The frozen star model of a black hole

In earlier blog posts we have claimed that a Schwarzschild black hole is essentially a "frozen star".

The frozen star model of a Schwarzschild black hole

Near the forming horizon of a Schwarzschild black hole, clocks tick extremely slowly, and the outermost shell of falling matter never quite reaches the forming horizon.

Inside the outermost forming horizon there may be other forming horizons from earlier matter shells which were dropped into the black hole.

In the case of an Oppenheimer-Snyder (1939) collapse, the collapse inside the forming horizon happens extremely slowly, and the surface of the dust ball never reaches the forming horizon.

Frozen in time

The basic idea is that at a low gravitational potential, local time runs ever slower. A black hole essentially is a collapse frozen in time.

In the syrup model of gravity, the viscosity of the syrup increases without bounds.

A global time in the system

The Schwarzschild external metric is static and it is easy to define a global time. Static observers at various places send light signals to each other, and use them to synchronize their clocks. Local time runs the slower, the closer we are to the horizon. The global time is the Schwarzschild coordinate time.

Inside the forming black hole, defining a global time is harder, and we do not have a mathematical formula for it. Since observers very close to each other can still send light signals to each other, we hope that a global time can be defined transitively: if adjacent observers 

       Aₙ and Aₙ₊₁

can synchronize their clocks, then we may be able to define a global time for a chain of observers

       A₀, A₁, A₂, ...

We believe that at a fundamental level, the system resides in Minkowski space, and distortions in clock speeds and distances are an artefact of the (newtonian) gravity field. If our hypothesis is true, then a global time can be defined: take any global time in the Minkowski space.

We conjecture that in the global time, the system essentially freezes.

Avoiding a singularity

An electron and a positron cannot fall into an "electric singularity" because they annihilate.

Classic newtonian gravity does allow a singularity to form. But if we add more complex effects of the gravity field, then we end up with the frozen star model, and avoid a singularity. General relativity can be regarded as nature's way of avoiding singularities – even though Roger Penrose and many other researchers in the 1960s believed that general relativity produces singularities.

Disassembling a black hole

If a black hole is a frozen star, it might be possible to disassemble it. Disassembling a singularity sounds like a hard thing – melting a frozen star sounds much easier.

So far, it looks like overspinning a black hole will disassemble it. The (fictitious) centrifugal force wins newtonian gravity. Matter inside the black hole will possess a lot of angular momentum and will be ejected away. The disassembling process might be quite quick, just like a collapse is.

We believe that we can overspin a black hole, using the rope pulling trick which we sketched on July 8, 2023.

Conclusions

Defining a global time is an essential step in proving that the frozen star model can work in general relativity. We will look at that problem in the future.

If we can disassemble a black hole, then a black hole is not an exotic object. A black hole is like a lump of syrup, nothing more. Exotic hypotheses about black hole thermodynamics, entropy, wormholes, white holes, string theories, and so on will be refuted; or, at least, those hypotheses become very improbable.

The black hole information paradox becomes a non-paradox: we can recover the fallen information from a black hole.

Effective potential in Schwarzschild metric is newtonian for reduced inertia

In this blog we have been claiming that general relativity is actually newtonian gravity if we take into account all effects of the newtonian gravity field.

Clocks slow down in a deep gravitational potential. We explained that by claiming that all particles acquire a "package" of extra inertia from the newtonian gravity field. The extra inertia might be explained by "field energy" moving around.

The gravity field of the central mass M is static. When a charge moves in a static electric field, we can ignore the magnetic field. That suggests that we can ignore "magnetic gravity" effects in the treatment below.

A modified newtonian effective potential

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

The effective potential in the Schwarzschild metric is

Let us assume that the mass m of the test particle is very small compared to the mass M of the central mass. Then

       μ = m

and

       M + m = M.

We can write

       L = p r,

where p is the momentum of the particle in the tangential direction.

Then the centrifugal potential in newtonian gravity is

       L² / (2 m r²)  =  p² / (2 m).

A faraway observer sees the particle to carry an inertia of

       m / sqrt(1 - r_s / r),

where r_s is the Schwarzschild radius.

Let us assume that only the momentum p' associated with the original inertia m "takes part" in the generation of the fictitious centrifugal force:

       p' = p * sqrt(1 - r_s / r).

