- #1
Poirot
- 94
- 3
Homework Statement
My end goal is to plot null geodesics around a black hole with realistic representations within the horizon (r<2GM, with c=1) using Mathematica. I've done this for outside the horizon using normal Schwarzschild coordinates and gained equation (1) below, and then used this with equation (3) and converted to u=1/r to find equation (4) in which I can plot u as a function of Φ. I then used NDSolve to solve this with varying impact parameters (D) but I know once it crosses the horizon I cannot justify the plot as being meaningful. I'm therefore converting to Eddington-Finkelstein coordinates using equation (5) to go from t -> v so the geodesic can cross the horizon fine. This gained the new metric equation (6). I can't find any help on how to do this properly for any geodesics that aren't radial so I've tried doing it myself a few different ways but all along the same logic as I did before. Every way I've tried it I get the same equation as (1) but with a different constant (Which I set as I plot so is meaningless to changing how the geodesic behaves at the horizon).
Homework Equations
\begin{array}
.Basic \ Schwarzschild \ metric \ in \ Lagrange \ form: \\
L = -\dot{t}^2 + \dot{r}^2 + r^2(\dot{\theta}^2+Sin^2{\theta}\dot{\phi}^2)\\
Relavent \ Equations:\\
\dot{r}^2 = E^2 - \frac{l^2}{r^2}(1-\frac{2GM}{r}) \ (1)\\
where,\ E = \dot{t}(1-\frac{2GM}{r}) \ (2)\\ and\ l = \dot{\phi}r^2 \ (3)\\
(\frac{du}{d\phi})^2 = 2Mu^3 - u^2 + 1/D^2\ (4)\\
where, \ D=\frac{l}{E} \\
t = v -r -2GMln|\frac{r}{2GM}-1| \ (5)\\
L = -(1-\frac{2GM}{r})\dot{v}^2 + 2\dot{v}\dot{r} + r^2\dot{\phi}^2\ (6) \\
Note:\ \theta = \frac{\pi}{2} \ so \ \dot{\theta}=0 \ and \ Sin^2{\theta}=1
\end{array}
The Attempt at a Solution
Attempted method 1:
I tried subbing back in for v and v(dot) so I could use equation (2) so I can use the same conserved quantities but everything just cancels back down to equation (1) like I had before.
Attempted method 2:
Finding a conserved quantity for v(dot) and calling the relevant constant say E' which once running all the maths through gains the same equation as (1) but with E' where E would be.