Chapter 2 Radiation Fundamentals

Think of the ray as a beam of photons, or light particles, some of which may be absorbed by the medium and vanish. The infinitesimal probability $dP$ of a photon being absorbed (e.g., by hitting an absorbing particle) in a thin slab of thickness $ds$ is directly proportional to $ds$: $dP=\kappa ds$, where the constant of proportionality

$$\boxed{\kappa \equiv \frac{dP}{ds}}$$

(2.18)

is called the linear absorption coefficient, and its dimension is inverse length. A value $\kappa =1{\mathrm{m}}^{-1}$ means that over some small distance $\mathrm{\Delta }s\ll {\kappa }^{-1}$ along the ray, $\mathrm{\Delta }s={10}^{-3}$ m for example, the small fraction $\kappa \mathrm{\Delta }s=1{\mathrm{m}}^{-1}\times {10}^{-3}\mathrm{m}={10}^{-3}$ of the photons will be absorbed. Why consider such thin layers rather than the whole absorber at once? Only if the layer is thin enough that $dP$ is infinitesimal does the probability of absorption within the layer increase linearly with thickness $ds$ so that $\kappa$ is a constant independent of $ds$. In a layer thick enough to absorb a significant fraction of the photons, fewer photons are absorbed near the far end of the layer simply because fewer photons have survived to be absorbed there, so the probability of absorption increases nonlinearly with thickness.

This simple macroscopic model doesn’t depend on the microscopic physical processes by which photons are absorbed. It is like the ideal gas model, which relates the macroscopic properties of a gas (temperature, pressure, and density) without considering the microscopic details about how individual gas molecules behave.

The fraction of the specific intensity lost to absorption in the infinitesimal distance $ds$ along the ray is

$$\frac{d{I}_{\nu }}{I_{\nu }}=-\kappa ds\mathit{\hspace{1em}}\text{(absorption only).}$$

(2.19)

Integrating both sides of Equation 2.19 along the absorbing path gives the output specific intensity as a fraction of the input specific intensity:

	$${\int }_{s_{\mathrm{in}}}^{s_{\mathrm{out}}}\frac{d{I}_{\nu }}{I_{\nu }}=-{\int }_{s_{\mathrm{in}}}^{s_{\mathrm{out}}}\kappa (s^{\prime })𝑑s^{\prime }={\ln I_{\nu }|}_{s_{\mathrm{in}}}^{s_{\mathrm{out}}},$$		(2.20)
	$$\ln [I_{\nu }(s_{\mathrm{out}})]-\ln [I_{\nu }(s_{\mathrm{in}})]=-{\int }_{s_{\mathrm{in}}}^{s_{\mathrm{out}}}\kappa (s^{\prime })𝑑s^{\prime },$$		(2.21)
	$$\frac{I_{\nu }(s_{\mathrm{out}})}{I_{\nu }(s_{\mathrm{in}})}=\exp [-{\int }_{s_{\mathrm{in}}}^{s_{\mathrm{out}}}\kappa (s^{\prime })d{s}^{\prime }].$$		(2.22)

The dimensionless quantity

$$\boxed{\tau \equiv -{\int }_{s_{\mathrm{out}}}^{s_{\mathrm{in}}}\kappa (s^{\prime })𝑑s^{\prime }}$$

(2.23)

is called the optical depth or opacity of the absorber. Note that $d\tau =-\kappa ds$. The “backward” direction of the path integration along the line of sight was chosen in Equation 2.23 to make $\tau >0$ and increasing as you look deeper into an absorber. Thus

$$\frac{I_{\nu }(s_{\mathrm{out}})}{I_{\nu }(s_{\mathrm{in}})}=\exp (-\tau )\mathit{\hspace{1em}}\text{(absorption only).}$$

(2.24)

If $\tau \ll 1$, the absorber is said to be optically thin; if $\tau \gg 1$, it is optically thick.

The intervening medium may also emit photons, again by some unspecified microscopic process. In any infinitesimal volume ($dsd\sigma$) of thickness $ds$ and cross section $d\sigma$, the probability per unit time that an isotropic source will emit a photon into the solid angle $d\mathrm{\Omega }$ is directly proportional to the volume and solid angle:

$${\dot{P}}_{\mathrm{em}}\propto dsd\sigma d\mathrm{\Omega }.$$

(2.25)

The emission coefficient $j_{\nu }$ is defined so that

$$\boxed{j_{\nu }\equiv \frac{d{I}_{\nu }}{ds}}$$

(2.26)

if there is no absorption. The dimensions of $j_{\nu }$ follow from its definition; they are power per unit volume per unit frequency per unit solid angle, and the corresponding MKS units are $\mathrm{W}{\mathrm{m}}^{-3}{\mathrm{Hz}}^{-1}{\mathrm{sr}}^{-1}$. Combining the effects of absorption (Equation 2.19) and emission (Equation 2.26) yields the equation of radiative transfer

$$\boxed{\frac{d{I}_{\nu }}{ds}=-\kappa I_{\nu }+j_{\nu }.}$$

(2.27)

Because the absorption and emission processes have not been specified, $\kappa$ and $j_{\nu }$ seem to be independent. However, they are not independent in full thermodynamic equilibrium (TE). In TE, matter and radiation are in equilibrium at the same temperature $T$. Cavity radiation from a container bounded by opaque walls pierced by a small hole (Figure 2.8) is equilibrium radiation. The intensity and spectrum of the radiation emerging from the hole is independent of the wall material (e.g., wood painted green, shiny copper, gray concrete, etc.) and any absorbing material (e.g., gas, dust, fog, etc.) that may be inside the cavity.

