Page:Elektrische und Optische Erscheinungen (Lorentz) 078.jpg

always remains the same vector; thus $F_{1}({\mathfrak {M}})$ must remain unchanged, which is only possible when this vector has the direction of ${\mathfrak {M}}$. With respect to the linear character of the sought relation, we consequently have to set

$F_{1}({\mathfrak {M}})=\sigma {\mathfrak {M}}$,

(65)

where $\sigma$ is a scalar constant.

The second vector $F_{2}({\mathfrak {{\dot {M}},p}})$ occurring in (62), has the following properties. First, its components are homogeneous, linear functions of ${\dot {\mathfrak {M}}}_{x},\ {\dot {\mathfrak {M}}}_{y},\ {\dot {\mathfrak {M}}}_{z}$ as well as from ${\mathfrak {p}}_{x},{\mathfrak {p}}_{y},{\mathfrak {p}}_{z}$. Second, after an arbitrary rotation of the figure consisting of the three vectors ${\mathfrak {{\dot {M}},p}}$ and $F_{2}({\mathfrak {{\dot {M}},p}})$, $F_{2}({\mathfrak {{\dot {M}},p}})$ must still fit to ${\dot {\mathfrak {M}}}$ and ${\mathfrak {p}}$. By that we derive^[1]

$F_{2}({\mathfrak {{\dot {M}},p}})=k[{\mathfrak {{\dot {M}}.p}}]$,

(66)

where k is a positive or negative constant, which by the way, as $\sigma$ above, can also depend on the oscillation period T.

1. If we decompose ${\mathfrak {p}}$ into two components ${\mathfrak {p}}_{1}$ and ${\mathfrak {p}}_{2}$, then it follows from the first mentioned property of $F_{2}({\mathfrak {{\dot {M}},p}})$
   $F_{2}({\mathfrak {{\dot {M}},p}})=F_{2}({\mathfrak {{\dot {M}},p_{1}}})+F_{2}({\mathfrak {{\dot {M}},p_{2}}})$

   It is assumed, that ${\mathfrak {p}}_{1}$ falls into the direction of ${\dot {\mathfrak {M}}}$, and ${\mathfrak {p}}_{2}$ is perpendicular to it. If we now rotate the figure (consisting of ${\dot {\mathfrak {M}}},{\mathfrak {p}}_{1}$ and $F_{2}({\mathfrak {{\dot {M}},{\mathfrak {p}}_{1}}})$) around an axis that falls into ${\dot {\mathfrak {M}}}$, ${\dot {\mathfrak {M}}}$ and ${\mathfrak {p}}_{1}$ stay were they are, and thus $F_{2}({\mathfrak {{\dot {M}},{\mathfrak {p}}_{1}}})$ may not change as well. Consequently, this vector must have the direction of ${\dot {\mathfrak {M}}}$ and ${\mathfrak {p}}_{1}$. That

   $F_{2}({\dot {\mathfrak {M}}},{\mathfrak {p}}_{1})=0$

   (67)

   can be shown in addition, by means of a rotation of 180° around an axis perpendicular to ${\dot {\mathfrak {M}}}$ and ${\mathfrak {p}}_{1}$. In the course of this rotation, the vector $F_{2}({\dot {\mathfrak {M}}},{\mathfrak {p}}_{1})$ would obtain the opposite direction; yet it shouldn't be changing, because both vectors ${\dot {\mathfrak {M}}}$ and ${\mathfrak {p}}_{1}$ change their sign.

   To find out the direction of $F_{2}({\dot {\mathfrak {M}}},{\mathfrak {p}}_{2})$, we turn the figure (which is formed by this vector with ${\dot {\mathfrak {M}}}$ and ${\mathfrak {p}}_{2}$) around an axis perpendicular to the plane $({\dot {\mathfrak {M}}},{\mathfrak {p}}_{2})$ or $({\mathfrak {{\dot {M}},p}})$, namely around 180°. Here, ${\dot {\mathfrak {M}}}$ and ${\mathfrak {p}}_{2}$ go over into $-{\dot {\mathfrak {M}}}$ and $-{\mathfrak {p}}_{2}$; the vector $F_{2}({\dot {\mathfrak {M}}},{\mathfrak {p}}_{2})$ thus may not be changed, which is only possible when it has the direction of the axis.

   Thus the vector $F_{2}({\dot {\mathfrak {M}}},{\mathfrak {p}}_{2})$ — and thus by (67) also the vector $F_{2}({\mathfrak {{\dot {M}},p}})$ — is perpendicular to the plane $({\mathfrak {{\dot {M}},p}})$; its magnitude is proportional to the values of ${\dot {\mathfrak {M}}}$ and ${\mathfrak {p}}_{2}$. Both we have expressed in (66).