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\markright{{\sf EM1 spring 96, Assignment 6, Problem 3}}
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\begin{document}

\chapter{Jackson problem 5.10}

\section{(a) Loop parallel to interface}

The force on a loop parallel to the interface is conveniently given by 
\begin{equation}
F_z=-\frac{2\pi aI}c\,B_\rho (a,2d)  \label{1}
\end{equation}
where the $z$ axis is perpendicular to the interface and $B_\rho $ is a
component of the field of the image loop. Using Jackson's (5.36), the vector
potential of the image loop is 
\[
A_\phi (\rho ,z)=\frac{I^{\prime }a}c\int_0^{2\pi }\frac{\cos \alpha
^{\prime }\;d\alpha ^{\prime }}{\sqrt{a^2+\rho ^2+z^2-2a\rho \cos \alpha
^{\prime }}} 
\]
where 
\[
I^{\prime }=\frac{\mu -1}{\mu +1}\,I 
\]
and 
\[
B_\rho (\rho ,z)=-\frac \partial {\partial z}\,A_\phi (\rho ,z)=\frac{%
I^{\prime }a}c\int_0^{2\pi }\frac{z\cos \alpha ^{\prime }\;d\alpha ^{\prime }%
}{\left( a^2+\rho ^2+z^2-2a\rho \cos \alpha ^{\prime }\right) ^{3/2}} 
\]
Thus $F_z$ is obtained directly as a single well-behaved integral, which can
be expressed as a hypergeometric function or as a sum of elliptic integrals
(see below). One can also obtain $F_z$ as a single integral over the product
of two Bessel functions, but such integrals are not easy to evaluate
numerically or to expand at large separations. Working with spherical
harmonics, it is possible to obtain for $F_z$ a power series in $a/\sqrt{%
a^2+4d^2}$ that is very useful for large separations. This is probably the
answer that Jackson wants (see below).

\subsubsection{Using elliptic integrals}

$B_\rho (\rho ,z)$ is given on page 11 of Lecture 12 in terms of elliptic
integrals: 
\[
B_\rho (\rho ,z)=\frac{2Iz}{c\rho \sqrt{(a+\rho )^2+z^2}}\left[ \frac{%
a^2+\rho ^2+z^2}{(a-\rho )^2+z^2}E(k^2)-K(k^2)\right] 
\]
with 
\[
k^2=\frac{4a\rho }{(a+\rho )^2+z^2} 
\]
Hence 
\begin{equation}
F_z=-\,\frac{\mu -1}{\mu +1}\,\frac{4\pi dI^2}{c^2\sqrt{a^2+d^2}}\left[ 
\frac{a^2+2d^2}{2d^2}\,E\left( \frac{a^2}{a^2+d^2}\right) -K\left( \frac{a^2%
}{a^2+d^2}\right) \right]  \label{2}
\end{equation}
For $a<<d,$ we can use the expansions (17.3.11 and 17.3.12 in Abramowitz and
Stegun) 
\begin{eqnarray*}
K(\alpha ) &=&\frac \pi 2\,\left( 1+\frac 14\,\alpha +\frac 9{64}\,\alpha
^2+\dots \right) \\
E(\alpha ) &=&\frac \pi 2\,\left( 1-\frac 14\,\alpha -\frac 3{64}\,\alpha
^2+\dots \right)
\end{eqnarray*}
and the square parenthesis in eq. (\ref{2}) reduces to 
\[
\frac \pi 2\left[ \left( 1+\frac{a^2}{2d^2}\right) \left( 1-\frac 14\,\frac{%
a^2}{a^2+d^2}-\frac 3{64}\,\frac{a^4}{d^4}\right) -1-\frac 14\,\frac{a^2}{%
a^2+d^2}-\frac 9{64}\,\frac{a^4}{d^4}\right] 
\]
which to leading order gives 
\[
\frac \pi 2\frac 3{16}\frac{a^4}{d^4} 
\]
so that, in agreement with part (c) here below, 
\begin{equation}
F_z\simeq -\frac{\mu -1}{\mu +1}\,\frac{3\pi ^2I^2}{8c^2}\frac{a^4}{d^4}
\label{3}
\end{equation}

