---
title: "Why hot surfaces glow, and why a good mirror glows weakly"
description: "Build Planck’s spectrum from mode energy and mode count, derive the Stefan–Boltzmann law, and use Kirchhoff’s law to explain the emission dip of SiC."
canonical_url: "https://sandboxphysics.com/stories/fluctuational-electrodynamics/thermal-emission-and-kirchhoff/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/fluctuational-electrodynamics/thermal-emission-and-kirchhoff/"
last_modified: "2026-10-30"
---

[Fluctuational electrodynamics](https://sandboxphysics.com/stories/fluctuational-electrodynamics/)

L1 · Chapter 1

# Why hot surfaces glow, and why a good mirror glows weakly

Sandbox Physics · 2026-10-30

An object at room temperature glows mostly in the infrared, near a wavelength of 10 µm. How much it emits at each frequency depends on two things: how much energy one thermal mode of the field holds, and how readily the surface lets that mode out. A black surface lets everything out. A surface that reflects strongly at some frequency emits weakly there, because emission and absorption are the same transmission run in opposite directions.

## Before you begin

Plane waves and the wave vector; reflection of light at an interface; the Boltzmann factor and the mean energy of a harmonic oscillator; integrals over frequency.

- Build the Planck spectrum from the mean energy of one mode times the number of modes per unit volume.
- Derive the Stefan–Boltzmann law and the flux between two black plates from that spectrum.
- Use Kirchhoff’s law to explain why silicon carbide emits weakly near a wavelength of 11 µm.

## The energy of one mode and the number of modes

Picture a cavity whose walls are at temperature $T$. The field inside splits into standing waves, and each one is a harmonic oscillator of angular frequency $\omega$. Quantum mechanics gives its mean energy above the ground state, $\Theta(\omega,T)$. At low frequency, $\hbar\omega\ll k_BT$, this is the classical $k_BT$. At high frequency, $\hbar\omega\gg k_BT$, it falls off exponentially.

Now count modes. With both polarisations, a unit volume holds $\omega^2d\omega/(\pi^2c^3)$ modes between $\omega$ and $\omega+d\omega$. The product is the energy per unit volume and per unit angular frequency inside the cavity, $u_0$.

Every later chapter keeps this split. The mean energy of a mode depends only on temperature and frequency; how many modes there are, and whether they can get out, depends on the materials and the geometry.

$$
\Theta(\omega,T)=\frac{\hbar\omega}{e^{\hbar\omega/k_BT}-1},\qquad u_0(\omega,T)=\Theta(\omega,T)\,\frac{\omega^{2}}{\pi^{2}c^{3}}
$$

Eq. (1) of Shchegrov et al. and Eq. (47) of Joulain et al. The unit of $u_0$ is $\mathrm{J\,s\,m^{-3}}$: energy density per unit angular frequency.

Source: A. V. Shchegrov · K. Joulain

## Integrate over frequency to get the Stefan–Boltzmann law

With $x=\hbar\omega/k_BT$ the integral becomes $\int_0^\infty x^3/(e^x-1)\,dx=\pi^4/15$, so the total energy density in the cavity grows as $T^4$.

A black surface emits into half of space. The energy crossing a plane per unit time is the energy density times the speed of light times the average normal velocity component over the outgoing half of all directions, which is $1/4$. So a black surface radiates $\sigma T^4$.

Where the spectrum peaks depends on the variable. Per unit angular frequency the peak sits at $\hbar\omega\approx2.82\,k_BT$, which is $1.11\times10^{14}\,\mathrm{rad\,s^{-1}}$ at 300 K. Per unit wavelength it sits near 10 µm, as Wien’s law says. The two peaks differ because the two spectra are densities in different variables.

$$
\int_0^{\infty}u_0\,d\omega=\frac{\pi^{2}k_B^{4}}{15\hbar^{3}c^{3}}\,T^{4},\qquad q_{\mathrm{black}}=\frac{c}{4}\int_0^{\infty}u_0\,d\omega=\sigma T^{4},\qquad \sigma=\frac{\pi^{2}k_B^{4}}{60\hbar^{3}c^{2}}
$$

With the SI constants that are exact since 2019, $\sigma=5.670374419\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$, the NIST value to all ten digits. The course benchmark script also integrates $u_0$ numerically and gets the same number.

Source: NIST Reference on Constants · K. Joulain

## Kirchhoff’s law: emission is absorption in reverse

A real surface reflects part of every wave. When Joulain and co-workers compute the energy density emitted by a half-space, each direction carries one factor: half the sum of the transmissions for the two polarisations. The material is full of blackbody radiation, and thermal emission is that radiation transmitted out through the interface.

Run the same transmission backwards and you have light from outside entering the material and being absorbed. The two transmissions are equal, so at every frequency, angle and polarisation the emissivity equals the absorptivity. That is Kirchhoff’s law, and here it follows directly from the reciprocity of the transmission.

