# How to derive exterior and/or interior Schwarzschild solution using Feynman excess radius equation?

Feynman excess radius equation for a uniformly dense spherical body is $R-\sqrt{\frac{A}{4\pi } } = \frac{G}{3c^{2}} M$

Where $R$ is the radius directly measured by digging a hole in the body, $A$ is the surface area of the body and $M$ is its mass.

I tried to use this equation, coupled with the fact that exterior gravitation due to a uniformly dense spherical body does not depend upon the radius of the body, to derive schwarzschild solution.

This is what I did:

$R-x = M$ (where $x= \sqrt{\frac{A}{4\pi } }$ ($x$ is the predicted radius); and $\frac{G}{3c^{2}}=1$.)

$\Rightarrow R-x=g\int 4\pi x^{2} dR,$ (where $g$ is the density of the body)

$\Rightarrow 1-\frac{dx}{dR} =4\pi gx^{2}$

$\Rightarrow \frac{dR}{dx} =\frac{1}{1-4\pi gx^{2}}$

My idea was to calculate, for every value of $x$, the value of $\frac{dR}{dx}$, for different values of density $g$, such that for every value of $x=a$, $g.\int ^{a}_{0}4\pi x^{2}dR=M=constant$. So i would get $\frac{dR}{dx}$ in the form of $x$ for every value of $x=a$, and integrate it to get the solution.

But many disasters happened in this scheme. Let me summarize by saying that, according to my understanding of Feynman excess radius equation, one needs infinite mass to create an event horizon around a uniformly dense spherical body. In other words, $\frac{dx}{dR}=0$ if and only if $R=\infty$. Which does not makes sense.

So, the question is, how will you derive the exterior and/or interior schwarzschild solution using excess radius equation?

• I am not sure if it is actually possible to get to the interior Schwarzschild solution starting only from the expression of Feynman excess radius equation. Just looking at how it can be derived from the interieror solution for a sphere of uniform density I have my doubts. I see two major problems: 1. the final result is given only by the two leading terms of a much more complex expression $L=\sqrt{R^3/2M}\arcsin\sqrt{2M/R}$. 2. This exact result for L is the solution of an integral over $\sqrt{g_{rr}}$ during that step you kind of lose the information about the functional form of $g_{rr}$.
– N0va
Aug 9 '16 at 20:16
• @M.J.Steil Unfortunately I tend to agree with your last point; I say unfortunately because Raja's ideas are a thoroughly lovely way to present GR if one could pull it off. Aug 10 '16 at 6:48

In addition to my comments and the answer of @Raja. There is actually a way to derive parts of the Schwarzschild solution using the exact expression for the excess: $$S=a \arcsin(\frac ar).$$ If you think of the excess as a distance measured instant in time and radially inward ($\Rightarrow 0=dt=d\theta=d\phi$) the equation for $S$ in general becomes: $$S=\int_0^R \frac {dr}{g_{rr}}.$$ This means if you take the derivative of the exact expression for $S$ you can extract $g_{rr}$ of the interior Schwarzschild solution: $$\frac{dS}{dr}\equiv\frac{1}{g_{rr}}=\frac{1}{\sqrt{1-r^2/a^2}}.$$

This only works with the exact expression and only gives you one metric potential, since $g_{tt}$ has nothing to do with the excess you can not extract it form it.

That all being said I am not to happy with this "derivation": I mean you get back to the interior Schwarzschild solution but only to one part of it and by using a very special equation.

I derived the interior Schwarzschild solution recently because I needed it to benchmark numerical code, which solves the TOV equation. One can derive the TOV equation without the Schwarzschild solution because of that I think a nice way to derive the exterior and interior Schwarzschild solution is to do it starting from the TOV equation. Getting the exterior solution is trivial and getting the interior solution can be done by integrating the remaing equation for $dP/dr$, once one puts in $M(r)=4\pi/3\rho_c r^3$. Or one could follow Schwarzschild's original way, which does not use the TOV equation at all. However the original derivation is much more involved then just integrating the TOV equation.

Personally I see the exterior and interior Schwarzschild solution as special, analytical solutions of the underling TOV equation anyway.

I found a solution to this question. Feynman's excess radius equation is wrong. I found the proof of this in Hartle's book 'gravity' (example 7.6, page no.147).

This example in Hartle's book describes the spacetime curvature of a uniformly dense star using following polar equation:

$dS^{2}=\frac{{dr}^{2}}{1-\frac{{r}^{2}}{{a} ^{2}}} +r^{2}\left( d\theta ^{2}+\sin \nolimits^{2}\theta d\phi ^{2}\right)$

Where $a$ is a constant depending only on the density of the star.

Now, following this, the book gives the formula of the actual radius $S$ of the star, measured directly by digging a hole to the its center:

$S=a\arcsin (\frac{r}{a} )$

Also, volume of the star is given in the book by:

$V=2\pi a^{2}[a\arcsin (\frac{r}{a} )-r\frac{(a^{2}-r^{2})}{a} ^{\frac{1}{2} }]$

$\Rightarrow V\propto [S-r\frac{(a^{2}-r^{2})}{a} ^{\frac{1}{2} }]$

Now for a star of constant density, $Volume\propto Mass$

$\Rightarrow Mass\propto [S-r\frac{(a^{2}-r^{2})}{a} ^{\frac{1}{2} }]$ ...(1)

Now, according to feynman's excess radius equation:

$Mass\propto S-r$ ...(2)

We notice that (1) and (2) cannot both be right.

This suggests that Feynman's excess radius equation is wrong. But actually it is not, as explained by M. J. Steil in the comments to this answer...

• Feynman's excess radius equation is not wrong. The expression of Feynman holds for $R>R_s$ it is the smal angle approximation of the expression given by Harle. I mentioned that in my comment on the original question.
– N0va
Aug 11 '16 at 22:51
• @M. J. Steil , by saying that 'it is the small angle approximation given by hartle' do you mean the equations given in hartle's book are approximate? It is not clear from your comment how otherwise both hartle and feynman can be right.... Please explain Sep 12 '16 at 22:11
• The expression given by Hartle is not an approximation. Feynman's excess equation is, but that does not mean it is wrong. If you take the expression for $S=a \arcsin(r/a)$ and assume $r<a\Leftrightarrow R>R_s$ and then make a series expansion of the $\arcsin$ one gets $S=r(1+\frac{r^2}{6a^2})+ O[\frac{r}{a}]^5$. So Feynman's excess equation is right up to the order $O[\frac{r}{a}]^5$. And if you make a series expansion of your equation (1). (1) and (2) are in agrement up to $O[[\frac{r}{a}]]^3$. The claim that "Feynman's excess radius equation is wrong" is not correct.
– N0va
Sep 12 '16 at 22:44