Do Black Holes End up as Quark Stars? and quantum gravity

[stevebd1]

The support mechanism in a star is pressure, not energy.

The gravitational field in GR is a product of two properties, the stress-energy tensor and the metric tensor. This can be represented by the following equation-

$$g=\frac{Gm}{r^2}\frac{1}{\sqrt{1-\frac{2Gm}{rc^2}}}$$

where the first part of the equation $(Gm/r^2)$ is an approximation of the stress energy tensor and the second part is the radial component from the Schwarzschild metric tensor. As you can see, gravity becomes infinite at 2M (the event horizon or the Schwarzschild radius). The stress energy tensor '..describes the density and flux of energy and momentum in spacetime' and the metric tensor '..captures all the geometric and causal structure of spacetime'. You can't look at gravity in GR without considering both. Looking at the Schwarzschild Metric-

$$c^2 {d \tau}^{2} =-\left(1-\frac{2Gm}{rc^2} \right) c^2 dt^2 + \left(1-\frac{2Gm}{rc^2}\right)^{-1} dr^2$$

In this form, the time component is negative and temporal for r>2M but when r<2M, the signs flip and r component becomes negative and temporal and there is no stable r, irrespective of pressure, ticking down towards r=0 hence the singularity. This is explained in more and better detail here in the The Schwarzschild Metric and Inside the Black Hole sections-

Spacetime Geometry Inside a Black Hole

Again, if we look at the time component from the Schwarzschild interior metric tensor (note- interior in this case means interior spacetime of an object of mass)-

$$c\ d\tau=\left( \frac{3}{2}\sqrt{1-\frac{2M}{r_0}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{r_0^{3}}}\right)c\ dt$$

where $r_0$ is the radius of the star, $M=Gm/c^2$ and $r$ is the radius of the star where you want to calculate the time dilation. Irrespective of pressure, type of mass or density, if r₀ collapses to 2.25M, then $\tau$ will equal zero at $r=0$ (i.e. the centre), there will be a runaway effect and a black hole will form. When considering the halting of collapse within a black hole, you need to consider both the mass and its effect on spacetime.

For the full Schwarzschild interior metric, see this post

 

[Bernie G]

“As you can see, gravity becomes infinite at 2M (the event horizon or the Schwarzschild radius).”

I think the Tolman-Volkoff equation and the equations above are wrong and result in the mis-analysis of black holes. For example, at 0.9 SR the gravitational acceleration would be about 1.23 that at 1.0 SR.

 

[PAllen]

“As you can see, gravity becomes infinite at 2M (the event horizon or the Schwarzschild radius).”

I think the Tolman-Volkoff equation and the equations above are wrong and result in the mis-analysis of black holes. For example, at 0.9 SR the gravitational acceleration would be about 1.23 that at 1.0 SR.

The Tolman-Volkoff equation is an exact consequence of GR given spherical symmetry and stasis. It is obviously inaccurate to the extent that a real world body is rotating and not strictly static - but only to that extent, unless you reject GR. The other statements you make (about gravitational acceleration inside the SR) are trivially wrong. If you can provide a reference for some way your claims can be considered correct, please do so, or desist from stating personal theories at odds with accepted physics as fact.

 

[PeterDonis]

Are you using the Tolman-Volkoff equation for that conclusion? This equation gives bad results for neutron star, let alone a black hole. If a star was significantly smaller than 1.0 SR, I think at 0.9 SR the gravitational acceleration would be about 1/0.8 that at 1.0 SR, a large number but still very finite.

The TOV equation assumes static equilibrium. But static equilibrium is not possible for an object with a surface radius less than 9/8 SR = 1.125 SR, by Buchdahl's Theorem. Any object with a surface radius smaller than this must be collapsing; it cannot be static. So the TOV equation does not apply. (Also, the "gravitational acceleration" goes to infinity at 1.0 SR; the TOV equation is not the equation that determines the gravitational acceleration.)

