\documentstyle[12pt,psfig]{article}
\message{
 	Copyright, 1993, all rights reserved, Charlie Ebner (Dept.of Physics, 
 The Ohio State University, Columbus OH 43210) and Mark Jarrell (Dept.of 
 Physics, The University of Cincinnati, Cincinnati, OH 45221-0011).  This 
 material may not be reproduced for profit, modified or published in any 
 form (this includes electronic redistribution) without the prior written 
 permission of the authors listed above.	
}
\title{Particle and Field Dynamics}
\author{Comte Joseph Louis Lagrange\\(1736 - 1813)}
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\tableofcontents

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In this chapter we shall study the dynamics of particles and fields. 
For a particle, the relativistically correct equation of motion is 
\beq
\der\p t=\F
\eeq
where $\p=m\ga\u$; the corresponding equation for the time rate of change of 
the particle's energy is
\beq
\der Et=\F\cdot\u.
\eeq
The dynamics of the electromagnetic field is given by the Maxwell equations,
\beqa
\div\Ext=4\pi\rhxt\hsph\hsph\div\Bxt=0\nonumber\\
\curl\Ext+\frac1c\pde\Bxt t=0\hsph\curl\Bxt-\frac1c\pde\Ext t=\frac{4\pi}c\Jxt.
\eeqa
These are tied together by the Lorentz force which gives $\F$ in terms of 
the electromagnetic fields
\beq
\F=q\leb\E+\frac1c(\u\times\B)\rib
\eeq
and by the expressions for $\rhxt$ and $\Jxt$ in terms of the particles' 
coordinates and velocities
\beq\barr{c}
\rhxt=\sum_iq_i\de(\x-\x_i(t))\\
\Jxt=\sum_iq_i\u_i(t)\de(\x-\x_i(t)).
\earr\eeq

In view of the fact that we already know all of this, what further do we 
want to do? Two things: (1) Formulate appropriate covariant Lagrangians and 
Hamiltonians from which covariant dynamical equations can be derived; 
and (2) applications.
\section{Lagrangian and Hamiltonian of a Charged Particle in an External Field}
We want to devise a Lagrangian for a charged particle in the presence of 
given applied fields which are treated as parameters and not as dynamic 
variables. This Lagrangian is to yield the equations of motion \eqs{1}{2} 
with $\F$ given by the Lorentz force. These equations can be written as
\beq
\der{p^\al}t=\frac q{mc\ga}F^{\al\be}p_\be
\eeq
which is almost in Lorentz covariant form. A more obviously Lorentz 
covariant form can be obtained by using the fact that the infinitesimal 
time element $dt$ can be related to an infinitesimal proper time element $d
\ta$ by $dt=\ga d\ta$. Then we have
\beq
\der{p^\al}\ta=\frac q{mc}F^{\al\be}p_\be.
\eeq

To get a Lagrangian from which these equations follow, we postulate the 
existence of the action $A$ which may be expressed as an integral,
\beq
A=\int_a^bdA,
\eeq 
over possible ``paths'' from $a$ to $b$.  The action is an extremum for 
the actual motion of the system. 

\centerline{\psfig{figure=paths.ps,height=2.0in,width=8.5in}}

\noindent In this case, the system consists of a single 
particle. The paths have the constraint that they start at given $(\x_a,
t_a)$ and end at $(\x_b,t_b)$.

Next comes a delicate point. We could say that the first postulate of 
relativity requires that $A$ be the same in all inertial frames\footnote{
This argument is (elegantly) made in {\em{The Classical Theory of
Fields}}}, which 
elevates the action and its consequences to ``law of nature'' status. It 
seems better to regard the invariance of $A$ as an assumption or postulate 
in its own right and to see where that leads us. 

Rewrite \eq{8} as follows:
\beq
A=\int_a^bdA=\int_{t_a}^{t_b}\der Atdt\equiv\int_{t_a}^{t_b}Ldt.
\eeq
This equation expresses nothing more than the parametrization of the 
integral using the time and the definition of the {\em Lagrangian} $L$ as
the derivative of $A$ with respect to $t$. Let us further parametrize the 
integral using the proper time $\ta$ of the particle,
\beq
A=\int_{\ta_a}^{\ta_b}L\ga d\ta
\eeq
where we use $dt=\ga d\ta$, $\ga=1/\sqrt{1-u^2/c^2}$, $\u$ being the 
particle's velocity as measured in the lab frame or the frame in which the 
time $t$ is measured. {\em The proper time is an invariant, so if we believe 
that $A$ is one also, we have to conclude that $L\ga$ is an invariant.}
This statement of invariance greatly limits the possible forms of $L$.

\subsection{Lagrangian of a Free Particle}

Consider first the case of a free particle. What invariants may we 
construct from the properties of a free particle? We have only the 
four-vectors $\pb$ and $\xb$. The presumed translational invariance of 
space rules out the use of the latter. That leaves only the 
four-momentum and the single invariant $p^\al p_\al=m^2c^2$ which is a 
constant. Hence we are led to $L\ga=C$ where $C$ is a constant. Hence, $L=C
/\ga$ and
\beq
A=C\int_{\ta_a}^{\ta_b}d\ta=C\int_{t_a}^{t_b}dt\,\sqrt{1-u^2/c^2}.
\eeq
We may find the constant $C$ by appealing to the nonrelativistic limit and 
expanding in powers of $u^2/c^2$. 
\beq
A \approx C\int_{t_a}^{t_b}dt\,\lep 1-\frac{u^2}{2c^2}
+\cdots \rip.
\eeq
The term proportional to $u^2$ should be 
the usual nonrelativistic Lagrangian of a free particle, $mu^2/2$. This 
condition leads to 
\beq
C=-mc^2
\eeq
and so
\beq
L_f=-mc^2\sqrt{1-u^2/c^2}
\eeq
is the free-particle Lagrangian.

