# What is the mass of a black hole made of?

anorlunda
Staff Emeritus
It certainly seems likely that whatever is there is broken down into elementary particles but we don't know that and likely won't until there evolves a provable theory of quantum gravity that may (or may not) let us figure out what is likely to be happening at the quantum level.
Do the possibilities include the matter being converted to massless particles such as photons? Or perhaps that the energy is stored in a field with no particles at all? Perhaps other types of conservation (charge?, baryon number? ...) forbids those scenarios.

It always seemed to me that science should discuss the energy inside the BH based on the observation that it gravitates, and avoid any presumption about the form of that energy since we don't know for sure.

kent davidge
PeterDonis
Mentor
2019 Award
Do the possibilities include the matter being converted to massless particles such as photons? Or perhaps that the energy is stored in a field with no particles at all?
Classically, whatever falls into the singularity at ##r = 0## just disappears; it doesn't get converted to something else. The mass (meaning the ##M## that appears in the Schwarzschild line element) is not "stored" anywhere except in the geometry of the spacetime as a whole.

When we add quantum gravity to the mix, our current best guess is probably that whatever falls into the black hole eventually gets converted into Hawking radiation and is radiated back out. But that's only a best guess; we won't know until we have a good theory of quantum gravity. And even the best guess I just described isn't backed up by anything very strong; there are plenty of speculations in this area that can't be tested against each other experimentally, now or in the foreseeable future.

Perhaps other types of conservation (charge?, baryon number? ...) forbids those scenarios.
If the best guess I gave above is correct, then baryon number isn't actually conserved; it's only approximately conserved, and deep inside a black hole is one of the places the approximation breaks down. (Actually, it probably breaks down in the early universe as well; the fact that our universe contains more matter than antimatter, when it probably started from a state at the very end of inflation that was matter-antimatter symmetric, indicates that baryon number was not conserved back then. But that's another area where we don't really understand what's going on.)

Charge is conserved, but any black hole of significant size is probably going to be electrically neutral, just because if it happens to get some charge, by having charged matter fall into it, it will attract opposite charges which will fall in and neutralize it again.

kent davidge
Classically, whatever falls into the singularity at ##r = 0## just disappears; it doesn't get converted to something else. The mass (meaning the ##M## that appears in the Schwarzschild line element) is not "stored" anywhere except in the geometry of the spacetime as a whole.

When we add quantum gravity to the mix, our current best guess is probably that whatever falls into the black hole eventually gets converted into Hawking radiation and is radiated back out. But that's only a best guess; we won't know until we have a good theory of quantum gravity. And even the best guess I just described isn't backed up by anything very strong; there are plenty of speculations in this area that can't be tested against each other experimentally, now or in the foreseeable future.

If the best guess I gave above is correct, then baryon number isn't actually conserved; it's only approximately conserved, and deep inside a black hole is one of the places the approximation breaks down. (Actually, it probably breaks down in the early universe as well; the fact that our universe contains more matter than antimatter, when it probably started from a state at the very end of inflation that was matter-antimatter symmetric, indicates that baryon number was not conserved back then. But that's another area where we don't really understand what's going on.)

Charge is conserved, but any black hole of significant size is probably going to be electrically neutral, just because if it happens to get some charge, by having charged matter fall into it, it will attract opposite charges which will fall in and neutralize it again.
An interesting popular physics account dealing with this phenomena

https://www.amazon.com/dp/0316016411/?tag=pfamazon01-20&tag=pfamazon01-20

Last edited by a moderator:
I we can only guess right? Since it's impossible/hard(maybe hawing radiation?) to get any information from a black hole.

phinds
Gold Member
2019 Award
I we can only guess right? Since it's impossible/hard(maybe hawing radiation?) to get any information from a black hole.
Hawking radiation doesn't tell you anything about what the stuff inside the BH looks like.

Last edited by a moderator:
The second statement of yours quoted above is easier to answer, so I'll start with that. "Mass" in the sense of what particles get via interaction with the Higgs field is not the same concept as "mass" in the sense of what appears as the constant ##M## in the spacetime metric of a black hole. The first concept is a property of particle interactions. The second concept is a property of the spacetime geometry. So it's important to keep the two concepts separate.

Given that, it should now be easier to see the answer to your first question. The mass of the black hole is a property of the spacetime geometry of the hole. It is not "made of" anything other than that. A better question would be, what causes that property of the spacetime geometry to be what it is? The answer to that question is: the matter and energy that formed the object that originally collapsed to form the hole.

According to classical GR, they hit the singularity at the center of the hole and are destroyed. Most physicists believe that is not what actually happens, because the laws of classical GR no longer work when the spacetime curvature gets strong enough, as it does close to the singularity. But we are not sure at this time what actually does happen in this regime.
It is possible to extrapolate some classical behaviors into the boundary of a black hole. In the case of supermassive black holes, the event horizon is so far from the center of mass that the gradient of the gravitational field is relatively small, A human astronaut could cross that boundary without being aware of the difference of gravitational attraction between his head and his feet. Near a much smaller black hole, the tidal effects could pull him apart. The same processes would hold for smaller objects, even down to the size of atoms. Eventually, the gradient approaching the singularity would pull everything apart. The question remains: how long does an astronaut have until that happens? With his speed of approach to the singularity increasing all the time, it should be expected that his perception of time, relative to an outside observer, should be decreasing. The details may be impossible to work out with confidence, but he could quite possibly spend a subjective eternity approaching the singularity. This is essentially the premise of Tipler's book, The Physics of Immortality.

