I'm sorry, I don't understand what you are asking for. Could you please clarify?

[PeterDonis]

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.

Looking into this further, I don't think $\Gamma^{r}_{rr}$ has to match at the boundary; and in this particular model, I would not expect it to, since there is a discontinuity in the density at the boundary--it goes from some constant, positive value inside the planet, to zero outside. That's why $\Gamma^{r}_{rr}$ jumps in value.

There is at least one Gamma that does need to match at the boundary: $\Gamma^{r}_{tt}$; that is proportional to the "acceleration due to gravity", which has to be continuous at the boundary.

 

[ApplePion]

"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."

Typically they do, because if they don't, then there is an infinite derivative of a gamma, which would lead to an infinite Riemann Tensor unless strangely canceled out.

 

[PeterDonis]

unless strangely canceled out.

I believe that's what happens with the Gammas that are discontinuous across the boundary; the terms they would contribute to the Riemann tensor end up cancelling out.

 

[GRstudent]

Peter, I calculated Einstein Tensor, please check if it is correct:

$G_{\mu \nu}=\left [ \begin{matrix} \dfrac{6 G M (2 M - r)}{r R^3}& 0 & 0 & 0 \\ 0 & \dfrac{2 M (-G r^3 + R^3)}{(2 M - r) r^2 (2 G M r^2 - R^3)} & 0 & 0 \\ 0 & 0 & \dfrac{-((M (M - r) (2 G (3 M - r) r^2 - R^3))}{((-2 M + r)^2 R^3)} & 0\\ 0 & 0 & 0 & \dfrac{-((M (M - r) (2 G (3 M - r) r^2 - R^3) sin^2 \theta}{((-2 M + r)-((M (M - r) (2 G (3 M - r) r^2 - R^3) sin^2 \theta R^3))} \end{matrix}\right ]$



I know that there is something wrong in this matrix. Check it please.

 

[ApplePion]

"I believe that's what happens with the Gammas that are discontinuous across the boundary; the terms they would contribute to the Riemann tensor end up cancelling out."

That would be a very bizarre situation.

Consider, for example, R1010 in spherical coordinates. Ignoring terms not linear in gamma. it is the derivative of [1,0,0] with respect to r, minus the derivative of [1,0,1] with respect to t. If [1,0,0] is discontinuous along r, then the first term in the Riemann tensor is infinite, and thus [1,0,1] would have to explode in time to keep the Riemann tensor finite.

Other components of the Riemann tensor also behave very implausibly if there are discontinuities in a gamma.

 

[GRstudent]

^
It would have been much clearer had you learned Latex. I myself didn't know it too. But trust me--it is very simple.

 

[Dale]

I have made that suggestion to ApplePion multiple times too.

 

[ApplePion]

"I have made that suggestion to ApplePion multiple times too."

Hi DaleSpam, did you get my email?

Best wishes to you.

 

[PeterDonis]

That would be a very bizarre situation.

I wasn't saying that *any* Gamma could be discontinous across the boundary. I posted previously that $\Gamma^{r}_{tt} = \Gamma^{1}_{00}$ does have to be continuous, since it's proportional to the "acceleration due to gravity", which is an observable. As you correctly point out, that Gamma being discontinuous would also introduce an infinite term into $R^{1}_{010}$.

But *some* Gammas could possibly be discontinuous without doing that. The one in question here is $\Gamma^{r}_{rr} = \Gamma^{1}_{11}$. That one, IIRC, would only contribute to components of the Riemann tensor that are identically zero by symmetry, meaning that any terms involving that Gamma that arise from the formulas cancel each other out. I haven't checked that explicitly, but it seems to me that it ought to work out that way, since as I pointed out before, in this particular solution there is a discontinuity in the density across the boundary, which should show up in that particular Gamma.

 

[GRstudent]

^
Peter,

Please take a look at post #109 (it's very hard to miss) where I posted Einstein Tensor matrix.

 

[PeterDonis]

Please take a look at post #109 (it's very hard to miss) where I posted Einstein Tensor matrix.

I saw it, it doesn't look like what I would have expected at a quick glance, but I haven't had a chance to check it in detail.

 

[GRstudent]

Yeah, I know that it is wrong. Please correct me when able.

 

[GRstudent]

So far we have computed only 4 Gammas:

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

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

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

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

Another one:

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

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

 

[Dale]

Those agree with my Mathematica output, except that I got the opposite sign for $\Gamma^{r}_{\theta \theta}$

 

[GRstudent]

^
Those Gammas weren't calculated by Mathematica--I calculated them by hand. On the contrary, I used Mathematica in Post #109, please check it.

 

[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:
$${\Gamma^r}_{tt}=\frac{1}{4} B r \left(3 \sqrt{1-A} \sqrt{1-B r^2}+B r^2-1\right)$$