Lattice Waves


Sound travels through crystalline solids via movements of the atoms that make up the crystal. Unlike air, where the molecules are free to go anywhere, the atoms in a crystal are restricted to small movements around the points they occupy in the crystal lattice. These movements can be analyzed in the form of "lattice vibrations," motions of the entire framework of the atoms. Due to the highly ordered nature of a crystal, many features of lattice vibrations can be deduced from a simple one-dimensional model.

In this section we develop the classical solution for a longitudinal harmonic wave on a lattice of this type. The wave carries energy, but in contrast to sound waves in air (or light waves), it is concluded that the wave does not carry momentum. What this means in regard to phonon momentum is discussed below.

Problem Formulation

The simple lattice model consists of a chain of N particles of equal mass M attached to N+1 springs. The end springs are connected to boundary points, at x=0 and x=L. The springs are assumed to be identical, massless, lossless, and perfectly linear. With all particles at rest, all springs have equal length, and the equilibrium positions are

where n=0 and n=N+1 represent the positions of the boundary points. Let dn(t) be the displacement from the nth equilibrium position. The displacements are assumed to be small compared to the equilibrium distance between particles x1. The force on the nth particle is

where γ is the spring constant. As a trial solution, suppose the displacements are given by

We impose as boundary conditions the displacements defined by equation (3) at the two end points. Then

The equation of motion is

This is the equation for a harmonic oscillator, and equation (3) is a solution iff

The wave phase velocity vp and group velocity vg are given by



The wavelength is


We assume fixed lattice points, so the frequency must be greater than zero. There is an upper limit on ω which occurs when the wavelength equals twice the distance between the equilibrium particle spacing x1. At this frequency adjacent particles move in exactly opposing directions, and the group velocity equals zero. For positive (negative) values of k the wave propagates in the positive (negative) x-direction.

Modal Solutions

If the boundary conditions are changed to impose zero displacements at each end point, then pairs of equal amplitude waves propagating in opposite directions are the solutions for this case. The mode frequencies are restricted to discrete values where there are an integral number of half-wavelengths between boundary points. These are standing wave solutions. Alternatively, a periodic boundary condition gives rise to traveling wave modes. The modes for the simple lattice model are all "acoustic" modes. For a slightly more complicated lattice with alternating masses of different size there are additional "optical" modes.

The Quantum Mechanics solution for these modes differs from the classical solution in that there is a non-zero minimum energy for each mode, and mode energy can only increase or decrease by quanta equal to the mode frequency times Planck's constant. These quanta are acoustic or optical phonons corresponding to their respective modes.

Power Flow

If the behavior of a single particle is observed, there is no way of knowing which way the wave is propagating. But if we observe two adjacent particles, the motion becomes evident. The force exerted on particle "n" by the spring between particle "n-1" and particle "n" is

 Substituting equation (3) we get

The motion of particle "n" in the time interval dt is

Therefore the incremental work done on particle "n" by the spring is

The total work integrated over a full cycle then yields an average power flow

For particle "n-1", the n in equation (11) becomes n-1, but this doesn't change the time-independent term in equation (12). The sign of the force is changed, and the total work over a cycle is the negative of equation (13). The end result is that, for positive k, there is a time-average flow of power from particle "n-1" to particle "n" given by equation (13).

Conceptually, there is an engine moving the boundary point at x=0 with the force given by equation (10), with n=0. There is power flow along the lattice, and the power is absorbed by a perfect load at x=L.

Wave Energy and Power Flow

Using equations (6) and (7) we can express the power flow as

At the group velocity vg the wave will travel from one particle to the next in time x1/vg, and in this time interval the energy transmitted is


This is equal to the peak kinetic energy per particle. Thus the power flow is equal to this energy flowing at the group velocity.

Momentum and Mass Flow

Obviously, since the particles are anchored to lattice points, there is no time-average transfer of particle mass (but there is mass flow - see Einstein's box below). From equation (10) the time average force on any particle, from any spring, is zero. The same is true for a wave reflected from a boundary. The easiest example is a pair of waves

This represents the case of a wave incident from x=L, perfectly reflected from a rigid boundary at x=0. The force exerted by the spring attached to the boundary at x=0 is

A litmus test for wave momentum is to bounce the wave off of something; if there is momentum a force will be exerted. Equation (17) shows the lattice wave momentum is purely oscillatory. This is quite different than the time-average momentum carried by a light wave. It is also different than the time-average momentum carried by a sound wave in air, even though the time-average particle mass flow in this case is also zero. The result for air is derived in another section based on the statistics of molecular motion. The wave momentum manifests itself by both an oscillatory pressure, and a non-zero time-average increase in pressure, when a wave is reflected at a boundary.

Goodstein (page 158) states that a phonon has momentum equal to E/vp. This is one form of the standard Quantum Mechanics relation that states that energy and momentum come in quanta equal to Planck's constant times frequency, and divided by wavelength, respectively. For high energy levels the Quantum Mechanics solution must agree with the classical solution, and the result here indicates that the time-average momentum of a lattice phonon is zero. Shortly after I originally posted this section I attended a physics class at UCSB to see how this apparent contradiction is resolved. What I learned is that phonons do not carry real momentum, consistent with the analysis here. But they do carry "quasimomentum" or "crystal momentum," meaning that in certain interactions they behave as if they carry momentum.

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