I Question regarding the nature of mass inside a black hole

[Buzz Bloom]

TL;DR Summary

If there is no matter inside a black hole (BH) horizon, what is the nature of the mass of the BH?

I was reading another thread which has been closed, so I cannot ask this question there. The question is about a post by @PeterDonis (post #21).

https://www.physicsforums.com/threa...a-of-a-mass-falling-into-a-black-hole.970627/​

There are no internal pressures inside a black hole. A black hole is vacuum; there is no matter inside. It is not an ordinary object with matter inside that is supported against gravity by pressure.​

If there is no matter inside the event horizon (EH) , what is the mass made of? The Friedmann equation gives four kinds of mass contributing to the manner in which the universe scale changes:

(1) Radiation, (2) Matter, (3) Curvature, and (4) Cosmological Constant="Dark Energy".​

If the matter inside the EH is not (2), is it one or a combination of the other three, or something else?

 

[PeterDonis]

If there is no matter inside the event horizon (EH) , what is the mass made of?

The mass of a black hole is a property of the spacetime geometry. It is not associated with any matter.

Any real black hole will have originally formed by the collapse of some massive object containing matter, but that collapse process leaves behind a spacetime geometry with a property that is called "mass" even though it is not associated with any matter.

The Friedmann equation

...is irrelevant to black holes. It describes a different kind of spacetime geometry.

 

[Ibix]

When you drop something into a black hole there is nothing to stop it. It crashes into the singularity, and our models do not work there. What happens, we do not know - until we have a theory of quantum gravity.

In terms of the Friedman equation, as I understand it you can treat a black hole as matter. The FLRW model doesn't care about non-uniformity in such small scales and from a distance a black hole just looks like a star, gravitationally speaking.

 

[Dale]

Summary: If there is no matter inside a black hole (BH) horizon, what is the nature of the mass of the BH?

If there is no matter inside the event horizon (EH) , what is the mass made of?

If you solve the EFE for a spherically symmetric vacuum then your solution has two properties, spherical symmetry and the stress energy tensor is 0 everywhere. That latter property means that there is no mass anywhere in the solution.

When you solve the EFE under those conditions using natural units and the usual coordinates you get a solution of the form $ds^2 = -(1-\frac{R}{r})dt^2+(1-\frac{R}{r})^{-1} dr^2+r^2(d\theta^2+sin^2\theta \; d\phi^2)$. This solution has one free parameter R, which is the Schwarzschild radius. It is a free parameter which can be adjusted so that the solution matches the curvature boundary condition just outside any spherically symmetric matter distribution .

Now, because we are working in natural units, you can replace R with M, which gives you a mass that is equivalent to the Schwarzschild radius in natural units. However, I personally would prefer that people did not make that substitution, because it leads to exactly your confusion. Since the solution is a vacuum solution the free parameter is not a mass, it is a characteristic length, the Schwarzschild radius, which sets the curvature of the vacuum on a given boundary. It is not coincidence that the R is equal to the amount of mass (in geometric units) that is inside the vacuum region, but it is important to understand that such mass is outside of the manifold (or at least outside of the portion of the manifold described by the above metric). In the manifold there is no mass, but there is curvature given by the characteristic length of the Schwarzschild radius.

 

[Buzz Bloom]

spacetime geometry with a property that is called "mass"

Hi Peter:

As matter falls from outside the EH and passes to inside, during it's time of continuing to fall towards the center, is it still matter? If so , when does it stop being matter? What is the process by which it stops being matter?

Just wondering. Is it possible that the spacetime geometry involves curvature with very large local value of Ω_k?

Regards,
Buzz

 

[PeterDonis]

As matter falls from outside the EH and passes to inside, during it's time of continuing to fall towards the center, is it still matter?

Yes.

when does it stop being matter?

When it hits the singularity. At least, that's the classical GR answer, but it's probably not the final answer. See below.

What is the process by which it stops being matter?

We don't know because we don't have a good theory of quantum gravity, and we expect that, at least close enough to the point where classical GR says there is a singularity, quantum gravity effects will be strong enough to significantly affect the physics. So this will probably remain an open question until we have a good theory of quantum gravity and can use it to tell us what happens in this regime.

Is it possible that the spacetime geometry involves curvature with very large local value of Ωk?

$\Omega_k$ only is meaningful in the FLRW spacetime geometry. It's not meaningful in the Schwarzschild spacetime geometry.

 

[Buzz Bloom]

but there is curvature given by the characteristic length of the Schwarzschild radius.

Hi Dale:

You seems to have answered a question I asked while you wee posting. Thank you.

Is it possible that the spacetime geometry involves curvature with very large local value of Ω_k?​

I understand from Peter's post #6 that the link between mass and curvature may be something other than Ω_k.

