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%TCIDATA{Created=Tue Nov 19 17:48:57 1996}
%TCIDATA{LastRevised=Tue Nov 19 18:02:36 1996}
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\markright{{\sf Conservation Laws    }}
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\begin{document}


\section{12.10\quad Conservation Laws.}

\subsection{12.10.1}

\quad \thinspace \textbf{a. }Conserved quantities in a relativistic field
theory obey equations such as $\partial ^{\alpha }J_{\alpha }\,=0$ (current
conservation). When such an equation is integrated over space, one finds 
\begin{equation}
-\int d^{3}x\,\mathbf{\nabla \cdot J}=\frac{\partial }{c\partial t}\int
J^{0}d^{3}x.  \label{1}
\end{equation}
The left-hand side can be transformed to a surface integral which represents
the net outflow of the conserved quantity. When this vanishes, the integral
on the r.h.s. is a constant of the motion.

\smallskip\ 

\textbf{b}. More generally, the conserved quantity could be a tensor of rank
higher than one. All field Lagrangians without external sources lead
naturally to the construction of a canonical \textsl{stress tensor}, $%
T^{\alpha \beta },$ such that $\partial _\alpha T^{\alpha \beta }=0.$

\smallskip\ 

\textbf{c. }For the free e.m. Lagrangian, 
\[
\frak{L}=-\frac 1{16\pi }F_{\mu \nu }F^{\mu \nu } 
\]
the canonical stress tensor is 
\[
T^{\alpha \beta }=\frac{\partial \frak{L}}{\partial \left( \partial _\alpha
A^\lambda \right) }\partial ^\beta A^\lambda -g^{\alpha \beta }\frak{L}%
=-\frac 1{4\pi }g^{\alpha \mu }F_{\mu \lambda }\partial ^\beta A^\lambda
-g^{\alpha \beta }\frak{L\,.} 
\]
The conservation law $\partial _\alpha T^{\alpha \beta }=0$ just follows
from the Euler-Lagrange equations for the \textsl{free} field.

\smallskip\ 

\textbf{d. }One can always add to $T^{\alpha \beta }$ a ``trivial''
conserved quantity. This is any tensor $t^{\alpha \beta }$ such that $%
\partial _\alpha t^{\alpha \beta }=0$ and that $\int t^{0\beta }d^3x=0.$ It
is trivial because the r.h.s. of the relation corresponding to (\ref{1})
vanishes identically when the integration is over all space. Often $%
t^{0\beta }$ is itself a divergence.

\smallskip\ 

\textbf{e. }In the e.m. case it is convenient for many purposes to subtract
from $T^{\alpha \beta }$ the ``trivial'' tensor $t^{\alpha \beta }=\left(
1/4\pi \right) \partial _\lambda \left( F^{\lambda \alpha }A^\beta \right) .$
One can easily check that $\partial _\alpha t^{\alpha \beta }=0.$

\newpage

\subsection{12.10.2}

\quad \quad \textbf{a. }The tensor 
\[
\Theta ^{\alpha \beta }=T^{\alpha \beta }-t^{\alpha \beta }=\frac{1}{4\pi }%
\left( g^{\alpha \mu }F_{\mu \lambda }F^{\lambda \beta }+\frac{1}{4}%
g^{\alpha \beta }F_{\mu \lambda }F^{\mu \lambda }\right) 
\]
is conserved, symmetric and manifestly gauge invariant (while $T^{\alpha
\beta }$ is not symmetric or gauge invariant). $\Theta ^{\alpha \beta }$ is
called the \textsl{symmetric} \textsl{stress tensor.} In terms of
three-dimensional quantities we have 
\begin{equation}
\Theta ^{\alpha \beta }=\left( 
\begin{tabular}{ll}
$u$ & $c\mathbf{g}$ \\ 
$c\mathbf{g}$ & $\Theta ^{ab}$%
\end{tabular}
\right)   \label{2}
\end{equation}
where 
\begin{eqnarray*}
u &=&\Theta ^{00}=\frac{1}{8\pi }\left( E^{2}+B^{2}\right) \,,\quad \quad
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{energy
density} \\
c\mathbf{g} &=&\left[ cg^{a}=\Theta ^{0a}\right] =\mathbf{S=}\frac{\mathbf{%
E\times B}}{4\pi }\,,\quad \mathbf{S}\text{:\textbf{\ }Poynting vector, }%
\mathbf{g}\text{: momentum density} \\
-\Theta ^{ab} &=&\frac{1}{4\pi }\left[ E_{a}E_{b}+B_{a}B_{b}-\frac{1}{2}%
\delta _{ab}\,(E^{2}+B^{2})\right] =T_{ab}^{(M)}\,,\quad \text{Maxwell
stress tensor}
\end{eqnarray*}