Kirchhoff derived the relation between $\kappa$ and $j_{\nu }$ in TE using the thought experiment illustrated in Figure 2.9. Two cavities made of different materials and containing different absorbers are connected through a passive (meaning, no energy is needed for it to work) filter transparent only in the narrow frequency range $\nu$ to $\nu +d\nu$. In equilibrium at any temperature $T$, radiation can transfer no net power from one cavity to the other, lest one cavity cool down and the other heat up. That would violate the second law of thermodynamics because the two cavities at different temperatures could be used to drive a heat engine.

In (full) thermodynamic equilibrium (TE) at temperature $T$,

$$\frac{d{I}_{\nu }}{ds}=0\mathit{\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}}{I}_{\nu }=B_{\nu }(T),$$

(2.28)

where $B_{\nu }(T)$ is the spectrum of equilibrium radiation at temperature $T$. The symbol $B_{\nu }(T)$ is used because equilibrium cavity radiation is the same as the blackbody radiation for a perfect absorber at temperature $T$, even if it is not in a cavity. Rearranging the equation of radiative transfer (Equation 2.27)

$$\frac{d{I}_{\nu }}{ds}=0=-\kappa B_{\nu }(T)+j_{\nu }$$

(2.29)

yields Kirchhoff’s law for a system in TE:

$$\boxed{\frac{j_{\nu }(T)}{\kappa (T)}=B_{\nu }(T),}$$

(2.30)

which is valid at any frequency $\nu$. Equation 2.30 is remarkable because it connects the properties $j_{\nu }(T)$ and $\kappa (T)$ of any kind of matter to the single universal spectrum $B_{\nu }(T)$ of equilibrium radiation. Note also that Kirchhoff’s argument is independent of the shapes of the cavities, so cavity radiation must be isotropic.

Although Kirchhoff’s law was derived for a system in thermodynamic equilibrium, its applicability is not limited to radiation in full thermodynamic equilibrium with its material environment. Kirchhoff’s law also applies whenever the radiating/absorbing material is in thermal equilibrium, in any radiation field. If the emitting/absorbing material is in thermal equilibrium at a well-defined temperature $T$, it is said to be in local thermodynamic equilibrium (LTE) even if it is not in equilibrium with the radiation field. For example, gas molecules in the Earth’s lower atmosphere have a Maxwellian speed distribution (Equation 4.34), so the gas is in LTE. The gas has a well-defined kinetic temperature $T\sim 300$ K measurable with an ordinary thermometer, but it is in a distinctly nonequilibrium radiation field (anisotropic $T\sim 5800$ K sunlight during the day, the cold dark sky at night, plus anisotropic emission from the ground).

Kirchhoff’s law applies in LTE as well as in TE. To show this, recall that $B_{\nu }(T)$ is independent of the properties of the radiating/absorbing material. In contrast, both $j_{\nu }(T)$ and $\kappa (T)$ depend only on the materials in the cavity (e.g., whether the walls are made of copper or concrete) and on the temperature of that material; they do not depend on the ambient radiation field or its spectrum.

This generalized version of Kirchhoff’s law is an exceptionally valuable tool for calculating the emission coefficient from the absorption coefficient or vice versa. For example, in Section 4.3 Kirchhoff’s law gives $\kappa (\nu )$ immediately following a calculation of $j_{\nu }(\nu )$ for free–free emission by an Hii region (a cloud of ionized hydrogen) in LTE.

The blackbody spectrum $B_{\nu }(T)$ (Equation 2.86) falls exponentially at frequencies above the peak, but only as a power law at lower frequencies (solid curve in Figure 2.20). Kirchhoff’s law is not intuitively obvious from experience because room-temperature ($T\sim 300$ K) objects in our environment are much too cold ($h\nu /kT\gg 1$) to emit detectable amounts of visible light. A glass of water might absorb 10% of the sunlight passing through it, but we do not see any emitted light because 10% (or even 100%) of the blackbody radiation emitted by a room-temperature object is very nearly zero. One familiar example of Kirchhoff’s law is a charcoal fire with flames and glowing coals. The infrared (heat) radiation from barely glowing black coals in LTE is much more intense than that from the hotter but nearly transparent visible flames, as you can verify by using a shield to cover either the coals or the flames. In contrast, the radio emission implied by Kirchhoff’s law is always significant because $h\nu /kT\ll 1$ when $T\sim 300\mathrm{K}$ and the blackbody spectrum below the peak falls off only as ${\nu }^{-2}$. This point is illustrated by radio emission and absorption in the Earth’s atmosphere (Section 2.2.3).

