L3 · Chapter 3

Heat in thin films: between ballistic and diffusive

Sandbox Physics · 2026-10-30

Between the two limits of the previous chapter lies the range where most nanoscale devices sit: the film is about one mean free path thick, so phonons neither fly straight across nor fully diffuse. This chapter solves for the steady heat flux in such a film exactly and finds a result that depends on the Knudsen number alone. A 40 nm silicon film carries only 43% of what Fourier’s law predicts. Where does the rest go? Mostly near the two walls, where the film next to each wall is not at the wall’s temperature.

A film between two black walls

Keep the setting of Chapter 2: a film of thickness $L$ between two black walls, the hot wall $\Delta T$ above the cold one, and one speed and one mean free path for all phonons. Dividing the steady flux by the ballistic bound $Cv\Delta T/4$ gives a dimensionless flux $\Psi$ . It depends only on the Knudsen number and runs from 0 in the diffusive limit to 1 in the ballistic limit.

Fourier’s law gives $\Psi_F=4\,\mathrm{Kn}/3$ , which exceeds 1 once the Knudsen number passes $3/4$ ; the previous chapter showed that this cannot happen. The true $\Psi$ has to respect both limits.

\Psi=\frac{q}{Cv\Delta T/4},\qquad \Psi_F=\frac{\kappa\Delta T/L}{Cv\Delta T/4}=\frac{4}{3}\,\mathrm{Kn}

The effective film conductivity

\kappa_{\mathrm{eff}}=qL/\Delta T

over the bulk value is

\Psi/\Psi_F

Source: A. A. Maznev

The exact steady solution

Measure lengths in mean free paths: the position in the film is $s=x/\Lambda$ and the thickness is $1/\mathrm{Kn}$ . Integrating the steady Boltzmann equation along each flight path and averaging over directions leaves an integral equation for the temperature alone. The temperature at a point is set partly by phonons arriving straight from the hot wall and partly by phonons arriving from wherever in the film they last scattered. The longer the path, the likelier a phonon scatters on the way, and the exponential integrals $E_n$ describe exactly that.

Write the dimensionless temperature as $\theta=(T-T_c)/\Delta T$ . Once $\theta(s)$ is known, the flux reaching the cold wall also has two parts: phonons from the hot wall that never scatter, and phonons that last scattered somewhere in the film.

The green settled value in the Lab comes from cutting the film into a few hundred cells, integrating this equation exactly over each cell and solving; it uses neither simulation on the stage. Computing the flux from the hot-wall and from the cold-wall formula gives the same number to numerical precision, an internal check.

\theta(s)=\tfrac12E_2(s)+\tfrac12\int_0^{1/\mathrm{Kn}}\theta(s')\,E_1(|s-s'|)\,ds'

Here

E_n(z)=\int_1^{\infty}e^{-zw}\,w^{-n}\,dw

. The first term comes from the hot wall, the second from the film itself.

\Psi=2E_3\!\left(\frac{1}{\mathrm{Kn}}\right)+2\int_0^{1/\mathrm{Kn}}\theta(s)\,E_2\!\left(\frac{1}{\mathrm{Kn}}-s\right)ds

The first term,

2E_3(1/\mathrm{Kn})

, is the share of the hot wall’s emission that reaches the cold wall without scattering.

Source: A. A. Maznev

Series resistance: an approximation worth remembering

Think of two resistances in series: the two walls together act like a ballistic resistance $4/(Cv)$ , and the film interior like the Fourier resistance $L/\kappa$ . In units of the ballistic resistance the total is $1+3/(4\,\mathrm{Kn})$ , and its inverse is the approximation below.

It is right in both limits and not bad in between. Test it against the exact solution by treating $1/\Psi-3/(4\,\mathrm{Kn})$ as an extra wall resistance, which the approximation sets to 1. The exact solution gives about 1.06 for Knudsen numbers from 0.2 to 1, falling to 1.03 for very thin films. So the approximation always slightly overestimates the flux, by a few percent at most.

Chapter 6 arrives at this 1 from a different direction: it is what remains of the boundary resistance of two black walls once the asymmetry of the phonon distribution inside the film is accounted for.

\Psi\approx\frac{1}{1+\dfrac{3}{4\,\mathrm{Kn}}}

It tends to 1 for large Knudsen numbers and to Fourier’s

4\,\mathrm{Kn}/3

for small ones.

Source: A. A. Maznev · Ioffe Institute

Temperature jumps at the walls

In the Lab’s temperature profile the film temperature does not run from the hot-wall to the cold-wall value. Next to the hot wall the film is clearly cooler than the wall; next to the cold wall it is clearly warmer. There is a jump at each end. The thinner the film, the bigger the jumps; in the ballistic limit the whole film sits near the average of the two wall temperatures.

The reason is that the temperature at a point is set by all the phonons arriving there. Right next to the hot wall, phonons heading towards the cold wall come from the hot wall, but phonons heading back come from cooler parts of the film, and the average lies below the hot-wall temperature. The jump is not a sign of a poor contact; even an ideal black wall has it.

These jumps are the extra wall resistance in the series picture. Because of them the effective film conductivity $\kappa_{\mathrm{eff}}$ falls below the bulk value even when phonons inside the film scatter exactly as in bulk.

