--- title: "Why near-field heat transfer grows as the inverse square of the gap" description: "Count the tunnelling channels between two SiC plates, show why the heat transfer coefficient grows as one over the gap squared, and test damping and materials." canonical_url: "https://sandboxphysics.com/stories/fluctuational-electrodynamics/tunnelling-channels-and-the-inverse-square-law/" language: "en" page_type: "story" alternate_languages: zh-CN: "https://sandboxphysics.com/zh/stories/fluctuational-electrodynamics/tunnelling-channels-and-the-inverse-square-law/" last_modified: "2026-10-30" --- [Fluctuational electrodynamics](https://sandboxphysics.com/stories/fluctuational-electrodynamics/) L3 · Chapter 5 # Why near-field heat transfer grows as the inverse square of the gap Sandbox Physics · 2026-10-30 Shrink the gap between two SiC plates from 20 nm to 10 nm and the model’s heat transfer coefficient rises from 2380 to 9340 $\mathrm{W\,m^{-2}\,K^{-1}}$, almost exactly four times. This chapter shows where the factor of four comes from: each channel carries at most one share, and the number of tunnelling channels that can open scales as the inverse square of the gap. ## Before you begin The channel picture and transmissions of chapter 4; the quasi-static reflection coefficient of chapter 2; logarithms and exponentials. - In the quasi-static limit, write the number of p tunnelling channels in a form that depends only on the permittivity and scales as the inverse square of the gap. - Estimate the number of channels at the line frequency for a 10 nm gap and compare with two black plates. - Explain why more damping increases the heat flow at 10 nm, and why gold paired with SiC transfers so little. ## Tunnelling channels form a ring in wavevector space When the gap is far below the wavelength, all tunnelling wavevectors are much larger than $k_0$. As in chapter 3, the p reflection coefficient then equals its quasi-static value $r=(\varepsilon-1)/(\varepsilon+1)$, independent of $\beta$, and s polarisation can be neglected. With $x=2\beta d$, the transmission depends only on $x$ and $r$. Near the line, $r$ is almost purely imaginary with a large modulus. The transmission is then a function of $y=|r|^2e^{-x}$ that reaches exactly 1 at $y=1$, that is at $\beta=\ln|r|/d$, and falls off on both sides. So the tunnelling channels do not fill a disc from zero up to some limit; they sit on a ring of radius about $\ln|r|/d$. At too small a wavevector the two surfaces reflect so strongly that energy bounces inside the gap without getting into the other plate; at too large a wavevector the wave decays too fast to get across. $$ \tau_p=\frac{4\,(\operatorname{Im}r)^{2}e^{-x}}{|1-r^{2}e^{-x}|^{2}},\qquad x=2\beta d,\qquad r=\frac{\varepsilon-1}{\varepsilon+1} $$ Two plates of the same material, p polarisation, quasi-static limit. Source: K. Joulain ## The number of channels scales as the inverse square of the gap Change the integration variable from $\beta$ to $x$: $\beta\,d\beta=x\,dx/(4d^2)$. All of $d$ moves outside the integral, and what remains depends only on $\varepsilon$. This step is where the inverse-square law comes from. The remaining integral has a closed form. Expand the transmission as a geometric series and integrate term by term to get the complex dilogarithm $\operatorname{Li}_2$. The course benchmark implements this function independently and compares it with the model’s numerical integral at a 1 nm gap and four frequencies; the largest difference is $1.5\times10^{-4}$. When $r$ is close to purely imaginary with a large modulus, the closed form reduces to $\ln|r|/(\pi d^2)$: the number of channels grows only logarithmically with the strength of the resonance. The heat transfer coefficient is the ratio of flux to temperature difference as that difference goes to zero. It weights the channel count at each frequency by $\partial\Theta/\partial T$ and integrates. Since the channel count at every frequency scales as $d^{-2}$, so does the coefficient. Eq. (93) of Joulain et al. states the same conclusion. $$ \Phi_p=\int_0^{\infty}\frac{\beta\,d\beta}{2\pi}\,\tau_p=\frac{1}{8\pi d^{2}}\int_0^{\infty}x\,\tau_p(x)\,dx=\frac{(\operatorname{Im}r)^{2}}{2\pi