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

and furthermore

$4\pi V^{2}{\frac {n}{m}}=\left\{\sigma \pm i(jm-kn{\mathfrak {p}}_{x})\right\}\left(V^{2}{\frac {m}{n}}-{\frac {n}{m}}-2{\mathfrak {p}}_{x}\right)$.

(112)

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

$n=in',\ m=im'$,

by that, equation (112) is transformed into

$4\pi V^{2}{\frac {n'}{m'}}=\left\{\sigma \mp (jm'-kn'{\mathfrak {p}}_{x})\right\}\left(V^{2}{\frac {m'}{n'}}-{\frac {n'}{m'}}-2{\mathfrak {p}}_{x}\right)$.

(113)

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

${\mathfrak {H}}_{y}=a\ \cos(n't-m'_{1}x),\ {\mathfrak {H}}_{z}=-a\ \sin(n't-m'_{1}x)$,

(114)

${\mathfrak {H}}_{y}=a\ \cos(n't-m'_{2}x),\ {\mathfrak {H}}_{z}=a\ \sin(n't-m'_{2}x)$,

(115)

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

${\mathfrak {H}}_{y}=2a\ \cos {\frac {1}{2}}(m'_{1}-m'_{2})x\ \cos \left\{n't-{\frac {1}{2}}(m'_{1}+m'_{2})x\ \right\}$, ${\mathfrak {H}}_{z}=2a\ \sin {\frac {1}{2}}(m'_{1}-m'_{2})x\ \cos \left\{n't-{\frac {1}{2}}(m'_{1}+m'_{2})x\ \right\}$.

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

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