Page:EB1911 - Volume 25.djvu/674

applying the theorem (7) to evaluate this expression, we have

$${\begin{aligned}{\frac {(-1)^{n}r^{n+1}}{n!}}{\frac {\partial ^{n}}{\partial z^{n}}}{\frac {1}{r}}&={\frac {(2n)!}{2^{n}n!n!}}{\frac {1}{r^{n}}}\left\{1-{\frac {r^{2}\nabla ^{2}}{2.2n-1}}+{\frac {r^{4}\nabla ^{4}}{2.4.2n-1.2n-3}}-\right\}z^{n}\\&={\frac {(2n)!}{2^{n}n!n!}}\left\{\mu ^{n}-{\frac {n(n-1)}{2.2n-1}}\mu ^{n-2}+\dots \right\},\end{aligned}}$$

the expression on the right side is $P^{n}(\mu )$, the zonal surface harmonic; we have therefore

$$P_{n}(\mu )={\frac {(-1)^{n}r^{n+1}}{n!}}{\frac {\partial ^{n}}{\partial z^{n}}}{\frac {1}{r}}.$$

The zonal harmonic has therefore all its poles coincident with the z axis. Next, suppose n - m axes coincide with the z axis, and that the remaining m axes are distributed symmetrically in the plane of x, y at intervals $\pi /m$, the direction cosines of one of them being $\cos \alpha ,\sin \alpha ,0$. We have

$${\begin{aligned}\prod _{0}^{m-1}\left\{\cos \left(\alpha +{\frac {r\pi }{m}}\right){\frac {\partial }{\partial x}}+\sin \left(\alpha +{\frac {r\pi }{m}}\right){\frac {\partial }{\partial y}}\right\}&={\frac {1}{2^{m}}}\prod {\Bigl \{}e^{i\left(\alpha +{\frac {r\pi }{m}}\right)}\left({\frac {\partial }{\partial x}}-i{\frac {\partial }{\partial y}}\right)\\&+e^{-i\left(\alpha +{\frac {r\pi }{m}}\right)}\left({\frac {\partial }{\partial x}}+i{\frac {\partial }{\partial y}}\right){\Bigr \}}.\end{aligned}}$$

Let $\xi =x+iy,\eta =x-iy$, the above product becomes

$$\prod _{0}^{m-1}\left\{e^{i\left(\alpha +{\frac {r\pi }{m}}\right)}{\frac {\partial }{\partial \xi }}+e^{-i\left(\alpha +{\frac {r\pi }{m}}\right)}{\frac {\partial }{\partial \eta }}\right\}$$

which is equal to

$e^{(m-1){\frac {i\pi }{2}}}\prod \left\{e^{mi\alpha }\left({\frac {\partial }{\partial \xi }}\right)^{m}-e^{-mi\alpha }\left({\frac {\partial }{\partial \eta }}\right)^{m}\right\}$; when $\alpha =0,{\frac {\pi }{2m}}$,

this becomes

$e^{(m-1){\frac {i\pi }{2}}}\left\{\left({\frac {\partial }{\partial \xi }}\right)^{m}-(-1)^{m}\left({\frac {\partial }{\partial \eta }}\right)^{m}\right\}$ and

$e^{(m-1){\frac {i\pi }{2}}}\left[\left({\frac {\partial }{\partial \xi }}\right)^{m}+(-1)^{m}\left({\frac {\partial }{\partial \eta }}\right)^{m}\right]$.

From (7), we find

$${\begin{aligned}{\frac {\partial ^{n-m}}{\partial z^{n-m}}}\left({\frac {\partial }{\partial x}}\pm i{\frac {\partial }{\partial y}}\right)^{m}{\frac {1}{r}}&={\frac {(2n)!}{2^{n}n!}}{\frac {1}{r^{2n+1}}}\left[1-{\frac {r^{2}\Delta ^{2}}{2.2n-1}}+\dots \right]z^{n-m}(x\pm iy)^{m}\\&=(-1)^{n}{\frac {(2n)!}{2^{n}n!}}{\frac {1}{r^{n+1}}}(\cos m\phi \pm i\sin m\phi )\sin ^{m}\theta {\Bigl \{}\cos ^{n-m}\theta \\&-{\frac {(n-m)(n-m-1)}{2.2n-1}}\cos ^{n-m-2}+\dots \theta {\Bigr \}}\end{aligned}}$$

hence

$${\frac {\partial ^{n-m}}{\partial z^{n-m}}}\left({\frac {\partial }{\partial x}}\pm i{\frac {\partial }{\partial y}}\right)^{m}{\frac {1}{r}}=(-1)^{n}{\frac {(n-m)!}{r^{n+1}}}(\cos m\phi \pm i\sin m\phi )P_{n}^{m}(\cos \theta )$$

