[Click here for a PDF of this post with nicer formatting]

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.