# Stress-Energy Tensor

1. Feb 18, 2006

### touqra

Can the energy density of an empty but expanding spacetime decrease with expansion?
Does a black hole have a stress-energy tensor?

2. Feb 18, 2006

### pmb_phy

Classically? Not really. If the cosmological constant of the universe is zero then the energy density is zero in empty spacetime.

Pete

3. Feb 18, 2006

### touqra

What about a non-zero cosmological constant?
Without a non-zero cosmological constant, an empty spacetime wouldn't expand, isn't it?

4. Feb 19, 2006

### pmb_phy

If the cosmological constant is non-zero then spacetime may still be either expanding or contracting.

And yes - A Black hole does have a stress-energy-momentum-tensor.

Pete

5. Feb 19, 2006

### touqra

I found out that the energy density of a Schwarzschild black hole, $$T_{00} = 0$$.
How is that possible if the mass of the BH contributes to a non-zero total energy and the volume of the BH can be taken as a spherical volume with a total area equals to that of its event horizon?

6. Feb 19, 2006

### pmb_phy

That is incorrect. That expression holds everywhere except at the origin, i.e. where the black hole is. The mass distriution is describribed by a delta function.

Pete

7. Feb 19, 2006

### Stingray

That's not true. It's possible to show rigorously that there are no solutions to Einstein's equations for 0+1 or 1+1 dimensional stress-energy tensors (within the context of standard distribution theory).

Commonly discussed black holes do not have stress-energy tensors (other than charged ones of course). They are instead vacuum solutions with nontrivial topology. So you could call it "topological mass."

There are no completely local notions of energy density in full GR. You could use $T_{00}$ of course, but this is rarely useful. It would imply that gravitational waves do not carry energy for example.

8. Feb 19, 2006

### pmb_phy

Of course it is.
So what? We're not talking about those tensors regarding a black hole, we're talkikng about a 2nd rank tensor (1+#). Otherwise please site proof.
Not from what I've seen.

Pete

9. Feb 19, 2006

### pervect

Staff Emeritus
In general it isn't possible to "localize" mass in GR. For the case of a static black hole, though, it is reasonable (though sloppy) to say that the mass is all concentrated in the central singularity, i.e. that it is a delta function of zero volume. While this answer is sloppy and Stingray's answer is more correct, I think the sloppy answer will probably serve tougra better.

If you take the Komar mass (surface intergal form) to represent "the" mass of the black hole, then you should get an answer that is zero if the surface does not surround the singularity, and non-zero if the surface does surround the singularity. (This is for an uncharged Schwarzschild hole). Which is consistent with the delta-function interpretation.

It turns out that different observers happen to agree about "where" the Komar mass is located, IIRC, this is one of the nice features of static metrics that can't be made more general.

I'm not sure where 1space + 1time entered the discussion, I think that part of the thread may have been deleted? Anyway I don't see it, and I'm only talking about 3space+1time.

10. Feb 19, 2006

### Stingray

Huh? By "0+1 dimensional," I meant something with 0+1 dimensional support; e.g. a delta-function on a worldline. The proof that Einstein's equations do not admit point particle or stringy stress-energy tensors in a reasonable sense is contained in Geroch and Traschen, Phys. Rev. D 36, 1017 (1987).

But this excludes things like Schwarzschild a priori (with some discussion). The reason is that we should have objects important to the theory - in particular the Riemann tensor - be physically interpretable. They should at least exist as distributions, which means that they must be locally integrable. Now, the Schwarzschild metric scales as 1/r, so the Riemann tensor must go as 1/r^4. But the volume element only grows like r^2, so the Schwarzschild Riemann tensor fails to be integrable in any neighborhood of the r=0. Differential geometry therefore fails to be terribly meaningful there. Who knows what it even means to talk about a stress-energy tensor in such a region.

You could still go on and say that only the Einstein tensor should be a distribution. I don't know for sure that this cannot be made to work, but as far as I know, all attempts have failed. In any case, whatever you'd have to do would be such a ridiculous kludge as to make it pointless I think. It's hard to even say what specific form a distributional stress-energy tensor should take (it's easy if you assume it is not self-gravitating, but that's not what we're talking about).

Perhaps you'd like to point to somebody talking about the stress-energy tensor of a black hole? Nobody talks about such things as far as I've ever seen.

11. Feb 19, 2006

### Stingray

Yes, sorry for derailing the thread a bit.

12. Feb 19, 2006

### Chronos

It's not meaningful to talk about stress energy tensors at points, AFAIK. But point particles are unphysical in our observable universe. Even black holes are now believed [at least by many theorists] to have a finite volume, hence are integrable.

13. Feb 20, 2006

### Stingray

Sort of. But that's in the quantum theory.

14. Feb 21, 2006

### pmb_phy

Why not? That's how it is actually defined. E.g. T^00 is energy density and that means its a point function. T^0i is ith component of momontum density which is also a point function. Same thing with all the other components.

Pete

15. Feb 21, 2006

### Stingray

From what I gathered, he meant that it's not meaningful to talk about stress-energy tensors of points rather than at points. Maybe not, though.

16. May 15, 2007

### MeJennifer

It's pointless to talk about a point particle with a non-vanishing stress-energy tensor since this means there is a singularity.

17. May 15, 2007

### pmb_phy

That's true. However in real life one considers the situation and determines how small the particle must be and how low the mass must be so as to let the geodesic approach that of a ideal test particle. In any case there is nothing wrong with a singularity. A micro black hole has a miniscule mass compared with the Earth (its on the order of mass of Mt. Everest). But there's nothing wrong with thinking of such a micro black hole as a test particle so long as one limits its mass to be so small as to not contribute significantly to the stress-energy-momentum tensor of the spacetime its moving in for regions of several Schwarzchild radii.

