Blackbody Radiation

The intensity and spectrum of blackbody radiation depends only on the temperature of the black body or cavity.  The same is true for the electrical noise generated by a warm resistor, which plays an important role in radio astronomy.  You may already be familiar with the standard derivation of the Rayleigh-Jeans and Planck radiation laws, but it is worth repeating because blackbody radiation is so fundamental and because its one-dimensional analog yields the spectrum of electrical noise generated by a resistor.

The Rayleigh-Jeans Law

Consider a large (meaning side length $a \gg \lambda$, where $\lambda$ is the longest wavelength of interest) cubical cavity filled with radiation in thermodynamic equilibrium.  The purpose of the cavity is to confine the radiation long enough for it to reach equilibrium.  The walls of a cavity capable of confining radiation must have nonzero conductivity because walls of zero conductivity would be transparent; they would generate no currents in response to the electric field of the incoming radiation, so the radiation would pass through unaffected. For walls having nonzero conductivity, the equilibrium electric field strength at the walls is $E = 0$ because modes with $E \ne 0$ at the walls are lossy.  Only those standing waves with $E = 0$ at the walls will persist after some time $t \gg a/c$.

We can enumerate all possible standing-wave modes in the cavity. For example, consider all standing waves whose wave normals point in the $x$ direction.

The boundary conditions $E = 0$ at $x = 0$ and at $x = a$ mean that only the waves having the discrete wavelengths
$${\lambda_{\rm x} \over 2} = a, \qquad {2 \lambda_{\rm x} \over 2} = a, \qquad {3 \lambda_{\rm x} \over 2} = a, ..., $$
can have nonzero amplitudes.  Thus
 $${n \lambda_{\rm x} \over 2} = a$$
where $n = 1$, 2, 3, $...$. Similarly,
$${n _{\rm x} \lambda_{\rm x} \over 2} = {n_{\rm y} \lambda_{\rm y} \over 2} = {n_{\rm z} \lambda_{\rm z} \over 2} = a$$
for wave normals in the $x,y,z$ directions.

What about a wave whose normal is in some arbitrary direction? Let $\alpha,\,\beta,\,\gamma$ be the angles between the wave normal and the $x,\,y,\,z$ axes, respectively.

From the figure it is clear that
$$\lambda = \lambda_{\rm x} \cos\, \alpha$$
where $\lambda$ is the wavelength measured in the direction of the wave normal and $\lambda_{\rm x} \geq \lambda$ is the spacing between the wave nodes measured along the $x$-axis. Thus
$$\lambda_{\rm x} = {\lambda \over \cos\,\alpha}$$
Similarly,
$$\lambda_{\rm y} = {\lambda \over \cos\,\beta}
{\rm ~~~and~~~}
\lambda_{\rm z} = {\lambda \over \cos\,\gamma}$$
Our boundary conditions become
$$n_{\rm x} = {2 a \over \lambda_{\rm x}},    n_{\rm y} = {2 a \over \lambda_{\rm y}},    n_{\rm z} = {2 a \over \lambda_{\rm z}}$$
$$n_{\rm x} = {2 a \cos\,\alpha \over \lambda},    n_{\rm y} = {2 a \cos\,\beta \over \lambda},    n _{\rm z} = {2 a \cos\,\gamma \over \lambda}$$
Squaring and summing the last three equations gives
$$n_{\rm x}^2 + n_{\rm y}^2 + n_{\rm z}^2 = \biggl({2 a \over \lambda}\biggr)^2 (\cos^2\,\alpha + \cos^2\,\beta + \cos^2\,\gamma)~.$$
The Pythagorean theorem implies $(\cos^2\,\alpha + \cos^2\,\beta + \cos^2\,\gamma) = 1$ so
$$n_{\rm x}^2 + n_{\rm y}^2 + n_{\rm z}^2 = \biggl({2 a \over \lambda}\biggr)^2$$
The permitted frequencies $\nu = c / \lambda$ of these standing waves are:
$$\nu = {c \over 2 a }\sqrt{n_{\rm x}^2 + n_{\rm y}^2 + n_{\rm z}^2}$$
for all positive integers $n_{ \rm x}$, $n_{\rm y}$, and $n_{\rm z}$.

