### The Sound of One Physicist Wailing

One of the delights of teaching elementary physics is discovering some basic thing that you thought you understood, but actually didn’t. Usually, this occurs late at night, while preparing your lecture for the next morning. And you wonder whether you’ll be able to keep a straight face, the next morning, as you say words you’re no longer quite so sure are true.

I’ve been teaching about waves in a non-technical course. One of the points I like to emphasize is that the energy density, or the intensity, of the wave is *quadratic* in the amplitude. There are lots of examples of that, with which you are doubtless familiar: electromagnetic waves, transverse waves on a stretched string, …

But we’re studying sound, now. So I thought I would reassure myself that the same is true of sound waves …

Where to start — not for the kids, mind you, but for my own satisfaction?

Clearly, we should start with the Navier-Stokes equations. Neglecting viscosity, these equations have five locally-conserved quantities: mass, energy, and (three components of) momentum. The equations we are dealing with are …

Navier-Stokes:

Conservation of mass:

and the equation of state:

It’s easy to write down the aforementioned conservation equations^{1}. Indeed, we already wrote one of them down, (2). The momentum density in the fluid is

and the stress tensor is

By virtue of (1),(2),(3), these satisfy $0 = \frac{\partial \vec{\mathcal{P}} }{\partial t} + \vec{\nabla}\cdot \overset{\leftrightarrow}{\sigma}$ The energy density in the fluid is

and the energy flux, or intensity

Again, by virtue of (1),(2),(3), these satisfy

Now, to write down a sound wave. Let’s work in rest frame of the fluid, and expand $\begin{split} p(\vec{x},t) &= p_0 + p_1(\vec{x},t) + \dots \\ \rho(\vec{x},t) &= \rho_0 + \rho_1(\vec{x},t) + \dots \\ \vec{v}(\vec{x},t) &= \vec{v}_1(\vec{x},t) + \dots \end{split}$ The general solution of the linearized equations is a (superposition of) plane wave(s)

where $v_s = \sqrt{\gamma p_o/\rho_0}$ is the sound speed, and $u^2= v_s^2$.

So far, so good. But, here comes the puzzle: if we plug (9) into (6) and (7), we find that the energy density and the flux have terms *linear* in $f$! Surely, they should be quadratic^{2}.

What the heck!?!

Turns out that the resolution is remarkably simple. We can modify the energy conservation equation (8), by adding a multiple of the mass conservation equation (2). If we choose astutely, we can kill the unwanted linear terms. Define

These still satisfy the same conservation equation (8), as before, but now, when we plug in (9), we find^{3}
$\begin{split}
\mathcal{E}' &= \frac{1}{\gamma p_0} f^2 \\
\vec{\mathcal{I}}' &= \frac{\vec{u}}{\gamma p_0} f^2
\end{split}$

Whew! That’s a relief.

Not that we’re going to ever discuss it in some Freshman physics class, but we should have some word to say about the interpretation of what we’ve done. The first term in (6) is the kinetic energy density in the fluid; the second term has the interpretation of a potential energy density. What we’ve done, in (10), is redefine the zero of the potential energy density in some peculiar, position-dependent, way. (We also added a constant, to make homogeneous solution have zero potential.) Is there a more insightful explanation of the modification we’ve made?

Also, you might try do something similar for the momentum density (4). But, I think, you will search in vain for a suitable modification. I think the momentum density really is linear in the fluctuations.

^{1} Many fluid dynamicists like to write conservation laws, using the convective derivative,
$\frac{D}{D t} = \frac{\partial}{\partial t} + \vec{v}\cdot \vec{\nabla}$
instead of $\frac{\partial}{\partial t}$. In this context, that would be a wacky thing to do. If I’m an observer at some fixed location, $\vec{x}$, I want to know how much *stuff* is flowing past my location. Perhaps if I were interested in bulk fluid motion, I might be interested in the observations of observers co-moving with the fluid. But, for present purposes, fixed observers are more natural.

^{2} There’s also a constant term in $\mathcal{E}$, but that’s harmless. We can just subtract it off, and do so in writing down (10).

^{3} There’s actually a little bit of trickiness involved. Naïvely, it appears that we need to evaluate the $\frac{1}{\gamma-1}(p-v_s^2\rho)$ term in $\mathcal{E}'$ to second order in the fluctuations. It would be rather ugly if we had to go back and solve Navier-Stokes to second order. Fortunately, expanding (3) to second order, we find
$\begin{split}
p_1 &= v_s^2 \rho_1\\
p_2 &= v_s^2 \left( \frac{\gamma-1}{2} \frac{\rho_1^2}{\rho_0} + \rho_2\right)
\end{split}$
which is *just* what is needed to evaluate (10).

## Re: The Sound of One Physicist Wailing

Fun stuff. It might also be fun to convert these equations from vector calculus to differential forms and see if the conservation laws can be expressed in terms of adjoint nilpotent operators $d^2 = 0$ and $\delta^2 = 0$ in analogy with Maxwell’s equations.

I used to the think we could start with a 0-form connection $A$ and compute the curvature $F = dA$, then follow the usual prescription from Maxwell, i.e.

$\mathcal{L} = \frac{1}{2} \int_{\mathcal{M}} (F,F) vol$

leading to

$dF = 0\quad\text{and}\quad \delta F = 0$

except $F$ is a 1-form. The constitutive equation (3) as well as the non-relativistic metric can be encoded in the Hodge star defining $\delta$.

Then we dig up some of Urs’ old notes on Hamiltonian evolution and Killing vector fields, etc.

There is probably some neat cohomology buried here too. For example, Equation (8) can be written as

$\delta \alpha = 0$

for some 1-form $\alpha$ and in going to Equation (10), you’re writing

$\alpha' = \alpha + \delta \beta$

for some 2-form $\beta$ so that obviously

$\delta\alpha' = \delta\alpha.$

Another way to maybe look at it is that you’re performing a gauge transformation, but here (unlike Maxwell) the “amplitude” is actually the gauge field.

Sorry for thinking out loud. It’s been ages since I’ve thought about this stuff and I wasn’t exactly an expert back then either :)

Anyway, thanks for writing this. It was a welcome distraction.