Mathematical foundations
of the entropy of radiation

Fundamentals

BlackBody radiation

Nonblackbody

Radiative Transfer

Entropy Radiative Transfer

Exercises
Fundamentals of Radiation
Solid Angle
Radiation description
Solid angle
In concept, the solid angle, $\Omega$, is a measure of how large an object appears to an observer. A small object nearby may subtend the same solid angle as a larger object farther away. The classical example here is the Moon. It is much smaller than the Sun but it is much closer to the Earth, and then it has the same solid angle, as can be perfectly appreciated in a solar eclipse.
Geometrically, the solid angle is the twodimensional angle in threedimensional space that an object subtends at a point, i.e., it is the 3D quantity corresponding to a 2D angle. Solid angle is expressed in steradian $($sr$)$, a dimensionless unit in 3D equivalent to the radian $($rad$)$ in 2D.
Mathematically, the solid angle of an object is equal to the area of the segment of a unit sphere that the object covers, centered at the angle's vertex. Continuing with the 2D analogy, the solid angle $($in steradians$)$ equals the area of a segment of a unit sphere in the same way a planar angle $($in radians$)$ equals the length of an arc of a unit circle.
The solid angle of a sphere measured from any point in its interior is 4$\pi$ sr. Solid angles can also be measured in square degrees, 1sr = $(\frac{180}{\pi})^2$ square degrees.
Shall we remember the spherical coordinates here, prior to the mathematical description of the solid angle. In physics, the spherical coordinates are usually denoted by the letters $r$ or $\rho$ $($radial distance$)$, $\theta$ $($polar angle$)$ and $\phi$ or $\psi$ $($azimuthal angle$)$. The position of a point is then determined by those three points, the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuth angle of its orthogonal projection on a reference plane that passes through the origin.
Commonly, the ranges of variation of the variables are: $r \geq 0$, $0^\circ \leq \theta \leq 180^\circ (\pi\text{ rad})$ and $0^\circ \leq \phi < 360^\circ (2\pi\text{ rad})$.
Spherical coordinates can be related to cartesian coordinates through the following relations:
\begin{equation} \hat{\textbf{r}} = \sin\theta \cos\phi \cdot\hat{\textbf{x}} + \sin\theta \sin\phi \cdot\hat{\textbf{y}} + \cos\theta \cdot\hat{\textbf{z}} \end{equation} \begin{equation} \hat{\theta} = \cos\theta \cos\phi \cdot\hat{\textbf{x}} + \cos\theta \sin\phi \cdot\hat{\textbf{y}}  \sin\theta \cdot\hat{\textbf{z}} \end{equation} \begin{equation} \hat{\phi} =  \sin\phi \cdot\hat{\textbf{x}} + \cos\phi \cdot\hat{\textbf{y}} \end{equation}The line element of an infinitesimal displacement from $(r, \theta, \phi)$ to $(r + dr, \theta + d\theta, \phi+d\phi)$ is:
\begin{equation} dr = dr\cdot\hat{r} + rd\theta \cdot\hat{\theta} + r\sin\theta d\phi \cdot\hat{\phi} \end{equation}The surface element on a spherical surface at constant radius $r$, spanning from $\theta + d\theta$ and to $\phi + d\phi$ is:
\begin{equation} dS_r = (rd\theta)(r\sin\theta d\phi) = r^2\sin\theta d\theta d\phi \end{equation}Thus, the differential solid angle is:
\begin{equation} d\Omega = \frac{dS_r}{r^2} = \sin\theta d\theta d\phi \end{equation}Radiation description
Now consider an infinitesimal pencil of the rays that pass through a point in a radiation field, and consider the emission of thermal radiation from an ideal solid surface. The considered radiation field will have energy as well as entropy, which will pass through the considered area. Lets do a general description of the radiation field, useful for both entropy and energy, and independent of the source.
Lets introduce the notation $dE_\lambda$ as the amount of radiant energy and $dS_\lambda$ as the amount of radiant entropy. The reader should keep in mind that both quantities have different units, but the following description will be similar.
The differential amount of radiation energy or entropy, $dE_\lambda$ or $dS_\lambda$, in a time interval $dt$ and in a specific wavelength interval $\lambda$ to $\lambda + d\lambda$, which crosses an element of area $dS_r$, and in directions confined to a differential solid angle, which is oriented at an angle $\theta$ to the normal $dS_r$, is expressed in terms of the specific energy intensity $K_\lambda$ and the specific entropy intensity $L_\lambda$ as:
\begin{equation} K_\lambda = \frac{dE_\lambda}{\cos\theta d\Omega d\lambda dt dS_r} \label{eqk} \end{equation} \begin{equation} L_\lambda = \frac{dS_\lambda}{\cos\theta d\Omega d\lambda dt dS_r} \end{equation}$K_\lambda$ represent the monochromatic energy intensity or spectral energy radiance $(J/m^2\cdot sr)$, and $L_\lambda$ the monochromatic entropy intensity or spectral entropy radiance $(J/m^2 \cdot sr\cdot K)$.
Blackbody Radiation
The Mode
This section is based on DelgadoBonal $($2017$)$^{1}. The mean of the radiation entropy function can be determined solving the equation:
\begin{equation} \left\{ m\cdot x + (m  1)\cdot \log \frac{1}{\left( e^x  1 \right)} \right\} \left( e^x  1 \right)^2 + \left\{ m \cdot x x e^x \right\}\left( e^x  1 \right)  x^2 \left( e^x\right)^2 = 0 \end{equation}$\vartheta$  B$_\vartheta$(T)d$\vartheta$  Dispersion rule  m  Energy  Entropy 

