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\markright{{\sf Lecture 7 (Ch 3 part 1) Legendre functions, spherical harmonics}}
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\chapter{Ch. 3, part 1}

\bigskip
\bigskip

\ \textsc{Lecture 7 - Legendre Functions, Spherical Harmonics}

\newpage 

The Laplace equation can be separated in several coordinate systems, the
most important of which are: Cartesian $x,y,z$, circular cylindrical $\rho
,\theta ,z$ and spherical polar $r,\theta ,\phi $. Confocal elliptical
coordinate systems are also useful, but are not in Jackson's book. In all
these systems the one-dimensional equations to be solved are similar to the
familiar 
\begin{equation}
\frac{d^2f}{dx^2}+k^2f=0  \label{C}
\end{equation}
but usually more complicated. For instance, in polar coordinates we write a
solution of Laplace's equation as $Q(\phi )P(\theta )U(r)/r$ and we find
that the equation in the azimuthal angle $\phi $ is the simple one 
\begin{equation}
\frac{d^2Q}{d\phi ^2}+m^2Q=0  \label{A}
\end{equation}
and the radial equation is almost as simple: 
\[
\frac{d^2U}{dr^2}-\lambda ^2\frac U{r^2}=0 
\]
but the equation in the polar angle $\theta $ is more complicated. If there
is no $\phi $ dependence, i.e. if $m=0,$ it reduces to 
\begin{equation}
\frac 1{\sin \theta }\frac d{d\theta }\left( \sin \theta \,\frac{dP}{d\theta 
}\right) +\lambda ^2P=0  \label{P}
\end{equation}
The ``separation constant'' $\lambda ^2$ appears as an eigenvalue in the
one-dimensional equations, and is to be determined with the help of the
boundary conditions. The solutions of this and similar eigenvalue problems
share many of the properties of the well-known solutions of the simple
equation (\ref{C}), namely:

\begin{itemize}
\item  A discrete set of eigenvalues

\item  Series expansions in the appropriate variables

\item  Recursion relations

\item  Existence of a generating function

These properties can be used to compute

\item  The functions and their derivatives

\item  Normalization integrals

\item  Three-function integrals, and others
\end{itemize}

\section{Azimuthal eigenfunctions}

All the above properties are quite trivial in the case of the eigenfunctions
of eq. (\ref{A}) with the boundary condition of periodicity in $(0,2\pi ).$
The (unnormalized) solutions are 
\[
U_m(\phi )=\exp (im\phi ) 
\]

\begin{itemize}
\item  The eigenvalues are $m^2$ with $m$ integer

\item  The series expansion in $\phi $ is 
\[
U_m(\phi )=\sum_{n=0}^\infty \frac{\left( im\phi \right) ^n}{n!}
\]
but it is more convenient to use the variable $X=e^{i\phi },$ in terms of
which we simply have 
\[
U_m(X)=X^m
\]

\item  There is a very simple one-step recursion relation 
\[
U_m(X)=XU_{m-1}(X)
\]
\end{itemize}

and the derivative is also given by a simple step-down relation 
\[
dU_m/dX=mU_{m-1} 
\]

\begin{itemize}
\item  The series expansion 
\[
\frac 1{1-aX}=\sum_{m=0}^\infty a^mX^m=\sum_{m=0}^\infty a^mU_m(X)
\]
shows what is meant by a ``generating function''
\end{itemize}

As for the use of these relations: if we take $U_0=1,$we can trivially
iterate $U_m(X)=XU_{m-1}(X)$ to get $U_m(X)=X^m,$ and the derivative is
explicitly given. Normalization is trivial because $\int \left| U_m\right|
^2d\phi =\int d\phi =2\pi .$ The three-function integrals 
\[
\int U_lU_mU_nd\phi 
\]
are also computed directly in this case. They vanish unless $l+m+n=0,$ and
have then the value $2\pi .$

