202004240406 Homework 1 (Q6)

Derive the Planck function $B_\nu$ .

HINTS: You can make the following steps.

(1) Show that the density of states for photons is given by

$g(E)=\displaystyle{\frac{\mathrm{d}N}{\mathrm{d}E}=\frac{8\pi VE^2}{h^3c^3}}$ ,

(2) the photon density per unit frequency is given by

$\displaystyle{\frac{\mathrm{d}n}{\mathrm{d}\nu}}=\displaystyle{\frac{8\pi \nu^2}{c^3(e^{h\nu /kT+1})}}$ , and

(3) the definition of the specific intensity $B_\nu =I_\nu$ is the energy flux per unit frequency per unit solid angle.

Note: the energy and momentum of photons are related as

$E=h\nu =pc$ ,

the energy flux per unit frequency is given by

$h\nu \displaystyle{\frac{\mathrm{d}n}{\mathrm{d}\nu}}c$ ,

and total solid angle for isotropic emission is $4\pi$ .

Attempts.

In attempting to derive the Planck function, extensive reference was made to the following several sources found on the Internet:

i. G. B. Rybicki & A. P. Lightman. (1979). Radiative Processes in Astrophysics. John Wiley & Sons, Inc.

ii. https://edisciplinas.usp.br/pluginfile.php/48089/course/section/16461/qsp_chapter10-plank(DOT)pdf

iii. https://en.wikipedia(DOT)org/wiki/Planck\%27s_law#Derivation

iv. http://web.phys.ntnu.no/~stovneng/TFY4165_2013/BlackbodyRadiation(DOT)pdf

Here I follow Rybicki and Lightman’s derivation (pp. 20–22, 1979) direct, and have changed only some wordings.

Solution.

The wave vector of the photon of frequency $\nu$ propagating in direction $\mathbf{n}$ is $\mathbf{k}=(2\pi /\lambda )\,\mathbf{n}=(2\pi\nu /c)\,\mathbf{n}$ . For a photon gas in a box of dimension $(L_x,L_y,L_z)$ , the number of nodes of the standing wave is $n_i=k_iL_i/2\pi$ for each direction $i\in\{ x,y,z\}$ and for some wave number $k_i$ . For all directions, write the variation of the number of nodes with the wave number

$\Delta n_i=\displaystyle{\frac{L_i\Delta k_i}{2\pi}}$

The three-dimensional wave vector element $\Delta k_x\Delta k_y\Delta k_z\equiv \mathrm{d}^3k$ has the number of states to be

$\Delta N=\Delta n_x\Delta n_y\Delta n_z=\displaystyle{\frac{L_xL_yL_z\,\mathrm{d}^3k}{(2\pi )^3}}$

By the fact that $L_xL_yL_z=V$ is the volume of the box and the fact that photons have two independent polarizations (i.e., two states per wave vector $\mathbf{k}$ ), for each 3D-wave vector, the number of states for every unit volume is

$\displaystyle{\frac{\Delta N}{V\,\mathrm{d}^3k}}=2/(2\pi )^3$ .

Using solid angle as in spherical coordinates, rewrite

$\mathrm{d}^3k=k^2\,\mathrm{d}k\,\mathrm{d}\Omega =\displaystyle{\frac{(2\pi )^3\nu^2\,\mathrm{d}\nu\,\mathrm{d}\Omega}{c^3}}$ .

Then, the density of states, i.e., the number of states per solid angle per volume per frequency, is given by

$\rho_s\stackrel{\textrm{def}}{=}\displaystyle{\frac{\mathrm{d}N}{\mathrm{d}\Omega\,\mathrm{d}V\,\mathrm{d}\nu}}=\displaystyle{\frac{2\nu^2}{c^3}}$ .

As each state of $n$ photons each of energy $h\nu$ is of energy $E_n=nh\nu$ , and according to statistical mechanics the probability of a state of energy $E_n$ is proportional to $\mathrm{exp}(-\beta E_n)$ (where $\beta =1/kT$ and $k=\textrm{ the Boltzmann's constant}$ ), the average energy of each state is

$\overline{E}=\displaystyle{\frac{\sum_{n=0}^\infty E_n\mathrm{exp}(-\beta E_n)}{\sum_{n=0}^\infty \mathrm{exp}(-\beta E_n)}}=-\displaystyle{\frac{\mathrm{\partial}}{\mathrm{\partial}\beta}\ln \bigg( \sum_{n=0}^\infty \mathrm{exp}(-\beta E_n)\bigg)}$ .

Recall the formula for the sum of a geometric series:

$\displaystyle{\sum_{n=0}^\infty}\mathrm{exp}(-\beta E_n)=\displaystyle{\sum_{n=0}^\infty}\mathrm{exp}(-nh\nu\beta )=(1-\mathrm{exp}(-\beta h\nu ))^{-1}$ ,

one has therefore the result (in Bose-Einstein statistics)

$\overline{E}=\displaystyle{\frac{h\nu\,\mathrm{exp}(-\beta h\nu )}{1-\mathrm{exp}(-\beta h\nu )}=\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}}$ .

where it can be seen that the occupation number

$n_\nu =\bigg[ \mathrm{exp}\bigg( \displaystyle{\frac{h\nu}{kT}}\bigg) -1\bigg]^{-1}$

means the average number of photons in some frequency $\nu$ , as one energy $h\nu$ corresponds to one frequency $\nu$ .

The energy per solid angle per volume per frequency is the product of i. $\overline{E}$ , the average energy of each state, and ii. $\rho_s$ , the density of states.

I.e., $\overline{E}\cdot \rho_s = \bigg( \displaystyle{\frac{2\nu^2}{c^3}}\bigg) \displaystyle{\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}}$

Define the specific energy density $u_\nu$ the energy per unit volume per unit frequency range. Then the energy density per unit solid angle can be expressed as

$\begin{aligned} u_\nu (\Omega ) & =\displaystyle{\frac{\mathrm{d}E}{\mathrm{d}V\,\mathrm{d}\Omega\,\mathrm{d}\nu}} \\ \dots \textrm{arranging } & \textrm{and equating the previous two equations}\dots \\ u_\nu (\Omega )\,\mathrm{d}V\,\mathrm{d}\nu\,\mathrm{d}\Omega & =\bigg( \displaystyle{\frac{2\nu^2}{c^3}}\bigg) \displaystyle{\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}}\,\mathrm{d}V\,\mathrm{d}\nu\,\mathrm{d}\Omega \\ \dots \textrm{comparing} & \dots \\ u_\nu (\Omega ) & = \bigg( \displaystyle{\frac{2\nu^2}{c^3}}\bigg) \displaystyle{\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}} \end{aligned}$

Observe that the specific energy density $u_\nu$ and the specific intensity (or brightness) $I_\nu$ are related by

$u_\nu (\Omega )=\displaystyle{\frac{I_\nu}{c}}$ .

Now that the specific intensity, $I_\nu$ , is the same as the source function (of specific emission mechanism), $B_\nu$ .
The frequency distribution is said to be a blackbody form, i.e.,

The Planck function is expressed as

$\boxed{I_\nu =B_\nu (T)=\displaystyle{\frac{2h\nu^3}{c^2}}\bigg[ \mathrm{exp}\bigg( \displaystyle{\frac{h\nu}{kT}} \bigg) -1\bigg]^{-1}}$

(Units: $\mathrm{erg\cdot s^{-1}\cdot Hz^{-1}\cdot cm^{-2}\cdot ster^{-1}}$ )

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