Chapter III

A System of Oscillators

179. Let us suppose that a large number $N$ of similar oscillators with parallel axes, acting quite independently of one another, are distributed irregularly in a volume-element of the field of radiation, the dimensions of which are so small that within it the intensities of radiation $K$ do not vary appreciably. We shall investigate the mutual action between the oscillators and the radiation which is propagated freely in space.

As before, the state of the field of radiation may be given by the magnitude and the azimuth of vibration $ψ$ of the principal intensities $K_{ν}$ and $K'$ of the pencils which strike the system of oscillators, where $K_{ν}$ and $K'$ depend in an arbitrary way on the direction angles $θ$ and $φ$ . On the other hand, let the state of the system of oscillators be given by the densities of distribution $w_{1}$ , $w_{2}$ , $w_{3}, \dots$ (166), with which the oscillators are distributed among the different region elements, $w_{1}$ , $w_{2}$ , $w_{3}, \dots$ being any proper fractions whose sum is $1$ . Herein, as always, the $n$ th region element is supposed to contain the oscillators with energies between $(n - 1) h ν$ and $n h ν$ .

The energy absorbed by the system in the time $d t$ within the conical element $d Ω$ is, according to (321),

(322)

$\frac{π N d t}{c L} {sin}^{2} θ (K {cos}^{2} ψ + K' {sin}^{2} ψ) d Ω .$

Let us now calculate also the energy emitted within the same conical element.

180. The total energy emitted in the time element $d t$ by all $N$ oscillators is found from the consideration that a single oscillator, according to (249), takes up an energy element $h ν$ during the time

(323)

$\frac{4 h ν L}{I} = τ,$

and hence has a chance to emit once, the probability being $η$ . We shall assume that the intensity $I$ of the exciting vibration does not change appreciably in the time $τ$ . Of the $N w_{n}$ oscillators which at the time $t$ are in the $n$ th region element a number $N w_{n} η$ will emit during the time $τ$ , the energy emitted by each being $n h ν$ . From (323) we see that the energy emitted by all oscillators during the time element $d t$ is

$\sum N w_{n} η n h ν \frac{d t}{τ} = \frac{N η I d t}{4 L} \sum n w_{n},$

or, according to (265),

(324)

$\frac{N (1 - η) d t}{4 p L} \sum n w_{n} .$

From this the energy emitted within the conical element $d Ω$ may be calculated by considering that, in the state of thermodynamic equilibrium, the energy emitted in every conical element is equal to the energy absorbed and that, in the general case, the energy emitted in a certain direction is independent of the energy simultaneously absorbed. For the stationary state we have from (160) and (265)

(325)

$K = K' = \frac{3 c}{32 π^{2}} I = \frac{3 c}{32 π^{2}} \frac{1 - η}{p η}$

and further from (271) and (265)

(326)

$w_{n} = \frac{1}{p I} {(\frac{p I}{1 + p I})}^{n} = η {(1 - η)}^{n - 1},$

and hence

(327)

$\sum n w_{n} = η \sum n {(1 - η)}^{n - 1} = \frac{1}{η} .$

Thus the energy emitted (324) becomes

(328)

$\frac{N (1 - η) d t}{4 L p η} .$

This is, in fact, equal to the total energy absorbed, as may be found by integrating the expression (322) over all conical elements $d Ω$ and taking account of (325).

Within the conical element $d Ω$ the energy emitted or absorbed will then be

$\frac{π N d t}{c} {sin}^{2} θ K d Ω,$

or, from (325), (327) and (268),

(329)

$\frac{π h ν^{3} (1 - η) N}{c^{3} L} \sum n w_{n} {sin}^{2} θ d Ω d t,$

and this is the general expression for the energy emitted by the system of oscillators in the time element $d t$ within the conical element $d Ω$ , as is seen by comparison with (324).

