A change in the metric can propagate faster than a photon in general relativity

We have treated this question in earlier blog posts. On February 25, 2023 we suggested that no change in the metric can overtake a photon. It turns out that if the change in the metric is such that it cannot be used to carry a signal, then it may propagate even at an infinite speed.

A spherically symmetric shell of matter

                   _____

                 /           \

               |      o       |    clock

                 \______/

           shell of matter

Let us have a spherically symmetric shell of matter which is kept static with its tangential pressure. We can make the shell to expand or contract. There is a clock at the center. It will tick faster or slower, depending on the radius of the sphere.

In general relativity, the spatial metric inside a static shell is flat and the metric of time is a constant.

Does the speed of the clock change immediately when we expand the radius or is there a delay which would depend on the speed of light?

Suppose that the clock would still tick slower for a while. Then the local speed of light would be slower at the center (slower in global coordinates) than at the edges of the volume inside the shell. Such a metric can focus parallel rays of light. Focusing requires positive Ricci curvature, but the volume is empty and the Ricci curvature has to be zero.

We conclude that the speed of the clock changes immediately when we change the radius of the shell.

                 ____

               /         \  +

                \____/

This is analogous to the electric potential inside a charged shell. We believe that the electric potential changes immediately when we adjust the radius of the charged shell.

Hypothesis. A change in the metric can propagate even infinitely fast if the change does not allow one to send a signal faster than the local speed of light. This is often case in spherically symmetric configurations.

Question. Is it possible to send a signal faster than the local speed of light?

Taking apart a compact object "quickly"

Suppose that our matter shell is so massive that the local speed of light (measured in global coordinates) is very slow.

Let us expand the shell quickly. The local speed of light becomes much faster and we can send a signal to an observer at the center of the shell in a reasonable time.

We were able to "melt" a frozen object quickly and then we can take it apart in a short time (time measured by a distant observer).

However, if we have very many shells nested inside each other, and each is close to being a black hole, then the method we used above will not work quickly.

Disassembling a black hole through overspinning

What about disassembling a Schwarzschild black hole by overspinning it?

Suppose that we can speed up the local flow of time inside the black hole by rotating it fast. That does not necessarily involve any signal which would travel faster than the local speed of light. Once time flows fast inside the black hole, it may be possible to take it apart.

However, spinning is not a spherically symmetric operation. An observer inside the black hole might receive a signal faster than what is the local speed of light.

The centripetal force versus newtonian gravity

On July 13, 2023 we suggested that the "true" radius (in global coordinates) of an extremal spinning black hole is

        R = sqrt(2) G M / c²,

where M is the total energy of the black hole, including the kinetic energy in rotation.

Furthermore, we suggested that the horizon at the equator is moving at a speed

       v = c / sqrt(2).

Let us have a particle, for example, a photon rotating along the horizon. We calculate the centripetal force and compare that to the newtonian gravity force.

https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit

According to Wikipedia, the innermost stable orbit and the photon sphere are located at the horizon for an extremal spinnning black hole.

The centripetal acceleration in special relativity is the same as in newtonian mechanics:

       v² / R = 1 / (2 sqrt(2)) * c⁴ / (G M).

Let us assume that only the irreducible mass of the spinning black hole generates newtonian gravity. Then the acceleration of gravity is

      1 / sqrt(2) G M / R²

      = 1 / (2 sqrt(2)) * c⁴ / (G M).

The accelerations match. Our newtonian model explains why there is a stable orbit at the event horizon.

Conclusions

We showed that a change in the metric of time can propagate at an infinite speed. But that cannot carry any signal.

We are not sure if general relativity allows a signal to propagate faster than the local speed of light, if the speed of light is slowed down by a deep gravitational potential.

Next we will study orbits around an overspun black hole. Does the matter fly out?

Does a rapidly spinning black hole feel a "centrifugal force"?

We are being hopeful that overspinning a black hole would disassemble it. A "centrifugal force" would throw away the matter which fell in.

If no signal can overtake a photon, then disassembling a Schwarzschild black hole does not seem to succeed

Suppose that we send a symmetric photon sphere down a Schwarzschild black hole. Let us then overspin the black hole. It looks like that no change in the metric can overtake the photons. The metric below the forming horizon, and all the matter there remains exactly like in the Schwarzschild black hole. Nothing can ever come out.

