## September 16, 2009

### 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:

(1)$\rho\left(\frac{\partial \vec{v}}{\partial t}+ (\vec{v}\cdot\vec{\nabla})\vec{v}\right) = - \vec{\nabla} p$

Conservation of mass:

(2)$0 =\frac{\partial\rho}{\partial t} + \vec{\nabla} \cdot(\rho\vec{v})$

and the equation of state:

(3)$p = c \rho^\gamma$

It’s easy to write down the aforementioned conservation equations1. Indeed, we already wrote one of them down, (2). The momentum density in the fluid is

(4)$\vec{\mathcal{P}} = \rho \vec{v}$

and the stress tensor is

(5)$\sigma_{i j} = p \delta_{i j} + \rho v_i v_j$

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

(6)$\mathcal{E} = \frac{1}{2} \rho v^2 +\frac{1}{\gamma-1} p$

and the energy flux, or intensity

(7)$\vec{\mathcal{I}} = \left(\frac{1}{2}\rho v^2 + \frac{\gamma}{\gamma-1} p \right) \vec{v}$

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

(8)$0 = \frac{\partial \mathcal{E}}{\partial t} + \vec{\nabla}\cdot \vec{\mathcal{I}}$

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)

(9)$\begin{split} p_1 &= f(\vec{x}-\vec{u} t)\\ \rho_1 &= \frac{1}{v_s^2}f(\vec{x}-\vec{u} t)\\ \vec{v}_1 &= \frac{\vec{u}}{\rho_0 v_s^2} f(\vec{x}-\vec{u} t) \end{split}$

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 quadratic2.

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

(10)$\begin{split} \mathcal{E}' &= \frac{1}{2} \rho v^2 +\frac{1}{\gamma-1} (p - v_s^2 \rho) + p_0 \\ \vec{\mathcal{I}}' &= \left(\frac{1}{2}\rho v^2 + \frac{1}{\gamma-1}(\gamma p- v_s^2\rho) \right) \vec{v} \end{split}$

These still satisfy the same conservation equation (8), as before, but now, when we plug in (9), we find3 $\begin{split} \mathcal{E}' &= \frac{1}{\gamma p_0} f^2 \\ \vec{\mathcal{I}}' &= \frac{\vec{u}}{\gamma p_0} f^2 \end{split}$

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).

Posted by distler at September 16, 2009 1:44 AM

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### 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.

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

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

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.

Posted by: Eric Forgy on September 16, 2009 2:48 AM | Permalink | Reply to this

### Re: The Sound of One Physicist Wailing

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 don’t see any such simplification. Incompressible fluid flow, in $n$ spatial dimensions, is simplified by writing the $(n-1)$-form, $\mathbf{v}$ as $\mathbf{v}= d\phi$. That’s particularly useful for $n=2$, where the vorticity, $\phi$, is a scalar.

But we’re not doing incompressible fluid flow (a limit in which the sound speed, $v_s\to\infty$) …

and in going to Equation (10), you’re writing $\alpha′=\alpha+\delta\beta$ for some 2-form $\beta$.

No, sorry, I don’t see that.

If that were true, the conservation equation for the 1-form “$\delta\beta$” would be an identity. But the mass conservation equation, (2), is not an identity.

Put a different way, $\mathcal{E}$ and $\mathcal{E}'$ are not gauge-equivalent. They have different physical meanings. It just turns out that $\mathcal{E}'$ is the relevant notion of energy density for the sound-wave.

Posted by: Jacques Distler on September 16, 2009 8:37 AM | Permalink | PGP Sig | Reply to this

### Re: The Sound of One Physicist Wailing

I don’t see any such simplification.

Thanks Jacques. I played around with it a little bit last night and things definitely are not as simple as I was hoping they would be.

Incidentally, I did find a paper that reformulates Navier-Stokes via vector-valued forms.

NAVIER-STOKES DYNAMICS ON A DIFFERENTIAL ONE-FORM

If one were inclined to pursue it (which is beyond me and probably misguided anyway), I wonder if this might help with a “gauge theoretic” reinterpretation. Instead of a $u(1)$-valued 0-form “connection” (and corresponding 1-form “curvature”) you might have higher dimensional Lie algebra (?)

No, sorry, I don’t see that.

[snip]

Put a different way, ℰ and ℰ′ are not gauge-equivalent. They have different physical meanings. It just turns out that ℰ′ is the relevant notion of energy density for the sound-wave.