At radio frequencies higher than $\nu \sim 1$ GHz, absorption by the Earth’s atmosphere may be large enough to affect the accuracy of flux-density measurements and atmospheric emission can increase noise errors. Radio astronomers can determine the amount of atmospheric absorption by measuring the amount of atmospheric emission as a function of zenith angle, or angle from the vertical, and using Kirchhoff’s law to calculate the zenith opacity of the (roughly isothermal) atmosphere at frequency $\nu$ from its kinetic (thermometer) temperature $T_{\mathrm{atm}}\sim 300\mathrm{K}$ and its radio brightness. The celestial sky above the atmosphere is much colder ($T\sim 3\mathrm{K}$) at high frequencies, so the “background” emission above the atmosphere can usually be ignored.

This is done by tilting the radio telescope and measuring $I_{\nu }$ as a function of the zenith angle $z$ as shown in Figure 2.10. There are two practical complications: (1) The atmospheric signal is noise indistinguishable from other noise sources, especially noise generated in the receiver itself. The output voltage of the receiver is proportional to the sum of these input noise powers. The amount of receiver noise power may not be well known, so only the change in $I_{\nu }$ with zenith angle $z$ can be measured, not the absolute value of $I_{\nu }$ at any zenith angle. (2) The power gain of the receiver system is difficult to calculate from first principles, so most measurements are made relative to some calibration source. For example, the receiver might be calibrated by having the feed look alternately at two absorbing plates with different known temperatures.

Because $B_{\nu }$ is directly proportional to $T$ in the Rayleigh–Jeans low-frequency approximation

$$B_{\nu }\approx \frac{2kT{\nu }^2}{c^2},$$

(2.31)

radio astronomers often find it convenient to specify the spectral brightness $I_{\nu }$, even if $I_{\nu }\ne B_{\nu }$, in terms of the equivalent Rayleigh–Jeans brightness temperature $T_{\mathrm{b}}$ defined by the equation

$$I_{\nu }=\frac{2k{T}_{\mathrm{b}}{\nu }^2}{c^2}.$$

(2.32)

Thus for any $I_{\nu }$,

$$\boxed{T_{\mathrm{b}}(\nu )\equiv \frac{I_{\nu }{c}^2}{2k{\nu }^2}.}$$

(2.33)

Equation 2.33 explicitly indicates that $T_{\mathrm{b}}$ can vary with frequency. Brightness temperature is just another way to specify power per unit solid angle per unit bandwidth in terms of the Rayleigh–Jeans approximation. It is convenient because radio telescopes are often calibrated by absorbers or “loads” of known temperature and because the temperature of a radio source is frequently a quantity of physical interest. Beware that brightness temperature is not the same as physical temperature. Nonthermal sources have frequency-dependent brightness temperatures but they do not have well-defined physical temperatures. Thermal sources have brightness temperatures lower than their physical temperatures if they are semitransparent (e.g., the atmosphere) or partially reflecting (e.g., the Moon). Even true blackbody radiators have $T_{\mathrm{b}}=T$ only in the low-frequency limit $h\nu \ll kT$.

The observed output from the radio telescope might vary with $z$ as shown in Figure 2.11.

To understand and interpret these observations, start with the radiative transfer equation (2.27),

$$\frac{d{I}_{\nu }}{ds}=-\kappa I_{\nu }+j_{\nu }.$$

(2.34)

Kirchhoff’s law (Equation 2.30) can eliminate the unknown $j_{\nu }$ because the atmosphere is in LTE:

$$j_{\nu }=\kappa B_{\nu }(T_{\mathrm{atm}}),$$

(2.35)

where $T_{\mathrm{atm}}$ is the kinetic temperature of the atmosphere as measured by an ordinary thermometer. Dividing by ${\kappa }_{\nu }$ yields

$$\frac{1}{\kappa }\frac{d{I}_{\nu }}{ds}=\frac{-d{I}_{\nu }}{d\tau }=-I_{\nu }+B_{\nu }(T_{\mathrm{atm}}).$$

(2.36)

Multiply both sides of this differential equation by $\exp (-\tau )$ and integrate along the ray in the telescope beam from the top of the atmosphere to the ground. Let ${\tau }_{\mathrm{A}}(z)$ be the total optical depth of the atmosphere along the ray at zenith angle $z$. Then

$${\int }_0^{{\tau }_{\mathrm{A}}}{e}^{-\tau }\frac{d{I}_{\nu }}{d\tau }𝑑\tau ={\int }_0^{{\tau }_{\mathrm{A}}}[I_{\nu }-B_{\nu }(T_{\mathrm{atm}})]e^{-\tau }𝑑\tau .$$

(2.37)