\subsubsection{Remark}

This case is overly simple, because the eigenfunctions obey in effect a
first-order differential equation. More typical is the case of eq.(\ref{C})
with zero boundary conditions at $x=\pm a/2$ (particle in a box). If we put $%
\phi =\pi x/a,$ this becomes eq. (\ref{A}) for $\phi $ in $(-\pi /2,\pi /2).$
Now the eigenfunctions are $U_m=\cos (m\phi )$ with positive half-integer $%
m. $ For any eigenvalue $m^2$ there is a ``second solution'' $\sin (m\phi )$
that does not satisfy the boundary conditions. The recursion relations are
now two-step, or involve derivatives. For instance 
\begin{eqnarray*}
\cos (m+1)\phi &=&\cos m\phi \cos \phi -\sin m\phi \sin \phi \\
&=&\cos m\phi \cos \phi -\frac{\sin ^2\phi }m\frac d{d\cos \phi }\cos m\phi
\end{eqnarray*}
The expansion of $\cos m\phi $ in the variable $X=\cos \phi $ is just a
polynomial of degree $m,$ the standard Chebyshev polynomial $T_m(X)$
(Arfken, page 741). This is just mentioned here as an example to make more
palatable the Legendre functions (coming up).

\section{Radial eigenfunctions}

The solutions of the equation 
\[
\frac{d^2U}{dr^2}-\lambda ^2\frac U{r^2}=0 
\]
are manifestly of two types: $r^{-l},$ which for $l>0$ is well behaved at $%
\infty $ but not at the origin, and $r^{l+1}$, which is well-behaved at the
origin but not at $\infty $ . In either case the eigenvalue is $\lambda
^2=l(l+1).$

\section{Polar eigenfunctions}

\subsection{Legendre functions}

The differential equation in $\theta $ is conveniently studied in the
variable $x=\cos \theta ,$ in the interval (-1,1). For $m=0$ it is the
Legendre equation: 
\[
\frac d{dx}\left[ \left( 1-x^2\right) \frac{dP}{dx}\right] +l\left(
l+1\right) P=0 
\]

\begin{itemize}
\item  The usual boundary conditions are that $P$ should remain finite at
the end points $x=1$ and $x=-1$ (corresponding to $\theta =0$ and $\theta
=\pi )$. They can be satisfied only if $l$ is a positive integer or zero.

\item  The coefficients $a_l$ of the series solution obey the recursion
relation 
\[
a_{j+2}=\frac{j(j+1)-l(l+1)}{\left( j+1\right) \left( j+2\right) }\,a_j
\]

\item  There are two solutions, one even in $x$, one odd.

\begin{itemize}
\item  If $l$ is an even integer or zero, the even solution reduces to the 
\emph{Legendre polynomial }of order $l$ and is well-behaved at $x=\pm 1$.
The ``other solution'' is not well behaved.

\item  If $l$ is an odd integer, the odd solution reduces to the \emph{%
Legendre polynomial }of order $l$ and is well-behaved at $x=\pm 1$. The
``other solution'' is not well behaved.
\end{itemize}

\item  The first few Legendre polynomials are 
\begin{eqnarray*}
P_0(x)=1 \\
P_1(x)=x \\
P_2(x)=\frac 12\left( 3x^2-1\right)  \\
P_3(x)=\frac 12\left( 5x^3-3x\right)  \\
P_4(x)=\frac 18\left( 35x^4-30x^2+3\right) 
\end{eqnarray*}

\item  To prove some properties of the $P_l$ (especially to evaluate
integrals) it is useful to know the \emph{Rodrigues formula} 
\[
P_l(x)=\frac 1{2^ll!}\frac{d^l}{dx^l}\left( x^2-1\right) ^l
\]

\item  Even more useful, especially in electrostatics, is the expansion (%
\emph{generating function}), valid for $\left| a\right| $ $<1,$%
\begin{equation}
\frac 1{\sqrt{1+a^2-2ax}}=\sum_{l=0}^\infty a^lP_l(x)  \label{G}
\end{equation}

If we go back to the variable $\theta $ and also put $a=r^{\prime }/r,$ this
becomes 
\[
\frac r{\sqrt{r^2+r^{\prime 2}-2rr^{\prime }\cos \theta }}=\sum_{l=0}^\infty
\left( \frac{r^{\prime }}r\right) ^lP_l(\cos \theta )
\]
or 
\[
\frac 1{\left| \mathbf{x-x}^{\prime }\right| }=\frac 1r\sum_{l=0}^\infty
\left( \frac{r^{\prime }}r\right) ^lP_l(\cos \theta )
\]
We recognize an expansion of the Coulomb potential, valid for $r^{\prime }<r$%
.

