The Matter-Vacuum Boundary in General Relativity

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

A perhaps naive answer: shouldn't the case you describe be physically non feasible? In a real physical scenario there must be a smooth interface between the vaccuum region and the planet's interior...