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

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 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

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).

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