Schwarzschild's null-geodesic new form or an error? My question is whether or not this form (radial acceleration of a photon)
$$\ddot{r}=\frac{L^2}{r^4}(r-3M)$$
is correct ?
Recall the standard set of second-order ODE for the Schwarzschild metric (for a massless particle)
$$\ddot{t}=-\frac{A'}{A}\dot{t}\dot{r}$$
$$\ddot{r}=-\frac{1}{2}AA'\dot{t}^2+\frac{1}{2}\frac{A'}{A}\dot{r}^2+Ar(\dot{\theta}^2+\sin^2{\theta}\dot{\phi}^2)$$
$$\ddot{\theta}=-\frac{2}{r}\dot{r}\dot{\theta}+\sin{\theta}\cos{\theta}\dot{\phi}^2$$
$$\ddot{\phi}=-\frac{2}{r}\dot{r}\dot{\phi}-2\frac{\cos{\theta}}{\sin{\theta}}\dot{\theta}\dot{\phi}$$
$$A=1-\frac{2M}{r}$$
$$A'=\frac{2M}{r^2}$$
Additionally, we have
$$E^2-\dot{r}^2=\frac{L^2}{r^2}A$$
$$E=A\dot{t}$$
$$L^2=r^4(\dot{\theta}^2+\sin^2{\theta}\dot{\phi}^2)$$
All of this is well know, see for example https://arxiv.org/abs/1511.06025.
$$\ddot{r}=-\frac{1}{2}AA'\dot{t}^2+\frac{1}{2}\frac{A'} {A}\dot{r}^2+Ar(\dot{\theta}^2+\sin^2{\theta}\dot{\phi}^2)$$
My focus is on $\ddot{r}$, let's rewrite it this way (taking advantage of $\dot{r}^2$ and the constants of motion $E$, $L^2$)
$$\ddot{r}=-\frac{1}{2}AA'\dot{t}^2+\frac{1}{2}\frac{A'}{A}\dot{r}^2+A\frac{L^2}{r^3}$$
$$\ddot{r}=-\frac{1}{2}\frac{A'}{A}E^2+\frac{1}{2}\frac{A'}{A}\dot{r}^2+A\frac{L^2}{r^3}$$
$$\ddot{r}=-\frac{1}{2}\frac{A'}{A}(E^2-\dot{r}^2)+A\frac{L^2}{r^3}$$
$$\ddot{r}=-\frac{1}{2}\frac{A'}{A}\frac{L^2}{r^2}A+A\frac{L^2}{r^3}$$
$$\ddot{r}=-\frac{1}{2}\frac{2M}{r^2}\frac{L^2}{r^2}+A\frac{L^2}{r^3}$$
$$\ddot{r}=-\frac{L^2}{r^4}M+A\frac{L^2}{r^3}$$
$$\ddot{r}=\frac{L^2}{r^3}\left(-\frac{M}{r} + A\right)$$
$$\ddot{r}=\frac{L^2}{r^3}\left(1-\frac{3M}{r}\right)$$
$$\ddot{r}=\frac{L^2}{r^4}(r-3M)$$
So finally, I get this form, which I've never seen in the literature, leading me to suspect it's incorrect and I made an error somewhere. If so what did I missed ?


**Not really important comments but this new form for $\ddot{r}$, "linearizes" the equation $dr/d\phi$.
Also the appearance of $3M$ is interesting because it's the location of the circular orbit for a trapped particle, and remplacing $r$ by $u=1/r$ get's us to $d^2u/d\phi^2=3Mu^2-u$ which is well known.
Edit: I did find one "paper" with this result https://www.astro.umd.edu/~miller/teaching/astr498/lecture10.pdf (eq. 2), the additionnal $\frac{M}{r^2}$ is because the author used a particle with a mass.
 A: That's a correct equation, which follows from the following 'conservation of mechanical energy' type of equation
\begin{align}
\frac{1}{2}\dot{r}^2+\underbrace{\frac{1}{2}\left(1-\frac{2m}{r}\right)\left(\frac{L^2}{r^2}+\kappa\right)}_{:=V_{\kappa}(r)}&=\frac{E^2}{2}.
\end{align}
Here, we set $\kappa=0$ for null geodesics and $\kappa=+1$ for timelike geodesics (your edit).
Differentiate this with respect to the affine parameter to get the equation you wrote. And this equation itself is obtained by using that:

*

*you're considering normalized geodesics $\gamma$, $g(\dot{\gamma},\dot{\gamma})=-\kappa$.

*$E=g(\dot{\gamma},\partial_t)$ is constant along $\gamma$ (since $\gamma$ is geodesic and $\partial_t$ is Killing... or as is immediately seen from the Euler-Lagrange equations)

*$L=g(\dot{\gamma},\partial_{\phi})$ is constant along $\gamma$ (similar to above)

*Without loss of generality we may consider motion in the $\theta\equiv\frac{\pi}{2}$ 'equatorial plane'.


Edit:

*

*This is also explained in Wald's GR text, in Section 6.3 (equation 6.3.14 to be specific).

*Also, see Tobias Osborne's lecture 22,23 on the geodesics of Schwarzschild if you like lectures.

