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 [sect:17.] Sec. 17.

Let us first calculate the total intensity $J=\overline{{⋿}_z^2\mathrm{\backslash Strut}}$ 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, $d\Omega$, is given by [eqn:(5)] (5), where the angles $\theta$ and $\varphi$, 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 ${𝖪}^{\prime }$ ([sect:17.] 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 $\psi$, no matter in which quadrant it lies, then, according to [eqn:(8)] (8), the specific intensity of the monochromatic pencil may be resolved into the two plane polarized components at right angles with each other,

$$\begin{gathered} 𝖪{\cos }^2\psi \&+{𝖪}^{\prime }{\sin }^2\psi \\ 𝖪{\sin }^2\psi \&+{𝖪}^{\prime }{\cos }^2\psi , \end{gathered}$$

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 ${⋿}_z^2$, 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 ${\displaystyle \frac{\pi }{2}}-\theta$ 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 ${\displaystyle \frac{c}{4\pi }}$ and the mean square of the electric field-strength. Hence the mean square of the electric field-strength of the pencil here considered is $\frac{4\pi }{c}(𝖪{\cos }^2\psi +{𝖪}^{\prime }{\sin }^2\psi )d\nu d\Omega ,$

and the mean square of its component in the direction of the $z$-axis is $\frac{4\pi }{c}(𝖪{\cos }^2\psi +{𝖪}^{\prime }{\sin }^2\psi ){\sin }^2\theta d\nu d\Omega .\mathrm{\backslash Label}[eqn](317)\text{upshape (317)}$ By integration over all frequencies and all solid angles we then obtain the value required $\overline{{⋿}_z^2\mathrm{\backslash Strut}}=\frac{4\pi }{c}\int {\sin }^2\theta d\Omega \int d\nu ({𝖪}_{\nu }{\cos }^2\psi +{𝖪}_{\nu }^{\prime }{\sin }^2\psi )=J.\mathrm{\backslash Label}[eqn](318)\text{upshape (318)}$

The space density u of the electromagnetic energy at a point of the field is u = 1 8 π ( ⋿ x 2 \Strut ― + ⋿ y 2 \Strut ― + ⋿ z 2 \Strut ― + 𝖧 x 2 \Strut ― + 𝖧 y 2 \Strut ― + 𝖧 z 2 \Strut ― ) , where ⋿ x 2 , ⋿ y 2 , ⋿ 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