Regards,
Buzz

 

[Dale]

I am not a cosmological expert, so perhaps I should defer, ... but to my understanding there is no local value for those parameters

 

[PeterDonis]

I understand from Peter's post #6 that the link between mass and curvature may be something other than $\Omega_k$.

$\Omega_kl$ in FLRW spacetime is not a "link between mass and curvature". It is a convenience for analyzing certain aspects of the dynamics of such geometries by putting the effect of spatial curvature on the same scale as the effects of the various types of stress-energy that can be present. None of this has anything to do with mass or spacetime curvature, or the relationship between the two, in Schwarzschild spacetime.

 

[Heikki Tuuri]

Suppose that you let a dust shell collapse into a black hole. A distant observer will continue "seeing" all the collapsed matter outside the Schwarzschild radius for ever. That is, at any time t in the future, a light speed signal can arrive from the falling dust particle if sent at a suitable time t'.

For the distant observer, all mass will look like frozen, floating above the forming event horizon.

The "foliation" of spacetime which is made according to the clock time of the distant observer always will have the matter outside the event horizon.

https://www.sciencemag.org/news/2007/06/no-more-black-holes
If there is some process which destroys a black hole in a finite time as measured by the distant observer, then the black hole will disappear before any event horizon will form. This means that no particle will ever observe itself to cross the Schwarzschild radius of the mass, and will not fall to a singularity.

Where is the matter content? The distant observer may imagine that the matter content is floating above the forming event horizon. It does not make much sense for him to think that it is past the event horizon, because he does not know if black holes are eternal or will be destroyed.

If black holes are eternal, then a falling observer does see the dust shell fall past the Schwarzschild radius. Some or all the matter will form a singularity of some kind, by the singularity theorem of Roger Penrose. What happens after that, we do not know because General Relativity does not work with singularities.

The time of a falling observer who is past the event horizon is "after" all the times of the distant observer, because no signal can go to the distant observer. The concept of simultaneity is not very sensible under these conditions. It would be strange to say that on June 1, 2019, some mass has "now" fallen behind an event horizon.

 

[PeterDonis]

The "foliation" of spacetime which is made according to the clock time of the distant observer always will have the matter outside the event horizon.

But this foliation does not cover the entire spacetime. I discuss this in my Insights articles about the Schwarzschild spacetime geometry:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

If there is some process which destroys a black hole in a finite time as measured by the distant observer, then the black hole will disappear before any event horizon will form.

This is not correct. We've had plenty of previous threads about this. The reference you give does not give a link to the actual paper, and I can't find it online.

The time of a falling observer who is past the event horizon is "after" all the times of the distant observer, because no signal can go to the distant observer.

This is not correct either. There are at least two known coordinate charts--Painleve and Eddington-Finkelstein--which have surfaces of simultaneity that extend from outside to inside the horizon.

 

[Heikki Tuuri]

https://arxiv.org/abs/1409.0187
Hawking Evaporation is Inconsistent with a Classical Event Horizon at r=2M

Borun D. Chowdhury (1), Lawrence M. Krauss (1,2)

Many people have observed that an event horizon cannot form if the black hole evaporates in a finite time. We assume here that the physics stays the same way past the Planck scale. A distant observer will see the matter come to a really infinitesimal distance from the Schwarzschild radius.

If there are quantum gravity effects at the horizon, then we do not know what happens.

 

[PeterDonis]

Many people have observed that an event horizon cannot form if the black hole evaporates in a finite time.

And all of those observations are obviously inconsistent with the fact that there are known spacetime geometries that describe a black hole that forms an event horizon and then evaporates by Hawking radiation. Hawking himself constructed the first of them back in the 1970s when he discovered Hawking radiation. There is an ongoing debate about how physically realistic those spacetime geometries are, but they are certainly mathematically consistent.

I'll take a look at the paper you linked to (thanks for finding the link), but at first glance it seems to be making the same error as above. A thing can't be impossible if someone has already done it.

 

[PeterDonis]

Many people have observed that an event horizon cannot form if the black hole evaporates in a finite time.

The standard rebuttal to this (based on the known spacetime geometries I referred to in my previous post just now), which you will find in plenty of threads here at PF, is that the statement that it takes infinite time, according to an observer at infinity, for the event horizon to form, is based on the "eternal" black hole spacetime geometry of Schwarzschild, which does not evaporate. Adding Hawking radiation and black hole evaporation changes the spacetime geometry, and that in turn changes what the observer at infinity observes. Instead of observing it to take an infinite time for anything to fall to the horizon, he now observes infalling objects crossing the horizon at the same instant that the hole evaporates--or, to put it in less misleading terms, any light emitted radially outward at the horizon reaches the distant observer along with light from the hole's final evaporation. But this in no way prevents there from being an event horizon and a region of spacetime inside it that cannot send light signals to infinity at all, and into which things can fall.

