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Next: Recurrence formulae for Rn,m Sn,m Up: Applications of Fast Multipole Previous: Series expansion of

   
Relations between solid harmonics Rn,m and Sn,m

Noting the following formulae (See Hobson[43] or Wallace[80]):
 
$\displaystyle S_{n,m}(\overrightarrow{O\mbox{\boldmath$ x $ }}) = (-1)^{n}\part...
...c{1}{\vert\overrightarrow{O\mbox{\boldmath$ x $ }}\vert}\right)\quad (m \ge 0),$     (B.1)


 
$\displaystyle S_{n,-m}(\overrightarrow{O\mbox{\boldmath$ x $ }}) = (-1)^m \over...
...c{1}{\vert\overrightarrow{O\mbox{\boldmath$ x $ }}\vert}
\right) \quad (m > 0),$     (B.2)


 
$\displaystyle \partial_{x\pm} = \left(\frac{\partial}{\partial x_1}
\pm i \frac{\partial}{\partial x_2} \right),$     (B.3)


 
$\displaystyle \partial_{x+}\partial_{x-} (\mbox{Harmonics})
= - \partial^2_{x_3} (\mbox{Harmonics})$     (B.4)

one can transform $S_{n,m}(\overrightarrow{O\mbox{\boldmath$\space x $ }})$ as follows:

\begin{eqnarray*}S_{n,m}(\overrightarrow{O\mbox{\boldmath$ x $ }}) &=& (-1)^{n}\...
...(\overrightarrow{\mbox{\boldmath$ y $ }\mbox{\boldmath$ x $ }}),
\end{eqnarray*}


where we have used (A.5) and the chain rule:

\begin{eqnarray*}\frac{\partial}{\partial X_i} = \frac{\partial}{\partial x_k}
\frac{\partial x_k}{\partial X_i} = \frac{\partial}{\partial x_i} .
\end{eqnarray*}


The underlined part (a) can be rewritten as follows:
In the case where $ m \ge m'$

\begin{eqnarray*}&& R_{n',m'}(\overrightarrow{\mbox{\boldmath$ y $ }O})(-1)^{n'+...
...(\overrightarrow{\mbox{\boldmath$ y $ }\mbox{\boldmath$ x $ }}).
\end{eqnarray*}


In the case where m' < m

\begin{eqnarray*}&& R_{n',m'}(\overrightarrow{\mbox{\boldmath$ y $ }O})(-1)^{n'+...
...(\overrightarrow{\mbox{\boldmath$ y $ }\mbox{\boldmath$ x $ }}).
\end{eqnarray*}


As a result, the underlined part (a) can be rewritten as

\begin{eqnarray*}(a)=\overline{R_{n',-m'}}(\overrightarrow{\mbox{\boldmath$ y $ ...
...(\overrightarrow{\mbox{\boldmath$ y $ }\mbox{\boldmath$ x $ }}).
\end{eqnarray*}


Hence one obtains
 
$\displaystyle S_{n,m}(\overrightarrow{O\mbox{\boldmath$ x $ }})$ = $\displaystyle \sum_{n'=0}^{\infty}\sum_{m'= 0 }^{n'}
\overline{R_{n',-m'}}(\ove...
...}) S_{n+n',m+m'}(\overrightarrow{\mbox{\boldmath$ y $ }\mbox{\boldmath$ x $ }})$  
  = $\displaystyle \sum_{n'=0}^{\infty}\sum_{m'= -n'}^{0}
\overline{R_{n',m'}}(\over...
...}) S_{n+n',m+m'}(\overrightarrow{\mbox{\boldmath$ y $ }\mbox{\boldmath$ x $ }})$  
  = $\displaystyle \sum_{n'=0}^{\infty}\sum_{m'= -n'}^{n'}
\overline{R_{n',m'}}(\ove...
...ox{\boldmath$ x $ }}\vert > \vert\overrightarrow{\mbox{\boldmath$ y $ }O}\vert.$ (B.5)