Next integrate the left side by parts and take $B_{\nu }(T_{\mathrm{atm}})$ outside the integral (i.e., make the approximation that the atmosphere below altitude $h$ is nearly isothermal):

	$${e^{-\tau }{I}_{\nu }|}_0^{{\tau }_{\mathrm{A}}}-{\int }_0^{{\tau }_{\mathrm{A}}}-e^{-\tau }{I}_{\nu }d\tau ={\int }_0^{{\tau }_{\mathrm{A}}}{I}_{\nu }{e}^{-\tau }𝑑\tau -B_{\nu }(T_{\mathrm{atm}}){\int }_0^{{\tau }_{\mathrm{A}}}{e}^{-\tau }𝑑\tau ,$$		(2.38)
	$$I_{\nu }(\tau ={\tau }_{\mathrm{A}})e^{-{\tau }_{\mathrm{A}}}-I_{\nu }(\tau =0)=B_{\nu }(T_{\mathrm{atm}})(e^{-{\tau }_{\mathrm{A}}}-1).$$		(2.39)

The term $I_{\nu }(\tau ={\tau }_{\mathrm{A}})\approx 0$ can be neglected because the brightness of emission above the atmosphere is small and doesn’t depend on zenith angle at high radio frequencies, so that

$$I_{\nu }(\tau =0)=(1-e^{-{\tau }_{\mathrm{A}}})B_{\nu }(T_{\mathrm{atm}}).$$

(2.40)

The path length through a plane-parallel atmosphere is proportional to $\sec z$ so at any zenith angle $z$,

$${\tau }_{\mathrm{A}}\approx {\tau }_{\mathrm{Z}}\sec z,$$

(2.41)

where ${\tau }_{\mathrm{Z}}\equiv {\tau }_{\mathrm{A}}(z=0)$ is the zenith opacity of the atmosphere above the radio telescope. Inserting the Rayleigh–Jeans approximation for $B_{\nu }(T_{\mathrm{atm}})$ gives

$$I_{\nu }=[1-\exp (-{\tau }_{\mathrm{Z}}\sec z)]\frac{2k{T}_{\mathrm{atm}}{\nu }^2}{c^2}.$$

(2.42)

Thus the brightness temperature of the atmospheric emission as a function of zenith angle is

$$T_{\mathrm{b}}=\frac{I_{\nu }{c}^2}{2k{\nu }^2}=T_{\mathrm{atm}}\left[1-\exp (-{\tau }_{\mathrm{Z}}\sec z)\right].$$

(2.43)

The variation of $T_{\mathrm{b}}$ with zenith angle is shown for different zenith opacities in Figure 2.12. By fitting observed data to such curves, radio astronomers can estimate the zenith opacity and correct the measured flux densities of celestial sources for atmospheric absorption.

If (1) the zenith opacity is low (${\tau }_{\mathrm{z}}\ll 1$), (2) the frequency is high enough ($\nu >1\mathrm{GHz}$) that the radio-source background is weak, and (3) there is little pickup of “spillover” radiation from the ground, then a good approximation to the observed system noise temperature $T_{\mathrm{s}}$ (Equation 3.150) is

$$T_{\mathrm{s}}\approx [T_{\mathrm{r}}+T_{\mathrm{cmb}}]+T_{\mathrm{atm}}{\tau }_{\mathrm{z}}\sec (z),$$

(2.44)

where $T_{\mathrm{r}}$ is the noise temperature of the receiver alone and $T_{\mathrm{cmb}}\approx 2.7\mathrm{K}$ is the brightness temperature of the cosmic microwave background. The quantity in square brackets is independent of $z$ and $T_{\mathrm{atm}}$ is close to the surface air temperature, which can easily be measured with a thermometer. Equation 2.44 implies that the zenith opacity

$${\tau }_{\mathrm{z}}\approx \frac{\mathrm{\Delta }{T}_{\mathrm{s}}/T_{\mathrm{atm}}}{\mathrm{\Delta }\sec (z)}$$

(2.45)

can be read directly from the slope of the linear tipping-curve plot of $T_{\mathrm{s}}/T_{\mathrm{atm}}$ versus $\sec (z)$. The solid line in Figure 2.13 shows an example in which ${\tau }_{\mathrm{z}}\approx 0.04$. Extrapolating the data to (the geometrically impossible) $\sec (z)=0$ yields $[T_{\mathrm{r}}+T_{\mathrm{cmb}}]/T_{\mathrm{atm}}$. If $T_{\mathrm{atm}}=280\mathrm{K}$, the example $[T_{\mathrm{r}}+T_{\mathrm{cmb}}]/T_{\mathrm{atm}}=0.14$ implies $[T_{\mathrm{r}}+T_{\mathrm{cmb}}]\approx 39.2\mathrm{K}$ and thus $T_{\mathrm{r}}\approx 36.5\mathrm{K}$. Alternatively, if $T_{\mathrm{r}}$ is measured in the laboratory, the tipping curve can be used to determine $T_{\mathrm{cmb}}$, the temperature of the cosmic microwave background, corrected for atmospheric emission. This is essentially how Penzias and Wilson [81] discovered the $T_{\mathrm{cmb}}\approx 3\mathrm{K}$ CMB at $\nu \approx 4\mathrm{GHz}$ using the low-spillover Bell Labs horn antenna (Figure 3.19).