 

[PeterDonis]

A Penrose diagram describing an evaporating black hole spacetime (basically the one Hawking found) is shown here:

https://backreaction.blogspot.com/2009/11/causal-diagram-of-black-hole.html

 

[Dale]

The "foliation" of spacetime which is made according to the clock time of the distant observer always will have the matter outside the event horizon.

Not really.

The vacuum region, which is described by the Schwarzschild metric, will not contain matter by definition. Within the non-vacuum region the metric is not the Schwarzschild metric, so the foliation defined by the Schwarzschild coordinates is not particularly natural any more. Furthermore, that region is also not static, so the Schwarzschild coordinate based foliation doesn’t make sense there since it is static.

A sensible foliation of the non-vacuum region requires attention to these issues. Making conclusions about the non-vacuum non-static region based on the naive foliation of the static vacuum region is a non-starter.

This means that no particle will ever observe itself to cross the Schwarzschild radius of the mass, and will not fall to a singularity.

This does not follow.

 

[Dale]

https://arxiv.org/abs/1409.0187
Hawking Evaporation is Inconsistent with a Classical Event Horizon at r=2M

Borun D. Chowdhury (1), Lawrence M. Krauss (1,2)

I have a serious objection to the papers claim in section 1 that “We make no attempt to rely upon any possible global description of a black hole encompassing both the points of view of the in falling and dis- tant observers, because, as is well known, no such global description exists.”

There are many coordinate charts that cover a distant observer and an infaller! I wonder if this is fixed in the published version.

 

[PeterDonis]

I have a serious objection to the papers claim in section 1 that “We make no attempt to rely upon any possible global description of a black hole encompassing both the points of view of the in falling and dis- tant observers, because, as is well known, no such global description exists.”

Yes, I'm surprised to see that coming from Krauss. But he wouldn't be the first physicist to misunderstand the Schwarzschild geometry.

Btw, reference [4] in the paper is this:

https://arxiv.org/pdf/gr-qc/0609024.pdf
Here is the abstract:

"We study the formation of black holes by spherical domain wall collapse as seen by an asymptotic observer, using the functional Schrodinger formalism. To explore what signals such observers will see, we study radiation of a scalar quantum field in the collapsing domain wall background. The total energy flux radiated diverges when backreaction of the radiation on the collapsing wall is ignored, and the domain wall is seen by the asymptotic observer to evaporate by non-thermal “pre-Hawking radiation” during the collapse process. Evaporation by pre-Hawking radiation implies thatan asymptotic observer can never lose objects down a black hole. Together with the non-thermalnature of the radiation, this may resolve the black hole information loss problem."

From this it seems to me like the sort of model Krauss actually has in mind is a model where quantum effects stop gravitational collapse before a horizon can form. Classically, such models would have a stress-energy tensor inside the collapsing matter that strongly violates the weak energy condition due to quantum field effects as the collapse approaches horizon formation. In other words, this looks like a kind of "bounce" model.

If that's where Krauss is coming from, his error in the 2014 paper is simply to confuse the "bounce" type of model in the 2006 paper with the Schwarzschild vacuum geometry--or, perhaps more realistically, with collapse models such as the original Oppenheimer-Snyder model, which assumed that the collapsing matter would satisfy the energy conditions everywhere. The argument in the 2014 paper, if it were correct, would imply that models like the original Oppenheimer-Snyder model simply cannot exist, which is obviously false.

 

[PeterDonis]

Another note: both the 2014 Krauss paper and the 2006 Krauss paper were published in Physical Review D, which as I understand it publishes much more speculative and less well checked material. I doubt these papers would have made it past peer review in one of the more rigorous GR journals.

 

[Nugatory]

For the distant observer, all mass will look like frozen, floating above the forming event horizon.

That's not right, although it's a very common misunderstanding. This "frozen mass" conclusion comes from misapplying the Schwarzschild solution, which is only valid if the infalling mass is negligible compared to the mass of the black hole so that the size of the horizon doesn't change (and note that even in that situation, Schwarzschild coordinates don't work at the horizon - you have to use some other coordinates that are not singular at the horizon to describe the Schwarzschild spacetime there).

