# I Does a black hole have a maximum density?

1. Aug 15, 2016

### GregoryC

If not why does the event horizon grow as more matter falls in?

2. Aug 15, 2016

### Grinkle

The space inside the event horizon will have exactly the average density that causes the event horizon, it cannot have a higher density or the event horizon would become larger.

The following analogy is not very good, but it paints a picture - if you blow up a balloon, the density of air inside the balloon is such that there is sufficient pressure to balance the elastic force of the balloon skin. Bad analogy, its all I can think of.

All of the matter inside the event horizon occupies (non-uniformly I suppose but I really don't know) a volume sufficient to cause an event horizon of the observed size. If more matter falls in, the event horizon grows.

3. Aug 15, 2016

### jbriggs444

As you add matter to a black hole, the radius of the event horizon grows linearly with the enclosed mass. This is actually an over-simplification -- a black hole does not actually have a radius in this sense. But it does have a surface area which we can pretend gives us an associated radius.

The volume of the black hole computed in this naive manner ($\frac{4}{3}\pi r^3$) increases more rapidly than the enclosed mass. So its density actually decreases as matter is added. If you want to maximize this notion of density then you need small black holes. Large ones will not do.

4. Aug 15, 2016

### Orodruin

Staff Emeritus
A black hole does not have any density or average density. It is not very meaningful to talk about the volume of a black hole. The very geometry of space-time is very different near and inside the black hole.

5. Aug 15, 2016

### Staff: Mentor

The short answer is that a black hole does not have a well-defined density, as Orodruin has pointed out.

Is not the "density" of the hole in any meaningful sense, because, as you note, the hole does not have a "radius"; it only has a surface area of the horizon. The spatial geometry inside the horizon is not Euclidean--in fact it's not even well-defined, because it's coordinate dependent. (In fact, you can choose coordinates in which the spatial volume inside the horizon is infinite.)

Because as more matter falls in the hole's effect on light becomes stronger, so the boundary of the region which cannot send light signals to infinity (which is the event horizon) moves outward.

6. Aug 15, 2016

### PAllen

You can give some physical meaning to the question of density by looking at BH formation rather than an 'old' BH (about whose interior little can be said; even classically; a realistic old interior for a collapse with rotation is unknown - Kerr Interior is perturbatively unstable; then there is the quantum gravity issue). Then, when the matter is first inside the horizon, it has as well defined a density as it did a moment earlier. This is strictly speaking coordinate dependent, but we have no trouble speaking of the density of neutron stars, which are also coordinate dependent in the strict sense. We adopt a standard space-time slicing which is natural to the symmetry of the problem. Then, you can say that the more massive the BH, the less dense it was as the horizon crossed the outer matter of the collapse. You can also ask about a matter infall across the horizon before any singularity has formed, and conclude that this will decrease the overall density inside the horizon (in any given, reasonable, foliation).

7. Aug 15, 2016

### Staff: Mentor

Not necessarily. You can define a local energy density as measured by an observer who is comoving with the infalling matter; that's an invariant with a well-defined physical meaning.

The problem is that the region of spacetime occupied by the infalling matter is only a very small part of the total region inside the event horizon, and any energy density defined inside the infalling matter does not apply to the black hole as a whole.

I would not say this. I would say that the more massive the hole, the less dense the infalling matter that formed it was when it crossed the horizon (or the horizon crossed it, whichever way you want to put it). I would not describe that density as the "density of the black hole", because the infalling matter is not static; it only occupies "all of the volume inside the horizon" (which is a problematic statement anyway) for a single instant.

8. Aug 15, 2016

### PAllen

Initially, it occupies essentially all the interior, as you note below.
Since the density of the BH, in general, is not defined, as explained by you and others, I am simply proposing a definition of something that is well defined and gets at the (IMO important, initially surprising) intuition that the more massive the BH, the less dense it was at formation.
I am simply saying that while the interior is everywhere non-singular, there is a well defined interior volume. It is foliation dependent, but I am claiming that for a given reasonable foliation, the growth of interior volume as a new infalling piece of matter is engulfed by the horizon is such that:

< original collapsing mass crossing horizon> / <interior volume at horizon crossing> is greater than
(<original collapsing mass + mass of new piece>) / <interior volume when new piece crosses growing horizon>

I do not have a reference for this, so consider this claim poorly justified unless I add more information later.

9. Aug 15, 2016

### Staff: Mentor

Not only the interior volume is foliation dependent; how long the interior is "everywhere non-singular" is foliation dependent. If we consider Painleve coordinates in the vacuum region to be a "reasonable foliation", then the coordinate time from the initial collapsing matter surface just crossing the horizon to the formation of the singularity is very short compared to the time it took for the collapsing matter to reach the horizon from its initial point (assuming the initial object was something like a star). So unless another object falls in before that short coordinate time elapses, the interior will be singular and your definition of "interior volume" will no longer hold.

10. Aug 16, 2016

### PAllen

I completely agree that the window for another object to fall in with my scenario is very short in typical coordinates. I didn't state that explicitly, but certainly understood it. In fact, I agree with everything you write, but consider it non inconsistent with what I wrote.

I guess a point mentioned in passing early in this thread is worth emphasizing: the apparent visual angular radius of a BH viewed from afar increases linearly with mass, as mass is added. This is wholly unexpected compared to other astronomic bodies, even including neutron stars.

11. Aug 20, 2016

### MattRob

With the disclaimer that I'm only a newbie undergrad student in the field, if I may, I think that while PeterDonis is almost certainly right more fundamentally speaking, and in many senses coordinate-invariant truths are really the only important thing, I think it's also useful to sometimes look at things in a more forgiving sort of intuitive way. Not missing the forest for the trees, so to speak.

Example: The Schwarzchild solution is only one of many, and not necessarily the best way to describe a black hole by any stretch - but it's also one of the simplest and most intuitive. If you take the coordinate radius as though it were a physical radius, then in this sort of intuitive way, strictly by the Schwarzchild solution alone there isn't a maximum "density" to a black hole - the less the mass, the smaller this "reduced radius" in a linear relationship and thus the greater the density.

Of course things get a lot more complicated when you consider what the "radius" really means in the metric, and even more so when you consider more realistic black hole models or quantum-scale effects, and, as added by PeterDonis, far more complex when you look for statements regarding the "volume" that are true in every coordinate system.

But at least the appeal to the Schwarzchild offers some sort of naive answer, at least, and points at how fascinating and strange the "volume" of a black hole can be with regards to its mass.

In any case, offering an answer to OP with the Schwarzchild seems like a reasonable enough thing to do, mentioning the caveat that things get a lot more complicated if one digs further into the fundamentals behind the mathematics of the Schwarzchild itself. But I don't think someone asking if a black hole has a maximum density is looking for tensor equations and coordinate invariant statements.

So, I guess the simplest answer to OP is "no". At least in Schwarzchild coordinates and neglecting everything else, no, because the "density" (defined as m/V where V = 4/3 π r2) is a function of mass, and the more massive the black hole, the less "dense" it is by this definition.

But there's a lot more depth to it, since asking how much "volume" a black hole has isn't as straightforward as it sounds when spacetime is distorted immensely and lengths depend on how you measure them.

I'm very thankful for all the insight provided, though, with regards to the question of "volume" as approached from many different coordinate systems and the various answers they provide. And somewhat excited that after many years of trying to get the subject I can finally understand most of what people are saying, haha.