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Feb25-12, 06:35 PM
Sci Advisor
P: 7,661
At the risk of hijacking the thread (which seems hopelessly confused anyway), I do have a specific problem along the lines of boundary conditions - one I think I solved correctly, that I presented earlier. I think there was some questions raised about it, but I didn't follow the questions.

If we consider one of the simplest possible forms for the interior metric of a photon star,


eq (1)

[tex]\frac{7}{4}\, dr^2 + r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \sqrt{\frac{7}{3}}\,r\,dt^2[/tex]

we might ask how do we go about enclosing said interior metric in a thin, massless shell, so we get the exterior Schwarzschild metric. I.e. how do we match up the exterior and interior Schwarzschild soultions at the boundary so that we have a solution for light in a spherical "box" by matching the interior solution given by (1) to some exterioor Schwarzschild solution.

I started with the line element from Wald for the spherically symmetric metric:


-f(r)\,dt^2 + h(r)\,dr^2 + r^2 \left(d \theta ^2 + sin \, \theta \: d\phi^2 \right)

Einsteins' equations give via equations 6.2.3 and 6.2.4 from Wald, General Relativity

[tex]8 \, \pi \rho = \frac{ \left( dh/dr \right) }{r \, h^2} + \frac{1}{r^2} \left( 1 - \frac{1}{h} \right) \; = \; \frac{1}{r^2} \frac{d}{dr} \left[r \, \left(1 - \frac{1}{h} \right) \right]

[tex]8 \, \pi \, P = \frac{ \left( df/dr \right) } {r \, f \, h} - \frac{1}{r^2} \left( 1 - \frac{1}{h} \right) [/tex]

Here [itex]\rho[/itex] and P are the density and pressure in the spherical shell.

Setting [itex]\rho[/itex] to zero and using 6.2.3 immediately tells us that r (1 - 1/h) is constant through the shell. For a thin shell, this means that h is the same inside the shell and outside the shell, because r is the same at the interior of the shell and the exterior of the shell, so h-, h inside the shell, equals h+, h outside the shell.

We can add 6.2.3 and 6.2.4 together to get

[tex]8 \pi \left(\rho + P \right) \; = \; \frac{ \left(dh/dr \right) } {r h^2 }+ \frac{ \left( df/dr \right) } {r \, f \, h} \; = \; \left( \frac{1}{ r \, f \, h^2 } \right) \, \frac{d}{dr} \left[ f \, h \right] [/tex]

So we can see that the product (f * h) can't be constant through the shell. So, we known that the right boundary conditions are that h is constant and f varies. In a shell of finite thickness, f will increase continuously throughout the shell. As we shrink it to zero width, f jumps discontinuously.

Simply put, for a _massless_ shell, we can say that the spatial curvature coefficient, h, is the same inside the shell and outside. This is a consequence of Einstein's equations. While h is constant, f, the time dilation metric coefficient, is NOT constant. This also follows from Einstein's equations.

We can do some more computation and find the exterior metric if we assume that the boundary of the shell is located at r=1. (It turns out we can place it wherever we like).

Then, the metric previously given in (1) is used for r<1, and for r> 1, we use

\frac{dr^2}{1-\frac{3}{7r}}+ r^2 \,d \theta^2 + r^2 sin^2 \theta \, d\phi^2 - \left( 1-\frac{3}{7\,r} \right) dt^2

We can do some more interesting stuff along the lines of comparing the Komar mass to the Schwarzschild mass parameter, but I think it suffices to say that the two agree for the total mass M as judged by the observer in asymptotically flat space-time, but are distrubuted differently in the interior.