Eight Lectures on Theoretical Physics/V

Last week I endeavored to point out that we find in the atomic theory a complete explanation for the whole content of the two laws of thermodynamics, if we, with Boltzmann, define the entropy by the probability, and I have further shown, in the example of an ideal monatomic gas, how the calculation of the probability, without any additional special hypothesis, enables us not only to find the properties of gases known from thermodynamics, but also to reach conclusions which lie essentially beyond those of pure thermodynamics. Thus, e. g., the law of Avogadro in pure thermodynamics is only a definition, while in the kinetic theory it is a necessary consequence; furthermore, the value of $$c_{v}$$, the mol-heat of a gas, is completely undetermined by pure thermodynamics, but from the kinetic theory it is of equal magnitude for all monatomic gases and, in fact, equal to $$3$$, corresponding to our experimental knowledge. Today and tomorrow we shall be occupied with the application of the theory to radiant heat, and it will appear that we reach in this apparently quite isolated domain conclusions which a thorough test shows are compatible with experience. Naturally, we take as a basis the electro-magnetic theory of heat radiation, which regards the rays as electro-magnetic waves of the same kind as light rays.

We shall utilize the time today in developing in bold outline the important consequences which follow from the electro-magnetic theory for the characteristic quantities of heat radiation, and tomorrow seek to answer, through the calculation of the entropy, the question concerning the dependence of these quantities upon the temperature, as was done last week for ideal gases. Above all, we are concerned here with the determination of those quantities which at any place in a medium traversed by heat rays determine the state of the radiant heat. The state of radiation at a given place will not be represented by a vector which is determined by three components; for the energy flowing in a given direction is quite independent of that flowing in any other direction. In order to know the state of radiation, we must be able to specify, moreover, the energy which in the time $$dt$$ flows through a surface element $$d\sigma$$ for every direction in space. This will be proportional to the magnitude of $$d\sigma$$, to the time $$dt$$, and to the cosine of the angle $$\theta$$ which the direction considered makes with the normal to $$d\sigma$$. But the quantity to be multiplied by $$d\sigma \cdot dt \cdot \cos \theta$$ will not be a finite quantity; for since the radiation through any point of $$d\sigma$$ passes in all directions, therefore the quantity will also depend upon the magnitude of the solid angle $$d\Omega$$, which we shall assume as the same for all points of $$d\sigma$$. In this manner we obtain for the energy which in the time $$dt$$ flows through the surface element $$d\sigma$$ in the direction of the elementary cone $$d\Omega$$, the expression:
 * $$\begin{align}&(28){\color{White}.}\qquad&&

K d\sigma dt \cdot \cos \theta \cdot d\Omega. \end{align}$$ $$K$$ is a positive function of place, of time and of direction, and is for unpolarized light of the following form:
 * $$\begin{align}&(29){\color{White}.}\qquad&&

K = 2 \int_{0}^{\infty} \mathfrak{K}_{\nu} d\nu \end{align}$$ where $$\nu$$ denotes the frequency of a color of wave length $$\lambda$$ and whose velocity of propagation is $$q$$:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\nu = \frac{q}{\lambda}, \end{align}$$ and $$\mathfrak{K}_{\nu}$$ denotes the corresponding intensity of spectral radiation of the plane polarized light.

From the value of $$K$$ is to be found the space density of radiation $$\epsilon$$, i. e., the energy of radiation contained in unit volume. The point $$0$$ in question forms the centre of a sphere whose radius $$r$$ we take so small that in the distance $$r$$ no appreciable absorption of radiation takes place. Then each element $$d\sigma$$ of the surface of the sphere furnishes, by virtue of the radiation traversing the same, the following contribution to the radiation density at $$0$$:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\frac{d\sigma \cdot dt \cdot K \cdot d\Omega}{r^{2} d\Omega \cdot q dt} = \frac{d\sigma \cdot K}{r^{2} q}. \end{align}$$ For the radiation cone of solid angle $$d\Omega$$ proceeding from a point of $$d\sigma$$ in the direction toward $$0$$ has at the distance $$r$$ from $$d\sigma$$ the cross-section $$r^{2} d\Omega$$ and the energy passing in the time $$dt$$ through this cross-section distributes itself along the distance $$q dt$$. By integration over all of the surface elements $$d\sigma$$ we obtain the total space density of radiation at $$0$$:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\epsilon = \int \frac{d\sigma K}{r^{2} q} = \frac{1}{q} \int K d\Omega, \end{align}$$ wherein $$d\Omega$$ denotes the solid angle of an elementary cone whose vertex is $$0$$. For uniform radiation we obtain:
 * $$\begin{align}&(30){\color{White}.}\qquad&&