Back up for a moment and consider a black hole of mass $M$, surrounded by a spherically symmetrical shell of dust with total mass $m$ at a great distance from the black hole and falling/collapsing into the black hole. Consider $R'$, the Schwarzschild radius of a hypotherical black hole of mass $M+m$ and $R$, the Schwarzschild radius of the black hole. Clearly $R'$ is greater than $R$ so is outside the event horizon; thus the infalling shell of dust will reach $R'$ in finite time according to you and other external observers. But once it gets there... we have a spherically symmetric mass distribution all inside its Schwarzschild radius $R'$, and that is a black hole with radius $R'$. So our initial state is a black hole of radius $R$ and our final state is a black hole of radius $R'$, and we get from one to the other in a finite external time. We just can't use the Schwarzschild solution to describe what's happening in between.

 

[PeterDonis]

Clearly $R'$ is greater than $R$ so is outside the event horizon; thus the infalling shell of dust will reach $R'$ in finite time according to you and other external observers.

Careful. Once the infalling shell of matter is inside the radial coordinate of the external observer (which by hypothesis is distant from the hole), the geometry to the future of the external observer is Schwarzschild with mass $M + m$, and therefore with event horizon at $R'$, not $R$. That means this external observer will not, in fact, see the shell reach $R'$. (We are assuming a purely classical model here with no Hawking radiation.)

Another way of seeing this is to note that once the infalling shell reaches $R'$, it is at the new horizon, so light emitted radially outward by the shell at $R'$ will stay at $R'$ forever. So such light can't possibly get back out to the external observer.

What will be true is that an observer at radial coordinate $R'$ before the shell falls in will see the shell reach $R'$, even though, since they are outside $R$, they never see anything fall through the horizon of the hole before the shell arrives.

So our initial state is a black hole of radius $R$ and our final state is a black hole of radius $R'$, and we get from one to the other in a finite external time.

No, we don't, if "external time" means a Schwarzschild time coordinate in the exterior region after the shell passes. See above.

We just can't use the Schwarzschild solution to describe what's happening in between.

We can still use the Schwarzschild solution for most of this spacetime; we just have to be precise about specifying which Schwarzschild solution applies where.

In 4-D spacetime geometry terms, what we have is the following: A region with vacuum Schwarzschild geometry with mass $M$ (region A), separated by a timelike thin shell region (region S) from a region with vacuum Schwarzschild geometry with mass $M + m$ (region B).

If the shell starts at rest at a finite radius, then an observer sufficiently far away (farther than the finite radius the shell starts at) will be in region B the whole time, and will never see the shell reach $R'$, as above. Also, the term "external time" is best used to refer to the Schwarzschild time coordinate in region B for this case. An observer inside the radius the shell starts at will start out in region A and will not see objects reach the original horizon radius at $R$; at some point they will pass through region S (when the shell falls past them) and will then be in region B. Whether or not they see the shell reach $R'$ will depend on whether they started out at a radius larger than $R'$, or between $R$ and $R'$ (assuming they stay at the same radius throughout).

If the shell falls in "from rest at infinity", then as we go farther into the past, observers at larger and larger radius will be in region A rather than B; there will be no finite radius at which an observer will be in region B the whole time. But for any finite time in the past, there will also be some sufficiently large finite radius which is still in region B; so even for this case, the best meaning for the term "external time" is the Schwarzschild time coordinate in region B.

 

[Heikki Tuuri]

If one draws the worldline of a hypothetical Hawking radiation quantum, the worldline cannot appear from a genuine event horizon. That is by definition: no worldline can climb up through the event horizon.

Therefore, all the energy that will ever radiate away, is already above the hypothetical event horizon, in any spacetime foliation, at every "moment" of global time.

It is hard to find a model where a genuine event horizon could exist under these conditions. If a mass M has fallen inside its Schwarzschild radius r = 2M, and we have the same mass M flowing outward above the radius 2M, then the event horizon should be at r = 4M, which makes no sense.

 

[PeterDonis]

If one draws the worldline of a hypothetical Hawking radiation quantum, the worldline cannot appear from a genuine event horizon.

That's right. But nobody claims that Hawking radiation comes from the event horizon. It comes from just above the event horizon. That does not mean there cannot be an event horizon in a spacetime in which there is Hawking radiation.

all the energy that will ever radiate away, is already above the hypothetical event horizon, in any spacetime foliation, at every "moment" of global time.

If you view the energy as being "stored" in the spacetime curvature outside the horizon, this is fine. But there is no way to localize where this energy is stored.

It is hard to find a model where a genuine event horizon could exist under these conditions.

It might be hard for you, but that just means you are not sufficiently familiar with the literature, in which, as I've already said, such models have been published. I suggest that you take some time to look through it. Start with Hawking's original papers on Hawking radiation.

If a mass M has fallen inside its Schwarzschild radius r = 2M, and we have the same mass M flowing outward above the radius 2M, then the event horizon should be at r = 4M

Huh? The total energy contained in the black hole plus Hawking radiation is constant. As the radiation is emitted, the mass of the hole decreases. There is no "double counting" of the mass.

