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}{\pi^2c^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}{4...
...2c^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\thet...
...{e^{\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{$\displaystyle
{\rm d}E_{\rm total\ emitted} = A \si...
...\mbox{5.67~10$\POW9,{-8}$\ W/m$\POW9,{2}$\ K$\POW9,{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.