# Integrating the Schwarzschild geodesics equations

I am trying to make a graph similar to this one from the Wikipedia article about Schwarzschild geodesics: There is an equation like this: $$\varphi = \int\frac{dr}{r^2\sqrt{\frac{1}{b^2}-(1-\frac{r_s}{r})\frac{1}{r^2}}}.$$

I did not get how they manage to draw this picture from that equation. Particular questions:

1. If it is a polar plot, usually it is opposite, like $r(\varphi)$, but not $\varphi(r)$. How can I draw the polar plot if I have $\varphi(r)$ function?

2. What is integration interval here? $\varphi = \int_{?}^{?}...$

• @Countto10 If I will integrate from -Infinity to +Infinity the result will not depend on r. It will be φ(rs, b), but I expect something like φ(rs, b, R). I put capital R to avoid confusion with r inside integral. – Zlelik Feb 24 '17 at 8:50
• My apologies Zlelik, I read the post too fast, (as usual) – user146020 Feb 24 '17 at 11:58

To your first question: Using a parametric plot in $r$ or $\phi$ and plotting $\left\{r\cos(\phi),r\sin(\phi)\right\}$.

To your second question: that is where things become tricky. Depending on what kind of geodesics comes out the radius shrinks or increases and the angle is also kind of problematic for closed orbits. A formulation with proper time as parameter is better suited for implementation.

This Wolfram Demonstration Geodesics in Schwarzschild Space of Niels Walet has a simple implementation of an integration over proper time.

That being said; with a bit of tweaking one can of course implement the differential equation for $\phi$. J. B. Hartle provides a Mathematica Notebook here as supplement to his book Gravity: An Introduction to Einstein's General Relativity.

• Thanks for the really good answer. Mathematica example is very useful. I use Mathematica as well, but I do not have access to Mathematica right now and I will be able to check this example later. What will be integration interval in the most simple case, just to draw the same picture as I attached in the question? I just need a picture, where I can change some parameters like b and rs. The proper time I do not really care. Or it is not possible without proper time? – Zlelik Feb 24 '17 at 8:48
• Well one can just implement the equation in $\phi$ chose a $\phi(r_0)$ and an $r_0$ and integrate to an $r_1$ for geodesics which fall directly into the black hole that works fine. But for unbound and bound curves one will run into some problems with the integration domain, since $r$ can get smaller first, then bigger and so on. The code of Hartle deals with that. – N0va Feb 24 '17 at 9:59

If I use this simple equation with integration interval from 0 to Rmax

$\Large \varphi(r_s,b,R_{max}) = \Huge\int_{0}^{R_{max}}\frac{dr}{r^2\sqrt{\frac{1}{b^2}-(1-\frac{r_s}{r})\frac{1}{r^2}}}$

I can get only red lines from the picture in my original question, when photon falls in the Black Hole, as M. J. Steil told. In order to get green lines this equation will not work and solution of differential equation is needed. Good example is here provided by M. J. Steil.

From this exampe I could get this picture where photon makes a loop around black hole and goes back. 