Werner Israel theorem (1967): nonrotating black hole must have a spherical horizon

Let us try to find an intuitive explanation for the theorem of Werner Israel (1967). The theorem states that a nonrotating static black hole in a vacuum must have a perfectly spherical horizon. It does not matter if the mass distribution inside the black hole is spherically symmetric or not.

Close to an event horizon of a static black hole, the potential is very low. If we send a photon from there, its redshift is infinite. Thus, the form of equipotential surfaces tells us something about the form of the horizon.

Making a nonspherical horizon with two masses?

Suppose that we try to make a nonspherical horizon. The obvious strategy is to put two masses M at some distance s from each other. We increase the masses gradually and watch what happens to the equipotential surfaces.

           field lines bend

                /              \

               |                 |

              ●                 ●

               <----- s ----->

              M               M

Let us describe the gravity field by "field lines". They are like lines of force for an electric field, but must take into account the nonlinear nature of gravity. We do not try to define the field line concept exactly here, but keep it intentionally vague. For weak gravity fields, the field lines are exactly like the lines of force of an electric field.

Each mass M distorts the metric of spacetime and makes the field lines of the other mass to bend toward M. The combined field of the masses is more spherically symmetric than the corresponding field of two electric charges.

As we increase the masses M, the equipotential surfaces become more and more spherical. It is not that surprising that the horizon is spherical when it ultimately forms.

What is the stress-energy tensor of a hypothetical nonspherical horizon?

Let us approach the problem from another direction. Let us assume that we have a static black hole in a vacuum and its horizon is elongated. What is the stress-energy tensor of the metric?

Our trick of putting two masses M at a distance s from each other did not succeed. What could we do to deform the horizon from a sphere?

There is no obvious way. We could try to add masses outside the event horizon, but that would breach the condition that the black hole sits in a vacuum.

The question is analogous to the following: suppose that we have a set of electric charges within a volume V of a diameter s. Far away, their field is essentially spherically symmetric. Can we break the spherical symmetry? Yes, but then we must put sources of the field outside the volume V. It is no longer a vacuum.

Dynamic configurations

The Israel theorem is formulated for a static metric. This leaves open a possibility that a dynamic system would have a nonspherical event horizon for a very long time.

Our previous blog post suggested that merging black holes are radially squeezed in the global pseudo-Schwarzschild metric around their common center of mass. Thus, it is possible that they can quickly fall inside their common Schwarzschild diameter. However, this not a mathematical proof.

https://en.wikipedia.org/wiki/No-hair_theorem

Wikipedia states that there exists no mathematical proof for the no-hair theorem. It is a conjecture.

Conclusion

There are various (partially heuristic?) proofs that a static or stationary black hole is either the Schwarzschild black hole or the Kerr black hole, depending on if its angular momentum is zero or not.

Such idealized black holes cannot exist in nature, though.

We have to check what is known about the end state of a black hole which is formed from realistic matter configurations. Can we show that it very quickly converges toward a stationary Kerr black hole?

Merger of two black holes

Our previous blog post suggests that a Schwarzschild black hole is a perfectly rigid object and its movement has to happen through "frame dragging". Observers very close to the horizon, or inside the horizon, will know of a merger process at a very late time, if at all.

The rigidity might cause problems in a merger. What guarantees that we can fit the two black holes inside the new horizon? The black holes lose some of the mass energy in the merger. The Schwarzschild radius of the new black hole is less than the sum of the radii of the merging black holes.

The gravity field of a Schwarzschild black hole M plus an external mass m

             "line of force"

           ________

               ___     \_______    • m

              /       \    

               \___/

                 M

Here we have a problem. If the "lines of force" of the gravity field of m go extremely close to the horizon of M, then the event horizon should effectively move farther behind M?

That probably is true. If m comes very close to the horizon, we expect it to increase the Schwarzschild radius by

       2 G m / c².

It makes sense that even before that, there is an effect on the radius of the event horizon.

