# Find resultant vector of light ray under the influence of a Schwarzschild black hole?

I'm working on implementing a GR (general relativity) raytracer for the purposes of displaying realistic visuals of a black hole against some fixed imagery (pictures of stars, for example.) I don't understand a lot of the language in general relativity, so I would like some help with the mathematics of it. Please bear with me, I am a programmer not a physicist.

Suppose I have three vectors (in 3-dimensional space), p (black hole position), p0 (light ray origin), v0 (light ray direction) and the mass M of the Schwarzschild black hole, how do I find the resultant direction of the light ray v1 after it has been fully influenced by the black hole? I understand that this might involve using some form of the geodesic equation:

$$$$\frac{d^2x^\mu}{ds^2} + \Gamma^\mu_{\nu\lambda}\frac{dx^\nu}{ds}\frac{dx^\lambda}{ds} = 0$$$$

I believe my use case requires a formulation of this equation integrated at infinity (to find vector v1.) I understand if the problem is non-trivial and cannot be explained or simplified to a single equation, and in this case I would highly appreciate pointers for better understanding of the topic. Numerical methods are more than acceptable.

Notes:

• During research I saw that proper time could be taken to be the time of an observer positioned at infinity, for simplicity. I don't know how to do this in practice.
• The black hole in question is not rotating.
• The black hole does not have charge. (?)
• your project is very cool! I have a bachelor in physics and im doing a master in theoretical physics atm, i have a passion for coding since a long time. I feel the question doesnt belong here too well, but do you feel like collaborating on this? Feb 24, 2022 at 18:16
• I think unfortunately this question might be too broad, but I don't know of any resource for this that isn't just a general relativity book. Not saying that one can't exist, only that I don't know of any. Feb 24, 2022 at 22:10

Indeed you need the geodesic equation, though you do not need its time dependence, but only its shape. This simplifies the problem a lot and you end up having to integrate an ODE to get the solution you want.

I will do the steps needed to obtain this equation. Start from the Schwartzchild metric $$$$ds^2 = \left(1 - \frac{2MG}r \right)dt^2 - \left(1 - \frac{2MG}r \right)^{-1}dr^2 - r^2(d\theta^2 + \sin^2\theta \ d\phi^2)$$$$ Now we consider the symmetries of the action $$$$S[x] = \frac1 2 \int d\lambda \ g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$$$$ The coordinates $$t$$ and $$\phi$$ are cyclic and the corresponding conserved Noether charges are energy and angular momentum. (There is an additional symmetry under reparametrizations $$\lambda^\prime = \lambda + \epsilon(\lambda)$$ but we do not need it). The expressions for energy and angular momentum are $$$$E = g_{tt}\frac{dt}{d\lambda} = \left(1 - \frac{2MG}r\right)\frac{dt}{d\lambda}$$$$

$$$$L = - g_{\phi\phi}\frac{d\phi}{d\lambda} = r^2 \frac{d\phi}{d\lambda}$$$$

Light-like geodesics are given by $$$$g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda} = 0$$$$ also your points p, p0 and the vector v0 together identify a plane which after a rotation we can say it is the plane $$\theta = \frac \pi 2$$ (I already considered this is the expression of $$L$$).

So we can write the equation for the light geodesics as $$$$\frac{E^2}{1- 2MG/r} - \frac 1 {1 - 2MG/r}\dot{r}^2 - \frac{L^2}{r^2} = 0$$$$ ($$\dot{r} = dr/d\lambda$$). Now we eliminate $$\lambda$$ in the equation, so we have an equation for the function $$r(\phi)$$, the light ray path on the plane defined above.

$$$$\frac{E^2}{1- 2MG/r} - \frac 1 {1 - 2MG/r}\frac{L^2}{r^4}\left(\frac{dr}{d\phi}\right)^2 - \frac{L^2}{r^2} = 0$$$$

You can than manipulate this equation a bit and end up with a second order equation. First we introduce the variable $$u = 1/r$$, then $$du = -r^{-2}dr$$ and we have hence $$$$\frac{E^2}{1-2MGu} - \frac {L^2} {1- 2MGu}\left(\frac{du}{d\phi}\right)^2 - L^2u^2 = 0$$$$

Then we cast it into $$$$\frac{E^2}{2L^2} = \frac 1 2 \left(\frac{du}{d\phi}\right)^2 + \frac {u^2} {2} (1 - 2MGu)$$$$

This looks like the energy of a particle of unit mass in the potential $$V(u) = \frac {u^2} {2} (1 - 2MGu)$$ and of energy $$E^2/2L^2$$ (with time replaced by $$\phi$$). Deriving with respect to $$\phi$$ we obtain the equation $$$$\frac{d^2u}{d\phi^2} = - \frac{dV}{du}$$$$

which you can integrate via numerical methods

• Thanks a lot for the insight. I would love to collaborate but this is for my final year thesis and the work must largely be my own. I will gladly include your contributions as reference in the final paper, though (this is actually mandated by my University.) This has been helpful but I'm not sure how I would calculate E, L or r in this equation, and I'm guessing this will need integrating to remove the derivative? I have sent you an email (the one on your GitHub) if this makes communication easier. Feb 25, 2022 at 13:52
• $E$, $L$ need to be calculated via the initial conditions, I will workout the formulas in terms of the data you have when I have time. $r$ is just the Schwarzchild radius wich is related to the radial distance from the black hole Feb 25, 2022 at 16:18
• $L/E$ is the impact parameter $b$. The deflection of light only depends on this, not $E$ and $L$ separately. Feb 25, 2022 at 20:49
• This looks promising, but I don't quite understand. Should I plug in V'(u) for dV? If this is the final equation to integrate, shouldn't it increase inversely proportional to the distance of the photon from the black hole? I know u=1/r, but the Schwarzchild radius (r) from the black hole is a fixed value dependent on its mass. Since we see d^2u/dϕ^2 on the left side, this would imply that the Schwarzchild radius changes over time if I'm not mistaken. Perhaps I am misunderstanding the definition for r? Feb 28, 2022 at 15:42