A body is opaque if its opacity is so high that photons cannot pass through it. The absorption coefficient $a(\nu )$ of an opaque body is the probability that an incident photon of frequency $\nu$ will be absorbed, and the reflection coefficient $r(\nu )$ is the probability that it will be reflected. Thus $a=1$ for a blackbody and $r=1$ for a perfect reflector. Each photon is either absorbed or reflected by any opaque body so

$$a(\nu )+r(\nu )=1.$$

(2.46)

The emission coefficient $e(\nu )$ of an opaque body is defined as the ratio of the spectral power it emits per unit area at frequency $\nu$ to that of a blackbody of the same temperature.

The values of $a(\nu )$ and $e(\nu )$ for any opaque body in local thermodynamic equilibrium are not independent. Imagine an opaque body surrounded by a filter passing frequencies in the range $\nu$ to $\nu +d\nu$ inside a cavity (Figure 2.14) in thermodynamic equilibrium at some temperature $T$. In equilibrium, the spectral power absorbed by the opaque body must equal the spectral power emitted, so

$$\boxed{e(\nu )=a(\nu )=1-r(\nu ).}$$

(2.47)

This is Kirchhoff’s law for opaque bodies. It implies that the brightness temperature $T_{\mathrm{b}}$ of an opaque body in LTE at physical temperature $T$ is given by

$$T_{\mathrm{b}}(\nu )=a(\nu )T=[1-r(\nu )]T.$$

(2.48)

Because $a\le 1$, then $T_{\mathrm{b}}\le T$. A perfect reflector ($r=1$) will always have $T_{\mathrm{b}}=0$; that is, it does not emit.

Example. Temperature differences across the GBT reflector (Figure 8.1) caused by differential solar heating can deform the surface and degrade its performance. A special paint that is white at visible wavelengths, black in the mid-infrared, and transparent at radio wavelengths keeps the surface cool and does not harm performance at radio wavelengths by absorbing incoming radio waves or emitting radio noise. The special paint exploits Kirchhoff’s law to perform three separate functions simultaneously: It is opaque and white in the visible portion of the spectrum to reflect sunlight ($T\approx 5800$ K). It is black in the mid-infrared so that the GBT ($T\approx 300$ K) can cool itself efficiently by reradiation. It is transparent at radio wavelengths so that it neither absorbs incoming radio waves nor emits thermal noise at radio wavelengths.

Consider a large (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 generate radiation and to confine the radiation long enough for it to reach equilibrium. The radiation must be generated by thermal accelerations of charged particles in the walls of any cavity with $T>0$. The walls have must have nonzero conductivity because walls of zero conductivity would be transparent—they would generate no currents in response to the electric fields of incoming radiation and the radiation would simply escape. The equilibrium radiation is otherwise independent of the wall material. For walls having nonzero conductivity, the transverse electric field strength at the walls is $E=0$ in equilibrium because fields with $E\ne 0$ induce currents in the walls and lose energy. Only those standing waves with $E=0$ at the walls will persist after some time $t\gg a/c$.

All possible standing-wave modes (a mode is a field pattern that oscillates sinusoidally with a fixed frequency and phase) in the cavity can be enumerated. For example, consider all standing waves whose wave vectors point in the $x$-direction (Figure 2.16). The boundary conditions $E=0$ at $x=0$ and at $x=a$ mean that only those waves having the discrete wavelengths

$$\frac{{\lambda }_x}{2}=a,\frac{2{\lambda }_x}{2}=a,\frac{3{\lambda }_x}{2}=a,\mathrm{\dots }$$

(2.59)

can have nonzero amplitudes. They all satisfy

$$\frac{n_x{\lambda }_x}{2}=a,$$

(2.60)

where $n=1$, 2, 3, $\mathrm{\dots }.$ Likewise,

$$\frac{n_y{\lambda }_y}{2}=\frac{n_z{\lambda }_z}{2}=a$$

(2.61)

for wave vectors pointing in the $y$- and $z$-directions, respectively.

What about a wave vector pointing in some arbitrary direction? Let $\alpha ,\beta ,\gamma$ be the angles between the wave vector and the $x$-, $y$-, $z$-axes, respectively. From Figure 2.18 it is clear that

$$\lambda ={\lambda }_x\cos \alpha ,$$

(2.62)

where $\lambda$ is the wavelength measured in the direction of the wave vector and ${\lambda }_x\ge \lambda$ is the component of $\lambda$ projected onto the $x$-axis. Thus

$${\lambda }_x=\frac{\lambda }{\cos \alpha }.$$

(2.63)

Notice that ${\lambda }_x>\lambda$; it is $\lambda$ divided by $\cos \alpha$, not multiplied by $\cos \alpha$. Simultaneously, the standing waves must also satisfy

$${\lambda }_y=\frac{\lambda }{\cos \beta }\text{ and }{\lambda }_z=\frac{\lambda }{\cos \gamma }.$$

(2.64)

Equations 2.60 through 2.64 imply that standing waves of all orientations must satisfy

$$n_x=\frac{2a}{{\lambda }_x},n_y=\frac{2a}{{\lambda }_y},n_z=\frac{2a}{{\lambda }_z},$$

(2.65)

$$n_x=\frac{2a\cos \alpha }{\lambda },n_y=\frac{2a\cos \beta }{\lambda },n_z=\frac{2a\cos \gamma }{\lambda }.$$