as we see on referring to (4); we thus obtain the formulae

$${\frac {\partial ^{n-m}}{\partial z^{n-m}}}\left\{\left({\frac {\partial }{\partial \xi }}\right)^{m}+\left({\frac {\partial }{\partial \eta }}\right)^{m}\right\}{\frac {1}{r}}=(-1)^{n}{\frac {(n-m)!}{2^{m-1}r^{n+1}}}\cos m\phi P_{n}^{m}(\cos \theta )$$

$${\frac {\partial ^{n-m}}{\partial z^{n-m}}}\left\{\left({\frac {\partial }{\partial \xi }}\right)^{m}-\left({\frac {\partial }{\partial \eta }}\right)^{m}\right\}{\frac {1}{r}}=(-1)^{n}{\frac {(n-m)!}{2^{m-1}r^{n+1}}}\sin m\phi P_{n}^{m}(\cos \theta )$$

It is thus seen that the tesseral harmonics of degree n and order m are those which have $n-m$ axes coincident with the z axis, and the other m axis distributed in the equatorial plane, at angular intervals $\pi /m$. The sectorial harmonics have all their axes in the equatorial plane.

8. Determination of the Poles of a given Harmonic.—It has been shown that a spherical harmonic $Y_{n}(x,y,z)$ can be generated by means of an operator

$f_{n}\left({\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}}\right)$ acting upon ${\frac {1}{r}}$,

the function $f_{n}$ being so chosen that

$$Y_{n}(x,y,z)=(-1)^{n}{\frac {(2n)!}{2^{n}n!}}\left\{1-{\frac {r^{2}\nabla ^{2}}{2.2n-1}}+\dots \right\}f_{n}(x,y,z);$$

this relation shows that if an expression of the form

$(x^{2}+y^{2}+^{2})f_{n-2}(x,y,z)$

is added to $f_{n}(x,y,z)$, the harmonic $Y_{n}(x,y,z)$ is unaltered; thus if $Y_{n}$ be regarded as given, $f_{n}(x,y,z)=0$, is not uniquely determined, but has an indefinite number of values differing by multiples of $x^{2}+y^{2}+z^{2}$. In order to determine the poles of a given harmonic, $f_{n}$ must be so chosen that it is resolvable into linear factors; it will be shown that this can be done in one, and only one, way, so that the poles are all real.

If x, y, z are such as to satisfy the two equations $Y_{n}(x,y,z)=0$, $x^{2}+y^{2}+z^{2}=0$, the equation $f_{n}(x,y,z)$ is also satisfied; the problem of determining the poles is therefore equivalent to the algebraical one of reducing $Y_{n}$ to the product of linear factors by means of the relation $x^{2}+y^{2}+z^{2}=0$, between the variables. Suppose

$$Y_{n}(x,y,z)=A\prod _{s=1}^{m}(l_{s}x+m_{s}y+n_{s}z)+(x^{2}+y^{2}+z^{2})V_{n-2}(x,y,z),$$

we see that the plane $l_{s}x+m_{s}y+n_{s}z=0$ passes through two of the 2n generating lines of the imaginary cone $x^{2}+y^{2}+z^{2}=0,$ in which that cone is intersected by the cone $Y_{n}(x,y,z)=0$. Thus a pole $(l_{s},m_{s},n_{s})$ is the pole with respect to the cone $x^{2}+y^{2}+z^{2}=0$, of a plane passing through two of the generating lines; the number of systems of poles is therefore $n(2n-1)$, the number of ways of taking the 2n generating lines in pairs. Of these systems of poles, however, only one is real, viz. that in which the lines in each pair correspond to conjugate complex roots of the equations $Y_{n}=0$, $x^{2}+y^{2}+z^{2}=0$. Suppose

$${\frac {x}{\alpha _{1}+i\beta _{1}}}={\frac {y}{\alpha _{2}+i\beta _{2}}}={\frac {z}{\alpha _{2}+i\beta _{2}}}$$

gives one generating line, then the conjugate one is given by

$${\frac {x}{\alpha _{1}-i\beta _{1}}}={\frac {y}{\alpha _{2}-i\beta _{2}}}={\frac {z}{\alpha _{2}-i\beta _{2}}}$$

and the corresponding factor $lx+my+nz$ is

$${\begin{vmatrix}x&y&z\\\alpha _{1}+i\beta _{1}&\alpha _{2}+i\beta _{2}&\alpha _{3}+i\beta _{3}\\\alpha _{1}-i\beta _{1}&\alpha _{2}-i\beta _{2}&\alpha _{3}-i\beta _{3}\\\end{vmatrix}}$$

which is real. It is obvious that if any non-conjugate pair of roots is taken, the corresponding factor, and therefore the pole, is imaginary. There is therefore only one system of real poles of a given harmonic, and its determination requires the solution of an equation of degree 2n. This, theorem is due to Sylvester (Phil. Mag. (1876), 5th series, vol. ii., "A Note on Spherical Harmonics").