Movement by a "translation"

If no change in the gravity field can overtake a photon, and the local speed of light (in Schwarzschild coordinates) is very slow inside a heavy neutron star or a black hole, then it may take very long for anything to change inside the neutron star or black hole.

But we must be able to move the neutron star or a black hole in a binary star. We have suggested that a "translation" implements the movement. The idea is that in the Einstein-Hilbert action we accept paths where a part of the system moves as a whole. Moving as a whole means that all matter and the metric move as is. An observer inside the translated volume does not notice any change.

Let us start to rotate a Schwarzschild black hole. Then the inside of the black hole and a thin layer above the forming horizon would move through a translation.

A translation can be understood as "total" frame dragging.

A translation in the context of electric fields

Suppose that we have a block of an electric insulator. We assume that the speed of light is very slow inside the insulator, but the speed of sound in the material is close to the speed of light in vacuum. The block is electrically charged.

                   |

                   | field line

                 ___

               /  +    \  _____

               \____/                      ● -   Q charge

            insulator

The electric field of the charge Q interacts of with the electric field outside the insulator and pulls the block as a whole. It is not necessary that the field of Q meets with the charge inside the insulator at all.

The insulator is very much electrically polarized. The charge Q actually interacts with the induced charge close to the surface of the insulator.

Here we have a model where the center of the insulator remains oblivious of the field of Q, but the center is pulled along as the block starts to move.

The center is pulled by stress forces (pressure) within the insulator. In this model, there are faster than light signals.

Stress forces (sound) propagate much faster than electromagnetic fields inside the insulator.

What if the speed of sound is slow, too?

Then the inside of the block stays static while the surfaces start to move. The charge and the matter inside the block is pressed against one side of the block.

The movement of the block would appear strange to an external observer.

An aside: interpreting the Schwarzschild gravity through polarization

In our example of an electrical insulator, polarization has two effects:

1. it slows down light,

2. it reduces the electric field inside the insulator.

For gravity, we have item 1, of course.

The gravity potential in the Schwarzschild solution is steeper than in newtonian gravity. Hypothetical polarization draws mass-energy in the surrounding empty space closer to the mass. The polarization would then make gravity stronger.

A translation in our own Minkowski & newtonian gravity model

In general relativity, gravitational forces and fictitious acceleration forces, like the centrifugal force, are treated as a whole, in the metric. The Moon orbits Earth in the Schwarzschild metric. There is no separate gravity force and centrifugal force.

Our own gravity model claims that gravity is an ordinary force and does not have a special status. There have to exist separate fictitious acceleration forces.

                        ___

                      /       \

                      \____/

             compact, massive

                      object

Hypothesis. In a compact massive object, all parts acquire a large amount of inertia from other parts. It is like a gearbox where turning any cogwheel involves inertia from other cogwheels. Different parts of the object can communicate through the inertia mechanism with the "combined" field of the "whole object". The communication speed is the speed of light in the spatial volume surrounding the object.

If we have a compact object, we can make it to move as a whole by exerting an external force. This is obvious from binary neutron stars and black holes.

The idea in the hypothesis is that the gravity of the object itself cannot slow down communication with the "combined" field of the "whole object".

But if the object is in the gravity field of another object, then the communication is slowed down by the field of the other object. This preserves an equivalence principle.

An object cannot use a "Baron Munchausen" trick of slowing down its reaction to external forces. Such a trick might break conservation laws. But the object can slow down its internal processes through the inertia which it imposes on its parts.

Our hypothesis is vague. We have to find a precise mathematical formulation.

No communication is allowed to happen faster than the speed of light of the surrounding empty space. In this way we avoid time travel paradoxes.

Accelerating a compact object in the Minkowski & newtonian model

Let us then assume that we, by some means, are able to accelerate (linearly) a compact object. A crucial question is if an observer inside the object can feel an acceleration, and how quickly.

                        ___

                      / • m\    --->  acceleration

                      \____/

             compact, massive

                      object

Let us have a test mass m inside the compact object. The test mass m has acquired a lot of inertia from the field of the object. It sounds natural that the acquired inertia tends to move with the compact object.

What about the inertia of the mass m itself? That inertia would probably cause m to lag behind. But since all the surrounding matter behaves in the exact same way, an observer riding on m would not notice anything.