The centrifugal potential in this modified newtonian gravity is

       p² / (2 m)  *  (1 - r_s / r)

       = L² / (2 m r²)  *  (1 - r_s / r)

       = L² / (2 m r²)  -  L² * 2 G M / (2 m r² c² r)

       = L² / (2 m r²)  -  G M L² / (c² m r³).

The formula for the centrifugal potential agrees with the one for the Schwarzschild metric.

The tangential velocity of the particle, as seen by a faraway observer

The above formula gives a constant of motion for the particle.

Let us assume that the particle is on a very eccentric orbit. What is its Schwarzschild coordinate velocity at the perihelion?

We assume that m is much smaller than M, which implies μ = m.

The proper time dτ is measured by an observer riding on the particle. To get the proper time dτ from the coordinate time dt we have to multiply dt by the factor

        1 / γ = sqrt(1 - v² / c²),

where v is the velocity measured by a local static observer. We also have to multiply dt by

        sqrt(|g₀₀|) = sqrt(1 - r_s / r),

where r_s is the Schwarzschild radius.

We have

      L = m r² dφ / dt  * 1 / sqrt(1 - r_s / r)

                                   * 1 / sqrt(1 - v² / c²).

Let us try to interpret the formula. The coefficient

       1 / sqrt(1 - r_s / r)

we explained with the model where the extra inertia acquired by the particle steals its fair share of the angular momentum.

The Lorentz factor is easy to understand:

        γ = 1 / sqrt(1 - v² / c²).

The kinetic energy of the particle gets its fair share of the angular momentum.

Question. Why the Lorentz factor does not show up in the effective potential formula? Could it be a result of the stretching of the radial metric? The stretching of the metric balances the fact that the tangential speed of the particle is reduced when kinetic energy takes its fair share of the tangential momentum?

The centrifugal force and newtonian gravity in circular orbits

Above we calculated that effective potential is like in a modified newtonian gravity where extra inertia has stolen some momentum from the particle, and only the momentum associated with the particle mass m takes part in the fictitious centrifugal force.

https://physics.stackexchange.com/questions/761407/orbital-velocity-formula-in-schwarzschild-metric

The user "ProfRob" calculates the coordinate speed for a circular orbit in the Schwarzschild metric. It agrees with newtonian gravity.

Conclusions

The above results support our view that general relativity is actually newtonian gravity where we take into account also complex, non-intuitive effects of the newtonian gravity field – particularly the extra inertia that a test body acquires in a low potential.

For example, a black hole captures any particle which endeavours below 1.5 Schwarzschild radii. This is because the extra inertia of the particle takes its fair share of the momentum of the particle, and the particle moves then so slowly (measured in Schwarzschild global coordinates) that the centrifugal force can no more win the newtonian gravity attraction.

We may imagine that there is "syrup" in low potentials. The particle is slowed down by the syrup, so that newtonian gravity wins the centrifugal force.

This may help us in analyzing what happens when we overspin a black hole. Newtonian gravity is much easier to analyze than general relativity.

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.

Perturbation from a small mass falling into a black hole

In our blog posts on June 11 and 16, 2023 we analyzed an electric charge close to the horizon of a Schwarzschild black hole, and tried to study a small mass falling into the horizon.

https://journals.aps.org/prd/abstract/10.1103/PhysRevD.8.3259

Hanni and Ruffini (1973) calculated the electric lines of force for a charge close to the horizon. The result is that the closer the charge is to the horizon, the more spherically symmetric is the electric field, where the origin is the center of the black hole.

Richard H. Price's thesis (1971)

https://thesis.library.caltech.edu/5526/

https://journals.aps.org/prd/abstract/10.1103/PhysRevD.5.2439

https://blackholes.tecnico.ulisboa.pt/gritting/pdf/black_holes/Richard-Price_Nonspherical-Perturbations-of-Relativistic-Gravitational-Collapse-I-Scalar-and-Gravitational-Perturbations.pdf

Richard H. Price (1971, 1972) calculated perturbations of a Schwarzschild black hole using a wave equation for a scalar field. He assumes that such a field gives us a clue of what happens with an electric charge or a small mass falling into the horizon.

His thesis claims that only the monopole field of an electric charge can survive. The monopole field is the spherical harmonic (0, 0), centered at the center of the black hole. All the other spherical harmonics are "reflected" back by the metric.

