy if the two fly apart at more than about half the speed of light. Bardeen, William Press and Saul Teukolsky had found the same bound in 1972.

What the Blandford-Znajek field does instead

The field needs no particle to break up. The turning horizon drags the field lines round, and a conductor turning in a magnetic field drives a current, as in Faraday's disc dynamo. The current flows out along the axis and carries energy and angular momentum away. In Gaussian units the power is

$P \approx \frac{k}{4\pi c}\,\Phi^2\,\Omega_H^2$

where $\Phi$ is the magnetic flux through one hemisphere of the horizon and $k$ is a number near 0.05 that depends on the shape of the field. Blandford and Znajek's own result went as $a^2$ and was worked out for slow spin. Alexander Tchekhovskoy, Ramesh Narayan and Jonathan McKinney found in 2010, in simulations of jets held in by a thin disc with the flux fixed, that writing it with $\Omega_H^2$ keeps it accurate to spins far above Vesper's; their 2011 paper gives it in the form above. For the example field, the estimate uses $\Phi = B\,\pi r_+^2$ . The example assumes that flux geometry.

Has anyone seen this happen, and what is still argued?Jets, a lab echo, a test of the area law, and the open questions

Jets that look like Blandford-Znajek

The strongest sign that holes give up spin is in jets, from the brightest active galaxies and from some stellar-mass holes. The Blandford-Znajek mechanism is the leading explanation for the most powerful of them frontier. In 2011 Tchekhovskoy, Narayan and McKinney simulated a disc that piles so much magnetic field onto a fast-spinning hole that the field holds the gas back, and found more energy flowing out than flowed in. The excess energy comes from the hole's rotation. In 2014 Jean-Pierre Lasota and colleagues showed that this is a Penrose process of a general kind: the field carries negative energy and angular momentum through the horizon, and in such simulations the energy out reaches about 300 % of the energy of the gas falling in.

By the 2020s no one had measured the energy a particular hole gives up in a jet and checked it against that hole's spin. The match between theory and the sky is a match of patterns, and spins themselves are measured imprecisely.

The same effect in a laboratory

Yakov Zel'dovich predicted in 1971 that a wave bouncing off a spinning absorber comes back stronger, if the absorber turns faster than the wave oscillates. In 2020 Marion Cromb and colleagues in Glasgow sent twisted sound waves through a spinning absorbing disc and measured them amplified by up to 30 %. The effect is called superradiance, and it is the wave version of the Penrose process: the amplified wave takes its energy from the rotation. Press and Teukolsky pointed out in 1972 that a mirror around a spinning hole would make the amplified wave grow without limit, a black-hole bomb. In nature the mirror can be the mass of a light particle, which is why spinning holes are used to search for new ones frontier.

Is the area law true?

Most of this page rests on the area theorem, so its test matters. In 2021 Maximiliano Isi and colleagues analysed the gravitational waves from GW150914, the first merger of two black holes ever detected, and compared the horizon areas of the two holes that went in with the area of the one that came out. The data agree with Hawking's theorem with 97 % probability, or 95 % if the ringdown is modelled without overtones. The measurement provides a test with substantial uncertainty. Vesper has no partner hole to merge with (A pair too wide to hear), so for the Vigil the theorem works as a ledger rule: whatever is fed in or extracted, the horizon area never falls.

Open questions

Quantum effects. The area theorem is classical. Hawking radiation (Black holes glow) makes the area shrink, very slowly, and the generalised second law, which counts the entropy of the radiation too, takes its place. The spin cools the glow too: Vesper radiates at 4.3 nanokelvin, against 5.1 for a hole of its mass without spin (The black hole is not your freezer). Neither touches the spin reserve over any time the Vigil plans for.
The field. Whether a hole of Vesper's size can hold a field of the strength the engine needs, and for how long, is not known design.
Real spins. Vesper's 0.7 is a choice for this world. The spins of real holes, found from their X-rays and gravitational waves, carry large errors, and how large is argued.