Schwarzschild black holes which collide head-on

https://arxiv.org/abs/2006.03939

Daniel Pook-Kolb et al. (2020) used numerical relativity to calculate various horizons in a head-on collision of two non-rotating black holes.

Unfortunately, their Figure 1 does not seem to be in scale. The sum of the individual radii (purple, red) is much less than the final radius (blue).

A crucial question is how close the two black holes can come before the local speed of light in the area slows down to a crawl, as measured by a faraway observer. If the two black holes would freeze in their places, their common event horizon could become elongated.

The Werner Israel theorem (1967)

https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-8/issue-3/Event-horizons-in-static-electrovac-space-times/cmp/1103840605.pdf

The Israel theorem speaks about a "static" solution. A black hole probably never becomes completely static. In the Oppenheimer-Snyder (1939) collapse, the surface keeps approaching the event horizon for ever.

Anyway, a static non-rotating system must always have a perfectly spherical event horizon. There cannot be any bulges in it. If we assume that the system quickly approaches a static solution, then this rules away the possibility that a common event horizon can form while the two merging black holes are not yet inside their common Schwarzschild radius.

The merger of a large black hole and a small black hole

In the Schwarzschild metric, the radial metric is stretched by the factor

       1 / sqrt(1  -  r_s / r),

where r_s is the Schwarzschild radius, and r is the Schwarzschild radial coordinate. Let us assume than the large black hole has

       M = 10

and the small

       m = 1.

Let us calculate the stretching of the radial metric of the large black hole at

       r = 1.2 r_s.

Then

       1 / sqrt(1 - 1 / 1.2)

       = 1 / sqrt( 1/6 )

       = 2.5.

Let us assume that the event horizon of the falling small black hole m remains as is in comoving coordinates, and that the event horizon of M stays as is.

                         M                             m

                          ●                         <---•

                                                      <---> 0.08 r_s

        <-----------------------------><----------->

                         2 r_s                0.2 r_s

           

We see that when the small black hole m falls toward M, it is squeezed into 40% or less in the radial direction in the Schwarzschild coordinate r. It fits easily within the diameter 2.2 r_s of the new, forming black hole.

The "original" event horizons of M and m probably do not touch yet when the common event horizon of the combined system M & m forms.

The event horizon is defined as the border from where a light signal can still reach infinity. It is obvious that the individual, up-to-date, event horizons of M and m have to touch before the common event horizon forms and grows to form a perfect sphere.

https://youtu.be/gmmD72cFOU4

Above we have two screenshots from a video by Teresita Ramirez and Geoffrey Lovelace (2018). In the video, both black holes seem to be squeezed in the radial direction from their common center of mass. They suddenly obtain a common event horizon when they are close enough. 

Could we "freeze" a non-spherically-symmetric field when two black holes meet?

               ___

             /        \    M

             \____/           

               _O_      O = small black hole

             /        \

             \____/     M

Let us try to build a "bridge" between the event horizons of two black holes which are not yet inside their common Schwarzschild diameter (diameter = 2 * radius). If we succeed, then the speed of light inside the bridge is essentially zero, and the bridge will "freeze" the black holes to their positions.

We put two black holes of a mass M close to each other, but not within the Schwarzschild diameter of 2 M. The letter O denotes a small black hole which we try to use as a bridge between the two larger ones.

Note a black hole M vertically squeezes the other black hole M if we use global (pseudo-"Schwarzschild") coordinates centered at O.

We can roughly add the perturbations of "weak" gravity fields in the Minkowski metric η. Both black holes M squeeze O vertically.

Let M = 10 and the mass of O be 1. Since all black holes are at a low potential, their Komar mass is considerably less.

Let the Schwarzschild radius for a mass 1 be 1.

Let the potential of O be -0.5. Then O only contributes 0.5 to the common Komar mass of the system. From the Schwarzschild metric formula we see that O is squeezed very roughly to 0.5 vertically.

If the potential of the lower big black hole in the field of the upper one is -1, then it contributes to the Komar mass 9.5, and is squeezed vertically to 9.

