---
title: "Normal scattering and second sound: why heat can slosh back and forth like a wave"
description: "Separate momentum-conserving normal scattering from resistive scattering, derive the damped heat-wave equation, and compute the second-sound speed and window."
canonical_url: "https://sandboxphysics.com/stories/phonon-heat-transport/normal-scattering-and-second-sound/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/phonon-heat-transport/normal-scattering-and-second-sound/"
last_modified: "2026-10-30"
---

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

L3 · Chapter 5

# Normal scattering and second sound: why heat can slosh back and forth like a wave

Sandbox Physics · 2026-10-30

In diffusion, heat only flows from hot to cold. But if most phonon collisions leave the total momentum unchanged, the heat flux itself gains inertia: phonons in a hot stripe surge together towards the cold stripe, overshoot and come back. The temperature then oscillates like a sound wave, and this is called second sound. This chapter derives the equation of second sound from a gray model with two scattering rates, works out its speed and the conditions for seeing it, and uses the thermal-grating Lab to watch hot and cold stripes trade places.

## Before you begin

The phonon Boltzmann equation from Chapter 2, the thermal grating from Chapter 4, and the solution of a linear second-order equation with constant coefficients, the damped oscillator. Crystal momentum is explained in the text.

- State the difference between normal and resistive scattering, and why only the second causes thermal resistance.
- Derive the damped heat-wave equation from the two-rate gray model and explain why the second-sound speed is the sound speed over the square root of three.
- Decide whether a grating oscillates for a given period and scattering times, and compute when its signal is lowest.

## Two kinds of scattering: conserving momentum or not

Lee and colleagues, discussing phonon hydrodynamics in graphene, split phonon scattering into two kinds. Normal scattering shares energy and momentum among phonons but keeps the total crystal momentum. Umklapp, impurity and boundary scattering change the total momentum and are together called resistive.

Only resistive scattering makes the heat flux decay. When normal scattering is fast, the phonons quickly settle into an equilibrium that drifts as a whole: phonon numbers are no longer equal in all directions but lean one way and carry a heat flux. Normal scattering preserves that flux; only resistive scattering can use it up.

Lee and colleagues point out that phonon hydrodynamics needs normal scattering to be much more frequent than resistive scattering. For graphene near 100 K they compute normal scattering rates of order ten to the ten per second, about two orders of magnitude higher than in diamond, which is why graphene and graphite are good places to look for second sound.

Source: S. Lee

## A gray model with two scattering rates

As in Chapter 4, take one spatial Fourier component $e^{iqx}$. Let $F(\mu,t)$ be the phonon energy in direction cosine $\mu$, in temperature units; $\theta$ is its average over directions, the local temperature, and $J$ its first moment, the heat flux divided by $Cv$.

Resistive scattering pulls the distribution towards the resting equilibrium $\theta$ in a time $\tau_R$. Normal scattering pulls it towards the drifting equilibrium $\theta+3\mu J$ in a time $\tau_N$; the drifting equilibrium carries the same flux as the current distribution, so normal scattering leaves $J$ alone. The Lab’s solver integrates this equation exactly on 64 directions.

At long periods the model reduces to diffusion with $\alpha=v^2\tau_R/3$, which contains only the resistive time.

$$
\frac{\partial F}{\partial t}+iqv\mu F=-\frac{F-\theta}{\tau_R}-\frac{F-\theta-3\mu J}{\tau_N},\qquad \theta=\tfrac12\int_{-1}^{1}F\,d\mu,\quad J=\tfrac12\int_{-1}^{1}\mu F\,d\mu
$$

Unlike a real material, every phonon here has one speed and two scattering times. It is the simplest model that separates what the two kinds of scattering do.

Source: S. Lee · A. A. Maznev

## Very fast normal scattering: the damped heat-wave equation

Take the zeroth and first moments of the equation. The zeroth gives energy conservation, $\dot\theta=-iqvJ$. The first involves the average of $\mu^2$; when normal scattering is very fast the distribution is the drifting equilibrium $\theta+3\mu J$, that average is $\theta/3$, and $\dot J=-iqv\theta/3-J/\tau_R$.

Eliminating $J$ gives a damped oscillator, the telegrapher equation, with wave speed $c=v/\sqrt3$. Lee and colleagues also write that in a three-dimensional Debye solid the second-sound speed is close to the sound speed over the square root of three.