\subsubsection{Equations of Motion}

The equations of motion are found by requiring that $A$ be an extremum,
\beq
\de A=\de\lep\int_{t_a}^{t_b}dt\,L\rip=0.
\eeq
The path $\x(t)$ is to be fixed at the end points $t_a$ and $t_b$,
$\de\x(t_a)=\de\x(t_b)=0$. Writing $L$ as a function of the Cartesian 
components of the position and velocity, we have, allowing for possible 
position-dependence which will appear if the particle is not free, 
$L=L(x_i,u_i,t)$, and
\beq
\de A=\sum_i\int_{t_a}^{t_b}dt\,\leb\lep\pde L{x_i}\rip\de x_i+\lep
\pde L{u_i}\rip\de u_i\rib.
\eeq
But $\de u_i$ is related to $\de x_i$ through $u_i=dx_i/dt$, so
\beqa
\de A=\sum_i\int_{t_a}^{t_b}dt\,\leb\lep\pde L{x_i}\rip\de x_i+\lep\pde 
L{u_i}\rip\de\lep\der{x_i}t\rip\rib\nonumber\\
=\left.\lep\pde L{u_i}\rip\de x_i\ril_{t_a}^{t_b}+\int_{t_a}^{t_b}dt\,
\leb\pde L{x_i}-\frac d{dt}\lep\pde L{u_i}\rip\rib\de x_i(t)
\eeqa
where we have integrated by parts to achieve the last step. The first term 
in the final expression vanishes because $\de x_i=0$ at the endpoints of 
the interval of integration. Arguing that $\de x_i(t)$ is arbitrary 
elsewhere, we conclude that the factor [...] in the final expression must 
vanish everywhere,
\beq
\frac d{dt}\lep\pde L{u_i}\rip-\pde L{x_i}=0
\eeq
for each $i=1,2,3$.

These are the {\em Euler-Lagrange} equations of motion. Let's apply them to 
the free-particle Lagrangian $L_f$,
\beq
\pde{L_f}{x_i}=0\andh\pde{L_f}{u_i}=m\ga u_i,
\eeq
so
\beq
\der{}{t}(m\ga\u)=0
\eeq
is the equation of motion. It is the same as 
\beq
\der\p t=0
\eeq
which is correct for a free particle.

\subsection{Lagrangian of a Charged Particle in Fields}

Next suppose that there are electric and magnetic fields  of roughly
the same order of magnitude present so that the particle experiences 
some force and acceleration. Then $L=L_f+L_{int}$ where $L_{int}$ is the 
``interaction'' Lagrangian and contains the information about the fields 
and forces. For the action to be an invariant, it must be the case 
that
\beq
A_{int}\equiv\int_{t_a}^{t_b}L_{int}dt
\eeq
is an invariant which means $L_{int}\ga$ has to be an invariant. Now, in the 
nonrelativistic limit one has, to lowest order, $L=T-V$ with $V=q\Ph$, so we 
have in this limit $L_{int}\ga=-q\Ph\ga=-qE\Ph/mc^2=-(q/mc)p_0A^0$. This is not
an invariant but can be made one by including the rest of 
$\pb\cdot\Ab$\footnote{We have little choice other than this form since we
only have the $p$, $x$ and $A$ four-vectors to work with}, 
and we expect that the result is valid not just in the nonrelativistic 
limit but in general:
\beq
L_{int}\ga=-\lep\frac q{mc}\rip p_\al A^\al.
\eeq
This choice of $L_{int}$ gives the desired invariant and reduces to the 
correct static limit. It is the simplest choice of the interaction 
Lagrangian with the following properties:
\ben
\item Translationally invariant (in the sense that it is independent of 
explicit dependence on $\x$; the potentials do depend on $\x$)
\item Linear in the charge (as are the forces on the particle)
\item Linear in the momenta (as are the forces)
\item Linear in the fields (as are the equations of motion of the particle)
\item A function of no time derivatives of $p^\al$ (appropriate for the 
equations of motion)
\een

\subsubsection{Equations of Motion}

Let us proceed to the Euler-Lagrange equations of motion. The total 
Lagrangian is
\beq
L=-mc^2\sqrt{1-u^2/c^2}+\frac qc\u\cdot\A-q\Ph;
\eeq
\beqa
\pde L{x_i}=\frac qc\u\cdot\pde\A{x_i}-q\pde\Ph{x_i}\nonumber\\
\pde L{u_i}=\frac{mc^2}{\sqrt{1-u^2/c^2}}\frac{u_i}{c^2}+\frac qc A_i,
\eeqa
so 
\beq
\frac d{dt}\lep\pde L{u_i}\rip=\frac d{dt}(m\ga u_i)+\frac qc\pde{A_i}t
+\frac qc(\u\cdot\grad)A_i.
\eeq
Notice the last term on the right-hand side of this equation. It is there 
because when we take the total time derivative, we must remember that the 
position variable $\x$ on which $\A$ depends is really the position of the 
particle at time $t$, so, by application of the chain rule, we pick up a 
sum of terms, each of which is the 
derivative of $\A$ with respect to a component of $\x$ times the derivative 
of that component of $\x$ with respect to $t$; the last is a component of the 
velocity of the particle.

