= Solution
The <four-velocity> and particle energy are
$$
u^\mu=\gamma(c,\mathbf v),
\qquad
\mathcal E=\gamma mc^2,
\qquad
p^0=mu^0=\frac{\mathcal E}{c}.
$$
With the conventions for the given metric, $F^{0i}=E_i/c$. The temporal component of the covariant <Lorentz force> is therefore
$$
\frac{d(\mathcal E/c)}{d\tau}
=qF^{0i}u_i
=\frac q c\gamma\mathbf E\mathbin{\cdot}\mathbf v.
$$
Since $dt/d\tau=\gamma$, cancellation of $\gamma/c$ gives the <relativistic electromagnetic work-energy theorem>
$$
\boxed{
\frac{d\mathcal E}{dt}=q\mathbf E\mathbin{\cdot}\mathbf v
}.
$$
The absence of $\mathbf B$ reflects that magnetic force is perpendicular to the velocity and does no work.
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