Blackbody Radiation

Classically, the radiation inside a box is a combination of standing-wave patterns, each behaving as a harmonic oscillator with frequency \(f = c/\lambda\). Each oscillator has two quadratic degrees of freedom, so equipartition gives it an average energy \(\bar{E} = k_{\mathrm{B}}T\). But there are infinitely many modes, most of them at very short wavelengths, so the total thermal energy would be infinite. That cannot be physically true.

Planck’s solution: a quantum oscillator has energies \(E_n = 0, hf, 2hf, \ldots\) (dropping the zero-point energy, which only multiplies \(Z\) by a constant). The partition function of a single oscillator is a geometric series,

and the average number of energy units in the oscillator is the Planck distribution:

Modes with \(hf \gg k_{\mathrm{B}}T\) are exponentially suppressed, “frozen out,” and the catastrophe disappears. Slide the mode frequency to compare.

Now take the units of field energy seriously as particles called photons. They are bosons and should follow the Bose–Einstein distribution with \(\varepsilon = hf\). Photons can be created or destroyed in any quantity, so their number is not conserved: \(N\) takes whatever value minimizes \(F\), which means \((\partial F/\partial N)_{T,V} = \mu = 0\). Then \(\bar{n}_{\mathrm{BE}}\) is exactly \(\bar{n}_{\mathrm{Pl}}\).

Photons are ultrarelativistic, \(\varepsilon = pc = hcn/2L\) with \(n = |\vec{n}|\), and come in two independent polarizations (playing the role of the electrons’ two spin states). Summing \(\varepsilon\,\bar{n}_{\mathrm{Pl}}\) over all modes in the eighth-sphere of \(n\)-space gives

The integrand is the spectrum: the energy density per unit photon energy,

With \(x = \varepsilon/k_{\mathrm{B}}T\) its shape is \(x^3/(e^x - 1)\), which peaks at \(x = 2.82\). Higher temperature means higher photon energies, larger frequency, and shorter wavelength: Wien’s law.

Problem 7.39. Changing variables to \(\lambda = hc/\varepsilon\), with \(d\varepsilon = -hc\,d\lambda/\lambda^2\), gives the energy density per unit wavelength:

The two peaks disagree because \(\lambda \propto 1/\varepsilon\) is nonlinear: equal intervals \(d\varepsilon\) correspond to unequal intervals \(d\lambda \propto d\varepsilon/\varepsilon^2\). Compare the last two rows of the panel. For sunlight, \(u(\varepsilon)\) peaks at 1.41 eV, which corresponds to 880 nm in the near infrared, yet \(u(\lambda)\) peaks near 500 nm.

Integrating the spectrum with \(\int_0^\infty x^3/(e^x-1)\,dx = \pi^4/15\) gives the total energy, and the rest follow:

Doubling the temperature multiplies the energy by 16. Dimensional analysis already suggests this: each photon carries something of order \(k_{\mathrm{B}}T\), and the only relevant length is the photons’ typical wavelength \(hc/k_{\mathrm{B}}T\). The density of states (Problem 7.40) is \(g(\varepsilon) = 8\pi V\varepsilon^2/(hc)^3\).

with \(a = 8\pi^5 k_{\mathrm{B}}^4 V/15(hc)^3\) and \(\int_0^\infty x^2/(e^x-1)\,dx = 2\zeta(3) \approx 2.404\) (Problem 7.44). \(S\) and \(N\) share the same dependence on \(V(k_{\mathrm{B}}T/hc)^3\), so the entropy per photon is a constant, \(S/N \approx 3.6\,k_{\mathrm{B}}\). For the pressure (Problem 7.45) and free energy (Problem 7.46),

and \(F\) can also be found mode by mode as \(-k_{\mathrm{B}}T\ln Z\), with the same result. Pick a temperature to see the numbers.

The radiation that fills the entire observable universe has an almost perfect thermal spectrum at 2.73 K. Its \(u(f)\) peaks at \(\varepsilon = 2.82\,k_{\mathrm{B}}T = 6.6 \times 10^{-4}\ \mathrm{eV}\), corresponding to a wavelength of about 1 mm, in the far infrared. There are about \(4.1 \times 10^8\) of these photons in every cubic meter of space. Part of the “snow” on an old analog television tuned between channels came from this leftover glow of the Big Bang. Choose Cosmic background in the spectrum above to see it.

Open a small hole of area \(A\) in a box of radiation. All photons travel at the same speed \(c\), regardless of wavelength, so the light that escapes has the same spectrum as the light inside. To find how much escapes in a time \(dt\), consider a chunk of the box at distance \(R\) and angle \(\theta\) from the hole, inside a hemispherical shell of thickness \(c\,dt\). Only photons in that shell can reach the hole during \(dt\).

The chunk holds energy \((U/V)\,R^2\sin\theta\,d\theta\,d\phi\,c\,dt\), and a fraction \(A\cos\theta/4\pi R^2\) of it heads for the hole; \(R\) cancels. Integrating over \(\phi\) from 0 to \(2\pi\) and \(\theta\) from 0 to \(\pi/2\),

The factor of \(\tfrac14\) is easy to check by brute force. Below, 2000 photons fly around a box at speed \(c\), bouncing off the walls; any that hit the hole escape. To keep the radiation inside in equilibrium, each is replaced by a photon entering through the hole, as if the hole opened onto more radiation at the same temperature. Compare the counted escape rate with \(\tfrac14 (N/V)\,c\,A\), and the escape angles with \(\sin\theta\cos\theta\).

This is Stefan’s law, discovered empirically in 1879 (remember “5–6–7–8”). Any nonreflecting (“black”) surface at the same temperature must emit the same power, and indeed the same entire spectrum, as the hole; otherwise placing it next to the hole, with a wavelength filter in between, would violate the second law. A surface that reflects some photons emits correspondingly less: power \(= \sigma e A T^4\), where \(e\) is the emissivity.

The Earth receives 1370 W/m² of sunlight (the solar constant) at a distance of \(1.5 \times 10^{8}\ \mathrm{km}\), so the Sun’s total output is \(4\pi (1.5 \times 10^{11}\ \mathrm{m})^2 \times 1370\ \mathrm{W/m^2} \approx 3.9 \times 10^{26}\ \mathrm{W}\). Its radius is about \(7.0 \times 10^8\ \mathrm{m}\), so with emissivity 1, \(T_\odot = (L/\sigma A)^{1/4} \approx 5800\ \mathrm{K}\).

The Earth absorbs sunlight across its disk, \(\pi r_E^2\), but radiates from its whole surface, \(4\pi r_E^2\). Balancing the two gives the Earth’s temperature:

Model the atmosphere as a single layer, transparent to visible light but opaque to infrared. In equilibrium the surface then receives twice as much energy as from sunlight alone, raising its temperature by \(2^{1/4}\): the greenhouse effect.