The Komar mass of the whole system is 2 * 9.5 + 0.5 = 19.5 and the vertical diameter only 2 * 9 + 0.5 = 18.5. The system is inside its combined Schwarzschild diameter. We failed. Our goal was to keep the system too wide to be inside its Schwarzschild diameter.

Building a bridge between the event horizons of the two black holes M seems to fail because everything is squeezed vertically. The whole system slips inside its common Schwarzschild diameter before a bridge is built.

Our analysis suggests that when the two black holes M merge, clocks in the gap between the black holes do not slow down too much, relative to the global "Schwarzschild" coordinate time. This means that the local speed of light in the gap stays relatively fast. The two black holes can approach each other rapidly, measured in the global "Schwarzschild" coordinates.

Only after a common, spherical event horizon forms around the two black holes, does the local speed of light in the gap go to almost zero. The system freezes when the two black holes are almost entirely inside their common Schwarzschild diameter. This is much like the Oppenheimer-Snyder (1939) collapse where only the surface of the dust ball lingers above the forming event horizon.

https://arxiv.org/abs/1110.1668

Cohen, Kaplan, and Scheel (2011) have made computer simulations of merging black holes. Their paper contains pictures of a thin bridge forming between the individual event horizons of the two black holes.

A microscopic black hole or a particle falling into a black hole

If a microscopic black hole falls into a large black hole, that is still a merger of two black holes. The analysis above suggests that a common event horizon quickly encloses both black holes.

The same is probably true of a particle falling into a black hole. If we use the Schwarzschild time coordinate, the particle quickly reaches a radius r which is the new Schwarzschild radius of the system.

What about a large mass (not a black hole) falling into a black hole? The mass consists of particles. If the hypothesis above is true, then a new horizon quickly encloses all the particles.

Conclusions

LIGO observations show that (spinning) black holes merge quickly. The ringdown phase lasts only for about 10 milliseconds. It cannot be that the merging black holes would freeze in a "dipole" formation in which they would produce large gravitational waves for a long time.

Our analysis above suggests that the merging black holes have separate event horizons until they are inside the Schwarzschild diameter of the entire system. Once they are inside that diameter, a common, spherical event horizon forms very quickly.

In our Minkowski & newtonian gravity model, the two black holes are essentially frozen behind the new common event horizon. Observers very close to the individual event horizons of each black hole never get a signal that anything has changed. This is because the local speed of light is very slow close to the horizon. The individual black holes, in this sense, are infinitely rigid. In our model, there is no further development behind the common event horizon. No collapse toward a singularity happens.

In our next blog posts we should analyze the proof of the Werner Israel theorem (1967), and analyze what happens in a merger of spinning black holes.

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

Also, we need to check what Richard H. Price wrote in his 1971 thesis about a mass falling into a spinning black hole.

https://nms.kcl.ac.uk/david.robinson/web_page/blackholes.pdf

D. C. Robinson (2012) wrote an overview of black hole uniqueness theorems.

Moving a very heavy neutron star in general relativity

Suppose that we have a Schwarzschild black hole. We have observers very close to the event horizon. In the Schwarzschild time coordinate, it takes a very long time to send a signal to such an observer.

Suppose then that we want to "bend" or otherwise change the metric of the black hole. This might happen in a black hole merger.

We assume that a "causal" change in the metric can only propagate at the local speed of light. A merger is a causal process, since we can either let the black hole to stand alone, or we can merge it with another black hole.

If the local speed of light, as seen by a faraway observer, stays extremely slow close to the horizon, then the merger process cannot change the metric very close to the horizon.

But an infinitely rigid object involves several paradoxes. One could send a superluminal signal by nudging such an object. How to resolve the paradoxes?

How to move a neutron star?

A milder version of the problem occurs with a neutron star. If we start to move a neutron star, it takes some time for the surface and the interior to "know" that it is moving.

We wrote about this problem also on March 17, 2023.

                      ___

                    /       \          <-- ●  mass M

                     \___/

              neutron star

Let us assume that we suddenly move a mass M close to the neutron star. It starts pulling on the star.

