Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/207

Performing the differentiations on $Q_{i}$ as given in equation (29), we obtain

$\vartheta _{i}^{(\sigma )}=\nu ^{\sigma }\sum \left\{(-1)^{n}{\frac {|{\underline {i-\sigma }}\ |{\underline {\sigma }}\ |{\underline {2i-2n}}}{2^{i-\sigma }|{\underline {i+\sigma }}\ |{\underline {n}}\ |{\underline {i-n}}\ |{\underline {i-\sigma -2n}}}}\mu ^{i-\sigma -2n}\right\}.$

(39)

We may also express it as a homogeneous function of $\mu$ and $\nu$,

$\vartheta _{i}^{(\sigma )}=\nu ^{\sigma }\sum \left\{(-1)^{n}{\frac {|{\underline {i-\sigma }}\ |{\underline {\sigma }}}{2^{2\sigma }|{\underline {n}}\ |{\underline {\sigma +n}}\ |{\underline {i-\sigma -2n}}}}\mu ^{i-\sigma -2n}\nu ^{2n}\right\}.$

(40)

In this expression the coefficient of the first term is unity, and the others may be written down in order by the application of Laplace’s equation.

The following relations will be found useful in Electrodynamics. They may be deduced at once from the expansion of $Q_{i}$.

$\mu Q_{i}-Q_{i+1}={\frac {1}{i+1}}\nu ^{2}{\frac {dQ_{i}}{d\mu }}={\frac {i}{2}}\nu \vartheta _{i}^{1},$

(41)

$Q_{i-1}-\mu Q_{i}={\frac {1}{i}}\nu ^{2}{\frac {dQ_{i}}{d\mu }}={\frac {i+1}{2}}\nu \vartheta _{i}^{1}.$

(42)

On Harmonics of Positive Degree.

133.] We have hitherto considered the spherical surface harmonic $Y_{i}$ as derived from the solid harmonic

$V_{i}=|{\underline {i}}\ M_{i}{\frac {Y_{i}}{r^{i+1}}}.$

This solid harmonic is a homogeneous function of the coordinates of the negative degree $-(i+1)$. Its values vanish at an infinite distance and become infinite at the origin.

We shall now shew that to every such function there corresponds another which vanishes at the origin and has infinite values at an infinite distance, and is the corresponding solid harmonic of positive degree $i$.

A solid harmonic in general may be defined as a homogeneous function of x, y, and z, which satisfies Laplace’s equation

${\frac {d^{2}V}{dx^{2}}}+{\frac {d^{2}V}{dy^{2}}}+{\frac {d^{2}V}{dz^{2}}}=0.$

Let $H_{i}$ be a homogeneous function of the degree $i$, such that

$H_{1}|{\underline {i}}\ M_{i}r^{i}Y_{i}=r^{2i+1}V_{i}.$

(43)

Then

${\frac {dH_{i}}{dx}}=(2i+1)r^{2i-1}xV_{i}+r^{2i+1}{\frac {dV_{i}}{dx}},$ ${\frac {d^{2}H_{i}}{dx^{2}}}=(2i+1)\left((2i-1)x^{2}+r^{2}\right)r^{2i-3}V_{i}+2(2i+1)r^{2i-1}x{\frac {dV_{i}}{dx}}+r^{2i+1}{\frac {d^{2}V_{i}}{dx^{2}}}.$