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Thytanium

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Thytanium

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- #2

ergospherical

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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?

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

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- #3

Thytanium

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

- #4

Thytanium

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Excuse me Ergospherical. Thanks for answering. Ergospherical.

- #5

Thytanium

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

- #6

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The Schwarzschild metric is the spherically symmetric vacuum solution. The stress energy tensor is zero.Excuse me again. I know the metric of Schwarzschild. This is the metric of perfect fluid?

- #7

Thytanium

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OK. Thanks for answering Orodruin.

- #8

ergospherical

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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

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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- #9

Thytanium

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Thanks Ergospherical. Very thankful,

- #10

Thytanium

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- #11

Thytanium

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- #12

ergospherical

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

- #13

Thytanium

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- #14

Thytanium

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Thakn you. Grateful Ergosferical.

- #15

Thytanium

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- #16

ergospherical

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\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*}

- #17

Thytanium

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