# Interior metric solution

1. Mar 14, 2010

### Orion1

$$g_{tt} = \left\{ \begin{array}{rcl} \frac{3}{2} \left( 1 - \frac{2GM(r)}{c^2 R} \right)^{\frac{1}{2}} - \frac{1}{2} \left( 1 - \frac{2 G M(r) r^2}{c^2 R^3} \right)^{\frac{1}{2}} \; \; \text{for} \; \; 0 \leq r \leq R \; \text{(interior)} \\ \left( 1 - \frac{2GM(r)}{c^2 R} \right) \; \; \text{for} \; \; r > R \; \text{(Schwarzchild)} \\ \end{array} \right.$$

$$g_{rr} = \left\{ \begin{array}{rcl} \left( 1 - \frac{2G}{c^2 r} \frac{4 \pi r^3}{3} \rho_0 \right)^{-1} \; \; \text{for} \; \; 0 \leq r \leq R \; \text{(interior)} \\ \left( 1 - \frac{2G M(r)}{c^2 r} \right)^{-1} \; \; \text{for} \; \; r > R \; \text{(Schwarzchild)} \\ \end{array} \right.$$

Please examine the derivation from General Relativity in reference 3.

My question is theoretical, why would the relativistic Equation of State for hydrostatic equilibrium which is the Tolman-Oppenheimer-Volkoff (TOV) equation, be based on the exterior metric as opposed to the interior metric?

What are the formal equation definitions for $$g_{\theta \theta}$$ and $$g_{\phi \phi}$$ for the interior metric?

Reference:
http://www.infn.it/thesis/PDF/getfile.php?filename=3852-Mana-specialistica.pdf"
http://en.wikipedia.org/wiki/Birkhoff%27s_theorem_%28relativity%29" [Broken]
http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_equation#Derivation_from_General_Relativity"

Last edited by a moderator: May 4, 2017
2. Mar 14, 2010

### bcrowell

Staff Emeritus
Based on a quick read-through of ref. 1, it looks to me like they're just using the Schwarzschild metric as a way of setting a boundary condition for the interior solution. I don't think the equation of state has anything to do with the exterior metric. The equation of state sets up the differential equations given on p. 16, and the exterior solution sets the boundary conditions for those differential equations.

3. Mar 16, 2010

### Altabeh

That is, as bcrowell said, just because on the boundary the interior solution reduces to the exterior solution with only one difference arising from the way the mass of the gravitating body affects the field and how it varies by $$r$$ and $$R$$.

They don't undergo any change from the spherically symmetric line-element setup! All changes from the exterior metric will be of the time and radial components and the reason is simple: we don't have any rotation or extra effects on the $$\theta$$ and $$\phi$$ coordinates when transitioning into the interior of the star. All we take into account in the interior solution involves the radius $$r$$ through pressure and mass density and the time-component of metric due to the connection the mass density has with $$G_{00}$$.

AB

4. Mar 17, 2010

### Orion1

Last edited by a moderator: Apr 24, 2017
5. Mar 17, 2010

### George Jones

Staff Emeritus
There is no known "stellar interior" that can be matched to an exterior Kerr metric; this is an important unsolved problem.

Last edited by a moderator: Apr 24, 2017
6. Mar 18, 2010

### Orion1

How was the formal definition for $$g_{tt}$$ derived for the non-rotating interior metric?

$$g_{tt} = \frac{3}{2} \left( 1 - \frac{r_s}{R} \right)^{\frac{1}{2}} - \frac{1}{2} \left( 1 - \frac{r_s r^2}{R^3} \right)^{\frac{1}{2}} \; \; \text{for} \; \; 0 \leq r \leq R \; \text{(interior)}$$

$$g_{tt} = \frac{3}{2} \left( 1 - \frac{2GM(r)}{c^2 R} \right)^{\frac{1}{2}} - \frac{1}{2} \left( 1 - \frac{2 G M(r) r^2}{c^2 R^3} \right)^{\frac{1}{2}} \; \; \text{for} \; \; 0 \leq r \leq R \; \text{(interior)}$$

7. Mar 18, 2010

### Altabeh

See Schutz B.F. A first course in general relativity, page 261.

AB

8. Mar 22, 2010

### stevebd1

Last edited by a moderator: Apr 24, 2017
9. Mar 22, 2010

### Orion1

The rotating interior metric solution must be capable of describing real physical systems in General Relativity in order to build stellar models upon to compare with real physical observations, therefore this cannot be the solution for the interior Kerr metric.

Reference:
http://www.csun.edu/~vcphy00d/PDFPublications/1976%20Collas-Lawrence.pdf" [Broken]

Last edited by a moderator: May 4, 2017
10. Mar 22, 2010

### JesseM

Is it possible to do a numerical simulation to get an approximate idea of what the curvature in the interior would look like (and perhaps fine-tune it until the exterior looks like the Kerr metric), or would this approach not be helpful in looking for an exact solution?

11. Mar 23, 2010

### Orion1

It is an interesting point that with all the computational power available to modern scientists, that none have yet discovered a solution to this important problem.

In my opinion, it is only a question of time before a solution is discovered, probably by an extremely powerful supercomputer.

12. Mar 24, 2010

### Altabeh

The method of 'fine tunning' is not applicable in this case because the interior of BH is a region not a point so imagine that you want to fine tune the local curvature, is it really possible to correspond to any point in the small region a number by hand in such a way that they all would form the "local curvature"? When talking of this method for the Lagrangian of an n-particle system, this can work because the Lagrangian has to be used to predict the dynamical behaviour of system at any time and point so if I found the right values for the coupling constants at time t and position x, the values work well for other points in spacetime! But finding the real value of Riemann curvature tensor at some point (assuming that the interior of BH has axial symmmety) means that one has to solve at least 20 nonlinear equations with no degrees of freedom to get the numerical values for tens of the first derivatives of metric tensor at only one given point, then predict what components generate these values and this by itself should be in agreement with what you get by doing fine tunning at some other point in the interior of BH.

This is only possible if technology reaches its infinite borders.

AB

13. Mar 25, 2010

### stevebd1

Last edited: Mar 25, 2010
14. Mar 25, 2010

### Frame Dragger

Hmmm, I know a Numerical Relativist working at the API, I could ask him if this sounds possible? He works on the 2-body problem, but he knows his computers and his SR/GR. That said, given what Altabeh has said, I'm guessing you've already gotten the answer he'd give.