$\nu^2$  $ 2\nu B_{\nu^2}(T)d\nu$  frequencysquared  2  $\frac{hc}{k_B (1.593624 ...)}$  $ \frac{hc}{k_B (1.178179641 ...)}$ 
$\nu$  $B_{\nu}(T)d\nu$  linear frequency  3  $ \frac{hc}{k_B (2.821439 ...)}$  $ \frac{hc}{k_B (2.538231893 ...)}$ 
$\sqrt{\nu}$  $\frac{1}{2\sqrt{\nu}}B_{\sqrt{\nu}}(T)d\nu$  square root frequency  7/2  $ \frac{hc}{k_B (3.380946 ...)}$  $ \frac{hc}{k_B (3.137016422 ...)}$ 
$\log \nu$  $\frac{1}{\nu}B_{\log \nu}(T)d\nu$  logarithmic frequency  4  $ \frac{hc}{k_B (3.920690...)}$  $ \frac{hc}{k_B (3.706085183 ...)}$ 
$\log \lambda$  $\frac{1}{\lambda}B_{\log \lambda}(T)d\lambda$  logarithmic wavelength  4  $ \frac{hc}{k_B (3.920690...)}$  $ \frac{hc}{k_B (3.706085183 ...)}$ 
$\sqrt{\lambda}$  $\frac{1}{2\sqrt{\lambda}}B_{\sqrt{\lambda}}(T)d\lambda$  square root wavelength  9/2  $ \frac{hc}{k_B (4.447304 ...)}$  $ \frac{hc}{k_B (4.255382544 ...)}$ 
$\lambda$  $B_{\lambda}(T)d\lambda$  linear wavelength  5  $ \frac{hc}{k_B (4.965114 ...)}$  $ \frac{hc}{k_B (4.791267357 ...)}$ 
$\lambda^2$  $2 \lambda B_{\lambda^2}(T)d\lambda$  wavelengthsquared  6  $ \frac{hc}{k_B (5.984901...)}$  $ \frac{hc}{k_B (5.838126229 ...)}$ 
The Median
The entropy fractional emissive power of isotropic radiation is determined as:
\begin{equation} \Im_{S} = \frac{15 }{\pi^4} \sigma T^3 \left\{ x^3Li_1(e^{x}) + 4x^2 Li_2(e^{x}) + 8 x Li_3(e^{x}) + 8Li_4(e^{x}) \right\} \end{equation}The total flux is obtained when the whole spectrum is considered. In such situation, the polylogarithmic term is reduced to $\frac{4\pi^4}{45}$ and the total entropy flux is given by the equivalent StefanBoltzmann's law $\frac{4}{3}\sigma T^3$. The emissive power normalized to the total entropy flux is therefore given by:
\begin{equation} \Im_{S,norm} = \frac{45 }{4 \pi^4} \left\{ x^3Li_1(e^{x}) + 4x^2 Li_2(e^{x}) + 8 x Li_3(e^{x}) + 8Li_4(e^{x}) \right\} \end{equation}The Mean
For an absolutely continuous distribution, the Mean is defined as the moment of order one:
\begin{equation} \int_0^\infty x\cdot S_x dx \end{equation}In the case of the entropy of radiation, the integral can be divided:
\begin{equation} \int_0^\infty x\cdot S_x dx = \frac{45}{4\pi^4} \left\{ \int_0^\infty x^4\frac{1}{e^x1} dx + \int_0^\infty \left(x^4 + x^3 \log \left( \frac{1}{e^x  1} \right) \right) dx \right\} \end{equation}Which can be solved as
\begin{equation} \int_0^\infty x\cdot S_x dx = \frac{45}{4\pi^4} \left[ 6\zeta(5) + 24\zeta(5) \right] \simeq 3.59272 \end{equation}being $\zeta(n)$ the Riemann zeta function. Undoing the change of variable, the Mean of the entropy of radiation in the $\lambda T$ product is $\simeq$ 4.00477 $\times 10^6$ nm K.