Energy of circle orbit in Schwarzschild metric

Question

For Schwarzschild metric we have invariants: \begin{equation} \begin{aligned} & r^2 \frac{d \varphi}{d \tau}=\frac{L}{m} \\ & \left(1-\frac{r_s}{r}\right) \frac{d t}{d \tau}=\frac{E}{m c^2} . \end{aligned} \end{equation} Which, after being put into metric give us orbit equation: \begin{equation} \left(\frac{d r}{d \varphi}\right)^2=\frac{r^4}{b^2}-\left(1-\frac{r_s}{r}\right)\left(\frac{r^4}{a^2}+r^2\right), \end{equation} If I set left part as zero for circle orbit, I still have no particular solution for E and L, the same orbit allows different energies and angular momentums. Such indetermenism confuses me, what am I doin wrong?

Answer 1

Requiring that $\frac{dr}{d\varphi}=0$ at a particular value of $r$ is not sufficient to guarantee that the orbit is circular. It only tells you that the orbit has a turning point at $r$. For each $r>r_s$ there is a one parameter family of orbits that have a turning point at that $r$.

To ensure that orbit is circular you need to also require that $\frac{d^2r}{d\varphi^2}=0$.