It takes some time for the interior of the neutron star to know that the mass M came nearby. The metric (in the comoving frame) inside the neutron star must remain constant during that period. But the neutron star, obviously, starts moving toward M immediately.

This means that the metric in the comoving frame of the interior really describes a "perfectly rigid object" during that short time.

The matter in a neutron star is not perfectly rigid. It is possible that the outer parts of the star start moving before the central part.

Let us then assume that the local speed of light (in global Schwarzschild coordinates) inside the star is extremely slow. We may be able to move the neutron star several diameters before its center knows that anything is happening. In this case, the central part must start moving before light has time to reach it.

The solution in this case might be frame dragging. The central part moves along with the outer parts "automatically". One cannot use frame dragging to communicate a signal to an observer at the center of the star, because the observer has no means of detecting frame dragging.

If we use comoving coordinates inside the neutron star, then we have to make the metric to change outside the star, to make the star to move.

If we use static coordinates inside the neutron star, then we have to let the mass start moving inside the star, and this process propagates faster than the local speed of light inside the star. If we have banned "causal" processes which happen faster than light, then we cannot use static coordinates.

Frame dragging in our Minkowski & newtonian model of gravity

In our own gravity model, a slow speed of light comes from the extra inertia which surrounding masses give to the photon.

If we move the surrounding masses, then the photon must move along. Frame dragging is a very natural process.

Also, the mutual inertia couples masses together. If we move the outer part of a neutron star, then the inner part must move along. The speed of this process exceeds the local speed of light. The speed is probably the global speed of light in the underlying Minkowski space.

Conclusions

We want to understand how black holes move, to figure out how a merger of two black holes technically happens.

Our analysis above suggests that we have to move the black hole with a frame dragging process - and the black hole really is a perfectly rigid object.

We do not know if numerical relativity uses frame dragging to move a black hole. And if not, what implications that has. Do the calculations allow superluminal causal effects?

A mass approaches a black hole: where is the horizon?

Suppose that we have a black hole in the process of an Oppenheimer-Snyder (1939) collapse. The (forming) event horizon is at the distance of the Schwarzschild radius of the collapsing mass-energy M. The event horizon is perfectly spherical.

Let an infinitesimal test mass m approach the black hole. Where is the event horizon in the new system?

                        •  test mass m

                      ___

                    /       \        

                     \___/

            Oppenheimer-

           Snyder collapse M

If the test mass would carry an electric charge, we know how the electric field would settle:

The diagram is from the paper by Hanni and Ruffini (1973):

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

Some lines of force would go extremely close to the horizon. From far away, it would look like that there is a negative induced charge at the locations where the lines of force approach the horizon, and a positive induced charge at the locations where the lines of force leave the horizon.

Perturbation of the event horizon

Suppose that an infinitesimal test mass dm approaches the horizon of a Schwarzschild black hole. What is the perturbation of the gravity field like?

It might be that the perturbation caused by dm is somewhat similar to the field lines of the electric test charge. Let us assume that dm falls on one side of the black hole. Let us study what happens to the field on the other side.

We assume that "causal" changes into a gravity field can propagate only at the local speed of light.

Suppose that dm would make the horizon to move dr farther. If dm is very small, it takes a very long time (in the global Schwarzschild time coordinate) for the metric close to the horizon to "know" that the horizon should be moved dr farther.

Is it so that in the Schwarzschild time coordinate, it takes an infinite time for the horizon to become "complete", in the sense that the metric of time at the horizon is zero?

Let us assume that the metric is "updated" to a new version in the order of a descending r coordinate. This is a natural assumption because it takes a very long time for light to reach points whose r coordinate is close to the old horizon.

A faraway observer sends down light signals which propagate a little behind the updating process. Let us assume that the metric is updated to a new Schwarzschild metric whose Schwarzschild radius is dr larger. It takes a long Schwarzschild time interval T for the light signal to come very close to the new Schwarzschild radius. If we assume that an observer sitting close to the new radius immediately sends the signal back, it takes another T for the faraway observer to receive the reply.