\item  There are many \emph{recursion relations}. The most useful are
probably

\begin{itemize}
\item  The step-up and step-down recursions, involving the derivative 
\[
\left( l+1\right) P_{l+1}=\left( l+1\right) xP_l+(x^2-1)P_l^{^{\prime }}
\]
\[
lP_{l-1}=lxP_l-(x^2-1)P_l^{^{\prime }}
\]

\item  The two-step recursion 
\[
\left( l+1\right) P_{l+1}=\left( 2l+1\right) xP_l-lP_{l-1}
\]
which is the most efficient numerical way to generate the $P_l$
\end{itemize}

\item  The orthogonality and normalization integrals are 
\[
\int_{-1}^1P_l(x)P_{l^{\prime }}(x)\,dx=\frac 2{2l+1}\,\delta _{ll^{\prime }}
\]

\item  The integrals 
\[
\int_{-1}^1P_l(x)P_m(x)P_n(x)\,dx
\]
are also useful, especially in atomic and nuclear physics. It can be shown
that they vanish unless

\begin{itemize}
\item  $l+m+n$ is even

\item  $l,m,n$ satisfy ``triangle'' inequalities, i.e. $\left| m-n\right|
\leq l\leq m+n$
\end{itemize}

We probably need them only for $n=1$ and $n=2$. See (J3.31-32).

\item  By writing $x^2-1=\left( x+1\right) \left( x-1\right) $ in the
Rodrigues formula it is easy to see that 
\begin{eqnarray*}
P_l(1)=1 \\
P_l(-1)=(-1)^l
\end{eqnarray*}
One also can find, using the binomial expansion of $\left( 1-x^2\right) ^l,$
that $P_l(0)$ vanishes for $l$ odd (obviously, by parity) and that for even $%
l$%
\[
P_l(0)=(-1)^l\frac{\left( 2l+1\right) !!}{2^ll!}
\]
\end{itemize}

\subsection{Linear multipoles}

We now run some checks of the above relations, starting from the generating
function expansion (\ref{G}), and describe the physical meaning of some of
the quantities.

We have already seen that for $r^{\prime }<r$%
\[
\frac 1{\left| \mathbf{x-x}^{\prime }\right| }=\frac 1r\sum_{l=0}^\infty
\left( \frac{r^{\prime }}r\right) ^lP_l(\cos \theta ) 
\]
Similarly, by putting $a=r/r^{\prime }$ in (\ref{G}), we obtain for $%
r<r^{\prime }$%
\[
\frac 1{\left| \mathbf{x-x}^{\prime }\right| }=\frac 1{r^{\prime
}}\sum_{l=0}^\infty \left( \frac r{r^{\prime }}\right) ^lP_l(\cos \theta ) 
\]
For $r=r^{\prime }$ both these expansions converge, except at $\theta =0,$
which of course corresponds to $\mathbf{x=x}^{\prime }.$ We note that, where
the series converges, both sides of these relations satisfy Laplace's
equation, because

\[
\frac 1{r^{l+1}}P_l(\cos \theta )\qquad \text{and \qquad }r^lP_l(\cos \theta
) 
\]
are the basic separable solutions when there is no dependence on $\phi .$

Generally, a $\phi $-independent solution of Laplace's equation in the whole
region \emph{exterior} to the sphere $r=r^{\prime }$ has the Legendre
expansion 
\[
\Phi (\mathbf{x)=}\frac 1r\sum_{l=0}^\infty C_l\left( \frac{r^{\prime }}%
r\right) ^lP_l(\cos \theta ) 
\]
and the coefficients $C_l$ can be found by equating the two sides for $%
\theta =0,$ where $P_l(1)=1,$ or for some other convenient $\theta .$ This
is a very useful trick. When $\Phi (\mathbf{x)=}1\mathbf{/}\left| \mathbf{x-x%
}^{\prime }\right| $ one quickly finds $C_l=1$ by this method and one can
also turn the argument around to obtain the values of $P_l(-1)$ and $P_l(0)$
already quoted.