Ibix
The question remains: how long does an astronaut have until that happens? With his speed of approach to the singularity increasing all the time, it should be expected that his perception of time, relative to an outside observer, should be decreasing. The details may be impossible to work out with confidence, but he could quite possibly spend a subjective eternity approaching the singularity. This is essentially the premise of Tipler's book, The Physics of Immortality.
I think you've misunderstood something, or I've misunderstood you. The infalling astronaut has a very short lifetime after crossing the horizon - less than a second, if memory serves. Viewed from a distance, though, the astronaut never reaches the horizon. She can exploit this to return in the far future by dropping arbitrarily close to the horizon and spending a short (to her) time there then returning (assuming a impossibly powerful rocket to hover and return).

Edit: also, you appear to be invoking special relativity's concept of kinematic time dilation, which isn't appropriate here. You need to look at spacetime intervals to get the elapsed times for two observers.

As the tittle says, what is the mass of a black hole made of? Would it be made of the neutrons (and protons, electrons) that formed the earlier star? If no, what happens to those particles? Where do they go to after the black hole is formed?
The question of the mass has been adequately answered, but maybe not the matter. Current thinking says a star of 10 to 29 solar masses will collapse into a neutron star, with protons and electrons compressed into neutrons, and the neutrons closely packed. Larger stars with more gravity become black holes. The neutrons may become crushed, so that the quarks are then packed closely. This is all conjecture, but someone said that the matter was destroyed. We don't know if the quarks are destroyed too. Maybe someone with high math skills could answer that. We know that black hole temperatures are on the order of a millionth of a degree. There seem to be different laws of physics for things this cold. Anyone want to elaborate?

PeterDonis
Mentor
2019 Award
With his speed of approach to the singularity increasing all the time, it should be expected that his perception of time, relative to an outside observer, should be decreasing.
No. First, the concept of "speed of approach to the singularity" has no meaning. The singularity is a moment of time, not a place in space. Spacetime geometry inside the horizon does not work the way you are thinking it does.

Second, there is no way to compare the "rate of time flow" of an astronaut inside the horizon with that of an outside observer. So the intuitive reasoning you are using here simply doesn't work for this case.

The details may be impossible to work out with confidence, but he could quite possibly spend a subjective eternity approaching the singularity.
No, that's not correct. The astronaut's experienced time from horizon to singularity is of the same order of magnitude as the time it would take light to travel a distance equal to the hole's Schwarzschild radius, i.e., ##2 GM / c^3##.

This is essentially the premise of Tipler's book, The Physics of Immortality.
Tipler was talking about a "Big Crunch" singularity in a closed universe (and his own particular kind of closed universe, too, not the standard kind that you see in cosmology textbooks), not a black hole singularity. These are very different spacetime geometries.

It is possible to extrapolate some classical behaviors into the boundary of a black hole. In the case of supermassive black holes, the event horizon is so far from the center of mass that the gradient of the gravitational field is relatively small, A human astronaut could cross that boundary without being aware of the difference of gravitational attraction between his head and his feet. Near a much smaller black hole, the tidal effects could pull him apart. The same processes would hold for smaller objects, even down to the size of atoms. Eventually, the gradient approaching the singularity would pull everything apart. The question remains: how long does an astronaut have until that happens? With his speed of approach to the singularity increasing all the time, it should be expected that his perception of time, relative to an outside observer, should be decreasing. The details may be impossible to work out with confidence, but he could quite possibly spend a subjective eternity approaching the singularity. This is essentially the premise of Tipler's book, The Physics of Immortality.

lets say we have three observers, a) one in moving around a static (Schwarzschild) black hole at a radius lets say r=10M b) one that is at infinity and gets light signals from the first observer and c) an observer that is somehow standing still at radius r=10M and also sees the light signals from the first observer.

Now for the first observer if we say that ##\theta=\pi /2## at a certain level and also his r is fixed we get

##-d\tau^2=ds^2 = - \bigg(1- \frac{2M}{r} \bigg) dt^2 + r^2 d\phi^2## from the metric element

and that would be ##\Delta\tau=20\sqrt{7}\pi M##

for the second observer the time that he would get between those signals are ##\Delta t=20\sqrt{10} \pi M##

and the third and static observer the proper time between those signals would be ##\Delta \tau' = 20\sqrt{8}\pi M##

So we get ##\Delta \tau < \Delta \tau' < \Delta t ## because for the observer far away measuring time t, the static observer feels only the gravitational expansion of time whereas the one going around the black hole feels both the gravitational (GR) and the kinetic (SR) contraction of time.