You also appear to think that the mass of a black hole is "stored" somewhere inside the horizon. It isn't. It's a global property of the spacetime geometry. As I said above, one way to think about Hawking radiation is that it is just a quantum effect that converts spacetime curvature, i.e., "mass" as a property of the spacetime geometry, to outgoing radiation.

 

[Heikki Tuuri]

Above, I assume that a "quantum" of Hawking radiation exists, and we can follow the worldline of the quantum backward in time.

There is no consensus in physics from where the hypothetical Hawking radiation originates in spacetime.

In Hawking's original paper, 1975, the "quanta" are the negative frequencies that appear in a hypothetical photon wave packet. The derivation assumes that we can calculate the wave packet backward in time, all the way through the collapsing star and to outer space. The whole derivation is based on this.

Thus, in Hawking's original paper, the outflowing energy is always outside the hypothetical event horizon.
https://www.iflscience.com/physics/...oven-mathematically-black-holes-do-not-exist/
The fuss around Laura Mersini-Houghton's calculation back in 2014 highlights the fact that nothing is settled science when we talk about black hole evaporation.

 

[PeterDonis]

In Hawking's original paper, 1975, the "quanta" are the negative frequencies that appear in a hypothetical photon wave packet.

If you mean the quanta of Hawking radiation that are observed far away, no, this is not correct.

The argument in Hawking's original paper ("Particle Creation by Black Holes", 1975) goes like this: he starts with a quantum field state that, in the asymptotic past, is a vacuum state. He then shows that this state, when it is evolved into the asymptotic future on a background Schwarzschild spacetime geometry, is not a vacuum state: it contains field quanta with positive energy propagating to infinity. (Note, btw, that he uses a scalar field, not photons.) This is possible because the decomposition of the field into positive and negative frequency parts is different in the far past vs. the far future. So a field state that is vacuum in the far past can, when viewed in the far future, be composed of a positive frequency part that propagates to infinity, and a negative frequency part that propagates into the black hole. The negative frequency part propagating into the hole is what decreases the hole's mass.

in Hawking's original paper, the outflowing energy is always outside the hypothetical event horizon.

If you call the positive frequency part of the field in the far future the "outflowing energy", yes, this is true. But there is also a part of the field that "flows" into the hole (below the horizon) and it is this part that reduces the hole's mass. See above.

Also, I don't know why you call the event horizon "hypothetical" in Hawking's model. He uses the Schwarzschild geometry as the background spacetime, and the Schwarzschild geometry contains a real event horizon, not a hypothetical one.

The fuss around Laura Mersini-Houghton's calculation back in 2014 highlights the fact that nothing is settled science when we talk about black hole evaporation.

Did you read the quoted comments from William Unruh in that article? They express the general opinion of physicists concerning this paper. We had a discussion of it on PF here:

https://www.physicsforums.com/threa...ay-prevent-formation-of-event-horizon.772816/

 

[Heikki Tuuri]

I did read Bill Unruh's comments back in 2014. The problem is that I am not aware of any reasonable calculation of the backreaction of Hawking radiation on the system. A part of this problem is the famous information paradox of black holes: if we would know where and how the "quanta" of Hawking radiation are born, then there would be no paradox - just calculations.

We do understand some radiation emitted by black holes - the accretion disk glows bright, even as a quasar in young galaxies. The photons are are born from collisions and interactions of particles in the accretion disk. For Hawking radiation, no such clear picture exists.

What do we really know about black holes? We know the empirical things - the accretion disk shines bright but we do not see radiation from the matter hitting any "surface" of a black hole. The binary pulsar observation of 1979 as well the waves deteted by LIGO prove that the geometry is roughly what general relativity predicts.

It is way too optimistic to believe we can have any settled science about black hole evaporation without quantum gravity.

 

[PeterDonis]

I am not aware of any reasonable calculation of the backreaction of Hawking radiation on the system.

It depends on your definition of "reasonable", I suppose. Until we have actually observed some Hawking radiation, I don't see how we will be able to actually test any such calculation, so it's a matter of everyone's individual opinion on reasonableness. It certainly seems reasonable to me that the energy observed to be carried away to infinity by Hawking radiation should be equal to the reduction in mass of the black hole.

A part of this problem is the famous information paradox of black holes

Yes. Some physicists, like Susskind, are claiming this has been resolved, but not all physicists agree. I don't think it has myself.

It is way too optimistic to believe we can have any settled science about black hole evaporation without quantum gravity.

Who here is claiming that we have "settled science" about black hole evaporation?

The fact that we don't know for sure what the right model is does not prevent us from pointing out obviously wrong things in some proposed models, or from making reasonable hypotheses about general characteristics like the energy balance I described above.

