Page:Elektrische und Optische Erscheinungen (Lorentz) 020.jpg

 By equation (4) we have

$$\bar{\mathfrak{S}}=\overline{\rho\mathfrak{v}}+\dot{\bar{\mathfrak{d}}}.$$

If the state of flow is stationary, then the observable magnitudes and also the averages are independent of time. Thus it will be

$$\bar{\mathfrak{S}}=\overline{\rho\mathfrak{v}}{,}$$

i.e. only the convection currents cause the action into the outside.

By the definition given in § 4, the components of $$\overline{\rho\mathfrak{v}}$$ are

$$\frac{1}{I}\int\rho\mathfrak{v}_{x}\ d\ \rho\text{, u. s. w.,}$$

or, when $$\rho$$ is different from zero only within the ions, and any ion is displaced without rotation

$$\frac{1}{I}\sum e\mathfrak{v}\text{, etc.,}$$

where e is the charge of an ion, and the sum is related to all charged particles contained in sphere I. We can easily see, that the result can be summarized in the formula

$$\bar{\mathfrak{S}}=\frac{1}{I}\sum e\mathfrak{v}$$

and this remains valid, when we don't interpret I just as a sphere, but as an arbitrary space, whose dimensions (albeit very small) are nevertheless much greater than the average distance of the ions. Of course, then the sum must be extended over the chosen space as well.

If there is a current within a lead wire with cross-section $$\omega$$, then we can take for I the part, that lies between two cross-sections which are mutually distant by ds. Since the magnitude of current will be determined by:

$$i=\omega\bar{\mathfrak{S}}{,}$$

and $$I=\omega\ d\ s$$, thus we obtain