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

\begin{equation}\label{eqn:magneticLorentzForceLagrangian:20}

\LL \rightarrow \LL + \frac{e}{c} \Bv \cdot \BA.

\end{equation}

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

\begin{equation}\label{eqn:magneticLorentzForceLagrangian:40}

S = \int \LL(x, \xdot, t) dt,

\end{equation}

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

\begin{equation}\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}

\end{equation}

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

\begin{equation}\label{eqn:magneticLorentzForceLagrangian:80}

\boxed{

\PD{x}{\LL} = \ddt{}\lr{\PD{\xdot}{\LL}}.

}

\end{equation}

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

\begin{equation}\label{eqn:magneticLorentzForceLagrangian:100}

\PD{x_i}{\LL}

=

\frac{e}{c} v_j \PD{x_i}{A_j}.

\end{equation}

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

\begin{equation}\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}

\end{equation}

Assembling the results, we’ve got

\begin{equation}\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}

\end{equation}

or

\begin{equation}\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}

\end{equation}

In vector form that is

\begin{equation}\label{eqn:magneticLorentzForceLagrangian:180}

m \ddot{\Bx}

=

\frac{e}{c} \Bv \cross \BB.

\end{equation}

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.