 

[member 656954]

https://www.sciencemag.org/news/2007/06/no-more-black-holes

I note that this reference is over a decade old. Has there been any update to Physicist Lawrence Krauss and Case Western Reserve colleagues' assertion that BHs can't form?

I'd guess the LIGO data says their 'constructed mathematical formulas' are incorrect, but the universe is a surprising beast (to me at least) and it seems possible elaborations on the premise may have yielded subtler outcomes than merely wrong/right.

 

[PAllen]

I did read Bill Unruh's comments back in 2014. The problem is that I am not aware of any reasonable calculation of the backreaction of Hawking radiation on the system. A part of this problem is the famous information paradox of black holes: if we would know where and how the "quanta" of Hawking radiation are born, then there would be no paradox - just calculations.

Whether you think they are reasonable or not, there are in fact many detailed calculations of backreaction of Hawking radiation on BH formation that claim it does not stop horizon formation. For example, a moderately recent classic is:

https://arxiv.org/abs/0906.1768
This was partly in response to the claims of Krauss, et. al. in some earlier papers.

 

[PAllen]

And here is a more formal response by Unruh to this line of work:

https://arxiv.org/abs/1709.00115

 

[Heikki Tuuri]

PAllen, thank you for the link to the

https://arxiv.org/abs/1709.00115

Arderucio-Costa & Unruh paper. I had not seen that paper. The paper is very technical and I did not yet check the math in it.

Let us think about the backreaction. A consensus is that a freely falling observer sees no drama at the forming horizon. The energy flux of hypothetical Hawking radiation is almost invisible for the freely falling observer.

On the other hand, static observers close to the horizon do see the Hawking energy flux. According to the 1975 paper by Hawking, the energy flux comes from negative frequencies which are generated by the rapidly changing gravitational field in a hypothetical wave packet which has passed through the star just before the formation of the horizon.

Static observers do see the entire mass M of the collapsing star converted to Hawking radiation. The radiation is moving up just above the forming horizon. It may take a photon 10^67 years to climb up. That is the reason why the evaporation is so slow.

A freely falling observer, on the other hand, sees almost no Hawking radiation. He sees the entire mass M falling ahead of him, first toward the forming horizon, and then toward the singularity.

Now we have a problem because static observers do see the falling mass M. It never disappears from their incoming light cone. Static observers also see the upcoming Hawking radiation. It contains the mass M, too. The total mass they see is 2M, which makes no sense.

Some of the mass M is converted to outgoing radiation through conventional processes in the accretion disk: friction and collisions. There is no paradox in that radiation. The paradox is in the bulk of matter M which seems to become duplicated at the horizon.

Leonard Susskind has tried to fix the paradox with something he calls black hole complementarity. The firewall paradox of the AMPS paper was a serious blow to complementarity.

My current view is that Hawking radiation does not exist. That would solve most paradoxes. I need to check what implications that would have for AdS/CFT and black hole thermodynamics.

 

[PeterDonis]

A consensus is that a freely falling observer sees no drama at the forming horizon.

Except for proponents of firewall models.

static observers close to the horizon do see the Hawking energy flux.

But they interpret this as Unruh radiation; the black hole horizon is also the Rindler horizon for these observers. They have no way of knowing, locally, that this radiation will over a very long period of time decrease the mass of the hole; for all they know it could be entirely due to their own acceleration, as Unruh radiation is.

static observers do see the falling mass M. It never disappears from their incoming light cone.

This is not correct. Once the falling mass is inside the horizon, it is outside the past light cone of any observer (static or not) outside the horizon.

The total mass they see is 2M

This is not correct. See above.

What the static observer actually sees is an infalling mass M disappearing from his past light cone, and then, much, much later, the same total amount of mass appearing over a long, long time as Hawking radiation. There is no issue with energy conservation at all.

 

[Heikki Tuuri]

PeterDonis,

https://phys.libretexts.org/Bookshelves/Relativity/Book:_General_Relativity_(Crowell)/7:_Symmetries/7.3:_Penrose_Diagrams_and_Causality

if you draw the Penrose diagram of a collapsing star, you notice that the falling matter remains in the light cone of a far-away observer. That is, for every future time t of the outside observer, there is a straight line 45 degrees up to the right, such that the line intersects the worldline of the observer at his proper time t. The line, of course, is outside the forming horizon. This means that the mass remains in his light cone.

If the mass inside the horizon keeps decreasing, then the line is not straight, but it nevertheless is a geodesic, a path of light.

The information problem of black holes is this paradox: how can the same matter and information fall behind the horizon in the Penrose diagram and be duplicated in the hypothetical Hawking radiation outside the horizon?

