# Trajectory of a photon around a Schwarzschild black hole?

Consider a photon coming from the infinity in a unbounded orbit to a Schwarzschild black hole (Schwarzschild radius $r_{s}$) (see this for illustration). Its impact parameter is $b$ and its distance of closest approach is $r_{0}$ with $$b^2=\frac{r_{0}^{3}}{r_{0}-r_{s}}$$.

Then its trajectory in polar coordinates is defined by :

$$\frac{d\varphi}{dr} = \frac{1}{r^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{r}\right)\frac{1}{r^2}}}$$

Consequently : $$\varphi\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}$$

and one can compute the total deviation using : $\Delta\varphi = 2\times\left(\lim_{r\to+\infty}\varphi\left(r\right)-\frac{\pi}{2}\right)$

But my question is : how can I plot/draw the trajectory using the integral expression of $\varphi\left(r\right)$ ?

Because if I compute : $$f\left(r\right) = 2\times\left(\int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}-\frac{\pi}{2}\right)$$ I obtain $f\left(r_{0}\right) = -\pi$, and then $f$ increases up to zero, crosses zero, and tends to its positive value at infinity $\Delta\varphi$. It does not make sense for me and I do not understand how to compute the trajectory from that.

If I compute : $$g\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}$$ it starts from $0$, and increase up to $\frac{\pi}{2}+\frac{\Delta\varphi}{2}$.

I would like to compute the trajectory in the $\left(x, y\right)$ plane, so how to use the values of $f\left(r\right)$ or $g\left(r\right)$ to compute the function $y\left(x\right)$ ?

## 1 Answer

In fact, the problem was that computing the deflection is not very intuitive. So the trajectory in polar coordinates is : $$\varphi\left(r\right) = \int_{r_{0}}^{r} \frac{dp}{p^2\sqrt{\frac{1}{b^2}-\left(1-\frac{r_s}{p}\right)\frac{1}{p^2}}}$$ and in cartesian coordinates it is nothing else than :

• $x=r\cos\left(\varphi\left(r\right)\right)$
• $y=r\sin\left(\varphi\left(r\right)\right)$

and it represents a photon starting from $\left(x_0, y_0\right) = \left(r_{0}, 0\right)$ and going up.

But what is non-intuitive (a drawing helps a lot), it that the half deflection $\alpha/2$ is in fact :

$$\frac{\alpha}{2} = \varphi\left(r\right)-\cos^{-1}\left(\frac{r_{0}}{r}\right)$$

Problem solved...