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 \bgroup\color{black}$ \sigma$\egroup along the geodesic. The Euler-Lagrange equation gives essentially the same result since the Lagrangian is independent of both \bgroup\color{black}$ t$\egroup and \bgroup\color{black}$ \varphi$\egroup, there are conserved quantities related to the energy ( \bgroup\color{black}$ \equiv\epsilon$\egroup) and angular momentum ( \bgroup\color{black}$ \equiv\ell$\egroup). The photon has a null geodesic \bgroup\color{black}$ ds=0$\egroup and we may use that equation directly.

$\displaystyle {dt\over d\sigma}$ $\displaystyle =\dot{t}={\epsilon\over\left(1-{r_s\over r}\right)}$    
$\displaystyle \dot{\varphi}$ $\displaystyle ={\ell\over r^2}$    
$\displaystyle ds^2$ $\displaystyle =0=\left(1 - \frac{r_s}{r} \right) c^2 \dot{t}^2 - \frac{\dot{r}^2}{1-\frac{r_s}{r}} - r^2 \dot{\varphi}^2$    
0 $\displaystyle =\left(1 - \frac{r_s}{r} \right) c^2 {\epsilon^2\over\left(1-{r_s\over r}\right)^2} - \frac{\dot{r}^2}{1-\frac{r_s}{r}} - {\ell^2\over r^2}$    
0 $\displaystyle = c^2 {\epsilon^2\over\left(1-{r_s\over r}\right)} - \frac{\dot{r}^2}{1-\frac{r_s}{r}} - {\ell^2\over r^2}$    
$\displaystyle \dot{r}^2$ $\displaystyle = c^2 \epsilon^2 - {\ell^2\over r^2}\left(1-{r_s\over r}\right)$    

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

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

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

\bgroup\color{black}$\displaystyle \dot{r}={dr\over d\sigma}={dr\over d\varphi}\dot{\varphi}={\ell\over r^2}{dr\over d\varphi} $\egroup

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

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

\bgroup\color{black}$ \displaystyle u'' +u-{3r_s\over 2}u^2 = 0 $\egroup



Subsections
Jim Branson 2012-10-21