\epsilon = \frac{4\pi K}{q} = \frac{8\pi}{q} \cdot \int_{0}^{\infty} \mathfrak{K}_{\nu} d\nu. \end{align}$$

The production of radiant heat is a consequence of the act of emission, and its destruction is the result of absorption. Both processes, emission and absorption, have their origin only in material particles, atoms or electrons, not at the geometrical bounding surface; although one frequently says, for the sake of brevity, that a surface element emits or absorbs. In reality a surface element of a body is a place of entrance for the radiation falling upon the body from without and which is to be absorbed; or a place of exit for the radiation emitted from within the body and passing through the surface in the outward direction. The capacity for emission and the capacity for absorption of an element of a body depend only upon its own condition and not upon that of the surrounding elements. If, therefore, as we shall assume in what follows, the state of the body varies only with the temperature, then the capacity for emission and the capacity for absorption of the body will also vary only with the temperature. The dependence upon the temperature can of course be different for each wave length.

We shall now introduce that result following from the second law of thermodynamics which will serve us as a basis in all subsequent considerations: “a system of bodies at rest of arbitrary nature, form and position, which is surrounded by a fixed shell impervious to heat, passes in the course of time from an arbitrarily chosen initial state to a permanent state in which the temperature of all bodies of the system is the same.” This is the thermodynamic state of equilibrium in which the entropy of the system, among all those values which it may assume compatible with the total energy specified by the initial conditions, has a maximum value. Let us now apply this law to a single homogeneous isotropic medium which is of great extent in all directions of space and which, as in all cases subsequently considered, is surrounded by a fixed shell, perfectly reflecting as regards heat rays. The medium possesses for each frequency $$\nu$$ of the heat rays a finite capacity for emission and a finite capacity for absorption. Let us consider, now, such regions of the medium as are very far removed from the surface. Here the influence of the surface will be in any case vanishingly small, because no rays from the surface reach these regions, and on account of the homogeneity and isotropy of the medium we must conclude that the heat radiation is in thermodynamic equilibrium everywhere and has the same properties in all directions, so that $$\mathfrak{K}_{\nu}$$, the specific intensity of radiation of a plane polarized ray, is independent of the frequency $$\nu$$, of the azimuth of polarization, of the direction of the ray, and of location. Thus, there will correspond to each diverging bundle of rays in an elementary cone $$d\Omega$$, proceeding from a surface element $$d\sigma$$, an exactly equal bundle oppositely directed, within the same elemental cone converging toward the surface element. This law retains its validity, as a simple consideration shows, right up to the surface of the medium. For in thermodynamic equilibrium each ray must possess exactly the same intensity as that of the directly opposite ray, otherwise, more energy would flow in one direction than in the opposite direction. Let us fix our attention upon a ray proceeding inwards from the surface, this must have the same intensity as that of the directly opposite ray coming from within, and from this it follows immediately that the state of radiation of the medium at all points on the surface is the same as that within. The nature of the bounding surface and the spacial extent of the medium are immaterial, and in a stationary state of radiation $$\mathfrak{K}_{\nu}$$ is completely determined by the nature of the medium for each temperature.