Finally, the equations of motion are
\beqa
\frac d{dt}(m\ga u_i)=-\frac qc\pde{A_i}t-\frac qc(\u\cdot\grad)A_i-q\pde\Ph
{x_i}+\frac qc\u\cdot\pde\A{x_i} \nonumber\\
=qE_i+\frac qc\u\cdot\pde\A{x_i}-\frac qc(\u\cdot\grad)A_i.
\eeqa

These are supposed to be familiar; consider 
\beq
(\u\times\B)_i=[\u\times(\curl\A)]_i=[\grad(\u\cdot\A)-(\u\cdot\grad)\A]_i
=\u\cdot\pde\A{x_i}-(\u\cdot\grad)A_i.
\eeq
Comparison of this expansion with \eq{27} demonstrates that the latter can be 
written as
\beq
\der{p_i}t=qE_i+\frac qc(\u\times\B)_i.
\eeq

\subsection{Hamiltonian of a Charged Particle}

One can also make a Hamiltonian description of the system. Introduce the 
canonical three-momentum $\vpi$ with components
\beq
\pi_i\equiv\pde L{u_i}=\ga mu_i+\frac qcA_i=p_i+\frac qcA_i.
\eeq
Then the Hamiltonian\footnote{The hamiltonian $H(q,p)$ obtained from
the Lagrangian through a Legendre transformation 
$H=\sum_i p_i\dot{q}_i - L(q,\dot{q})$} is
\beq
H=\vpi\cdot\u-L=\vpi\cdot\u+mc^2\sqrt{1-u^2/c^2}+q\Ph-\frac qc(\u\cdot\A).
\eeq
We want $H$ to depend on $\x$ and $\vpi$ but not on $\u$. To this end 
consider how to write $\u$ in terms of $\vpi$,
\beq
\vpi=\frac{m\u}{\sqrt{1-u^2/c^2}}+\frac qc\A,
\eeq
or
\beq
\lep\vpi-\frac qc\A\rip^2\lep1-\frac{u^2}{c^2}\rip=m^2u^2
\eeq
which may be solved for $\u$ to give
\beq
\u=c\frac{c\vpi-q\A}{\sqrt{m^2c^4+(c\vpi-q\A)^2}}
\eeq
Use of this result in the expression for the Hamiltonian leads to 
\beq
H=\sqrt{m^2c^4+(c\vpi-q\A)^2}+q\Ph.
\eeq
The development of Hamilton's equations will be left as an exercise.

The Hamiltonian is the 0$^{th}$ component of a four-vector. Notice, from 
\eq{35}, that
\beq
(H-q\Ph)^2-(c\vpi-q\A)^2=m^2c^4,
\eeq
is an invariant. This invariant is the inner product of a four-vector with 
itself. The spacelike components are $c\vpi-q\A=c\p$, and the timelike 
component is $H-q\Ph$. The vector is just the energy-momentum four-vector 
in the presence of fields,
\beq
p^\al=(E/c,\p)=\lep\frac1c(H-q\Ph),\vpi-\frac ec\A\rip.
\eeq

\subsection{Invariant Forms}

Next we are going to repeat everything that we have just done, but in a 
manner that is ``manifestly'' covariant. That is, we want to rewrite
the Lagrangian in terms of invariant 4-vector products.
We can write the free-particle Lagrangian as
\beq
L_f=-mc^2\sqrt{1-u^2/c^2}=-\frac1\ga\sqrt{E^2-p^2c^2}=-\frac c\ga\sqrt{
p_\al p^\al}=-\frac{mc}\ga\sqrt{U_\al U^\al}
\eeq
where 
\beq
U^\al=(E/mc,\p/m)\equiv dx^\al/d\ta
\eeq
is a four-vector we shall call the {\em four-velocity}. The action of the 
free particle is
\beq
A=-\int_{t_a}^{t_b}dt\,\ga^{-1}mc\sqrt{U^\al U_\al}=-mc\int^{\ta_b}_{\ta_a}
d\ta\,\sqrt{U^\al U_\al}
\eeq
Note the manifestly invariant form.
However, note that we must also impose the constraint $U^\al U_\al=c^2$.
Thus we may not freely vary this action to find the equations of motion.
There are two ways to overcome this.  First, we could introduce a Lagrange
multiplier to impose the constraint\footnote{This approach is discussed
in {\bf Electrodynamics and Classical Theory of Fields and Particles}
by A.O. Barut, Dover, page 65}, or we could introduce an additional degree
of freedom into our equations, and use it to impose the constraint 
{\em a posteriori}.  Following Jackson, we will follow the later
(less conventional) route.  To this end we rewrite the action, and introduce
$s$.
\beq
A=-mc\int_{\ta_a}^{\ta_b}\sqrt{dx_\al dx^
\al}=-mc\int_{s_a}^{s_b}ds\,\sqrt{g_{\al\be}\der{x^\al}s\der{x^\be}s}
\eeq
where the path of integration has been parametrized using some (invariant) 
$s$. We shall now {\bf treat} each $dx^\al/ds$ as an independent 
generalized velocity, and the Lagrangian takes on the functional form
$L(x^\al,dx^\al/ds,s)$.
This (more general) parametrization of the action integral is just as good 
as the standard one using the time; the Euler-Lagrange equations of motion, 
found by demanding that $A$ be an extremum, are familiar in appearance,
\beq
\frac d{ds}\lep\pde{\tilde{L}}{(dx_\al/ds)}\rip-\pde{\tilde{L}}{x_\al}=0
\eeq
In this case, we obtain the equation of motion
\beq
mc\der{}{s}\leb\frac{dx^\al/ds}
{\sqrt{\frac{dx^\beta}{ds} \frac{dx_\beta}{ds}}}\rib=0
\eeq
These velocities are constrained by the condition
\beq
\sqrt{g_{\al\be}\der{x^\al}s\der{x^\be}s}ds=cd\ta
\eeq
because there are really only three independent generalized velocities,
so that
\beq
m\frac{d^2x^\al}{d\tau^2}=0
\eeq

