In [1] it is claimed in an Aharonov-Bohm discussion that a Lagrangian modification to include electromagnetism is

\label{eqn:magneticLorentzForceLagrangian:20}
\LL \rightarrow \LL + \frac{e}{c} \Bv \cdot \BA.

That can’t be the full Lagrangian since there is no $$\phi$$ term, so what exactly do we get?

If you have somehow, like I did, forgot the exact form of the Euler-Lagrange equations (i.e. where do the dots go), then the derivation of those equations can come to your rescue. The starting point is the action

\label{eqn:magneticLorentzForceLagrangian:40}
S = \int \LL(x, \xdot, t) dt,

where the end points of the integral are fixed, and we assume we have no variation at the end points. The variational calculation is

\label{eqn:magneticLorentzForceLagrangian:60}
\begin{aligned}
\delta S
&= \int \delta \LL(x, \xdot, t) dt \\
&= \int \lr{ \PD{x}{\LL} \delta x + \PD{\xdot}{\LL} \delta \xdot } dt \\
&= \int \lr{ \PD{x}{\LL} \delta x + \PD{\xdot}{\LL} \delta \ddt{x} } dt \\
&= \int \lr{ \PD{x}{\LL} – \ddt{}\lr{\PD{\xdot}{\LL}} } \delta x dt
+ \delta x \PD{\xdot}{\LL}.
\end{aligned}

The boundary term is killed after evaluation at the end points where the variation is zero. For the result to hold for all variations $$\delta x$$, we must have

\label{eqn:magneticLorentzForceLagrangian:80}
\boxed{
\PD{x}{\LL} = \ddt{}\lr{\PD{\xdot}{\LL}}.
}

Now lets apply this to the Lagrangian at hand. For the position derivative we have

\label{eqn:magneticLorentzForceLagrangian:100}
\PD{x_i}{\LL}
=
\frac{e}{c} v_j \PD{x_i}{A_j}.

For the canonical momentum term, assuming $$\BA = \BA(\Bx)$$ we have

\label{eqn:magneticLorentzForceLagrangian:120}
\begin{aligned}
\ddt{} \PD{\xdot_i}{\LL}
&=
\ddt{}
\lr{ m \xdot_i
+
\frac{e}{c} A_i
} \\
&=
m \ddot{x}_i
+
\frac{e}{c}
\ddt{A_i} \\
&=
m \ddot{x}_i
+
\frac{e}{c}
\PD{x_j}{A_i} \frac{dx_j}{dt}.
\end{aligned}

Assembling the results, we’ve got

\label{eqn:magneticLorentzForceLagrangian:140}
\begin{aligned}
0
&=
\ddt{} \PD{\xdot_i}{\LL}

\PD{x_i}{\LL} \\
&=
m \ddot{x}_i
+
\frac{e}{c}
\PD{x_j}{A_i} \frac{dx_j}{dt}

\frac{e}{c} v_j \PD{x_i}{A_j},
\end{aligned}

or
\label{eqn:magneticLorentzForceLagrangian:160}
\begin{aligned}
m \ddot{x}_i
&=
\frac{e}{c} v_j \PD{x_i}{A_j}

\frac{e}{c}
\PD{x_j}{A_i} v_j \\
&=
\frac{e}{c} v_j
\lr{
\PD{x_i}{A_j}

\PD{x_j}{A_i}
} \\
&=
\frac{e}{c} v_j B_k \epsilon_{i j k}.
\end{aligned}

In vector form that is

\label{eqn:magneticLorentzForceLagrangian:180}
m \ddot{\Bx}
=
\frac{e}{c} \Bv \cross \BB.

So, we get the magnetic term of the Lorentz force. Also note that this shows the Lagrangian (and the end result), was not in SI units. The $$1/c$$ term would have to be dropped for SI.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.