The initial stripes carry no flux, so $\dot\theta(0)=0$. When $cq>1/(2\tau_R)$ the solution oscillates under an envelope $e^{-t/(2\tau_R)}$ and reaches its lowest, negative value after half an oscillation, at $t=\pi/\omega'$.

$$
\ddot\theta+\frac{\dot\theta}{\tau_R}+c^{2}q^{2}\theta=0,\qquad c=\frac{v}{\sqrt3}
$$

When $\tau_R$ is short the inertial term drops out and diffusion returns, with $\alpha=c^2\tau_R=v^2\tau_R/3$.

$$
\theta(t)=e^{-t/(2\tau_R)}\left[\cos\omega' t+\frac{\sin\omega' t}{2\tau_R\,\omega'}\right],\qquad \omega'=\sqrt{c^{2}q^{2}-\frac{1}{4\tau_R^{2}}}
$$

Its derivative is proportional to $-\sin\omega' t$, so the first minimum falls exactly at $t=\pi/\omega'$, with the value $-e^{-\pi/(2\tau_R\omega')}$.

Source: S. Lee

## The second-sound window

Lee and colleagues give the condition for second sound: the oscillation frequency $\Omega$ must lie above the resistive rate and below the normal rate, $1/\tau_R<\Omega<1/\tau_N$. Too low, and resistive scattering drains the flux within one period, so diffusion returns. Too high, and normal scattering cannot build the drifting equilibrium in time, so phonons fly independently.

In a thermal grating $\Omega=cq=2\pi c/L$, so the window is a range of periods. There is also a finer line: for the telegrapher equation to oscillate at all needs $cq>1/(2\tau_R)$, a lower bound half as large as the window’s.

The gray model has a flaw that must be stated. With no scattering at all the amplitude follows $\sin(qvt)/(qvt)$, which also dips below zero, to about −0.217, because every phonon has the same speed and they fly in step. When Huberman and colleagues switched off all scattering in their 50 K graphite calculation, the dip disappeared, because real phonon speeds spread widely. So in a gray model a negative signal alone does not prove second sound; what matters is how it changes with period and scattering times.

Source: S. Lee · S. Huberman

## Worked example: the second-sound preset

Problem: the Lab’s “Second sound: stripes swap” preset uses a phonon speed $v=10\,\mu\mathrm m/\mathrm{ns}$, a period $L=10\,\mu\mathrm m$, $\tau_R=5\,\mathrm{ns}$ and $\tau_N=0.01\,\mathrm{ns}$. These numbers are chosen to show the effect clearly and describe no particular material. Find the wave speed, the oscillation frequency, the window, and the time and value of the telegrapher minimum.

The wave speed is $c=10/\sqrt3=5.774\,\mu\mathrm m/\mathrm{ns}$, the wavenumber $q=2\pi/10=0.6283\,\mu\mathrm m^{-1}$, and $\Omega=cq=3.628\,\mathrm{ns^{-1}}$. The window is $0.2<\Omega<100\,\mathrm{ns^{-1}}$, and $\Omega$ sits inside it.

With $\omega'=\sqrt{3.628^2-0.1^2}=3.626\,\mathrm{ns^{-1}}$, the minimum comes at $t=\pi/\omega'=0.866\,\mathrm{ns}$ with the value $-e^{-0.1\times0.866}=-0.917$. Hot and cold stripes almost fully swap.

The Lab’s 64-direction solver keeps the finite $\tau_N$ and finds a minimum of −0.876 at 0.871 ns. It is a little shallower than the telegrapher value because normal scattering at 0.01 ns is fast but not infinitely fast.

The difference can be computed. With normal scattering finite, the distribution keeps a little of its second moment (the $\mu^2$ term), which scattering pulls back in a time $\tau$ with $1/\tau=1/\tau_N+1/\tau_R$. Keeping that term to lowest order adds damping that grows as $q^2$: $\gamma=1/(2\tau_R)+\tfrac{2}{15}v^2\tau q^2$. Here the extra part is $0.0525\,\mathrm{ns^{-1}}$, so $\gamma=0.1525\,\mathrm{ns^{-1}}$ and the minimum becomes $-e^{-0.1525\times0.867}=-0.876$, as the solver finds. This damping matters more at shorter periods, so the Lab’s damped-wave fit includes it.