Oops! Sorry about that. It was wishful thinking. That makes your note even more interesting than I already thought it was!

How was the lecture?

Posted by: Eric Forgy on September 16, 2009 11:05 AM | Permalink | Reply to this

### Re: The Sound of One Physicist Wailing

I don’t think we ought to force models into the mold of our favorite mathematical frameworks just because we can. We also have to consider the physical properties of the model. For the Navier-Stokes equation, this means coming up with a mathematical framework which makes the Galilean symmetry of the system explicit.

Posted by: Grapes on October 18, 2009 9:41 PM | Permalink | Reply to this

### Re: The Sound of One Physicist Wailing

The reason this works is that in non-rel hydrodynamics mass is conserved, so we can always add a term to the energy density which is proportional to the mass density.

In terms of a more physical picture I would argue the following: In a density wave there is a non-zero “background” energy density, and part of the local energy density of the wave is a periodic change in this background energy. This change is linear in the amplitude, but it integrates to zero over a period of the wave, and we therefore do not include it in the total energy of the wave. The effect is nevertheless real – I could try to build a little powerplant that extracts energy from this term (or from the analogous potential energy term in a surface wave).

Posted by: Thomas on September 16, 2009 11:25 AM | Permalink | Reply to this

### Re: The Sound of One Physicist Wailing

This change is linear in the amplitude, but it integrates to zero over a period of the wave, and we therefore do not include it in the total energy of the wave. The effect is nevertheless real…

It’s most definitely real. And it would be a mistake to focus only on time-averaged quantities (for which that term integrates to zero). We are, after all, interested in sending signals with our sound waves. So we really do care about the time-dependence.

… or from the analogous potential energy term in a surface wave

I should probably make sure I understand the surface wave case, too …

Posted by: Jacques Distler on September 16, 2009 12:01 PM | Permalink | PGP Sig | Reply to this

### Surface waves

I could try to build a little powerplant that extracts energy from this term (or from the analogous potential energy term in a surface wave).

The analogous potential energy term for the surface wave is

1. everwhere positive

So it doesn’t pose the same puzzles that this sinusoidal (and linear in the amplitude) term, for the sound wave, poses.

I’m not sure you can build a device to extract energy from the latter, though you certainly can extract energy from the former.

Posted by: Jacques Distler on September 17, 2009 12:01 PM | Permalink | PGP Sig | Reply to this

### Re: Surface waves

Are you sure? The motion of the fluid in a gravity wave can be described as individual fluid particles performing a circular motion with constant angular velocity (and an amplitude that decreases exponentially as you go away from the surface). This would seem to give a gravitational potential energy which is proportional to the density of the fluid, linear in the amplitude and alternating in sign.

Posted by: Thomas on September 17, 2009 1:08 PM | Permalink | Reply to this

### Re: Surface waves

I should have been more precise: It would seem to give a linear and a quadratic term, where the linear term averages to zero, and the quadratic term is one half of the total energy of the wave (the other half residing in kinetic energy).

Posted by: Thomas on September 17, 2009 1:12 PM | Permalink | Reply to this

### Re: Surface waves

Yes, I’m sure. The potential energy density of the surface wave is $\begin{split} \mathcal{U}(x,y,t) &= \int_0^{h(x,y,t)} \rho g z d z\\ &= \tfrac{1}{2} \rho g {h(x,y,t)}^2 \end{split}$ This is

2. non-negative, regardless of the sign of $h$.
Posted by: Jacques Distler on September 17, 2009 1:54 PM | Permalink | PGP Sig | Reply to this

### Re: Surface waves

It’s a small amplitude wave, isn’t it? So h=h_0+a*Cos(k*x-w*t), where a is the amplitude of the wave.

Posted by: Thomas on September 17, 2009 2:28 PM | Permalink | Reply to this

### Re: Surface waves

Obviously, we want to normalize so that the potential energy density vanishes for vanishing displacement from equilibrium. That corresponds to taking $h_0=0$ in your notation.

Posted by: Jacques Distler on September 17, 2009 2:49 PM | Permalink | PGP Sig | Reply to this

### Re: Surface waves

You do more than that, don’t you? You change the sign of the gravitational force at h_0, so that a particle acquires positive gravitational potential energy no matter if it is raised or lowered.

Posted by: Thomas on September 17, 2009 3:11 PM | Permalink | Reply to this

### Re: Surface waves

No.

