---
title: "The phonon Boltzmann equation: from flights and collisions to Fourier’s law"
description: "Derive phonon conductivity and the ballistic limit from the Boltzmann equation with one relaxation time, and see why Fourier’s law overshoots in 40 nm silicon."
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  zh-CN: "https://sandboxphysics.com/zh/stories/phonon-heat-transport/phonon-boltzmann-equation/"
last_modified: "2026-10-30"
---

[Phonon heat transport](https://sandboxphysics.com/stories/phonon-heat-transport/)

L2 · Chapter 2

# The phonon Boltzmann equation: from flights and collisions to Fourier’s law

Sandbox Physics · 2026-10-30

Fourier’s law says that heat flux is proportional to the temperature gradient. It is not a fundamental law but an approximation for phonons that collide often. This chapter keeps track of phonon energy direction by direction and writes the simplest phonon Boltzmann equation: free flight plus collisions with a single relaxation time. When collisions are frequent it gives Fourier’s law and the kinetic formula of Chapter 1. With no collisions at all it gives a different bound, the most heat phonons can carry between two black walls. Comparing the two tells you how thin a film must be before Fourier’s law goes wrong.

## Before you begin

The kinetic formula and Knudsen number from Chapter 1, partial derivatives, and integrals of simple polynomials from −1 to 1. No partial differential equations need to be solved.

- Write the gray phonon Boltzmann equation and say what its flight and collision terms mean.
- Derive Fourier’s law in the limit of frequent collisions, including the factor of one third in the conductivity.
- Compute the ballistic bound on heat flux between two black walls and compare it with Fourier’s law.

## Bookkeeping phonon energy by direction

Take one-dimensional heat flow along $x$. Phonons fly in every direction; call the cosine of the angle between a phonon’s velocity and the $x$ axis $\mu$, from −1 to 1. Let $e(x,\mu,t)$ be the energy per volume carried by phonons heading in direction $\mu$. Its average over directions, divided by the heat capacity, is the local temperature.

In a gray model every phonon has speed $v$, so phonons heading in direction $\mu$ move along $x$ at $v\mu$. The heat flux is energy times that speed, averaged over directions.

$$
C\,T(x,t)=\tfrac12\int_{-1}^{1}e\,d\mu,\qquad q(x,t)=\tfrac12\int_{-1}^{1}v\mu\,e\,d\mu
$$

The factor $\tfrac12$ is the averaging weight for direction cosines from −1 to 1. Temperature and flux are measured from a reference temperature.

Source: A. A. Maznev

## Flight plus collisions: the relaxation-time approximation

Without collisions the energy in each direction simply drifts at speed $v\mu$. Collisions share energy between directions and push the distribution towards local equilibrium, which is the same in every direction and equal to $C\,T$. The simplest way to write this is to let the departure from equilibrium decay exponentially with one time constant $\tau$: the relaxation-time approximation.

The temperature in the equilibrium distribution must be the actual local temperature, so that collisions neither create nor destroy energy. Maznev and colleagues use the same single-mode relaxation-time approximation for thermal gratings, with a different $\tau$ for each phonon frequency. This course uses one $\tau$ for all.

$$
\frac{\partial e}{\partial t}+v\mu\,\frac{\partial e}{\partial x}=-\frac{e-C\,T}{\tau}
$$

The second term on the left is flight; the right side is collisions. The mean free path is $\Lambda=v\tau$.

Source: A. A. Maznev

## Frequent collisions: Fourier’s law appears

If the temperature hardly changes over one mean free path, the distribution is only slightly off equilibrium. Treat the right side as dominant: replace $e$ on the left by its equilibrium value and, for slow changes, drop the time derivative. The first correction says that slightly more phonons head down the temperature gradient than up it.

Put the corrected distribution into the definition of the flux. The isotropic part carries no flux; only the term proportional to $\mu^2$ survives, and $\tfrac12\int_{-1}^{1}\mu^2\,d\mu=\tfrac13$. That is where the one third in the kinetic formula comes from.

The same derivation says when Fourier’s law fails: when the temperature changes noticeably within one mean free path, that is, when the Knudsen number is no longer small, the first correction is not small and the expansion breaks down.

$$
e\approx C\,T-\tau v\mu\,C\,\frac{\partial T}{\partial x}\quad\Longrightarrow\quad q=-\frac{C v^{2}\tau}{3}\,\frac{\partial T}{\partial x}=-\kappa\,\frac{\partial T}{\partial x}
$$

So $\kappa=Cv^2\tau/3=Cv\Lambda/3$, the kinetic formula of Chapter 1.

Source: A. A. Maznev · Ioffe Institute

## No collisions: the ballistic bound between black walls

The opposite limit has no collisions at all. Put a film between two walls that absorb every phonon arriving and emit phonons at their own temperature; such walls are called black. The hot wall is $\Delta T$ warmer than the cold one.