Source: S. Lee · S. Huberman

## Try it in the Lab

### [Make hot and cold stripes swap](https://sandboxphysics.com/labs/transient-thermal-grating/)

1. Choose “Second sound: stripes swap”. Watch the stripes on the stage swap near 1 ns and read the lowest grating amplitude.
2. In the direction chart, the orange curve between stripes leans clearly to one side: the phonons there drift together.
3. Shorten the resistive scattering time step by step until the oscillation disappears and the minimum is no longer negative; compare with this chapter’s damping condition.

### [Tell second sound from free flight](https://sandboxphysics.com/labs/transient-thermal-grating/)

1. Choose “Free flight: a small dip” and note the minimum; then choose “Too much resistance: no wave” and note its minimum.
2. Explain in this chapter’s terms why free flight in a gray model also pushes the signal slightly negative, and why that is not second sound.

## Exercises with explained answers

1. Take $v=10\,\mu\mathrm m/\mathrm{ns}$, $\tau_R=5\,\mathrm{ns}$ and $\tau_N=0.01\,\mathrm{ns}$. Which range of grating periods does the window correspond to?

   Show the answer

   With $\Omega=2\pi c/L$, the lower bound $\Omega=1/\tau_R$ gives $L=2\pi c\tau_R=181\,\mu\mathrm m$ and the upper bound $\Omega=1/\tau_N$ gives $L=0.363\,\mu\mathrm m$. Second sound can be seen at periods in between.

2. With the same numbers, below what period does the telegrapher equation actually oscillate?

   Show the answer

   It needs $cq>1/(2\tau_R)$, that is $L<4\pi c\tau_R=363\,\mu\mathrm m$, twice the period at the lower edge of the window.

3. A three-dimensional Debye solid has a sound speed of 6351 m/s. What second-sound speed does the approximation of Lee and colleagues give?

   Show the answer

   $6351/\sqrt3=3667\,\mathrm{m/s}$. This is only the Debye estimate; in graphite the second-sound speed lies between the slow and fast transverse speeds and cannot be estimated this way.

4. Why does normal scattering cause no thermal resistance of its own, yet is necessary for second sound?

   Show the answer

   Normal scattering conserves crystal momentum and pulls the distribution towards a drifting equilibrium that carries the same flux, so it does not use up the flux and adds no resistance. But it is what makes the phonons drift together, which gives the flux inertia and lets the temperature oscillate like a wave. Without it, phonons fly independently and only diffusion or free flight remains.

[Previous: Transient thermal gratings: measuring heat diffusion with interference fringes](https://sandboxphysics.com/stories/phonon-heat-transport/transient-thermal-grating/) [Next: Interface thermal resistance: why the temperature jumps where two materials meet](https://sandboxphysics.com/stories/phonon-heat-transport/interface-thermal-resistance/)

## Sources and what we recomputed

1. [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.

2. [S. Lee, D. Broido, K. Esfarjani and G. Chen, Hydrodynamic phonon transport in suspended graphene, Nature Communications 6, 6290 (2015), doi:10.1038/ncomms7290](https://doi.org/10.1038/ncomms7290)

   Main text and Methods read; the Supplementary Information was not opened. Used here: normal scattering conserves crystal momentum while umklapp and other resistive processes do not; second sound needs a frequency window between the resistive and the normal scattering rates; in a three-dimensional Debye solid the second-sound speed is close to the sound speed divided by the square root of three; and the predicted second-sound speeds in graphene of roughly 2000 to 3000 m/s. The course quotes no equation numbers from this paper.

3. [S. Huberman, R. A. Duncan, K. Chen, B. Song, V. Chiloyan, Z. Ding, A. A. Maznev, G. Chen and K. A. Nelson, Observation of second sound in graphite at temperatures above 100 K, Science (2019), doi:10.1126/science.aav3548, arXiv:1901.09160v1](https://arxiv.org/abs/1901.09160)

   Read in full from arXiv v1, including the supplementary materials. Used here: the transient-grating setup with a 515 nm pump and a 532 nm probe in heterodyne detection; the diffusion time constant written as the period squared over four pi squared times the diffusivity; 11 cm²/s at a 37.5 µm period at 300 K and the slower-than-diffusive decay at short periods (Fig. 1); the sign flip at 85 K and the velocity of about 3200 m/s from the first minimum (Fig. 2); oscillations at 104 K and 125 K that vanish at 150 K and at 50 K (Fig. 3); the predicted window between 50 and 250 K (Fig. 4); the first-principles value of 3650 m/s; and, from the supplement, the dip minimum near 1.2 ns for the 7.5 µm trace and the grating amplitude of about 0.2 K during the oscillation. The published Science version was not compared line by line with arXiv v1.