 Solution

If $L$ is independent of $q_1$, the Euler-Lagrange equation makes the generalized momentum $p_1=\partial L/\partial {\dot{q}}_1$ constant. If $L$ is independent of time, the energy function $E={\sum }_a{\dot{q}}_a(\partial L/\partial {\dot{q}}_a)-L$ is constant; differentiating it gives $\dot{E}=-L_t$ along solutions.

Set $A(r)=r^2/(r-2m{)}^2$ and $B(r)=r^3/(r-2m)$. Their derivatives are $A^{\prime }=-4mr/(r-2m{)}^3$ and $B^{\prime }=2r^2(r-3m)/(r-2m{)}^2$. The radial and angular equations areEquivalently,Two conserved quantities areThe second equals $L$ here because the Lagrangian is homogeneous quadratic in velocities. For a constant-radius orbit with nonzero angular motion, the radial equation requires $r-3m=0$. ThusIt lies outside $2m$ and is the circular photon orbit. If both velocities were zero, the differential equations would also admit static constant configurations at any radius; these are degenerate zero-energy paths rather than the intended moving photon trajectory.