Page:SearleEllipsoid.djvu/12

When $h$ is large the quantity in [ ]

${=}h-l\left({\frac {h^{2}}{l^{2}}}+1\right)\left({\frac {l}{h}}+{\frac {l^{2}}{3h^{3}}}\dots \right)$

vanishing when $h=\infty$.

Thus, making use of $\mu \mathrm {K} v^{2}=1$ we have for the magnetic energy

$\mathrm {T} {=}{\frac {q^{2}u^{2}}{4l\mathrm {K} v^{2}}}\left\{{\frac {a^{2}-l^{2}}{2l^{2}}}\log {\frac {a+l}{a-l}}-{\frac {a}{l}}\right\}$.

Now by (17) we have at the surface of the ellipsoid

$\mathbf {\Psi } _{0}{=}{\frac {q\alpha }{\mathrm {K} }}\int _{a}^{\infty }{\frac {dh}{h^{2}-l^{2}}}{=}{\frac {q\alpha }{2\mathrm {K} l}}\log {\frac {a+l}{a-l}}$.

Hence the total electromagnetic energy of the ellipsoid is

$\mathrm {W} ={\frac {1}{2}}q\mathbf {\Psi } _{0}+2\mathrm {T} ={\frac {q^{2}}{4\mathrm {K} l}}\left\{\left(1+{\frac {u^{2}a^{2}}{v^{2}l^{2}}}\right)\log {\frac {a+l}{a-l}}-2{\frac {u^{2}a}{v^{2}l}}\right\}$.

(23)

Here we must remember that $l^{2}=a^{2}-\alpha b^{2}$.

(A) Energy of Heaviside Ellipsoid. If we put $a/l$= S and make S large we have

${\begin{aligned}\mathrm {W} &={\frac {q^{2}\mathrm {S} }{2\mathrm {K} a}}\left\{\left(1+{\frac {u^{2}\mathrm {S} ^{2}}{v^{2}}}\right)\left({\frac {1}{\mathrm {S} }}+{\frac {1}{3\mathrm {S} ^{3}}}+\cdot \cdot \cdot \right)-{\frac {u^{2}\mathrm {S} }{v^{2}}}\right\}\\&={\frac {q^{2}}{2\mathrm {K} a}}\left(1+{\frac {1}{3}}{\frac {u^{2}}{v^{2}}}\right)\ \mathrm {when} \ \mathrm {S} =\infty \end{aligned}}$

(24)

This corresponds to the Heaviside ellipsoid, for when $S=\infty$, $a^{2}=ab^{2}$. The energy of the same ellipsoid at rest is

${\frac {q^{2}{\sqrt {\alpha }}}{2\mathrm {K} a}}\cdot {\frac {v}{u}}\sin ^{-1}{\frac {u}{v}}$.

(B) Energy of a Sphere. Putting $b=a$ we have $l=au/v$, and thus

$\mathrm {W} ={\frac {q^{2}}{2\mathrm {K} a}}\left({\frac {v}{u}}\log {\frac {v+u}{v-u}}-1\right)$.

(25)

If $u$ is small compared with $v$ we have

$\mathrm {W} {=}{\frac {q^{2}}{2\mathrm {K} a}}\left(1+{\frac {2}{3}}{\frac {u^{2}}{v^{2}}}+\cdot \cdot \cdot \right)$.

It will be found that as far as $u^{2}/v^{2}$ the magnetic energy is

${\frac {q^{2}u^{2}}{3\mathrm {K} av^{2}}}{=}{\frac {\mu q^{2}u^{2}}{3a}}$

as has been found by Mr. Heaviside.^[1] It follows from this

1. 'Electrical Papers' vol. ii. p. 505.