The explicit expressions of $P_l(\cos \theta )/r^{l+1}$ for small $l$ are 
\begin{eqnarray*}
&&\frac 1r \\
\frac{\cos \theta }{r^2} &=&\frac z{r^3} \\
\frac 1{2r^3}\left( 3\cos ^2\theta -1\right) &=&\frac 1{2r^5}\left(
3z^2-r^2\right)
\end{eqnarray*}
We recognize the potential of a dipole oriented along the $z$ axis and of a
quadrupole, and we also note that 
\begin{eqnarray*}
\frac z{r^3} &=&-\frac \partial {\partial z}\frac 1r \\
\frac 1{r^5}\left( 3z^2-r^2\right) &=&-\frac \partial {\partial z}\frac
z{r^3}
\end{eqnarray*}
We can show that in general 
\begin{equation}
-\frac \partial {\partial z}\left[ \frac 1{r^{l+1}}P_l(\cos \theta )\right] =%
\frac{l+1}{r^{l+2}}P_{l+1}(\cos \theta )  \label{EM}
\end{equation}
so that the successive application of $-\partial /\partial z$ (at constant $%
x $ and $y$) generates the fields of \emph{linear multipoles} consisting of
charges aligned along $z.$ As already stated, these are the basic fields for
an exterior problem.

Similarly for an \emph{interior} problem the basic fields are $r^lP_l(\cos
\theta ).$ Their explicit expressions for small $l$ are 
\begin{eqnarray*}
&&1 \\
r\cos \theta &=&z \\
\frac 12r^2\left( 3\cos ^2\theta -1\right) &=&\frac 12\left( 3z^2-r^2\right)
\end{eqnarray*}
and we have the ``step-down'' relations 
\begin{equation}
\frac \partial {\partial z}\left[ r^lP_l(\cos \theta )\right]
=lr^{l-1}P_{l-1}(\cos \theta )  \label{IM}
\end{equation}

\subsection{Associated Legendre functions}

For any $m,$ the differential equation in $\theta $ is best studied using
the variable $x=\cos \theta ,$ in the interval $(-1,1)$. It is the
associated Legendre equation: 
\[
\frac d{dx}\left[ \left( 1-x^2\right) \frac{dP}{dx}\right] +l\left(
l+1\right) P-\frac{m^2}{1-x^2}\,P=0 
\]
We consider only the case of integer $m$, and we can take $m\geq 0.$ Then:

\begin{itemize}
\item  The usual boundary conditions are that $P$ should remain finite at
the end points $x=1$ and $x=-1$ (corresponding to $\theta =0$ and $\theta
=\pi $). They can be satisfied only if $l$ is a positive integer or zero,
and $m$ $\leq l$

\item  The solutions can be written as power series, but it is easier to
derive them from the Legendre functions $P_l(x)$ by the formula 
\[
P_l^m(x)=\left( -1\right) ^m\left( 1-x^2\right) ^{m/2}\frac{d^m}{dx^m}P_l(x)
\]

\item  For any $l,m$ pair there are two solutions, one even in $x$, one odd.
If $l$ is an integer, one solution is a polynomial of order $l-m$ in $x=\cos
\theta $ times the factor $\sin ^m\theta $. This solution, which has parity $%
l-m,$ is well-behaved at $x=\pm 1$. The ``other solution'' is not well
behaved.

\item  From the Rodriguez formula for $P_l$ one obtains at once the \emph{%
Rodrigues formula} for $P_l^m$ 
\[
P_l^m(x)=\frac{\left( -1\right) ^m}{2^ll!}\left( 1-x^2\right) ^{m/2}\frac{%
d^{l+m}}{dx^{l+m}}\left( x^2-1\right) ^l
\]

\item  Clearly $P_l^0(x)=P_l(x).$ Also, from the Rodrigues formula: 
\begin{eqnarray*}
P_l^l(\cos \theta )=\left( -1\right) ^l\frac{\left( 2l\right) !}{2^l\,l!}%
\sin ^l\theta  \\
P_l^{l-1}(\cos \theta )=\left( -1\right) ^{l-1}\frac{\left( 2l\right) !}{%
2^l\,l!}\sin ^l\theta \cos \theta =-\frac d{d\theta }P_l^l(\cos \theta )
\end{eqnarray*}

