= Solution
Take a variation $y_\varepsilon=y+\varepsilon\eta$ with $\eta(a)=\eta(b)=0$. Differentiating the <functional> at zero and using <integration by parts> gives
$$
\delta F=\int_b^a(f_y\eta+f_{y'}\eta')dx=\left[f_{y'}\eta\right]_b^a+\int_b^a\left(f_y-\frac d{dx}f_{y'}\right)\eta\,dx.
$$
The boundary term vanishes. Since the remaining <integral> must vanish for every smooth endpoint-zero variation, the <fundamental lemma of the calculus of variations> gives the <Euler-Lagrange equation> $d(f_{y'})/dx=f_y$.
For the mechanical problem put $A(r)=1-2m/r>0$. The <Lagrangian> is independent of $\theta$, a <cyclic coordinate>, so its <conjugate momentum> $\partial L/\partial\dot\theta=2r^2\dot\theta=2J$ is conserved. Therefore $J$ is constant, including the factor of two between $J$ and the canonical momentum for this normalization of $L$.
For the radial coordinate, $\partial L/\partial\dot r=2\dot r/A$ and $\partial L/\partial r=-\dot r^2A'/A^2+2r\dot\theta^2-\mu A'/A^2$. Its <Euler-Lagrange equation> is
$$
\frac{2\ddot r}{A}-\frac{A'\dot r^2}{A^2}-2r\dot\theta^2+\frac{\mu A'}{A^2}=0.
$$
For a <circular orbit> $r=R>2m$, the radial <velocity> and <acceleration> vanish. Using $A'(R)=2m/R^2$ and $\dot\theta=J/R^2$ leaves $2J^2/R^3=2m\mu/[R^2A(R)^2]$, hence
$$
\boxed{J^2=\frac{m\mu R}{(1-2m/R)^2}.}
$$
This is the <circular orbit condition for a radial kinetic Lagrangian> applied to the coefficients here.
Back to article page
