Here's a question inspired by Edward's answer to this question.

It's my understanding that the average energy density of a black hole in its rest frame is $\rho_\text{BH}(A)$, a function of surface area. I calculated $3c^2/2GA$ for a Schwarzschild black hole, but that's presumably not applicable here since I'm talking about an extended energy distribution. Anyway, suppose you are in a space filled with some sort of energy, matter, or whatever, which produces a potentially time-dependent stress-energy tensor. And further suppose that there is some finite, spherical region of surface area $A$ in this space, over which you measure the average energy density to be $\rho_\text{BH}(A)$, the net charge to be zero, and the total angular momentum of the matter within the region to be zero. (I'm assuming there is a measurement procedure available which can be carried out without entering the region, if it matters.)

Now, Edward's argument in the other question shows that there are at least two ways to produce that (average) energy density:

  1. The energy density arises from a fluid or other material at rest, such that the region is a black hole and there are no timelike geodesics exiting it
  2. The energy density has been "augmented" by a Lorentz boost from some other frame, implying that there are timelike geodesics exiting the region

Is it possible, in general, to distinguish between case 1 and case 2 by only looking at the other components of the stress energy tensor, without actually calculating the geodesics? If so, how? What would be the "signature" of a black hole in the components of $T^{\mu\nu}$ (or rather, their averages over the region in question)?

  • 1
    $\begingroup$ Hopefully this isn't the case, but I believe currently there is no known simple answer. What you essentially want is a full proof of the hoop conjecture which there is none that I know of. While it makes intuitive sense that if we shove enough mass into a region a black hole forms, it is difficult to discuss precisely. It is even difficult to define what we mean by mass contained in a local region in a completely general case en.wikipedia.org/wiki/Mass_in_general_relativity $\endgroup$
    – Edward
    Commented Mar 30, 2011 at 3:41
  • $\begingroup$ @Edward: thanks, I figured there was a decent chance that I was asking something which can't be answered. I'll see what people have to say, though. $\endgroup$
    – David Z
    Commented Mar 30, 2011 at 4:28
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    $\begingroup$ Sorry, David, but the energy density $T_{00}$ in the Schwarzschild black hole solution is zero everywhere except for the singularity where it's ill-defined. Your disagreement with this simple fact makes your question confusing, to say the least. Are you actually asking about the stars that are going to collapse into black holes sometime in the future? $\endgroup$ Commented Mar 30, 2011 at 8:02
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    $\begingroup$ @Luboš: Yes, I know the stress-energy tensor itself is singular, but I was talking about the average energy density, which (as I tried to make clear) could be defined as the mass of the black hole divided by the volume of a Euclidean sphere with the same surface area as the event horizon. I would be very surprised to hear that either mass or surface area is ill-defined for a black hole. I've tried to formulate the question in a way that doesn't rely on measuring the value of $T^{00}$ at an individual point, only on its average over a region. $\endgroup$
    – David Z
    Commented Mar 30, 2011 at 18:12
  • $\begingroup$ the asymptotic structure of the spacetime will define you a covariant asymptotic 4-momentum vector, which will have magnitude $M$, so the mass is indeed well-defined. But the volume enclosed by the black hole area is not an invariant. You can get a bit tricky with this, as the $r=0$ singularity has a similar structure to the $r=0$ singularity of a point charge, so you, in some sense, can think of $\rho$ as a delta function, but I dont' think this generalizes to even the Kerr metric, so it's probably not useful. $\endgroup$ Commented Mar 31, 2011 at 3:30

3 Answers 3



As you can see from Lubos's comment there is a serious flaw in the formulation of this question, so I will use this answer simply to explain some aspects of the topic. First the flaw:

the energy density of a Schwarzchild black hole

The Schwarzchild solution is a solution of the Vacuum Einstein equations ie $R^{u,v}=0$ and so $T^{u,v}=0$ ie the Stress Energy Tensor is zero throughout the Schwarzchild solution (removing the origin from this manifold avoids the undefinedness there). With zero stress-energy there are no fluids or local energy densities to measure or examine.

What is meant in the question (as it arises from the earlier Stack questions) has some relationships with the Hoop conjecture as Edward points out, but just for clarification I shall add more.

Let us not consider any more an Einstein vacuum and assume that some form of matter (or radiation - I will just say matter below) must have been present "originally". This matter is part of a non-vacuum solution of the Einstein equations, and so there will be a corresponding non-zero Stress-Energy Tensor whose matter is destined to become the Black Hole. So now the question begins to make sense.

So the question is really about what determines whether a given non-vacuum solution of Einstein's equations forms a Black Hole and whether this fact is measurable locally. The sentence that asks this is:

What would be the "signature" of the Black Hole in the components of $T^{u,v}$?$

I dont believe that this answer is known, partly because the space of all solutions of Einstein's equations is not yet known. If one considers the points made below, one might also conclude that GR alone was inadequate to predict the BH formation - matter properties are central too.

There are the classical Hawking-Penrose theorems which give a topological-geometric answer to this question by positing the existence of "closed trapped surfaces", along with certain properties of $T^{u,v}$. In that sense there is an answer, but it doesnt tell us when the closed trapped surfaces will form (generically).

