The Matter-Vacuum Boundary in General Relativity

+ 4 like - 0 dislike
491 views

A previous Stack question (before I joined) asking about continuity in GR received replies which suggested that Curvature would be discontinuous at say a planetary boundary (assume no atmosphere for simplicity). I will analyse some basics of this and then return to that question.

It is true that the Stress-Energy Tensor $T{_a}{_b}=0$ outside the body and is nonzero in the interior resulting in a discontinuity at the surface. This would imply that the Ricci Tensor $R{_a}{_b}$ is also discontinous at the boundary, and zero in the vacuum part as expected from the Einstein equations. However the Riemann Curvature Tensor $R{_a}{_b}{_c}{_d}$ (which generates the physically measurable accelerations) has contributions from the Weyl Curvature Tensor $C{_a}{_b}{_c}{_d}$ as well. In fact the Ricci Tensor "hands over" to the Weyl Tensor at the boundary: thus the Riemann Tensor stays non-zero there. However this "hand over" does not imply continuity, unless there is some GR Theorem which says that the Riemann Tensor stays continuous in this region.

Also in the Newtonian approximation the analogous role is played by the gravitational potential $\phi$ in the Poisson equation $\nabla^2 \phi = 4 \pi G\rho$. Clearly this shows a discontinuity too as the density $\rho$ suddenly drops off at the boundary. However the discontinuity is in the second derivative of the potential: the potential itself is continuous. This means that in exiting a planetary cave or mine one does not suddenly meet a change in Gravitational potential.

However I do not know any theorem in GR which guarantees such continuity. The applicable in-the-large scenario might be the surface of a neutron star; there may be in-the-small particle models too.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Roy Simpson
asked Jan 24, 2011

4 Answers

+ 4 like - 0 dislike

Roy, your wishful thinking is manifestly impossible. If the tensor $T_{\mu\nu}$ is discontinuous, and it surely is on the surface of a solid, then Einstein's equations guarantee that the Einstein tensor $G_{ab}$ is discontinuous as well - up to a normalization, it's the same tensor. It follows that the Ricci tensor and Riemann tensor, $R_{\mu\nu}$ and $R_{\kappa\lambda\mu\nu}$, must also be discontinuous because the Einstein tensor $G_{\mu\nu}$ can be easily calculated both from the Ricci tensor as well as from the Riemann tensor, so if the Ricci or Riemann tensor were continuous, the Einstein tensor would have to be continuous, which is an obvious contradiction.

I just proved the opposite theorem that the Riemann tensor is discontinuous.

You should realize that the Riemann tensor has a higher number of components than the Ricci (or Einstein) tensor, so its continuity - which means the continuity of all of its components - is an even stronger condition than the continuity of the Ricci (or Einstein) tensor. The argument above proves that none of these tensors is continuous in the presence of solids - which is why there can't be any theorem saying the opposite thing (it would be wrong). Another question is whether the Weyl tensor is continuous near such boundaries. I don't know the answer. The answer could be easily calculated from the very formula for the Weyl tensor.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Luboš Motl
answered Jan 25, 2011 by (10,278 points)
Very simple, but interesting Lubos! I had not realised before this Question that GR would have problems with solids. If the main Tensors are discontinuous, then (as QGR below also points out) its derivatives will go singular/distributional. My GR texts just assume "continuous matter distributions" - and continue from there...

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Roy Simpson
Dear @Roy Simpson, thanks for your answer but I didn't write that GR has problems with solids because it doesn't have any problems. There is absolutely nothing wrong with the curvature tensors' being discontinuous. It just means that some of the derivatives of the curvature tensor would include delta-functions. That's not a problem by itself. Derivatives of curvature are third derivatives of the metric - very far on the line of derivatives. The only "naturally occurring" derivatives of the curvature tensor is $\nabla_\mu R^{\mu\nu}$ which vanishes identically, even when $R$ is discontinuous.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Luboš Motl
The Riemann curvature Tensor is what is physically measurable however (relative accelerations). If it is OK to have GR solutions consisting of patches with discontinuities of measurable properties, then I have learned something today...

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Roy Simpson
Yes, @Roy Simpson, absolutely, it is OK to find solutions composed out of patches. Clearly, if the right-hand side of the equation is discontinuous and composed of "patches", something about the left-hand side must have the same property. After all, this was true already in classical physics, wasn't it? The Laplacian of the gravitational potential $\phi$ was exactly as discontinuous as the curvature tensor.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Luboš Motl
Incidentally you have not yet proven that the metric itself is continuous in this case.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Roy Simpson
+ 1 like - 0 dislike