Analyzing and including the interaction Lagrangian in the same manner leads 
to a total Lagrangian and an action which is
\beq
A=-\int_{s_a}^{s_b}ds\,\leb mc\sqrt{g^{\al\be}\der{x_\al}s\der{x_\be}s}
+\frac ec\der{x_\al}sA^\al(x)\rib\equiv-\int_{s_a}^{s_b}ds\,\tilde{L}.
\eeq
The equation of motion may be found in the same manner
and in the present application these turn out to be
\beq
m\sde{x^\al}\ta=\frac ec(\partial^\al A^\be-\partial^\be A^\al)\der{x_\be}
\ta, 
\eeq
and they are correct, as one may show by comparing them with the standard 
forms.
The corresponding canonical momenta are
\beq
\pi^\al=\pde{\tilde{L}}{(dx_\al/ds)}=mU^\al+\frac ec A^\al.
\eeq
Hence the Hamiltonian is
\beq
\tilde{H}=\pi_\al U^\al-\tilde{L}=\frac 1{2m}\lep\pi_\al-\frac{eA_\al}c\rip
\lep\pi^\al-\frac{eA^\al}c\rip-\frac12mc^2.
\eeq
Hamilton's equations of motion\footnote{For $H(p,q)$ Hamiltons equations
are $\dot{q}_i=\pde{H}{p_i}$, are $\dot{p}_i=-\pde{H}{q_i}$, and
$\pde{L}{t}=-\pde{H}{t}$} are
\beqa
\der{x^\al}\ta=\pde{\tilde{H}}{\pi_\al}=\frac1m\lep\pi^\al-\frac ecA^\al
\rip\nonumber\\\der{\pi^\al}\ta=-\pde{\tilde{H}}{x_\al}=\frac e{mc}\lep\pi_\be-
\frac{eA_\be}c\rip\partial^\al A^\be
\eeqa

\section{Lagrangian for the Electromagnetic Field}

The electromagnetic field and fields in general have continuous degrees of 
freedom. The analog of a generalized coordinate $q_i$ is the value of a 
field $\ph_k$ at a point $\xb$. There are an infinite number of such points 
and so we have an infinite number of generalized coordinates. The 
corresponding ``generalized velocities'' are derivatives of the field with 
respect to the variables, $\partial \ph_k(\xb)/\partial x^\al$ or $\partial 
\ph_k(\xb)\partial x_\al$ with $\al=0,1,2,3.$ 
\beqa
q_i&\to&\ph_k(\xb) \nonumber\\
\dot{q}_i&\to&\pde{\ph_k(\xb)}{x^\al}
\eeqa
Instead of a Lagrangian $L$ 
which depends on the coordinates and velocities $q_i$ and $\dot{q}_i$, one now 
has a {\em Lagrangian density} ${\cal L}$, and the Lagrangian is obtained by 
integrating this density over position space,
\beq
L=\iniv\,{\cal L}(\ph_k(\xb),\partial^\al\ph_k(\xb));
\eeq
The action is the integral of this over time, or
\beq
A=\int d^4x\,{\cal L}(\ph_k(\xb)\,,\,\partial^\al\ph_k(\xb)).
\eeq
Given that $A$ and $d^4x$ are invariants, ${\cal L}$ must also be an 
invariant.

The Euler-Lagrange equations of motion are obtained as usual by demanding 
that $A$ be an extremum with respect to variation of the fields, or
\beq
\de A/\de\ph_k(\xb)=0
\eeq
for each field $\ph_k$. The resulting equations are, explicitly, 
\beq
\partial^\be\lep\frac{\partial{\cal L}}{\partial(\partial^\be\ph_k)}\rip-
\frac{\partial{\cal L}}{\partial\ph_k}=0.
\eeq

Now let's turn to the question of an appropriate Lagrangian density for the 
electromagnetic field. The things we have to work with are $F^{\al\be}$, 
$A^\al$, and $J^\al$, if we rule out explicit dependence on space and time 
(a translationally invariant universe). We must make an invariant out of 
these. One which practically suggests itself is
\beq
{\cal L}=-\frac1{16\pi}F_{\al\be}F^{\al\be}-\frac1cJ_\al A^\al.
\eeq
The various constants are a matter of definition; otherwise we have 
something which is linear in components of $\Ab$ and of $\Jb$, and bilinear 
derivatives of components of $\Ab$. Let's write it out in detail:
\beqa
{\cal L}&=&-\frac1{16\pi}(\partial_\al A_\be-\partial_\be A_\al)(\partial^\al
A^\be-\partial^\be A^\al)-\frac1cJ_\al A^\al\nonumber\\
&=&-\frac1{16\pi}g_{\al
\ga}g_{\be\de}(\partial^\ga A^\de-\partial^\de A^\ga)(\partial^\al A^\be-
\partial^\be A^\al)-\frac1cJ_\al A^\al
\eeqa

The generalized fields (called $\ph_k$ above) are the components of $\Ab$. 
Hence the functional derivatives of ${\cal L}$ which enter the Euler-Lagrange 
equations are
\beq
\frac{\partial{\cal L}}{\partial(\partial^\be A^\al)}=
\frac1{4\pi}F_{\al\be}\andh
\frac{\partial{\cal L}}{\partial A^\al}=-\frac1c J_\al
\eeq
and so the equations of motion are
\beq
\frac1{4\pi}\partial^\be F_{\be\al}=\frac1cJ_\al.
\eeq
These are indeed the four\footnote{Four equations come from the one scalar
and one vector inhomogeneous Maxwell's equations} inhomogeneous Maxwell
equations. The homogeneous 
equations are automatically satisfied because we have constructed the 
Lagrangian in terms of the potentials. The charge continuity equation 
follows from taking the contravariant derivative of the equation above,
\beq
\frac1{4\pi}\partial^\al\partial^\be F_{\be\al}=\frac1c\partial^\al J_\al;
\eeq
the left-hand side is zero when summed because $F_{\al\be}=-F_{\be\al}$ 
and so we have
\beq
\partial^\al J_\al=0.
\eeq
\section{Stress Tensors and Conservation Laws}
Conservation of energy emerges from the usual Lagrangian formulation if $L$ 
has no explicit dependence on the time; then $dH/dt=0$ which means that the 
Hamiltonian is a constant of the motion. If we want to carry this sort of 
thing over to our field theory, we need to construct a {\em Hamiltonian density 
} ${\cal H}$ whose integral over all position space, $H$, is interpreted as the 
energy. If one proceeds in analogy with the particle case, he would take a 
Lagrangian density
\beq
{\cal L}={\cal L}(\ph_k(\xb),\partial^\al\ph_k(\xb));
\eeq
introduce momentum fields 
\beq
\Pi_k(\xb)\equiv\partial{\cal L}/\partial(\partial\ph_k/\partial t);
\eeq
and a Hamiltonian density
\beq
{\cal H}=\sum_k\Pi_k(\xb)(\partial\ph_k(\xb)/\partial t)-{\cal L}.
\eeq
We are going to generalize this procedure by introducing a rank-two tensor 
instead of a simple Hamiltionian density. The reason is that if one has a 
simple density and introduces $H$ as
\beq
H=\iniv{\cal H}=\int dx_0\frac{d^3x}{dx_0}{\cal H},
\eeq
and if one wants this to be an energy, which, as we have seen, transforms 
as the $0^{th}$ component of a four-vector, then ${\cal H}$ should be the 
(0,0) component of a rank-two tensor. To this end, let us introduce
\beq
\ps_k^\al(\xb)\equiv\partial{\cal L}/\partial(\partial_\al\ph_k)
\eeq
and
\beq
T^{\al\be}\equiv\sum_k\ps_k^\al(\xb)\partial^\be\ph_k-g^{\al\be}{\cal L}.
\eeq
This rank-two tensor is called the {\em canonical stress tensor}. 

