6.8 Blackbody Radiation

The Bose-Einstein distribution of the previous section can also be used for understanding the emission of light and other electromagnetic radiation. If you turn on an electric stove, the stove plate heats up until it becomes red hot. The red glow that you see consists of photons with energies in the visible red range. When the stove plate was cold, it also emitted photons, but those were of too low energy to be seen by our unaided eyes.

The radiation system that is easiest to analyze is the inside of an empty box. Empty should here be read as devoid of matter. For if the temperature inside the box is above absolute zero, then the inside of the box will still be filled with the electromagnetic radiation that the atoms in the box surfaces emit. This radiation is representative of the radiation that truly black surfaces emit. Therefore, the radiation inside the box is called “blackbody radiation.”

Before the advent of quantum mechanics, Rayleigh and Jeans had computed using classical physics that the energy of the radiation in the box would vary with electromagnetic frequency $\omega$ and temperature $T$ as

\begin{displaymath}
\rho(\omega)= \frac{\omega^2}{\pi^2 c^3} k_{\rm B}T
\end{displaymath}

where $k_{\rm B}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.38 10$\POW9,{-23}$ J/K is the Boltzmann constant and $c$ the speed of light. That was clearly all wrong except at low frequencies. For one thing, the radiation energy would become infinite at infinite frequencies!

It was this very problem that led to the beginning of quantum mechanics. To fix the problem, in 1900 Planck made the unprecedented assumption that energy would not come in arbitrary amounts, but only in discrete chunks of size $\hbar\omega$. The constant $\hbar$ was a completely new physical constant whose value could be found by fitting theoretical radiation spectra to experimental ones. Planck’s assumption was however somewhat vague about exactly what these chunks of energy were physically. It was Einstein who proposed, in his 1905 explanation of the photoelectric effect, that $\hbar\omega$ gives the energy of photons, the particles of electromagnetic radiation.

Photons are bosons, relativistic ones, to be sure, but still bosons. Therefore the Bose-Einstein distribution should describe their statistics. More specifically, the average number of photons in each single-particle state should be

\begin{displaymath}
\iota^{\rm {b}}_\gamma=\frac{1}{e^{{\vphantom' E}^{\rm p}/{k_{\rm B}}T} - 1} %
\end{displaymath} (6.10)

where $\gamma$ is the standard symbol for a photon. Note the missing chemical potential. As discussed in chapter 11, the chemical potential is related to conservation of the number of particles. It does not apply to photons that are readily created out of nothing or absorbed by the atoms in the walls of the box. (One consequence is that Bose-Einstein condensation does not occur for photons, {N.21}.)

To get the energy of the photons in a small frequency range ${\rm d}\omega$, simply multiply the number of single particle states in that range, (6.7), by the number of photons per state $\iota^{\rm {b}}_\gamma$ above, and that by the single-photon energy ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\omega$.

That gives the radiation energy per unit volume of the box and per unit energy range as

\begin{displaymath}
\fbox{$\displaystyle
\rho(\omega) =
\frac{\omega^2}{\p...
...^3} \frac{\hbar\omega}{e^{\hbar\omega/{k_{\rm B}}T}-1}
$} %
\end{displaymath} (6.11)

This expression is known as “Planck’s blackbody spectrum.”

For low frequencies, the final ratio is about ${k_{\rm B}}T$, giving the Rayleigh-Jeans result. That is readily verified from writing a Taylor series for the exponential in the denominator. For high frequencies the energy is much less because of the rapid growth of the exponential for large values of its argument. In particular, the energy no longer becomes infinite at high frequencies. It becomes zero instead.

To rewrite the blackbody spectrum in terms of the frequency $f$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\omega$$\raisebox{.5pt}{$/$}$$2\pi$ in cycles per second, make sure to convert the actual energy in a frequency range, ${\rm d}{E}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\rho(\omega){\,\rm d}\omega$, to ${\rm d}{E}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\bar\rho(f){\,\rm d}{f}$. Merely trying to convert $\rho$ will get you into trouble. The same if you want to rewrite the blackbody spectrum in terms of the wave length $\lambda$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c$$\raisebox{.5pt}{$/$}$$f$.

For engineering purposes, what is often the most important is the amount of radiation emitted by a surface into its surroundings. Now it so happens that if you drill a little hole in the box, you get a perfect model for a truly black surface. An ideal black surface is defined as a surface that absorbs, rather than reflects, all radiation that hits it. If the hole in the box is small enough, any radiation that hits the hole enters the box and is never seen again. In that sense the hole is perfectly black.

And note that a black surface does not have to look black. If the black plate of your electric stove is hot enough, it will glow red. Similarly, if you would heat the inside of the box to the same temperature, the radiation inside the box would make the hole shine just as red. If you would heat the box to 6,000 K, about as hot as the surface of the sun, the hole would radiate sunlight.

