Since $eφ$ is a potential energy, with $H = ½mv^2 + eφ$ being the total energy, then this suggests treating $e$ as a potential momentum (especially since Maxwell, who invoked the idea of $$, called it the Electromagnetic Momentum) and $ = m + e$ the total momentum. So, if the total energy $H$ is also the Hamiltonian, then the Lagrangian is
$$L = · - H = mv^2 + e· - ½mv^2 - eφ = ½mv^2 + e(· - φ).$$
This has the form $L = T - U$, where $T() = ½mv^2$ is velocity dependent, as usual, but where
$$U(,t,) = e(φ(,t) - (,t)·)$$
is both position and velocity dependent. Correspondingly, the equations of motion
$$
\frac{d\left(½mv^2\right)}{dt} = e(·) = e·\left(-∇φ - \frac{∂}{∂t}\right), \\
\frac{d(m)}{dt} = e( + ×) = e\left(-∇φ - \frac{∂}{∂t} + ×(∇×)\right)
$$
can be rewritten, with the aid of vector algebra
$$×(∇×) = ∇(·) - ·∇$$
and the chain rule
$$\frac{dφ}{dt} = ·∇φ + \frac{∂φ}{∂t}, \hspace 1em \frac{d}{dt} = ·∇ + \frac{∂}{∂t},$$
as:
$$\frac{dH}{dt} = \frac{∂}{∂t}U(,t,), \hspace 1em \frac{d}{dt} = -∇U(,t,).$$
These two equations continue to hold in Special Relativity, but with
$$
T() = m f(), \hspace 1em U(,t,) = e(φ(,t) - (,t)·), \\
L = T - U = m f() + e(· - φ), \\
= M + e, \hspace 1em H = M f() + eφ.
$$
where
$$
f() = \frac{v^2}{1 + \sqrt{1 - (v/c)^2}}, \hspace 1em M = \frac{m}{\sqrt{1 - (v/c)^2}}.
$$
You may verify that
$$\frac{∂f()}{∂} = \frac{}{\sqrt{1 - (v/c)^2}}, \hspace 1em ·\frac{∂f()}{∂} - f() = \frac{f()}{\sqrt{1 - (v/c)^2}},$$
so that the relations
$$ = \frac{∂L}{∂}, \hspace 1em H = · - L,$$
therefore still hold.
Note, also, that
$$f() - c^2 = -\sqrt{1 - (v/c)^2}, \hspace 1em \frac{f()}{\sqrt{1 - (v/c)^2}} + c^2 = \frac{c^2}{\sqrt{1 - (v/c)^2}}
$$
so that with the choice of $T() = -mc^2$, instead of the more natural $T() = 0$, you could instead write:
$$
T() = -m c^2\sqrt{1 - (v/c)^2}, \hspace 1em U(,t,) = e(φ(,t) - (,t)·), \\
L = T - U = -m c^2\sqrt{1 - (v/c)^2} + e(· - φ), \\
= M + e, \hspace 1em H = Mc^2 + eφ.
$$