A linear translation in the Minkowski & newtonian model looks very much like the one in general relativity. However, there is slight squeezing and compression because the amount of inertia depends on how far the test mass m is from the center.

If we start to rotate the object, there is an associated fictitious centrifugal force. If the object is strong enough not to come apart, an observer feels a centripetal acceleration. If the object starts to disintegrate, then an observer sees radial distances of particles start to grow.

Conclusions

If we assume that in general relativity no change in the metric can overtake a photon, then it looks like one cannot disassemble a Schwarzschild black hole by overspinning it.

Textbooks on general relativity are vague about if a change of the metric can spread faster than a photon.

In our own Minkowski & newtonian gravity model, it might be possible to disassemble the black hole. This requires that the inertia of the field can communicate at the light speed of the surrounding space. It must be able to communicate faster than the local speed of light.

We have to look further into the Kerr metric of extremal spinning black holes. That may offer us clues if an overspun black hole really comes apart.

Overspinning an electrically charged sphere versus a black hole

Suppose that we have a sphere with a very large electric charge Q. The sphere is an electric insulator, so that the charge must rotate along the sphere.

Overspinning an electrically charged sphere

                        |

                        |

                      ___

          ____   /       \ Q   _____ E electric field

                    \____/

                     <---  rotation

                        |

                        |

Let us assume that the electric field outside the sphere contains mass-energy and inertial mass such that the energy density D is

       D = 1/2 ε₀ E²,

where E is the electric field strength and ε₀ is vacuum permittivity.

It is like a carousel where a substantial mass is outside the sphere. The far field of the sphere will move at the speed of light, or almost at the speed of light. There is a substantial "centrifugal" force which tries to pull the charge Q out from the sphere.

We can increase the centifugal force by adding more energy into the rotation of the field, since then the mass-energy of the field will be larger. Eventually, the centrifugal force will pull the charge Q out from the sphere.

If our sphere is analogous to a black hole in gravity, then over-spinning a black hole is a way to disassemble the black hole. 

A spinning black hole

https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit

In an extremal black hole, orbits to the same direction as the black hole spins (prograde orbits) can touch the horizon. A "centrifugal" force seems to be at play.

https://en.wikipedia.org/wiki/Kerr_metric

The Kerr metric in the usual Boyer-Lindquist oblate spheroidal coordinates is:

We can move to a rotating coordinate system to diagonalize the metric:

       φ' = φ - Ω t.

Let us assume that c = G = M = 1, and calculate the values for an extremal spinning black hole. Then r_s = 2, the horizon is at r = 1, and a = 1.

The angular speed in the equatorial plane (θ = 90 degrees), relative to the coordinate time t is then

       Ω = 1 / 2.

We can map the oblate spheroidal coordinates to cartesian coordinates:

The cartesian (x, y, z) coordinate radius of r = 1 is

       sqrt(r² + a²) = sqrt(2).

The cartesian coordinate speed of the rotation at r = 1 is then

       sqrt(2) / 2.

It is 71% of the speed of light, as seen by a faraway observer.

Let us then slow down the black hole to a standstill, so that a = 0. The new mass is only sqrt(2) / 2, because 29% was "reducible" mass. The cartesian radius of the event horizon is then

       2 * sqrt(2) / 2 = sqrt(2).

The radius stayed the same when we slowed down the rotation.

The circumference of an extremal black hole

https://arxiv.org/abs/2112.15441

Francesco Sorge (2021) calculated the rotating, diagonalized metric explicitly:

Let us determine the equatorial circumference of the event horizon of the extremal black hole in these rotating coordinates.

In the metric, the ratio

     A / Σ = 4

holds the usual place of the square of the radius. The radius, if defined from the circumference, is 2.

The radius is 2 in the nonrotating Boyer-Lindquist coordinates, too.

But in the cartesian coordinates, the radius is only sqrt(2).

Which is the "true" radius? We can stop the rotation, and the black hole becomes a Schwarzschild black hole whose radius is sqrt(2) in the Schwarzschild coordinates.

https://en.wikipedia.org/wiki/Ehrenfest_paradox

The Ehrenfest paradox concerns the circumference of a fast rotating cylinder.

The speed of the horizon is c / sqrt(2) in the cartesian coordinates. The length contraction factor is

       1 / γ = sqrt(1 - v² / c²)

                = 1 / sqrt(2).