\smallskip\ 

\textbf{b.} In the presence of sources, i.e. with $-(e/c)A_{\alpha
}J^{\alpha }$ in $\frak{L},$ the symmetric stress tensor can be shown to
obey 
\begin{equation}
\partial _{\alpha }\Theta ^{\alpha \beta }=-\frac{1}{c}F^{\beta \lambda
}J_{\lambda \,}.  \label{3}
\end{equation}
The r.h.s. of this equation is the negative of the Lorentz force density 
\[
f^{\beta }=\frac{1}{c}F^{\beta \lambda }J_{\lambda }=\left( \frac{1}{c}%
\mathbf{J\cdot E,\;}\rho \mathbf{E+}\frac{1}{c}\mathbf{J\times B}\right) .
\]
Therefore (\ref{3}) can be written 
\[
-\int d^{3}x\,\partial _{a}\Theta ^{a\beta }=\frac{dP^{\beta }}{dt}\,,
\]
where 
\[
\dfrac{dP^{\beta }}{dt}=\frac{d}{cdt}\int d^{3}x\,\Theta ^{0\beta }+\int
f^{\beta }d^{3}x.
\]
Clearly $P^{\beta }$ is the total 4-momentum of particles \textsl{and}
fields.

\newpage 

\subsection{12.10.3}

\quad \quad \textbf{a. }A simple derivation of the symmetric form of the
stress tensor is obtained by analogy with the formula that gives the elastic
stress in a crystal as the derivative of the elastic energy density with
respect to the strain. The analogue of the strain is the metric tensor $%
g_{\alpha \beta }.$ In any set of coordinates (curvilinear, accelerating,
rotating), $g_{\alpha \beta }$ is defined by 
\begin{equation}
g_{\alpha \beta }=\frac{\partial x^{\gamma }}{\partial \xi ^{\alpha }}\frac{%
\partial x^{\delta }}{\partial \xi ^{\beta }}\eta _{\gamma \delta }
\label{4}
\end{equation}
where $\xi ^{\alpha }$ are inertial cartesian coordinates and $\eta _{\gamma
\delta }$ is the tensor with $(-1,1,1,1)$ on the diagonal, i.e. the tensor $%
g_{\alpha \beta }$ in the coordinates $\xi ^{\alpha }.$ Then 
\begin{equation}
\Theta ^{\alpha \beta }=\frac{2}{\sqrt{\left| g\right| }}\frac{\partial
\left( \mathcal{L}\sqrt{\left| g\right| }\right) }{\partial g_{\alpha \beta }%
}  \label{5}
\end{equation}
where $g$ is the  determinant of $g_{\alpha \beta }.$ (After
differentiation, one can set again $g_{\alpha \beta }=\eta _{\alpha \beta },$
$g=-1).$ The reason why $\sqrt{\left| g\right| }$ appears in (\ref{5}) is
that the action in general coordinates is 
\begin{equation}
\mathcal{A}=\int \mathcal{L}\sqrt{\left| g\right| }d^{4}x  \label{6}
\end{equation}
where the extra factor $\sqrt{\left| g\right| }$ is nothing but the Jacobian.

\smallskip\ 

\emph{Details of algebra.}

\smallskip \ 

With  
\[
\mathcal{L}=-\frac{1}{16\pi }F_{\alpha \beta }F^{\alpha \beta }=-\frac{1}{%
16\pi }g_{\mu \mu ^{\prime }}g_{\nu \nu ^{\prime }}F^{\mu \nu }F^{\mu
^{\prime }\nu ^{\prime }}\,,
\]
it follows that\emph{\ } 
\[
\frac{\partial \mathcal{L}}{\partial g_{\alpha \beta }}=-\frac{1}{16\pi }%
\left[ \delta _{\mu }^{\,\,\,\,\alpha }\delta _{\mu ^{\prime
}}^{\,\,\,\,\beta }g_{\nu \nu ^{\prime }}+g_{\mu \mu ^{\prime }}\delta _{\nu
}^{\,\,\,\,\alpha }\delta _{\nu ^{\prime }}^{\,\,\,\,\beta }\right] F^{\mu
\nu }F^{\mu ^{\prime }\nu ^{\prime }}=-\frac{1}{8\pi }g_{\nu \nu ^{\prime
}}F^{\alpha \nu }F^{\beta \nu ^{\prime }}\,.
\]
Also, $\partial g/\partial g_{\alpha \beta }=gg^{\alpha \beta }$\ (=
algebraic complement, or cofactor, of $g_{\alpha \beta }$\ in the
determinant $g),$\ and of course 
\[
\frac{\partial \sqrt{\left| g\right| }}{\partial g_{\alpha \beta }}=-\frac{1%
}{2\sqrt{\left| g\right| }}\frac{\partial g}{\partial g_{\alpha \beta }}\,.
\]
With these results, (\ref{2}) follows from (\ref{5}).

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