Page:Elektrische und Optische Erscheinungen (Lorentz) 117.jpg

 and furthermore

Now, if $$\sigma, j, k$$ and n are given, we can determine m from this equation, namely we obtain two values, depending on whether we apply the above, or the below sign.

§ 86. We put

$$n = i n',\ m = i m'$$,

by that, equation (112) is transformed into

from which two real values are given from $$m'$$, which we want to denote by $$m'_1$$ and $$m'_2$$.

For $$\nu=+i,\ m'=m'_1$$, it becomes now

$$\mathfrak{H}_{y}=ae^{i(n't-m'_{1}x)},\ \mathfrak{H}_{z}=iae^{i(n't-m'_{1}x)}$$,

and for $$\nu=-i,\ m'=m'_2$$,

$$\mathfrak{H}_{y}=ae^{i(n't-m'_{2}x)},\ \mathfrak{H}_{z}=-iae^{i(n't-m'_{2}x)}$$.

If we eventually take the real parts, we arrive at the following two particular solutions

which obviously represent two opposite, circular-polarized light beams of propagation velocities $$n'/m'_1$$ and $$n'/m'_2$$.

The composition of these states of motion leads in a known way to a beam of linear-polarized light, whose oscillation direction gets rotated. Namely, addition of the values (114) and (115) gives the solution

The rotation $$\omega$$ of the polarisation plane related to unit volume, consequently amounts

$$\omega=\frac{1}{2}(m'_{1}-m'_{2})$$.