## Problem: Lorentz force gauge transformation.

Show that the gauge transformation $$A \rightarrow A + \grad \psi$$ applied to the Lorentz force Lagrangian
\begin{equation}\label{eqn:gaugeLorentzSTA:20}
L = \inv{2} m v^2 + q A \cdot v/c,
\end{equation}
does not change the equations of motion.

The gauge transformed Lagrangian is
\begin{equation}\label{eqn:gaugeLorentzSTA:40}
L = \inv{2} m v^2 + q A \cdot v/c + \frac{q v}{c} \cdot \grad \phi.
\end{equation}
We know that the Lorentz force equations are obtained from the first two terms, so need only consider the effects of the new $$\phi$$ dependent term on the action. First observe that
\begin{equation}\label{eqn:gaugeLorentzSTA:60}
=
\frac{dx^\mu}{d\tau} \PD{x^\mu}{\phi}
=
\frac{d \phi}{d\tau}.
\end{equation}
This means that the action is transformed to
\begin{equation}\label{eqn:gaugeLorentzSTA:80}
S
\rightarrow S + \frac{q}{c} \int d\tau \frac{d\phi}{d\tau}
= S + \frac{q}{c} \evalbar{\phi}{\Delta \tau}.
\end{equation}
As the action is evaluated over a fixed interval, the gauge transformation only changes the action by a constant, so the equations of motion are unchanged.