Removing a particle from below the equilibrium height makes a positive contribution to $\mathcal{U}$, just as surely as adding a particle above the equilibrium height.

Posted by: Jacques Distler on September 17, 2009 3:16 PM | Permalink | PGP Sig | Reply to this

### Re: Surface waves

My definition of the potential energy of a fluid column and yours differ by a constant and a term proportional to the mass in the column. Since total mass is conserved your definition is indeed as good as mine.

I would argue that my expression is the one that follows from the textbook definition of the energy density of a fluid. In that sense, the situation with density waves and surface waves is indeed exactly the same: If the text book definition is adopted we find a term in the local energy density which is linear in the amplitude, but it disappears if one computes the total energy of the wave, or if one redefines the energy density by a suitable multiple of the mass density.

The exercise does shed some light on the meaning of the extra term: The fluid motion involves local mass transport, and the extra term in the energy density accounts (to leading order) for the energy required to take a mass element and put it somewhere else.

Posted by: Thomas on September 17, 2009 9:04 PM | Permalink | Reply to this

### Re: Surface waves

If I wanted the gravitational potential energy corresponding to the total mass in the column (of equilibrium depth, $D$), I would compute $\begin{split} \mathcal{U}'(x,y,t) &= \int_{-D}^{h(x,y,t)} \rho g z d z \\ &= \tfrac{1}{2} \rho g (h^2 - D^2) \end{split}$ This differs by a constant ($\tfrac{1}{2} \rho g D^2$) from the expression I wrote before.

As always, I think I am justified in defining $h(x,y,t)$ as the deviation from the equilibrium height of the column. Again, up to the additive constant, the potential energy density is a quadratic function of that deviation if we choose the surface of the water (in equilibrium) as the zero of the gravitational potential (as above).

This seems like quite a natural choice.

Moreover, if you want to study surface waves, where the channel has a varying depth (waves breaking at the beach!), it’s the only choice that makes sense.

Posted by: Jacques Distler on September 17, 2009 11:26 PM | Permalink | PGP Sig | Reply to this

### Re: The Sound of One Physicist Wailing

“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 quadratic2.”

Silly question from ignoramus here but why is it necessary to have quadratic terms instead of linear here, what problems exactly would this cause?

Posted by: TinyGrasshopper on September 16, 2009 6:10 PM | Permalink | Reply to this

### Re: The Sound of One Physicist Wailing

[W]hy is it necessary to have quadratic terms instead of linear here, what problems exactly would this cause?

Among other weirdnesses, the latter is not positive-definite…

Posted by: Jacques Distler on September 17, 2009 12:04 PM | Permalink | PGP Sig | Reply to this
Weblog: Musings
Excerpt: More elementary hydrodynamics. Why? 'Cuz it's fun.
Tracked: September 18, 2009 12:29 PM

### Re: The Sound of One Physicist Wailing

Just out of curiosity, why is there no viscosity and why is the equation of state a power law? Not that it affects your main argument at all.

Posted by: Grapes on October 18, 2009 7:29 PM | Permalink | Reply to this

### Re: The Sound of One Physicist Wailing

Just out of curiosity, why is there no viscosity

Simplicity.

Putting viscosity into the linearized equations, I was aiming for, just dissipates the plane wave solutions that we found. And, in so doing, it dissipates the energy that — in the absence of viscosity — is a conserved quantity.

Since the point was to discuss the latter, it makes sense to ignore viscosity for the purposes of this discussion.

and why is the equation of state a power law?

Again, the departures from a perfect fluid are irrelevant to the discussion.

Posted by: Jacques Distler on October 18, 2009 11:24 PM | Permalink | PGP Sig | Reply to this

### Re: The Sound of One Physicist Wailing

The true root of the problem is that zero point we are expanding about doesn’t have zero compressibility. We have exactly the same situation for electromagnetic waves propagating over a nonzero electromagnetic background field, or sound waves over a solid medium. What we ought to do instead is to average the energy density over several wavelengths.

Posted by: Grapes on October 18, 2009 9:27 PM | Permalink | Reply to this

### Re: The Sound of One Physicist Wailing

In fact, at least up to the quadratic level and for small enough perturbations, it’s possible to show that provided $\rho=\rho_0$ and $v=0$ at the boundaries, even though the energy density at any given point might be below the zero point energy density, the total energy has to be greater than or equal to the zero point energy.
Posted by: Grapes on October 18, 2009 9:35 PM | Permalink | Reply to this

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