\subsection{Free Field Lagrangian and Hamiltonian Densities}

Let's 
see what form the Lagrangian density and canonical stress tensor take
in the absence of any sources 
$\J^\al$.  In this case the Lagrangian density becomes ${\cal L}_{ff}$, 
the free-field Lagrangian 
density. 
\beq
{\cal L}_{ff} = -\frac1{16\pi}F_{\al\be}F^{\al\be}
\eeq
By carrying out the implied manipulations we find
\beq
T^{\al\be}=-\frac1{4\pi}g^{\al\ga}F_{\ga\de}\partial^\be A^\de-g^{\al\be}{
\cal L}_{ff}.
\eeq
Look in particular at $T^{00}$:
\beqa
T^{00}=-\frac1{4\pi}(F_{0\ga}\partial^0A^\ga)-\frac1{8\pi}(E^2-B^2)\nonumber\\
=-\frac1{4\pi}\leb E_x\frac1c\pde{A_x}t+E_y\frac1c\pde{A_y}t+E_z\frac1c
\pde{A_z}t\rib-\frac1{8\pi}(E^2-B^2)\nonumber\\=\frac1{4\pi}E^2+\E\cdot
\grad\Ph-\frac1{8\pi}(E^2-B^2)=\frac1{8\pi}(E^2+B^2)+\frac{\E\cdot\grad
\Ph}{4\pi}.
\eeqa
This contains the expected and desired term $(E^2+B^2)/8\pi$, which is
the feild energy density, but there is 
an additional term $\E\cdot\grad\Ph$. Because $\div\E=0$ for free fields, 
it is the case that $\E\cdot\grad\Ph=\div(\E\Ph)$ and so the integral over 
all space of this part of $T^{00}$ will vanish for a localized field 
distribution. Hence we find that
\beq
\iniv T^{00}=\frac1{8\pi}\iniv(E^2+B^2)
\eeq
is indeed the field energy.

And what of the other components of the stress tensor? These too have some 
unexpected properties. For example, one can show that 
\beq
T^{0i}=\frac1{4\pi}(\E\times\B)_i+\frac1{4\pi}\div(A_i\E)
\eeq
and
\beq
T^{i0}=\frac1{4\pi}(\E\times\B)_i+\frac1{4\pi}\leb(\curl(\Ph\B))_i-\frac
\partial{\partial x_0}(\Ph E_i)\rib.
\eeq
Evidently, this tensor is not symmetric. Also, one would have hoped that 
these components of the tensor would have turned out to be components of 
the Poynting vector, with  appropriate scaling, so that we would have found 
an equation $0=\partial_\al T^{0\al}$ which would have been equivalent to 
the Poynting  theorem,
\beq
\pde ut+\div\S=0.
\eeq
Although this is not going to happen, there is some sort of conservation 
law contained in our stress tensor. One can show that 
\beq
\partial_\al T^{\al\be}=0
\eeq
which gives not one but four conservation laws. To demonstrate this 
equation, consider the following:
\beqa
\pop_\al T^{\al\be}=\sum_k\pop_\al\leb\pde\scl{(\pop_\al\ph_k)}\pop^\be
\ph_k\rib-\pop^\be\scl\nonumber\\
=\sum_k\leb\pop_\al\lep\pde\scl{(\pop_\al\ph_k)}\rip\pop^\be\ph_k+
\lep\pde\scl{(\pop_\al\ph_k)}\rip\pop^\be(\pop_\al\ph_k)\rib-\pop^\be\scl
\nonumber\\
=\sum_k\leb\lep\pde\scl{\ph_k}\rip\pop^\be\ph_k+\lep\pde\scl{(\pop_\al\ph_k)}
\rip\pop^\be(\pop_\al\ph_k)\rib-\pop^\be\scl
\eeqa
where we have used the Euler-Lagrange equations of motion \eq{55} 
on the first term 
in the middle line. Now we can recognize that the terms summed over $k$ in the 
last line are $\pop^\be\scl$ since $\scl$ is a function of the fields 
$\ph_k$ and their derivatives $\pop_\al\ph_k$. Hence we have demonstrated 
that
\beq
\pop_\al T^{\al\be}=\pop^\be\scl-\pop^\be\scl=0
\eeq

These give familiar global conservation laws when integrated over all space 
for a localized set of fields. Consider
\beq
0=\iniv\pop_\al T^{\al\be}=\frac\pop{\pop x^0}\lep\iniv T^{0\be}\rip+\iniv
\frac\pop{\pop x^i}(T^{i\be}).
\eeq
The last term on the right-hand side is zero as one shows by integrating 
over that coordinate with respect to which the derivative is taken and 
appealing to the fact that we have localized fields which vanish as $|x^i|$ 
becomes large. Hence our conclusion is that
\beq
\frac d{dt}\lep\iniv T^{0\be}\rip=0
\eeq
If one looks at the explicit components of the tensor, one finds that these 
simply say the total energy and total momentum are constant, using our 
identifications (from chapter 6) of $u$ and $\g$ as the energy and momentum 
density.
\beq
u=\frac{1}{8\pi}\lep E^2+B^2\rip\;\;\;\g=\frac{1}{4\pi c}\lep\E\times\B\rip
\eeq

