Page:Elektrische und Optische Erscheinungen (Lorentz) 062.jpg

§ 41. The rule contained in the formula

${\overline {\epsilon }}={\frac {1}{l}}\int \epsilon d{\mathfrak {r}}$

can be express somewhat differently. Namely we shall choose for the point P an infinite amount, we want to say k, locations that are uniformly distributed over the sphere I, and take the arithmetic mean of the values of $\epsilon$ that are valid for these locations, i.e. we put

${\overline {\epsilon }}={\frac {1}{k}}\Sigma \epsilon$

(49)

Any ion, that has its equilibrium position in the interior of I, will (during its displacement) now pass through some positions that are connected with the element $d\sigma$ and thus add some terms to the sum $\Sigma \epsilon$. We obtain the whole sum, if we at first add to one another the terms that stem from a certain ion, and than sum over all ions.

Let Q be the equilibrium position of the considered ions, and Q the new position; so $QQ'={\mathfrak {q}}$. The length and the direction of this line are given, as well as the direction and magnitude of $d\sigma$. Whether the particle hits the surface element and provides the part e for the sought sum, only depends on the relative positions of P and Q. Thus we can, instead of giving k positions in the sphere l to P, also let remain the point at its position and lead point Q around a sphere I. As $QQ'$ now hits the fixed element $d\sigma$, when Q lies in a certain, easily specifiable cylinder of area ${\mathfrak {q}}_{n}d\sigma$, then the number of "relevant" positions is related to the integer k, as the area of that cylinder is related to the area of sphere I. This number is thus

${\frac {k}{I}}{\mathfrak {q}}_{n}d\sigma$,

and the sum $\Sigma \epsilon$, as far it is caused by the ion Q,

${\frac {k}{I}}e{\mathfrak {q}}_{n}d\sigma$.