Phonons leaving the hot wall travel only with $\mu>0$ and carry $C\Delta T$ more energy than those from the cold wall. Integrating over that half of the directions, $\tfrac12\int_0^1\mu\,d\mu=\tfrac14$, gives a net flux of $Cv\Delta T/4$. It does not depend on the thickness: it is the most heat phonons can carry between two black walls.

Maznev and colleagues note that this slab between two black walls is the model theorists favour, and that it is hard to build in the laboratory because every real interface has its own thermal resistance. Chapter 6 deals with real interfaces.

$$
q_{B}=\frac{C v}{4}\,\Delta T
$$

The ballistic limit. The same one quarter appears in the kinetic theory of gases, in the rate at which molecules strike a wall.

Source: A. A. Maznev · Ioffe Institute

## Worked example: when Fourier’s law asks for too much

Problem: with the gray silicon numbers of Chapter 1 ($C=1.63\times10^{6}\,\mathrm{J\,m^{-3}\,K^{-1}}$, $v=6351\,\mathrm{m/s}$, $\tau=5.93\,\mathrm{ps}$), check the conductivity formula, then compare Fourier’s flux across a 40 nm silicon film with 1 K across it against the ballistic bound.

The conductivity is $Cv^2\tau/3=130\,\mathrm{W\,m^{-1}\,K^{-1}}$, matching the 1.3 W/(cm K) of the tables. This is a consistency check, since $\tau$ was inferred from that value.

Fourier’s law gives $\kappa\,\Delta T/L=130/(40\times10^{-9})=3250\,\mathrm{MW/m^2}$; the ballistic bound is only $Cv\Delta T/4=2589\,\mathrm{MW/m^2}$. Fourier’s law asks for more heat than phonons could carry if they never collided at all, which is impossible.

The two are equal at $L=4\kappa/(Cv)=4\Lambda/3=50.2\,\mathrm{nm}$. Thinner than that, Fourier’s law must overestimate. Somewhat thicker, it still overestimates, because the real flux is limited both by diffusion inside the film and by ballistic flight near the walls. Chapter 3 computes the overestimate exactly.

Source: Ioffe Institute · A. A. Maznev

## Try it in the Lab

### [Watch the cold-wall flux approach the ballistic bound](https://sandboxphysics.com/labs/phonon-ballistic-diffusive-film/)

1. Choose “Very thin: phonons fly across” (8 nm). The heat leaving the cold wall, over the ballistic limit, rises from zero; note where it settles.
2. Read the settled film conductivity over the bulk value. Convert it back to the flux over the ballistic bound, $\Psi$, with $\kappa_{\mathrm{eff}}/\kappa=\Psi\cdot3/(4\,\mathrm{Kn})$, and compare with the previous step.

## Exercises with explained answers

1. Compute $\tfrac12\int_{-1}^{1}\mu^2\,d\mu$ and $\tfrac12\int_0^1\mu\,d\mu$, and say which result each one enters.

   Show the answer

   The first is $\tfrac13$, the one third in the kinetic formula $\kappa=Cv\Lambda/3$. The second is $\tfrac14$, the one quarter in the ballistic bound $Cv\Delta T/4$.

2. A 100 nm silicon film has 1 K across it. What flux does Fourier’s law give, and what fraction of the ballistic bound is that?

   Show the answer

   $\kappa\,\Delta T/L=1300\,\mathrm{MW/m^2}$, which is $0.502$ of the 2589 MW/m² bound, equal to $4\,\mathrm{Kn}/3$. Fourier’s law no longer exceeds the bound here, but Chapter 3 shows that it still overestimates.

3. How thin must a silicon film be for Fourier’s flux to be twice the ballistic bound?

   Show the answer

   Setting $\kappa/L=2\cdot Cv/4$ gives $L=2\kappa/(Cv)=2\Lambda/3=25.1\,\mathrm{nm}$.

4. The “very thin” preset has a Knudsen number of 4.71, and the Lab reports a settled film conductivity of 0.134 of bulk. What is the cold-wall flux over the ballistic bound?

   Show the answer

   $\Psi=0.134\times4\times4.71/3=0.842$. A film one fifth of a mean free path thick already carries close to the ballistic bound, while Fourier’s law would ask for 6.28 times that bound.

[Previous: What carries heat in a crystal, and how far a phonon flies](https://sandboxphysics.com/stories/phonon-heat-transport/heat-carriers-and-mean-free-path/) [Next: Heat in thin films: between ballistic and diffusive](https://sandboxphysics.com/stories/phonon-heat-transport/ballistic-and-diffusive-films/)

## Sources and what we recomputed

1. [Ioffe Institute, NSM Archive: Thermal properties of Silicon (Si) and Mechanical properties, elastic constants, lattice vibrations of Silicon (Si)](http://www.ioffe.ru/SVA/NSM/Semicond/Si/thermal.html)

   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.

2. [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](https://arxiv.org/abs/1108.3770)

   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.