L  &  =T-U\label{p}\\
&  =\frac{1}{2}m\overset{.}{x}^{2}+qxE_{0}. \label{p1}$$ The Lagrangian equation of motion is$$\frac{d}{dt}\frac{\partial L}{\partial\overset{.}{x}}-\frac{\partial
L}{\partial x}=m\overset{..}{x}-qE_{0}=0, \label{q}$$ which is just the elementary Newtonian equation$$m\overset{..}{x}=qE_{0}. \label{r}$$ The simplest initial conditions for this problem – initial position and velocity set to zero – lead to the solution$$x=\frac{qE_{0}}{2m}t^{2}. \label{s}$$ From Eqs. (\[p\]) and (\[p1\]), it follows that$$T=\frac{1}{2m}\left(  qE_{0}t\right)  ^{2},\qquad U=-\frac{1}{2m}\left(
qE_{0}t\right)  ^{2},\qquad T+U=0. \label{t}$$ The anticipated conservation of energy holds true.

Despite the uniqueness of the potentials of Eq. (\[o\]), there exists an apparently proper gauge transformation generated by the function$$\Lambda=ctxE_{0}. \label{u}$$ The gauge-transformed potentials are$$\widetilde{\phi}=0,\qquad\widetilde{A}=-ctE_{0}, \label{v}$$ and the Lagrangian function is [@hrjmo]$$\widetilde{L}=\frac{1}{2}m\overset{.}{x}^{2}-qtE_{0}\overset{.}{x}. \label{w}$$ The kinetic energy is unaltered ($\widetilde{T}=T$), but the new potential energy is$$\widetilde{U}=qtE_{0}\overset{.}{x}, \label{x}$$ which is explicitly time-dependent. The new equation of motion is$$\frac{d}{dt}\left(  \frac{\partial\widetilde{L}}{\partial\overset{.}{x}}\right)  -\frac{\partial\widetilde{L}}{\partial x}=m\overset{..}{x}-qE_{0}=0,
\label{y}$$ which is identical to that found in the original gauge, so that the solution is the same as Eq. (\[s\]). However, the altered gauge has introduced a fundamental change. The gauge-transformed potential energy is evaluated as$$\widetilde{U}=\frac{1}{m}\left(  qE_{0}t\right)  ^{2}, \label{z}$$ so that $$\widetilde{T}+\widetilde{U}=\frac{3}{2m}\left(  qE_{0}t\right)  ^{2}.
\label{aa}$$ The total energy is not conserved, as was presaged by the explicit time dependence of the gauge-transformed Lagrangian (\[w\]).

How did this happen? One constraint placed on gauge 