= Solution
{c}
On the equatorial path, the <Lagrangian> per unit <mass> is
$$
L=\frac12\left(-F\dot t^2+F^{-1}\dot r^2+r^2\dot\phi^2\right)-\frac{eQ}{mr}\dot t.
$$
Both $t$ and $\phi$ are <cyclic coordinates>. The <Euler-Lagrange equations> therefore conserve their <canonical momenta>:
$$
p_t=\frac{\partial L}{\partial\dot t}=-F\dot t-\frac{eQ}{mr},\qquad
p_\phi=\frac{\partial L}{\partial\dot\phi}=r^2\dot\phi.
$$
Consequently
$$
\boxed{E=-mp_t=mF\dot t+\frac{eQ}{r},\qquad l=p_\phi=r^2\dot\phi,\qquad \dot E=\dot l=0.}
$$
Here $E$ is the total conserved <energy> in the specified gauge of the <electromagnetic four-potential>. The paper's $l$ is the <angular momentum> per unit <mass>; the physical <angular momentum> is $ml$. The equatorial choice is consistent with the $\theta$ <Euler-Lagrange equation>, since its angular force term contains $\sin\theta\cos\theta=0$ at $\theta=\pi/2$ and the <electromagnetic four-potential> has no angular component.
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