I don't know of a theorem of the kind you want, but the standard treatment of the matter-vacuum boundary in general relativity is given by the Israel junction conditions. Consider an infinitely thin shell of matter embedded in a manifold. The shell has two sides (assuming the shell is an orientable manifold) $S^+$ and $S^-$. The embedding of each of these surfaces in the background manifold is given by an extrinsic curvature $K^{\pm}_{ij}$. The difference in the extrinsic curvature between the two sides is related to the intrinsic curvature of the shell:

$$8 \pi \left( R_{ij} - \frac{1}{2} g_{ij} R \right) = K^-_{ij} - K^+_{ij}$$

The left hand side is also known as the Einstein tensor $G_{ij}$, i.e. the traceless part of the Ricci curvature $R_{ij}$. And this in turn is related to the energy-momentum tensor $T_{ij}$ of the matter constituting the shell by Einstein's equation:

$$G_{ij} = 8\pi T_{ij}$$

For now this should be enough to get you going. I'll add more as I learn more. A good reference for this and other topics is the set of notes on advanced GR by Eric Poisson which you can find here

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user user346
answered Jan 24, 2011 by (1,985 points)
+ 1 like - 0 dislike

The question touches on Birkhoff’s theorem. A spherical distribution of matter has a gravity field entirely equivalent to a black hole of the same mass. So if we were to ignore the matter in a star the interior configuration of spacetime must behave as if it were a Schwarzschild metric (or Reissnor-Norstrom, Kerr, etc). Suppose we have a metric of with $g_{tt}~=~F(r)$ and $g_{rr}~=~1/g_{tt}$, and within a surface there is the $T^{ab}$ for material inside the body. The Bianchi identity ${T^{ab}}_{;b}~=~0$ gives $$p_{,r}~+~(m~+~p)\Big(\frac{F_{,rr}}{F_{,r}}~-~\frac{2F_{,r}}{F}\Big)~=~0$$ where now one matches the condition on the vacuum metric with this condition. From there the metric $F(r)$ inside the body can be computed.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Lawrence B. Crowell
answered Jan 24, 2011 by (590 points)
Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be stationary and asymptotically flat. ref: Wikipedia. I don't directly see it as saying anything about the question of describing the matter-vacuum boundary for a general mass distribution.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user user346
Birkhoff's theorem was proven with the intention of extending a vacuum solution across a matter boundary. The properties of the spherically symmetric solution is what permits this. Appealing to the continuation of the vacuum solution permits one to work the appropriate interior metric, with various simplifying assumptions. This was used to work out the physics of stellar gravity and collapse That is half the point of it all. As for a general matter distribution, that is something which has no general theorem as far as I know.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Lawrence B. Crowell
I have just studied a little of the stellar modelling theory and I think that the idea there is to match interior-exterior GR solutions up without discontinuities. Hence Birkhoff plays a role in what the exterior solution is. Stellar interior models are some kind of Fluid to allow this. I assume the p term in your equation is a variable pressure or density. But here we deal with Solids and it looks like we are forced into discontinuity. See LM's Answer.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Roy Simpson
The curvature is matched across the boundary, and the source is discontinuous. In the elementary model the pressure is constant. If you want to work more realistic models it varies. The purpose is to get curvature which matches across the surface. In the case of a S-schild solution the neck up to the horizon is replaced by a "bowl." Nothing is automatic, and it can take a lot of fiddle work to get curvatures to match, to get $C^k$ $K > 2$ continuity and differentiability across the boundary, and so forth.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Lawrence B. Crowell
Thanks. I suppose the issue I need to understand - in the light of comments with LM above - is why the curvatures need to match?

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Roy Simpson
+ 1 like - 0 dislike

If there is a discontinuity in the Ricci tensor at the boundary, this would show up as a Dirac delta contribution to the covariant derivative of the Ricci tensor there.

Consider the Bianchi identity $R_{\mu\nu\rho\sigma;\tau}+R_{\mu\nu\sigma\tau;\rho}+R_{\mu\nu\tau\rho;\sigma}=0$.

Decompose the Riemann tensor as the sum of the Weyl tensor plus Ricci contributions. After this decomposition, group the terms in the Bianchi identity into Weyl terms, and Ricci terms. Generically, the Ricci terms have a Dirac delta contribution. So, in order to sum up to zero, so must some terms of the covariant derivative of the Weyl tensor.

In other words, we generically have a discontinuity in the Weyl tensor.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user QGR
answered Jan 25, 2011 by (250 points)
Thanks, yes I am mulling over the implications of these Answers, as I am not familiar with discontinuities (or dirac-distributions) in GR (except at BH Singularities).

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Roy Simpson
With ${\bf Riemann}~=~{\bf Ricci}~+~{\bf Weyl}$ one has a jump in $\bf Ricci$ and $\bf Weyl$, but if one models things right the Riemannian curvature should be continuous. With the Einstein field equation the “turning on” of a $T_{\mu\nu}$ at the surface should counter the jump in the Ricci curvature. All of this has to be done “by hand,” and I don’t think there is a formal procedure for doing it. Physically it makes sense that we do not expect gravity to discontinuously jump if we climb down a mine shaft.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Lawrence B. Crowell
@Crowell: The gravitational field is given by the Christoffel symbol, not the Riemann tensor.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user QGR
The Christoffel symbol gives the Connexion does it not?

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user Roy Simpson
@Roy: Yes, it gives the affine connection.

This post imported from StackExchange Physics at 2014-04-01 16:28 (UCT), posted by SE-user QGR

Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:$\varnothing\hbar$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.