Silicon carbide (SiC) is the main material of this course. Its permittivity comes from lattice vibrations. Between the transverse optical phonon frequency $\omega_T$ and the longitudinal one $\omega_L$ the real part of $\varepsilon$ is negative: the lattice responds against the field, and the surface reflects almost everything. This band is called the reststrahlen band.

$$
e(\omega,\theta)=\tfrac{1}{2}\left[\left(1-|r_s|^{2}\right)+\left(1-|r_p|^{2}\right)\right]
$$

The emissivity inside Eq. (46) of Joulain et al. Here $r_s$ and $r_p$ are the Fresnel reflection amplitudes for the two polarisations at angle $\theta$. A black surface has $r=0$ and $e=1$.

$$
\varepsilon(\omega)=\varepsilon_\infty\,\frac{\omega_L^{2}-\omega^{2}-i\gamma\omega}{\omega_T^{2}-\omega^{2}-i\gamma\omega}
$$

The SiC permittivity printed by Shchegrov et al. on p. 1549, the same oscillator as Eq. (22) of Joulain et al.: $\varepsilon_\infty=6.7$, $\omega_L=182.7$, $\omega_T=149.5$ and $\gamma=0.9$, with frequencies in $10^{12}\,\mathrm{rad\,s^{-1}}$.

Source: K. Joulain · A. V. Shchegrov

## Worked example: two black plates and a dip in SiC

Problem. Two black plates 1 cm apart are held at 400 K and 300 K. Find the net heat flux. Then find the normal emissivity of SiC at $\omega=166$ and $\omega=200$ (in $10^{12}\,\mathrm{rad\,s^{-1}}$) and say what fraction of a black surface’s emission SiC at 300 K gives off at 166.

Model. A black plate absorbs everything that reaches it, so each plate emits $\sigma T^4$ and absorbs all of the other’s emission. One centimetre is far larger than the 10 µm thermal wavelength, so the gap does not matter. The net flux is $q=\sigma(T_1^4-T_2^4)$.

Numbers. $\sigma T_1^4=1452\,\mathrm{W\,m^{-2}}$ and $\sigma T_2^4=459.3\,\mathrm{W\,m^{-2}}$, so $q=992\,\mathrm{W\,m^{-2}}$. The linear form $4\sigma T^3\Delta T$ with $T=300\,\mathrm{K}$ gives $6.12\times100=612\,\mathrm{W\,m^{-2}}$, 38% too low: for a 100 K difference the curvature of $T^4$ matters.

Now SiC. At 166, $\varepsilon=-7.48+0.41i$, and the normal-incidence formula gives $|r|^2=0.966$, an emissivity of only 0.034. At 200, $\varepsilon=2.51+0.04i$ and the emissivity is 0.949. So at 166 SiC emits 3.4% of what a black surface does, and at 200 it emits 95%. The far-field spectrum has a broad dip between 150 and 180, the band Shchegrov and co-workers show in their Fig. 1(a).

$$
e_\perp(\omega)=1-\left|\frac{1-\sqrt{\varepsilon(\omega)}}{1+\sqrt{\varepsilon(\omega)}}\right|^{2}
$$

At normal incidence the two polarisations coincide. Take the branch of $\sqrt{\varepsilon}$ with a positive imaginary part.

Source: NIST Reference on Constants · A. V. Shchegrov

## Try it in the Lab

### [See the emission dip of SiC from far away](https://sandboxphysics.com/labs/thermal-near-field-probe/)

1. Predict first: with the probe 1 mm above SiC at 300 K, in which band will the measured spectrum fall below the dashed black-surface reference?
2. Choose the Far-field spectrum preset. Compare the solid curve with the black-surface reference between 150 and 180 (in $10^{12}\,\mathrm{rad\,s^{-1}}$), then read the strongest frequency.
3. Raise the sample temperature to 600 K. Does the dip move? Explain with the permittivity, which in this model does not depend on temperature.

### [Move the Planck peak](https://sandboxphysics.com/labs/blackbody-radiation/)

1. Predict first: when the cavity goes from 300 K to 600 K, by what factor does the peak wavelength change?
2. In the blackbody Lab, set the two temperatures in turn and read the peak wavelength. Then set the gray emissivity to 0.5.
3. Explain why a gray body only scales the spectrum down and leaves the peak in place, while SiC cuts a hole into one band.

## Exercises with explained answers

1. Two black plates are at 310 K and 300 K. Find the net flux with the exact formula and with the linear form.

   Show the answer

   Exactly, $\sigma(310^4-300^4)=5.670\times10^{-8}\times1.135\times10^{9}=64.4\,\mathrm{W\,m^{-2}}$. The linear form at 300 K gives $4\sigma T^3\Delta T=61.2\,\mathrm{W\,m^{-2}}$; at the mean temperature of 305 K it gives $64.4\,\mathrm{W\,m^{-2}}$. For a small difference evaluated at the mean temperature, the linear form is nearly exact.

