On a Heuristic Point of View Concerning the Generation and Transformation of Light

§ 3. On the entropy of radiation.Show the German source

The following consideration is contained in a famous paper by Mr. W. Wien and is given a place here only for the sake of completeness.Show the German source

Let there be given a radiation that occupies the volume $v$.Show the German source We assume that the perceptible properties of the radiation in question are completely determined when the radiation density $\varrho(\nu)$ is given for all frequencies.^[1)]Show the German source Since radiations of different frequencies are to be regarded as separable from one another without the performance of work and without the supply of heat, the entropy of the radiation can be represented in the formShow the German source $$S = v \int_0^\infty \varphi(\varrho, \nu) d \nu$$$S$Radiation entropy$v$Volume$\phi$Spectral entropy density$\varrho$Spectral energy density$\nu$Frequency of the lightExplain this equationShow the German source where $\varphi$ denotes a function of the variables $\varrho$ and $\nu$.Show the German source $\varphi$ can be reduced to a function of only one variable by formulating the statement that the entropy of a radiation is not changed by its adiabatic compression between reflecting walls.Show the German source We shall not go into this, however, but shall at once investigate how the function $\varphi$ can be determined from the radiation law of the black body.Show the German source

In “black radiation”, $\varrho$ is such a function of $\nu$ that the entropy is a maximum for a given energy, i.e., thatShow the German source $$\delta \int_0^\infty \varphi(\varrho, \nu) d \nu = 0,$$$\phi$Spectral entropy density$\varrho$Spectral energy density$\nu$Frequency of the lightExplain this equationShow the German source ifShow the German source $$\delta \int_0^\infty \varrho d \nu = 0.$$$\varrho$Spectral energy density$\nu$Frequency of the lightExplain this equationShow the German source From this it follows that for every choice of $\delta \varrho$ as a function of $\nu$Show the German source $$\int_0^\infty \left( \frac{\partial \varphi}{\partial \varrho} - \lambda \right) \delta \varrho d \nu = 0,$$$\phi$Spectral entropy density$\varrho$Spectral energy density$\lambda$Lagrange multiplier$\nu$Frequency of the lightExplain this equationShow the German source where $\lambda$ is independent of $\nu$.Show the German source In black radiation, therefore, $\partial \varphi / \partial \varrho$ is independent of $\nu$.Show the German source

For the increase in temperature of a black radiation of volume $v = 1$ by $d T$, the following equation holds:Show the German source $$d S = \int_{\nu = 0}^{\nu = \infty} \frac{\partial \varphi}{\partial \varrho} d \varrho d \nu,$$$S$Radiation entropy$\nu$Frequency of the light$\phi$Spectral entropy density$\varrho$Spectral energy densityExplain this equationShow the German source or, since $\partial \varphi / \partial \varrho$ is independent of $\nu$:Show the German source $$d S = \frac{\partial \varphi}{\partial \varrho} d E.$$$S$Radiation entropy$\phi$Spectral entropy density$\varrho$Spectral energy density$E$Radiation energy in the bandExplain this equationShow the German source Since $d E$ is equal to the heat supplied and the process is reversible, we also have:Show the German source $$d S = \frac{1}{T} d E.$$$S$Radiation entropy$T$Temperature of the radiation$E$Radiation energy in the bandExplain this equationShow the German source By comparison one obtains:Show the German source $$\frac{\partial \varphi}{\partial \varrho} = \frac{1}{T}.$$$\phi$Spectral entropy density$\varrho$Spectral energy density$T$Temperature of the radiationExplain this equationShow the German source This is the law of black radiation.Show the German source One can therefore determine the law of black radiation from the function $\varphi$ and, conversely, the function $\varphi$ from the latter by integration, bearing in mind that $\varphi$ vanishes for $\varrho = 0$.Show the German source

Bern, 17 March 1905.Show the German source

(Received 18 March 1905.)Show the German source