In this sense, we could say that the new Schwarzschild metric is never completed, from the point of view of a faraway observer.

Conclusion

In this blog we believe that the metric of spacetime near a mass is simulated by the gravity force through adding inertia to particles. Clocks tick slower when inertia is larger. The "true" underlying metric is the Minkowski metric - that never changes.

A question is if the inertia can become infinite. Then clocks would stop entirely. This does not sound right. Our hypothesis is that clocks can slow down arbitrarily much, but never stop entirely.

At the Schwarzschild event horizon, the metric of time would be zero. We try to show that such horizon can never form.

Since the speed of light (in the global Schwarzschild coordinates) is extremely slow inside a forming event horizon, does all movement essentially stop there? Is the frozen star model right? In a merger of two black holes, how do both objects get inside a common, spherical horizon, if nothing can move inside the horizon? We will look at this in the next blog post.

Black hole no-hair theorems: an electric charge close to the horizon

Our previous blog post asked what happens if the collapsing dust ball is not symmetric. Furthermore, what happens if two black holes merge.

https://en.wikipedia.org/wiki/No-hair_theorem

Werner Israel proved in 1967 that a static, non-spinning black hole must have the Schwarzschild metric outside its event horizon, assuming some sanity conditions about the event horizon. That is, the event horizon has to be a perfect sphere.

Brandon Carter, David C. Robinson, and Stephen Hawking proved further results which say that a spinning black hole must have the must have the Kerr-Newman metric outside it.

Why is the Werner Israel 1967 result true?

Suppose that we have a black hole which satisfies the Scwarzschild metric. Let us drop a test mass m into it.

The black hole bends the field lines of the gravity field of the test mass m in such a way that when the test mass is relatively close to the horizon, its field seem to originate from a point which is very close to the center of the black hole.

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

Richard Hanni and Remo Ruffini calculated this for electric field lines in 1973.

https://physics.stackexchange.com/questions/416413/what-happens-when-an-electric-charge-crosses-the-event-horizon

This suggests that even if we try to make the gravity field of a black hole deformed by dropping a large number of test masses to one side of the black hole, the event horizon will stay perfectly spherical.

It is probably an "optical" phenomenon: the field of the black hole distorts the field of the test mass in such a way that the location of the test mass is hidden.

An electric charge close to the event horizon

              ___

            /       \  ●   charge Q

            \____/

          horizon

Suppose that we have an electric charge Q very close to the horizon of a black hole.

The charge is a source of the electric field E and there are no other sources or sinks of the field. The lines of force must extend to infinity.

https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Book%3A_Applications_of_Maxwells_Equations_(Cochran_and_Heinrich)/03%3A_Electrostatic_Field_II/3.04%3A_The_Field_Energy_as_Minimum

Let ε₀ be vacuum permittivity. We assume that the electric field E becomes such that it minimizes the total field energy

       W = ∫     1/2 ε₀ E²  * U dV,

           entire

           space

where we integrate over the entire space, and U is the gravitational potential at the volume element dV. Close to the horizon, U is very close to zero.

How should we draw the lines of force, so that the field energy would be the smallest possible?

Can we let lines of force go very close to the horizon? Let us calculate the field energy for such a configuration.

The Schwarzschild metric is (Wikipedia):

where r is the Schwarzschild radial coordinate and r_s is the Schwarzschild radius of the spherically symmetric mass M.

Let us denote

       r = r_s + r'

where r' is a small positive distance.

                   r'    dr

               |        |||

               |        |||  field lines

               |        |||

                    

  horizon         layer

What is the energy of a configuration where we draw all the field lines from the charge Q very close along the horizon?

We assume that c = 1 and r_s = 1. The metric of time at a distance r' from the horizon is

        sqrt(1  -  1 / (1 + r'))

        = sqrt(r')

        = U.

Let us pack the bunch of field lines into a "layer" whose thickness in the Schwarzschild radial coordinate is dr << r', as in the diagram.

The radial metric tells us the proper thickness of that dr. It is

       dr / sqrt(r').

