Page:Elektrische und Optische Erscheinungen (Lorentz) 052.jpg

where all occurring $\rho {\mathfrak {v}}_{x}$ are related to the same instant, namely to the instant when

$t'={\frac {r_{0}}{V}}$

is the local time of $Q_{0}$.

Since ${\mathfrak {v}}_{x}$ is equal for all points of an ion, then, if we write e for the charge of such a particle, the last integral transforms into

$\Sigma e{\mathfrak {v}}_{x}.$

The sum is extending over all ions of the molecule.

Furthermore, if ${\mathfrak {q}}$ is now the displacement of an ion from its equilibrium position, then

${\mathfrak {v}}_{x}={\frac {d{\mathfrak {q}}_{x}}{dt}}{,}$

and

$\Sigma e{\mathfrak {v}}_{x}={\frac {d}{dt}}\Sigma e{\mathfrak {q}}_{x}.$

This has a simple meaning. We can conveniently call the vector $\Sigma e{\mathfrak {q}}$ the electric moment of the molecule and denote it by ${\mathfrak {m}}$. Then it is

$\Sigma e{\mathfrak {q}}_{x}={\mathfrak {m}}_{x}{,}$

$\psi _{x}=-{\frac {1}{r_{0}}}{\frac {d{\mathfrak {m}}_{x}}{dt}}=-{\frac {\partial }{\partial t}}\left({\frac {{\mathfrak {m}}_{x}}{r_{0}}}\right);$

after the things said here, we have to take the value of the derivative for the instant when the local time in $Q_{0}$ is $t'={\tfrac {r_{0}}{V}}$. Obviously we can also write

$\psi _{x}=-{\frac {\partial }{\partial t'}}\left({\frac {{\mathfrak {m}}_{x}}{r_{0}}}\right){,}$

where ${\mathfrak {m}}_{x}$ means the first component of the electric moment in that very instant. After (by that and by two equations of the same from) we have found $\psi _{x}$, $\psi _{y}$, $\psi _{z}$ for the point (x, y, z) and the local time $t'$ at this place, the study of the propagating oscillations is very simple. The equations (37) give

${\mathfrak {H'}}_{x}={\frac {\partial }{\partial t'}}\left({\frac {\partial }{\partial y}}\right)'\left({\frac {{\mathfrak {m}}_{x}}{r_{0}}}\right)-{\frac {\partial }{\partial t'}}\left({\frac {\partial }{\partial z}}\right)'\left({\frac {{\mathfrak {m}}_{y}}{r_{0}}}\right){\text{, etc.,}}$

(39)