 

[PeterDonis]

According to Schwarzschild metric, world lines within 2M (or the Schwarzschild radius) are spacelike

More precisely, curves of constant $r$ in the region $r < 2M$ are spacelike. This makes it impossible, as you say, for any object to "hover" at fixed $r$ in the region $r < 2M$.

which means r is temporal, just as t is temporal outside 2M

More precisely, $r$ becomes a timelike coordinate for $r < 2M$ in Schwarzschild coordinates. There are other coordinate charts describing the same geometry (such as Painleve or Eddington-Finkelstein) for which $r$ is spacelike everywhere. A better way of stating the coordinate-independent feature of this spacetime for $r < 2M$ is that all timelike worldlines must have decreasing $r$.

according to the the Schwarzschild interior metric (which looks at the curvature of spacetime within a spherical mass)

The metric you refer to is only valid for the interior of a spherically symmetric mass of uniform density. There is no known closed-form solution for the (much more realistic) case where the density varies with radius; that case has to be solved numerically. The uniform density metric is still useful as an idealized example to illustrate qualitative features, but it's good to be aware of its limitations.

the minimum stable radius for a spherical mass is 9/4M (or 2.25M)

This is actually true regardless of the behavior of the density as a function of radius; the general theorem that demonstrates this for any spherically symmetric, static mass distribution, as I noted in my previous post, is Buchdahl's Theorem.

 

[Bernie G]

Probably the sentence should have been "I think at 0.9 SR the gravitational acceleration would be about 1.23 that at 1.0 SR."

 

[Bernie G]

The TOV equation assumes static equilibrium. But static equilibrium is not possible for an object with a surface radius less than 9/8 SR = 1.125 SR, by Buchdahl's Theorem. Any object with a surface radius smaller than this must be collapsing; it cannot be static. So the TOV equation does not apply. (Also, the "gravitational acceleration" goes to infinity at 1.0 SR; the TOV equation is not the equation that determines the gravitational acceleration.)

I think "the gravitational acceleration goes to infinity at 1.0 SR" is just plain wrong. Gravitational acceleration = force/mass = F/m. Why should a contracting object exert infinite force at some point? That would be like creating energy for free.

 

[PeterDonis]

I think "the gravitational acceleration goes to infinity at 1.0 SR" is just plain wrong.

Have you actually done the math? This is a common homework problem for physics students studying GR; it is not at all contentious.

Gravitational acceleration = force/mass = F/m.

In GR, gravity is not a "force", and you can't use Newtonian reasoning like this. The proper acceleration required to maintain a constant altitude goes to infinity at the horizon because of the geometry of spacetime, not because of any "force" becoming infinite.

 

[Bernie G]

Have you actually done the math? This is a common homework problem for physics students studying GR; it is not at all contentious.

Please give an internet source with the solution.

 

[PAllen]

Please give an internet source with the solution.

With your rude tone, you don't deserve this, however here is such a link:

http://en.wikipedia.org/wiki/Proper_acceleration#Four-vector_derivations

Note the formula for acceleration needed maintain hovering positions, measured locally by an accelerometer:

This becomes infinite as r decreases to r_s.

 

[PeterDonis]

Please give an internet source with the solution.

You are the one making the extraordinary claim (that GR is wrong), so it's up to you to give references. However, PAllen was nice and gave you the formula. Please bear in mind the PF rules about personal theories and making extraordinary claims without references to back them up. If you continue to repeat erroneous claims, you will receive a warning.

 

[Bernie G]

"With your rude tone, you don't deserve this, however here is such a link"

Please accept my apologies if I am coming across as rude. Perhaps you are interpreting tenacity for rudeness. I'll be posting one last question about black hole spin rates shortly. Do not feel obligated to answer it if I'm trying your patience.

 

[Bernie G]

Don't the rather slow and finite spin rates of the fastest spinning black holes imply a large object within the black hole? Its logical that the equatorial velocity of the fastest spinning black holes would be a large fraction of the speed of light. If an equator has a radius of 10 - 25 km and a spin rate of 1000 revolutions per second, its tangental velocity would be 0.2c - 0.5c, which seem reasonable. But if the equator only had a radius of 1 km and a spin rate of 1000 RPS the equatorial velocity would only be 0.02c, which seems illogical for the fastest spinning black holes.

