Page:Motion of Electrification through a Dielectric.djvu/7

distribution of electric force, which has to be found by the additional datum

$\scriptstyle {div\ c\mathbf {E} =4\pi \rho ,\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(17)}}}$

where E is the real force.

9. "Symbolically" expressed, the solution of (16) is

$\scriptstyle {\mathbf {A} ={\frac {4\pi \rho \mathbf {u} }{p^{2}/v^{2}-\nabla ^{2}}}={\frac {-4\pi \rho \mathbf {u} /\nabla ^{2}}{1-p^{2}/v^{2}\nabla ^{2}}}.\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(18)}}}$

Here the numerator of the fraction to the right is the vector-potential of the convection-current. Calling it A₀, we have

$\scriptstyle {\mathbf {A} _{0}={\frac {4\pi \rho \mathbf {u} }{-\nabla ^{2}}}=\sum {\frac {\rho \mathbf {u} }{r}}.\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(19)}}}$

Inserting in (18) and expanding, we have

$\scriptstyle {\mathbf {A} =\left\{1+(p/v\nabla )^{2}+(p/v\nabla )^{4}+\cdots \right\}\mathbf {A} _{0}.\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(20)}}}$

Given then ρu as a function of position and time, A₀ is known by (19), and (20) finds A, whilst (15) finds H.

Complete Solution in the Case of Steady Rectilinear Motion.⁠Physical Inanity of Ψ.

10. When the motion of the electrification is all in one direction, say parallel to the z-axis, u, A₀, and A are all parallel to this axis, so that we need only consider their tensors. When there is simply one charge q at a point, we have

$\scriptstyle {A_{0}=qu/r,}$

and (20) becomes

$\scriptstyle {A=q\left\{1+(p/v\nabla )^{2}+(p/v\nabla )^{4}+\cdots \right\}(u/r)\quad \cdots \cdots \cdots \cdots \cdots \cdots {\text{(21)}}}$

at distance r from q. When the motion is steady, and the whole electromagnetic field is ultimately steady with respect to the moving charge, we shall have, taking it as origin,

$\scriptstyle {p=-u(d/dz)=-uD\,}$

for brevity; so that

$\scriptstyle {A=qu\left\{1+(uD/v\nabla )^{2}+(uD/v\nabla )^{4}+\cdots \right\}r^{-1}.\quad \cdots \cdots \cdots \cdots {\text{(22)}}}$

Now the property

$\scriptstyle {\nabla ^{2}r^{n+2}=(n+2)(n+3)r^{n}\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(23)}}}$

brings (22) to

$\scriptstyle {A=qu\left\{{\frac {1}{r}}+{\frac {u^{2}}{v^{2}}}D^{2}{\frac {r}{2!}}+{\frac {u^{4}}{v^{4}}}D^{4}{\frac {r^{3}}{4!}}+\cdots \right\};\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(24)}}}$

and the property

$\scriptstyle {D^{2n}r^{2n-1}=1^{2}.3^{2}.5^{2}\cdots (2n-1)^{2}\nu ^{2n}/r,\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(25)}}}$

where ν=sin θ, θ being the angle between r and the axis, brings (24) to

$\scriptstyle {A={\frac {qu}{r}}\left\{1+{\frac {u^{2}}{v^{2}}}{\frac {\nu ^{2}}{2}}(1+{\frac {u^{2}}{v^{2}}}{\frac {3}{4}}\nu ^{2}(1+{\frac {u^{2}}{v^{2}}}{\frac {5}{6}}\nu ^{2}(1+\cdots \right\};\quad \cdots \cdots {\text{(26)}}}$