Simple explanation of relation between speed of sound and r.m.s. speed?

In an ideal gas, the speed of sound $v_s$ is related to the r.m.s. molecular speed $v_m$ by

$$\frac{v_s}{v_m}=\sqrt{\frac{\gamma}{3}} \qquad ,$$

where $\gamma=C_p/C_v=7/5$ for a diatomic gas. I understand how to prove this relation from first principles. However, it seems mysterious to me that it pops out like this in the end.

I suppose it's inevitable for dimensional reasons that there will be some relation of this form, since if we want to describe a sample of an ideal gas, a sufficient set of unitful parameters is $m$ and $kT$, and there is a unique way of combining these to give units of velocity. (In a solid or liquid, we have other parameters such as the Young's modulus.) However, this type of dimensional argument doesn't prove that the ratio of the speeds is of order unity.

Is there any straightforward physical plausibility argument for the fact that this ratio of speeds is constant, and for the fact that the ratio is of order unity? I guess it's implausible to have $v_m \gg v_s$, since it seems like then a sound wave would sort of get scrambled and lose its identity because of molecular motion, and this scrambling would take place in much less than a period.

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Should they not be coupled through the mean time between collisions/mean-free-path? At the hand-waving level that alone would satisfy me as far as expecting a order unity coefficient. – dmckee Sep 28 '13 at 15:00
@dmckee: Should they not be coupled through the mean time between collisions/mean-free-path? That's not immediately obvious to me, but if you can expand it into an answer, I'd love to read it. By reducing the pressure while maintaining constant temperature, you can make the mean free path go to infinity while the speed of sound is unchanged. – Ben Crowell Sep 28 '13 at 17:33
@user23660: The ratio is constant and depends only on $\gamma$. This relation is valid only for ideal gases. – Ben Crowell Sep 28 '13 at 17:34
Ben, On thinking a little I'd same I was way off the mark with that comment, and it should have been obvious: the ideal gas is nominally non-interacting. A better explanation for what I thought I was getting at is that pressure is intimately coupled to mean velocity in a gas (even a non-ideal one) because it is the expression of momentum transfer to a impinging object when the gas molecules collided with the surface. Sound being a variation in pressure there must be a relationship. If I get done with my work this weekend I'll work that up into a proper answer. – dmckee Sep 28 '13 at 20:24
The simplest model that includes both sound wave propagation and molecule speeds would be kinetic theory. Consider Boltzmann equation $\partial_t f + (v \cdot \nabla) f = St$. Density function that corresponds to sound impulse propagating in direction x has the form f(x-v_s t, y, z, v_x, v_y, v_z) (with the only t dependence in the first argument). So, we will then have $(-v_s + v )\partial_x f = St$. So there, if we assume that rms speed $v_m$ is increased $\alpha$ times, $v_s$ should also be increased accordingly. – user23660 Sep 29 '13 at 7:56

Let us recap how we obtain the quantities in question.

The easiest way to obtain the speed of sound is from continuous mechanics. Assuming isentropic flow and small perturbations of speed and density and linearizing Euler (or Navier-Stokes) and continuity equations we obtain $$v_s^2=\left(\frac{\partial p}{\partial \rho}\right)_s,$$ where the derivative of pressure is taken at constant entropy. For an ideal gas that has adiabatic equation in the form $p/\rho^\gamma = \mathrm{const}$ that means $$v_s = \sqrt{\frac {\gamma p}{\rho}} = \sqrt{\frac{\gamma\, k \, T}{m}},$$ where $k$ is Boltzmann constant, $T$ temperature and $m$ is mass of molecule.

The rms speed is obtained from statistical mechanics of gases using Maxwell–Boltzmann distribution: $$v_m = \sqrt{\frac{3 \,k \, T}{m}},$$ here the factor 3 under the square root is consequence of 3-dimensionality of our world.

In order to clarify relationship between those two speeds we can consider the theory that includes both the sound wave propagation and statistical distributions of molecular speeds. That would be the kinetic theory and in particular Boltzmann equation: $$\frac{\partial}{\partial t} f+(v\cdot \nabla) f=\mathrm{St},$$ where $f=f(\vec{r},\vec{v},t)$ is the density function, and $\mathrm{St}$ in the rhs (from Stosszahlansatz) is the integral collision operator.

If we consider the sound wave propagating along the direction $x$, then its density function would have the form $f(x-v_s\,t, v_x, v_y, v_z)$ (with the only $t$ dependence in the first argument and without dependence on $y$ and $z$). Substituting this into Boltzmann equation we obtain $$- v_s\frac{\partial }{\partial x} f + v_x \frac{\partial}{\partial x} f=\mathrm{St}.$$

There, just from the form of equation we could see, that if we assume that rms speed $v_m$ is increased $\alpha$ times (by rescaling the density function $f' =\alpha^{-3} \cdot f(x-v'_s t,\alpha \vec{v})$), $v_s$ should also be increased accordingly to maintain the solution. So we can conclude that $v_s / v_m = \mathrm{const}$.

This is, of course, valid if we can assume that the collision term in rhs also transforms accordingly or if we could simply neglect the rhs at a first approximation, which should hold for rarefied gases. Another case is rigid-sphere approximation for the collisions that provides appropriate transformation, so $v_s/v_m$ should also be independent on temperature there (but, obviously, if the density is large enough in this case it will be dependent on it).

Here is random example of paper that considers sound-wave propagation in the framework of Boltzmann equation: R.D.M. Garcia, C.E. Siewert, 'The linearized Boltzmann equation: sound-wave propagation in a rarefied gas', DOI:10.1007/s00033-005-0007-8 (just the first from Google scholar results that has online source).

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