Black Holes arose as physically plausible solutions to Einstein's equations because of the early work of Oppenheimer et al. Here there are two metrics combined to form the matter leading to a Black Hole:

Friedmann Dust (interior) + Schwarzchild (exterior)

(a clever trick to consider "massless dust" allows one to use only the Friedmann Dust solution). These two solutions need to be "glued together" to form the surface of the star. The $T^{u,v}$ (in the comoving frame and its translation into other frames) for this was given in Edward's earlier answer, and is non-zero in the star interior.

What causes another layer of complication in discussing the formation of Black Holes and Event Horizons is the teleological nature of their formation in General Relativity. This arises because "time" is just a parameter in the theory and the formation of the Black Hole is determined by the overall solution (thus in a time independent way). Now it has been concluded that for stellar objects of mass > TOV limit a Black Hole will form. But as remarked above translating this condition into a condition on the Stress-Energy Tensor alone may not be possible.

Physically one might expect that there is some local condition despite all these issues, such as the Hoop conjecture which includes the object's mass in its formulation. There are several subtleties connected with this unproven conjecture and one problem here is that "mass" is not a local property in GR (because mass = energy and the gravitational field contributes energy too, not just the Stress-Energy Tensor - hence we again may need the full solution of $G^{u,v}=T^{u,v}$.)

I shall add this link from Willie Wong for anyone interested in the latest on the Hoop Conjecture.

Finally my answer to the linked question might be of interest.

  • $\begingroup$ Nice answer :) I will have to mull this over a bit. Incidentally, my $\rho_\text{BH}(A)$ is in fact the same thing you labeled as $D_R$ in your answer to the other question. $\endgroup$
    – David Z
    Commented Mar 31, 2011 at 0:25

Also, note that there are exact solutions of Einstein's equation that correspond to naked singularities. For an easy example, if you have a charged black hole with $Q>M$, there is no event horizon at all, but you still have a mass $M$ measured by an observer at asymptotic infinity. So, these solutions are a class of solutions that have "infinite" density at the singularity, but at the same time, aren't black holes.

  • $\begingroup$ Right. But then again you have the cosmic censorship conjecture which says that naked singularities cannot form. And even if they do, the existence of a singularity in any form is an indication of a breakdown of the existing theory. In any theory the values of various fields and observables where the theory develops singularities are de facto limiting values. $\endgroup$
    – user346
    Commented Mar 30, 2011 at 23:54
  • $\begingroup$ @Jerry: Interesting, I didn't think about that... thanks for pointing it out. So I guess the question is a moot point for black holes with sufficient charge or angular momentum, but I had in mind a situation where $Q = 0$ and $J = 0$ so I will update the question to make that clear, in case it changes anything. +1 for your insight ;) $\endgroup$
    – David Z
    Commented Mar 30, 2011 at 23:55
  • $\begingroup$ @Deepak: but the cosmic censorship conjecture has been proven false on at least a the complement of a dense subset of initially physically reasonable initial states of GR. Regardless, this shows that the notion of a 'universaal black hole energy density' isn't going to work out. $\endgroup$ Commented Mar 30, 2011 at 23:56
  • $\begingroup$ Also, there are more exotic naked singularities corresponding to sheets of or lines of mass for which no horizon forms, even when you get infinite mass. You can get these for (very contrived) physically reasonable (but not asymptotically flat) initial conditions. $\endgroup$ Commented Mar 30, 2011 at 23:59
  • $\begingroup$ @Jerry, you say proven false on a complement of a dense subset ... wouldn't the complement of a dense subset be a non-dense subset of the solution space? Can you explain that statement in somewhat less technical language? $\endgroup$
    – user346
    Commented Mar 31, 2011 at 0:13

It really is not a density. The critical spatial quantity is the Schwarzschild radius $R~=~2GM/c^2$, where the mass $M$ is contained in a volume with a radius $R$. The critical factor is not the amount of matter per unit volume, or density. A super massive black hole has a mass of 10 billion solar masses, and a radius 10 billion times the solar mass radius of 1.5 km. The volume in standard coordinates is $3.4\times 10^{30}km^3$. The sun has a volume of $1.4\times 10^{18}km^3$. !0 billion suns have a volume of $1.4\times 10^{28}km^3$. So 10 billion suns could be packed into the volume of a 10 billion solar mass black hole with room to spare.

  • $\begingroup$ Right, and the critical average density for a black hole of that size would be 0.7 kg/m^3, which is less than the Sun's density by a factor of nearly 2000. I don't really see what you're getting at... $\endgroup$
    – David Z
    Commented Mar 31, 2011 at 0:02
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    $\begingroup$ Absolutely @David, and this is what suggests that the line of inquiry based on volume densities might be incorrect. Regardless of the density of the interior bulk - which may be large or very small as you point out - one common characteristic of all black holes is that their bounding surface (the "horizon") satisfies $S=A/4$. Therefore IMHO it is the entropy density per unit area on the boundary of a region with any given field configuration which will ultimately determine whether or not that region will undergo grav. collapse. This also sidesteps all issues about the definition of energy. $\endgroup$
    – user346
    Commented Mar 31, 2011 at 0:17

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