The "backreaction" to Hawking radiation should explain how the mass-energy of the black hole decreases. I will look at the AdS/CFT black hole and try to decipher the backreaction there.

UPDATE: In the Penrose diagram, let us draw the worldlines of baryons B1 and B2 falling toward the horizon. The lines go up left at an angle of, say, 50 degrees from the horizontal line.

horizon
______/ photon
_____/_/_____ F4 folio
_\_\/_/______ F3 folio
__\/\/_______ F2 folio
__/\_\_______ F1 folio
_/ B1 B2

If the baryons happen to collide, then a free falling observer sees them emit a photon and he sees the combined mass-energy of B1 & B2 decrease. A late time observer sees the same thing: he receives the photon and sees the system B1 & B2 lose some kinetic energy or heat.

The geometry of spacetime at the observer is determined from the information within his light cone. The geometry correctly reflects the fact that energy was conserved in the system B1 & B2 & photon.

If B1 and B2 fall uneventfully to the horizon, then the geometry is the classical black hole.

Let us add an energy flux of Hawking radiation. We may imagine a shell around the black hole close to the horizon. The shell collects all positive energy from Hawking radiation.

The observer sees the mass-energy M fall uneventfully to the horizon. He also sees the same mass-energy M collected by the shell. Free falling observers see these same things: the mass M is in the falling matter and the mass M is also in the shell.

General relativity calculates the geometry in the next folio of spacetime from the information in the previous folio. The input is the geometry in the previous folio plus the energy density at each point, measured by a free falling observer.

In the folio F3 of the diagram, the geometry reflects both the mass M which falls to the horizon, and also the same mass M contained in the shell. This contradicts the conservation of energy.

 

[PeterDonis]

for every future time t of the outside observer, there is a straight line 45 degrees up to the right, such that the line intersects the worldline of the observer at his proper time t. The line, of course, is outside the forming horizon. This means that the mass remains in his light cone.

This is correct, but it does not mean what you claim it means.

Here is what it does mean: (1) For every event on the worldline of the outside observer, there is infalling matter present somewhere in the past light cone of that event (though you might have to go back very, very far in time to reach it).

Here is what it does not mean: (2) For every event in the spacetime region occupied by the infalling matter, there is an outgoing null worldline that originates at that event and reaches the worldline of the outside observer. This is obviously false for the portion of the infalling matter region that is at or below the horizon.

You are pointing out that statement #1 above is true (which it is), but the argument you are trying to make requires statement #2 above to be true (and it isn't).

If the mass inside the horizon keeps decreasing

This is not a correct description of what is happening. The region occupied by the infalling matter inside the horizon is inside the horizon; in a Penrose diagram of a spacetime with a black hole that evaporates by Hawking radiation (such as the one appearing in Hawking's original paper), statement #2 above is still false, and the Hawking radiation therefore cannot be coming from the region occupied by infalling matter inside the horizon. That matter region ends at the future singularity, just as it does in the Penrose diagrams shown in the article you linked to.

However, the "mass" of the black hole at times long after the original gravitational collapse is not obtained by looking at the infalling matter region. It's obtained by looking at the vacuum region outside the horizon and measuring its spacetime geometry. In other words, "mass" is a global property of the geometry. And so is "decreasing mass due to Hawking radiation" for that case.

The information problem of black holes is this paradox: how can the same matter and information fall behind the horizon in the Penrose diagram and be duplicated in the hypothetical Hawking radiation outside the horizon?

And one answer (the answer Hawking originally gave, although his opinion changed much later on) is that the Hawking radiation does not contain the information in the infalling matter: that the information in the infalling matter is lost when it hits the singularity. Many physicists don't like this answer because it violates unitarity, but that just means they prefer one counterintuitive thing (figuring out how unitarity is preserved) to another (figuring out how unitarity can be violated).

 

[PeterDonis]

General relativity calculates the geometry in the next folio of spacetime from the information in the previous folio.

This is one way to find a global solution in GR, but by no means the only one.

 

[PeterDonis]

The observer sees the mass-energy M fall uneventfully to the horizon. He also sees the same mass-energy M collected by the shell.

Not at the same time. First the observer sees M fall uneventfully to the horizon, Then, much later, he sees it collected by the shell. He never sees it in both places at once.

 

[PeterDonis]

In the folio F3 of the diagram, the geometry reflects both the mass M which falls to the horizon, and also the same mass M contained in the shell. This contradicts the conservation of energy.

First, there is no global conservation of energy in GR. Energy conservation is local: the covariant divergence of the stress-energy tensor is zero at every event. That is satisfied by the solution you are describing.