                         ___

                       /        \   black hole

                       \____/

                          •  •

                      +Q   -Q

What does this reflection mean? We use as an example a pair of charges, +Q and -Q placed close to each other near the horizon. Let us use these charges as a dipole antenna. We try to send very long radio waves, and move the charges extremely slowly.

The diagrams of Hanni and Ruffini (1973) show that the electric fields of the charges cancel each other out almost entirely at some distance from the horizon. Both fields are almost perfectly spherically symmetric.

We conclude that long radio waves cannot get very far: they are "reflected" back. If a very long wave is reflected from a potential wall with a 180 degree phase shift, then destructive interference cancels the wave close to the wall.

Note that if we move the charges very quickly, generating very short waves, our geometric argument of canceling of spherically symmetric electric fields does not work: the fields get substantially distorted.

Did we "prove" that the electric field of a black hole is the simple spherically symmetric field? No, because the interplay of gravity and electromagnetism could produce a surprise. There is no proof that the Einstein field equations have a solution at all for such a complex setup. Solutions for nonlinear equations could diverge.

However, if we assume that the charges and their electric field does not affect the Schwarzschild metric in any way, then it is obvious that the result of Hanni and Ruffini shows that the field is almost symmetric for a finite number of charges falling into the black hole.

Using the merger of two black holes as a model

In our June 26, 2023 blog post we studied the merger of two black holes of comparable sizes. Let us then assume that one of the black holes is microscopic and the other is a large Schwarzschild black hole.

                  ___

                /        \    <--  • microscopic black hole

                \____/         

            black hole

The merger process probably is qualitatively similar to one with two large black holes. In the Schwarzschild coordinates, the small black hole is very much squeezed in the radial direction. The two event horizons will remain separate until the microscopic black hole is within the Schwarzschild diameter of the combined system. Once within, a common event horizon quickly grows around the whole system.

What about a (classical) elementary particle falling in? A classical particle, actually, is a microscopic black hole.

Let us then consider a small body of "continuous" matter, with no point particles. In the Oppenheimer-Snyder (1939) collapse we have an analogous process. The surface of the dust ball never falls within the Schwarzschild radius of the whole system. Any internal point at a reasonable distance > 0 from the surface quickly is enclosed inside the forming event horizon.

                    ___

                  /        \   <-- ●P   continuous body 

                  \____/         

               black hole

We can guess that the same is true for the small body of matter. Let P be the point on the surface of the body, such that its radial distance is the largest. We expect P to remain outside the new forming horizon for ever.

Sometimes it is claimed that a falling particle can never cross the event horizon. That is true for a zero-mass idealized particle, but seems not to be true for a real particle.

Perturbation of the event horizon

https://arxiv.org/abs/2109.09514

Riccardo Bonetto, Adam Pound, and Zeyd Sam (2021) calculated the "tidal effect" on the event horizon of a Schwarzschild black hole M when a smaller black hole, 0.1 M is orbiting it.

                         <------ counterclockwise orbit

They try to explain why the bulge of the event horizon of the large black hole is ahead of the the orbiting small black hole. In the diagram (Figure 2) in the paper, the small black hole orbits counterclockwise. The figure uses the advanced time coordinate. The dotted line points to the small black hole. The dashed line points to the maximum deformation.

We saw a similar effect in the animation made by Teresita Ramirez and Geoffrey Lovelace (2018). We interpreted that the squeezing of the large black hole happens in the global pseudo-"Schwarzschild" coordinates whose origin is the center of mass of the system. The small black hole causes the radial metric to stretch (that is, g₁₁ becomes larger, if we use the (- + + +) signature).

In the coordinate time of the large Schwarzschild black hole, the trip of light from the orbiting black hole takes a very long time. Presumably, the horizon never "knows" that the small black hole has started to orbit the large black hole. If we assume that there are no superluminal signals, then the metric close to the horizon can never change. Thus, the form of the horizon cannot change for local observers at all. The horizon is perfectly rigid. There cannot be any bulge in the horizon rotating around.

However, in global coordinates the horizon can appear squeezed. We have to check if the authors have made some conceptual error when they think that a bulge exists.

Conclusions

In general relativity objects slip almost entirely behind an event horizon, and quite quickly. The prominent example is the Oppenheimer-Snyder (1939) collapse of a dust ball.

We have to check in more detail the paper by Bonetto, Pound, Sam (2021).

We will also look at the Kerr metric, the ergosphere, as well as various horizons in a rotating black hole.