181. Let us now, as a preparation for the following deductions, consider more closely the properties of the different pencils passing the system of oscillators. From all directions rays strike the volume-element that contains the oscillators; if we again consider those which come toward it in the direction $(θ, φ)$ within the conical element $d Ω$ , the vertex of which lies in the volume-element, we may in the first place think of them as being resolved into their monochromatic constituents, and then we need consider further only that one of these constituents which corresponds to the frequency $ν$ of the oscillators; for all other rays simply pass the oscillators without influencing them or being influenced by them. The specific intensity of a monochromatic ray of frequency $ν$ is

$K + K'$

where $K$ and $K'$ represent the principal intensities which we assume as non-coherent. This ray is now resolved into two components according to the directions of its principal planes of vibration (Sec. 176).

The first component,

$K {sin}^{2} ψ + K' {cos}^{2} ψ,$

passes by the oscillators and emerges on the other side with no change whatever. Hence it gives a plane polarized ray, which starts from the system of oscillators in the direction $(θ, φ)$ within the solid angle $d Ω$ and whose vibrations are perpendicular to the axis of the oscillators and whose intensity is

(330)

$K {sin}^{2} ψ + K' {cos}^{2} ψ = K'' .$

The second component,

$K {cos}^{2} ψ + K' {sin}^{2} ψ,$

polarized at right angles to the first consists again, according to Sec. 176, of two parts

(331)

$\begin{array}{l} (K {cos}^{2} ψ + K' {sin}^{2} ψ) {cos}^{2} θ \end{array}$

(332)

$a n d (K {cos}^{2} ψ + K' {sin}^{2} ψ) {sin}^{2} θ,$

of which the first passes by the system without any change, since its direction of vibration is at right angles to the axes of the oscillators, while the second is weakened by absorption, say by the small fraction $β$ . Hence on emergence this component has only the intensity

(333)

$(1 - β) (K {cos}^{2} ψ + K' {sin}^{2} ψ) {sin}^{2} θ .$

It is, however, strengthened by the radiation emitted by the system of oscillators (329), which has the value

(334)

$β' (1 - η) \sum n w_{n} {sin}^{2} θ,$

where $β'$ denotes a certain other constant, which depends only on the nature of the system and whose value is obtained at once from the condition that, in the state of thermodynamic equilibrium, the loss is just compensated by the gain.

For this purpose we make use of the relations (325) and (327) corresponding to the stationary state, and thus find that the sum of the expressions (333) and (334) becomes just equal to (332); and thus for the constant $β'$ the following value is found:

$β' = β \frac{3 c}{32 π^{2} p} = β \frac{h ν^{3}}{c^{2}} .$

Then by addition of (331), (333) and (334) the total specific intensity of the radiation which emanates from the system of oscillators within the conical element $d Ω$ , and whose plane of vibration is parallel to the axes of the oscillators, is found to be

(335)

$K''' = K {cos}^{2} ψ + K' {sin}^{2} ψ + β {sin}^{2} θ (K_{e} - (K {cos}^{2} ψ + K' {sin}^{2} ψ))$

where for the sake of brevity the term referring to the emission is written

(336)

$\frac{h ν^{3}}{c^{2}} (1 - η) \sum n w_{n} = K_{e} .$

Thus we finally have a ray starting from the system of oscillators in the direction $(θ, φ)$ within the conical element $d Ω$ and consisting of two components $K''$ and $K'''$ polarized perpendicularly to each other, the first component vibrating at right angles to the axes of the oscillators.

In the state of thermodynamic equilibrium

$K = K' = K'' = K''' = K_{e},$

a result which follows in several ways from the last equations.

182. The constant $β$ introduced above, a small positive number, is determined by the spacial and spectral limits of the radiation influenced by the system of oscillators. If $q$ denotes the cross-section at right angles to the direction of the ray, $Δ ν$ the spectral width of the pencil cut out of the total incident radiation by the system, the energy which is capable of absorption and which is brought to the system of oscillators within the conical element $d Ω$ in the time $d t$ is, according to (332) and (11),

(337)

$q Δ ν (K {cos}^{2} ψ + K' {sin}^{2} ψ) {sin}^{2} θ d Ω d t .$

Hence the energy actually absorbed is the fraction $β$ of this value. Comparing this with (322) we get

(338)

$β = \frac{π N}{q \cdot Δ ν \cdot c L} .$