9. Expression for the Zonal Harmonic with any Axis.—The zonal surface harmonic, whose axis is in the direction

${\frac {x'}{r'}},{\frac {y'}{r'}},{\frac {z'}{r'}}$, is $P_{n}\left({\frac {xx'+yy'+zz'}{rr'}}\right)$

or $P_{n}(\cos \theta \cos \theta '+\sin \theta \sin \theta '\cos \phi -\phi ')$; this is expressible as a linear function of the system of zonal, tesseral, and sectorial harmonics already found. It will be observed that it is symmetrical with respect to $(x,y,z)$ and $(x',y',z')$, and must thus be capable of being expressed in the form

$$a_{0}P_{n}(\cos \theta )P_{n}(\cos \theta ')+\sum _{1}^{n}a_{m}P_{n}^{m}(\cos \theta )P_{n}^{m}(\cos \theta ')\cos m(\phi -\phi ')$$

and it only remains to determine the co-efficients $a_{0},a_{1},\dots ,a_{m},\dots a_{n}$ To find this expression, we transform $(x'x+y'y+z'z)^{n}$, where x, y, z satisfy the condition $x^{2}+y^{2}+z^{2}=0$; writing $\xi =x+iy,\eta =x-iy,\xi '=x'+iy',\eta '=x'-iy'$, we have

$$(xx'+yy'+zz')^{m}=({\frac {1}{2}}\eta '\xi +{\frac {1}{2}}\xi '\eta +zz')^{n}$$

which equals

$$(xx'+yy'+zz')^{n}=(zz')^{n}+\sum \sum {\frac {n!}{a!b!(n-a-b)!}}\left\{{\frac {\eta '^{a}\xi '^{b}\eta ^{b}\xi ^{b}+\eta '^{b}\xi '^{a}\xi ^{b}\eta ^{a}}{2^{a+b}}}\right\}(zz')^{n-a-b}$$

the summation being taken for all values of a and b, such that $a+b\leq n$, $a>b$; the values $a=0,b=0$ corresponding to the term $(zz')^{n}$. Using the relation $\xi \eta =-z^{2}$, this becomes

$${\begin{aligned}(xx'+yy'+zz')^{n}=(zz')^{n}+\sum \sum &{\frac {(-1)^{b}}{2^{a+b}}}{\frac {n!}{a!b!(n-a-b)!}}(\eta '\xi ')^{b}z'^{n-a-b}\\&\{(\eta '\xi )^{a-b}+(\xi '\eta )^{a-b}\}z^{n-a+b},\end{aligned}}$$

putting $a-b=m$, the coefficient of $\xi ^{m}z^{n-m}$ on the right side is

$$\sum {\frac {(-1)^{b}}{2^{m+2b}}}{\frac {n!}{b!(m+b)!(n-m-2b)!}}(\xi '\eta ')^{b}\eta '^{m}z'^{n-m-2b},$$

from $b=0$ to $b={\frac {1}{2}}(n-m)$, or ${\frac {1}{2}}(n-m-1)$, according as $n-m$ is even or odd. This coefficient is equal to

$${\begin{aligned}&{\frac {n!}{2^{m}m!(n-m)!}}(x'-iy')^{m}{\Bigl \{}z'^{n-m}-{\frac {(n-m)(n-m-1)}{2.2m+2}}z'^{n-m-2}(x'^{2}+y'^{2})\\&+{\frac {(n-m)(n-m-1)(n-m-2)(n-m-3)}{2.4.2m+2.2m+3}}z'^{n-m-4}(x'^{2}+y'^{2})^{2}-{\Bigr \}}\end{aligned}}$$

in order to evaluate this coefficient, put $z=1$, $x'=i\cos \alpha$, $y'=i\sin \alpha$, then this coefficient is that of $(i\cos \alpha +\sin \alpha )^{m}$, or of $i^{m}e^{-mi\alpha }$ in the expansion of $(z'+ix'\cos \alpha +iy'\sin \alpha )^{n}$ in powers of $e^{-i\alpha }$ and $e^{i\alpha }$, this has been already found, thus the coefficient is

$${\frac {n!}{(n+m)!}}e^{-im\phi '}P_{n}^{m}(\cos \theta ')r'^{n}.$$

Similarly the coefficient of $\eta ^{m}z^{n-m}$ is

$${\frac {n!}{(n+m!)}}e^{+im\phi '}P_{n}^{m}(\cos \theta ')r'^{n};$$

hence we have

$${\begin{aligned}{\frac {1}{r'^{n}}}(xx'+yy'+zz')^{n}=z^{n}P_{n}^{m}(\cos \theta ')+n!\sum _{1}^{n}&P_{n}^{m}(\cos \theta ')\{\cos m\phi '(\xi ^{m}+\eta ^{m})\\&+i\sin m\phi '(\eta ^{m}-\xi ^{m})\}{\frac {z^{n-m}}{(n+m!)}}\end{aligned}}$$

In this result, change x, y, z into

$${\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial z}}$$