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%TCIDATA{Created=Wed Nov 13 08:45:07 1996}
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\markright{{\sf Lagrangian of a charged particle      }}
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\begin{document}


\section{12.1\quad Lagrangian of a Charged Particle.}

\subsection{12.1.1}

\quad \thinspace \textbf{a. }Although the Einstein-Lorentz equation 
\[
\frac{dp^\alpha }{d\tau }=eF^{\alpha \beta }\frac{u_\beta }c\quad \quad 
\text{or\quad \quad }\frac{d\mathbf{p}}{dt}=e\left( \mathbf{E+}\frac{\mathbf{%
v}}c\times \mathbf{B}\right) 
\]
is sufficient for the discussion of the dynamics of charged particles, it is
instructive to give a Lagrangian formulation of the problem and the related
action principle. The Hamiltonian formulation is also useful for the
connection to quantum mechanics, but in fact relativistic hamiltonians are
rather unnatural and the Lagrangian formulation is preferred.

\smallskip\ 

\textbf{b}. We recall that the nonrelativistic Lagrangian of a charged
particle is 
\[
L=\frac 12mv^2+\frac ec\mathbf{v\cdot A-\,}e\phi , 
\]
that the canonical momentum is 
\[
\mathbf{P}=\frac{\partial L}{\partial \mathbf{v}}=m\mathbf{v+}\frac ec%
\mathbf{A} 
\]
and that the Euler-Lagrange equations 
\[
\frac{d\mathbf{P}}{dt}-\frac{\partial L}{\partial \mathbf{x}}=0 
\]
are equivalent to the Newton-Lorentz equation.

\smallskip\ 

\textbf{c. }Further, the Euler-Lagrange equations can be derived from the
action principle $\delta \mathcal{A}=0,$ where $\mathcal{A}%
=\int_{t_{1}}^{t_{2}}Ldt.$is the action. The variation is over all paths $%
\mathbf{x(}t)$ between assigned end points $\mathbf{x(}t_{1})=\mathbf{x}_{1},
$ $\mathbf{x(}t_{2})=\mathbf{x}_{2}.$

\smallskip\ 

\textbf{d. }The easiest way to generalize this to relativistic velocities is
to require that $\mathcal{A}$ be a Lorentz-invariant, so that $\delta 
\mathcal{A}=0$ is an invariant statement. Hence $Ldt$ must be invariant and
since $dt=\gamma d\tau $ $(d\tau =$ proper time), $\gamma L$ must be
invariant.

\smallskip\ 

\textbf{f. }For the free-particle part of the Lagrangian, the choice 
\begin{equation}
\gamma L_{free}=-mc^{2}  \label{1}
\end{equation}
gives $L_{free}=-mc^{2}\sqrt{1-v^{2}/c^{2}}\,.$ For small $v/c,$ $%
L_{free}\rightarrow -mc^{2}+(1/2)mv^{2}$; thus, apart from an additive
constant, (\ref{1}) reduces to the correct Newtonian limit.

\newpage

\subsection{12.1.2}

\quad \quad \textbf{a. }For the interacting part, $(e/c)\mathbf{v\cdot A-\,}%
e\phi $ is already a proper relativistic $L_{int},$ because then $\gamma
L_{int}=-(e/c)u^\alpha A_\alpha ,$ with $u^\alpha =dx^\alpha /d\tau =(\gamma
c,\gamma \mathbf{v),}$ is manifestly invariant.

\smallskip\ 

\textbf{b.} The total relativistic Lagrangian is then 
\begin{equation}
L=L_{free}+L_{int}=-\left( mc^2\sqrt{1-\frac{v^2}{c^2}}+e\phi -e\frac{%
\mathbf{v}}c\cdot \mathbf{A}\right)  \label{2}
\end{equation}

\smallskip\ 

\textbf{c. }It can be checked that the Euler-Lagrange equations for this $L$
give indeed the Einstein-Lorentz equation in the form $d\mathbf{p}%
/dt=e\left( \mathbf{E+(v}/c)\times \mathbf{B}\right) $ However, this
formulation is not manifestly covariant, since the path is given by $\mathbf{%
x}(t),$ i.e. $t$ acts as a parameter and $\mathbf{x}$ is a dependent
variable, as in Newtonian mechanics. The gist of relativity is that $\mathbf{%
x}$ and $t$ should be treated on the same footing.