We pack, say, Q field lines into the layer. The field strength

       E ~ sqrt(r') / dr,

and the (locally measured) energy density

       E² ~ r' / dr².

The locally measured total energy of the field is ~ proper thickness * E², or

       ~ dr / sqrt(r')  *  r' / dr²

       = sqrt(r') / dr.

Since U ~  sqrt(r'), we have that the total energy, as measured by an observer far away, is

       ~ r' / dr.

Let us choose

       dr = b r',

where b is a small constant > 0. For example, b could be 0.01. The energy associated to putting Q lines of force to that layer is

       constant * 100.

We can now divide the range r' > 0 into an infinite number of layers which converge to r' = 0.

 horizon

              |     |       |             |                   |

              |     |       |             |                   |

              ...  layer   layer        layer

Suppose that we want to have Q lines of force going along the horizon, close to the horizon. We divide Q lines into N layers. The field energy in one layer is

       ~ E²

       ~ Q² / N².

The total field energy in N layers is

        ~ Q² / N.

By choosing N large we can make the field energy as small as we like.

If we have a charge very close to the horizon, then we can lead the lines of force around the horizon, and the associated field energy of the field close to the horizon is negligible, as measured by a distant observer.

In order to minimize the field energy in the entire space, we need to draw the lines of force from the horizon to infinity. How can we minimize the energy? Intuitively, we have to minimize the field energy associated with the lines of force which leave the horizon.

                  lines of force

                            |

                         ___  +

               ____ /       \ ____      lines of force

                        \___/

                                 metal sphere

                           |

If we have a charge Q on a metal sphere, the minimum energy electric field is the spherically symmetric uniform field. The field lines are normal to the surface of the sphere. This is the solution for a black hole, too.

Our proof is not rigorous, though. Since the charge Q is not exactly at the horizon, drawing the field lines downward almost to touch the horizon does involve some field energy. Anyway, we have now an intuitive explanation for the fact that a black hole hides the electric charge distribution in it, so that the field looks as if the charge would be uniformly distributed over the horizon.

Conclusion

We found a simple heuristic argument which explains the no-hair theorem for an electric charge in a black hole.

Let us next try to extend this to the gravity field. Is the gravity field of a non-rotating black hole spherically symmetric, even though the masses would not be? This is the claim of the Werner Israel 1967 theorem.

The Schwarzschild metric behind the horizon never forms?

In our blog we have held the view that there cannot exist "infinite" forces. An example of an infinite force is gravity at the horizon of the Schwarzschild metric at the event horizon, or inside the horizon.

The metric inside the horizon would make matter to move toward the center regardless of how strong a force opposes the movement.

A collapsing symmetric spherical dust ball

General relativity says that the metric outside the ball is the Schwarzschild metric. Outside the event horizon, we can define simultaneous events in a "sensible" way. The standard Schwarzschild coordinates give us a sensible definition of simultaneity.

In these coordinates, the outer surface of the shell moves ever slower, never reaching the Schwarzschild radius R of the collapsing mass M.

The natural spacelike foliation is given by the Schwarzschild standard coordinates t and r.

                                           R

           t = ∞                        ●

               .

               .                           

               .

                                      / 

            t = 2    ----------------

                                   /

            t = 1    ----------------

                               /

            t = 0    ----------------|

                          /            

                     surface        R

The surface does reach R in a finite proper time of someone standing on the surface. Also, the proper distance of the horizon is finite from a static observer outside the horizon. This implies that the time for surface to descend to the horizon is finite, if measure the time by summing proper time intervals measured by static observers on the path.

But our opinion in this blog is that proper time is nothing more than what a clock shows. And if a clock runs ever slower, it may never progress to a specific time t₀. There is no reason why we should be able to extend proper time beyond that reading of the clock.

Question. Is there a spacelike foliation where the spatial metric is static in the vacuum area, simultaneousness is defined in a "sensible" way, and where the surface would reach R?

