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

[ApplePion]

<<It will if g_rr is fixed>>

The g_tt metric component looks wrong, too. It is a bit difficult to understand what he is saying because he made a typing mistake. Note in the second radical for the g_tt formula he has a closed parenthesis without an open parenthesis. Regardless, it does not appear that you get the required g_tt = 1 - 2M/R, for r=R. (I'm setting G and c equal to 1, for simplicity)

 

[PeterDonis]

The g_tt metric component looks wrong, too. It is a bit difficult to understand what he is saying because he made a typing mistake. Note in the second radical for the g_tt formula he has a closed parenthesis without an open parenthesis.

The open parenthesis is at the beginning of the entire thing, before the first radical. The whole expression adding both radicals together is inside the parentheses; then that entire expression gets squared to give the final g_tt coefficient. It's the same as what George Jones quoted in his post.

Regardless, it does not appear that you get the required g_tt = 1 - 2M/R, for r=R. (I'm setting G and c equal to 1, for simplicity)

It does when read correctly; see above. You have 3/2 times sqrt(1 - 2M/R) minus 1/2 times sqrt(1 - 2M/R), which gives sqrt(1 - 2M/R); then you square that, which gives the correct g_tt at r = R to match up with the exterior vacuum metric.

 

[ApplePion]

OK, thanks Peter. I did not notice the open parenthesis.

 

[GRstudent]

OK, here I modified it (please confirm that it is correct so I can go and calculate Einstein Tensor):$g_{tt}=(\dfrac{3}{2}\sqrt{1-\dfrac{2MG}{Rc^2}}-\dfrac{1}{2}\sqrt{1-\dfrac{2MGr^{2}}{R^{3}c^2}})^{2}$

$g_{rr}=\dfrac{1}{(1-\dfrac{2MGr^2}{R^3c^2})}$

$g_{\theta \theta}=r^2$

$g_{\phi \phi}=r^2sin^2\theta$

 

[PeterDonis]

OK, here I modified it (please confirm that it is correct so I can go and calculate Einstein Tensor

This looks correct to me.

 

[Mark M]

Yes, it's correct.

EDIT: Oh, Peter beat me to it.

 

[PeterDonis]

This looks correct to me.

Oops, there is one other thing: either there needs to be a minus sign in front of g_tt, or there need to be minus signs in front of g_rr, g_theta_theta, and g_phi_phi. The former sign convention is more common and is probably easier to work with.

 

[GRstudent]

what's $\dfrac{\partial g_{rr}}{\partial r}$?

 

[GRstudent]

^^

Right.

$R_{tt}=-\dfrac{\partial^2_{r} g_{tt}}{2g_{rr}}+\dfrac{\partial_r g_{tt}}{4g_{rr}}(\dfrac{\partial_{r} g_{tt}}{g_{tt}}+\dfrac{\partial_{r}g_{rr}}{g_{rr}}) - \dfrac{\partial_{r}g_{tt}}{r g_{rr}}$

Can someone show me how to solve above equation? I have some trouble with differentiating those messy terms.

Can anyone calculate this? I am getting some crazy mess on my Ti-89.

 

[ApplePion]

GRstudent, why are you interested in knowing R_tt explicitly?

 

[GRstudent]

^
All I want that someone puts my metric into Mathematica and calculates Gammas, Riemanns, Ricci, and Einstein Tensor. I have mathematica but I don't know why GR package doesn't work on my PC.

 

[PeterDonis]

what's $\dfrac{\partial g_{rr}}{\partial r}$?

If you're not comfortable taking the derivative of that g_rr expression with respect to r, I would suggest that you need a calculus review before tackling GR. Andy by "taking the derivative" I mean doing it by hand, not with computer assistance.

Can someone show me how to solve above equation?

How are you coming up with that equation for R_tt?

 

[GRstudent]

^
When I wanted to calculate the derivative of g_rr with respect to r, my Ti-89 gave me some crazy answers.

