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

[PeterDonis]

^
Don't understand what you wrote? Can you express your equation in terms of A and B?

$$g_{rr} = \frac{1}{F(r)} = \frac{1}{- \left(1 - Br^{2} \right)}$$

 

[GRstudent]

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

$\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-\dfrac{2MG}{R^3c^2}r^2))\dfrac{1}{- \left(1 - \dfrac{2MG}{R^3c^2}r^{2} \right)}$

Is this correct? What can I do with it further?

Wait...What?

Gamma is 1/2? With c=G=1 and r=R it should reduce to $\dfrac{M}{2MR-R^2}$

Something is wrong...

 

[PeterDonis]

Is this correct?

No. The gammas involve the *derivative* of g_rr, not g_rr itself. I was pointing out that you took the derivative of g_rr wrong; you need to go back and correct that before going further.

 

[GRstudent]

^
What do you mean? You seem very confused:

$\Gamma^{r}_{rr}=\dfrac{1}{2}g^{rr}(\dfrac{\partial g_{rr}}{\partial r} +\dfrac{\partial g_{rr}}{\partial r}-\dfrac{\partial g_{rr}}{\partial r})$

After cancelations:

$\Gamma^{r}_{rr}=\dfrac{1}{2}g^{rr}(\dfrac{\partial g_{rr}}{\partial r})$

$g^{rr}$ IS INVERSE OF $g_{rr}$. You said that derivative of g_{rr} what you wrote in terms A and B.

 

[PeterDonis]

I said that you computed

$$\dfrac{\partial g_{rr}}{\partial r}$$

wrong. I didn't say anything about $g^{rr}$.

 

[GRstudent]

^
Can you RE-write the derivative of g_{rr} in terms of A and B correctly?

 

[PeterDonis]

Can you RE-write the derivative of g_{rr} in terms of A and B correctly?

I could, but I'd prefer that you did it yourself, using the hints I gave you in posts #89 and #91.

 

[GRstudent]

Is this one correct? I re-did my earlier calculations:

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

$\Gamma^{r}_{rr}=\dfrac{1}{2} (-(1-\dfrac{2MG}{R^3c^2}r^2)) \dfrac{-2r\dfrac{2MG}{R^3c^2}}{(\dfrac{2MG}{R^3c^2}r^2 -1)^2}$

 

[PeterDonis]

Is this one correct?

Looks OK to me.

 

[GRstudent]

^
Nope, there is a mistake in factor somewhere. When reduced to ordinary Schwarzschild metric (c=G=1 and r=R) it should be

$\dfrac{M}{2MR-R^2}$

Instead I got $\dfrac{-2M}{2MR-R^2}$

Factor (-2) is somewhere unnecessary, where?

 

[PeterDonis]

^
Nope, there is a mistake in factor somewhere. When reduced to ordinary Schwarzschild metric (c=G=1 and r=R) it should be

$\dfrac{M}{2MR-R^2}$

The Gamma symbols are not the same as the metric. Why are you trying to compare them?

 

[GRstudent]

^
I was trying to compare $\Gamma^{r}_{rr}$ in Schwarzschild Interior to that in ordinary Schwarzschild.

 

[PeterDonis]

^
I was trying to compare $\Gamma^{r}_{rr}$ in Schwarzschild Interior to that in ordinary Schwarzschild.

What do you think is the general formula for $\Gamma^{r}_{rr}$, as a function of r, in the Schwarzschild exterior vacuum metric (I assume that's what you mean by "ordinary Schwarzschild")?

 

[PeterDonis]

^
I was trying to compare $\Gamma^{r}_{rr}$ in Schwarzschild Interior to that in ordinary Schwarzschild.

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.

 

[GRstudent]

I have GREAT.m package for Mathematica successfully installed. Can someone show me how to put matrix into it?

 

[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:

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



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.