**Sound Wave Attenuation Due to Loss**

A sound wave in air has an attenuation factor due
to a transfer of kinetic energy from the systematic mean molecular velocity
of the wave, to random molecular motion (i.e. heat). This is analogous to frictional
loss. Since the theory on this web site is __based__ on molecular velocity
statistics, both systematic and random, one would expect the results to include
this attenuation - but in fact the plane wave solutions are completely lossless!
How can this be?

This section reexamines the physics, at a molecular level, to answer this question. The actual attenuation in air is complicated, due to the mixture of various gases and water vapor. The goal here is to shed light on the basic physics, and not to accurately compute the attenuation in real air (however the answer is pretty close).

The basic assumption in the derivation in Section 1 is: at any point there exists a cubic region, small compared to a wavelength, containing many molecules, whose velocity statistics are same as a gas at equilibrium. Specifically, within the cube: (1) there is equipartition of energy between all degrees of freedom; and (2) the probability distribution of the x, y, and z velocity components is Gaussian (actually the only property used is that it is symmetric with respect to its mean value). The only additional assumption in the theory is to drop some second order terms in deriving the acoustic equations. This is a very minimal set of assumptions; yet buried in one of them is the cause of the missing attenuation.

For a plane wave the cubic region can be replaced with a thin slice perpendicular to the direction of propagation, and extending to infinity in the other two coordinates; therefore "containing many molecules" is not an issue. I initially suspected equipartition, or the elimination of the second order terms. To my surprise, it is a slight asymmetry of the velocity probability distribution that leads to the attenuation factor.

**Physical Source of an Asymmetric Probability Distribution**

The velocity statistics of a gas at equilibrium are
Gaussian, and remain Gaussian. A computer model was created to determine how
an asymmetry develops from a non-equilibrium condition. A collection of 2x10^{6}
molecules, centered around the origin, are given x-velocities with a Gaussian
distribution at t=0. The molecules subsequently travel in one dimension without
colliding. The x-velocity statistics for a slice of molecules near the origin
are sampled as a function of time.

Various non-equilibrium conditions were imposed at t=0. It was determined that linear or quadratic density variations did not introduce a significant asymmetry in the statistics. Neither did a linear gradient in mean velocity, or a quadratic gradient in variance. The two conditions which did produce an asymmetry were: (1) a quadratic gradient in mean velocity, and (2) a linear gradient in velocity variance. Under typical conditions the asymmetry is much greater for the variance gradient than for the velocity gradient. For a plane wave, the variance gradient is sinusoidal, and the asymmetry is in phase quadrature with it.

In the absence of collisions the asymmetry initially grows linearly with time. As discussed in another section, collisions cause any initial velocity distribution to move toward a Gaussian distribution. Therefore a balance is reached between the growth of asymmetry due to the non-equilibrium variance gradient, and the decay of asymmetry due to collisions. More on this below.

**Molecular Model Derivation Revisited**

The kinetic energy term in table
2 involves the expected value of the cube of v_{x}. For any probability
distribution symmetric around its mean, all odd moments are zero, including
the 3^{rd} moment

This factor was set to zero in computing the kinetic energy term in table 3. If this term is retained, it gets added onto equations (15) and (19). For the 1-dimensional case equation (19) becomes (dropping the subscript x)

The wave equations in 1-dimension are then

where

** ****Traveling Wave
Solution**

We look for a solution of the form

The wave equations are satisfied iff

and

The two equations above cannot be satisfied in general; however for α <<k we have

and if the relation between pressure and velocity is approximately the same as a wave without attenuation

Then the equations are approximately satisfied with

Thus the third moment is in phase quadrature with the variance gradient, consistent with the computer model. The attenuation coefficient is proportional to the peak value of the third moment

**Rates of Generation and Decay of the Third Moment**

As noted above, during wave propagation the third moment of the velocity probability distribution is generated by a gradient in the velocity variance, and decays as a result of collisions. The two effects are evaluated using two different computer models.

