
\documentclass[12pt]{article}
\pagestyle{empty}
\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}
\setlength{\textwidth}{6.5in}
\setlength{\parindent}{0in}
\setlength{\parskip}{.2in}
\setlength{\topmargin}{0in}
\setlength{\headsep}{0in}
\setlength{\headheight}{0in}
\setlength{\textheight}{9in}

\begin{document}
\begin{flushright}
\today
\end{flushright}
\begin{center}
\bf{PHY 711 -- Lecture notes on Lagrangian for Electric and
Magnetic Fields}
\end{center}

For simplicity, consider a  Lagrangian for a single particle
having the  form (in Cartesian coordinates)
$L(x,y,z,\dot{x},\dot{y},\dot{z},t) \equiv T - U$. The
Euler-Lagrange equations have the form:
\begin{equation}\label{el}
  \frac{d}{dt} \left( \frac{\partial{L}}{\partial{\dot{x}}}
  \right) -  \frac{\partial{L}}{\partial{{x}}} = 0,
\end{equation}
with similar equations for $y$ and $z$.  We can show that this
form is consistent with Newton's Laws if the potential function
$U$ takes the form:
\begin{equation}\label{U}
  U = U^0(x,y,z,t) + U^{EM}(x,y,z,\dot{x},\dot{y},\dot{z},t),
\end{equation}
where $U^{EM}$ represents the interaction of our particle (having
charge $q$) with an electric field $\mathbf{E}$ and magnetic field
$\mathbf{B}$ where we can represent the fields in terms of the
scalar and vector potentials:
\begin{equation}\label{field}
  \mathbf{E} = - \nabla \phi -
  \frac{1}{c}\frac{\partial{\mathbf{A}}}{\partial{t}} \;\;\;\;
  {\mbox{\rm{and}}} \;\;\;\;\mathbf{B} = \nabla \times \mathbf{A}.
\end{equation}
We must find $U^{EM}$ which is both consistent with the
Euler-Lagrange Eq.(\ref{el}) and with the Lorentz force (written
in the $x$ direction):
\begin{equation}\label{lor}
F_x = q (E_x + \frac{1}{c} \mathbf{(\dot{r}}\times
\mathbf{B})\rfloor_x ) = -\frac{\partial{U^{EM}}}{\partial{x}}
+\frac{d}{dt}\left(\frac{\partial{U^{EM}}}{\partial{\dot{x}}}
\right).
\end{equation}
We note that the magnetic field terms can be evaluated:
\begin{equation}\label{curl}
\mathbf{\dot{r}}\times (\mathbf{\nabla \times A})\rfloor_x =
\dot{y} \left(\frac{\partial{A_y}}{\partial{x}} -
\frac{\partial{A_x}}{\partial{y}} \right) - \dot{z}
\left(\frac{\partial{A_x}}{\partial{z}} -
\frac{\partial{A_z}}{\partial{x}} \right).
\end{equation}
The right hand side of Eq.(\ref{curl}) (with the addition and
subtraction of a convenient term) can be written:
\begin{equation}\label{expand}
 \underbrace{\dot{x}\frac{\partial{A_x}}{\partial{x}} +
 \dot{y}\frac{\partial{A_y}}{\partial{x}} +
 \dot{z}\frac{\partial{A_z}}{\partial{x}}}_{{\partial{({\mathbf{\dot{r}\cdot A}})}}/{\partial{x}}} -
 \underbrace{\dot{x}\frac{\partial{A_x}}{\partial{x}} -
 \dot{y}\frac{\partial{A_x}}{\partial{y}} -
 \dot{z}\frac{\partial{A_x}}{\partial{z}}}_{-{d A_x}/{dt} +
 {\partial{A_x}}/{\partial{t}}},
\end{equation}
where we are assuming that $A_x = A_x(x,y,z,t)$. Noting that $A_x
= \partial{({\mathbf{\dot{r}\cdot A}})}/\partial{\dot{x}}$, the
electromagnetic force can thus be written:
\begin{equation}\label{fem}
F_x = \underbrace{-q\frac{\partial{\phi}}{\partial{x}} -
\frac{q}{c} \frac{\partial{A_x}}{\partial{t}}}_{q E_x} +
\frac{q}{c} \left(
\underbrace{\frac{\partial{({\mathbf{\dot{r}\cdot
A}})}}{\partial{x}} -
\frac{d}{dt}\frac{\partial{({\mathbf{\dot{r}\cdot
A}})}}{\partial{\dot{x}}} +
\frac{\partial{A_x}}{\partial{t}}}_{\mathbf{(\dot{r}}\times
\mathbf{B})\rfloor_x} \right).
\end{equation}
Simplifying this equation, we obtain
\begin{equation}\label{simp}
F_x = -\frac{\partial}{\partial{x}} \left( q \phi -
\frac{q}{c}{\mathbf{\dot{r}\cdot A}} \right)  - \frac{d}{dt}
\frac{\partial}{\partial{\dot{x}}} \left( \frac{q}{c}
{\mathbf{\dot{r}\cdot A}} \right).
\end{equation}
Thus, we finally have the result
\begin{equation}\label{final}
  U^{EM} = q \phi - \frac{q}{c} {\mathbf{\dot{r}\cdot
A}}.
\end{equation}








\end{document}