 

[PAllen]

Don't the rather slow and finite spin rates of the fastest spinning black holes imply a large object within the black hole? Its logical that the equatorial velocity of the fastest spinning black holes would be a large fraction of the speed of light. If an equator has a radius of 10 - 25 km and a spin rate of 1000 revolutions per second, its tangental velocity would be 0.2c - 0.5c, which seem reasonable. But if the equator only had a radius of 1 km and a spin rate of 1000 RPS the equatorial velocity would only be 0.02c, which seems illogical for the fastest spinning black holes.

A black hole with spin can have any spin, and this is unrelated to any surface or object inside the horizon. The BH would end up with roughly the same spin as just before final collapse.The GR solution describing such objects is the Kerr solution:

See:

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

 

[PeterDonis]

Don't the rather slow and finite spin rates of the fastest spinning black holes imply a large object within the black hole?

No. I don't understand why you think it would.

Its logical that the equatorial velocity of the fastest spinning black holes would be a large fraction of the speed of light.

A rotating black hole doesn't have an "equatorial velocity". It has an angular momentum, and you can compute an angular velocity from this that is sometimes loosely interpreted as the "angular velocity of the horizon". But the hole doesn't have a surface, so the angular velocity doesn't translate into an ordinary speed. An object rotating around the hole with the angular velocity of the horizon, at the horizon, would have to be moving at the speed of light; no actual object can do this (just as no actual object can "hover" at the horizon of a non-rotating black hole).

In short, you are thinking of the hole as an ordinary rotating object. It isn't, and trying to reason as though it is will lead you to incorrect conclusions.

 

[stevebd1]

Don't the rather slow and finite spin rates of the fastest spinning black holes imply a large object within the black hole? Its logical that the equatorial velocity of the fastest spinning black holes would be a large fraction of the speed of light. If an equator has a radius of 10 - 25 km and a spin rate of 1000 revolutions per second, its tangental velocity would be 0.2c - 0.5c, which seem reasonable. But if the equator only had a radius of 1 km and a spin rate of 1000 RPS the equatorial velocity would only be 0.02c, which seems illogical for the fastest spinning black holes.

A couple of things here. firstly black holes are observed to spin slower than they are actually spinning due to gravitational and relativistic redshift. For instance, when an article says '..the black hole is spinning at close to the speed of light..' they should add 'as observed from infinity'. Also what is being measured isn't strictly speaking the BH itself, but the frame dragging rate of the spacetime being dragged around the BH. The tangential velocity of the frame dragging rate for a black hole (as observed from infinity) at various r is-

$v=\omega R$

where $\omega$ is the frame dragging rate (or angular velocity) and $R$ is the reduced circumference, but the actual local tangential velocity is-

$v=(\omega R)/\alpha$

where $\alpha$ is the reduction factor or redshift. For the full equations for $\omega, R$ and $\alpha$, see this old library post- What is frame dragging.

Secondly, if the BH has any kind of spin, no matter how small, it will have an ergosphere $(r_{e+})$ just outside the horizon $(r_+)$, and if it has an ergosphere, there will be a region where spacetime is rotating faster than c relative to infinity so technically, all BH's locally spin at c at the ergosphere boundary. For the equations for $r_{e+}$ and $r_+$, see this old library post- https://www.physicsforums.com/threads/radius-of-a-black-hole.762981/

When observed from infinity, this rapid rotating of the ergosphere will not be detected, and the event horizon will appear to spin at anything up to c (with c being maximal, i.e. a/M=1), whereas locally, the boundary of the ergosphere will always be spinning at c which will increase exponentially as you proceed through the ergosphere to the event horizon (it's also worth noting that a BH's spin is normally measured by the location of the marginally stable orbit (MSO) which is at 6M for a static BH (a/m=0) and at M for a BH at maximum spin (a/m=1).

 

[PeterDonis]

black holes are observed to spin slower than they are actually spinning due to gravitational and relativistic redshift.

Actually, you can't observe a black hole's spin directly at all. You can only observe the motion of objects close to its horizon, and note that, the closer to the horizon they get, the more constrained their angular velocity is. In the limit of an object at the horizon (which is not actually physically possible, as I've said before), the angular velocity of the object can only have one value, which is often (somewhat sloppily) called "the angular velocity of the horizon".

