Page:Elektrische und Optische Erscheinungen (Lorentz) 051.jpg

$\rho {\mathfrak {v}}_{x}=f(x,y,z,t')$

and thus have

$\psi _{x}=-\int {\frac {1}{r}}f\left(\xi ,\eta ,\zeta ,t'-{\frac {r}{V}}\right)d\tau$

(38)

a solution of (36)^[1]. By that we have to imagine two points; first, the fixed point (x, y, z), for which we want to calculate $\psi _{x}$ and which we call P; second, a moving point Q, which has to traverse the whole space, where $\rho {\mathfrak {v}}_{x}$ is different from zero. r represents the distance QP, and $t'$ the local time of P at the instant for which we wish to calculate $\psi _{x}$; furthermore we have to understand by ξ, η, ζ, the coordinates of Q, and by dτ an element of the just mentioned space. The function $\textstyle {f\left(\xi ,\eta ,\zeta ,t'-{\frac {r}{V}}\right)}$ is the value of $\rho {\mathfrak {v}}_{x}$ in this element, namely, if the local time that is valid at this place, is $\textstyle {t'-{\frac {r}{V}}}$.

A single luminous molecule.

§ 33. To excite electric oscillations, a single molecule with oscillating ions shall serve; let $Q_{0}$ be an arbitrary fixed point in it — for brevity, we say, "the molecule is present in $Q_{0}$" —, and for P a place is chosen, whose distance from $Q_{0}$ is much larger than the dimensions of the molecules. For distinction, $Q_{0}P=r_{0}$.

We now want to replace the various distances r, that are present in formula (38), by $r_{0}$ and also neglect the differences of local times at the various points of the molecule. In this way,

$\psi _{x}=-{\frac {1}{r_{0}}}\int \rho {\mathfrak {v}}_{x}d\tau {,}$

1. The proof for this can be found, for example, in my treatise: La théorie électromagnétique de Maxwell et son application aux corps mouvants.