= Solution
Choose <Minkowski metric> $(+,-,-,-)$, so $A_\mu=(\phi/c,-\mathbf A)$ and $d\tau=dt\sqrt{1-v^2/c^2}$. In laboratory time the <relativistic charged-particle action> has <Lagrangian>
$$
L=-mc^2\sqrt{1-v^2/c^2}-e\phi(\mathbf x,t)+e\mathbf A(\mathbf x,t)\cdot\mathbf v.
$$
Here $e$ denotes the signed particle charge in the action. The <canonical momentum> and the position derivative are
$$
\frac{\partial L}{\partial v_i}=m\gamma v_i+eA_i,\qquad
\frac{\partial L}{\partial x_i}=-e\partial_i\phi+e v_j\partial_iA_j,
\qquad \gamma=(1-v^2/c^2)^{-1/2}.
$$
The <Euler-Lagrange equation> therefore gives
$$
\frac d{dt}(m\gamma v_i)=e\left[-\partial_i\phi-\partial_t A_i+v_j(\partial_iA_j-\partial_jA_i)\right].
$$
The last term is $(\mathbf v\times\mathbf B)_i$, and the first two terms form the <electric field>. Thus the <relativistic Lorentz force> equation is
$$
\boxed{\frac d{dt}(m\gamma\mathbf v)=e(\mathbf E+\mathbf v\times\mathbf B).}
$$
Its power relation and acceleration form are
$$
\boxed{\frac d{dt}(\gamma mc^2)=e\mathbf E\cdot\mathbf v,\qquad
\dot{\mathbf v}=\frac e{m\gamma}\left[\mathbf E+\mathbf v\times\mathbf B-\frac{\mathbf v(\mathbf E\cdot\mathbf v)}{c^2}\right].}
$$
The power relation follows by taking the scalar product with $\mathbf v$ and using the <relativistic energy-momentum relation>. It also shows why the <magnetic field> alone cannot change the particle's energy.
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