Sound Waves and the Boltzmann Distribution

Question

Imagine a sound wave traveling linearly in a given direction through a monatomic ideal gas.

Based on gas laws and the wave equation, we have that the wave should travel, in that given direction, at a speed of $\sqrt{\frac{5k_bT}{3m}}$, where $k_b$ is Boltzmann's Constant, $T$ is the temperature, and $m$ is the molecular mass. However, we have, per the Boltzmann Distribution, that the root-mean-square molecules in that dimension is merely $\sqrt{\frac{k_bT}{m}}$. In other words, this means that the sound wave (and, consequently, the molecules carrying the sound wave) are moving faster than the air molecules are moving in general in a given dimension. Theoretically, this means that the kinetic energy of the gas molecules is vastly increasing in the presence of sound, which would in turn cause a vast temperature increase, something that does not match up with real-life experience in a noisy room (to be fair, air is mostly diatomic, but changing the coefficient from $\frac{5}{3}$ to $\frac{7}{5}$ doesn't fundamentally change things).

What am I missing?

Answer 1

You seem to be interested in reconciling the molecular view with the view of continuum theory.

The average velocity in the direction of sound propagation, which you estimate to be $$ \bar{v}_x=\sqrt{\langle v_x^2\rangle} = \sqrt{\frac{k_BT}{m}}$$ amounts to about 300 m/s, which is already quite close to the actual speed of sound. You may regard this number as an estimate for the speed at which pressure variations in the gas can be 'communicated' among gas molecules along the wave.

Note, however, that a sound wave is a collective excitation of gas molecules. Individual packets of a large number of air molecules oscillate back and forth with a typical speed much (i.e. orders of magnitude) smaller than the thermal motion of individual molecules or the speed of sound. For a sound wave of 60 dB SPL, I find a speed of these wave packets of 68 $\mu$m/s. This low speed (compared to the thermal molecular speed of about 500 m/s) does clearly not cause any significant temperature increase.

The continuum description is possible, because a large number of molecules within one packet exert an effective pressure on the neighboring packet by intermolecular collisions combined with their thermal speed.

Answer 2

I think the problem with your argument is that it is possible to define a whole series of "average speeds" with variations between them of several percent (most probable speed, rms speed, mean speed...). So, taking one of these speeds, you can believe that the sound wave has changed the speed of molecules of several percent, which is not the case.

What we can say is that the speed of sound is of the order of magnitude of the speed of the molecules and it is in agreement with the idea of ​​a very weak perturbation, propagated by the molecules.

(Sorry for my english)

Answer 3

The (phase) speed of the sound wave is not the speed of the particles in the wave nor the speed with with macroscopic parcels of the gas move around. The velocity of the macroscopic parcels of is is much lower than the phase speed of the wave. Both have no direct relation to the average thermal speed distribution. (Further, a sound wave is not an equilibrium phenomenon, so deviations from the equilibrium behaviour should not be surprising per se.)

The drift velocity (so to say, the macroscopic velocity of the gas) in a sound wave can be obtained by applying the continuity equation for the mass current: $$ \dot \rho + \nabla \cdot \vec j = 0. $$ The dependency of the mass current on the drift velocity is $$ \vec j = \rho \vec v_D. $$ The mass profile in a plane wave with wave number $k$ travelling with the speed of sound $c$ in $x$-direction is given by (where $\delta$ is the amplitude and assumed small compared to the density $\rho_0$ since otherwise we will get out of the validity of the assumption of a linear density-pressure response – remember that we use the adiabatic compressibility to derive the phase velocity): $$ \rho(\vec r, t) = \rho_0 + \delta \cos(ckt - kx). $$ Using the continuity equation we can determine the mass current, and from that the drift velocity, by calculating the derivative: $$ \partial_t \rho = -\nabla \cdot \vec j = -\delta c k \sin(ckt-kx) $$ and then the anti-derivative: $$ \vec j = \vec j_0 - \delta c \cos(ckt - kx). $$ The anti-derivative allows for an additional global mass current (a wind blowing through all our system), which we can just set zero (because it has nothing to do with the sound wave, but can be removed by Galilei transforming the system).

So the drift velocity of the gas at a fixed location (let's say zero, all locations should be roughly equivalent in a plane wave) is (remember: we assumed $\delta \ll \rho_0$): $$ \vec v_D(\vec 0, t) = \delta c \cos(ckt) / \rho(t) \approx \delta c \cos(ckt)/\rho_0 \approx \frac {cp} {p_0} \cos(ckt). $$ So the macroscopic parcels of the gas oscillate around with a maximal drift speed that is proportional to the relative amplitude of the wave and the speed of sound, so it is always lower than the speed of sound in the gas (and the validity of the linear wave equation is no longer given long before that – if the speed of the gas in the wave approaches the speed of sound there will be shock waves not sound waves).