This law suffers a modification, however, in the special case that the medium is absolutely diathermanous for a definite frequency $$\nu$$. It is then clear that the capacity for absorption and also that for emission must be zero, because otherwise no stationary state of radiation could exist, i. e., a medium emits no color which it does not absorb. But equilibrium can then obviously exist for every intensity of radiation of the frequency considered, i. e., $$\mathfrak{K}_{\nu}$$ is now undetermined and cannot be found without knowledge of the initial conditions. An important example of this is furnished by an absolute vacuum, which is diathermanous for all frequencies. In a complete vacuum thermodynamic equilibrium can therefore exist for each arbitrary intensity of radiation and for each frequency, i. e., for each arbitrary distribution of the spectral energy. From a general thermodynamic point of view this indeterminateness of the properties of thermodynamic states of equilibrium is explained through the presence of numerous different relative maxima of the entropy, as in the case of a vapor which is in a state of supersaturation. But among all the different maxima there is a special maximum, the absolute, which indicates stable equilibrium. In fact, we shall see that in a diathermanous medium for each temperature there exists a quite definite intensity of radiation, which is designated as the stable intensity of radiation of the frequency $$\nu$$ considered. But for the present we shall assume for all frequencies a finite capacity for absorption and for emission.

We consider now two homogeneous isotropic media in thermodynamic equilibrium separated from each other by a plane surface. Since the equilibrium will not be disturbed if one imagines for the moment the surface of separation between the two substances to be replaced by a surface quite non-transparent to heat radiation, all of the foregoing laws hold for each of the two substances individually. Let the specific intensity of radiation of frequency $$\nu$$, polarized in any arbitrary plane within the first substance (the upper in Fig. 1) , be $$\mathfrak{K}_{\nu}$$ and that within the second substance $${\mathfrak{K}_{\nu}}'$$ (we shall in general designate with a dash those quantities which refer to the second substance). Both quantities $$\mathfrak{K}_{\nu}$$ and $${\mathfrak{K}_{\nu}}'$$, besides depending upon the temperature and the frequency, depend only upon the nature of the two substances, and, in fact, these values of the intensity of radiation hold quite up to the boundary surface between the substances, and are therefore independent of the properties of this surface.



Each ray from the first medium is split into two rays at the boundary surface: the reflected and the transmitted. The directions of these two rays vary according to the angle of incidence and the color of the incident ray, and, in addition, the intensity varies according to its polarization. If we denote by $$\rho$$ (the reflection coefficient) the amount of the reflected energy of radiation and consequently by $$1 - \rho$$ the amount of transmitted energy with respect to the incident energy, then $$\rho$$ depends upon the angle of incidence, upon the frequency and upon the polarization of the incident ray. Similar remarks hold for $$\rho'$$, the reflection coefficient for a ray from the second medium, upon meeting the boundary surface.

Now the energy of a monochromatic plane polarized ray of frequency $$\nu$$ proceeding from an element $$d\sigma$$ of the boundary surface within the elementary cone $$d\Omega$$ in a direction toward the first medium (see the feathered arrow at the left in Fig. 1) is for the time $$dt$$, in accordance with $$(28)$$ and $$(29)$$:
 * $$\begin{align}&(31){\color{White}.}\qquad&&

dt \cdot d\sigma \cdot \cos \theta \cdot d\Omega \cdot \mathfrak{K}_{\nu} d\nu, \end{align}$$ where
 * $$\begin{align}&(32){\color{White}.}\qquad&&

d\Omega = \sin \theta d\theta d\varphi. \end{align}$$ This energy is furnished by the two rays which, approaching the surface from the first and the second medium respectively, are reflected and transmitted respectively at the surface element $$d\sigma$$ in the same direction. (See the unfeathered arrows. The surface element $$d\sigma$$ is indicated only by the point $$0$$.) The first ray proceeds in accordance with the law of reflection within the symmetrically drawn elementary cone $$d\Omega$$: the second approaches the surface within the elementary cone
 * $$\begin{align}&(33){\color{White}.}\qquad&&

d\Omega' = \sin \theta' d\theta' d\varphi', \end{align}$$ where, in accordance with the law of refraction,
 * $$\begin{align}&(34){\color{White}.}\qquad&&