The permitted standing-wave modes can be represented as a lattice in the positive octant of the space whose axes are $n_{\rm x}$, $n_{\rm y}$, and $n_{\rm z}$. Each point of the lattice represents one possible mode of electromagnetic radiation. The space density of points in this lattice is unity, so the number of points in any volume is equal to that volume.

The (x,y) plane in the imaginary space whose axes are $(n_{\rm x}, \, n_{\rm y}, \, n_{\rm z})$.  Permitted standing waves in the $(n_{\rm x}, \, n_{\rm y})$ plane are indicated by dots at positive integer values of these axes. 

Let $\rho$ be the radial coordinate in $(n_{\rm x}, \, n_{\rm y}, \, n_{\rm z})$-space. Then
$$\rho^2 = n_{\rm x}^2 + n_{\rm y}^2 + n_{\rm z}^2$$
and
$$\nu = {c \rho \over 2 a}$$
The number $N(\nu) d\nu$ of modes having frequencies in the range $\nu$ to $\nu + d\nu$ is the volume of the spherical octant shell between $\rho$ and $\rho + d \rho$, multiplied by 2 (two polarizations):
$$N(\nu) d\nu = {4 \pi \rho^2 d \rho \over 8} \times 2$$
$$N(\nu) d\nu = \pi \biggl({2 a \nu \over c}\biggr)^2 {2a \over c} d\nu$$
In thermodynamic equilibrium at temperature $T$, each mode has average energy $\langle E \rangle = kT$ according to the classical Boltzmann law, but not according to quantum mechanics.  If we proceed using $kT$ for the average energy per mode, the spectral energy density $u_\nu(T)$ of radiation in the frequency range $\nu$ to $\nu + d\nu$ is the total energy of all modes in that frequency range divided by the volume $a^3$ of the cavity:
$$ u_\nu(T) = {N(\nu) d\nu \over a^3} kT$$
$$u_\nu(T) = {8 \pi a^3 \over a^3} {\nu^2 \over c^3} kT = 8 \pi k T {\nu^2 \over c^3}$$
The spectral energy density of the radiation is spectral energy per unit volume.  It equals the total flow of spectral power per unit area divided by the flow speed $c$:
$$\bbox[border:3px blue solid,7pt]{u_\nu = {1 \over c} \int I_\nu d \Omega} \rlap{\quad \rm {(2C1)}}$$
Calling the specific intensity of blackbody radiation $B_\nu$ and making use of the fact the blackbody radiation is isotropic, we get for blackbody radiation:
$$u_\nu = {1 \over c}\int_{4\pi} B_\nu d \Omega =  {B_\nu \over c}
\int_{4 \pi} d \Omega = {4 \pi \over c} B_\nu$$
Thus
$${4 \pi \over c} B_\nu = {8 \pi k T \nu^2 \over c^3}$$
and we obtain the Rayleigh-Jeans law for the spectral brightness of blackbody radiation
$$\bbox[border:3px blue solid,7pt]{B_\nu = {2 k T \nu^2 \over c^2}}\rlap{\quad \rm {(2C2)}}$$
that is valid only in the low-frequency limit $h \nu \ll k T$.

Notice that:

1. Brightness is proportional to frequency squared because the volume of a spherical shell in three dimensions is proportional to frequency squared.
   
2. We assumed all modes have $\langle E \rangle = k T$, the  classical assumption that breaks down at high frequencies.
3. $B_\nu$ diverges at high $\nu$. This is called the "ultraviolet catastrophe".
4. $B_\nu$ is independent of direction.
   