The amount of radiation that is emitted by the hole can be found by simply multiplying Planck’s spectrum by one quarter of the speed of light $c$, {D.27}. That gives for the radiation energy emitted per unit area, per unit frequency range, and per unit time:

\begin{displaymath}
\fbox{$\displaystyle
{\cal I}(\omega) =
\frac{\omega^2...
...^2} \frac{\hbar\omega}{e^{\hbar\omega/{k_{\rm B}}T}-1}
$} %
\end{displaymath} (6.12)

A perfectly black surface area would radiate the same amount as the hole.

If you see the hole under an angle, it will look just as bright per unit area as when you see it straight on, but it will seem smaller. So your eyes will receive less radiation. More generally, if $A_e$ is a small black surface at temperature $T$ that emits radiation, then the amount of that radiation received by a small surface $A_r$ is given by

\begin{displaymath}
{\rm d}E =
\frac{\omega^2}{4\pi^3c^2}
\frac{A_e \cos\t...
...^{\hbar\omega/{k_{\rm B}}T}-1}
{\,\rm d}\omega {\,\rm d}t %
\end{displaymath} (6.13)

Here $r$ is the distance between the small surfaces, while $\theta_e$ and $\theta_r$ are the angles that the connecting line between the surfaces makes with the normals to the emitting and receiving surfaces respectively.

Often the total amount of energy radiated away by a black surface is of interest. To get it, simply integrate the emitted radiation (6.12) over all values of the frequency. You will want to make a change of integration variable to $\hbar\omega$$\raisebox{.5pt}{$/$}$${k_{\rm B}}T$ while doing this and then use a table book like [40, 18.80, p. 132]. The result is called the “Stefan-Boltzmann law:

\begin{displaymath}
\fbox{$\displaystyle
{\rm d}E_{\rm total\ emitted} = A \...
...r^3c^2} \approx
5.67\; 10^{-8}\mbox{ W/m$^2$\ K$^4$}
$} %
\end{displaymath} (6.14)

Since this is proportional to $T^4$, at 6,000 K 160,000 times as much radiation will be emitted as at room temperature. In addition, a much larger part of that radiation will be in the visible range. That is the reason you will see light coming from a hole in a box if it is at 6,000 K, but not when it is at room temperature.

A surface that is not perfectly black will absorb only a fraction of the radiation that hits it. The fraction is called the “absorptivity” $a$. Such a surface will also radiate less energy than a perfectly black one by a factor called the “emissivity” $e$. This assumes that the surface is in stable thermal equilibrium. More simply put, it assumes that no external source of energy is directed at the surface.

Helmholtz discovered that the absorptivity and emissivity of a surface are equal in thermal equilibrium, {D.28}. So poor absorbers are also poor emitters of radiation. That is why lightweight emergency blankets typically have reflective metallic coatings. You would think that they would want to absorb, rather than reflect, the heat of incoming radiation. But if they did, then according to Helmholtz they would also radiate precious body heat away to the surroundings.

Since a surface cannot absorb more radiation than hits it, the absorptivity cannot be greater than one, It follows that the emissivity cannot be greater than one either. No surface can absorb better or emit better than a perfectly black one. At least not when in thermodynamic equilibrium.

Note that absorptivity and emissivity typically depend on electromagnetic frequency. Substances that seem black to the eye may not be at invisible electromagnetic frequencies and vice-verse. It remains true for any given electromagnetic frequency that the absorptivity and emissivity at that frequency are equal. To soak up the heat of the sun in a solar energy application, you want your material to be black in the visible frequency range emitted by the 6,000 K surface of the sun. However, you want it to be white in the infrared range emitted at the operating temperature of the material, in order that it does not radiate the heat away again.

Absorptivity and emissivity may also depend on the direction of the radiation, polarization, temperature, pressure, etcetera. In thermodynamic equilibrium, absorptivity and emissivity must still be equal, but only at the same frequency and same directions of radiation and polarization.

For surfaces that are not black, formula (6.13) will need to be modified for the relevant emissivity. A simplifying grey body assumption is often made that the absorptivity, and so the emissivity, is constant. Absorptivity and emissivity are usually defined as material properties, cited for infinitely thick samples. For objects, the terms absorptance and emittance are used.

Fluorescence/phosphorescence and stimulated emission (lasers) are important examples of radiative processes that are not in thermal equilibrium. The above discussion simply does not apply to them.


Key Points
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Blackbody radiation is the radiation emitted by a black surface that is in thermal equilibrium.

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Planck’s blackbody spectrum determines how much is radiated at each frequency.

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Surfaces that are not black emit radiation that is less by a factor called the emissivity.

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Emissivity equals absorptivity for the same frequency and direction of radiation.

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If the material is not in thermal equilibrium, like energized materials, it is a completely different ball game.