\subsubsection{Using spherical harmonics}

$B_r$ and $B_\theta $ are given by Jackson in (5.48) and (5.49). Then 
\[
B_\rho =B_r\sin \theta +B_\theta \cos \theta 
\]
where in our case 
\[
\sin \theta =\frac a{\sqrt{a^2+4d^2}}\quad \quad \quad \quad \cos \theta =%
\frac{2d}{\sqrt{a^2+4d^2}} 
\]
and one obtains 
\begin{eqnarray}
F_z &=&-\frac{\mu -1}{\mu +1}\,\frac{2\pi ^2I^2}{c^2}\,\,\sum_{n=0}^\infty 
\frac{(-1)^n(2n+1)!!}{2^nn!}\left( \sin \theta \,\right) ^{2n+3}  \label{4}
\\
&&\times \left[ 2\sin \theta \,P_{2n+1}(\cos \theta )-\frac{\cos \theta }{n+1%
}\,P_{2n+1}^1(\cos \theta )\right]  \nonumber
\end{eqnarray}
For $a<<d,$ which is the same as $\theta <<1,$ the important term is $n=0.$
For this term 
\[
P_1=\cos \theta \quad \quad \quad \;P_1^1=-\sin \theta 
\]
and the square bracket in eq. (\ref{4}) reduces to $3\sin \theta $. Hence we
find again the result (\ref{3}).

\section{(b) Loop perpendicular to interface}

The force on the loop perpendicular to the interface can be computed
directly, but leads to a double integral, one over the loop and one over its
image. One of the integrals can be expressed as a sum of elliptic integrals,
but the other remains. Working with Bessel functions one can obtain $F$ as a
single integral over the product of three Bessel functions, not easy to
evaluate numerically or to expand at large separations. Working with
spherical harmonics, it is possible to obtain a straight power series in $%
a/d $ that is very useful for large separations. This is probably the answer
that Jackson wants.