d^{2}}\,\frac{\operatorname{Im}\operatorname{Li}_2(r^{2})}{\operatorname{Im}(r^{2})} $$ The second equality is a change of variable; the third uses $\operatorname{Im}[w/(1-w)]=\operatorname{Im}w/|1-w|^2$ and $\int_0^\infty x\,a e^{-x}/(1-a e^{-x})\,dx=\operatorname{Li}_2(a)$. $$ r\approx i|r|,\;|r|\gg1:\qquad \tau_p\approx\frac{4y}{(1+y)^{2}},\quad y=|r|^{2}e^{-x},\qquad \Phi_p\approx\frac{\ln|r|}{\pi d^{2}} $$ Rewrite the integral of $x\,\tau_p$ as one over $y$: the area under the transmission curve is about 4; multiply by the ring position $x=2\ln|r|$ and divide by $8\pi d^2$. $$ h(d)=\lim_{T_1\to T_2}\frac{q}{T_1-T_2}=\int_0^{\infty}\frac{d\omega}{2\pi}\,\frac{\partial\Theta}{\partial T}\,\Phi(\omega,d)\;\xrightarrow{\;d\to0\;}\;\propto d^{-2} $$ The definition of Eq. (92) of Joulain et al. The course model evaluates it at 300 K. Source: K. Joulain ## The heat transfer coefficient from the model The figure shows the heat transfer coefficient between two SiC plates at 300 K against the gap. Above about 10 µm it is nearly constant, about $3.3\,\mathrm{W\,m^{-2}\,K^{-1}}$, below the $6.12\,\mathrm{W\,m^{-2}\,K^{-1}}$ of two black plates for the reason given in chapter 4. Below 1 µm the curve steepens quickly. A fit between 10 and 50 nm gives an exponent of $-1.92$, close to $-2$. The product $h\,d^2$ is $9.34$, $9.51$ and $10.5\times10^{-13}\,\mathrm{W\,K^{-1}}$ at 10, 20 and 50 nm. As the gap closes, the share carried by p tunnelling rises from 92% to 99.6% and the curve approaches a pure inverse square. At 10 nm, $h=9.34\times10^{3}\,\mathrm{W\,m^{-2}\,K^{-1}}$: 1530 times two black plates and 2870 times its own value at 100 µm, about three and a half orders of magnitude. Joulain and co-workers write that at 10 nm the coefficient has grown by four orders of magnitude over its far-field value; their Fig. 17 is reproduced from another paper. We did not digitise that figure and have not found the cause of the half-order difference. The inverse square cannot be pushed to zero gap. Joulain and co-workers mention that it has been questioned because the flux diverges at zero gap; their answer is that the thermal resistance of a 10 nm vacuum gap remains far larger than that of conduction across a hundred nanometres of solid, so the plates can be treated as uniform in temperature. The course model also assumes a local permittivity, so it stops at 10 nm; non-local corrections at smaller gaps were not tested. Heat transfer coefficient between two SiC plates at 300 K against the gap, both axes logarithmic. Dots: the course model, from the benchmark record. Amber dashes: two black plates, $4\sigma T^3=6.12\,\mathrm{W\,m^{-2}\,K^{-1}}$. Grey dashes: an inverse-square line through the 10 nm point, for reference. Source: K. Joulain · NIST Reference on Constants ## Worked example: how many channels open at the line frequency at 10 nm Problem. Two SiC plates at 300 K are 10 nm apart. At $\omega=178.7\times10^{12}\,\mathrm{rad\,s^{-1}}$, estimate the number of p tunnelling channels per unit area and compare with two black plates at the same frequency. Then predict the 10 nm coefficient from the 20 nm one. Model. Use the quasi-static reflection coefficient and the logarithmic approximation, then compare with the closed form. Numbers. $\varepsilon=-1.008+0.129i$, so $r=(\varepsilon-1)/(\varepsilon+1)=2.01+15.39i$, $|r|=15.5$ and $\ln|r|=2.74$. The ring radius is $\beta\approx2.74/d=2.74\times10^{8}\,\mathrm{m^{-1}}$. The channel count is $\Phi_p\approx2.74/(\pi d^2)=8.72\times10^{15}\,\mathrm{m^{-2}}$; the closed form gives $8.69\times10^{15}\,\mathrm{m^{-2}}$, or 8690 per square micrometre. Comparison. Two black plates at this frequency have $k_0^2/(2\pi)=0.0565$ channels per square micrometre for both polarisations together. Tunnelling opens $1.5\times10^{5}$ times more. That many channels exist only in a narrow band around the line, so after integrating over the whole spectrum the coefficient is 1530 times that of black plates. Prediction. At 20 nm, $h=2377\,\mathrm{W\,m^{-2}\,K^{-1}}$; four times that is 9508, while the model gives 9342, so the prediction is 1.8% high. Part of the difference is the 1.5% of