I need someone to solve this with Mathematica and then, I will analyze the solution. Please! Help HS student!

 

[PeterDonis]

When I wanted to calculate the derivative of g_rr with respect to r, my Ti-89 gave me some crazy answers.

Did you read the part where I said you should be able to calculate that derivative yourself, without computer assistance? If you aren't comfortable doing that with your expression for g_rr, which is reasonably simple as such expressions go, then you are probably not ready to dig into the math of GR. You need to learn the requisite calculus first.

I need someone to solve this with Mathematica and then, I will analyze the solution.

If you are not capable of deriving the solution yourself, I don't think you are capable of "analyzing" it. Certainly not without the math skills I referred to above. The solution is just going to be more math expressions that you will need to be able to understand and manipulate.

If you are just trying to get some kind of basic physical understanding of what the Einstein Field Equation says, then I would not recommend trying to do so by just looking at equations and trying to "analyze" them. You might try reading this discussion by John Baez:

http://math.ucr.edu/home/baez/einstein/

 

[GRstudent]

^
Oh, not g_{rr}. I meant g_{tt} with respect to {r}. If my Ti-89 gives the messiest equation that I have ever seen in my life how would you expect me to solve it? I have taken Calculus courses and successfully competed Differentiation chapter.

 

[GRstudent]

Unless I am presented with some contradictory evidence, I think it is correct:

$R_{tt}=-\dfrac{\partial^2_{r} g_{tt}}{2g_{rr}}+\dfrac{\partial_r g_{tt}}{4g_{rr}}(\dfrac{\partial_{r} g_{tt}}{g_{tt}}+\dfrac{\partial_{r}g_{rr}}{g_{rr}} ) - \dfrac{\partial_{r}g_{tt}}{r g_{rr}}$

 

[PeterDonis]

Oh, not g_{rr}. I meant g_{tt} with respect to {r}.

That one is more complicated, yes, but you should still be able to differentiate it by hand if you've completed courses in calculus including differentiation. Hint: a lot of the symbols represent constants; to make the differentiation easier you might want to roll as many of the constants as possible into single symbols. When you do that you get an expression that looks like this:

$$g_{tt} = \left( \frac{3}{2} \sqrt{A} - \frac{1}{2} \sqrt{1 - B r^{2}} \right)^{2}$$

where A and B are constants. That should help in figuring out how to take its derivative.

If my Ti-89 gives the messiest equation that I have ever seen in my life how would you expect me to solve it?

The old-fashioned way, as I've said repeatedly. You should not be relying on your calculator to figure things out that you can't figure out without it. The calculator can help with speed and accuracy for computations that you already understand how to do; but if you don't understand how to do it without the calculator, you won't understand what the calculator is telling you.

Unless I am presented with some contradictory evidence, I think it is correct

How did you obtain this equation for R_tt?

 

[GRstudent]

The only way to compute these messy equations is to simplify them--establish new variables:

$A=\dfrac{2MG}{Rc^2}$

$B=\dfrac{2MG}{R^3c^2}$

$g_{\mu \nu}=\left [ \begin{matrix} (1.5\sqrt{1-A}-0.5\sqrt{1-Br^2})^2& 0 & 0 & 0 \\ 0 & \dfrac{1}{-(1-Br^2)}& 0 & 0 \\ 0 & 0 & -r^2 & 0\\ 0 & 0 & 0 & -r^2 sin^2 \theta \end{matrix} \right ]$

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

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

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

Please correct me if I am wrong!

Others:

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

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

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

 

[PeterDonis]

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

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

This doesn't look right. $g_{rr}$ is of the form

$$\frac{1}{F(r)}$$

where F(r) is a function of r; so its derivative should be

$$\frac{-1}{F(r)^{2}} \frac{dF}{dr}$$

Edit: Corrected typo in the last formula, both F's are capital F's.

 

[GRstudent]

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

 

[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?