\varphi' = \varphi\quad \text{and}\quad \frac{\sin \theta}{\sin \theta'} = \frac{q}{q'}. \end{align}$$ We now assume that the ray is either polarized in the plane of incidence or perpendicular to this plane, and likewise for the two radiations out of whose energies it is composed. The radiation coming from the first medium and reflected from $$d\sigma$$ contributes the energy:
 * $$\begin{align}&(35){\color{White}.}\qquad&&

\rho \cdot dt \cdot d\sigma \cos \theta \cdot d\Omega \cdot \mathfrak{K}_{\nu} d\nu, \end{align}$$ and the radiation coming from the second medium and transmitted through $$d\sigma$$ contributes the energy:
 * $$\begin{align}&(36){\color{White}.}\qquad&&

(1 - \rho') \cdot dt \cdot d\sigma \cos \theta' \cdot d\Omega' \cdot {\mathfrak{K}_{\nu}}' d\nu. \end{align}$$ The quantities $$dt$$, $$d\sigma$$, $$\nu$$, and $$d\nu$$ are here written without the accent, since they have the same values in both media.

Adding the expressions $$(35)$$ and $$(36)$$ and placing the sum equal to the expression $$(31)$$, we obtain:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\rho \cos \theta d\Omega \mathfrak{K}_{\nu} + (1 - \rho') \cos \theta' d\Omega' {\mathfrak{K}_{\nu}}' = \cos \theta d\Omega \mathfrak{K}_{\nu}. \end{align}$$ Now, in accordance with $$(34)$$:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\frac{\cos \theta d\theta}{q} = \frac{\cos \theta' d\theta'}{q'}, \end{align}$$ and further, taking note of $$(32)$$ and $$(33)$$:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

d\Omega' \cos \theta' = d\Omega \cos \theta \cdot \frac{q'^{2}}{q^{2}}, \end{align}$$ and it follows that:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\rho \mathfrak{K}_{\nu} + (1 - \rho') \frac{q'^{2}}{q^{2}} {\mathfrak{K}_{\nu}}' = \mathfrak{K}_{\nu} \end{align}$$ or:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\frac{\mathfrak{K}_{\nu}}{{\mathfrak{K}_{\nu}}'} \cdot \frac{q^{2}}{q'^{2}} = \frac{1 - \rho'}{1 - \rho}. \end{align}$$

In the last equation the quantity on the left is independent of the angle of incidence $$\theta$$ and of the kind of polarization, consequently the quantity upon the right side must also be independent of these quantities. If one knows the value of these quantities for a single angle of incidence and for a given kind of polarization, then this value is valid for all angles of incidence and for all polarizations. Now, in the particular case that the rays are polarized at right angles to the plane of incidence and meet the bounding surface at the angle of polarization,
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\rho = 0\quad \text{and}\quad \rho' = 0. \end{align}$$ Then the expression on the right will be equal to $$1$$, and therefore it is in general equal to $$1$$, and we have always:
 * $$\begin{align}&(37){\color{White}.}\qquad&&

\rho = \rho',\quad q^{2} \mathfrak{K}_{\nu} = q'^{2} {\mathfrak{K}_{\nu}}'. \end{align}$$ The first of these two relations, which asserts that the coefficient of reflection is the same for both sides of the boundary surface, constitutes the special expression of a general reciprocal law, first announced by Helmholz, whereby the loss of intensity which a ray of given color and polarization suffers on its path through any medium in consequence of reflection, refraction, absorption, and dispersion is exactly equal to the loss of intensity which a ray of corresponding intensity, color and polarization suffers in passing over the directly opposite path. It follows immediately from this that the radiation meeting a boundary surface between two media is transmitted or reflected equally well from both sides, for every color, direction and polarization.

The second relation, $$(37)$$, brings into connection the radiation intensities originating in both substances. It asserts that in thermodynamic equilibrium the specific intensities of radiation of a definite frequency in both media vary inversely as the square of the velocities of propagation, or directly as the squares of the refractive indices. We may therefore write
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

q^{2} \mathfrak{K}_{\nu} = F(\nu, T), \end{align}$$ wherein $$F$$ denotes a universal function depending only upon $$\nu$$ and $$T$$, the discovery of which is one of the chief problems of the theory.