As I said before, translating this angular velocity into a "speed" has a number of issues involved. See below.

what is being measured isn't strictly speaking the BH itself, but the frame dragging rate of the spacetime being dragged around the BH.

No, what is being observed, as above, is the motion of objects close to the horizon. The possible worldlines for such objects are affected by the frame dragging rate, so by measuring them, you can indirectly measure frame dragging. But what you measure, even indirectly, is not a velocity; at best, it's an angular velocity. See further comments below.

The tangential velocity of the frame dragging rate for a black hole (as observed from infinity) at various r is-

$$
v=\omega R$$

where $\omega$ is the frame dragging rate (or angular velocity) and $R$ is the reduced circumference, but the actual local tangential velocity is-

$v=(\omega R)/\alpha$

where $\alpha$ is the reduction factor or redshift.

What is the physical meaning of these velocities? How would you measure them? I think that, if you try to answer these questions, you will find a number of issues lurking that you haven't considered.

if the BH has any kind of spin, no matter how small, it will have an ergosphere $(r_{e+})$ just outside the horizon $(r_+)$, and if it has an ergosphere, there will be a region where spacetime is rotating faster than c relative to infinity

This is, at best, extremely sloppily phrased, and at worst, simply wrong. I strongly recommend taking some time to think through the physical meaning of what's going on. Spacetime does not "rotate", and you can't observe spacetime directly; you can only observe the motion of objects. An object inside the ergosphere cannot be static; that is, it cannot remain at constant spatial coordinates. In other words, it cannot have zero angular velocity. But that in no way translates to "rotating faster than c relative to infinity"; I don't know where you're getting that from.

locally, the boundary of the ergosphere will always be spinning at c

I don't know where you're getting this from either. What references are you using?

 

[stevebd1]

'At a point r_e it becomes impossible to counteract the rotational sweeping force. The particle is in a kind of space-time maelstrom. The surface determined by r_e is the static limit: from there in, you cannot avoid rotating. Space-time is rotating here in such a way that you cannot do anything in order to not co-rotate with it.'
Source- Introduction to Black Hole Astrophysics - Page 51

'..the (outer) event horizon of a Kerr black hole is surrounded by a second critical surface, the static limit, which has the shape of an oblate spheroid and touches the event horizon at the poles. The space between these two surfaces is called the ergosphere. Within the ergosphere the spacetime is dragged in the direction of the spinning black hole at a speed greater than c with respect to the outside universe at rest, while at the static limit this speed equals c.'
Source- Extreme Environment Astrophysics - Page 17

'In a rotating black hole, the ergosphere is associated with the stationary limit, the location at which space-time is flowing at the speed of light'
Source- http://www.astro.cornell.edu/academics/courses/astro201/ergosphere.htm (Cornell Centre for Astrophysics and planetary science)

'All objects in the ergosphere become dragged by a rotating spacetime.'
(the phrase rotating spacetime is highlighted and actually links to the frame-dragging page)
Source- Penrose process (Wikipedia)

While I understand that frame dragging cannot be measured directly at the event horizon or at the boundary of the ergosphere, it can be predicted by establishing where the marginally stable orbit is (which is what I state at the end of my post). A number of recent articles have stated that '..the black hole is spinning at half the speed of light.' or '..the black hole is spinning at close to the speed of light making it almost maximal'. I imagine these statements are based on where the MSO is which is normally defined by the inner edge of the accretion disk where there's plenty of matter sending pulses of radiation. For example, if a black hole has an MSO at 4M, then working backwards using the equations (eq 32, http://relativity.livingreviews.org/Articles/lrr-2013-1/articlese2.html ) to establish the MSO, the rate of spin can be established, in this case, the spin would be a/M=~0.56. Based on this, using $\omega R$, it can be predicted that the frame-dragging rate at the event horizon is 0.305c as observed from infinity, (the black hole is spinning at nearly a third the speed of light)

One source of $\omega R/\alpha$ is-

$$v_s=(\Omega_s-\omega)\frac{R}{\alpha}$$

where $v_s$ is the local velocity required for a stable orbit and $\Omega_s$ is the angular velocity required for stable orbit. This equation can be rewritten as-

$$v_s=(\Omega_s-\omega)\frac{\Sigma^2\sin\theta}{\rho^2\sqrt{\Delta}}$$

which can be seen in some form in this paper, equation at the top of page three. Remove $\Omega_s$ and you have the local tangential velocity of the frame-dragging rate. Remove the $\alpha$ component and you have the tangential velocity of the frame-dragging rate as observed from infinity. I've also seen the equation in various forms elsewhere.