It is true that, because of the way the geometry works inside the horizon, you can find spacelike hypersurfaces on which the infalling matter region appears (far inside the horizon) and a region containing Hawking radiation also appears (outside the horizon). However, you can also find other spacelike hypersurfaces for which this is not the case. As the article you linked to notes, there is no way to pick out any particular set of spacelike hypersurfaces as the "correct" or "true" one.

 

[Heikki Tuuri]

the forming
horizon
______/ quantum E
_____/_/_____ F4 folio
_\_\/_/______ F3 folio
__\/\/_______ F2 folio
__/\_\_______ F1 folio
_/ B1 B2
mass M

The diagram I drew in the previous message tells us what should happen, to keep the geometry consistent for a collapsing mass M.

Let us first think about a classic non-evaporating black hole. The baryons B1, B2 fall into the singularity. There are no outcoming photons. The horizon forms at the folio F3 as a perfect straight 45 degree line to northeast.

Let us then add a quantum of energy E of Hawking radiation. The quantum is present in the folio F4. How do we modify the diagram, so that the geometry for an outside observer stays the same? Birkhoff's theorem states that the gravitational field of the collapsing spherical mass M must stay constant at all times.

If we add another quantum with energy -E falling into the horizon, then Birkhoff will be happy:

-E \/ E

A problem in this is that no one has been able to formulate a mathematical model where quanta of negative energy pop up and are swallowed by the black hole. In Hawking's 1975 paper, the quanta E gain their energy at the forming horizon. The energy E is the negative frequencies in a hypothetical wave packet which travels through the collapsing mass. There is no negative energy quantum in the calculation.

Anyway, let us assume that there is a process which sends the energy -E down to the horizon.

If E gains its energy at the intersection of the lines B2 and E, then B2 is accompanied by an energy -E falling with B2 to the horizon. Actually, this description fits ordinary radiation emitted by B2. No horizon ever forms because all the mass-energy of B2 is canceled by the Hawking quanta -E.

For a horizon to form, the quantum E must gain its energy at a folio later than F2. In Hawking's 1975 paper, the energy E is gained at the folio F3, because the horizon has just formed there. But then the negative quanta -E would fall into the horizon in one huge batch, and the horizon would disappear at that instant.

A peaceful evaporation of a black hole is possible if the quanta E gain their energy at very late folios and send the negative energy -E down to the horizon. But in Hawking's 1975 paper, we do follow the path of E back in time all the way to the distant past. The path is extremely close to the forming horizon at the folio F3. That is the natural place where the energy E is gained.

The title of this thread is "The nature of mass inside a black hole". If there is no Hawking radiation, then the mass is in a singularity and we do not know anything about the nature there.

If Hawking radiation exists, then we face many additional problems. A few of those problems are presented in this message. The information loss problem and the firewall come on top of that.

The geometry outside a horizon is not affected in any way by what happens behind the horizon, because no event behind the horizon is in the light cone of an outside observer. The observer does not need to know anything about the nature behind the horizon. Currently, there is no consensus about what happens to an observer who jumps into the horizon.

I will look at the AdS/CFT black hole. As far as I know, no one has been able to decipher there what is the backreaction to hypothetical black hole evaporation.

There is one good reason to believe that Hawking radiation exists: thermodynamics. The "derivations" of Hawking radiation are riddled with problems.

...

UPDATE: It just occurred to me that if the quanta of negative energy -E are moving outward, just like the quanta E of positive energy, then the diagram would allow the creation of a horizon. Then there would be a positive mass M behind the horizon, a negative mass -M very slowly falling into the horizon, and a positive mass M slowly escaping the black hole.

This would also solve the problem of the conservation of momentum: if Hawking quanta exist, from where do they get the momentum p to the direction out of the horizon? The momenta of -E and E would cancel each other. The black hole would swallow the momentum -p as well as the energy -E.

The diagram which describes the creation of quanta -E and E should look like this:

-E // E

The new diagram stresses that the quanta move at the speed of light upward from the horizon. The quanta -E very slowly approach the singularity. Slowly, because they are moving at the speed of light upward. The quanta -E are below the Schwarzschild radius. Light moving straight out there will end up at the singularity.

An observer falling into the black hole would encounter a huge mass -M of negative energy quanta at the horizon. Does he see them? He does not see upcoming Hawking quanta.

 

[PeterDonis]

Birkhoff's theorem states that the gravitational field of the collapsing spherical mass M must stay constant at all times.

Not in this case, it doesn't. Birkhoff's theorem applies to a spherically symmetric vacuum region of spacetime. If there is Hawking radiation present, the region of spacetime exterior to the collapsing matter is not vacuum.

If we add another quantum with energy -E falling into the horizon

Real quanta can't have negative energy.

Please review the PF rules on personal speculation, which is what you are verging on here.

 

[PeterDonis]

The OP question has been addressed. Thread closed.