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\bce
SUMMARY PHYSICS 707\\
{\bf Electrostatics}
\ece
The basic differential equations of electrostatics are
\beq
\div\Ex=4\pi\rhx\hbox{ and }\curl\Ex=0
\eeq
where $\Ex$ is the electric field and $\rhx$ is the electric charge
density. The field is defined by the statement that a charge $q$ at point
$\x$ experiences a force $\F=q\Ex$ where $\Ex$ is the field produced by all
charge other than $q$ itself. These equations have integral equivalents,
\beq
\inac\Ex\cdot\nn=4\pi\inv\rhx=4\pi Q
\eeq
where $Q$ is the charge enclosed by the surface $S$ surrounding the domain
$V$ and $\nn$ is a unit outward (from $V$) normal vector at a point on $S$;
and
\beq
\invlc\cdot\Ex=0.
\eeq
Finally, if one applies these equations on the surface of a conductor
(inside of which $\E=0$), then one finds the surface charge density 
and (negative) pressure are
\beq
E_n(x)=4\pi\si(x)\;\;\; p=2\pi\si^2
\eeq

There is an integral solution for $\Ex$ if one knows $\rh$ everywhere,
\beq
\Ex=\inivp\frac{\rhxp(\x-\xp)}{\xxpa^3}.
\eeq
Introduce a scalar potential $\Phx$ such that $\Ex=-\grad\Phx$ (This can be
done because $\curl\Ex=0$ everywhere). Then
\beq
\lap\Phx=-4\pi\rhx
\eeq
which has the integral solution
\beq
\Phx=\inivp\frac\rhxp\xxpa.
\eeq
Note in particular that the solution for a unit point charge is $\xxpi$ and
that this function is such that
\beq
\lap\lep\xxpi\rip=-4\pi\de\xxp.
\eeq

The meaning of $\Phx$ is that $q\Phx$ is the energy of interaction of $q$,
located at $\x$, with the charges that produce the potential. The energy of a
localized charge distribution can be written as
\beq
W=\frac1{8\pi}\iniv\Ex\cdot\Ex=\frac12\iniv\rhx\Phx.
\eeq
Define the electrostatic energy density as
\beq
w(\x)=\frac1{8\pi}\Ex\cdot\Ex\ge0.
\eeq
\bce
{\bf Solution of Boundary Value Problems}
\ece
We learned how to solve boundary value problems by a variety of methods
(For the actual boundary conditions, see the section on macrostatic
electrostatics).
\ben
\item Image method
\item Green's functions
\item Conformal maps (two-dimensional systems)
\item Orthogonal functions and expansions
\ben
\item Cartesian coordinates (Fourier series)
\item Spherical coordinates (Spherical harmonics, Legendre polynomials)
\item Cylindrical coordinates (Bessel functions, Fourier-Bessel series)
\een
\een
In particular, in spherical coordinates a general solution is given by
\beq
\Phx=\sum_{l=0}^\infty\sum_{m=-l}^l\lep A_lr^l+B_l/r^{l+1}\rip\ylm;
\eeq
this expansion is crafted in such a way as to provide a complete orthonormal
set of functions, the spherical harmonics, on a spherical surface.
In cylindrical coordinates, one may write
\beq
\Phx=\sum_{m=-\infty}^\infty\sum_{n=1}^\infty\lep A_{mn}\sinh(x_{mn}z/a)
+B_{mn}\cosh(x_{mn}z/a)\rip J_m(x_{mn}\rh/a)e^{im\ph}
\eeq
which is constucted so as to provide a complete orthogonal set of
functions, the Bessel functions and exponentials with imaginary argument,
on n a disc of radius $a$; $x_{mn}$ is the $n^{th}$ zero of the Bessel
function of order $m$. Alternatively, one may write
\beq
\Phx=\sum_{m=-\infty}^\infty\sum_{n=1}^\infty\lep A_{mn}I_m(n\pi\rh/c)
+B_{mn}K_m(n\pi\rh/c)\rip\sin(n\pi z/c)e^{im\ph}
\eeq
which provides a complete orthogonal set of functions in the form of
Fourier series on a surface at constant $\rh$ with $0\le z\le c$.

