Chapter II

One Oscillator in the Field of Radiation

175. If in any field of radiation whatever we have an ideal oscillator of the kind assumed above (Sec. 135), there will take place between it and the radiation falling on it certain mutual actions, for which we shall again assume the validity of the elementary dynamical law introduced in the preceding section. The question is then, how the processes of emission and absorption will take place in the case now under consideration.

In the first place, as regards the emission of radiant energy by the oscillator, this takes place, as before, according to the hypothesis of emission of quanta (Sec. 147), where the probability quantity $η$ again depends on the corresponding spectral intensity $I$ through the relation (265).

On the other hand, the absorption is calculated, exactly as above, from (234), where the vibrations of the oscillator also take place according to the equation (233). In this way, by calculations analogous to those performed in the second chapter of the preceding part, with the difference only that instead of the Fourier’s series (235) the Fourier’s integral (311) is used, we obtain for the energy absorbed by the oscillator in the time $τ$ the expression

$\frac{τ}{4 L} \int d μ (A_{μ} cos 2 π μ t + B_{μ} sin 2 π μ t),$

where the constants $A_{μ}$ and $B_{μ}$ denote the mean values expressed in (316), taken for the spectral region in the neighborhood of the natural frequency $ν_{0}$ of the oscillator. Hence the law of absorption will again be given by equation (249), which now holds also for an intensity of vibration $I$ varying with the time.

176. There now remains the problem of deriving the expression for $I$ , the spectral intensity of the vibration exciting the oscillator, when the thermodynamic state of the field of radiation at the oscillator is given in accordance with the statements made in Sec. 17.

Let us first calculate the total intensity $J = \bar{E_{z}^{2}}$ of the vibration exciting an oscillator, from the intensities of the heat rays striking the oscillator from all directions.

For this purpose we must also allow for the polarization of the monochromatic rays which strike the oscillator. Let us begin by considering a pencil which strikes the oscillator within a conical element whose vertex lies in the oscillator and whose solid angle, $d Ω$ , is given by (5), where the angles $θ$ and $φ$ , polar coordinates, designate the direction of the propagation of the rays. The whole pencil consists of a set of monochromatic pencils, one of which may have the principal values of intensity $K$ and $K'$ (Sec. 17). If we now denote the angle which the plane of vibration belonging to the principal intensity $K$ makes with the plane through the direction of the ray and the $z$ -axis (the axis of the oscillator) by $ψ$ , no matter in which quadrant it lies, then, according to (8), the specific intensity of the monochromatic pencil may be resolved into the two plane polarized components at right angles with each other,

$\begin{array}{l} K {cos}^{2} ψ & + K' {sin}^{2} ψ \\ K {sin}^{2} ψ & + K' {cos}^{2} ψ, \end{array}$

the first of which vibrates in a plane passing through the $z$ -axis and the second in a plane perpendicular thereto.

The latter component does not contribute anything to the value of $E_{z}^{2}$ , since its electric field-strength is perpendicular to the axis of the oscillator. Hence there remains only the first component whose electric field-strength makes the angle $\frac{π}{2} - θ$ with the $z$ -axis. Now according to Poynting’s law the intensity of a plane polarized ray in a vacuum is equal to the product of $\frac{c}{4 π}$ and the mean square of the electric field-strength. Hence the mean square of the electric field-strength of the pencil here considered is

$\frac{4 π}{c} (K {cos}^{2} ψ + K' {sin}^{2} ψ) d ν d Ω,$

and the mean square of its component in the direction of the $z$ -axis is

(317)

$\frac{4 π}{c} (K {cos}^{2} ψ + K' {sin}^{2} ψ) {sin}^{2} θ d ν d Ω .$

By integration over all frequencies and all solid angles we then obtain the value required

(318)

$\bar{E_{z}^{2}} = \frac{4 π}{c} \int {sin}^{2} θ d Ω \int d ν (K_{ν} {cos}^{2} ψ + K' {sin}^{2} ψ) = J .$

The space density $u$ of the electromagnetic energy at a point of the field is

$u = \frac{1}{8 π} (\bar{E_{x}^{2}} + \bar{E_{y}^{2}} + \bar{E_{z}^{2}} + \bar{H_{x}^{2}} + \bar{H_{y}^{2}} + \bar{H_{z}^{2}}),$

where $E_{x}^{2}$ , $E_{y}^{2}$ , $E_{z}^{2}$ , $H_{x}^{2}$ , $H_{y}^{2}$ , $H_{z}^{2}$ denote the squares of the field-strengths, regarded as “slowly variable” quantities, and are hence supplied with the dash to denote their mean value. Since for every separate ray the mean electric and magnetic energies are equal, we may always write

(319)

$u = \frac{1}{4 π} = (\bar{E_{x}^{2}} + \bar{E_{y}^{2}} + \bar{E_{z}^{2}}) .$

If, in particular, all rays are unpolarized and if the intensity of radiation is constant in all directions, $K_{ν} = K'$ and, since

(319a)

$\begin{matrix} \int {sin}^{2} θ d Ω = \iint {sin}^{3} θ d θ d φ = \frac{8 π}{3} \\ \bar{E_{z}^{2}} = \frac{32 π^{2}}{3 c} \int K_{ν} d ν = \bar{E_{x}^{2}} = \bar{E_{y}^{2}} \end{matrix}$

and, by substitution in (319),

$u = \frac{8 π}{c} \int K_{ν} d ν,$

which agrees with (22) and (24).

177. Let us perform the spectral resolution of the intensity $J$ according to Sec. 174; namely,

$J = \int I_{ν} d ν .$

Then, by comparison with (318), we find for the intensity of a definite frequency $ν$ contained in the exciting vibration the value

(320)

$I = \frac{4 π}{c} \int {sin}^{2} θ d Ω (K_{ν} {cos}^{2} ψ + K' {sin}^{2} ψ) .$

For radiation which is unpolarized and uniform in all directions we obtain again, in agreement with (160),

$I = \frac{32 π^{2}}{3 c} K .$

178. With the value (320) obtained for $I$ the total energy absorbed by the oscillator in an element of time $d t$ from the radiation falling on it is found from (249) to be

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

Hence the oscillator absorbs in the time $d t$ from the pencil striking it within the conical element $d Ω$ an amount of energy equal to

(321)

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