Let us fix our attention again on the case of a diathermanous medium. We saw above that in a medium surrounded by a non-transparent shell which for a given color is diathermanous equilibrium can exist for any given intensity of radiation of this color. But it follows from the second law that, among all the intensities of radiation, a definite one, namely, that corresponding to the absolute maximum of the total entropy of the system, must exist, which characterizes the absolutely stable equilibrium of radiation. We now see that this indeterminateness is eliminated by the last equation, which asserts that in thermodynamic equilibrium the product $$q^{2}\mathfrak{K}_{\nu}$$ is a universal function. For it results immediately therefrom that there is a definite value of $$\mathfrak{K}_{\nu}$$ for every diathermanous medium which is thus differentiated from all other values. The physical meaning of this value is derived directly from a consideration of the way in which this equation was derived: it is that intensity of radiation which exists in the diathermanous medium when it is in thermodynamic equilibrium while in contact with a given absorbing and emitting medium. The volume and the form of the second medium is immaterial; in particular, the volume may be taken arbitrarily small.

For a vacuum, the most diathermanous of all media, in which the velocity of propagation $$q = c$$ is the same for all rays, we can therefore express the following law: The quantity
 * $$\begin{align}&(38){\color{White}.}\qquad&&

\mathfrak{K}_{\nu} = \frac{1}{c^{2}} F(\nu, T) \end{align}$$ denotes that intensity of radiation which exists in any complete vacuum when it is in a stationary state as regards exchange of radiation with any absorbing and emitting substance, whose amount may be arbitrarily small. This quantity $$\mathfrak{K}_{\nu}$$ regarded as a function of $$\nu$$ gives the so-called normal energy spectrum.

Let us consider, therefore, a vacuum surrounded by given emitting and absorbing bodies of uniform temperature. Then, in the course of time, there is established therein a normal energy radiation $$\mathfrak{K}_{\nu}$$ corresponding to this temperature. If now $$\rho_{\nu}$$ be the reflection coefficient of a wall for the frequency $$\nu$$, then of the radiation $$\mathfrak{K}_{\nu}$$ falling upon the wall, the part $$\rho_{\nu} \mathfrak{K}_{\nu}$$ will be reflected. On the other hand, if we designate by $$E_{\nu}$$ the emission coefficient of the wall for the same frequency $$\nu$$, the total radiation proceeding from the wall will be:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\rho_{\nu} \mathfrak{K}_{\nu} + E_{\nu} = \mathfrak{K}_{\nu}, \end{align}$$ since each bundle of rays possesses in a stationary state the intensity $$\mathfrak{K}_{\nu}$$. From this it follows that:
 * $$\begin{align}&(39){\color{White}.}\qquad&&

\mathfrak{K}_{\nu} = \frac{E_{\nu}}{1 - \rho_{\nu}}, \end{align}$$ i. e., the ratio of the emission coefficient $$E_{\nu}$$ to the capacity for absorption $$(1-\rho_{\nu})$$ of a given substance is the same for all substances and equal to the normal intensity of radiation for each frequency (Kirchoff). For the special case that $$\rho_{\nu}$$ is equal to $$0$$, i. e., that the wall shall be perfectly black, we have:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\mathfrak{K}_{\nu} = E_{\nu}, \end{align}$$ that is, the normal intensity of radiation is exactly equal to the emission coefficient of a black body. Therefore the normal radiation is also called “black radiation.” Again, for any given body, in accordance with $$(39)$$, we have:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

E_{\nu} < \mathfrak{K}_{\nu}, \end{align}$$ i. e., the emission coefficient of a body in general is smaller than that of a black body. Black radiation, thanks to W. Wien and O. Lummer, has been made possible of measurement, through a small hole bored in the wall bounding the space considered.

