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