 Solution

The Euler-Lagrange equations areUsing time independence of the vector potential gives $m{\ddot{r}}_i=(e/c){\dot{r}}_j({\partial }_i{A}_j-{\partial }_j{A}_i)$, which is the component form ofTaking its scalar product with the velocity shows $\dot{T}=m\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}=0$: the kinetic energy is constant, since the magnetic force does no work.

For $\mathbf{B}=(0,0,F^{\prime }(x))$, the equations give $m\ddot{y}=-(e/c)F^{\prime }(x)\dot{x}$ and $\ddot{z}=0$. Thus $mc\dot{y}+eF(x)=-C$ and $\dot{z}=v_z$ are constants of motion. Substitution into $2T/m={\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2$ yieldsThe constant $D$ in this representation is nonpositive. The gauge choice $\mathbf{A}=(0,F(x),0)$ also makes the conserved transverse canonical momenta explicit.