Talk:PlanetPhysics/Radiation at Thermodynamic Equilibrium Kirchhoffs Law Black Radiation

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\subsection{CHAPTER II} \subsection{Radiation at Thermodynamic Equilibrium. Kirchoff's Law. Black Radiation} From \htmladdnormallink{the theory of heat radiation}{http://planetphysics.us/encyclopedia/TheoryOfHeatRadiation.html} by Max Planck

{\bf 24.} We shall now apply the laws enunciated in the last chapter to the special case of \htmladdnormallink{thermodynamic equilibrium}{http://planetphysics.us/encyclopedia/ThermalEquilibrium.html}, and hence we begin our consideration by stating a certain consequence of the second principle of \htmladdnormallink{Thermodynamics}{http://planetphysics.us/encyclopedia/Thermodynamics.html}: A \htmladdnormallink{system}{http://planetphysics.us/encyclopedia/GenericityInOpenSystems.html} of bodies of arbitrary nature, shape, and \htmladdnormallink{position}{http://planetphysics.us/encyclopedia/Position.html} which is at rest and is surounded by a rigid cover impermeable to \htmladdnormallink{heat}{http://planetphysics.us/encyclopedia/Heat.html} will, no matter what its initial state may be, pass in the course of time into a permanent state, in which the \htmladdnormallink{temperature}{http://planetphysics.us/encyclopedia/BoltzmannConstant.html} of all bodies of the system is the same. This is the state of thermodynamic equilibrium, in which the \htmladdnormallink{entropy}{http://planetphysics.us/encyclopedia/ThermodynamicLaws.html} of the system has the maximum value compatible with the total \htmladdnormallink{energy}{http://planetphysics.us/encyclopedia/CosmologicalConstant2.html} of the system as fixed by the initial conditions. This state being reached, no further increase in entropy is possible.

In certain cases it may happen that, under the given conditions, the entropy can assume not only one but several maxima, of which one is the absolute one, the others having only a relative significance[1]. In these cases every state corresponding to a maximum value of the entropy represents a state of thermodynamic equilibrium of the system. But only one of them, the one corresponding to the absolute maximum of entropy, represents the absolutely stabe \htmladdnormallink{equilibrium}{http://planetphysics.us/encyclopedia/ThermalEquilibrium.html}. All the others are in a certain sense unstable, inasmuch as a sutiable, however small, disturbance may produce in the system a permanent change in the equilibrium in the direction of the absolutely stable equilibrium. An example of this is offered by supersaturated steam enclosed in a rigid vessel or by any explosive substance. We shall also meet such unstable equilibria in the case of \htmladdnormallink{radiation}{http://planetphysics.us/encyclopedia/Cyclotron.html} phenomena (Sec. 52).

{\bf 25.} We shall now, as in the previous chapter, assume that we are dealing with homogeneous isotropic media whose condition depends only on the temperature, and we shall inquire what laws the radiation phenomena in them must obey in order to be consistent with the deduction from the second principle mentioned in the preceding \htmladdnormallink{section}{http://planetphysics.us/encyclopedia/IsomorphicObjectsUnderAnIsomorphism.html}. The means of answering this inquiry is supplied by the investigation of the state of thermodynamic equilibrium of one or more of such media, this investigation to be conducted by applying the conceptions and laws established in the last chapter.

We shall begin with the simplest case, that of a single medium extending very far in all directions of space, and, like all systems we shall here consider, being surrounded by a rigid cover impermeable to heat. For the preseent we shall assume that the medium has finite coefficients of \htmladdnormallink{absorption}{http://planetphysics.us/encyclopedia/FluorescenceCrossCorrelationSpectroscopy.html}, emission, and scattering.

Let us consider, first, points of the medium that are far away from the surface. At such points the influence of the surface is, of course, vanishingly small and from the homogeneity and the isotropy of the medium it will follow that in a state of thermodynamic equilibrium the radiation of heat has everywhere and in all directions the same properties. Then $\bf{K}_\nu$, the specific intensity of radiation of a plane polarized ray of frequency $\nu$ (Sec. 17), must be independent of the azimuth of the plane of \htmladdnormallink{Polarization}{http://planetphysics.us/encyclopedia/FluorescenceCrossCorrelationSpectroscopy.html} as well as of position and direction of the ray. Hence to each pencil of rays starting at an element of area $d\sigma$ and diverging within a conical element $d\omega$ corresponds an exactly equal pencil of opposite direction converging within the same conical element toward the element of area.

