104.

In [sect:98.] Sec. 98 we have found the intensity of entropy radiation of a definite frequency in a definite direction by adding the entropy radiations of the two independent components $𝖪$ and ${𝖪}^{\prime }$, polarized at right angles to each other, or $𝖫(𝖪)+𝖫({𝖪}^{\prime }),\mathrm{\backslash Label}[eqn](141)\text{upshape (141)}$ where $𝖫$ denotes the function of $𝖪$ given in equation [eqn:(134)] (134). This method of procedure is based on the general law that the entropy of two mutually independent physical systems is equal to the sum of the entropies of the separate systems.

If, however, the two components of a ray, polarized at right angles to each other, are not independent of each other, this method of procedure no longer remains correct. This may be seen, e.g., on resolving the radiation intensity, not with reference to the two principal planes of polarization with the principal intensities $𝖪$ and ${𝖪}^{\prime }$, but with reference to any other two planes at right angles to each other, where, according to equation [eqn:(8)] (8), the intensities of the two components assume the following values ${3} &\mathsf{K} \cos^{2} \psi &&+ \mathsf{K}' \sin^{2} \psi &&= \mathsf{K}'' \ &\mathsf{K} \sin^{2} \psi &&+ \mathsf{K}' \cos^{2} \psi &&= \mathsf{K}'''.

\Label[eqn]{(142)}\tag*{\upshape (142)}$

In that case, of course, the entropy radiation is not equal to $𝖫({𝖪}^{″})+𝖫({𝖪}^{‴})$.

Thus, while the energy radiation is always obtained by the summation of any two components which are polarized at right angles to each other, no matter according to which azimuth the resolution is performed, since always ${𝖪}^{″}+{𝖪}^{‴}=𝖪+{𝖪}^{\prime },\mathrm{\backslash Label}[eqn](143)\text{upshape (143)}$ a corresponding equation does not hold in general for the entropy radiation. The cause of this is that the two components, the intensities of which we have denoted by ${𝖪}^{″}$ and ${𝖪}^{‴}$, are, unlike $𝖪$ and ${𝖪}^{\prime }$, not independent or non-coherent in the optic sense. In such a case $𝖫({𝖪}^{″})+𝖫({𝖪}^{‴})>𝖫(𝖪)+𝖫({𝖪}^{\prime }),\mathrm{\backslash Label}[eqn](144)\text{upshape (144)}$ as is shown by the following consideration.

Since in the state of thermodynamic equilibrium all rays of the same frequency have the same intensity of radiation, the intensities of radiation of any two plane polarized rays will tend to become equal, i.e., the passage of energy between them will be accompanied by an increase of entropy, when it takes place in the direction from the ray of greater intensity toward that of smaller intensity. Now the left side of the inequality [eqn:(144)] (144) represents the entropy radiation of two non-coherent plane polarized rays with the intensities ${𝖪}^{″}$ and ${𝖪}^{‴}$, and the right side the entropy radiation of two non-coherent plane polarized rays with the intensities $𝖪$ and ${𝖪}^{\prime }$. But, according to [eqn:(142)] (142), the values of ${𝖪}^{″}$ and ${𝖪}^{‴}$ lie between $𝖪$ and ${𝖪}^{\prime }$; therefore the inequality [eqn:(144)] (144) holds.

At the same time it is apparent that the error committed, when the entropy of two coherent rays is calculated as if they were non-coherent, is always in such a sense that the entropy found is too large. The radiations ${𝖪}^{″}$ and ${𝖪}^{‴}$ are called "partially coherent," since they have some terms in common. In the special case when one of the two principal intensities $𝖪$ and ${𝖪}^{\prime }$ vanishes entirely, the radiations ${𝖪}^{″}$ and ${𝖪}^{‴}$ are said to be "completely coherent," since in that case the expression for one radiation may be completely reduced to that for the other. The entropy of two completely coherent plane polarized rays is equal