\item  The first few associated Legendre functions with $m>0$ are 
\begin{eqnarray*}
P_1^1(x)=-\left( 1-x^2\right) ^{1/2}=-\sin \theta  \\
P_2^1(x)=3x\left( 1-x^2\right) ^{1/2}=-3\cos \theta \,\sin \theta  \\
P_2^2(x)=3\left( 1-x^2\right) =3\sin ^2\theta  \\
P_3^1(x)=-\frac 32\left( 5x^2-1\right) \left( 1-x^2\right) ^{1/2}\ =-\frac
32\left( 5\cos ^2\theta -1\right) \sin \theta  \\
P_3^2(x)=15x\left( 1-x^2\right) =15\cos \theta \,\sin ^2\theta  \\
P_3^3(x)=-15\left( 1-x^2\right) ^{3/2}=-15\sin ^3\theta 
\end{eqnarray*}

\item  Also useful is the expansion (\emph{generating function}), valid for $%
\left| a\right| $ $<1,$%
\[
\frac{\left( 2m\right) !}{2^mm!}\,\frac{\left( 1-x^2\right) ^{m/2}}{\left(
1+a^2-2ax\right) ^{m+1/2}}=\sum_{l=0}^\infty a^lP_{l+m}^m(x)
\]

However, it does not have a direct physical meaning.

\item  There are very many \emph{recursion relations}. Recursions in $m$ are
considered more fully in the next section. We list here only

\begin{itemize}
\item  the two-step recursion in $l$, which is the efficient numerical way
to generate $P_{l+m}^m$ moving up from $P_m^m$ and $P_{m+1}^m$%
\[
\left( l-m+1\right) P_{l+1}^m=\left( 2l+1\right) xP_l^m-(l+m)P_{l-1}^m
\]

\item  the two-step recursion in $m$, which can generate $P_l^m$ moving down
from $P_l^l$ and $P_l^{l-1}$%
\[
P_l^{m+1}+\frac{2mx}{\sqrt{1-x^2}}P_l^m+\left[ l(l+1)-m(m+1)\right]
P_l^{m-1}=0
\]
\end{itemize}

\item  The orthogonality and normalization integrals are 
\[
\int_{-1}^1P_l^m(x)P_{l^{\prime }}^m(x)\,dx=\frac 2{2l+1}\,\frac{\left(
l+m\right) !}{\left( l-m\right) !}\,\delta _{ll^{\prime }}
\]
Three-function integrals are considered in the next section.

\item  Associated Legendre functions for a negative upper index are given by 
\[
P_l^{-m}(x)=\left( -1\right) ^m\frac{\left( l+m\right) !}{\left( l-m\right) !%
}P_l^m(x)
\]

With this definition, the Rodrigues formula, recursion relations, and
normalization integral are valid for all integer values of $m.$ Note that we
have not introduced new eigenfunctions, simply relabeled the ones we had.
Why we want to do this is discussed next.
\end{itemize}

\subsection{Spherical harmonics}

It is often better to deal with the $\theta $ and $\phi $ dependence at the
same time by introducing the normalized eigenfunctions 
\[
Y_{lm}(\theta ,\phi )=\sqrt{\frac{2l+1}{4\pi }\frac{\left( l-m\right) !}{%
\left( l+m\right) !}\,}P_l^m(\cos \theta )\,e^{im\phi } 
\]
Note that $Y_{lm}$ and $Y_{l,-m}$ are different eigenfunctions, related by 
\[
Y_{l,-m}=\left( -1\right) ^mY_{l,m}^{*} 
\]
These \emph{spherical harmonics} form an orthogonal and complete set. The
first few should be memorized, at least in their $\theta ,\phi $ dependence.