Now the condition of thermodynamic equilibrium requires that the temperature shall be everywhere the same and shall not vary in time. Therfore in any given arbitrary time just as much radiant heat must be absorbed as is emitted in each volume-element of the medium. For the heat of the body depends only on the heat radiation, since, on account of the uniformity in temperature, no \htmladdnormallink{conduction}{http://planetphysics.us/encyclopedia/Conduction.html} of heat takes place. This condition is not influenced by the phenomenon of scattering, because sacttering refers only to a change in direction of the energy radiated, not to a creation or destruction of it. We shall, therefore, calculate the energy emmitted and absorbed in the time $dt$ by a volume-element $v$.

According to equation (2) the energy emitted has the value

$$ dt \, v \cdot 8 \pi \int_0^{\infty}{ \epsilon_\nu d\nu } $$

where $\epsilon_\nu$, the coefficient of emission of the medium, depends only on the frequency $\nu$ and on the temperature in addition to the chemical nature of the medium.

{\bf 26.} For the calculation of the energy absorbed we shall employ the same reasoning as was illustrated by Fig. 1 (Sec. 22) and shall retain the notation there used. The radiant energy absorbed by the volume-element $v$ in the time $dt$ is found by considering the intensities of all the rays passing through the element $v$ and taking that fraction of each of these rays which is absorbed in $v$. Now, according to (19), the conical element that starts from $d\sigma$ and cuts out of the \htmladdnormallink{volume}{http://planetphysics.us/encyclopedia/Volume.html} $v$ a part equal to $fs$ has the intensity (energy radiated per unit time)

$$ d\sigma \cdot \frac{f}{r^2} \cdot K $$

or according to (12), by considering the different parts of the \htmladdnormallink{spectrum}{http://planetphysics.us/encyclopedia/CohomologyTheoryOnCWComplexes.html} separately:

$$ 2 \, d\sigma \frac{f}{r^2} \int_0^{\infty}{ K_{\nu} \, dv}.$$

Hence the intensity of a monochromatic ray is:

$$ 2 \, d\sigma \frac{f}{r^2} K_{\nu} \, dv. $$

The amount of energy of this ray absorbed in the distance $s$ in the time $dt$ is, according to (r),

$$ dt \, \alpha_{\nu} s \, 2 \, d\sigma \frac{f}{r^2} K_{\nu} \, dv.$$

Hence the absorbed part of the energy of this small cone of rays, as found by integrating over all frequencies, is:

$$dt \, 2 \, d\sigma \frac{fs}{r^2} \int_0^{\infty}{ \alpha_{\nu} K_{\nu} \, dv}.$$

When this expression is summed up over all the different crossections $f$ of the concial elements starting at $d\sigma$ and passing through $v$, it is evident that $\sum \,fs = v$, and when we sum up over all elements $d\sigma$ of the spherical surface of radius $r$ we have

$$ \int{ \frac{d\sigma}{r^2} = 4 \pi}. $$

Thus for the total radiant energy absorbed in the time $dt$ by the volume-element $v$ the following expression is found:

\begin{equation} \tag{25} dt \, v \, 8\pi \, \int_0^{\infty}{ \alpha_{\nu} K_{\nu} \, dv}. \end{equation}

By equating the emitted and absorbed energy we obtain:

$$ \int_0^{\infty}{\epsilon_{\nu} d\nu} = \int_0^{\infty}{\alpha_{\nu} K_{\nu} d\nu}$$