I've left out many math steps but that would give you an idea although this case is different .

Another thing is while approaching the black hole and for r<2GM time t from timelike becomes spacelike. so when you are travelling towards a Black Hole from a point and on it's like moving forward in time, like going to the future, yes I say what happens when you travel in space by say what will happen from the time point of view.
"Thus you can no more stop moving toward the singularity than you can stop getting older. Since proper time is maximized along a geodesic, you will live the longest if you don't struggle. As you fall to the singularity your feet and head will be pulled apart from each other while your torso is squeezed to infinitesimal thinness" as detailed in the book "Gravitation" by Misner, Thorne and Wheeler.

Last edited:
kent davidge
Nugatory
Mentor
We know that black hole temperatures are on the order of a millionth of a degree. There seem to be different laws of physics for things this cold. Anyone want to elaborate?
Temperatures much colder than that have been achieved in labs without any "different laws of physics" being seen. In fact, it often works the other way: random thermal motion overwhelms some of the more subtle effects we're looking for, so many physics experiments must be done at the coldest possible temperatures just so that we can see the laws of physics that we already know at work.

But all of this is a complete red herring because the low temperatures that you're talking about are the equilibrium temperatures at the event horizon, while this thread is about conditions well inside the horizon.

I think you've misunderstood something, or I've misunderstood you. The infalling astronaut has a very short lifetime after crossing the horizon - less than a second, if memory serves. Viewed from a distance, though, the astronaut never reaches the horizon. She can exploit this to return in the far future by dropping arbitrarily close to the horizon and spending a short (to her) time there then returning (assuming a impossibly powerful rocket to hover and return).

Edit: also, you appear to be invoking special relativity's concept of kinematic time dilation, which isn't appropriate here. You need to look at spacetime intervals to get the elapsed times for two observers.
Yes, I was invoking time dilation, but fairly reasonably so. Gravity also makes clocks run slower. As for the astronaut's lifetime, it all does depend on your frame of reference, doesn't it?

phinds
Gold Member
2019 Award
As for the astronaut's lifetime, it all does depend on your frame of reference, doesn't it?
Yeah, but the astronaut isn't going to care for very long about anyone else's FOR.

Do the possibilities include the matter being converted to massless particles such as photons? Or perhaps that the energy is stored in a field with no particles at all? Perhaps other types of conservation (charge?, baryon number? ...) forbids those scenarios.
It does not seem to make sense to have several laws of nature being violated inside a black hole. If that was the case, then such a object would be unlike everything physicists ever dealt with, either theoretically or experimentally. To me, it is more probable that something familiar is happening there, that is, something one can explain in part using the laws already known, perhaps with the help of a Quantum Gravity theory.

Does anyone think so?

Ibix
Yes, I was invoking time dilation, but fairly reasonably so. Gravity also makes clocks run slower.
But gravitational time dilation doesn't have anything to do with speed, which was what you were talking about. Particularly not speed with respect to a black hole singularity, which is problematic in a number of ways as PeterDonis pointed out.

As for the astronaut's lifetime, it all does depend on your frame of reference, doesn't it?
Indeed. But you said that the infalling astronaut could quite possibly spend a subjective eternity approaching the singularity. I took "subjective" in this context to mean that you were referring to the astronaut's proper time - which is very short, and in no way a subjective eternity.

PeterDonis
Mentor
2019 Award
for the first observer if we say that ##\theta=\pi /2## at a certain level and also his r is fixed we get

$$-d\tau^2=ds^2 = - \bigg(1- \frac{2M}{r} \bigg) dt^2 + r^2 d\phi^2$$

from the metric element
This is correct as far as it goes, but it's not sufficient by itself to calculate ##\Delta \tau##. You need to know what ##dt## and ##d\phi## are. You haven't said anything about them at all. Can you give more details about what specific orbit you are assuming and how you are calculating your numbers?

This is correct as far as it goes, but it's not sufficient by itself to calculate ##\Delta \tau##. You need to know what ##dt## and ##d\phi## are. You haven't said anything about them at all. Can you give more details about what specific orbit you are assuming and how you are calculating your numbers?
##-d \tau^2=ds^2=-(1-\frac{2M}{r})dt^2+
r^2 d\phi ^2## →##d \tau^2 = [(1-\frac{2M}{r})(\frac{dt}{d \phi})^2-r^2]dφ^2## and for circular orbits like the one assumed around the static black hole ##Ω^2=(\frac{d φ}{dt})^2=\frac{M}{r^3}## so you have

##Δτ=\int_{0}^{2π} \sqrt{\frac{r}{M}-3} rdφ=2πr\sqrt{ \frac{r}{M}-3}## and for r=10M you get ##Δτ=20√7πM##

PeterDonis
Mentor
2019 Award
for circular orbits like the one assumed around the static black hole ##Ω^2=(\frac{d φ}{dt})^2=\frac{M}{r^3}##
Ah, ok, so you were assuming a free-fall orbit.

I was thinking of buying this book. Is it good? Has anyone of you read it?
I read it and like it.