The proper length of the spinning horizon would correspond to a radius 2. This matches the value in the Boyer-Lindquist coordinates. Maybe we should call sqrt(2) the "true" radius?

https://www.eftaylor.com/pub/SpinNEW.pdf

This paper by Edwin F. Taylor gives the speed of prograde light in "bookkeeper" coordinates as 1.0, which presumably means the speed of light in those coordinates.

Accelerating a Schwarzschild black hole: a new Ehrenfest paradox?

The speed of light close to the horizon of a Schwarzschild black hole is extremely slow. If we start rotating the black hole, observers close to the horizon do not learn about the spinning until much later. How can the metric stay unchanged close to the horizon, while the metric is length-contracted a little farther away? Do we have here a new Ehrenfest paradox?

A similar process happens when we linearly accelerate a black hole. Points close to the event horizon do not know about the acceleration until much later – but a faraway observer sees the entire black hole to length-contract. How is this possible?

If we accelerate a ruler, observers sitting on the ruler feel an acceleration. The spatial metric measured by the observers does not contract significantly, while static observers see it to contract. There is no paradox there.

Let us use comoving coordinates inside the black hole, and static coordinates at some distance from it. There does not need to be any paradox. Observers inside the black hole are "frame-dragged", and do not feel any acceleration. Their comoving spatial metric does not change in any way.

Conclusions

We now have a grasp of what happens when we start to rotate a Schwarzschild black hole.

We still need to analyze if we really can over-spin the black hole using ropes. Then analyze if the black hole starts to disassemble through over-spinning.

Also: how fast does the new metric propagate toward the horizon? If the propagation never reaches the falling matter, then we cannot eject the matter out. However, this would be strange, since how could the black hole store excess angular momentum in that case?

If a neutron star contains too much angular momentum to form a black hole, a computer simulation showed that it will form a torus. The crucial question is if an Oppenheimer-Snyder collapse is irreversible. There is no proof that it would be. We conjecture that over-spinning will reverse an Oppenheimer-Snyder collapse.

https://www.jstor.org/stable/1968902

Albert Einstein's 1939 paper on rapidly spinning Einstein clusters may offer a clue. Over-spinning a black hole may turn it into an Einstein cluster.

https://arxiv.org/abs/1006.1764

Ted Jacobson and Thomas P. Sotiriou (2010) write about destroying a black hole by over-spinning it. They correctly note that it is hard to deduce what is the dynamical process like if we start to over-spin an extremal Kerr black hole.

Spontaneous superradiance does not exist

Yakov Zeldovich conjectured in 1971 that a rotating body can excite "vacuum fluctuations", giving them energy, and in that way create real quanta. Zeldovich discussed the idea with Stephen Hawking, who was inspired to develop the hypothesis of Hawking radiation (1974).

https://arxiv.org/abs/1609.06723

Solomon Endlich and Riccardo Penco (2016) write about a superradiant spontaneous emission.

There is no reason why a certain phase of a probability amplitude would be preferred: the sum is zero

The mechanism would work if we would have an individual "quantum fluctuation". Imagine a pair of photons, one of which has positive energy and the other negative energy. Such photons are allowed in Feynman diagrams and they can "exist" for a short time.

                                            real photon

                  virtual              ^

                  pair                 /

                    -----------------> --------> real photon

                     \_________/

                           ___        interaction

                         /       \     

                         \____/

                           <---

                 

             rotating metal cylinder

The rotating object then interacts with the positive energy photon, giving it more energy. The pair recombines. Since the pair has now more energy, it can become a real pair of photons.

However, it is not right to calculate the mechanism for an individual pair. We have to sum the probability amplitudes (or photon phases) for all histories. Let us consider a photon leaving at a certain time t to a certain direction p.

There is no reason why a certain phase φ of the probability amplitude should be preferred. Probability amplitudes with different phases cancel each other out. The sum is zero. There is no radiation.

We are allowed to sum the probability amplitudes of different histories if the macroscopic part of the system does not retain information of which history happened. The classic example is the double slit experiment, where the walls of the slits do not preserve information about which slit the photon passed through. That is why we can sum the probability amplitudes on the screen.

In the case of a rotating cylinder there is no clear mechanism which would tell us which of various histories happened.