The Planck Radiation Law

Planck postulated that possible mode energies are not continuously distributed, but rather they are quantized and must satisfy
$$E = n h \nu, \qquad n = 1,~2,~3,~...$$
where $h \approx 6.63 \times 10^{-27}$ erg s is Planck's constant and $n$ is the number of photons (particles of light each having energy $E = h \nu$) in that mode. Then
$$P(E) = P(n h \nu) \propto \exp\biggl({ - n h \nu \over k T}\biggr)$$
and the average energy per mode is calculated by summing over only the discrete energies permitted instead integrating over all energies.
$$\langle E \rangle = {\sum_{n = 0}^\infty n h \nu P(n h \nu) \over \sum_{n = 0}^\infty P(n h \nu)} =
{\sum_{n = 0}^\infty n h \nu \exp\bigl({ - n h \nu \over k T}\bigr) \over
\sum_{n = 0}^\infty \exp\bigl({ - n h \nu \over k T}\bigr)}$$
Evaluating this sum yields
$$ \langle E \rangle = {h \nu \over {\exp\bigl({h \nu \over kT}\bigr) -1}}$$
Thus the correct radiation law becomes
$$B_\nu = {2 k T \nu^2 \over c^2} \biggl[{h \nu / ( k T) \over \exp [h \nu / ( kT)]  -1} \biggr]~,$$
where the first factor is the Rayleigh-Jeans law and the quantity in brackets is the quantum correction factor.
Planck's law for blackbody radiation is usually written in the form
$$\bbox[border:3px blue
solid,7pt]{B_\nu = {2 h \nu^3 \over c^2}
{1 \over \exp\bigl({h \nu \over k T}\bigr) -1}}
\rlap{\quad \rm {(2C3)}}$$
This eliminates the ultraviolet catastrophe because integrating over all frequencies gives a finite integrated brightness a blackbody radiator at temperature $T$:
$$\bbox[border:3px blue
solid,7pt]{B(T) \equiv \int_0^\infty B_\nu(T) d\nu = {\sigma T^4 \over \pi}}
\rlap{\quad \rm {(2C4)}}$$
where
$$\sigma \equiv {2 \pi^5 k^4 \over 15 c^2 h^3} \approx 5.67 \times 10^{-5} {{\rm erg} \over {\rm cm}^2 {\rm ~s~K}^4 {\rm ~sr}}$$
is called the Stefan-Boltzmann constant.  Beware that the dimensionless sr in the denominator is dropped by many authors.

The spectral energy density per unit solid angle of radiation is just the specific intensity $I_\nu$ divided by the flow rate $c$.  For isotropic radiation, the spectral energy density is
$$u_\nu = 4 \pi I_\nu / c$$ 
so the total radiation energy density $u = \int_0^\infty u_\nu d \nu$ of blackbody radiation is
$$\bbox[border:3px blue solid,7pt]{u =  {4 \sigma T^4\over c}}
\rlap{\quad \rm {(2C5)}}$$

The frequency $\nu_{\rm max}$ at which $B_\nu$, the brightness per unit frequency of a blackbody, is maximum is the solution of
$$ {\partial B_\nu \over \partial \nu} = 0 ; $$
it is
$$ \nu_{\rm max} \approx {59 {\rm ~GHz} \over T {\rm~(K)}}.$$
Note that $\nu_{\rm max}$ isn't quite the same as $c / \lambda_{\rm max} \approx 103 {\rm ~GHz}$, where
$$\lambda_{\rm max} \approx {0.29{\rm ~cm} \over T {\rm~(K)}}$$
is the wavelength at which $B_\lambda$, the brightness per unit wavelength, is highest.  The latter equation is the familiar form of Wien's Law used by optical astronomers, whose spectrometers measure wavelengths instead of frequencies.