A useful formula (the addition theorem):
\beq
P_l(\cos\ga)=\frac{4\pi}{2l+1}\sum_{m=-l}^l\ylmsp\ylm
\eeq
where $\ga$ is the angle between $\x$ and $\xp$.\newline
\bce
{\bf Electric multipoles; Macroscopic Electrostatics}
\ece
The potential of a localized charge distribution can be expanded as
\beq
\Phx=\frac Qr+\frac{\edm\cdot\x}{r^3}+\frac12\sum_{i,j=1}^3\frac{x_iQ_{ij}
x_j}{r^5}+...
\eeq
where
\beq
Q=\inivp\rhxp,
\eeq
\beq
\edm=\inivp\xp\rhxp,
\eeq
and
\beq
Q_{ij}=\inivp(3x_i'x_j'-r'^2\de_{ij})\rhxp.
\eeq
The energy of a localized charge distribution ``centered'' at the origin
and subjected to an external applied field $\Ex$ is
\beq
W_e=Q\Ph(0)-\edm\cdot\E(0)-\frac16\sum_{i,j=1}^3Q_{ij}\left.\pde{E_j}{x_i}\ril_0
+...
\eeq
from which one can find the dipole-dipole interaction
\beq
W_{d-d}=\frac{\edm_1\cdot\edm_2-3(\edm_1\cdot\nn)(\edm_2\cdot\nn)}{r^3}.
\eeq
The electric field of a point dipole at the origin is
\beq
\E_d(\x)=\frac{3(\edm\cdot\nn)\nn-\edm}{r^3}-\frac{4\pi}3\edm\de(\x).
\eeq

In macroscopic electrostatics the sources are the macroscopic charge
density $\rhx$ and the polarization or electric dipole moment per unit
volume, $\Px$. The macroscopic electric field is written as $\Ex$ and one
defines the electric displacement $\Dx=\Ex+4\pi\Px$. The differential
equations for the fields are
\beq
\div\Dx=4\pi\rh\hbox{ and }\curl\Ex=0;
\eeq
the latter implies that there is a scalar potential for $\Ex$. There is a
real charge density associated with the polarization, $\rh_p=-\div\P$; at
a boundary between two materials, this has a surface-charge component,
$\si_p=-(\P_2-\P_1)\cdot\nn$ where $\nn$ is the unit normal to the boundary
pointing into material 2. Boundary conditions on the fields are
\beq
(\D_2-\D_1)\cdot\nn=4\pi\si\hbox{ and }\nn\times(\E_2-\E_1)=0.
\eeq
Here, $\si$ is the macroscopic surface-charge density; it does not include
the polarization surface-charge density. In order to solve for the fields, one
must have a constitutive relation involving (at least) two of $\D$, $\E$, and
$\P$. The simplest such relation, valid for uniform, linear, isotropic
materials, is $\P=\ch_e\E$ in which case one can write $\D=\ep\E$ with the
dielectric constant given by $\ep=1+4\pi\ch_e$.

The energy change accompanying an infinitesimal change $\de\Dx$ in the
displacement can be written quite generally as
\beq
\de W=\frac1{4\pi}\iniv\Ex\cdot\de\Dx;
\eeq
if the materials are linear, then one can further write
\beq
W=\frac1{8\pi}\iniv\Ex\cdot\Dx=\frac12\iniv\Phx\rhx.
\eeq
Finally, the force on a piece of dielectric can be determined from
energy-conservation-based arguments to be
\beq
F_\et=-\lep\pde W\et\rip_Q=\lep\pde W\et\rip_V.
\eeq
We also discussed the local field, the connection between the dielectric
constant and the molecular polarizability, and the physical origins of the
polarizability.