A rotating cylinder does add energy to a laser beam which is directed toward it in a suitable angle. Superradiance, in that sense, does exist.

Conclusions

Spontaneous superradiance has never been observed, and it should be very weak if it exists. Our argument shows that the process cannot exist with a perfectly round rotating cylinder.

What happens if the cylinder is not perfectly round?

We cannot over-charge a Schwarzschild black hole

In the previous blog post we found a way to add angular momentum to a black hole by "grabbing" its gravity field.

Can we figure out a similar trick with which to over-charge a black hole which is extremal with respect to an electric charge?

The Reissner-Nordström metric of an electrically charged black hole

https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric

Here M is the mass-energy of the black hole, Q is its electric charge, G is the gravitational constant, c is the speed of light, and ε₀ is vacuum permittivity.

The event horizons are where the metric for dr diverges:

For an extremal black hole, the formula within the square root is zero. We see that the radius of the event horizon is exactly a half of the corresponding uncharged black hole with the same mass-energy M.

The mass-energy of the static electric field

https://journals.aps.org/prd/abstract/10.1103/PhysRevD.4.3552

Demetrios Christodoulou and Remo Ruffini (1971) state in the abstract that 50% of the energy of an extremal charged black hole can be extracted.

                 _____

               /           \

             |      ● black hole Q

               \______/

               shell -Q

Let the charge of the extremal black hole be Q. We put a zero mass charged spherical shell with the opposite charge -Q around it and lower the shell slowly down. In this way we can extract the energy of the electric field of Q.

What is the energy of the electric field, from the viewpoint of a faraway observer?

The time component g₀₀ of the metric tensor is:

At the coordinate radius r, the time runs slower, causing a "redshift of energy" by the factor

       sqrt(g₀₀),

but that is compensated by the stretching of the radial metric of r, which makes a volume element larger by the factor:

       1 / sqrt(g₀₀).

The energy of the electric field, as seen by a faraway observer, is as if the metric on the time t and the radial distance r would be Minkowski.

Let us calculate the energy of the static electric field for an extremal black hole. An extremal black hole has 2 r_Q = r_s, which implies

       Q = sqrt(G / k) * M.

We denote Coulomb's constant by

       k = 1 / (4 π ε₀).

The electric field strength E at a distance r is

       E = k Q / r².

The energy density D is

       D = 1/2 ε₀ E²

           = 1 / (8 π k) * E²

           = k / (8 π) * Q² * 1 / r⁴.

The electric field energy W from r to infinity is

       W = k / (8 π) * Q²

                     ∞

               *  ∫ 1 / r⁴  * 4 π r² dr

                  r

            = k / 2 * Q² * 1 / r.

For an extremal black hole, we have

       Q² = G / k * M²,

and the event horizon is at r = G M / c².

       W = k / 2 * G / k * M² * c² / (G M)

            = 1/2 M c².

We conclude that the static electric field contains exactly 1/2 of the whole mass-energy of the extremal black hole.

Let us then calculate the energy of the electric field per a short distance dr:

       dW / dr = k / 2 * Q² * 1 / r²

                      = k / 2 * G / k * M² * 1 / r²

                      = 1/2 G M² / r².

At the extremal black hole event horizon r = G M / c², we have

       dW / dr = 1/2 c⁴ / G,

or in terms of the mass of the electric field,

       dm / dr = 1/2 c² / G.

For an uncharged black hole, the Schwarzschild radius is

       r = 2 G M / c²,

or

       M / r = 1/2 c² / G.

The ratio M / r tells us how much mass we have to add if we want increase the Schwarzschild radius by a unit length.

We see that the ratio dm / dr at the horizon of an extremal black hole is almost enough to increase the Schwarzschild radius.

We get the following diagram for the extremal black hole whose total mass is M:

         |           M / 2           |                 M / 2 ...

      r = 0                     r = G M / c²

          irreducible mass    mass of electric field

The horizon r = GM / c² is at the location where it would be if we would remove the charge and the associated electric field energy, leaving just 1/2 M of the mass in the system.

Can we use over-charging to shrink the horizon of an uncharged Schwarzschild black hole?

Short answer: no.

Let us start from a neutral Schwarzschild black hole whose Schwarzschild radius is R. We assume that we have a shell of matter whose entire mass-energy is in its static electric field. Its "intrinsic" mass is zero.

