L3 · Chapter 4

Radiative heat between two plates, counted channel by channel

Sandbox Physics · 2026-10-30

With only vacuum between two plates, heat can cross only as radiation. When the gap is much larger than the thermal wavelength, this is the textbook blackbody exchange and the result does not depend on the gap. As the gap shrinks, the evanescent fields of the two surfaces start to overlap; waves that could never reach the other side now tunnel across and the flux grows. This chapter sets up the calculation and uses it to recover the familiar far-field result.

Each channel carries at most one share

Two half-infinite plates at temperatures $T_1$ and $T_2$ face each other across a vacuum gap of width $d$ . Each emits thermal radiation as in chapter 3 and absorbs what the other sends.

Split the field by frequency $\omega$ , in-plane wavevector $\beta$ and polarisation. The plates are uniform in the plane, so each $(\omega,\beta,\text{polarisation})$ is independent and can be treated as one channel. Plate 1 feeds $\Theta(\omega,T_1)$ into the channel, plate 2 feeds back $\Theta(\omega,T_2)$ , and the energy transmission $\tau$ of the channel sets how much gets across.

The net flux is the sum over channels of the difference in mode energy, times the transmission, times the number of channels per unit area. This is the Polder–Van Hove form. Eqs. (90) and (91) of Joulain et al. split the same result into a travelling term and an evanescent term.

q=\int_0^{\infty}\frac{d\omega}{2\pi}\,\bigl[\Theta(\omega,T_1)-\Theta(\omega,T_2)\bigr]\,\Phi(\omega,d),\qquad \Phi(\omega,d)=\sum_{s,p}\int_0^{\infty}\frac{\beta\,d\beta}{2\pi}\,\tau(\omega,\beta,d)

\Phi

is the effective number of channels per unit area and per unit angular frequency, in

\mathrm{m^{-2}}

. The zero-point energy

\hbar\omega/2

is the same on both sides and drops out of the difference.

Two plates at temperatures

T_1

and

T_2

across a vacuum gap

d

. A travelling wave (amber) crosses with an in-plane wavevector below

k_0

; an evanescent wave (cyan) decays from both surfaces and tunnels across only when

d

is comparable to its decay length. Original schematic, not to scale.

Source: K. Joulain

Transmission of travelling and tunnelling channels

Channels with in-plane wavevector below $k_0$ are travelling waves. A wave leaves plate 1 and bounces between the plates, interfering as in a Fabry–Perot cavity. The numerator holds the transmission of each interface and the denominator keeps track of the repeated reflections.

Channels with in-plane wavevector above $k_0$ are evanescent. At a single interface they carry no energy away; with a second interface opposite, the two decaying fields overlap and energy tunnels across. The transmission is proportional to $e^{-2\kappa d}$ and grows as the gap shrinks. The numerator holds the imaginary parts of the two reflection coefficients, which chapter 2 showed are large near the surface wave.

Neither transmission can exceed one. The course benchmark samples many frequencies, wavevectors and gaps; the largest value it finds is 0.999999. A channel can pass at most the whole difference in mode energy between the two sides, never more.

\tau=\begin{cases}\dfrac{(1-|r_1|^{2})(1-|r_2|^{2})}{|1-r_1r_2e^{2ik_zd}|^{2}}, & \beta<k_0\\[2ex]\dfrac{4\operatorname{Im}r_1\,\operatorname{Im}r_2\,e^{-2\kappa d}}{|1-r_1r_2e^{-2\kappa d}|^{2}}, & \beta>k_0\end{cases}

The transmission factors in Eqs. (90) and (91) of Joulain et al. Here

r_1

and

r_2

are the reflection coefficients of the two plates for the same polarisation and the same

\beta

Source: K. Joulain

Black plates: counting channels gives the Stefan–Boltzmann law

Two black plates reflect nothing, $r=0$ . Travelling channels transmit fully and evanescent channels not at all. With both polarisations, $\Phi=k_0^2/(2\pi)$ : the number of channels per unit area grows as the square of the frequency.

Put this into the flux and the integrand becomes exactly $(c/4)\,[u_0(T_1)-u_0(T_2)]$ , which integrates to $\sigma(T_1^4-T_2^4)$ . What chapter 1 found by averaging energy density times the speed of light becomes, here, a count of channels.

The course model checks this limit at gaps of 10 nm, 1 µm and 100 µm; the relative error never exceeds $2\times10^{-10}$ and does not depend on the gap. Black plates show no near-field growth, because a black surface holds no evanescent field.

Real materials are grey in the far field. Averaged over the interference fringes within a frequency interval, the travelling term returns to the classical grey-plate formula without interference. The benchmark compares the two at a 100 µm gap and finds them within 1% at three frequencies.