\subsection{Symmetric Stress Tensor}

It is troubling that the canonical stress tensor is not symmetric. This 
becomes a serious problem when one examines the angular momentum. Consider 
the rank-three tensor
\beq
M^{\al\be\ga}\equiv T^{\al\be}x^\ga-T^{\al\ga}x^\be.
\eeq
If this is to represent the angular momentum in some way we would like it 
to provide a conservation law in the form $\pop_\al M^{\al\be\ga}=0$. But 
that doesn't happen. Rather,
\beq
\pop_\al M^{\al\be\ga}=T^{\ga\be}-T^{\be\ga}+(\pop_\al T^{\al\be})x^\ga
-(\pop_\al T^{\al\ga})x^\be=T^{\ga\be}-T^{\be\ga}
\eeq
which doesn't vanish because $\Tb$ is not symmetric. 

The standard way out of this and other difficulties associated with the 
asymmetry of the canonical stress tensor is to define a different stress 
tensor which works. By regrouping terms in the canonical stress tensor one 
can write
\beq
T^{\al\be}=\frac1{4\pi}\lep g^{\al\ga}F_{\ga\de}F^{\de\be}+\frac14g^{\al\be}
F_{\ga\de}F^{\ga\de}\rip-\frac1{4\pi}g^{\al\ga}F_{\ga\de}\pop^\de A^\be.
\eeq
Now, the second term is 
\beqa
-\frac1{4\pi}g^{\al\ga}F_{\ga\de}\pop^\de A^\be=-\frac1{4\pi}F^{\al\de}
\pop_\de A^\be=\frac1{4\pi}F^{\de\al}\pop_\de A^\be=\nonumber\\
\frac1{4\pi}(F^{\de\al}\pop_\de A^\be+A^\be\pop_\de F^{\de\al})=\frac1{4\pi}
\pop_\de(F^{\de\al} A^\be).
\eeqa
This is a four-divergence, so for fields of finite extent, it must be the 
case that
\beq
\iniv\pop_\de(F^{\de0}A^\be)=0.
\eeq
Moreover, it has a vanishing four-divergence,
\beq
\pop_\al\pop_\de(F^{\de\al}A^\be)=0, 
\eeq
which follows from the antisymmetric character of the field tensor. Hence, 
if we simply remove this piece from the stress tensor, leaving a new tensor
$\thb$, known as the {\em symmetric stress tensor},
\beq
\th^{\al\be}\equiv\frac1{4\pi}(g^{\al\ga}F_{\ga\de}F^{\de\be}+\frac14g^{\al
\be}F_{\ga\de}F^{\ga\de}),
\eeq
then this tensor is such that
\beq
\der{}t\lep\iniv\th^{0\be}\rip=0\andh\pop_\al\th^{\al\be}=0.
\eeq

It is easy to work out the components of this tensor; they are $(i,j=1,2,3)$
\beq\barr{l}
\th^{00}=\frac1{8\pi}(E^2+B^2)\\
\th^{i0}=\th^{0i}=\frac1{4\pi}(\E\times\B)\\
\th^{ij}=-\frac1{4\pi}[E_iE_j+B_iB_j-\frac12\de_{ij}(E^2+B^2)].
\earr\eeq
Hence in block matrix form,

\beq
\th^{\al\be}=\lep\barr{cc}u&c\g\\c\g&-T^{(M)}_{ij}\earr\rip
\eeq
where $T^{(M)}_{ij}$ is the $ij$ component of the Maxwell stress tensor. 
The conservation laws 
\beq
\pop_\al\th^{\al\be}=0
\eeq
are well-known to us. They are the Poynting theorem, for $\be=0$, and the 
momentum conservation laws
\beq
\pde{g_i}t-\sum_j\pde{T^{(M)}_{ij}}{x_j}=0.
\eeq
when $\be=i$

Now consider once again the question of angular momentum. Define
\beq
M^{\al\be\ga}\equiv\th^{\al\be}x^\ga-\th^{\al\ga}x^\be.
\eeq
Then the equations
\beq
\pop_\al M^{\al\be\ga}=0
\eeq
express angular momentum conservation as well as some other things.

\subsection{Conservation Laws in the Presence of Sources}

Finally, what happens if there are sources? Then we won't find the same 
form for the conservation laws. Consider
\beqa
\pop_\al\th^{\al\be}=\frac1{4\pi}\leb\pop^\ga(F_{\ga\de}F^{\de\be})+
\frac1{4}\pop^\be(F_{\ga\de}F^{\ga\de})\rib\nonumber\\
=\frac1{4\pi}\leb(\pop^\ga F_{\ga\de}F^{\de\be}+F_{\ga\de}(\pop^\ga F^{\ga
\be})+\frac12F_{\ga\de}(\pop^\be F^{\ga\de})\rib.
\eeqa
Making use of the Maxwell equations $\pop^\ga F_{\ga\de}=\frac{4\pi}{c}J_\de$, 
we can rewrite this as
\beq
\pop_\al\th^{\al\be}+\frac1cF^{\be\de}J_\de=\frac1{8\pi}\leb F_{\ga\de}(
\pop^\ga F^{\de\be}+\pop^\ga F^{\de\be}+\pop^\be F^{\ga\de})\rib.
\eeq
Now recall that (these are the homogeneous Maxwell's equations)
\beq
\pop^\ga F^{\de\be}+\pop^\be F^{\ga\de}+\pop^\de F^{\be\ga}=0,
\eeq
so \eq{90} may be written as
\beq
\pop_\al\th^{\al\be}+\frac1cF^{\be\de}J_{\de}=\frac1{8\pi}F_{\ga\de}(\pop^
\ga F^{\de\be}-\pop^\de F^{\be\ga}).
\eeq
However,
\beq
(\pop^\ga F^{\de\be}-\pop^\de F^{\be\ga})F_{\ga\de}=(\pop^\ga F^{\de\be}+
\pop^\de F^{\ga\be})F_{\ga\de}
\eeq
is a contraction of an object symmetric in the indices $\ga$ and $\de$ and 
one which is antisymmetric; therefore it is zero. 
Hence we conclude that
\beq
\pop_\al\th^{\al\be}=-\frac1cF^{\be\de}J_\de.
\eeq
The four equations contained in this conservation law are the familiar ones
\beq
\pde ut+\div\S=-\J\cdot\E\;\;\;{\rm{when}}\; \be=0
\eeq
and
\beq
\pde{g_i}t-\sum_j\frac\pop{\pop x^j}T^{(M)}_{ij}=-[\rh E_i+\frac1c(\J\times
\B)_i]\;\;\;{\rm{when}}\; \be=i \,.
\eeq