We charge the black hole by pushing in these charged shells. The black hole under the newly built electric field does not change.

Let us assume that we are able to over-charge the black hole. We assume that we have used concentric shells whose charges are Q and -Q, to add the new electric field only to the radii

       R ... R + dR,

where dR is small. The configuration is like this:

               Q             -Q

               | |              |

               R              R + dR

           event

         horizon

But now the black hole and the static electric field between R and R + dR contain too much mass-energy, as seen by a distant observer. The event horizon has advanced to R + dR or beyond.

If we would have the electric field also from R + dR to infinity, that does not change the fact that the event horizon has risen up. An event horizon in a spherically symmetric system is not affected if we add mass outside the sphere, at least as long as the horizon does not rise outside the spherically symmetric system.

Our Minkowski & newtonian viewpoint on overcharging

Let us analyze the problem using our own model of gravity. We ask if we can destroy a Schwarzschild black hole by pushing electric charge in.

The answer is obviously no. The volume inside and very close to the forming event horizon is essentially "frozen". Clocks tick extremely slow there, and the local speed of light is very slow, as seen by a faraway observer.

Pushing electric charge to the surface does not produce any force inside the black hole and cannot change anything. One cannot use that method to shrink the event horizon.

Conclusions

Stephen Hawking and other researchers claimed in the early 1970s that it is impossible to shrink the event horizon of a black hole. They, certainly, looked at over-charging a black hole, too.

Our previous blog post suggests that one may be able to disassemble a black hole by over-spinning it. A "centrifugal force" may be the way to reverse the direction of the force close to the horizon and make objects, which were falling in, to fly out. If this is the case, then it is a big step forward in showing that black holes do not contain a one-way membrane, and behave like any normal system in everyday physics.

We can overspin a black hole by pulling ropes at almost the speed of light around it?

How to "overspin" a rotating black hole? If the value of the parameter

       a = J / M

in the Kerr metric exceeds a certain threshold, then the event horizons disappear. In the formula, M is the ADM mass of the black hole and J is its angular momentum.

https://en.wikipedia.org/wiki/Kerr_metric

Various authors have tried to prove that you cannot over-spin a black hole by throwing light or other material into it.

https://www.eftaylor.com/pub/SpinNEW.pdf

Let us have a black hole for which the quantity J / M is at the threshold value. Such a black hole is called extremal.

In the link we have a paper, probably written by Edwin F. Taylor, where Figure 2 seems to indicate that the tangential speed of light to the spinning direction of an extremal black hole is less than the speed of light far away.

Pulling a Schwarzschild black hole with a rope

             ___

           /       \  __________  rope

            \___/                -->

       black hole           F

Let us lower a strong rope very close to the horizon of a Schwarzschild black hole.

The radial metric is stretched, but that does not affect our reasoning. The speed of light is very slow at the left end of the rope, as measured by a faraway observer. The end of the rope is "stuck in the syrup" of the gravity field of the black hole.

Let us pull the rope with a large force F for a time t. We give to the system the black hole & the rope a momentum

       p = F t.

We give it the energy

       E = F v t,

where v is the average speed of the right end of the rope. How is the momentum distributed between the black hole and the rope?

Let the mass of the rope be m. Since the black hole weighs a lot, the rope receives almost all of the energy E.

Let v' be the final velocity of the rope when it is completely far away from the black hole. We have

       E = F v t  >  1/2 m v'²,

because the rope had some negative potential energy.

       p' = m v'  <  sqrt( 2 F v t m ).

If v is "small", then the black hole receives almost all of the momentum p.

We get another estimate from the fact that v' < c. The final momentum of the rope is at most

        p'  <  m c  +  F t v / c.

If m is very small, then p' / p < v / c. If such a lightweight, very strong rope exists, we can use it as a tether to pull a black hole around.

Let us analyze what is the fundamental mechanism of pulling the black hole. The end of the rope acquires a lot of inertia from the black hole. When we pull the rope, we pull that inertia, too. When the rope moves away, the momentum absorbed by that inertia remains with the black hole.

Accelerating the spin by pulling on ropes whose end is close to the black hole

Let us put ropes around the black hole and pull them very fast, at almost the speed of light (the light speed far away).