\subsubsection{Using spherical harmonics}

Put the loop at $\frac 12\mathbf{R+a},$ where $\mathbf{R}=(2d,0,0)$ and $%
\mathbf{a}=a(\cos \alpha ,\,\sin \alpha ,\,0),$ and put the image loop at $%
\,-\frac 12\mathbf{R+a}^{\prime },$ where $\mathbf{a}^{\prime }=a(\cos
\alpha ^{\prime },\,\sin \alpha ^{\prime },\,0).$ The force on the loop is 
\[
\mathbf{F}=-II^{\prime }\int d\mathbf{l}\cdot d\mathbf{l}^{\prime }\,\frac{%
\mathbf{R+a-a}^{\prime }}{\left| \mathbf{R+a-a}^{\prime }\right| ^3}%
=II^{\prime }\nabla _{\mathbf{R}}M(R\mathbf{)}
\]
where in our case 
\[
I^{\prime }=-I\,\frac{\mu -1}{\mu +1}
\]
and $M(R\mathbf{)}$ is the mutual inductance 
\[
M(R)=\int \,\frac{d\mathbf{l}\cdot d\mathbf{l}^{\prime }}{\left| \mathbf{%
R+a-a}^{\prime }\right| }
\]
Since $d\mathbf{l}$ is perpendicular to $\mathbf{a},$ and $d\mathbf{l}%
^{\prime }$ to $\mathbf{a}^{\prime },$ we have $d\mathbf{l}\cdot d\mathbf{l}%
^{\prime }=a^2\cos (\alpha -\alpha ^{\prime })d\alpha \,d\alpha ^{\prime }.$
It is convenient to put $\mathbf{a}^{\prime }\mathbf{-a}=\rho (\cos \phi
,\,\sin \phi ,\,0)$ and to expand 
\[
\frac 1{\left| \mathbf{R+a-a}^{\prime }\right| }=\sum_{l=0}^\infty
\sum_{m=0}^l\frac{\rho ^l}{R^{l+1}}\frac{(l-m)!}{(l+m)!}\left[
P_l^m(0)\right] ^2e^{im\phi }
\]
We can evaluate $\rho ^le^{im\phi }$ with the help of 
\[
\rho \cos \phi =a(\cos \alpha ^{\prime }-\cos \alpha )\quad \quad \quad \rho
\sin \phi =a(\sin \alpha ^{\prime }-\sin \alpha )
\]
These equations give 
\[
\rho e^{i\phi }=a\left( e^{i\alpha ^{\prime }}-e^{i\alpha }\right)
=-2iae^{i(\alpha ^{\prime }+\alpha )/2}\sin \beta 
\]
\[
\rho =2a\,\sin \beta 
\]
with $\beta =\frac 12\left( \alpha -\alpha ^{\prime }\right) ,$ when $0\leq
\beta \leq \pi .$ It follows that 
\[
\rho ^{l-m}\left( \rho e^{i\phi }\right) ^m=\left( -i\right) ^me^{im(\alpha
+\alpha ^{\prime })/2}\left( 2a\sin \beta \right) ^l
\]
and 
\[
M(R)=\sum_{l=0}^\infty \sum_{m=0}^l\frac{a^{l+2}}{R^{l+1}}\frac{(l-m)!}{%
(l+m)!}\left[ P_l^m(0)\right] ^2C_{lm}
\]
where 
\[
C_{lm}=\left( -i\right) ^m\int e^{im\alpha ^{\prime }}e^{im\beta }(2\sin
\beta )^l\,\cos 2\beta \,\,d\alpha \,d\alpha ^{\prime }
\]
Because the integrand is periodic, we can replace the integral over $\alpha $
with an integral over $\beta $ from 0 to $\pi $. The result is that $%
C_{lm}=\delta _{m0}C_{l0}$ with 
\[
C_{l0}=4\pi \int_0^\pi \left( 2\sin \beta \right) ^l\,(1-2\sin ^2\beta
)\,d\beta 
\]
At this point it is convenient to recall that $P_l^m(0)$ vanishes for odd $l,
$ so that for even $l$ we can write 
\[
C_{l0}=2\pi \int_0^{2\pi }\left[ \left( 2\sin \beta \right) ^l-\frac
12\,\left( 2\sin \beta \right) ^{l+2}\right] \,d\beta 
\]
and we can use 
\[
\int_0^{2\pi }\left( 2\sin \beta \right) ^{2n}\,d\beta =2\pi \binom{2n}%
n=2\pi \frac{\left( 2n\right) !}{\left( n!\right) ^2}
\]
to obtain $C_{00}=0$ and 
\[
C_{2n,0}=-4\pi ^2\frac{\left( 2n\right) !}{\left( n!\right) ^2}\frac
n{(n+1)}=-4\pi ^2\frac{\left( 2n\right) !}{(n-1)!(n+1)!}
\]
for $n>0.$ We have arrived at the desired result 
\[
M(R)=-4\pi ^2\sum_{n=1}^\infty \frac{a^{2n+2}}{R^{2n+1}}\left[
P_{2n}(0)\right] ^2\frac{\left( 2n\right) !}{(n-1)!(n+1)!}
\]
where we recall that 
\[
\left| P_{2n}(0)\right| =\frac{\left( 2n-1\right) !!}{2^nn!}=\frac{\left(
2n-1\right) !}{2^nn!\,2^{n-1}(n-1)!}=\frac{\left( 2n\right) !}{2^{2n}\left(
n!\right) ^2}=\left( \frac 12\right) ^{2n}\binom{2n}n
\]
and in partcular $P_{2n}(0)=\frac 12\,.$ Then $F=II^{\prime }\partial
M/\partial R$ gives 
\[
F=-4\pi ^2I^2\,\frac{\mu -1}{\mu +1}\sum_{n=1}^\infty \frac{a^{2n+2}}{\left(
2d\right) ^{2n+2}}\left[ P_{2n}(0)\right] ^2\frac{\left( 2n+1\right) !}{%
(n-1)!(n+1)!}
\]
and for $d>>a,$ in agreement with part (c) here below, 
\[
F\simeq -\frac{\mu -1}{\mu +1}\,\frac{3\pi ^2I^2}{16c^2}\frac{a^4}{d^4}
\]

