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Expression for the Lagrangian for a charged particle in electromagnetic field is given and the Euler Lagrange equations are shown to coincide with EOM with Lorentz force on the charged particle.
$\newcommand{\Label}[1]{\label{#1}} \newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}}
\newcommand{\dd}[2][]{\frac{d #1}{d#2}}$
Notation
We follow 123 notation for components of a vector. Thus we write the position vector as \(\vec{x} = (x_1,x_2,x_3)\).
The force on a charged particle in electric field \(\vec{E}\) and magentic field \(\vec{B}\) is given by
\begin{equation}\Label{EQ01} \vec{F} = q (\vec{E} + \vec{v}\times \vec{B}) \end{equation}
where \(q\) is the electric charge of of the charged particle. This expression \eqRef{EQ01} is known as Lorentz force. The equation of motion of the particle under influence of electric and magnetic field is given by
\begin{equation}\Label{EQ02} m \ddot{\vec{r}} = q (\vec{E} + \vec{v}\times \vec{B}) \end{equation}
Vector and Scalar Potentials
We now ask what is the Lagrangian for the charged particle so that the Euler Lagrange equations imply the above equation of motion. The Lagrangian cannot be written in terms of the fields \( \vec{E},\vec{B}\). We need to introduce the vector and scalar potentials \(\vec{A}, \phi\). Recall that the vector and scalar potentials are related to the fields by
\begin{equation}\label{EQ03} \vec{E} = - \nabla \phi - \pp[\vec{A}]{t}, \quad \vec{B} = \nabla \times \vec{A}. \end{equation}
\subsubsection{Lagrangian for a charged particle} In terms of the scalar and vector potentials the Lagrangian for a charged particle is give by
\begin{equation}\label{EQ04} L =\frac{1}{2} m \dot{\vec{x}}^{\,2} + q \dot{\vec{r}}\cdot\vec{A}(\vec{x},t) - q\phi(\vec{x},t) \end{equation}
How do we arrive at this Lagrangian? There will be several approaches.
- Start with d' Alembert's principle and follow the steps used to derive Euler Lagrange equations of motion.
- Note that the generalized potential \(U(\vec{r},t)\) \begin{equation} U(\vec{x},t) = - q\dot{\vec{x}}\cdot\vec{A}(\vec{x},t) + q\phi(\vec{x},t) \end{equation} gives rise to correct expression for force by means of the equation \begin{equation} F_k = \dd{t}\Big(\dd[U]{\dot{x}_k}\Big) - \pp[U]{x_k}. \end{equation}
We leave it as an exercise for reader to verify that Lagrangian in \eqref{EQ04} gives correct equations of motion \eqref{EQ02}.
Gauge Invariance
It is well known that the electric and magnetic fields do not determine the potentials uniquely. Two sets of different potentials \(\vec{A}, \vec{\phi}\) and \(\vec{A}^\prime, \vec{\phi}^\prime\) give rise to the same fields and hence same EOM if they are related by a gauge transformation
\begin{eqnarray}
\vec{A}^\prime(\vec{x},t) &=& \vec{A}(\vec{x},t) + \nabla
\Lambda(\vec{x},t),\\
\phi^\prime(\vec{x},t) &=& \vec{\phi}(\vec{x},t) - \pp[\Lambda(\vec{x},t)]{t}.
\end{eqnarray}
If we write Lagrangians \(L\) and \(L^\prime\) using the two sets of potentials related as above, the two Lgarangians should give the same Euler Lagrange equations of motion. We will now prove this by showing that \(L\) and \(L^\prime\) differ by a total time derivative.
Let us consider \(L^\prime\)
\begin{eqnarray}
L &=& \frac{1}{2} m \dot{\vec{x}}^{\,2} +
q \dot{\vec{r}}\cdot\vec{A}^\prime(\vec{x},t) -
q\phi^\prime(\vec{x},t)\\
&=& \frac{1}{2} m \dot{\vec{x}}^{\,2} +
q \dot{\vec{r}}\cdot\Big(\vec{A}(\vec{x},t) +\nabla \Lambda \Big) -
q\Big(\phi(\vec{x},t) - \pp[\Lambda]{t}\Big)\\
&=& L + \sum_k q \,\dot{x}_k\cdot \pp{x_k}\Lambda(\vec{x},t) + q
\pp[\Lambda(\vec{x},t)]{t} \\
&=& L + \dd[\Lambda(x_k,t)]{t}
\end{eqnarray}
This completes the proof that the Lagrangians \(L^\prime\) and \(L\) differ by a total time derivative. Therefore they will give rise to the same equations of motion.
