Page:CunninghamPrinciple.djvu/5

From Wikisource
Jump to: navigation, search
This page has been validated.
542
 
Mr. E Cunnigham on the

Hence

\frac{dW}{dv_{0}}f=+v_{0}K+\frac{v_{0}f}{c^{2}\beta}\int\rho_{0}x\left(F_{x}\right)_{0}d\tau_{0}

and therefore

f\left\{ \frac{1}{v_{0}}\frac{dW}{dv_{0}}-\frac{1}{c^{2}\beta}\int\rho_{0}x\left(F_{x}\right)_{0}d\tau_{0}\right\} =K,

so that the longitudinal mass is equal to

\frac{1}{v_{0}}\frac{dW}{dv_{0}}-\frac{1}{c^{2}\beta}\int\rho_{0}x\left(F_{x}\right)_{0}d\tau_{0}

It is the second term in this expression that is neglected by Abraham, and which he has to account for by assuming the energy of the electron to be made up of W together with a term not electromagnetic in origin.

We proceed to evaluate this expression in the two cases (i.) of a sphere with uniform volume density; (ii.) of a sphere with uniform surface charge.


(i.) Volume Distribution.

\begin{array}{l}
G=\frac{4}{5}\frac{e^{2}}{ac^{2}}\frac{v_{0}}{\beta};\\
\\W=\frac{3}{5}\frac{e^{2}}{a\beta}\left(1+\frac{v_{0}^{2}}{3c^{2}}\right).\end{array}

Hence

\begin{array}{rl}
\frac{dG}{dv_{0}} & =\frac{4}{5}\frac{e^{2}}{ac^{2}}\left\{ \frac{1}{\beta}+\frac{v_{0}^{2}/c^{2}}{\beta^{3}}\right\} \\
\\ & =\frac{4}{5}\frac{e^{2}}{ac^{2}\beta^{3}},\\
\\\frac{1}{v_{0}}\frac{dW}{dv_{0}} & =\frac{3}{5}\frac{e^{2}}{a\beta}\left\{ \frac{2}{3c^{2}}+\left(1+\frac{v_{0}^{2}}{3c^{2}}\right)\frac{1}{c^{2}\beta^{2}}\right\} \\
\\ & =\frac{1}{5}\frac{e^{2}}{ac^{2}\beta^{3}}\left\{ 5-\frac{v_{0}^{2}}{c^{2}}\right\} .\\
\\\int\rho_{0}x\left(F_{x}\right)_{0}d\tau_{0} & =\rho_{0}\int_{0}^{a}\int_{0}^{\pi}r\ \cos\theta\cdot\frac{4}{3}\pi r\rho_{0}\ \cos\theta\ 2\pi r^{2}\ \sin\theta\ d\theta\ dr\\
\\ & =\frac{8\pi^{2}\rho_{0}^{2}}{3}\cdot\frac{a^{3}}{5}\cdot\frac{2}{3}\\
\\ & =\frac{1}{5}\frac{e^{2}}{a}.\end{array}.