# Calculation of radial excess in General Relativity Theory

• A
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?

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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?

Staff Emeritus
Homework Helper
Gold Member
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.

vanhees71
Thytanium

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

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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.