542
Mr. E Cunnigham on the
Hence
d
W
d
v
0
f
=
+
v
0
K
+
v
0
f
c
2
β
∫
ρ
0
x
(
F
x
)
0
d
τ
0
{\displaystyle {\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
{
1
v
0
d
W
d
v
0
−
1
c
2
β
∫
ρ
0
x
(
F
x
)
0
d
τ
0
}
=
K
{\displaystyle 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
1
v
0
d
W
d
v
0
−
1
c
2
β
∫
ρ
0
x
(
F
x
)
0
d
τ
0
{\displaystyle {\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.
G
=
4
5
e
2
a
c
2
v
0
β
;
W
=
3
5
e
2
a
β
(
1
+
v
0
2
3
c
2
)
.
{\displaystyle {\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
d
G
d
v
0
=
4
5
e
2
a
c
2
{
1
β
+
v
0
2
/
c
2
β
3
}
=
4
5
e
2
a
c
2
β
3
,
1
v
0
d
W
d
v
0
=
3
5
e
2
a
β
{
2
3
c
2
+
(
1
+
v
0
2
3
c
2
)
1
c
2
β
2
}
=
1
5
e
2
a
c
2
β
3
{
5
−
v
0
2
c
2
}
.
∫
ρ
0
x
(
F
x
)
0
d
τ
0
=
ρ
0
∫
0
a
∫
0
π
r
cos
θ
⋅
4
3
π
r
ρ
0
cos
θ
2
π
r
2
sin
θ
d
θ
d
r
=
8
π
2
ρ
0
2
3
⋅
a
3
5
⋅
2
3
=
1
5
e
2
a
.
{\displaystyle {\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}}}