A similar \htmladdnormallink{relation}{http://planetphysics.us/encyclopedia/Bijective.html} may be obtained for the separate parts of the spectrum. For the energy emitted and the energy absorbed in the state of thermodynamic equilibrium are equal, not only for the entire radiation of the whole spectrum, but also for each monochromatic radiation. This is readily seen from the following. The \htmladdnormallink{magnitudes}{http://planetphysics.us/encyclopedia/AbsoluteMagnitude.html} of $\epsilon_{nu}$, $\alpha_{\nu}$, and $K_{\nu}$ are independent of position. Hence, if for any single color the absorbed were not equal to the emitted energy, there would be everywhere in the whole medium a continuous increase or decrease of the energy radiation of that particular color at the expense of the other colors. This would be contradictory to the condition that $K_{\nu}$ for each separate frequency does not change with time. We have therefore for each frequency the radiation:

\begin{equation} \tag{26} \epsilon_{\nu} = \alpha{\nu} K_\nu \end{equation}

\begin{equation} \tag{27} K_{\nu} = \frac{\epsilon_{\nu}}{\alpha_{\nu}} \end{equation}

\emph{i.e.: in the interior of a medium in a state of thermodynamic equilibrium the specific intensity of radiation of a certain frequency is equal to the coefficient of emission divided by the coefficient of absorption of the medium for this frequency.}

{\bf 27.} Since $\epsilon_{\nu}$ and $\alpha_{\nu}$ depend only on the nature of the medium, the temperature, and the frequency $\nu$, the intensity of radiation of a definite color in the state of thermodynamic equilibrium is completely defined by the nature of the medium and the temperature. An exceptional case is when $\alpha_{\nu} = 0$, that is, when the medium does not at all absorb the color in question. Since $K_{\nu}$ cannot become infinitely large, a first consequence of this that in that case $\epsilon_{\nu}= 0$ also, that is, a medium does not emit any color which it does not absorb. A second consequence is that if $\epsilon_{\nu}$ and $\alpha_{\nu}$ both vanish, equation (26) is satisfied by every value of $K_{\nu}$. \emph{In a medium which is diathermanous for a certain color thermodynamic equilibrium can exist for any intensity of radiation whatever of the color.}

This supplies an immediate illustration of the cases spoken before (Sec. 24), where, for a given value of the total energy of a system enclosed by a rigid cover impermeable to heat, several states of equilibrium can exist, corresponding to several relative maxima of the entropy. That is to say, since the intensity of radiation of the particular color in the state of thermodynamic equilibrium is quite independent of the temperature of a medium which is diathermanous for this color, the given total energy may be arbitrarily distributed between radiation of that color and the heat of the body, without making thermodynamic equilibrium impossible. Among all these distributions there is one particular one, corresponding to the absolute maximum of entropy, which represents absolutely stable equilibrium. This one, unlike all the others, which are in a certain sense unstable, has the property of not being appreciably affected by a small disturbance. Indeed we shall see later (Sec. 48) that among the infinite number of values, which the quotient $frac{\epsilon_{\nu}}{\alpha_{\nu}}$ can have, if numerator and denominator both vanish, there exists one particular one which depends in a definite way on the nature of the medium, the frequency $\nu$, and the temperature. This distinct value of the fraction is accordingly called the stable intensity of radiation $K_{\nu}$ in the medium, which at the temperature in question is diathermanous for rays of the frequency$\nu$.

Everything that has just been said of a medium which is diathermanous for a certain kind of rays holds true for an absolute vacuum, which is a medium diathermanous for rays of all kinds, the only difference being that one cannot speak of the heat and the temperature of an absolute vacuum in any definite sense.

For the present we again shot put the special case of diathermancy aside and assume that all the media considered have a finite coefficient of absorption.

{\bf 28.} Let us now consider briefly the phenomenon of scattering at thermodynamic equilibrium. Every ray meeting the volume element $\nu$ suffers there, apart from absorption, a certain weakening of its intensity because a certain fraction of its energy is diverted in different directions. The value of the total energy of scattered radiation received and diverted, in the time $dt$ by the volume-element $v$ in all directions, may be calculated from expression (3) in exactly the same way as the value of the absorbed energy was calculated in Sec. 26. Hence we get an expression similar to (25), namely,

\begin{equation} \tag{28} dt \, v\, 8 \pi \int_0^{\infty}{\beta_{\nu} K_{\nu} d{\nu}} \end{equation}