Source: A. A. Maznev

The transient: when the first phonons reach the cold wall

The Lab starts with the hot wall stepping up in temperature. Phonons flying straight across are fastest and still need $L/v$ to arrive, so the cold-wall flux is zero until then. It then rises and settles within a few to a few tens of flight times; the smaller the Knudsen number, the longer it takes, because diffusion has to build the temperature profile first.

Two independent simulations run on the stage. One splits directions into a set of discrete ordinates and advances them with a limited upwind scheme; the other is Monte Carlo, in which packets change direction at random after exponentially distributed free times. They agree within their statistical errors and both approach the green exact settled value.

Source: A. A. Maznev

Worked example: a 40 nm silicon film

Problem: take a gray silicon mean free path of 37.7 nm and a 40 nm film. Find the Knudsen number, Fourier’s dimensionless flux, the series approximation and the exact value, the effective conductivity ratio, the transit time, and the share of hot-wall emission that reaches the cold wall unscattered.

The Knudsen number is $37.7/40=0.943$ . Fourier’s law gives $\Psi_F=4\times0.943/3=1.26$ , above the ballistic bound. The series approximation gives $1/(1+0.796)=0.557$ and the exact solution $\Psi=0.540$ .

The effective conductivity ratio is $\Psi/\Psi_F=0.540/1.26=0.429$ : this film carries only 43% of the flux the bulk conductivity predicts. The extra wall resistance is $1/0.540-0.796=1.06$ , as in the previous section.

The transit time is $L/v=40\,\mathrm{nm}/(6351\,\mathrm{m/s})=6.30\,\mathrm{ps}$ . The share of hot-wall energy reaching the cold wall unscattered is $2E_3(1/0.943)=0.202$ ; the other four fifths scatter at least once.

Source: Ioffe Institute · A. A. Maznev

Try it in the Lab

Drag the cold wall between ballistic and diffusive

Choose “Thickness near the mean free path”. When the cold-wall flux has settled, find the current Knudsen number on the settled-flux curve in the data row.
Drag the blue cold wall to stretch the film from 8 nm to 800 nm and watch the jumps at both ends of the temperature profile shrink.
With “Thick film: Fourier returns”, compare the exact value with Fourier’s law and note the difference.

Compare the two simulations with the exact value

With “Very thin: phonons fly across”, compare the discrete-ordinates solver, the Monte Carlo run and the green exact value, and check that they agree within the error bars.
Reduce the number of packets in the advanced settings and watch the Monte Carlo error bars widen.

Exercises with explained answers

What is the Knudsen number of an 800 nm silicon film (mean free path 37.7 nm)? What dimensionless flux do Fourier’s law and the exact solution give, and what is the effective conductivity over bulk?
Show the answer
The Knudsen number is $0.0471$ ; Fourier’s law gives $\Psi_F=0.0628$ and the exact solution about $0.059$ , so the effective conductivity is about 0.94 of bulk. Even at 21 mean free paths thick, the extra wall resistance costs about 6% of the flux.
What does the series approximation give for an 8 nm silicon film? The exact value is 0.842; by how much does the approximation overestimate?
Show the answer
The Knudsen number is 4.71, and the approximation gives $1/(1+3/(4\times4.71))=0.863$ , 2.5% above the exact value.
The film is still 40 nm, but alloy scattering shortens the mean free path to 4 nm. The exact value is about 0.1168; what is the effective conductivity ratio?
Show the answer
The Knudsen number is 0.1 and Fourier’s law gives $0.133$ , so the ratio is $0.1168\times3/(4\times0.1)=0.876$ . With a shorter path the same film is close to Fourier’s law again, but the alloy’s bulk conductivity has itself fallen to about a tenth of pure silicon’s.
At a Knudsen number of 1, what share of the hot wall’s emission reaches the cold wall unscattered? Use $E_3(1)=0.1097$ .
Show the answer
$2E_3(1)=0.219$ . Even when the film is only one mean free path thick, about four fifths of the energy scatters at least once on the way.

Sources and what we recomputed

Ioffe Institute, NSM Archive: Thermal properties of Silicon (Si) and Mechanical properties, elastic constants, lattice vibrations of Silicon (Si)
Both pages read in full (the mechanical page is at mechanic.html in the same folder). Values used at 300 K: specific heat 0.7 J/(g K), thermal conductivity 1.3 W/(cm K), thermal diffusivity 0.8 cm²/s, density 2.329 g/cm³, and the sound speeds 8.43 km/s (longitudinal) and 5.84 km/s (transverse) along the [100] direction. The course combines the two speeds into one Debye average; that average is our step, not a value from the pages.
A. A. Maznev, J. A. Johnson and K. A. Nelson, Onset of nondiffusive phonon transport in transient thermal grating decay, Physical Review B 84, 195206 (2011), arXiv:1108.3770v2
Read in full (arXiv v2). Introduction: the kinetic-theory mean free path of tens of nanometres, the gray-body approximation as an oversimplification, the slab between two black walls favoured by theorists, the grating period set by the crossing angle of the pump beams, and grating decay times of about 0.3 to 30 ns in silicon for periods of 1 to 10 µm. The abstract states that the reduction of the effective conductivity in silicon is significant at periods as large as 10 µm. The course quotes no equation numbers from this paper.