For the standard Schwarzschild coordinates the following holds: if a static observer A sends a light signal to a nearby static observer B, and B immediately sends a signal back, then the time coordinate when the signal arrives at B is the average of the time coordinates of when A sent the signal and received the signal.

Suppose then that we have a folio F which in the diagram above contains the dot ● at R, at the external time ∞. Obviously, one cannot define simultaneousness in the above "sensible" way. A static observer A at the dot ● cannot send a signal at all to a static observer B at a slightly larger radius, and vice versa.

The Oppenheimer-Snyder (1939) collapsing dust ball

https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455

Inside the collapsing uniform dust ball, Oppenheimer and Snyder (1939) use comoving coordinates and solve the metric in the form above. The coordinate τ measures the proper time of a comoving observer.

The authors move to Schwarzschild-like coordinates which have the metric in the form above. They then match the metric to the static Schwarzschild metric outside the ball:

where r₀ is the Schwarzschild radius of the ball.

In the Schwarzschild-like metric and coordinates t and r, the tangential metric is 1. That is, the circumference around the center of the system, as measured by observers at r is 2 π r. In that sense, we may call an observer at a fixed radius r a "static" observer. It is reasonable to say that an observer at r = 0 is static - our definition agrees with that.

Is the Schwarzschild time t a reasonable global time in the system? So that it would be sensible to say that events having the same coordinate t are "simultaneous" relative to a faraway observer?

In a Schwarzschild-like metric tensor, the non-zero elements are on the diagonal. Observers which are initially static at a certain r and a fixed t will probably agree that the events with the fixed time coordinate t are simultaneous. At least this holds in the case of the Minkowski metric.

Oppenheimer and Snyder calculate the approximate mapping from the Schwarzschild global time t to the proper time τ of comoving observers. When t approaches infinity, the proper time τ approaches a finite value. In the formula above, R⁢_b is the comoving radial coordinate of the surface of the ball, and r₀ is the Schwarzschild radius, and R is the comoving radial coordinate of a dust particle inside the ball.

When t goes to infinity, the formula says that the proper time τ progresses slightly further when R is larger.

Interpreting the Oppenheimer-Snyder result in our Minkowski & newtonian gravity

In our own gravity model, slowdown of clocks is a result of the gravity interaction which gives more inertia to parts of the clock. The stretching of the radial metric is due to the inertia being larger in a radial movement compared to a tangential movement.

In our model the underlying metric of spacetime does not change. The underlying metric is always Minkowski. It is the paths of particles which change.

Let us introduce:

Hypothesis. The global time coordinate in the Schwarzschild-like metric matches the time coordinate of the underlying Minkowski coordinates and metric, where the dust ball is static in the Minkowski coordinates.

Then we have a very nice interpretation for the Oppenheimer-Snyder result. It describes how the dust ball freezes as it collapses.

There is an "event horizon" in the following sense: a photon sent from inside the ball will never get out of the ball. Our event horizon has no strange properties of the one in the standard Schwarzschild solution of general relativity. There is no infinite force at the horizon in our model. And the metric does not switch the roles of the radial coordinate and the time coordinate at the horizon.

The interpretation of the whole collapse process is very mundane in our model: a system of particles moves slower and slower, but never freezes totally. It resembles the "frozen star" hypothesis of general relativity.

The mass-energy concentrates itself close to the Schwarzschild radius

The center of the dust ball falls into an extremely low potential. Its Komar mass is negligible relative to the dust particles at surface of the ball which acquire a lot of kinetic energy, relative to a static coordinate system.

Could this explain why the gravity field of the resulting black hole is spherically symmetric, even if the dust cloud initially was not symmetric?

We need to figure out what happens with spinning dust balls, and how does a merger of two black holes happen.

Conclusions

Our Minkowski & newtonian model makes a collapsing dust ball well behaved. We avoid some strange properties of a standard Schwarzschild black hole.

We need to figure out what happens with rotating black holes, and what happens in a merger of two black holes.

We also need to consider quantum mechanical effects. There are extremely strong forces at the surface of a collapsing dust ball. Are these forces able to materialize new particles from "empty" space?