The question as to what becomes of this energy is readily answered. On account of the isotropy of the medium, the energy scattered in $v$ and given by (28) is radiated uniformly in all directions just as in the case of the energy entering $v$. Hence that part of the scattered energy received in $v$ which is radiated out in a cone of \htmladdnormallink{solid}{http://planetphysics.us/encyclopedia/CoIntersections.html} angle $d\Omega$ is obtained by multiplying the last expression by$\frac{d\Omega}{4\pi}$. This gives

$$2 dt \, v\, d\Omega \int_0^{\infty}{\beta_{\nu} K_{\nu} d{\nu}}$$

and, for monochromatic plane polarized radiation, \begin{equation} \tag{29} dt \, v\, d\Omega \beta_{\nu} K_{\nu} d{\nu} \end{equation}

Here it must be carefully kept in mind that this uniformity of radiation in all directions holds only for all rays striking the element $v$ taken together; a single ray, even in an isotropic medium, is scattered in different directions with different intensities and different directions of polarization. (See end of Sec. 8.)

(start page 30)
 * Missing pages 28 and 29, will get later *******

on the way. For $\beta_{\nu} = 0$ expressions (33) and (30) are identical, as may be seen by comparison with (27). Generally, however, (30) is larger than (33) because the energy reaching $d \sigma$ contains also some rays which were not at all emitted from elements inside of the pencil, but somewhere else, and have entered later on by scattering. In fact, the volume-elements of the pencil do not merely scatter outward the radiation which is being transmitted inside the pencil, but they also collect into the pencil rays coming from without. The radiation $E'$ thus collected by the volume-element at $r_0$ is found, by putting in (29),

$$dt = 1, v = dr_0 d\Omega r_0^2, d\Omega = \frac{d\sigma}{r_0^2}$$

to be

$$E' = dr_0 \, d\Omega \, d \sigma \beta_{\nu} K_{\nu} d\nu $$

This energy is to be added to the energy $E$ emitted by the volume-element, which we have calculated in (31). Thus for the total energy contributed to the pencil in the volume-element at $r_0$ we find:

$$E + E' = dr_0 d\Omega d\sigma (\epsilon_{\nu} + \beta_{\nu} K_{\nu}) d\nu$$

The part of this reaching $O$ is, similar to (32):

$$dr_0 d\Omega d\sigma (\epsilon_{\nu} + \beta_{nu} K_{\nu}) d\nu e^{-r_0\left ( \alpha_{\nu} + \beta_{\nu} \right ) } $$

Making due allowance for emission and collection of scattered rays entering on the way, as well as for losses by absorption and scattering, all volume-elements of the pencil combined give for the energy ultimately reaching $d\sigma$

$$d \Omega d \sigma \left (\epsilon_{\nu} + \beta_{\nu} K_{\nu} \right ) d\nu \int_0^{\infty} dr_0 e^{-r_0 \left (\alpha_{\nu} + \beta{\nu} \right )} = d\Omega d\sigma \frac{\epsilon_{\nu} + \beta_{\nu} K_{\nu}}{\alpha_{\nu} + \beta_{\nu}} \nu $$

and this expression is really exactly equal to that given by (30), as may be seen by comparison with (26).

{\bf 32.} The laws just derived for the state of radiation of a homogeneous isotropic medium when it is in thermodynamic equilibrium hold, so far as we have seen, only for parts of the medium which lie very far away from the surface, because for such parts only may the radiation be considered, by symmetry, as independent of position and direction. A simple consideration, however, shows that the value of $K_{\nu}$, which was already calculated and given by (27), and which depends only on the temperature and the nature of the medium, gives the correct value of the intensity of radiation of the frequency considered for all directions up to points directly below the surface of the medium. For in the state of thermodynamic equilibrium every ray must have just the same intensity as the one traveling in an exactly opposite direction, since otherwise the radiation would cause a unidirectional transport of energy. Consider then any ray coming from the surface of the medium and directed inward; it must have the same intensity as the opposite ray, coming from the interior. A further immediate consequence of this is \emph{that the total state of radiation of the medium is the same on the surface as in the interior.}