(2.66)

Squaring and summing the three parts of Equation 2.66 gives

$$n_x^2+n_y^2+n_z^2={\left(\frac{2a}{\lambda }\right)}^2({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma ).$$

(2.67)

The Pythagorean theorem implies $({\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma )=1$ so all standing waves must satisfy

$$n_x^2+n_y^2+n_z^2={\left(\frac{2a}{\lambda }\right)}^2.$$

(2.68)

The permitted frequencies $\nu =c/\lambda$ of the standing waves are

$$\nu =\frac{c}{2a}\sqrt{n_x^2+n_y^2+n_z^2}$$

(2.69)

for all positive integers $n_x$, $n_y$, and $n_z$.

The permitted standing waves can be represented as a lattice in the positive octant of the space whose axes are $n_x>0$, $n_y>0$, and $n_z>0$ (Figure 2.18). Each point of the lattice represents one possible mode of equilibrium cavity radiation. The space density of points in this lattice is unity, so the average number of points in any volume equals that volume.

Let $\rho$ be the radial coordinate in $(n_x,n_y,n_z)$-space. Then

$${\rho }^2=n_x^2+n_y^2+n_z^2$$

(2.70)

and

$$\nu =\frac{c}{\lambda }=\left(\frac{c}{2a}\right)\rho .$$

(2.71)

The number $N_{\nu }(\nu )d\nu$ of independent modes having frequencies in the range $\nu$ to $\nu +d\nu$ equals the volume of the spherical octant shell between $\rho$ and $\rho +d\rho$, multiplied by 2 to account for the two independent orthogonal polarizations (Section 2.3) of electromagnetic waves:

$$N_{\nu }(\nu )d\nu =\frac{4\pi {\rho }^{2}d\rho }{8}\times 2=\pi {\left(\frac{2a\nu }{c}\right)}^2\frac{2a}{c}d\nu .$$

(2.72)

Classically, each radiation mode can have any energy $E>0$. In thermodynamic equilibrium at temperature $T$, the probability $P(E)$ that any mode has energy $E$ is given by the continuous Boltzmann probability distribution

$$P(E)\propto \exp \left(-\frac{E}{kT}\right).$$

(2.73)

Then the average energy per mode is

$$⟨E⟩=\frac{{\int }_0^{\mathrm{\infty }}EP(E)𝑑E}{{\int }_0^{\mathrm{\infty }}P(E)𝑑E}=kT.$$

(2.74)

Thus each mode has the same average energy $⟨E⟩=kT$. The spectral energy density $u_{\nu }(T)$ of cavity radiation at frequency $\nu$ is the total energy of all $N_{\nu }$ modes (Equation 2.72) at that frequency divided by the volume $a^3$ of the cavity:

$$u_{\nu }(T)=\frac{N_{\nu }(\nu )kT}{a^3}=\frac{8\pi a^3}{a^3}\frac{{\nu }^2}{c^3}kT=8\pi kT\frac{{\nu }^2}{c^3}.$$

(2.75)

The spectral energy density of radiation (blackbody or not) is spectral energy per unit volume. It equals the total flow of spectral power per unit area divided by the flow speed $c$:

$$\boxed{u_{\nu }=\frac{1}{c}\int I_{\nu }d\mathrm{\Omega }.}$$

(2.76)

Calling the specific intensity of blackbody radiation $B_{\nu }$ and making use of the fact the blackbody radiation is isotropic yields, for blackbody radiation,

$$u_{\nu }=\frac{1}{c}{\int }_{4\pi }{B}_{\nu }𝑑\mathrm{\Omega }=\frac{4\pi }{c}{B}_{\nu }.$$

(2.77)

Combining Equations 2.75 and 2.77 gives

$$\frac{4\pi }{c}{B}_{\nu }=\frac{8\pi kT{\nu }^2}{c^3}.$$

(2.78)

Solving for $B_{\nu }$ yields the Rayleigh–Jeans approximation for the spectral brightness of blackbody radiation

$$\boxed{B_{\nu }=\frac{2kT{\nu }^2}{c^2}=\frac{2kT}{{\lambda }^2}}$$

(2.79)

that is valid only in the low-frequency limit $h\nu \ll kT$.

Notice that

1. 1.

   the spectral brightness $B_{\nu }$ is proportional to frequency squared because the volume of a spherical shell in three dimensions is proportional to frequency squared;

2. 2.

   all modes have $⟨E⟩=kT$, the classical assumption that breaks down at high frequencies;

3. 3.

   the Rayleigh–Jeans approximation implies that $B_{\nu }$ diverges at high $\nu$; this is called the ultraviolet catastrophe;

4. 4.

   $B_{\nu }$ is independent of direction; blackbody radiation is isotropic;

5. 5.

   blackbody radiation is unpolarized; you can show this with a thought experiment involving two cavities connected by a passive polarizing filter, as in Figure 2.9;

6. 6.

   this result derived for a cubical cavity applies to equilibrium radiation in a cavity of any shape or (large) size, which you can demonstrate by a thought experiment in which a cubical cavity is connected through a small filter window to the other cavity.