This is backed up to some extent when you look at the boundary for the ergosphere (or the static limit). If we look at the ergosphere of a 3 solar mass BH with a spin parameter of 0.95 (numbers rounded up). Note that r is the coordinate radius and R is the reduced circumference in the azimuth plane-

$$r_e=M+ \sqrt{M^2-a^2 \cos^2\theta}$$

See what is frame dragging to calculate $\omega, R$ and $\alpha$

For $\theta=90$, frame-dragging rate at $r_e$ (8861.099m) is 3.6937e-05 rads/s, R=10,674.7743m, $\alpha$=0.394296 which means v=1c

For $\theta=45$, frame-dragging rate at $r_e$ (7712.59695m) is 4.8909e-05 rads/s, R=6559.6475m, $\alpha$=0.320827 which means v=1c

For $\theta=5$, frame-dragging rate at $r_e$ (5861.7987m) is 8.06154e-05 rads/s, R=629.7636m, $\alpha$=0.050769 which means v=1c

which collaborates with the previous statements that spacetime is being dragged at 1c at the ergosphere boundary (or the static limit).

I'll admit, there appears to be a coordinate singularity at $\theta=0$ though that just might be the way I've got things set up.

$\omega R$ also features in this link (in this case, $R$ is written as $\varpi$), albeit in an algebraic sense. Wheeler also mentions 'tangential velocity $Rd\phi/dt$ as recorded by the Kerr bookkeeper' in Exploring Black Holes (where $d\phi/dt=\omega$).

 

[PeterDonis]

One source of ωR/α\omega R/\alpha

What are $R$ and $\alpha$?

$v_s$ is the local velocity required for a stable orbit

I assume that by "local velocity" you mean "velocity relative to a static observer at the same spatial coordinates". But in the ergosphere there are no static observers--that is, there are no timelike worldlines with constant spatial coordinates/zero angular velocity; all timelike worldlines have positive angular velocity. So "local velocity" as you are using the term has no meaning in the ergosphere. (Your second expression for $v_s$ makes this clear, since $\sqrt{\Delta} = 0$ at the static limit, and becomes imaginary inside the static limit since $\Delta$ itself is negative.)

As you get closer and closer to the horizon, the range of allowed angular velocities for timelike worldlines gets smaller and smaller; in the limit at the horizon, there are no timelike worldlines at all that do not fall into the hole, and the outgoing null worldlines that stay at the horizon all have one fixed positive angular velocity, which is sometimes referred to as the "angular velocity of the hole" (a more precise term would be "angular velocity of the horizon"). This is the correct way of saying what the various sources you quote say sloppily.

Remove $\Omega_s$ and you have the local tangential velocity of the frame-dragging rate.

What do you mean by "remove $\Omega_s$? And what do you mean by "the local tangential velocity of the frame-dragging rate"? Remember that, as above, there are no static observers inside the static limit, so there is no meaning to the concept of "local velocity" relative to such observers.

If you are not familiar with the concept of Zero Angular Momentum Observers (ZAMOs) in Kerr spacetime, I strongly recommend learning about it. This family of observers gives a much better way of physically realizing things like what you are calling the "frame dragging rate" without having to use the sort of sloppy terminology that the sources you refer to use. This paper has a good discussion of ZAMOs:

http://arxiv.org/abs/1408.6316

Another very good source is Visser's paper on Kerr spacetime:

http://arxiv.org/abs/0706.0622

Remove the $\alpha$ component and you have the tangential velocity of the frame-dragging rate as observed from infinity.