{\bf 33.} While the radiation that starts from a surface element and is directed toward the interior of the medium is in every respect equal to that emanating from an equally large parallel element of area in the interior, it nevertheless has a different history. That is to say, since the surface of the medium was assumed to be impermeable to heat, it is produced only by \htmladdnormallink{reflection}{http://planetphysics.us/encyclopedia/FluorescenceCrossCorrelationSpectroscopy.html} at the surface of radiation coming from the interior. So far as special details are concerned, this can happen in very different ways, depending on whether the surface is assumed to be smooth, \emph{i.e.}, in this case reflecting, or rough, \emph{e.g.}, white (Sec. 10). In the first case there corresponds to each pencil which strikes the surface another perfectly definite pencil, symmetrically situated and having the same intensity, while in the second case every incident pencil is broken up into an infinite number of reflected pencils, each having a different direction, intensity, and polarization. While this is the case, nevertheless the rays that strike a surface-element from all different directions with the same intensity $K_{\nu}$ also produce, all taken together, a uniform radiation of the same intensity $K_{\nu}$, directed toward the interior of the medium.

{\bf 34.} Hereafter there will not be the slightest difficulty in dispensing with the assumption made in Sec. 25 that the medium in question extends very far in all directions. For after thermodynamic equilibrium has been everywhere established in our medium, the equilibrium has been everywhere established in our medium, the equilibrium is, according to the results of the last paragraph, in no way disturbed, if we assume any number of rigid surfaces impermeable to heat and rough or smooth to be inserted in the medium. By means of these the whole system is divided into an arbitrary number of perfectly closed separate systems, each of which may be chosen as small as the general restrictions stated in Sec. 2 permit. It follows from this that the value of the specific intensity of radiation $K_{\nu}$ given in (27) remains valid for the thermodynamic equilibrium of a substance enclosed in a space as small as we please and of any shape whatever.

{\bf 35.} From the consideration of a system consisting of a single homogeneous isotropic substance we now pass on to the treatment of a system consisting of two different homogeneous isotropic substances contiguous to each other, the system being, as before, enclosed by a rigid cover impermeable to heat. We consider the state of radiation when thermodynamic equilibrium exists, at first, as before, with the assumption that the media are of considerable extent. Since the equilibrium is nowise disturbed, if we think of the surface separating the two media as being replaced for an instant by an area entirely impermeable to heat radiation, the laws of the last paragraphs must hold for each of the two substances separately. Let the specific intensity of radiation of frequency $\nu$ polarized in an arbitrary plane be $K_{\nu}$ in the first substance (the upper one in Fig. 3), and $K_{\nu}'$ in the second, and, in general, let all quantities referring to the second substance be indicated by the addition of an account. Both of the quantities $K_{\nu}$ and $K_{\nu}'$ depend, according to equation (27), only on the temperature, the frequency $\nu$, and the nature of the two substances, and these values of the intensities of radiation hold up to very small distances from the bounding surface of the substances, and hence are entirely independent of the properties of this surface.

{\bf 36.} We shall now suppose, to begin with, that the bounding surface of the media is smooth (Sec. 9). Then every ray coming from the first medium and falling on the bounding surface is divided into two rays, the reflected and the transmitted ray. The directions of these two rays vary with the angle of incidence and the color of the incident ray; the intensity also varies with its polarization. Let us denote by $\rho$ (coefficient of reflection) the fraction of the energy reflected, then the fraction transmitted is $(1 - \rho), \rho$ depending on the angle of incidence, the frequency, and the polarization of the incident ray. Similar remarks apply to $\rho'$ the coefficient of reflection of a ray coming from the second medium and falling on the bounding surface.

Now according to (11) we have for the monochromatic plane polarized radiation of frequency $\nu$, emitted in time $dt$ toward the first medium (in the direction of the feathered arrow upper left hand in Fig. 3)

(INSERT FIGURE 3)

from an element $d\sigma$

[1] See, e.g., M. Planck, Vorlesungen uber Thermodynamik, Leipzig, Vet and Comp., 1911 (or English Translation, Longmans Green \& Co.), Secs. 165 and 189, et seq.

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