The electrical power generated by a warm resistor

A resistor is any electrical device that absorbs all of the electrical power applied to it; it is the "black body" of electric circuits.  A warm resistor at temperature $T > 0$ K generates electrical noise.   The amount of power generated depends only on $T$, and the frequency spectrum of the noise is independent of the resistor material, obeying laws analogous to the blackbody radiation laws. Also, the electrical noise generated is indistinguishable from the noise coming from an antenna observing a blackbody radiator. Warm resistors are useful in radio astronomy as standards for calibrating receiver gains, and the power received by a radio telescope is often described in terms of the resistor temperature required to generate the same power thermally. Thus the gain of a radio receiver is normally calibrated by connecting it alternately to hot and cold loads having known temperatures.

The derivation of the electrical power per unit bandwidth $P_\nu$ generated by a resistor is essentially a one-dimensional version of the three-dimensional argument used for blackbody radiation [Nyquist, H. 1928, Phys. Rev. 32, 110; Barrett, A. H. 1970, Fundamentals of Radio Astronomy, p. 22-23 (unpublished)]. For radio frequencies, $h\nu \ll kT$, the Rayleigh-Jeans limit. Recall that the Rayleigh-Jeans derivation starts with a large cube of side length $a$ containing standing waves of thermal radiation. The average energy in each standing-wave mode is $\langle E \rangle = kT$, and the number of modes with frequency $\nu$ to $\nu + d\nu$ is proportional to $\nu^2$, so $B_\nu \propto \nu^2$. For the resistor, consider two identical resistors at temperature $T$ connected by a lossless transmission line (e.g., a pair of parallel wires) of length $a$ much larger than the longest wavelength of interest. 

We assume that the transmission line has a characteristic impedance matched to the resistance of the resistors so that power is efficiently coupled between them. In equilibrium, the transmission line can only support those wave modes satisfying
$$ a = {n \lambda \over 2}, \qquad n = {\rm 1,~2,~3,}...,$$
where $\lambda$ is the wavelength. Electrical signals do not travel at exactly the speed of light on a transmission line, but at some slightly lower velocity $v < c$. Thus
$$n = {2 a \nu \over v}$$
and, for $a \gg \lambda$, the number of modes per unit frequency is
$${\Delta n \over \Delta \nu} = {2 a \over v}~.$$
In thermal equilibrium, the classical Boltzmann law says that each mode has average energy $\langle E \rangle = kT$, so the average energy per unit frequency $E_\nu$ in the transmission line is
$$E_\nu = {\Delta n \over \Delta \nu} kT = {2akT \over v}$$
This energy takes a time $\Delta t = a / v$ to flow from one end of the transmission line to the other, so the classical power (energy per unit time) per unit frequency flowing on the transmission line is
$$ P_\nu = {E_\nu \over \Delta t} = {2kT}$$

Note that the velocity $v$ has dropped out. The spectral power $P_\nu$ generated by the two identical resistors must be $2kT$ and, by symmetry, the spectral power generated by each resistor is
$$\bbox[border:3px blue solid,7pt]{P_\nu = kT}
\rlap{\quad \rm {(2C6)}}$$
in the limit $h \nu \ll k T$.  This equation is called the Nyquist formula and is the electrical equivalent of the Rayleigh-Jeans law for radiation.  Since the "space" of the transmission line has only one dimension instead of three, the frequency dependence is $\nu^0$ instead of $\nu^2$. 

Still, the Nyquist formula suffers from an "ultraviolet catastrophe" of its own: the total power integrated over all frequencies diverges. Planck's quantization rule is the cure for this problem also: the electrical energy in each mode must be an integer multiple of $h \nu$, and the corrected version of the Nyquist formula is
$$P_\nu = kT \biggl[{h \nu /( k T) \over \exp [h \nu / (kT)] -1} \biggr]~,$$
where the quantity in large brackets is the same quantization correction previously found for blackbody radiation.  This equation is usually written in the form
$$\bbox[border:3px blue solid,7pt]{P_\nu = {h \nu \over {\exp\bigl({h \nu \over kT}\bigr) -1}}}
\rlap{\quad \rm {(2C7)}}$$
In many radio-astronomical applications, ${h \nu \ll kT}$ and the simpler $P_\nu = k T$ (Eq. 2C6) can be used.