\section{(c) The limit of large $d/a$}

For large $d/a,$ the loop and its image can be regarded as dipoles. The
force between two dipoles is $F=-\partial U/\partial r,$ where 
\[
U=\frac{\mathbf{m\cdot m}^{\prime }-3(\mathbf{m\cdot n})(\mathbf{m}^{\prime }%
\mathbf{\cdot n)}}{r^3}
\]
and $\mathbf{n}$ is the unit vector along the line joining them. We must
differentiate with respect to $r$ first, then put $r=2d.$ In our problem $%
\mathbf{n}$ is along the $z$ axis and 
\[
m=\frac{\pi Ia^2}c\quad \quad \quad m^{\prime }=\frac{\pi Ia^2}c\,\frac{\mu
-1}{\mu +1}
\]
In case (a) the dipoles are parallel to each other and to $\mathbf{n}$,
hence $U=-2mm^{\prime }/r^3.$ 
\[
F=-\,\frac{6mm^{\prime }}{r^4}=-\frac{3\pi ^2I^2}{8c^2}\,\frac{a^4}{d^4}%
\frac{\mu -1}{\mu +1}
\]
In case (b) the dipoles are opposite to each other and perpendicular to $%
\mathbf{n}$, hence $U=-mm^{\prime }/r^3$ and the force is half as large as
in case (a), $F=-3mm^{\prime }/(2d)^4.$

\newpage\ 

\subsubsection{Using Bessel functions}

\begin{equation}
\frac 1{\sqrt{\rho ^2+a^2+z^2-2a\rho \cos (\phi -\alpha ^{\prime })}%
}=\sum_{m=-\infty }^\infty e^{im(\phi -\alpha ^{\prime })}\int_0^\infty
dk\,e^{-k|z|}J_m(ka)J_m(k\rho )  \label{G3}
\end{equation}
which is a particular case of the Green function formula given by Jackson in
Problem 3.14(b). Using 
\[
\int_0^{2\pi }\cos (\alpha -\alpha ^{\prime })e^{im(\phi -\alpha ^{\prime
})}\,d\alpha ^{\prime }=\pi \left[ \delta _{m1}e^{i(\phi -\alpha )}+\delta
_{m,-1}e^{-i(\phi -\alpha )}\right] 
\]
we find 
\[
M(d)=2\pi a^2\int_0^{2\pi }d\alpha \,\cos (\phi -\alpha )\int_0^\infty
dkJ_1(ka)J_1(k\rho ) 
\]
We must now perform the integral over $\alpha ,$ keeping in mind that 
\[
\rho \cos \phi =d+a\cos \alpha \quad \quad \quad \quad \rho \sin \phi =a\sin
\alpha 
\]
\[
\cos (\phi -\alpha )=\frac 1\rho \,(a+d\cos \alpha ) 
\]
For this purpose we use another formula from problem 3.13: 
\begin{equation}
J_0(k\rho )=J_0(k\sqrt{d^2+a^2+2ad\cos \alpha })=\sum_{m=-\infty }^\infty
e^{im(\pi -\alpha )}J_m(ka)J_m(kd)  \label{J0}
\end{equation}
which gives by differentiation 
\[
\frac{a+d\cos \alpha }\rho J_1(k\rho )=-\sum_{m=-\infty }^\infty e^{im(\pi
-\alpha )}J_m^{\prime }(ka)J_m(kd) 
\]
We obtain 
\[
M(d)=-2\pi a^2\int_0^{2\pi }d\alpha \sum_{m=-\infty }^\infty e^{im(\pi
-\alpha )}\int_0^\infty dk\,J_1(ka)J_m^{\prime }(ka)J_m(kd) 
\]
and finally 
\[
M(d)=4\pi ^2a^2\int_0^\infty dk\left[ J_1(ka)\right] ^2J_0(kd) 
\]
\[
F=II^{\prime }\frac{\partial M}{\partial d}=-4\pi ^2a^2II^{\prime
}\int_0^\infty kdk\left[ J_1(ka)\right] ^2J_1(kd) 
\]
\[
\frac{ad\sin \alpha }\rho J_1(k\rho )=-\sum_{m=-\infty }^\infty ime^{im(\pi
-\alpha )}J_m(ka)J_m^{\prime }(kd) 
\]
It is convenient to perform first an integration by parts, using 
\[
-\frac \partial {\partial k}J_0(ka)=aJ_1(ka)\quad \quad \quad \quad \frac
\partial {\partial k}J_1(k\rho )=\frac \rho 2\left[ J_0(k\rho )-J_2(k\rho
)\right] 
\]
to obtain 
\[
\int_0^\infty dkJ_1(ka)J_1(k\rho )=\frac \rho {2a}\int_0^\infty
dkJ_0(ka)\left[ J_0(k\rho )-J_2(k\rho )\right] 
\]
\[
\frac 1{\left| \mathbf{R+a-a}^{\prime }\right| }=\sum_{l=0}^\infty
\sum_{m=-l}^le^{im(\phi -\alpha ^{\prime })}P_l^m(0)P_l^m(0)\,\frac a\rho 
\]
First, we need the result of Problem 5.4(b) for the vector potential of a
loop of radius $a$%
\[
A_\phi (\rho ,z)=\frac{2\pi Ia}c\int_0^\infty dk\,e^{-k|z|}J_1(ka)J_1(k\rho
) 
\]
This follows from Jackson's eq. (5.36) 
\[
A_\phi (\rho ,z)=\frac{Ia}c\int_0^{2\pi }\frac{\cos \phi \,\,d\phi }{\sqrt{%
\rho ^2+a^2+z^2-2a\rho \cos \phi }} 
\]
by using the Green function formula from Problem 3.14 
\begin{equation}
\frac 1{\sqrt{\rho ^2+a^2+z^2-2a\rho \cos \phi }}=\sum_{m=-\infty }^\infty
e^{im\phi }\int_0^\infty dk\,e^{-k|z|}J_m(ka)J_m(k\rho )  \label{G3}
\end{equation}

