Page:SearleEllipsoid.djvu/12

 When $$h$$ is large the quantity in [ ]

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

vanishing when $$h=\infty$$.

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

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

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

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

Hence the total electromagnetic energy of the ellipsoid is

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

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

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

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

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

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

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

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

as has been found by Mr. Heaviside. It follows from this