Pete

18. May 15, 2007

### Chris Hillman

As some respondents already said, in the context of relativistic cosmology (right?) "empty spacetime" usually means a vacuum solution of the Einstein field equation (EFE) of the general theory of relativity (gtr). By definition, in a vacuum solution, the stress-energy tensor vanishes identically. The energy-density is just one component (evaluated in an appropriate frame field) of this tensor, so it vanishes too.

The simplest and most often discussed models of black holes in gtr are exact vacuum solutions such as the Schwarzschild or Kerr vacuums, so these have stress-energy tensors which are well-defined but which vanish. (The equation x=0 has a real root, zero, whereas the equation x^2+1 has no real roots--- this analogy suggests why I would pedantically object to any suggestion that a vacuum solution might not possess a stress-energy tensor at all!)

However, there are other models of black holes (compact objects which possess event horizons) such as the ingoing Vaidya null dust (1951)
$$ds^2 = -\left( 1-\frac{2m(t)}{r} \right) \, dt^2 + \, 2 \, dt \, dr + \, r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),$$
$$-\infty < t < \infty, \; 0 < r < \infty, \; 0 < \theta < \pi, \; -\pi < \phi < \pi$$
where m(t) is an arbitrary non-negative non-decreasing smooth function, which models a black hole with infalling incoherent radiation, and the Reissner-Nordstrom electrovacuum, which models a static charged black hole). The stress-energy tensors of these solutions do not everywhere vanish.

For those of you learning about black hole formation, especially if you are worried about the odd global structure of the "eternal black hole", the Vaidya null dust is a terrific example. Consider the case of a smooth function m(t) which is initially zero and at some time begins to increase, then levels off at some positive value. This models a region of Minkowski spacetime, until along comes a contracting shell of incoherent radiation (see the discussion in Feynman's Lectures on Physics of a similar thought experiment), which collapses on a point forming a black hole there. Follow the evolution of the event horizon and consider the implications! (I can explain if this doesn't become clear after some thought.)

Another good example: the Schwarzschild-de Sitter lambdavacuum (1918)
$$ds^2 = -\left(1- \frac{2m}{r} + \frac{r^2}{a^2} \right) \, dt^2 + \, \frac{dr^2}{1- \frac{2m}{r} + \frac{r^2}{a^2} } + \, r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),$$
$$-\infty < t < \infty, \; r_0 < r < \infty, \; 0 < \theta < \pi, \; -\pi < \phi < \pi$$
where $r_0$ is the positive real root of $r^3 - a^2 \, r + 2 m \, a = 0$, models a static spherically symmetric black hole with a Lambda term (as I wrote the solution above, $\Lambda=\frac{3}{a^2}$, where a is the "radius" of the cosmological horizon). This term contributes a nonzero term to the stress-energy tensor.

For those of you learning about global structure (Carter-Penrose diagrams and all that), here's an exercise: find the Carter-Penrose diagram for the Schwarzschild-de Sitter lambdavacuum. (I can provide a citation to an arXiv eprint which gives the solution.)

The Kasner vacuum (1921)
$$ds^2 = -dt^2 + \, t^{2a} \, dx^2 + \, t^{2b} \, dy^2 + \, t^{2c} \, dz^2, \; \; 0 < t < \infty, \; -\infty < x, \, y, \, z < \infty$$
where
$$a+b+c = a^2+b^2+c^2 = 1$$
is a simple homogeneous anisotropic vacuum solution which expands from a Big Bang type singularity. You can take
$$a = \frac{-u}{1+u+u^2}, \; b = \frac{1+u}{1+u+u^2}, \; c = \frac{u \, (1+u)}{1+u+u^2}$$
so for example $a=-2/7, \; b=3/7, \; c=6/7$ is a rational solution of the constraint equations. This exact vacuum solution arises as a limiting case of the Kasner dust solution. (In gtr, a "dust" is a perfect fluid with vanishing pressure; the so-called "matter-dominated FRW models" in cosmology, or better, FRW dusts, are exact dust solutions.)

The Kasner vacuum is a terrific computational example for those of you learning about geodesic equations, or about computing curvature the Cartan way, or about the kinematic decomposition, BTW. It also provides an excellent example of a simple exact solution with extremely interesting "light cones in the large" (see the monograph by Hawking and Ellis, The Large Scale Structure of Space-Time).

Not sure I understand the question, but as pervect (and Stingray) said, in gtr, localizing the energy of the gravitational field itself, is quite tricky. As he said, for certain kinds of spacetimes which model "isolated objects", there are sensible methods of assigning a physically reasonable mass and angular momentum. These include black holes solutions like the Kerr vacuum.

No, that wouldn't be correct even for the simplest kind of stellar model, in which a static spherically symmetric perfect fluid ball (these are all known, and simple examples can even be found readily; see the arxiv for recent reviews coauthored by Matt Visser) is matched across the surface (where the pressure vanishes but the density typically does not) to an exterior Schwarschild vacuum region. The geometric significance of the Schwarzschild radius is that $A = 4 \pi \, r_0^2$ for the surface area of the sphere $t=t_0, \, r=r_0$ remains valid, but distance along a radius $t=t_0, \, \theta=\theta_0, \, \phi=\phi_0$ and volume are given by more elaborate formulas, in the case of such a stellar model.

In the case of a black hole, the best short answer is that the interior of an event horizon is nothing like the interior of a hollow beach ball; roughly, the spacetime is so curved that there is nothing inside which can be given a volume. Maybe the best short explanation of the reason why not would be to say that inside the horizon, no matter can remain static, so unlike the case of the static fluid ball, there is no sensible way of assigning a volume to some static ball-like region of "space" inside the horizon.

Now read again what I said above about the Vaidya solution modeling the black hole formed by collapse of a shell of radiation.