I'm not sure what you mean by "remove the $\alpha$ component", but if you mean removing the "time dilation factor" from the formula to adjust for the "rate of time flow" at infinity, it's important to recognize that the numbers you get that way have no physical meaning; they are just "bookkeeper" numbers, as the Wheeler reference you mention indicates. Calling such a number "the velocity observed at infinity" is, IMO, very misleading; it's the sort of thing that leads to misconceptions like thinking that objects "freeze" at a black hole's horizon and never fall inside.

 

[stevebd1]

$\sqrt{\Delta}=0$ occurs at the event horizon $(r_+)$, the static limit is defined by $g_{tt}=0$ where
$$g_{tt}=1-2Mr/\rho^2$$
$v_s=(\Omega_s-\omega)R/\alpha$ becomes meaningless at the photon sphere where $v_s=1$ which ranges from 3M for a static BH to 1M for a/M=1 (eq 30, http://relativity.livingreviews.org/Articles/lrr-2013-1/articlese2.html ) (though for an object of mass, the last stable orbit (MSO) ranges from 6M for a/M=0 to 1M for a/M=1). $\ v_s$ reduces to the static solution for a/M=0.

$R$ is the reduced circumference where $R=(\Sigma/\rho)\sin\theta$, $\ \alpha$ is the reduction factor (or redshift) where $\alpha=(\rho/\Sigma)\sqrt{\Delta}$. $\ R$ is equal to coordinate r in schwarzschild metric but in Kerr metric, $R$ becomes slightly greater than coordinate r. $\ \alpha$ reduces to $\sqrt{1-2M/r}$ when a/M=0.

(For the definitions of $\Sigma, \rho$ and $\Delta$ see this link).

 

[PeterDonis]

$\sqrt{\Delta}=0$ occurs at the event horizon $(r_+)$

Ah, yes, you're right, I was misremembering which coefficient is which.

the static limit is defined by $g_{tt}=0$

Yes, which means that an integral curve of $\partial / \partial t$ is null at the static limit and spacelike inside it. That is why there can't be any static observers (observers following integral curves of $\partial / \partial t$, i.e., who have constant spatial coordinates) inside the static limit.

$v_s=(\Omega_s-\omega)R/\alpha$ becomes meaningless at the photon sphere

Yes, I see, I was mixing up $v_s$ and what you were calling "frame dragging velocity"; the latter is what becomes meaningless at the static limit (since at and inside the static limit there are no static observers).

Let me rephrase some of this in the terminology I am used to seeing. The angular velocity of a ZAMO (zero angular momentum observer) is given by

$$
\Omega_Z = - \frac{g_{t \phi}}{g_{\phi \phi}} = \frac{2 M a r}{\rho^2 \left( r^2 + a^2 \right) + 2 M r a^2 \sin^2 \theta}
$$

This angular velocity is what you have been calling the "frame dragging rate" $\omega$. In the region outside the static limit, where there are static observers (observers with zero angular velocity, whose spatial coordinates do not change), we can convert $\Omega_Z$ to a velocity, the velocity of a ZAMO relative to a static observer at the same spatial location, by basically the same method you are using to convert $\Omega_s$, the angular velocity required for a stable orbit, to a velocity; that is, we have

$$
v_Z = \Omega_Z \frac{R}{\alpha}
$$

where $R$ and $\alpha$ are the reduced circumference and redshift factor as you have defined them. This could be called the "frame dragging velocity", provided we remember that it is only valid outside the static limit.

This also makes clear that what you are calling $v_s$, the "velocity" required for a stable orbit, is not velocity relative to a static observer; it is velocity relative to a ZAMO (since $\Omega_s - \omega = \Omega_s - \Omega_Z$ is what appears in the formula). Since there are ZAMOs all the way down to the horizon, $v_s$ will be meaningful all the way down to the horizon as you have defined it. But we would only be able to define a velocity for a stable orbit relative to a static observer outside the static limit; that velocity would be

$$
v_S = \Omega_s \frac{R}{\alpha}
$$

(note that $\Omega_s$ appears here instead of $\Omega_s - \omega$).