We proceed now to the treatment of the problem of determining the specific intensity $$\mathfrak{K}_{\nu}$$ of black radiation in a vacuum, as regards its dependence upon the frequency $$\nu$$ and the temperature $$T$$. In the treatment of this problem it will be necessary to go further than we have previously done into those processes which condition the production and destruction of heat rays; that is, into the question regarding the act of emission and that of absorption. On account of the complicated nature of these processes and the difficulty of bringing some of the details into connection with experience, it is certainly quite out of the question to obtain in this manner any reliable results if the following law cannot be utilized as a dependable guide in this domain: a vacuum surrounded by reflecting walls in which arbitrary emitting and absorbing bodies are distributed in any given arrangement assumes in the course of time the stationary state of black radiation, which is completely determined by a single parameter, the temperature, and which, in particular, does not depend upon the number, the properties and the arrangement of the bodies. In the investigation of the properties of the state of black radiation the nature of the bodies which are supposed to be in the vacuum is therefore quite immaterial, and it is certainly immaterial whether such bodies actually exist anywhere in nature, so long as their existence and their properties are compatible throughout with the laws of electrodynamics and of thermodynamics. As soon as it is possible to associate with any given special kind and arrangement of emitting and absorbing bodies a state of radiation in the surrounding vacuum which is characterized by absolute stability, then this state can be no other than that of black radiation. Making use of the freedom furnished by this law, we choose among all the emitting and absorbing systems conceivable, the most simple, namely, a single oscillator at rest, consisting of two poles charged with equal quantities of electricity of opposite sign which are movable relative to each other in a fixed straight line, the axis of the oscillator. The state of the oscillator is completely determined by its moment, $$f(t)$$; i. e., by the product of the electric charge of the pole on the positive side of the axis into the distance between the poles, and by its differential quotient with regard to the time:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\frac{df(t)}{dt} = \dot{f}(t). \end{align}$$ The energy of the oscillator is of the following simple form:
 * $$\begin{align}&(40){\color{White}.}\qquad&&

U = \tfrac{1}{2} Kf^{2} + \tfrac{1}{2} L \dot{f}^{2}, \end{align}$$ wherein $$K$$ and $$L$$ denote positive constants which depend upon the nature of the oscillator in some manner into which we need not go further at this time.

If, in the vibrations of the oscillator, the energy $$U$$ remain absolutely constant, we should have: $$dU = 0$$ or:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

K f(t) + L \ddot{f}(t) = 0, \end{align}$$ and from this there results, as a general solution of the differential equation, a pure periodic vibration:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

f = C \cos (2\pi \nu_{0} t - \theta), \end{align}$$ wherein $$C$$ and $$\theta$$ denote the integration constants and $$\nu_{0}$$ the number of vibrations per unit of time:
 * $$\begin{align}&(41){\color{White}.}\qquad&&

\nu_{0} = \frac{1}{2\pi} \sqrt{\frac{K}{L}}. \end{align}$$ Such an oscillator vibrating periodically with constant energy would neither be influenced by the electromagnetic field surrounding it, nor would it exert any external actions due to radiation. It could therefore have no sort of influence on the heat radiation in the surrounding vacuum.

In accordance with the theory of Maxwell, the energy of vibration $$U$$ of the oscillator by no means remains constant in general, but an oscillator by virtue of its vibrations sends out spherical waves in all directions into the surrounding field and, in accordance with the principle of conservation of energy, if no actions from without are exerted upon the oscillator, there must necessarily be a loss in the energy of vibration and, therefore, a damping of the amplitude of vibration is involved. In order to find the amount of this damping we calculate the quantity of energy which flows out through a spherical surface with the oscillator at the center, in accordance with the law of Poynting. However, we may not place the energy flowing outwards in accordance with this law through the spherical surface in an infinitely small interval of time $$dt$$ equal to the energy radiated in the same time interval from the oscillator. For, in general, the electromagnetic energy does not always flow in the outward direction, but flows alternately outwards and inwards, and we should obtain in this manner for the quantity of the radiation outwards, values which are alternately positive and negative, and which also depend essentially upon the radius of the supposed sphere in such manner that they increase toward infinity with decreasing radius—which is opposed to the fundamental conception of radiated energy. This energy will, moreover, be only found independent of the radius of the sphere when we calculate the total amount of energy flowing outwards through the surface of the sphere, not for the time element $$dt$$, but for a sufficiently large time. If the vibrations are purely periodic, we may choose for the time a period; if this is not the case, which for the sake of generality we must here assume, it is not possible to specify a priori any more general criterion for the least possible necessary magnitude of the time than that which makes the energy radiated essentially independent of the radius of the supposed sphere.

