Page:Motion of Electrification through a Dielectric.djvu/8

which, by the Binomial Theorem, is the same as

$\scriptstyle {A=(qu/r)\left\{1-u^{2}\nu ^{2}/v^{2}\right\}^{-{\frac {1}{2}}},\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(27)}}}$

the required solution.

11. To derive H, the tensor of the circular H, let rv=h, the distance from the axis. Then, by (15),

$\scriptstyle {H=-{\frac {dA}{dh}}=-\nu {\frac {dA}{dr}}+{\frac {\mu \nu }{r}}{\frac {dA}{d\mu }}={\frac {qu\nu }{r^{2}}}\left(1+\mu {\frac {d}{d\mu }}\right)\left(1-{\frac {u^{2}}{v^{2}}}\nu ^{2}\right)^{-{\frac {1}{2}}},\quad {\text{(28)}}}$

by (27), if μ=cos θ. Performing the differentiation, and also getting out E the tensor of the electric force, we have the final result that the electromagnetic field is fully given by^[1]

$\scriptstyle {cE={\frac {q}{r^{2}}}\cdot {\frac {1-u^{2}/v^{2}}{\left(1-u^{2}\nu ^{2}/v^{2}\right)^{\frac {3}{2}}}},\quad \quad H=cEuv,\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(29)}}}$

with the additional information that E is radial and H circular.

Now, as regards Ψ, if we bring it in, we have only got to take it out again. When the speed is very slow we may regard the electric field as given by $\scriptstyle {-\nabla \Psi }$ plus a small correcting vector, which we may call the electric force of inertia. But to show the physical inanity of Ψ, go to the other extreme, and let u nearly equal v. It is now the electric force of inertia (supposed) that equals $\scriptstyle {+\nabla \Psi }$ nearly (except about the equatorial plane), and its sole utility or function is to cancel the other $\scriptstyle {-\nabla \Psi }$ of the (supposed) electrostatic field. It is surely impossible to attach any physical meaning to Ψ and to propagate it, for we require two Ψ's, one to cancel the other, and both propagated infinitely rapidly.

As the speed increases, the electromagnetic field concentrates itself more and more about the equatorial plane, $\scriptstyle {\theta ={\frac {1}{2}}\pi .}$ To give an idea of the accumulation, let u²/v²=.99. Then cE is .01 of the normal value q/r² at the pole, and 10 times the normal value at the equator. The latitude where the value is normal is given by

$\scriptstyle {\nu =(v/u)\left[1-(1-u^{2}/v^{2})^{\frac {2}{3}}\right]^{\frac {1}{2}}.\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(30)}}}$

Limiting Case of Motion at the Speed of Light. Applications to a Telegraph Circuit.

12. When u=v, the solution (29) becomes a plane electromagnetic wave, E and H being zero everywhere except in the equatorial plane. As, however, the values of E and H are infinite, distribute the charge along a straight line moving in its own line, and let the linear-density be q. The solution is then^[2]

$\scriptstyle {H=Ecv=2qv/r\quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots {\text{(31)}}}$

at distance r from the line, between the two planes through the ends of the line perpendicular to it, and zero elsewhere.

To further realize, let the field terminate internally at r=a, giving a cylindrical-surface distribution of electrification, and terminate the tubes

1. The Electrician, Dec. 7, 1888, p. 148 [p. 495, vol. II].
2. Ibid. Nov. 23, 1888, p. 84. [p. 493, vol. II.].