Now take two points O and O' where two inequalities $\vert\overrightarrow{O\mbox{\boldmath$\space y $ }}\vert
< \vert\overrightarrow{O\mbox{\boldmath$\space x $ }}\vert$ and $\vert\overrightarrow{O'\mbox{\boldmath$\space y $ }}\vert < \vert\overrightarrow{O'\mbox{\boldmath$\space x $ }}\vert$ are valid and consider the following identity:
 
$\displaystyle \left(\frac{1}{\vert\mbox{\boldmath$ x $ }-\mbox{\boldmath$ y $ }...
...oldmath$ y $ }})
\overline{S_{n,m}}(\overrightarrow{O'\mbox{\boldmath$ x $ }}).$     (B.6)

From (B.5) we obtain the following formula:
 
$\displaystyle \overline{S_{n,m}}(\overrightarrow{O\mbox{\boldmath$ x $ }})=\sum...
...arrow{O'O})\overline{S_{n+n',m+m'}}(\overrightarrow{O'\mbox{\boldmath$ x $ }}).$     (B.7)

Substituting (B.7) into the left-hand side of (B.6), one obtains
 
% latex2html id marker 35337
$\displaystyle \mbox{ LHS of (\ref{tako})} = \sum_{...
...arrow{O'O})\overline{S_{n+n',m+m'}}(\overrightarrow{O'\mbox{\boldmath$ x $ }}).$     (B.8)

Setting n+n'=n'',m+m'=m'' and deleting n in (B.8), one has
 
% latex2html id marker 35341
$\displaystyle \mbox{ LHS of (\ref{tako})}$ = $\displaystyle \sum_{n=0}^{\infty}\sum_{m=-n}^{n}\sum_{n'=0}^{\infty}\sum_{m'=-n...
...tarrow{O'O})\overline{S_{n+n',m+m'}}(\overrightarrow{O'\mbox{\boldmath$ x $ }})$  
  = $\displaystyle \sum_{n'=0}^{\infty}\sum_{m'=-n'}^{n'}
\sum_{n''=n'}^{\infty}\sum...
...tarrow{O'O})
\overline{S_{n'',m''}}(\overrightarrow{O'\mbox{\boldmath$ x $ }}).$ (B.9)

Exchanging the order of the summation in (B.9) ( $\displaystyle
\sum_{n'}\sum_{n''} \rightarrow \sum_{n''}\sum_{n'}$, See Fig.B), one obtains
 
% latex2html id marker 35346
$\displaystyle \mbox{ LHS side of (\ref{tako})}$ = $\displaystyle \sum_{n'=0}^{\infty}\sum_{m'=-n'}^{n'}
\sum_{n''=n'}^{\infty}\sum...
...htarrow{O'O})
\overline{S_{n'',m''}}(\overrightarrow{O'\mbox{\boldmath$ x $ }})$  
  = $\displaystyle \sum_{n''=0}^{\infty}\sum_{m''=-n''}^{n''}
\sum_{n'=0}^{n''}\sum_...
...htarrow{O'O})
\overline{S_{n'',m''}}(\overrightarrow{O'\mbox{\boldmath$ x $ }})$  
  = $\displaystyle \sum_{n=0}^{\infty}\sum_{m=-n}^{n}
\sum_{n'=0}^{n}\sum_{m'=-n'}^{...
...rightarrow{O'O})
\overline{S_{n,m}}(\overrightarrow{O'\mbox{\boldmath$ x $ }}).$ (B.10)

Finally, comparing the right-hand side of (B.6) with that of (B.10) one obtains the following formula:
 
$\displaystyle R_{n,m}(\overrightarrow{O'\mbox{\boldmath$ y $ }})=
\sum_{n'=0}^{...
...,m-m'}(\overrightarrow{O\mbox{\boldmath$ y $ }})R_{n',m'}(\overrightarrow{O'O})$     (B.11)


  
Figure B.1: Exchange of indices
\begin{figure}
\begin{center}
\leavevmode
\epsfile{file=FIG/fig4.1b.eps,scale=0.8}
\end{center}\end{figure}


next up previous contents
Next: Recurrence formulae for Rn,m Sn,m Up: Applications of Fast Multipole Previous: Series expansion of
Ken-ichi Yoshida
2001-07-28