Radial Excess Calc: GR, 4πGρ & R00

[Thytanium]

Which is the mathematical procedure to obtain $\delta r = \frac{GM}{3c^2}$ from $\nabla^2 V = R_{00} = 4\pi G\rho$ where $\nabla^2 V$ is volume contraction of a spherical mass of density $\rho$ and $R_{00}$ is the 00 component of Ricci tensor $R_{ij}$?

 

[ergospherical]

Assuming you mean the difference between the Schwarzschild coordinate $r_g$ at the surface and the physical distance $\displaystyle{\int_0^{r_g}} \sqrt{g_{rr}} dr$ for a spherical mass distribution of constant density...

...first of all, do you know what the metric of the interior solution for a spherically symmetric, perfect fluid is?

 

[Thytanium]

I don't know the metric of the interior solution for a spherically symmetric perfect fluid.

 

[Thytanium]

Excuse me Ergospherical. Thanks for answering. Ergospherical.

 

[Thytanium]

Excuse me again. I know the metric of Schwarzschild. This is the metric of perfect fluid?

 

[Orodruin]

Excuse me again. I know the metric of Schwarzschild. This is the metric of perfect fluid?

The Schwarzschild metric is the spherically symmetric vacuum solution. The stress energy tensor is zero.

 

[Thytanium]

OK. Thanks for answering Orodruin.

 

[ergospherical]

I mean, you can find it on Wikipedia:
https://en.wikipedia.org/wiki/Interior_Schwarzschild_metric
(there are derivations in Wald, etc.)

Specifically, with a (-+++) signature, the $rr$ component is $g_{rr} = \left( 1- \dfrac{2GMr^2}{c^2 r_g^3} \right)^{-1}$. That's all you need to do the integration

 

[Thytanium]

Thanks Ergospherical. Very thankful,

 

[Thytanium]

 

[Thytanium]

Good day friends. Thankful for your immense help because I already evaluated the integral and obtained the result of the physical distance. But now I have a drawback and it is that I do not know how to do the subtraction between the Euclidean radius and the physical distance that I calculate. I think it is not a simple subtraction because it is a curved space within the mass sphere M. I would appreciate your help again and excuse my ignorance about it.

 

[ergospherical]

Did you already compute the integral? It is a standard one of the form $\int (1-\alpha r^2)^{-\frac{1}{2}} dr$ which you can do with a trigonometric substitution.

The result you gave doesn't look exact, although is correct if you truncate the power series for $\arcsin$...

 

[Thytanium]

Well. I am going to do the serial development of the integration result and then I am going to try to do the subtraction: $r_{g}$ - physical distance. The physical distance is the result of the series expansion of what I obtained by solving the integral and $r_{g}$ is the Euclidean radius. Ok.

 

[Thytanium]

Thakn you. Grateful Ergosferical.

 

[Thytanium]

The physical distance gives me exactly $r_{g}$ after the calculations with the series expansion of the arcsin (z), leaving the first term "z" and neglecting the other higher order terms of the series. Due to this result I think I must have a misconception because the Euclidean radius should not be $r_{g}$ since the radial excess would give me "0". Please, if you can clarify these doubts, I thank you very much. And excuse my ignorance please.

 

[ergospherical]

Can you write out your work? You should end up with\begin{align*}
\int_0^{r_g} \sqrt{g_{rr}} dr &= \sqrt{\dfrac{c^2 r_g^3}{2GM}} \arcsin{\left(r_g \sqrt{\frac{2GM}{c^2 r_g^3}}\right)} = r_g + \dfrac{GM}{3c^2} + O(r_g^{-1})
\end{align*}

 

[Thytanium]

The result I made of my integration was correct but I had not included the second term of the series. Because of that I was wrong. A thousand apologies friend. Very grateful to you Ergospherical. Eternally grateful. I wish you a very good day. That is the correct calculation.