Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/256

equal to $Ce^{u}$ where $C$ is a numerical quantity which for simplicity we may make equal to unity.

We therefore put

$u=\log {\frac {O'P}{OP}},\ \alpha =\log {\frac {O'A}{OA}},\ \beta =\log {\frac {O'B}{OB}}.$

Let

$\beta -\alpha =\varpi ,\ u-a=\theta .$

Then all the successive images of $P$ will lie on the arc $OAPBO'$.

The position of the image of $P$ in $A$ is $Q_{0}$ where

$u\left(Q_{0}\right)=\log {\frac {O'Q}{OQ}}=2a-u.$

That of $Q_{0}$ in $B$ is $P_{1}$ where

$u\left(P_{1}\right)=\log {\frac {O'P_{1}}{OP_{1}}}=u+2\varpi .$

Similarly

$u\left(P_{s}\right)=u+2s\varpi ,\ u\left(Q_{s}\right)=2a-u-2s\varpi .$

In the same way if the successive images of $P$ in $B,A,B$, &c. are $Q_{0}',P_{1}',Q_{1}',$, &c.,

${\begin{array}{lll}u\left(Q_{0}'\right)=2\beta -u,&&u\left(P_{0}'\right)=u-2\varpi ;\\u\left(P_{s}'\right)=u-2s\varpi ,&&u\left(Q_{s}'\right)=2\beta -u+2s\varpi .\end{array}}$

To find the charge of any image $P_{s}$ we observe that in the inverted figure its charge is

$P{\sqrt {\frac {OP_{s}}{OP}}}$

In the original figure we must multiply this by $O'P_{s}$. Hence the charge of $P_{s}$ in the dipolar figure is

$P{\sqrt {\frac {OP_{s}\cdot O'P_{s}}{OP\cdot O'P}}}.$

If we make $\xi ={\sqrt {OP\cdot O'P}}$, and call $\xi$ the parameter of the point $P$, then we may write

$P_{s}={\frac {\xi _{s}}{\xi }}P,$

or the charge of any image is proportional to its parameter.

If we make use of the curvilinear coordinates $u$ and $v$, such that

$e^{u+{\sqrt {-1}}v}={\frac {x+{\sqrt {-1}}y-k}{x+{\sqrt {-1}}y+k}},$

then

$x=-{\frac {k\sin hu}{\cos hu-\cos v}},\ y={\frac {k\sin v}{\cos hu-\cos v}};$