%\section{Maxwell's Equations from Coulomb's Law? No.}

\section{Examples of Relativistic Particle Dynamics}

\subsection{Motion in a Constant Uniform Magnetic Induction}

Given an applied 
constant magnetic induction, the equations of motion for a particle of 
charge $q$ are
\beq
\der{E}{t}=\F\cdot\u=0 ,\;\;\;\;
\der\p t=\frac qc(\u\times\B)=m\ga\der\u t
\eeq
where the last step follows from the fact that $\p=m\ga\u$ and the fact 
that $\ga mc^2$, the particle's energy, is constant because magnetic forces do 
no work. Hence the equations reduce to
\beq
\der\u t=\u\times\om_B
\eeq
where $\om_B=q\B/m\ga c$. Notice that this frequency depends on the energy 
of the particle. For definiteness, let $\B=B\ept$. Also, write $\u=u\pll
\ept+\u\per$ where $\u\per\cdot\ept=0$. 

\centerline{\psfig{figure=fig2.ps,height=2.0in,width=8.5in}}

\noindent From the equations of motion, one 
can see that $u\pll$ is a constant while $\u\per$ obeys
\beq
\der{\u\per}t=\om_B(\u\per\times\ept),
\eeq
or 
\beq
\der{u_x}t=\om_Bu_y\andh\der{u_y}t=-\om_Bu_x.
\eeq
Combining these we find, e.g.,
\beq
\sde{u_x}t=-\om_B^2u_x
\eeq
with the general solution
\beq
u_x=u_0e^{-i\om_Bt}
\eeq
where $u_0$ is a complex constant. Further,
\beq
u_y=\frac1{\om_B}\der{u_x}t=-iu_x,
\eeq
so
\beq
\u\per=u_0(\epu-i\epd)e^{-i\om_Bt}.
\eeq
We may integrate over time to find the trajectory:
\beq
\der\x t=u\pll\ept+\u\per
\eeq
and so 
\beqa
\x(t)=\x(0)+\int_0^tdt'\,\leb u\pll\ept+u_0(\epu-i\epd)e^{-i\om_Bt'}\rib
\nonumber\\=\x(0)+u\pll t\ept+i\frac{u_0}{\om_B}(\epu-i\epd)(e^{-i\om_Bt}-1
).
\eeqa
The physical trajectory is the real part of this and is, for real $u_0$,
\beq
\x(t)=\x(0)+u\pll t\ept+\frac{u_0}{\om_B}[\sin(\om_Bt)\epu+(\cos(\om_Bt)-1)
\epd].
\eeq
This equation describes helical motion with the helix axis parallel to 
the $z$-axis. The radius of the axis is $a$, where $a=u_0/\om_B$.

It is worthwhile to establish the connection betwen $a$ and $|\p\per|$ 
where $\p\per=m\ga\u\per$ is the momentum in the plane perpendicular to the 
direction of $\B$.
\beq
p\per=m\ga u_0=m\ga\om_Ba=m\ga\frac{qB}{m\ga c}a=\frac{qBa}c.
\eeq

\centerline{\psfig{figure=fig3.ps,height=3.0in,width=8.5in}}

\noindent This relation, $p\per=qBa/c$, tells us the radius of curvature in 
the plane 
perpendicular to $\B$ (which is not the same as the radius of curvature of 
the orbit) is a linear function of $p\per$, and it suggests a simple way 
to select particles of a given momentum out of a beam containing particles 
with many momenta. One simply passes the beam through a region of space 
where there is some $\B$ applied transverse to the direction of the beam. 
The amount by which a particle is deflected will increase with decreasing
$p\per$ and so the beam is spread out much as a prism separates the 
different frequency components of a beam of light. The device is a momentum 
selector.

\subsection{Motion in crossed $\E$ and $\B$ fields, $E<B$.} 

For $\E\cdot\B=0$ 
in frame $K$, we can find a frame $K'$ where $\E'=0$, provided $E<B$. 
This may be seen from the form of the field transforms.
\beq\barr{cc}
\E\pll'=\E\pll&\E\per'=\ga[\E\per+(\vbe\times\B)]\\
\B\pll'=\B\pll&\B\per'=\ga[\B\per-(\vbe\times\E)]\earr
\eeq
In fact, we have already solved exactly this problem in chapter 11
where we found that $K'$ moves relative to $K$ with a velocity which 
is $\v=c(\E\times\B)/
B^2$. If we let $\E=E\epd$ and $\B=B\ept$, then $\v=c(E/B)\epu$, and 
$\B'=B\sqrt{1-E^2/B^2}\ept$. 

