# When the Schwarzschild radius is inside the star

1. Aug 23, 2011

### lavinia

What happens to the Schwarzschild metric for an isolated non-rotating body when the horizon radius is inside the body? As I remember from classical physics all of the gravitational pull on an object inside a shell cancels out so it would seem that the horizon radius can not include any mass outside of it.

2. Aug 23, 2011

### tiny-tim

hi lavinia!
yes

another way of looking at it is that the Schwarzschild metric is only valid in empty space …

find the Schwarzschild radius from the actual mass, then if that is less than the radius R of the body, the Schwarzschild metric is valid for r > R

(eg the metric in the empty space outside the Sun is the Schwarzschild metric for a black hole with the same mass as the Sun)

3. Aug 23, 2011

### K^2

Just to make sure it's clear, that is always the case for a star.

4. Aug 25, 2011

### lavinia

Thanks Tiny-tim! So the metric breaks down inside the star. That makes sense. What happens to the metric?

5. Aug 25, 2011

### K^2

There is a different metric for inside of a spherically symmetric body. It matches Swarzschild metric at the boundary (surface). There's a name for it, but I don't recall what that name is. I'm also not sure if there is an analytic solution for it, or if it's purely numerical.

6. Aug 25, 2011

### WannabeNewton

It just isn't valid in the interior of the star. If R > 2M and you come to r = R you won't really see anything happen in the metric but since this metric is a vacuum solution it can only be allowed to describe anything for r > R. If you want to describe the interior you will have to solve the EFEs, possibly with spherical symmetry conditions applied, for a non - vanishing Stress - Energy tensor (such as that of a static fluid to get the solution for the interior of a static, spherically symmetric star).

7. Aug 25, 2011

### WannabeNewton

Oh the Tolman - Oppenheimer equation? - $$\frac{\mathrm{d} p}{\mathrm{d} r} = \frac{{(\rho + p)(m(r) + 4\pi r^{3}p)}}{r(r - 2m(r))}$$ where $m(r) = \frac{1}{2}r(1 - e^{-2\Lambda })$ and you have the equation $$(\rho + p)\frac{\mathrm{d} \Phi }{\mathrm{d} r} = \frac{\mathrm{d} p}{\mathrm{d} r}$$ to get the metric. I also believe there is no analytic solution for it but I could be wrong.

8. Aug 25, 2011

### George Jones

Staff Emeritus
An equation of state $p = p \left(\rho \right)$ also is needed. For realistic models (eauations of state), this is done numerically, but Schwarzschild solved Einstein's equation analytically for a constant-density spherical star, with result
$$g=\left( \frac{3}{2}\sqrt{1-\frac{2M}{R}}-\frac{1}{2}\sqrt{1-\frac{2Mr^{2}}{R^{3}}}\right) ^{2}dt^{2}-\left( 1-\frac{2Mr^{2}}{R^{3}}\right) ^{-1}dr^{2}-r^{2}\left( d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right).$$

Unfortunately, the term "interior Schwarzschild solution" can refer to either of two very different metrics (solutions to Einstein's equation):

1) spacetime inside the event horizon of the Schwarzschild black hole solution, where there "is a singularity";

2) the non-singular solution inside a spherically symmetric, constant density massive object.