\subsubsection{Multipole fields and ``ladder'' operations}

It is actually easiest to memorize the basic interior solutions, $%
r^lY_{l,m}(\theta ,\phi ),$ as a function of the Cartesian coordinates $%
x,y,z.\,$ For $l=0$%
\[
\sqrt{4\pi }\,Y_{00}=1 
\]
For $l=1$%
\[
-\sqrt{4\pi }\,rY_{1,1}=\sqrt{\frac 32}\,r\sin \theta \,e^{i\phi }=\sqrt{%
\frac 32}\,(x+iy) 
\]
\[
\sqrt{4\pi }\,rY_{10}=\sqrt{3}r\cos \theta =\sqrt{3}\,z 
\]
For $l=2$%
\[
\sqrt{4\pi }\,r^2Y_{22}=\sqrt{\frac{15}8}\,r^2\sin ^2\theta \,e^{2i\phi }=%
\sqrt{\frac{15}8}\,\left( x+iy\right) ^2 
\]
\[
-\sqrt{4\pi }\,r^2Y_{21}=\sqrt{\frac{15}2}\,r^2\sin \theta \cos \theta
\,e^{i\phi }=\sqrt{\frac{15}2}\,\left( x+iy\right) z 
\]
\[
\sqrt{4\pi }\,r^2Y_{20}=\sqrt{\frac 54}\,r^2(3\cos ^2\theta \,-1)=\sqrt{%
\frac 54}\,\left( 3z^2-r^2\right) 
\]
For $l=3$%
\[
\,r^3Y_{33}\approx (x+iy)^3 
\]
\[
r^3\,Y_{32}\approx (x+iy)^2z 
\]
\[
\,r^3\,Y_{31}\approx (x+iy)\left( 5z^2-r^2\right) 
\]
\[
r^3\,Y_{30}\approx z(5z^2-3r^2) 
\]
A look at these expressions suggests that it is convenient to introduce 
\begin{eqnarray*}
x_{+} &=&x+iy \\
x_{-} &=&x-iy
\end{eqnarray*}
with the inverse transformation 
\begin{eqnarray*}
x &=&\frac 12\,(x_{+}+x_{-}) \\
y &=&\frac 1{2i}\,(x_{+}-x_{-})
\end{eqnarray*}
which gives 
\begin{eqnarray*}
\frac \partial {\partial x_{+}} &=&\frac 12\left( \frac \partial {\partial
x}-i\,\frac \partial {\partial y}\right) \\
\frac \partial {\partial x_{-}} &=&\frac 12\left( \frac \partial {\partial
x}+i\,\frac \partial {\partial y}\right)
\end{eqnarray*}
Then apart from factors, when applied to $r^lY_{l,m}(\theta ,\phi )$:

\begin{itemize}
\item  $\dfrac \partial {\partial z}\,$ steps down from $l,m\,$ to $l-1,m$

\item  $\dfrac \partial {\partial x_{+}}\,$ steps down from $l,m\,$ to $%
l-1,m-1$

\item  $\dfrac \partial {\partial x_{-}}\,$ steps from $l,m$ to $l-1,m+1$

\item  $z\dfrac \partial {\partial x_{+}}-x_{-}\dfrac \partial {\partial z}$
steps down from $l,m\,$ to $l,m-1$

\item  $x_{+}\dfrac \partial {\partial z}-z\dfrac \partial {\partial x_{-}}$
steps up from $l,m\,$ to $l,m+1$
\end{itemize}

As we did earlier in the case $m=0,$ we can also consider the multipole
fields $Y_{l,m}(\theta ,\phi )/r^{l+1},$ which are the basic exterior
solutions. The ``ladder'' operations for these fields are

\begin{itemize}
\item  $\dfrac \partial {\partial z}\,$ steps up from $l,m\,$ to $l+1,m$

\item  $\dfrac \partial {\partial x_{+}}\,$ steps from $l,m\,$ to $l+1,m-1$

\item  $\dfrac \partial {\partial x_{-}}\,$ steps up from $l,m\,$ to $l+1,m+1
$

\item  The other two operations, which are very familiar in the quantum
theory of angular momentum, have the same effect as before.
\end{itemize}

The integrals of three spherical harmonics 
\[
\int Y_{lm}Y_{l^{\prime }m^{\prime }}Y_{l"m"}\,d\Omega 
\]
are often useful, especially in radiation problems (next semester). It can
be shown that they vanish unless

\begin{itemize}
\begin{itemize}
\item  $m+m^{\prime }+m^{\prime \prime }=0$ and $l+l^{\prime }+l^{\prime
\prime }$ is even

\item  $l,l^{\prime },l^{\prime \prime }$ satisfy ``triangle'' inequalities,
i.e. $\left| l-l^{\prime }\right| \leq l^{\prime \prime }\leq l+l^{\prime }$
\end{itemize}
\end{itemize}