 

[Bernie G]

The detailed answers here mostly describe what happens in the vacuum outside the singularity or object in a black hole. Does anybody want to add something about the following statements from the original poster's paper: "As a result of the continual gravitational collapse of the black hole, a stage will be reached when ... the entire matter in the black hole will be converted into quark - gluon plasma permeated by leptons. . ... We have an ultrahigh energy particle accelerator in the form of a gravitationally collapsing black hole, which can, in the absence of any physical process inhibiting the collapse of the black hole to a singularity, accelerate particles to an arbitrarily high energy and momentum without any limit."

 

[PeterDonis]

Does anybody want to add something about the following statements from the original poster's paper: "As a result of the continual gravitational collapse of the black hole, a stage will be reached when ... the entire matter in the black hole will be converted into quark - gluon plasma permeated by leptons. . ... We have an ultrahigh energy particle accelerator in the form of a gravitationally collapsing black hole, which can, in the absence of any physical process inhibiting the collapse of the black hole to a singularity, accelerate particles to an arbitrarily high energy and momentum without any limit."

In the classical model of gravitational collapse to a black hole, yes, the collapsing matter will heat up, in principle to unbounded temperatures, and will pass through various states of matter appropriate to those temperatures (quark-gluon plasma permeated by leptons is just the state of matter appropriate to a very high temperature).

Defining the energy of individual particles in the collapsing matter is problematic, however, because the region of spacetime in which the collapse occurs is not stationary; in other words, there is no time translation or symmetry, so there is no natural definition of energy. All we can say is that the stress-energy tensor describing the collapsing matter obeys the local conservation law that its covariant divergence is zero.

 

[Bernie G]

In the classical model of gravitational collapse to a black hole, yes, the collapsing matter will heat up, in principle to unbounded temperatures, and will pass through various states of matter appropriate to those temperatures (quark-gluon plasma permeated by leptons is just the state of matter appropriate to a very high temperature).

Defining the energy of individual particles in the collapsing matter is problematic, however, because the region of spacetime in which the collapse occurs is not stationary; in other words, there is no time translation or symmetry, so there is no natural definition of energy. All we can say is that the stress-energy tensor describing the collapsing matter obeys the local conservation law that its covariant divergence is zero.

Even if your analysis has problems defining energy there should be conservation of energy in a collapsing black hole; energy is not created. If the collapsing contents can not be accelerated to infinite energy the resulting object should have a finite size.

 

[PeterDonis]

Even if your analysis has problems defining energy there should be conservation of energy in a collapsing black hole

If you can't define "energy", how can you check whether it's conserved? In a non-stationary spacetime, there is no global definition of "energy", so there is way to even given meaning to the question of whether energy is conserved.

Locally, energy is conserved--more precisely, as I said before, the stress-energy tensor obeys the local conservation law that its covariant divergence is zero. But that's only local; it doesn't give you any way to define a globally conserved "energy".

If the collapsing contents can not be accelerated to infinite energy

This has no meaning because there is no way to define what "accelerated to infinite energy" means. The only local notion of "energy" is the one I referred to above, that the stress-energy tensor obeys a local conservation law. If you actually look at the underlying math, instead of trying to reason using imprecise ordinary language, you will see why this is the case.

 

[Bernie G]

If you can't define "energy", how can you check whether it's conserved? In a non-stationary spacetime, there is no global definition of "energy", so there is way to even given meaning to the question of whether energy is conserved.

Locally, energy is conserved--more precisely, as I said before, the stress-energy tensor obeys the local conservation law that its covariant divergence is zero. But that's only local; it doesn't give you any way to define a globally conserved "energy".
This has no meaning because there is no way to define what "accelerated to infinite energy" means. The only local notion of "energy" is the one I referred to above, that the stress-energy tensor obeys a local conservation law. If you actually look at the underlying math, instead of trying to reason using imprecise ordinary language, you will see why this is the case.

 

[Bernie G]

No matter what math you use and even if your model can't define energy, there's a good argument that the total kinetic energy is less than or equal to Mc^2

 

[PeterDonis]

No matter what math you use and even if your model can't define energy, there's a good argument that the total kinetic energy is less than or equal to Mc^2

Ok, then please give the argument, in detail, with references. I think you will find that, since "kinetic energy" is frame-dependent, the argument does not prove what you think it does.