                                      F

                                     ---> pull

              -----------------------  rope

                             r

                

                             ●  rotating extremal black hole

                           <-- direction of rotation

      rope ------------------------

      pull <---

                F

Does the field of the black hole resist the pull? We are not sure. The radial metric is stretched by the black hole, and clocks tick slower. Does this contribute some resisting force for the pulling?

Let us assume that the black hole resists the pull of each rope with a force F.

Let us calculate the mass-energy and the angular momentum which we input to the system black hole & its gravity field.

The mass-energy m which we input is approximately

       m = 2 F c t / c² = 2 F t / c,

where c is the speed of light and t is the time which we pull.

The angular momentum j which we input is approximately

       j = 2 F t r,

where r is the distance of rope ends from the black hole. The ratio

       j / m = c r.

By choosing r large we can make the ratio as large as we like. 

Does this guarantee that we can push the ratio J / M of the whole system over the threshold?

Not necessarily. As we speed up the system, it may lose some angular momentum in gravitational waves. We need to analyze this in more detail.

Lowest orbits around an extremal spinning black hole touch the horizon: can we disassemble the black hole?

https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit

According to Wikipedia, the innermost stable circular orbit (ISCO) touches the horizon of an extremal black hole.

Maybe adding still more spin, making the system superextremal, then causes particles to fly away from the forming horizon?

We have been claiming that the event horizon of a Schwarzschild black hole is never "completed" because clock slow down so much when close to the forming horizon.

In principle, it should be possible to reverse the process and make the falling matter to fly away. Over-spinning an extremal black hole might do the trick.

Once the topmost matter flies away, lower layers of the forming black hole are exposed. If we continue over-spinning the system, we may be able to disassemble the forming black hole entirely.

Conclusions

We need to study overspinning in detail. Is the excess angular momentum carried away in gravitational waves?

https://en.wikipedia.org/wiki/Black_hole_thermodynamics

If over-spinning really is a mechanism to disassemble a black hole, that refutes black hole thermodynamics which was developed by Jacob Bekenstein and others in the early 1970s. The entropy of a black hole can be quite low if it is built from large chunks of matter.

A naked singularity? The gravity of a cylinder

In newtonian gravity, an axially symmetric cylindrical uniform dust cloud obviously collapses into a line segment singularity. If the cylinder does not contain too much mass per unit length, then it probably does not become a black hole in general relativity. Could it be that in general relativity, too, there is a singularity there, and the singularity is naked since it is not hidden behind an event horizon?

Shapiro and Teukolsky (1991)

https://core.ac.uk/download/pdf/216288565.pdf

Stewart Shapiro and Saul Teukolsky (1991) used a computer program to calculate a dust collapse where the dust spheroid is elongated (prolate). The results suggested that a "spindle" singularity forms without an apparent horizon.

The example is contrived, though. Nature does not contain infinitely fine dust particles. If we assume classical point particles, then each particle is a tiny black hole. In principle, we can order these tiny black holes on a straight line, and their horizons do not touch. Thus, at a microscopic level, our cylinder would not be a cylinder at all, but a string of tiny beads.

Division of classical matter into elementary particles prevents naked singularities?

We can extend our argument in the previous section to all naked singularities which are supposed to contain infinitely dense matter. If each elementary particle is a tiny black hole, then infinitely dense matter presumably contains many such black holes which have merged together. The singularity, if any, is behind their event horizon. It is not naked.

Is a naked singularity a problem at all in classical physics?

People have done a lot of general relativity calculations with "domain walls", which are assumed to be infinitely thin massive walls. Classically, there is nothing pathological in such a hypothetical structure.

Some people assume that the center of a (classical) Schwarzschild black hole contains a point which contains all the matter in the black hole. Is there anything wrong with such a model in classical physics? As long as it is a point which is solely characterized by it mass M, there should be no problem.

In quantum mechanics, infinitely dense objects may be problematic, though.

Conclusions

If classical matter consists of elementary particles, then there seem to be no classical naked singularities.

If there exists infinitely fine dust, then, for example, the Shapiro and Teukolsky setup may create a naked singularity. However, there probably is nothing contradictory or problematic in such an object.

In quantum mechanics, an electron is often thought of as a point particle which is not a black hole, though. Quantum mechanics does not play well together with objects which are infinitely dense. The existence of singularities or naked singularities in quantum gravity is a subject which we will not touch here.