Orbits of Photons

Since the photon has zero rest mass, we cannot use the proper time to parametrize its path. Nevertheless, we can use some increasing (affine) parameter $\sigma$ along the geodesic. The Euler-Lagrange equation gives essentially the same result since the Lagrangian is independent of both $t$ and $\varphi$, there are conserved quantities related to the energy ( $\equiv\epsilon$) and angular momentum ( $\equiv\ell$). The photon has a null geodesic $ds=0$ and we may use that equation directly.

$${dt\over d\sigma} =\dot{t}={\epsilon\over\left(1-{r_s\over r}\right)}$$

$$\dot{\varphi} ={\ell\over r^2}$$

$$ds^2 =0=\left(1 - \frac{r_s}{r} \right) c^2 \dot{t}^2 - \frac{\dot{r}^2}{1-\frac{r_s}{r}} - r^2 \dot{\varphi}^2$$

$$=\left(1 - \frac{r_s}{r} \right) c^2 {\epsilon^2\over\left(1-{r_s\over r}\right)^2} - \frac{\dot{r}^2}{1-\frac{r_s}{r}} - {\ell^2\over r^2}$$ 0

$$= c^2 {\epsilon^2\over\left(1-{r_s\over r}\right)} - \frac{\dot{r}^2}{1-\frac{r_s}{r}} - {\ell^2\over r^2}$$ 0

$$\dot{r}^2 = c^2 \epsilon^2 - {\ell^2\over r^2}\left(1-{r_s\over r}\right)$$

If we try an analysis with $V_{eff}={\ell^2\over r^2}\left(1-{r_s\over r}\right)$, we see that the potential reaches a maximum at $r={3\over 2} r_s$ and falls off in either direction from there, so there are no stable orbits, although there is in principle an unstable circular orbit for photons only at $r={3\over 2} r_s$. For smaller radii, the photon will be sucked in. For larger radii, it will be deflected.

To solve the equation of motion, it will be best to get to a linear differential equation, by differentiating the null geodesic equation above.

$$2\ddot{r}\dot{r} -{2\ell^2\over r^3}\dot{r}+{3r_s\ell^2\over r^4}\dot{r} = 0$$

$$\ddot{r} -{\ell^2\over r^3}+{3r_s\ell^2\over 2r^4} = 0$$

It will be easiest to analyze the deflection by parametrizing the trajectory in terms of the azimuthal angle $\varphi$, rather than using the general parameter $\sigma$.

$$\dot{r}={dr\over d\sigma}={dr\over d\varphi}\dot{\varphi}={\ell\over r^2}{dr\over d\varphi}$$

We can now transform the differential equation into one with $\varphi$ as the independent variable. At the same time lets make the standard transformation $u={1\over r}$. This means that $u'\equiv {du\over d\varphi}=-{1\over r^2}{dr\over d\varphi}=-{1\over r^2}{r^2\over\ell}\dot{r}=-{\dot{r}\over\ell}$ and $u''\equiv {d^2u\over d\varphi^2}=-{r^2\over\ell}{\ddot{r}\over\ell}=-{r^2\ddot{r}\over\ell^2}$.

$$\ddot{r} -{\ell^2\over r^3}+{3r_s\ell^2\over 2r^4} = 0$$

$$-{\ell^2\over r^2}u'' -{\ell^2\over r^2}u+{3r_s\ell^2\over 2r^2}u^2 = 0$$

$$u'' +u-{3r_s\over 2}u^2 = 0$$