I Question regarding the nature of mass inside a black hole

Buzz Bloom

Gold Member
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).
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?

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PeterDonis

Mentor
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

Mentor
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

Gold Member
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 Ωk?

Regards,
Buzz

PeterDonis

Mentor
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

Gold Member
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 Ωk?​
I understand from Peter's post #6 that the link between mass and curvature may be something other than Ωk.

Regards,
Buzz

Dale

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

PeterDonis

Mentor
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.

• Buzz Bloom

PeterDonis

Mentor
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:

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.

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

Mentor
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

Mentor
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

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

Dale

Mentor
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

Mentor
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

Mentor
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  in the paper is this:

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.

• Dale

PeterDonis

Mentor
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

Mentor
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

Mentor
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

Mentor
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/physicist-claims-have-proven-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

Mentor
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:

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