the 20 nm flux carried by travelling waves and s tunnelling; the rest is that even the p part is not yet fully quasi-static at 20 nm. Source: K. Joulain ## Try it in the Lab ### [Test the inverse square](https://sandboxphysics.com/labs/near-field-heat-transfer/) 1. Predict first: as the gap goes from 20 nm to 10 nm, will the coefficient times gap squared grow, shrink or stay about the same? 2. Choose the One-color transfer preset, open More numbers and read the coefficient times gap squared; read it again at 20 nm and 50 nm. Find where the Heat transfer coefficient against gap chart becomes a straight line of slope minus two. 3. Use how the p tunnelling share changes with gap to explain why the product is larger at 50 nm. ### [More damping, more heat?](https://sandboxphysics.com/labs/near-field-heat-transfer/) 1. Predict first: at a 10 nm gap, if the SiC damping doubles, will the flux grow or shrink? 2. In the One-color transfer preset, set the damping multiplier to 0.5, 1, 2 and 4 and note the net flux. Then choose the Gold plates and Mismatched plates presets and compare the s and p shares. 3. Explain the damping effect with a channel count that grows only logarithmically with resonance strength while the resonant band widens with damping. Explain the pairing effect by whether the two plates resonate at the same frequency. ## Exercises with explained answers 1. With the SiC damping scaled by 0.25, 1 and 4, the model’s 10 nm coefficient is 5008, 9342 and 12230 $\mathrm{W\,m^{-2}\,K^{-1}}$. Why does more loss give more heat flow? Show the answer At each frequency the channel count is about $\ln|r|/(\pi d^2)$, which changes only logarithmically with resonance strength: more damping lowers $|r|$ at the peak and removes only a few channels. But the band where $|r|$ is large widens in proportion to the damping, so more frequencies have many channels. The product grows. At larger damping the resonance smears out and the gain slows: going from 2 to 4 adds only 8%. 2. Two gold plates 100 nm apart have a coefficient of 73.2 $\mathrm{W\,m^{-2}\,K^{-1}}$, 99% of it carried by s tunnelling; SiC facing gold gives only 1.56 $\mathrm{W\,m^{-2}\,K^{-1}}$. Why does gold transfer almost nothing in p polarisation? Why is SiC with gold so poor? Show the answer At the frequencies of 300 K thermal radiation, gold has $|\varepsilon|$ above $10^{3}$, so its quasi-static $r_p$ is close to 1 with a tiny imaginary part, and its p tunnelling channels are nearly closed. The p transmission is proportional to the product of the imaginary parts of both plates’ reflection coefficients. SiC has a large imaginary part only near 178.7, where gold’s is small, so the two do not match and the channels do not open. What remains is mostly s channels, which need a highly conducting material like gold. 3. If $|r|$ rises from 15.5 to 31, by what factor does the 10 nm channel count change in the logarithmic approximation? What if the gap goes from 10 nm to 5 nm? Show the answer $\ln31/\ln15.5=3.43/2.74=1.25$, only 25% more. Halving the gap gives a factor of 4 from $d^{-2}$. Changing the gap is far more effective than strengthening the resonance. 4. Why can’t the inverse square be extrapolated to 1 nm to claim a coefficient one hundred times the 10 nm value? Show the answer At least three things are untested. At one or two nanometres the permittivity is no longer local; the response of electrons and lattice near the surface changes the reflection coefficient. At atomic gaps the macroscopic permittivity itself stops applying. And conduction inside the plates must still be able to hold the surface temperature. The course model stops at 10 nm and draws no conclusion below it. [Previous: Radiative heat between two plates, counted channel by channel](https://sandboxphysics.com/stories/fluctuational-electrodynamics/heat-across-a-vacuum-gap/) [Next: How the Casimir force gives a micromachined plate two swings](https://sandboxphysics.com/stories/fluctuational-electrodynamics/casimir-force-and-a-bistable-oscillator/) ## Sources and what we recomputed 1. [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. 2. [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.