\centerline{\psfig{figure=fig4.ps,height=2.5in,width=8.5in}}

\noindent Now imagine a particle is injected into this 
system with an initial velocity\footnote{We could be more general and 
include a component of $\u$ parallel to $\B$; this would not lead to 
anything significantly different from what we are about to find.}
$\u(0)=u_0\epu$. In the frame $K'$, its initial velocity is
\beq
\u'(0)=\frac{u_0-v}{1+u_0v/c^2}\epu.
\eeq

From our first example, we know that the particle will proceed to execute 
circular motion in this frame, always with the same speed $u'(0)$. What 
then is its motion in frame $K$? Superposed on the circular motion will be 
a drift velocity $\v$. If $q>0$ and $u_0>v$, we get the first motion shown 
below. But if $u_0<v$, we get the second motion. For the special case of 
$u_0=v$, the particle is at rest in $K'$ which means it moves in $K$ 
at a constant velocity $\u(t)=\v$. 

\centerline{\psfig{figure=fig5.ps,height=2.0in,width=8.5in}}

Such a device can be 
employed as a velocity selector and so it complements the device described 
in the first example which was a momentum selector. The idea is that a 
particle coming in with a speed $u_0$ greater than $v$ will experience a 
magnetic force greater than the electric force and so it will be deflected 
accordingly. But one coming in with a speed smaller than $v$ will 
experience an electric force greater than the magnetic one, and it will be 
deflected in the other direction.

The picture chages after a while, however, because the particle will speed 
up and slow down under the influence of the electric field. Suppose that 
initially $u_0>v$ ($u_0<v$). Then the $B$-field (the $E$-field) force 
dominates, and the particle is deflected in such a way that it moves 
against (with) the electric field. This causes it to slow down (speed up) 
so that after some time $u_0<v$ ($u_0>v$). Then the electric (magnetic) 
field force dominates, causing the particle to swing around so that it 
eventually moves with (against) the electric field force. And so on. The 
end result is a trajectory that produces a time-averaged velocity equal to 
$\v$ or $c(\E\times\B)/B^2$. This is called the {\em $E$ cross $B$ drift 
velocity}. It is in the direction of $\E\times\B$ no matter what is the 
sign of the charge.

\subsection{Motion in crossed $\E$ and $\B$ fields, $E>B$} 

This time we want 
to consider the motion in a frame $K'$ moving at velocity $\v=c(\E\times\B)
/E^2\rta c(B/E)\epu$, if we keep the same directions of the fields as in 
the preceding example. In this frame there is only an electric field
$\E'=\E\sqrt{1-B^2/E^2}$ which will cause the particle to move away in the 
direction of $\E'$. The equations of motion in $K'$ are
\beq
mc^2\der{\ga'}{t'}=qE'\der{y'}{t'}\andh\der{p_y'}{t'}=qE';
\eeq
the components of the momentum in the other directions are constant. One 
easily solves to find
\beq
\p'(t')=\p'(0)+qE't'\epd
\eeq
and we can then find $\ga'$ directly from the dispersion relation,
\beq
\ga'=\frac1{mc^2}\sqrt{m^2c^4+\p'(t')\cdot\p'(t')c^2}.
\eeq
The speed $u_y'$ is found easily from the equation of motion for $\ga'$ 
which integrates trivially to produce
\beqa
y'(t')=y'(0)+\frac{mc^2}{qE'}(\ga'(t')-\ga'(0))\nonumber\\
=y'(0)+\frac{\sqrt{m^2c^4+(\p'(t'))^2c^2}-\sqrt{m^2c^4+(\p'(0))^2c^2}}{qE'}.
\eeqa
Consider also $\x'\per$, the component of $\x'$ perpendicular to the 
electric field. Because $\p'\per/dt=0$, it is true that $\ga'\y'\per=\ga'(0)
\u'\per(0)$, a constant. Hence
\beq
\u'\per(t')=\u'\per(0)\sqrt{1+(\p'(0))^2/m^2c^2}/\sqrt{1+(\p'(t'))^2/
m^2c^2}.
\eeq
We can integrate the velocity over 
time to find the displacement of the particle. For the special case that 
there is no component of $\p'(0)$ in the direction of the field, one finds 
that
\beq
\x'\per(t)-\x'\per(0)=\frac{\p'\per(0)}{qE'}\ln\leb\frac{qE't'}{m\ga(0)}
+\sqrt{1+\lep\frac{qE't'}{m\ga(0)}\rip^2}\rib.
\eeq
We can combine \eqs{113}{115} to remove the time and so have an equation that 
determines the shape of the trajectory. For simplicity, let $\x'\per(0)=y'(
0)=0.$ Then one finds
\beq
\frac{x'\per qE'}{m\ga'(0)u'(0)}=\ln\leb\sqrt{\lep1+\frac{qE'y'}{m\ga'(0)}
\rip^2-1}+1+\frac{qE'y'}{m\ga'(0)}\rib.
\eeq
For short times satisfying the condition $|qE'y'/m\ga'(0)|<<1$, the 
trajectory is a parabola,
\beq
\frac{\x'\per qE'}{m\ga'(0)u'(0)}\approx\sqrt{\frac{2qE'y'}{m\ga'(0)}}
\eeq
or
\beq
y'=\frac{qE'x'^2\per}{2m\ga'(0)(u'(0))^2}.
\eeq
The long time behavior is displayed for $|qE'y'/m\ga'(0)|>>1$, and it is 
such that
\beq
y'=\frac{m\ga'(0)}{2qE'}\exp\lep\frac{x'\per qE'}{m\ga'(0)u'(0)}\rip.
\eeq

\subsection{Motion for general uniform $\E$ and $\B$.} 

Then we cannot find a 
frame where one of the fields can be made to vanish. But there is a frame 
where the electric field and magnetic induction are parallel; here the 
solution of the equations of motion is relatively simple and is left as an 
exercise.

\subsection{Motion in slowly spatially varying $\B(\x)$}. 

This problem is 
greatly simplified by (1) the fact that then energy, or $\ga$, is a 
constant and by (2) the assumption that $\Bx$ does not vary much relative 
to its magnitude over distances on the order of the radius of the 
particle's orbit.
\edo