\smallskip\ 

\textbf{d. }To achieve a manifestly covariant formulation we write, using $%
dt=\gamma d\tau $%
\begin{equation}
\mathcal{A}=\int Ldt=-\int \left( mc^{2}+\frac{e}{c}\gamma \phi -\frac{e}{c}%
\gamma \frac{\mathbf{v}}{c}\cdot \mathbf{A}\right) d\tau \,.  \label{3}
\end{equation}
Now the integral is carried out using any convenient parameter $\sigma $ to
specify the path: $\mathbf{x(}\sigma ),$ $t(\sigma ),$by writing $d\tau =$
or 
\[
d\tau =c\,\frac{d\tau }{d\sigma }\,d\sigma =c\sqrt{g_{\alpha \beta }\frac{%
dx^{\alpha }}{d\sigma }\frac{dx^{\beta }}{d\sigma }}\,d\sigma 
\]
\quad \quad and $\gamma L_{int}d\tau =-\,(e/c)\,dx^{\alpha }A_{\alpha }$
(see \textbf{a}).Hence 
\begin{equation}
\mathcal{A}=-\int_{\sigma _{1}}^{\sigma _{2}}\left[ mc\sqrt{g_{\alpha \beta }%
\frac{dx^{\alpha }}{d\sigma }\frac{dx^{\beta }}{d\sigma }}+\frac{e}{c}\frac{%
dx^{\alpha }}{d\sigma }A_{\alpha }\right] d\sigma =\int \frak{L}d\sigma 
\label{4}
\end{equation}

\smallskip\ 

\textbf{e. }Application of the Euler-Lagrange equations yields
\[
\frac{d}{d\sigma }\frac{\partial \frak{L}}{\partial \left( \dfrac{dx^{\alpha
}}{d\sigma }\right) }-\frac{\partial \frak{L}}{\partial x^{\alpha }}=0.
\]

\smallskip\ 

\textbf{f. }It is particularly instructive to note that for a free particle
the action principle $\delta \mathcal{A}=0$ is equivalent to the
``geodesic'' principle $\delta \int ds=0,.$where $ds=cd\tau .$

\newpage

\subsection{12.1.3}

\quad \quad \textbf{a. }A Hamiltonian formalism can be introduced by the
formulae 
\[
P_{\alpha }=-\,\frac{\partial \frak{L}}{\partial \left( \dfrac{dx^{\alpha }}{%
d\sigma }\right) }\quad \quad \quad \frak{H}=\frac{1}{2}\left( P_{\alpha }%
\dfrac{dx^{\alpha }}{d\sigma }+L\right) \,.
\]

\smallskip\ 

\textbf{b. }It is convenient to specialize $d\sigma $ to be $ds=cd\tau .$
Then 
\[
P_\alpha =mu_\alpha +\frac ecA_\alpha \quad \quad \text{and\quad \quad }%
\frak{H}=\frac 1{2m}\left( P_\alpha -\frac ecA_\alpha \right) \left(
P^\alpha -\frac ecA^\alpha \right) -\frac 12mc^2\,. 
\]

\smallskip\ 

\textbf{c. }The only problem with this ``Hamiltonian'' is that it is
identically zero. However, considered formally as a function of $\mathsf{P}$
and $\mathsf{x}$ (which is hidden in $\mathsf{A}$) it serves admirably.

\smallskip\ 

\textbf{d. }The main purpose of going through the derivation of $\frak{H}$
is to recall that the Klein-Gordon equation for a spinless particle in
quantum mechanics is simply $\frak{H}\psi =0.$ Of course $\mathbf{P}$
becomes $-i\hbar \mathbf{\nabla ,}$ $P_{0}$ becomes $i\hbar \partial
/c\partial t.$ Explicitly, for $A^{\alpha }=0,$ the Klein-Gordon equation is 
\[
\frac{\hbar ^{2}}{c^{2}}\frac{\partial ^{2}\psi }{\partial t^{2}}-\hbar
^{2}\nabla ^{2}\psi +m^{2}c^{2}\psi =0.
\]

\smallskip\ 

\textbf{e. }The Hamiltonian formulation is not unique. Another approach,
which stays close to the Newtonian model, is to define 
\[
\mathbf{P}=\frac{\partial L}{\partial \mathbf{v}}=m\gamma \mathbf{v-}\frac ec%
\mathbf{A\quad \quad \quad }H=\mathbf{P\cdot v-}L. 
\]

Then $H$ turns out to be the total energy $H=m\gamma c^{2}+e\phi $ or $H=%
\sqrt{\left( c\mathbf{P-}e\mathbf{A}\right) ^{2}+m^{2}c^{4}}+e\phi .$
Squaring this we have $(H-e\phi )^{2}-\left( c\mathbf{P-}e\mathbf{A}\right)
^{2}=m^{2}c^{2},$ which shows that $H/c$ should be identified with the
zeroth component of $P_{\alpha }$ as defined in \textbf{b. }This formulation
is not manifestly covariant, but one cannot have everything.

\end{document}