\bce
{\bf Magnetostatics}
\ece
The differential equations of magnetostatics are
\beq
\div\Bx=0\hbox{ and }\curl\Bx=\frac{4\pi}c\Jx;
\eeq
the magnetic induction $\Bx$ is defined by the torque produced on a test
dipole of moment $\mdm$ at point $\x$, $\N=\mdm\times\Bx$. In
magnetostatics, $\div\Jx=0$.

The integral forms of the preceding equations are
\beq
\inac\Bx\cdot\nn=0\hbox{ and }\invlc\cdot\Bx=\frac{4\pi}c\ina\Jx\cdot\nn=
\frac{4\pi}cI_S.
\eeq.
Given $\Jx$, the integral solution for $\Bx$ is
\beq
\Bx=\frac1c\inivp\frac{\Jxp\times\xxp}{\xxpa^3}.
\eeq

The force on a localized current distribution $\Jx$ produced by an applied
field $\Bx$ can be obtained by integrating over the force density which is
\beq
\f(\x)=\frac1c\Jx\times\Bx;
\eeq
the torque on the distributin is
\beq
\N=\frac1c\iniv[\x\times(\Jx\times\Bx)].
\eeq
Because the divergence of $\B$ is always zero, there is a vector potential
$\A$ such that $\B=\curl\A$. An integral solution for $\A$ is
\beq
\Ax=\frac1c\inivp\frac\Jxp\xxpa;
\eeq
the differential equation for $\A$ is
\beq
\lap\Ax-\grad(\div\Ax)=-\frac{4\pi}c\Jx.
\eeq
If one chooses $\A$ so that $\div\A=0$, which is always possible, then the
Cartesian components of $\A$ obey Poisson equations. More generally, one
can choose the divergence of the vector potential arbitrarily; doing so is
called choosing a gauge.\newline
\bce
{\bf Multipoles and Macroscopic Magnetostatics}
\ece
A localized current distribution may for many purposes be treated in the
dipole approximation. The magnetic dipole moment of the distribution is
defined as
\beq
\mdm=\frac1{2c}\iniv\x\times\Jx\equiv\iniv{\cal M}(\x)
\eeq
where we introduce also the magnetic moment density. In the dipole
approximation, the vector potential is
\beq
\A_d(\x)=\frac{\mdm\times\x}{r^3}
\eeq
and the field is
\beq
\B_d(\x)=\frac{3(\mdm\cdot)\nn-\mdm}{r^3}+\frac{8\pi}3\mdm\de(\x);
\eeq
the singular piece is strictly valid only for a point dipole.
The force on the distribution given an applied field $\Bx$ is, in this
approximation, $\F=\grad(\mdm\cdot\Bx)$; the torque is $\N=\mdm\times\B$.

The sources in macroscopic magnetostatics are the macroscopic or free
current density $\Jx$ and the magnetization or magnetic dipole moment
density $\Mx$ which comes from charges bound on molecules. The differential
field equations are
\beq
\div\Bx=0\hbox{ and }\curl\Hx=\frac{4\pi}c\Jx
\eeq
where the magnetic field is $\H=\B-4\pi\M$. For linear, isotropic, uniform
materials, $\M=\ch_m\H$ which gives $\B=\mu\H$; $\mu$ is called the
magnetic permeability. One introduces a magnetization current density
$J_M(\x)=c\,\curl\Mx$; at a boundary between two materials, there is a
surface-current density component in $J_M$; it is $K_M=\nn\times(\M_2-\M_1)
$ where $\nn$ is the unit normal into material 2 at the boundary.

The continuity conditions on the macroscopic fields are
\beq
(\B_2-\B_1)\cdot\nn=0\hbox{ and }\nn\times(\H_2-\H_1)=\frac{4\pi}c\K
\eeq
where $\K$ is the surface-current density at the boundary (it does not
include the magnetization surface-current density) and $\nn$ is as above.
\edo

