Page:Elektrische und Optische Erscheinungen (Lorentz) 116.jpg

Now, the equation ($IV_{e}$) will be satisfied by us, by putting

${\mathfrak {E}}_{x}=0,\ {\mathfrak {E}}_{y}={\frac {n}{m}}{\mathfrak {H}}_{z},\ {\mathfrak {E}}_{z}=-{\frac {n}{m}}{\mathfrak {H}}_{y}$,

and then it follows from ($V_{e}$), ($VI_{e}$) and ($III_{e}$), one after the other,

${\mathfrak {d}}_{x}=0,\ 4\pi V^{2}{\mathfrak {d}}_{y}=\left({\frac {n}{m}}+{\mathfrak {p}}_{x}\right){\mathfrak {H}}_{z},\ 4\pi V^{2}{\mathfrak {d}}_{z}=-\left({\frac {n}{m}}+{\mathfrak {p}}_{x}\right){\mathfrak {H}}_{y}$, ${\mathfrak {H}}'_{x}=0,\ {\mathfrak {H}}'_{y}=\left(1-{\frac {n}{m}}{\frac {{\mathfrak {p}}_{x}}{V^{2}}}\right){\mathfrak {H}}_{y},\ {\mathfrak {H}}'_{z}=\left(1-{\frac {n}{m}}{\frac {{\mathfrak {p}}_{x}}{V^{2}}}\right){\mathfrak {H}}_{z}$. ${\mathfrak {D}}_{x}=0,\ 4\pi {\mathfrak {D}}_{y}=\left({\frac {m}{n}}-{\frac {{\mathfrak {p}}_{x}}{V^{2}}}\right){\mathfrak {H}}_{z},\ 4\pi {\mathfrak {D}}_{z}=-\left({\frac {m}{n}}-{\frac {{\mathfrak {p}}_{x}}{V^{2}}}\right){\mathfrak {H}}_{y}$,

whose latter values are also in agreement with condition ($I_{c}$).

Eventually we derive from (X)

${\mathfrak {M}}_{x}=0,\ 4\pi V^{2}{\mathfrak {M}}_{y}=\left(V^{2}{\frac {m}{n}}-{\frac {n}{m}}-2{\mathfrak {p}}_{x}\right){\mathfrak {H}}_{z}$, $4\pi V^{2}{\mathfrak {M}}_{z}=-\left(V^{2}{\frac {m}{n}}-{\frac {n}{m}}-2{\mathfrak {p}}_{x}\right){\mathfrak {H}}_{y}$,

and then we only have to satisfy condition (XI).

The first one of the herein summarized relation gives nothing new, while the second and third ones read:

${\mathfrak {E}}_{y}=\sigma {\mathfrak {M}}_{y}-j\ {\frac {\partial {\mathfrak {M}}_{z}}{\partial x}}+k{\dot {\mathfrak {M}}}_{z}{\mathfrak {p}}_{x}$,

(110)

and

${\mathfrak {E}}_{z}=\sigma {\mathfrak {M}}_{z}+j\ {\frac {\partial {\mathfrak {M}}_{y}}{\partial x}}-k{\dot {\mathfrak {M}}}_{y}{\mathfrak {p}}_{x}$.

(111)

Now, since by the reported formulas

${\mathfrak {E}}_{y}=-\nu {\mathfrak {E}}_{z}$ und ${\mathfrak {M}}_{y}=-\nu {\mathfrak {M}}_{z}$

it can be written for (110) and (111)

$\nu ({\mathfrak {E}}_{z}-\sigma {\mathfrak {M}}_{z})=j\ {\frac {\partial {\mathfrak {M}}_{z}}{\partial x}}-k{\dot {\mathfrak {M}}}_{z}{\mathfrak {p}}_{x}$,

and

${\mathfrak {E}}_{z}-\sigma {\mathfrak {M}}_{z}=-\nu \left(j\ {\frac {\partial {\mathfrak {M}}_{z}}{\partial x}}-k{\dot {\mathfrak {M}}}_{z}{\mathfrak {p}}_{x}\right)$,

thus at first we find

$\nu ^{2}=-1,\ \nu =\pm i$,