In this way we succeed in finding for the energy emitted from the oscillator in the time from $$t$$ to $$t + \mathfrak{T}$$ the following expression:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\frac{2}{3c^{3}} \int_{t}^{t + \mathfrak{T}} \ddot{f}^{2}(t) dt. \end{align}$$ If now, the oscillator be in an electromagnetic field which has the electric component $$\mathfrak{E}_{z}$$ at the oscillator in the direction of its axis, then the energy absorbed by the oscillator in the same time is:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\int_{t}^{t + \mathfrak{T}} \mathfrak{E}_{z} \dot{f} \cdot dt. \end{align}$$ Hence, the principle of conservation of energy is expressed in the following form:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\int_{t}^{t + \mathfrak{T}} \left(\frac{dU}{dt} + \frac{2}{3c^{3}} \ddot{f}^{2} - \mathfrak{E}_{z} \dot{f}\right) dt = 0. \end{align}$$ This equation, together with the assumption that the constant
 * $$\begin{align}&(42){\color{White}.}\qquad&&

\frac{4\pi^{2} \nu_{0}}{3c^{3} L} = \sigma \end{align}$$ is a small number, leads to the following linear differential equation for the vibrations of the oscillator:
 * $$\begin{align}&(43){\color{White}.}\qquad&&

Kf + L\ddot{f} - \frac{2}{3c^{3}} \overset{\ldots}{f} = \mathfrak{E}_{z}. \end{align}$$ In accordance with what precedes, in so far as the oscillator is excited into vibrations by an external field $$\mathfrak{E}_{z}$$, one may designate it as a resonator which possesses the natural period $$\nu_{0}$$ and the small logarithmic decrement $$\sigma$$. The same equation may be obtained from the electron theory, but I have considered it an advantage to derive it in a manner independent of any hypothesis concerning the nature of the resonator.

Now, let the resonator be in a vacuum filled with stationary black radiation of specific intensity $$\mathfrak{K}_{\nu}$$. How, then, does the mean energy $$U$$ of the resonator in a state of stationary vibration depend upon the specific intensity of radiation $$\mathfrak{K}_{\nu_{0}}$$ with the natural period $$\nu_{0}$$ of the corresponding color? It is this question which we have still to consider today. Its answer will be found by expressing on the one hand the energy of the resonator $$U$$ and on the other hand the intensity of radiation $$\mathfrak{K}_{\nu_{0}}$$ by means of the component $$\mathfrak{E}_{z}$$ of the electric field exciting the resonator. Now however complicated this quantity may be, it is capable of development in any case for a very large time interval, from $$t = 0$$ to $$t = \mathfrak{T}$$, in the Fourier's series:
 * $$\begin{align}&(44){\color{White}.}\qquad&&

\mathfrak{E}_{z} = \sum\limits_{n = 1}^{n = \infty} C_{n} \cos \left(\frac{2\pi n t}{\mathfrak{T}} - \theta_{n}\right), \end{align}$$ and for this same time interval $$\mathfrak{T}$$ the moment of the resonator in the form of a Fourier's series may be calculated as a function of $$t$$ from the linear differential equation $$(43)$$. The initial condition of the resonator may be neglected if we only consider such times $$t$$ as are sufficiently far removed from the origin of time $$t = 0$$.

