Page:Lorentz Grav1900.djvu/10

 and those for the state $$(\mathfrak{d}', \mathfrak{H}')$$

{{MathForm2|(II)|$$\left.\begin{array}{c} Div\ \mathfrak{d}'=\varrho'\\ Div\ \mathfrak{H}'=0\\ Rot\ \mathfrak{H}'=4\pi\varrho'\mathfrak{v}'+4\pi\mathfrak{\dot{d}}'\\ 4\pi V^{2}Rot\ \mathfrak{d}'=-\mathfrak{\dot{H}}';\end{array}\right\} $$.}}

In the ordinary theory of electromagnetism, the force acting on a particle, moving with velocity $$\mathfrak{v}$$, is

$4\pi V^{2}\mathfrak{d}+\left[\mathfrak{v\ }.\ \mathfrak{H}\right]$

per unit charge.

In the modified theory, we shall suppose that a positively electrified particle with charge e experiences a force

on account of the field $$(\mathfrak{d}, \mathfrak{H})$$, and a force

on account of the field $$(\mathfrak{d}', \mathfrak{H}')$$, the positive coefficients $$\alpha$$ and $$\beta$$ having slightly different values.

For the forces, exerted on a negatively charged particle I shall write

and

expressing by these formulae that e is acted on by ($$\mathfrak{d}, \mathfrak{H}$$) in the same way as e'  by ($$\mathfrak{d}', \mathfrak{H}'$$), and vice versa.

§ 7. Let us next consider the actions exerted by a pair of oppositely charged ions, placed close to each other, and remaining so during their motion. For convenience of mathematical treatment, we may even reason as if the two charges penetrated each other, so that, if they are equal, $$\varrho'=-\varrho$$.