Does the (0,0) component of the Ricci tensor equal Poisson's equation?

[Dale]

As far as I can tell 109 isn't even close. However, it is unnecessarily cluttered with all of the G and c and M and R terms. You should just stick with A and B.

 

[GRstudent]

^
OK, I feel that there is something wrong with it. Can you check it with your Mathematica?

Also, there other non-zero Gammas which I haven't mentioned.

 

[Dale]

Also, there other non-zero Gammas which I haven't mentioned.

Well, there are the obvious symmetries:
$\Gamma^{\theta}_{r \theta}=\Gamma^{\theta}_{\theta r}$ and $\Gamma^{\phi}_{r \phi}=\Gamma^{\phi}_{\phi r}$ and $\Gamma^{\phi}_{\theta \phi}=\Gamma^{\phi}_{\phi \theta}$

Beyond those the following terms are also non-zero:
$\Gamma^{t}_{r t} = \Gamma^{t}_{t r}$
$\Gamma^{r}_{t t}$
$\Gamma^{r}_{\phi \phi}$

 

[pervect]

Also, I'm not sure that the Gammas have to match at the boundary. The metric components themselves do, but I'm not sure the Gammas do. I'll have to check further on that.

You'll definitely see jumps in the christoffel symbols, and in the metric, if you have pressurized shells - for instance, if if you have a ball or radiation trapped in a box. They'll be sudden jumps only in the limit of very thin shells IIRC.

I suspect you won't see any jumps matching interior to exterior solutions lacking such pressure differences, but it's worth checking to be sure.

It'd be wrong to try and match an interior solution to an exterior solution where the interior pressure wasn't equal to the exterior pressure (i.e. zero for a vacuum exterior) without adding some sort of shell to acount for the pressure difference.

 

[PeterDonis]

It'd be wrong to try and match an interior solution to an exterior solution where the interior pressure wasn't equal to the exterior pressure (i.e. zero for a vacuum exterior) without adding some sort of shell to acount for the pressure difference.

In the particular solution in question (constant density spherical massive body surrounded by vacuum--it's discussed, for example, in MTW), the pressure is continuous at the boundary (the surface of the body)--at least, I'm pretty sure that's right--but the density is not; it jumps from its interior value to zero at the boundary. The jump in density causes a jump in the radial derivative of g_rr (g_rr itself is continuous), which shows up as a jump in $\Gamma^{r}_{rr}$. As far as I can tell, that is the only Christoffel symbol that is affected.

 

[GRstudent]

Others so far:

$\Gamma^{r}_{\phi\phi}=\sin^2\theta r (Br^2-1)$

$\Gamma^{r}_{tt}=\dfrac{0.5r(\sqrt{1-r^2B}-3 \sqrt{1-A})B(1-Br^2)}{2 \sqrt{1-r^2B}}$

Please check them!

 

[Dale]

Please check them!

Those are correct.

 

[GRstudent]

List of Gammas:

$\Gamma^{r}_{rr}=\dfrac{-rB}{(Br^2 -1)}$

$\Gamma^{\theta}_{\theta r }=\dfrac{1}{r}$

$\Gamma^{\phi}_{\phi r}=\dfrac{1}{r}$

$\Gamma^{\phi}_{\phi \theta} = \dfrac{1}{\tan \theta}$

$\Gamma^{r}_{\theta \theta}=-r(Br^2-1)$

$\Gamma^{\theta}_{\phi\phi}=-\sin\theta \cos\theta$

$\Gamma^{r}_{\phi\phi}=\sin^2\theta r (Br^2-1)$

$\Gamma^{r}_{tt}=\dfrac{0.5r(\sqrt{1-r^2B}-3 \sqrt{1-A})B(1-Br^2)}{2 \sqrt{1-r^2B}}$

 

[Mentz114]

I think those are correct except for $\Gamma^r_{tt}$
$${\Gamma^r}_{tt}=-\frac{r\,B\,\sqrt{1-{r}^{2}\,B}\,\left( \sqrt{1-{r}^{2}\,B}-3\,\sqrt{A}\right) }{4}$$

and ${\Gamma^t}_{rt}$ is not in the list. Calculated by Maxima ctensor package.

 

[GRstudent]

^
Can you calculate Einstein Tensor of my Gammas?

 

[Dale]

I think those are correct except for $\Gamma^r_{tt}$
$${\Gamma^r}_{tt}=-\frac{r\,B\,\sqrt{1-{r}^{2}\,B}\,\left( \sqrt{1-{r}^{2}\,B}-3\,\sqrt{A}\right) }{4}$$

and ${\Gamma^t}_{rt}$ is not in the list. Calculated by Maxima ctensor package.

I think those are the same. The expression that Mathematica gives seems to be about halfway in-between both your expression and GRstudent's:
$${\Gamma^r}_{tt}=\frac{1}{4} B r \left(3 \sqrt{1-A} \sqrt{1-B r^2}+B r^2-1\right)$$