# A Israel junction conditions for simple metrics

1. Mar 2, 2016

### FunkyDwarf

Hello,

I am curious as to how one appropriately matches an interior and exterior solution in GR, i.e. where the interior corresponds to the field of some finite spherical mass (perfect fluid sphere, for the Schwarzschild interior solution). Specifically, looking at both the Schwarzschild interior, and the metrics given here

http://iopscience.iop.org/article/10.1088/0305-4470/10/4/017/meta
http://www.jstor.org/stable/78530?seq=1#page_scan_tab_contents

it seems that not only is the dt^2 coefficient continuous at the boundary of the object (r=R), but it is also differentiable (in the schwarzschild coordinates, i think?). This is in contrast to the dr^2 coefficient.

Is there some deeper requirement that forces the dt^2 to be smooth and continuous whereas the dr^2 coefficient does not? How to Israel's junction conditions translate into conditions on simple metrics? That is to say, if you know that you are matching an interior to the vacuum solution, what can you say/conditions can you impose on the value of the dr^2 and dt^2 interior coefficients (and their derivatives) at the boundary?

Thanks!
-FD

2. Mar 2, 2016

### bcrowell

Staff Emeritus
I would expect *all* components of the metric to be twice differentiable, because the curvature is computed from those second derivatives, and the curvature should be finite.

Too bad the articles are paywalled.

3. Mar 15, 2016

### FunkyDwarf

Okay so I've a better idea now: the requirement is that the induced metric and extrinsic curvature match at the boundary. The induced metric is easy enough to get my head around, but the extrinsic curvature is harder.

I thought i'd lucked out with these notes here
But their definition of the normal to the hypersurface is something I find a bit confusing. Specifically, I get a different answer for the K_tt term if I use it verbatim. It seems consistent with the definition given here
http://physics.stackexchange.com/questions/100975/hypersurface-normal

Specifically when f(x) = r = const = R (joining at the surface of an object is what I am interested in) you get
$n_{\alpha}^{\mu} = n_{r}^{\mu} = \frac{1}{\sqrt{g_{rr}}} \delta_{r}^{\mu}$
because unless alpha = r, the derivative kills the whole thing off, which is the same as in the pdf linked above. But this ends up giving
$K_{tt} = \frac{GM}{R^2} (1-\frac{2GM}{R})^{3/2}$
which is the wrong power.

I'm sure i'm missing something silly here, thanks in advance!

4. Mar 15, 2016

### PAllen

J.L.Synge has a detailed discussion of junction conditions in his 1960 GR text, which he states is modeled on the work of Israel. He shows, among other things, that if the metric is continuous and differentiable (but not necessarily second differentiable ) across the junction boundary, and is second differentiable elsewhere, then the junction conditions are satisfied. He doesn't argue the other direction of inference.

5. Mar 15, 2016

### FunkyDwarf

Well the junction conditions do seem to require continuity and differentiability of the g_tt component, but only continuity of g_rr, which is what i expect.

Also, I think I found the problem: in that L2.pdf the normal vector has g_rr in the denominator, but based on the definition in the other link it should be g^rr, in which case I get the same answer.