# I Schwarzschild interior boundary conditions

1. Jun 30, 2017

### Ibix

I've been playing around with Maxima and it's ctensor library for tensor manipulation. I decided to have a crack at deriving Schwarzschild's solution for the interior of a constant-density sphere.

I've managed to derive a static, spherically symmetric solution, but am struggling a bit with the boundary conditions. I've used a coordinate system based on the Schwarzschild exterior solution - $\theta$ and $\phi$ have the usual definitions and $r$ is defined in the same sense as the exterior ($\sqrt{A/4\pi}$). I ended up with three constants of integration, and a need to somehow associate the density used in the interior solution with the total mass that the exterior solution uses - four constraints needed. I've got three.

In the circumstances described above (static spherically symmetric solution), am I allowed to require $g^{\mathrm{exterior}}_{tt}=g^{\mathrm{interior}}_{tt}$ and $g^{\mathrm{exterior}}_{rr}=g^{\mathrm{interior}}_{rr}$? I think, given that I'm using the same coordinate basis vectors (give or take a scale factor), I'd simply be insisting that that scale factor is 1. I think that's OK, but am a bit worried that I'm basing something physical on coordinates.

2. Jun 30, 2017

### Staff: Mentor

The way I've done this derivation by hand in the past is to first show that, for a static, spherically symmetric solution, you can write the metric such that

$$g_{rr} = \frac{1}{1 - \frac{2m(r)}{r}}$$

where $m(r)$ is the mass inside radial coordinate $r$, and is in general a function of $r$, but of course it only varies with $r$ in the interior of the sphere--at the surface and in the exterior, it's just $M$, the total mass. This method makes it obvious that $g_{rr}$ matches across the boundary.

Doing the derivation this way also establishes that

$$\frac{dm}{dr} = 4 \pi \rho r^2$$

where $\rho$ is the (assumed known) constant density inside the sphere. Since $\rho = 0$ outside the sphere, this also makes it obvious why $m(r)$ is constant outside the sphere. Furthermore, it gives you a way to relate $\rho$ and $M$: just integrate the above from $r = 0$ to $r = R$, where $R$ is the radial coordinate of the surface of the sphere. The result will be $M$, the mass that appears in the exterior line element.

At this point the only undetermined function is $g_{tt}$, and you already know what that is in the exterior region, by Birkhoff's theorem (any spherically symmetric vacuum metric must be the Schwarzschild metric). Matching $g_{tt}$ at the boundary and using the EFE then gives you $g_{tt}$ in the interior.

More precisely, you can always choose coordinates such that this is the case. The method I described above for $g_{rr}$ amounts to enforcing this, by requiring that the integral of $m(r)$ over the interior equals the total mass $M$ that appears in the exterior line element. For $g_{tt}$, it amounts to choosing the time coordinate so that the time dilation factor for an observer on the surface of the sphere is the same in both the exterior and interior line elements. (Note that this also assumes that we have chosen coordinates in general so that the metric is diagonal--that plus the assumption of a static spacetime ensures that none of the metric coefficients are functions of $t$ and that there are no cross terms with $dt$ in the line element, so $g_{tt}$ is the only metric coefficient involved in the time dilation factor and it's constant in time for a given $r$.)

3. Jul 1, 2017

### Ibix

Thanks, Peter.

I had seen in http://eagle.phys.utk.edu/guidry/astro421/lectures/lecture490_ch10.pdf that the proof of $g_{rr}=(1-2m(r)/r)^{-1}$ is just a clever re-parameterisation of the ${}^0_0$ Einstein equation. I hadn't quite seen the value of it.

I did the derivation by modifying Chris Hillman's coframe ansatz (https://www.physicsforums.com/threads/brs-using-maxima-for-gtr-computations.378991/) from Painleve coordinates to Schwarzschild coordinates for the vacuum. Then I extended it to the non-vacuum case, arguing that Schwarzschild-like coordinates are sensible because hovering observers are at rest with respect to the matter (assuming equilibrium) and therefore the stress-energy tensor for pressurised dust is diagonal. So I can insist that the metric components match because I'm deriving them from the same interpretation of the same frame field at the surface. I think.

I'd been worrying about matching metric components because doing so if I'd used Painleve coordinates (for example) in the exterior and Schwarzschild in the interior I would produce a different result. But in that case I would be starting with the same frame field and explicitly using a different expression of them in coordinate terms, so I wouldn't expect the metric coefficients to match.

4. Jul 1, 2017

### Staff: Mentor

The SET being diagonal doesn't necessarily impose a diagonal metric. However, the fact that hovering observers are at rest with respect to the matter at equilibrium, plus the assumption that the matter itself is static (note that this assumption includes "not rotating"--see below) at equilibrium, implies that there is a timelike Killing vector field everywhere in the spacetime (not just in the vacuum exterior region, where Birkhoff's Theorem shows that such a KVF is present given the assumption of spherical symmetry) and that that KVF is hypersurface orthogonal (this is the part that requires the "not rotating" assumption). That is sufficient to ensure that you can always choose coordinates to make the metric completely diagonal.

Yes, you are basically saying that the coordinate description of the hovering observer at the surface has to match in both the exterior and interior metrics--or more precisely, that you can always choose the coordinates so that they match.

To put it another way, we know there is a timelike KVF everywhere, and we've chosen coordinates such that the basis vector $\partial / \partial t$ is that timelike KVF (which we can always do). That means the 4-velocity of a hovering observer is given by the vector field

$$U = \frac{1}{\sqrt{g_{tt}}} \frac{\partial}{\partial t}$$

Since this 4-velocity must be the same in the exterior and interior charts (because it's the same observer), and $\partial / \partial t$ is the same in both (because the timelike KVF is the same and we've defined the charts so that $\partial / \partial t$ is the timelike KVF), then $g_{tt}$ must be the same in both as well--i.e,. $g_{tt}$ must match at the boundary.

Actually, you wouldn't for $g_{tt}$--it's the same in both charts!

The problem with Painleve coordinates is actually $g_{rr}$; that chart does not have the property that $g_{rr} = 1 / 1 - 2m(r) / r$. This is a manifestation of the fact that the surfaces of constant Painleve coordinate time are not the same as the surfaces of constant Schwarzschild coordinate time, so a coordinate increment $dr$ in Painleve coordinates is in a different "direction" in spacetime than the same increment $dr$ in Schwarzschild coordinates. The presence of the $dt dr$ cross term in the Painleve chart also reflects this difference. So trying to match across the boundary would mean that surfaces of constant coordinate time would have a "corner" in them; they would not be smooth.