\Phi_{\mathrm{black}}=2\int_0^{k_0}\frac{\beta\,d\beta}{2\pi}=\frac{k_0^{2}}{2\pi},\qquad q=\int_0^{\infty}\frac{\Theta_1-\Theta_2}{4\pi^{2}c^{2}}\,\omega^{2}d\omega=\sigma\left(T_1^{4}-T_2^{4}\right)

For each polarisation, the channel count is the area of a disc of radius

k_0

divided by

(2\pi)^2

q_{\mathrm{grey}}=\frac{\sigma\left(T_1^{4}-T_2^{4}\right)}{1/e_1+1/e_2-1}

Far-field exchange between two opaque grey plates without interference, with emissivities

e_1

and

e_2

Source: K. Joulain · NIST Reference on Constants

Worked example: SiC plates 1 µm apart

Problem. Two SiC plates at 400 K and 300 K face each other across 1 µm. The model gives a net flux of $2.18\times10^{3}\,\mathrm{W\,m^{-2}}$ , 70.6% of it carried by tunnelling channels. How does this compare with two black plates at the same temperatures? What happens at a 20 µm gap?

Model. The black-plate result does not depend on the gap; it is the $992\,\mathrm{W\,m^{-2}}$ of the worked example in chapter 1. The SiC result comes from the full channel sum.

Numbers. At 1 µm, $2178/992.3=2.19$ : already more than twice the blackbody limit, with a gap only ten times smaller than the thermal wavelength. At 20 µm the model gives $566\,\mathrm{W\,m^{-2}}$ , 0.57 of the black plates, and tunnelling carries only 1.3%.

Explanation. At 20 µm almost only travelling channels remain, and SiC reflects almost everything in its reststrahlen band, so its emissivity is low and it falls short of black plates. At 1 µm, the surface waves in exactly that highly reflecting band open many tunnelling channels. What Kirchhoff’s law holds against SiC in the far field becomes its advantage in the near field.

Checks. Set both plates to 300 K and the flux is exactly zero. Swap SiC and gold and exchange their temperatures, and the flux changes sign with zero residual. Swap the two plates and every channel’s transmission stays the same. These are detailed balance and reciprocity, built directly into the model.

Source: K. Joulain

Try it in the Lab

From the blackbody regime to 1 µm

Predict first: as the gap shrinks from 20 µm to 1 µm, will the flux between SiC plates end up above or below that of two black plates?
Choose the Blackbody regime preset and read the ratio to two black plates and the share carried by tunnelling. Then bring the gap back to 1 µm.
Look at the Where in the spectrum the heat flows chart, find the band that carries the extra flux at 1 µm, and explain it with the surface wave of chapter 2.

Where the flux goes at equal temperatures

Predict first: with both plates at 300 K and a 100 nm gap, what is the net flux? What is the heat transfer coefficient?
Choose the Equal temperatures preset, read the net flux, then open More numbers and read the heat transfer coefficient.
Explain why the net flux is zero while the coefficient is not: the coefficient is the slope with which the flux grows for a small temperature difference.

Exercises with explained answers

At $178.7\times10^{12}\,\mathrm{rad\,s^{-1}}$ , how many channels per square micrometre and per polarisation connect two black plates, counted as $k_0^2/(4\pi)$ ?
Show the answer
$k_0=5.961\times10^{5}\,\mathrm{m^{-1}}$ , so $k_0^2/(4\pi)=2.83\times10^{10}\,\mathrm{m^{-2}}$ , or 0.0283 per square micrometre. The next chapter shows that a 10 nm gap opens more than four orders of magnitude more tunnelling channels.
Two grey plates both have emissivity 0.5 and sit at 400 K and 300 K, much farther apart than the thermal wavelength. What is the net flux?
Show the answer
The denominator is $1/0.5+1/0.5-1=3$ , so $q=992/3=331\,\mathrm{W\,m^{-2}}$ . Radiation bounces back and forth between the plates, so the result is smaller than half the black-plate value.
A SiC plate and a gold plate are both at 300 K, 10 nm apart, with strong tunnelling. Will net heat flow from SiC to gold because SiC is the better emitter?
Show the answer
No. The net flux in each channel is proportional to $\Theta(\omega,T_1)-\Theta(\omega,T_2)$ , which is zero in every channel at equal temperatures, whatever the transmission. Moreover, the transmission does not change when the plates are swapped: a channel that carries much from SiC to gold carries just as much from gold to SiC.
Why does only $\Theta$ appear in the flux formula and not the zero-point energy $\hbar\omega/2$ ?
Show the answer
The zero-point energy is the same on both sides. The net flux in each channel is the difference in mode energy times the transmission, so the zero-point part cancels. Joulain and co-workers make the same point in their Sec. 3.4: the vacuum energy cancels between emission and absorption when computing fluxes, but must be kept when computing the Casimir force. Chapter 6 uses this.

Sources and what we recomputed

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)
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.
NIST Reference on Constants, Units, and Uncertainty: speed of light in vacuum, reduced Planck constant, Boltzmann constant and Stefan–Boltzmann constant
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.