Second, we need a formula for $J_1(k\rho )$ that is the analog of the
formula 
\begin{equation}
J_0(k\sqrt{d^2+a^2-2ad\cos \phi })=\sum_{m=-\infty }^\infty e^{im\phi
}J_m(ka)J_m(kd)  \label{J0}
\end{equation}
again from Problem 3.14. To prove this formula, we follow three steps:

(i) Put $a=0$ in (\ref{G3}) and obtain 
\[
\frac 1{\sqrt{\rho ^2+z^2}}=\int_0^\infty dk\,e^{-k|z|}J_0(k\rho ) 
\]

(ii) Put $\rho ^2=d^2+a^2-2ad\cos \phi $ in the above and obtain 
\[
\frac 1{\sqrt{d^2+a^2-2ad\cos \phi +z^2}}=\int_0^\infty dk\,e^{-k|z|}J_0(k%
\sqrt{d^2+a^2-2ad\cos \phi }) 
\]

(iii) Comparing this result with eq. (\ref{G3}) for $\rho =d$, obtain the
Laplace transform of (\ref{J0}).

\noindent Now do the following:

(i) Take $\partial /\partial a$ of (\ref{G3}) and then put $a=0.$ Using $%
J_1^{\prime }(0)=\frac 12$ and $J_{-1}=-J_1$obtain 
\[
\frac \rho {\left( \rho ^2+z^2\right) ^{3/2}}=\int_0^\infty
kdk\,e^{-k|z|}J_1(k\rho ) 
\]

(ii) Put $\rho ^2=d^2+a^2-2ad\cos \phi $ in the above and obtain 
\[
\frac{\sqrt{d^2+a^2-2ad\cos \phi }}{\left( d^2+a^2+z^2-2ad\cos \phi \right)
^{3/2}}=\int_0^\infty kdk\,e^{-k|z|}J_1(k\sqrt{d^2+a^2-2ad\cos \phi }) 
\]

(iii) Compare this result with the derivative of eq. (\ref{G3}) for $\rho =d$%
, obtain the Laplace transform of (\ref{J0}).

we can also write 
\[
M(d)=-4\pi ^2\sum_{n=1}^\infty \frac{a^{2n+2}}{d^{2n+1}}\left|
P_{2n}(0)\right| ^3\frac n{n+1}
\]

\end{document}