2. A gray body has emissivity 0.5 at every frequency. How do its spectrum and total emission compare with a black body at the same temperature? Why is SiC not a gray body?

   Show the answer

   The gray body’s spectrum is halved at every frequency, its peak stays in place, and it emits $\sigma T^4/2$ in total. The emissivity of SiC changes sharply with frequency, from 0.95 at 200 to 0.03 at 166 (in $10^{12}\,\mathrm{rad\,s^{-1}}$). Its spectrum loses a piece inside the reststrahlen band and changes shape.

3. The SiC near-field line sits at $\omega=178.7\times10^{12}\,\mathrm{rad\,s^{-1}}$. Find $\hbar\omega/k_BT$ at 300 K and $\Theta$ in units of $k_BT$. By what factor does $\Theta$ grow at 600 K?

   Show the answer

   At 300 K, $\hbar\omega/k_BT=4.55$ and $\Theta/k_BT=4.55/(e^{4.55}-1)=0.0486$. At 600 K, $\hbar\omega/k_BT=2.275$ and $\Theta$ is 10.7 times its 300 K value. Doubling the temperature raises the mode energy of this line tenfold because the line sits on the exponential tail of the Planck spectrum.

4. A surface at 300 K sits inside a cavity at 300 K. What would happen if its emissivity at one frequency exceeded its absorptivity? What does the derivation of Joulain and co-workers add to this thermodynamic argument?

   Show the answer

   At that frequency the surface would emit more than it absorbs. It would cool while the cavity warms, creating a temperature difference without work, which the second law forbids; so the two must be equal. Thermodynamics only demands overall balance. Joulain and co-workers write the emissivity as a transmission, and the transmission is unchanged when source and detector are swapped, so the equality holds separately at each frequency, angle and polarisation.

[Next: Evanescent waves and the surface phonon polariton of silicon carbide](https://sandboxphysics.com/stories/fluctuational-electrodynamics/evanescent-waves-and-surface-polaritons/)

## Sources and what we recomputed

1. [A. V. Shchegrov, K. Joulain, R. Carminati and J.-J. Greffet, Near-field spectral effects due to electromagnetic surface excitations, Physical Review Letters 85, 1548–1551 (2000), HAL hal-01624761v1](https://hal.science/hal-01624761v1)

   Full text of the published letter in the HAL archive. Eq. (1): the Planck spectrum as mode energy times mode density. Eqs. (2) to (5): thermal currents from the fluctuation–dissipation theorem and the energy density of the emitted electric field. Eq. (6): the surface-wave dispersion relation. Eqs. (7) and (8): the near-field spectrum as mode energy times a local mode density that grows as one over height cubed. Parameters of SiC on p. 1549. Fig. 1: spectra at 1000, 2 and 0.1 µm above SiC at 300 K, with the line at 178.7 in units of 10 to the 12 rad/s. Fig. 3: decay with height at 166 and 210 in the same units. We recomputed the line position, the height law and the decay at 166; we did not digitise the figures.

2. [K. Joulain, J.-P. Mulet, F. Marquier, R. Carminati and J.-J. Greffet, Surface electromagnetic waves thermally excited: radiative heat transfer, coherence properties and Casimir forces revisited in the near field, arXiv:physics/0504068v2; Surface Science Reports 57, 59–112 (2005)](https://arxiv.org/abs/physics/0504068v2)

   Version 2, 24 April 2005. Sec. 2: surface waves, the Drude permittivity (Eq. 21) and the SiC oscillator (Eq. 22). Sec. 3.2: the fluctuation–dissipation theorem and the mean oscillator energy (Eqs. 25 to 30). Secs. 4.1 and 4.2: the emitted energy density (Eqs. 44 to 48), emissivity as the half-sum of the two transmission factors and its link to Kirchhoff’s law, and Fig. 6 at 100 µm, 1 µm and 100 nm above SiC. Sec. 7: heat flux between two half-spaces (Eqs. 89 to 93), the growth as the inverse square of the gap, and Figs. 17 and 18, which they reproduce from another paper. Sec. 8: the Casimir force from the same fluctuating fields. The evanescent term of Eq. 45 as printed carries an extra factor of two; we fixed the prefactor from Eq. 48. The prefactor of Eq. 93 is garbled in the text version we read, so we quote only its inverse-square dependence on the gap.

3. [NIST Reference on Constants, Units, and Uncertainty: speed of light in vacuum, reduced Planck constant, Boltzmann constant and Stefan–Boltzmann constant](https://physics.nist.gov/cuu/Constants/index.html)

   The value pages for the four constants. Since 2019 the speed of light, the Planck constant and the Boltzmann constant are exact in the SI, so the Stefan–Boltzmann constant computed from them agrees with the NIST value 5.670374419 to all ten digits listed.