\subsection{The addition theorem}

This is a generalization of the formula 
\[
\cos \gamma =\cos \theta \cos \theta ^{\prime }+\sin \theta \sin \theta
^{\prime }\cos \left( \phi -\phi ^{\prime }\right) 
\]
for the angle $\gamma $ between the vectors in the directions of $\mathbf{x}$
and $\mathbf{x}^{\prime }.$ It is 
\[
P_l(\cos \gamma )=\frac{4\pi }{2l+1}\sum_{m=-l}^lY_{lm}^{*}(\theta ^{\prime
},\phi ^{\prime })Y_{lm}(\theta ,\phi ) 
\]

When inserted in the generating function formula it gives the extremely
useful expansion 
\[
\frac 1{\left| \mathbf{x-x}^{\prime }\right| }=\sum_{l=0}^\infty
\sum_{m=-l}^l\ \frac{4\pi }{2l+1}\frac{r_{<}^l}{r_{>}^{l+1}}%
Y_{lm}^{*}(\theta ^{\prime },\phi ^{\prime })Y_{lm}(\theta ,\phi ) 
\]
where $r_{>}$ is the larger of $r,r^{\prime }.$

\section{Fields in a hole or near a point}

The field is very strong near a sharp conducting point and does not
penetrate well inside holes, with a power-law behavior similar to that near
edges and inside corners. Near edges and in corners (2-d) we found $\Phi
\propto \rho ^\alpha ,$ $E$ $\propto \rho ^{\alpha -1}$ with $\alpha =\pi
/\theta _0$; near conical points and in conical holes (3-d) one finds $\Phi
\propto r^\nu ,$ $E$ $\propto r^{\nu -1}$, but the relation between $\nu $
and the cone angle $\beta $ (see Fig. 3.5) is not as simple as $\nu =\pi
/2\beta ,$ where $2\beta $ in 3-d is the same as $\theta _0$ in 2-d.
However, for a hole $(0<\beta <\pi /2)$ there is the excellent approximation
(see Fig. 3.6) 
\[
v\simeq \frac{x_{01}}\beta -\frac 12-c_0\,\beta 
\]
where $x_{01}$ is the first zero of the Bessel function $J_0(x)$ and $c_0$
is chosen so that $\nu =1$ for a flat surface ($\beta =\pi /2$), $%
c_0=4x_{01}/\pi ^2-3/\pi .\,$ Numerically, $x_{01}\simeq 2.405$ and $%
c_0\simeq $ $0.01978.$ Actually, this approximation is also good for a point
that is not too sharp and fails only when $\pi -\beta <10\degree ,$ in which
case 
\[
\nu \simeq \left( 2\ln \frac 2{\pi -\beta }\right) ^{-1} 
\]
To obtain these results one must construct solutions of the Legendre
equation that are regular at the North pole ($\theta =0$) and are not
necessarily regular at the South pole ($\theta =\pi ).$ These solutions are
known as the \emph{Legendre functions} $P_\nu (x),$ where $x=\cos \theta $:
they are conventionally normalized by setting $P_\nu (1)=1$ and reduce to
the Legendre polynomials for integer $\nu .$ For a given $\beta ,$ one picks 
$\nu $ so that $\cos \beta $ is the first zero of $P_\nu (x)$. The dominant
behavior near the apex is then 
\[
\Phi =A_\nu r^\nu P_\nu (\cos \theta ) 
\]
and the field components are given by (J3.46). There is a simple expansion
of $P_\nu $ in powers of the variable $(1-x)/2=\sin ^2(\theta /2)$ that is
quite useful for the fields in shallow holes ($\nu $ near 1). The recursion
relations for the Legendre polynomials are also valid for the Legendre
functions, so in practice we need only tables of $P_\nu (x)$ for $0<\nu <1.$

\section{Other Legendre functions}

Jackson confines himself, wisely, to problems defined on the whole spherical
surface or on a spherical cap, with azimuthal symmetry. The most general
case of separable boundary value problem on the sphere involves the
functions $P_\nu ^\mu $ for non-integer $\mu $ and $\nu ,$ if the North pole
is included in the domain, and in addition the ``second'' Legendre functions 
$Q_\nu ^\mu $ if the North pole is excluded. For instance, problems on an
equatorial belt involve $P_\nu ^m$ and $Q_\nu ^m$ for $m$ integer, problems
in a watermelon slice involve $P_n^\mu $ for $n$ integer.

\end{document}