Third moment generation

In a plane wave, for the case of a diatomic molecule, the velocity variance is equal to

The value of this variance gradient at t=0 is used as the initial condition for the computer model. For small values of t the resulting third moment is well matched by the equation

A value of σ_{o}=290.7 is used in the
model.

Third moment decay

A computer model of collisions between monatomic
molecules is described in another section.
The results for diatomic molecules are assumed to be similar. The initial conditions
for x-velocities of this model are the final velocities of the "generation"
computer model. The y and z velocities are zero-mean Gaussian random variables
with the same value of σ_{o}. The model then collides particle
pairs in 3-dimensions, with a random selection of pairs, and a random geometrical
orientation. The moments of the resulting velocities are then sampled after
each round of collisions. There is a remarkably good fit to the results given
by

where nc = the number of collisions. In this model
every particle has exactly the same number of collisions; this should not greatly
alter the results as long as the number of collisions is greater than 10 or
so. For a diatomic molecule the time scale might differ, but not by a large
factor. In air the collision frequency is approximately 5x10^{9}. Using
this value we get

Balance of generation and decay

The generation and decay rates will be in balance for

the corresponding value of the attenuation constant is

The actual attenuation for dry air is proportional to the square of frequency above roughly 10 kHz. According to the reference cited below the value of the attenuation factor above 10kHz is

Attenuation values produced at this web site agree well with this equation. Below 10kHz the variation with frequency flattens out.

I am very happy with this level of agreement.

It is not difficult to generalize the above derivation for a gas at an arbitrary pressure; the result can be expressed as

where γ=5/3 for a monatomic molecule, and P_{0}
is the pressure. Experimental and theoretical values of attenuation for helium
over a wide range of pressure and frequency are given in a PDF document published
by NIST, which can be found by Googling for the title "Transmission of
Sound Waves in Gases at Low Pressures." For values of ω/P_{0}
less than 1.4x10^{4}, corresponding to attenuations less than 8 dB per
wavelength, the results agree well with the above equation. For higher values
the curve in the NIST document flattens out, reaching a peak attenuation of
approximately 16 dB per wavelength.

**Discussion**

The decay model should be reasonably accurate as
long as there are 10 or more collisions per molecule, or t>2 nanoseconds
for air. In the absence of third moment generation this would reduce the third
moment by a factor in excess of 100. The third moment generation model is reasonably
accurate for a time period of 2 nanoseconds for frequencies as high as 10^{7}
Hz. At this frequency the wavelength in air is 34 micrometers and the attenuation
is 0.5 dB per wavelength of travel. The agreement with the helium data indicates
that the model is accurate well beyond these estimates. At the 20kHz top end
of the audio spectrum the attenuation is about 0.1 dB per meter.

The classical method for calculating sound attenuation, developed by Stokes and Rayleigh, is to add terms to the wave equation representing viscosity and thermal conduction. There are additional terms for diatomic molecules representing "molecular relaxation," an energy exchange between translational and rotational energy, and molecular vibration. The analysis here is totally different; it should capture the same losses as the classical method, but not the relaxation loss. I am surely biased, but I really like the derivation based directly on molecular velocities a lot better than invoking viscosity and heat conduction.

To the physics table of contents

_______________________

Reference for the attenuation of air: Crocker, Malcom
J., *Handbook of Acoustics*, Wiley, N. Y., 1998.

This section was posted Feb 3, 2006. Generalization for attenuation as a function of pressure added July 2006. As far as I am aware the analysis in this section regarding the relationship between attenuation and asymmetry of the velocity statistics is original work. In August 2007 I went back and re-read Vincenti and Kruger; in their "Kinetic Theory" derivation of the equations of gas dynamics in chapter IX they do include a "heat flux" term which looks like my 3rd moment term. However they do not use these equations to derive the sound wave equations, which are instead derived from an earlier set of gas equations derived in Chapter VIII.