Derivation of Blackbody Radiation (No-Nonsense Version)
HaifeiWith nothing better to do, let’s review blackbody radiation~
Many articles on blackbody radiation start by telling you a long history lesson, which can easily get confusing.
This article gets straight to the point: shut up and calculate.
1. What Is the Blackbody Radiation Formula?
The blackbody radiation formula refers to the energy density radiated by a blackbody per unit frequency.
To calculate the blackbody radiation formula, we need to calculate how many quantum states there are between frequencies \(\nu\) and \(\nu + \mathrm{d}\nu\), as well as how many photons occupy each quantum state, and then multiply them by the photon energy \(h\nu\).
\[\varepsilon(\nu) \mathrm{d} \nu = [(\nu,\nu+\mathrm{d}\nu) 中有多少个量子态] \times [这些量子态上有多少个光子] \times h\nu\]2. Density of States
How many quantum states are there between frequencies \(\nu\) and \(\nu + \mathrm{d}\nu\)?
Consider a rectangular box with periodic boundary conditions, whose length, width, and height are \(L_{x,y,z}\). Then the wavenumber \(\kappa_{x,y,z}\) of the electromagnetic waves in the box satisfies:
\[\kappa_i = n_i/L_i, \quad(i=x,y,z)\]Note that \(\kappa\) is the wavenumber, not the wavevector \(k\). Their relation is \(k = 2\pi \kappa\).
Thus, the number of quantum states is:
\[\mathrm{d} N = \mathrm{d} n_x\mathrm{d}n_y\mathrm{d}n_z = L_xL_yL_z \mathrm{d}\kappa_x \mathrm{d}\kappa_y\mathrm{d}\kappa_z = V \mathrm{d}^3\vec{\kappa}\]Let the number of states between \(\nu\) and \(\nu + \mathrm{d}\nu\) be \(f(\nu)\). Then
\[f(\nu) \mathrm{d}\nu = \mathrm{d}N = V \mathrm{d}^3\vec{\kappa} = V \cdot4\pi \kappa^2 \mathrm{d}\kappa = V \cdot\frac{4 \pi \nu^2}{c^3} \mathrm{d} \nu\]where \(\kappa = |\vec{\kappa}|\), and \(\nu = c\kappa\).
Therefore, \(f(\nu) = V \cdot4 \pi \nu^2/c^3\).
Since an electromagnetic wave in a given spatial mode has two polarizations, we must also multiply by 2. In addition, since we seek the energy density rather than the energy, we can divide by the volume in advance:
\[f(\nu) = 8 \pi \nu^2/c^3\]This is the number of states (per unit volume) between \(\nu\) and \(\nu + \mathrm{d}\nu\).
The idea here is that in three-dimensional space, the density of states near frequency \(\nu\) is proportional to the square of \(\nu\), just as the area of a sphere is proportional to the square of its radius.
3. Number of Photons in a Quantum State
At thermal equilibrium, how many photons occupy a state of frequency \(\nu\)? The answer is \(1/(e^{h\nu/kT} - 1)\). Let us derive it below:
The partition function of a bosonic system is:
\[\begin{aligned} Z &= \sum_{n_1,n_2,\cdots} \exp[-\beta (n_1 \varepsilon_1 + n_2 \varepsilon_2 +\cdots)] \\ &= \left(\sum_{n_1} e^{-\beta n_1\varepsilon_1}\right)\left(\sum_{n_2} e^{-\beta n_2\varepsilon_2}\right)\cdots \\ &= \left(\frac{1}{1-e^{-\beta \varepsilon_1}}\right) \left(\frac{1}{1-e^{-\beta \varepsilon_2}}\right) \cdots \end{aligned}\]where \(\varepsilon_i\) denotes the photon energy in the i-th mode, and \(n_i\) denotes the number of photons in the i-th frequency mode.
Then the average photon number in the i-th mode is:
\[\begin{aligned} \langle n_i\rangle = -\frac{1}{\beta}\frac{\partial \ln Z}{\partial \varepsilon_i} = \frac{1}{e^{\beta \varepsilon_i} - 1} \end{aligned}\]That is: the thermal-equilibrium photon number in a state of frequency \(\nu\) is \(\langle n(\nu)\rangle = 1/(e^{h\nu/kT} - 1)\).
This is in fact the Bose-Einstein distribution without a chemical potential. It is also called the Planck distribution.
4. Result
\[\begin{aligned} \varepsilon(\nu) \mathrm{d} \nu &= [(\nu,\nu+\mathrm{d}\nu) 中有多少个量子态] \times [这些量子态上有多少个光子] \times h\nu \\ &=f(\nu) \mathrm{d} \nu \cdot\langle n(\nu)\rangle \cdot h\nu\\ &=\frac{8\pi \nu^2}{c^3} \cdot \frac{1}{e^{h\nu/kT}-1} \cdot h\nu \end{aligned}\]Viewed this way, the blackbody radiation formula is quite easy to remember: one only needs to remember the density of states \(f(\nu)\) and the photon number at thermal equilibrium \(\langle n(\nu)\rangle\). The density of states is proportional to the square of the frequency, while the thermal-equilibrium photon number follows the Bose-Einstein distribution.
At low frequencies, \(\varepsilon(\nu)\) is limited by the density of states and the energy of an individual photon; at high frequencies, \(\varepsilon(\nu)\) is limited by the average photon number.
Exercise 1: In a two-dimensional world, what does the density of states look like? What does the blackbody radiation formula look like?
Exercise 2: Which step above goes wrong in the ultraviolet catastrophe?
Exercise 3: At room temperature and visible-light frequencies ( \(\sim 5.5 \times 10^{14} \text{Hz}\) ), what is the approximate thermal-equilibrium photon number in a single mode?
Answer 1: In a two-dimensional world, the density of states is \(f(\nu)=2\pi\nu/c^2\), and the blackbody radiation formula is \(\varepsilon(\nu)=\frac{2\pi h\nu^2}{c^2(e^{h\nu/kT}-1)}\)
Answer 2: The ultraviolet catastrophe results from directly replacing \(\frac{h\nu}{e^{h\nu/kT}-1}\) with \(kT\) from the classical equipartition theorem. When \(\nu \ll 1\), \(\frac{h\nu}{e^{h\nu/kT}-1} \rightarrow kT\), but when \(\nu \gg 1\), \(\varepsilon(\nu)\) diverges; this is the ultraviolet catastrophe.
Answer 3: At room temperature and visible-light frequencies, \(h\nu /kT = \frac{6.6 × 10^{-34} \times 5.5 \times 10^{14}}{1.4 × 10^{-23} \times 300} = 86\), \(\frac{1}{e^{86}-1} \ll 1\). This is why we cannot see blackbody radiation in the visible-light band.