If it be now recalled that in a stationary state of vibration the mean energy $$U$$ of the resonator is given, in accordance with $$(40)$$, $$(41)$$ and $$(42)$$, by:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

U = K \bar{f}^{2} = \frac{16\pi^{4} \nu_{0}{}^{3}}{3 \sigma c^{3}} \bar{f}^{2}, \end{align}$$ it appears after substitution of the value of $$f$$ obtained from the differential equation $$(43)$$ that:
 * $$\begin{align}&(45){\color{White}.}\qquad&&

U = \frac{3 c^{3}}{64\pi^{2} \nu_{0}{}^{2}} \mathfrak{T} \bar{C}_{n0}{}^{2}, \end{align}$$ wherein $$\bar{C}_{n0}{}^{2}$$ denotes the mean value of $$C_{n}$$ for all the series of numbers $$n$$ which lie in the neighborhood of the value $$\nu_{0} \mathfrak{T}$$, i. e., for which $$\nu_{0} \mathfrak{T}$$ is approximately $$= 1$$.

Now let us consider on the other hand the intensity of black radiation, and for this purpose proceed from the space density of the total radiation. In accordance with $$(30)$$, this is:
 * $$\begin{align}&(46){\color{White}.}\qquad&&

\epsilon = \frac{8\pi}{c} \int_{0}^{\infty} \mathfrak{K}_{\nu} d\nu = \frac{1}{8\pi} (\bar{\mathfrak{E}}_{x}{}^{2} + \bar{\mathfrak{E}}_{y}{}^{2} + \bar{\mathfrak{E}}_{z}{}^{2}                 + \bar{\mathfrak{H}}_{x}{}^{2} + \bar{\mathfrak{H}}_{y}{}^{2} + \bar{\mathfrak{H}}_{z}{}^{2}), \end{align}$$ and therefore, since the radiation is isotropic, in accordance with $$(44)$$:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\frac{8\pi}{c} \int_{0}^{\infty} \mathfrak{K}_{\nu} d\nu = \frac{3}{4\pi} \bar{\mathfrak{E}}_{z}{}^{2} = \frac{3}{8\pi} \sum\limits_{n = 1}^{n = \infty} C_{n}{}^{2}. \end{align}$$ If we write $$\Delta n/\mathfrak{T}$$ on the left instead of $$d\nu$$, where $$\Delta n$$ is a large number, we get:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\frac{8\pi}{c} \sum\limits_{n = 1}^{n = \infty} \mathfrak{K}_{v} \frac{\Delta n}{\mathfrak{T}} = \frac{3}{8\pi} \sum\limits_{n = 1}^{n = \infty} C_{n}{}^{2}, \end{align}$$ and obtain then by “spectral” division of this equation:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\frac{8\pi}{c} \mathfrak{K}_{\nu_{0}} \frac{\Delta n}{\mathfrak{T}} = \frac{3}{8\pi} \sum\limits_{n_{0} - (\Delta n/2)}^{n_{0} + (\Delta n/2)} C_{n}{}^{2}, \end{align}$$ and, if we introduce again the mean value
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\frac{1}{\Delta n} \cdot \sum\limits_{n_{0} - (\Delta n/2)}^{n_{0} + (\Delta n/2)} C_{n}{}^{2} = \bar{C}_{n0}{}^{2}, \end{align}$$ we then get:
 * $$\begin{align}&{\color{White}.(00)}\qquad&&

\mathfrak{K}_{\nu_{0}} = \frac{3 c \mathfrak{T}}{64\pi^{2}} \cdot \bar{C}_{n 0}. \end{align}$$ By comparison with $$(45)$$ the relation sought is now found:
 * $$\begin{align}&(47){\color{White}.}\qquad&&

\mathfrak{K}_{\nu_{0}} = \frac{\nu_{0}{}^{2}}{c^{2}} U, \end{align}$$ which is striking on account of its simplicity and, in particular, because it is quite independent of the damping constant $$\sigma$$ of the resonator.

This relation, found in a purely electrodynamic manner, between the spectral intensity of black radiation and the energy of a vibrating resonator will furnish us in the next lecture, with the aid of thermodynamic considerations, the necessary means of attack in deriving the temperature of black radiation together with the distribution of energy in the normal spectrum.