The only flaw in the derivation of the Rayleigh–Jeans approximation is the classical assumption that each radiation mode can have any energy $E>0$. To eliminate the ultraviolet catastrophe, Planck postulated that the possible mode energies are not continuously distributed, but rather they are quantized and must satisfy the new constraint

$$E=nh\nu ,n=1,2,3,\mathrm{\dots },$$

(2.80)

where $h\approx 6.63\times {10}^{-27}$ erg s is Planck’s constant and $n$ is the number of photons, or discrete particles of light each having energy

$$\boxed{E=h\nu .}$$

(2.81)

Then

$$P(E)=P(nh\nu )\propto \exp \left(-\frac{nh\nu }{kT}\right)$$

(2.82)

and the average energy per mode is calculated by summing over only the discrete energies permitted instead of integrating over all energies. Planck’s quantized replacement for Equation 2.74 is

$$⟨E⟩=\frac{\sum _{n=0}^{\mathrm{\infty }}nh\nu P(nh\nu )}{\sum _{n=0}^{\mathrm{\infty }}P(nh\nu )}=\frac{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}nh\nu \exp \left(-{\displaystyle \frac{nh\nu }{kT}}\right)}{{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\exp \left(-{\displaystyle \frac{nh\nu }{kT}}\right)}.$$

(2.83)

Planck’s sum is evaluated in Appendix B.1; it is

$$⟨E⟩=\frac{h\nu }{\exp \left({\displaystyle \frac{h\nu }{kT}}\right)-1}=kT\left[\frac{{\displaystyle \frac{h\nu }{kT}}}{\exp \left({\displaystyle \frac{h\nu }{kT}}\right)-1}\right].$$

(2.84)

The quantity $kT$ is the classical mean energy per mode, and the quantity in square brackets is the quantum correction factor. In the limit $h\nu \ll kT$ the quantum correction factor is unity, but in the limit $h\nu \gg kT$ it falls exponentially to zero. That is, the high-frequency or “ultraviolet” modes may contain few or even zero photons, thereby avoiding the “ultraviolet catastrophe.”

Thus the correct blackbody radiation law becomes

$$B_{\nu }=\frac{2kT{\nu }^2}{c^2}\left[\frac{{\displaystyle \frac{h\nu }{kT}}}{\exp \left({\displaystyle \frac{h\nu }{kT}}\right)-1}\right],$$

(2.85)

where the first factor is the Rayleigh–Jeans approximation and the quantity in square brackets is the quantum correction factor. Planck’s equation for the spectral brightness $B_{\nu }$ of blackbody radiation is usually written in the simpler form

$$\boxed{B_{\nu }(\nu ,T)=\frac{2h{\nu }^3}{c^2}\frac{1}{\exp \left({\displaystyle \frac{h\nu }{kT}}\right)-1}.}$$

(2.86)

The corresponding brightness per unit wavelength $B_{\lambda }$ follows from Equation 2.5; it can be written either as a function of frequency:

$$B_{\lambda }(\nu ,T)=\frac{{\nu }^2}{c}{B}_{\nu }(\nu ,T)=\frac{2h{\nu }^5}{c^3}\frac{1}{\exp \left({\displaystyle \frac{h\nu }{kT}}\right)-1}$$

(2.87)

or as a function of wavelength:

$$B_{\lambda }(\lambda ,T)=\frac{2h{c}^2}{{\lambda }^5}\frac{1}{\exp \left({\displaystyle \frac{hc}{\lambda kT}}\right)-1}.$$

(2.88)

Integrating Planck’s law over all frequencies (see Appendix B.2) gives the Stefan–Boltzmann law specifying a finite integrated brightness for a blackbody radiator at temperature $T$:

$$\boxed{B(T)\equiv {\int }_0^{\mathrm{\infty }}{B}_{\nu }(T)d\nu =\frac{\sigma T^4}{\pi }.}$$

(2.89)

The quantity

$$\sigma \equiv \frac{2{\pi }^5{k}^4}{15c^2{h}^3}\approx 5.67\times {10}^{-5}\frac{\mathrm{erg}}{{\mathrm{cm}}^2\mathrm{s}{\mathrm{K}}^4(\mathrm{sr})}$$

(2.90)

is called the Stefan–Boltzmann constant and is also derived in Appendix B.2. The unit sr is dimensionless because solid angle has dimensions angle${}^2$ = (length/length)${}^2$. A dimensionless unit can be inserted or deleted without changing the dimensions of an equation—hence the parentheses around sr in Equation 2.90. For clarity it should be used where appropriate. For example, it is needed in the blackbody brightness (erg cm${}^{-2}$ s${}^{-1}$ K${}^{-4}$ sr${}^{-1}$) Equation 2.89 but not in the radiation energy density (erg cm${}^{-3}$) Equation 2.93 or in the flux (erg cm${}^{-2}$ s${}^{-1}$ K${}^{-4}$) Equation 2.111.

