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I am using the book "A first course in general relativity" by Bernard Schutz. On page 267 he derives equation 10.54 but leaves out some steps that I am trying to do myself. The following is a picture showing the page.

enter image description here

I tried solving it in MATLAB to avoid calculation error (and also it's faster). My progress so far is:

$$(\rho + p)\frac{d\phi}{dr}=-\frac{dp}{dr}\tag{10.27}$$ Using (10.48) and rearranging gives:

$$\frac{d\phi}{dr}=\frac{4\pi r}{3}\frac{\rho+3p}{1-8\pi r^2\rho /3}.$$

This can now be integrated:

$$ \int^{\phi(R)}_{\phi(0)}d\phi = \int^{R}_{0} \frac{4\pi r}{3}\frac{\rho+3p}{1-8\pi r^2\rho /3} \,dr$$

Where $p$ equals (10.52) note that there is a typo for equation (10.52) it should state "$p$" instead of "$p_c$".

Now, $\phi(R)$ one can easily get from the stated boundary condition $g_{00}(R)=-e^{2\phi(R)}=-(1-2M/R) \Rightarrow \phi(R)=\frac{1}{2}\log(1-\frac{2M}{R})$ (this step I feel confident about being correct since this was in my lecture). However I am a bit unsure on how to achieve $\phi(0)$. A cheaty way is to simply use the result (10.54) which gives (by setting $r=0$) $\phi(0)=\log(\frac{3}{2}(1-\frac{2M}{R})^{1/2}-\frac{1}{2})$. The next issue is that if one performs this integration then one notice that the integral on the right side of the equation gives problem due to solution containing $\log$ which gets negative if one inserts the lower limit $0$. I thought this could be solved by taking the exponent of both sides since the result we want is exp($\phi$) but this also did not work out for me.

I will include by MATLAB code below which I have been using to trying to solve this.

My question is the following. How do I derive the Schwarzschild constant-density interior solution (equation 10.54)?

clc  
clear

syms rho M r R phi p


g00 = -(1-2M/R);

p = rho((1-2Mr^2/R^3)^(1/2)-(1-2M/R)^(1/2))/(3(1-2M/R)^(1/2)-(1-2M*r^2  /R^3)^(1/2));


phiR = (1/2)log(1-2M/R);

phi0 = log((3/2)(1-2M/R)^(1/2)-1/2);


LS = 1;

RS = (4pir/3)(rho+3p)(1/(1-8pir^2rho/3)); %10.27


simplify(exp(int(LS,phi,phi0,phiR)))

simplify(exp(int(RS,r)))
ludz
  • 963

1 Answers1

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To get the equation (10.54) you should integrate from $\Phi(r)$ to $\Phi(R)$ and not from $\Phi(0)$ to $\Phi(R)$. The same relates to the integral on the right side of your equation which has to be from $r$ to $R$.

JanG
  • 1,831
  • Good point. That is probably correct. However, when changing the limits on the right side of the equation to "from r to R" then the intergral cannot be solved analytically which I believe should be possible. – ludz May 04 '22 at 08:32
  • Write $r^2=u$, $2rdr=du$, and put the corresponding integral function into https://www.wolframalpha.com/calculators/integral-calculator/. You will see that integral has analytical solution. – JanG May 04 '22 at 10:08
  • @ludz, there is a more simple derivation of $g_{00}$ that you can see on https://physics.stackexchange.com/a/679431/281096 . The expression in equation (9) for $p_{1}=0$ (zero pressure on the star surface) is identical with the result derived by Bernard Schulz in (10.54). – JanG May 04 '22 at 10:21
  • I tried the substitution you suggested and both wolframalpha and MATLAB failed to solve it analytically. Remember that p depends on r. The expression for $p$ is the large one seen in equation 10.52 (note there is a type in the book. Replace $p_c$ with $p$). – ludz May 04 '22 at 13:39
  • If you get it right you should come to the equation: $d\Phi=\alpha/2~(1-\alpha u)^{-1/2}~(3 (1-\alpha)^{1/2}-(1-\alpha u)^{1/2})^{-1} du$ that can be analytically integrated. By the way, $\alpha\equiv 2GM/(c^2 R^2)$. – JanG May 04 '22 at 14:47
  • Sorry, I do not see where you get that equation from. For example where is the $\frac{4\pi }{3}$ factor. – ludz May 05 '22 at 09:06
  • I suppose you still use Matlab for deriving the integration function. Can you write down explicitly the function you want to integrate? By the way, for constant density one can write: $(p+\rho) d\Phi/dr=-dp/dr \equiv -d(p+\rho)/dr$ . This can be re-written as $d\Phi=d(p+\rho)/(p+\rho)$ . The integration of the last equation results in $\Phi(R)-\Phi(r)=\ln{(p/\rho+1)}$. The last step you have to make alone. Using the pressure from equation (10.52) you will get the $\Phi(r)$ as in equation (10.54). – JanG May 05 '22 at 14:20
  • This is the integral we are talking about $\int_r^R \frac{4\pi r}{3}\frac{\rho+3p}{1-8\pi r^2\rho/3}dr$. Where $p = \rho \frac{(1-2Mr^2/R^3)^{1/2}-(1-2M/R)^{1/2}}{3(1-2M/R)^{1/2}-(1-2Mr^2/R^3)^{1/2}}$. With the substitution you proposed ($r^2=u$) this turns into $\int_{\sqrt u}^R \frac{2\pi}{3}\frac{\rho+3p}{1-8\pi r^2\rho/3}du$ where now $p = \rho \frac{(1-2Mu/R^3)^{1/2}-(1-2M/R)^{1/2}}{3(1-2M/R)^{1/2}-(1-2Mu/R^3)^{1/2}}$. – ludz May 05 '22 at 19:51
  • All right, forget the substitution. Assign $\beta\equiv 2M/R^3$ and notice that $\rho\equiv 3/(8\pi)\beta$ and $1-8\pi r^2\rho/3\equiv 1-\beta r^2$. Calculate $p+3\rho$ and set it into your function. The result is $f(r) = \beta~ r/[(3(1-\beta R^{2})^{1/2}- (1-\beta r^{2})^{1/2})\cdot (1-\beta r^{2})^{1/2}]$. This function can be easily integrated. – JanG May 06 '22 at 10:58
  • It should be "Calculate $\rho+3p$" and not "$p+3\rho$" as I have written, sorry! – JanG May 06 '22 at 18:01
  • I entered $\int_r^Rf(r)dr$ exactly as you wrote it into wolframalpha and it was not able to solve it. Did you try it? – ludz May 10 '22 at 21:32
  • Why don't you calculate it by yourself? Wolfram https://www.wolframalpha.com/calculators/integral-calculator/ has no problem with that integral: integral(b r (1 - b r^2)^(-1/2))/(3 sqrt(1 - b R^2) - sqrt(1 - b r^2)) dr = log(sqrt(1 - b r^2) - 3 sqrt(1 - b R^2)) + constant – JanG May 11 '22 at 08:35