From Section 9.1, in General Relativity by Woodhouse:

For a normal star, the Schwartzchild radius is well inside the star itself. As it is not in the vacuum region of space-time, the Ricci tensor does not vanish at $r=2m$, and so the Schwartzchild solution is not valid there. Inread the metric is that of an 'interior' Schwartzchild solution, found by solving Einstein's equations for a static spherically symmetric metric, with the energy-momentum tensor of an appropriate form of matter on the right hand side. In such metrics, generally nothing exceptional happens at the Schwartzchild radius. But in the extreme case, all of the body lies within its Schwartzchild radius and the vacuum solution extends down to $r=2m$. In this case, we have a spherical black hole.

1) What is the connection to the Ricci tensor here and why does this mean that the Schwarzschild solution is not valid there?

2) What is an $\textbf{interior}$ Schwarzschild solution?

3) Why does nothing exceptional happen here?

4) Why do we care that all of the body needs to be inside $2m$? I would have thought that the Schwarzschild solution just gives a problem at $r=2m$ where you are dividing by zero effectively. (This is probably more a mathematical problem).


Let $R_{ab}(x)$ be the Ricci tensor at a point $x$ in spacetime, let $T^{ab}(x)$ be the stress-energy tensor, and let $D$ denote the number of spacetime dimensions (so $D=4$ in the real world). Then, for $D\neq 2$, the Einstein field equation implies $$ \text{$R_{ab}(x)=0$ if and only if $T_{ab}(x)=0$} \tag{0} $$ at each point $x$. To deduce this, use the Einstein field equation $$ R_{ab}-\frac{1}{2}g_{ab}g^{cd}R_{cd}\propto T_{ab} \tag{1} $$ along with the fact that the same equation may also be written $$ R_{ab}\propto T_{ab}-\frac{1}{D-2}g_{ab}g^{cd}T_{cd} \tag{2} $$ where $g_{ab}$ is the metric tensor, with the usual summation convention for repeated indices. Equation (2) may be deduced from (1) by contracting both sides of (1) with $g^{ab}$, using the result to write the Ricci scalar $g^{ab}R_{ab}$ in terms of the trace $g^{ab}T_{ab}$ of the stress-energy tensor, and then substituting this expression for $g^{ab}R_{ab}$ back into (1). Equation (1) says that the condition $R_{ab}=0$ implies $T_{ab}=0$, and equation (2) says that the condition $T_{ab}=0$ implies $R_{ab}=0$, as claimed. In words:

  • The Ricci tensor must be zero in vacuum regions, such as the exterior of an idealized star. (But the curvature tensor $R_{abcd}$ can still be non-zero where $T_{ab}=0$.)

  • Inside the star, where we have a non-vacuum ($T_{ab}\neq 0$), the Ricci tensor must also be non-zero.

1) What is the connection to the Ricci tensor here and why does this mean that the Schwarzschild solution is not valid there?

The solution that is most often simply called the Schwarzschild solution is a solution that has $R_{ab}=0$ everywhere (because $T_{ab}=0$ everywhere) except at the singularity. It describes a black hole. Thanks to Birkhoff's theorem [1], this solution also applies outside any spherically symmetric, non-rotating distribution of matter, such as an idealized star, where $T_{ab}=0$. It doesn't apply inside the star, though, because $T_{ab}\neq 0$ (and therefore $R_{ab}\neq 0$) inside the star.

2) What is an interior Schwarzschild solution?

An interior Schwarzschild solution is a metric that solves the Einstein field equation inside a star and also matches the usual exterior Schwarzschild solution at the boundary between the vacuum and non-vacuum regions. The simplest example of an interior Schwarzschild solution is the one corresponding to having a constant density everywhere inside the star. This is described in section 12.3 of [2] and in section 2 of [3]. The constant-density assumption is not realistic, but it is relatively simple mathematically and it is good enough to answer the questions being asked here.

3) Why does nothing exceptional happen here?

For a typical star with mass $m$, the star's radius is much greater than the Schwarzschild radius $2m$. (For a neutron star, change "much greater" to "a little bit greater".) Nothing special happens at $r=2m$ because this is well inside the region where $T_{ab}\neq 0$. The event horizon associated with the usual empty-space Schwarzschild solution is not relevant in this case, because the usual empty-space Schwarzschild solution is valid only in regions where $T_{ab}=0$. For a star, we have a different overall solution that only matches the usual empty-space Schwarzschild solution outside the star. Inside, the metric is different. It has no event horizon and no singularity.

4) Why do we care that all of the body needs to be inside $2m$?

In order to get an event horizon, we need to have sufficient mass concentrated in less than its Schwarzschild radius. A typical star does not satisfy this condition; if it did, then it would collapse, and it wouldn't be a star any more.

In addition to searching with the keywords "interior Schwarzschild solution," you can probably find more information by searching for the keywords "Buchdal's theorem." This is how I found reference [3].


[1] https://en.wikipedia.org/wiki/Birkhoff%27s_theorem_(relativity)

[2] Hobson, Efstathiou, and Lasenby (2006), General Relativity: An Introduction for Physicists, Cambridge University Press

[3] Rezzolla, "An Introduction to Stellar Collapse to Black Holes ," https://www.researchgate.net/publication/239533143_An_Introduction_to_Stellar_Collapse_to_Black_Holes

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.