For isotropic radiation, the spectral energy density (Equation 2.76) is simply

$$u_{\nu }=\frac{4\pi I_{\nu }}{c},$$

(2.91)

so the total radiation energy density

$$u\equiv {\int }_0^{\mathrm{\infty }}{u}_{\nu }𝑑\nu =\frac{4\pi I}{c}$$

(2.92)

of blackbody radiation is

$$\boxed{u=\frac{4\sigma T^4}{c}=a{T}^4,}$$

(2.93)

where the quantity

$$a\equiv 4\sigma /c\approx 7.56577\times {10}^{-15}\mathrm{erg}{\mathrm{cm}}^{-3}{\mathrm{K}}^{-4}$$

(2.94)

is called the radiation constant.

In any narrow frequency range, the number density of photons is their spectral energy density divided by the energy per photon, $E=h\nu$. The total photon number density $n_{\gamma }$ is

$$n_{\gamma }={\int }_0^{\mathrm{\infty }}\frac{u_{\nu }}{h\nu }𝑑\nu .$$

(2.95)

The photon number density of blackbody radiation at temperature $T$ is

	$n_{\gamma }=$	${\displaystyle \frac{4\pi }{ch}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{B_{\nu }(T)}{\nu }}𝑑\nu$		(2.96)
	$=$	${\displaystyle \frac{8\pi }{c^3}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{{\nu }^{2}d\nu }{\exp \left(\frac{h\nu }{kT}\right)-1}}$		(2.97)
	$=$	${\displaystyle \frac{8\pi }{c^3}}{\left({\displaystyle \frac{kT}{h}}\right)}^3{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{x^{2}dx}{e^x-1}},$		(2.98)

where the integral

$${\int }_0^{\mathrm{\infty }}\frac{x^{2}dx}{e^x-1}\approx 2.404$$

(2.99)

is evaluated in Appendix B.2. Numerically, the photon number density of blackbody radiation is

$$\boxed{\left(\frac{n_{\gamma }}{{\mathrm{cm}}^{-3}}\right)\approx 20.3{\left(\frac{T}{\mathrm{K}}\right)}^3.}$$

(2.100)

The mean photon energy of blackbody radiation is

$$\boxed{⟨E_{\gamma }⟩=\frac{u}{n_{\gamma }}=\frac{4\sigma T^4}{c{n}_{\gamma }}=\frac{{\pi }^{4}kT}{15}{\left({\int }_0^{\mathrm{\infty }}\frac{x^{2}dx}{e^x-1}\right)}^{-1}\approx \mathrm{\hspace{0.17em}2.70}kT.}$$

(2.101)

The frequency $⟨\nu ⟩$ corresponding to this mean photon energy is

$$\left(\frac{⟨\nu ⟩}{\mathrm{GHz}}\right)=\frac{⟨E_{\gamma }⟩}{h}\approx 56\left(\frac{T}{\mathrm{K}}\right).$$

(2.102)

It is close to the frequency ${\nu }_{\max }$ at which $B_{\nu }$, the brightness per unit frequency of a blackbody, is maximum. Setting

$$\frac{\partial B_{\nu }(\nu )}{\partial \nu }=0$$

(2.103)

yields

$$\boxed{\left(\frac{{\nu }_{\max }}{\mathrm{GHz}}\right)\approx 59\left(\frac{T}{\mathrm{K}}\right).}$$

(2.104)

The wavelength ${\lambda }_{\max }$ at which

$$\frac{\partial B_{\lambda }(\lambda )}{\partial \lambda }=0$$

(2.105)

maximizes $B_{\lambda }$, the brightness per unit wavelength of a blackbody. It is

$$\boxed{\left(\frac{{\lambda }_{\max }}{\mathrm{cm}}\right)\approx 0.29{\left(\frac{T}{\mathrm{K}}\right)}^{-1}.}$$

(2.106)

Equation 2.106 is the familiar form of Wien’s displacement law used by optical astronomers, whose spectrometers measure wavelengths instead of frequencies. Note that the peak frequency $c/{\lambda }_{\max }\approx 103\mathrm{GHz}T(\mathrm{K})$ is much higher than ${\nu }_{\max }\approx 59\mathrm{GHz}T(\mathrm{K})$.

The power emitted per unit area per unit frequency by any opaque isotropic radiator of brightness $I_{\nu }$ is the spectral flux density

$$F_{\nu }=\int I_{\nu }\cos \theta d\mathrm{\Omega },$$

(2.107)

where the integration covers the hemisphere above the unit area:

$$F_{\nu }={\int }_0^{\pi /2}{I}_{\nu }\cos \theta \left({\int }_0^{2\pi }\sin \theta d\varphi \right)𝑑\theta =2\pi I_{\nu }{\int }_0^{\pi /2}\cos \theta \sin \theta d\theta ,$$

(2.108)

$$\boxed{F_{\nu }=\pi I_{\nu }.}$$

(2.109)

For a blackbody at temperature $T$, the spectral flux density is

$$F_{\nu }(T)=\pi B_{\nu }(T).$$

(2.110)

The corresponding total power per unit area integrated over all frequencies is

$$\boxed{F(T)=\pi B(T)=\sigma T^4.}$$

(2.111)