When the Schwarzschild radius is inside the star

[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.

 

[tiny-tim]

hi 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.

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)

 

[K^2]

When the Schwarzschild radius is inside the star.

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

 

[lavinia]

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)

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

 

[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.

 

[WannabeNewton]

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

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).

 

[WannabeNewton]

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.

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.

 

[George Jones]

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.

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.