4.8 Route 4 — from Hamilton’s principle

The first three routes all reached the wave equation by writing down a force balance — Newton’s law for a fluid slab (route 1), for a chain of oscillators (route 2), for the molecular momentum flux (route 3). This fourth route starts one level deeper, from a single scalar principle: of all the ways the field could evolve between two instants, nature picks the one that makes a quantity called the action stationary. The wave equation falls out as the condition for that stationarity, and — this is the payoff no force balance can give — the same principle hands us the conserved energy and momentum of the sound field almost for free, through Noether’s theorem.

This lesson is the one place in the chapter where every symbol needs to be pinned down carefully, because the objects (a field, its action, a Lagrangian density) are less familiar than a force. We go slowly.

What a variational principle says

In ordinary mechanics, a particle’s path $x(t)$ between a fixed start and end obeys Hamilton’s principle (full development: calculus of variations →): the path actually taken is the one for which the action

S \;=\; \int_{t_1}^{t_2} L\,\big(x, \dot x\big)\; dt, \qquad L \;=\; T - U,

is stationary — unchanged to first order under a small perturbation of the path. Here $L$ is the Lagrangian, kinetic energy $T$ minus potential energy $U$ . Requiring $\delta S = 0$ for every such perturbation $\delta x(t)$ produces the Euler–Lagrange equation, and for $L = \tfrac12 m\dot x^2 - U(x)$ that equation is exactly $m\ddot x = -U'(x)$ — Newton’s law. So $F = ma$ is not the foundation; it is a consequence of “the action is stationary.”

For a field — a quantity $\phi(\mathbf r, t)$ defined at every point of space and time rather than a single coordinate $x(t)$ — the same idea holds, with two changes: the Lagrangian becomes a Lagrangian density $\mathcal L$ (Lagrangian per unit volume), and the action integrates it over space as well as time,

S \;=\; \int dt \int d^3 r\;\; \mathcal L.

Everything below is this one principle, $\delta S = 0$ , applied to the sound field.

The field and its dictionary

We describe the sound field by a single scalar, the velocity potential $\phi(\mathbf r, t)$ . It is a bookkeeping device: instead of tracking the vector velocity and the pressure separately, we derive both from $\phi$ through a fixed dictionary,

\mathbf v' \;=\; \nabla\phi, \qquad p' \;=\; -\rho_0\,\partial_t\phi .

where

$\phi(\mathbf r, t)$: velocity potential — the single field we vary m^2/s
$\mathbf v' = \nabla\phi$: acoustic particle velocity (the fluid's small back-and-forth motion) m/s
$p' = -\rho_0\,\partial_t\phi$: acoustic pressure perturbation Pa
$\rho_0$: equilibrium (background) air density kg/m^3
$c$: speed of sound m/s

That a single scalar $\phi$ can carry the whole field is a real economy: a velocity potential exists precisely because sound is irrotational ( $\nabla\times\mathbf v' = \nabla\times\nabla\phi = 0$ ), which linear acoustics guarantees. The two relations above are the entire translation between $\phi$ and the physical quantities you can measure. Keep them in view — every term we write is one of them in disguise.

The acoustic Lagrangian density, term by term

Here is the Lagrangian density for the sound field:

\boxed{\;\mathcal L \;=\; \underbrace{\frac{\rho_0}{2c^2}\big(\partial_t\phi\big)^2}_{\text{compression energy}} \;-\; \underbrace{\tfrac12\,\rho_0\,\big(\nabla\phi\big)^2}_{\text{flow kinetic energy}}. \;}

Read each term through the dictionary above.

The gradient term $\tfrac12\rho_0(\nabla\phi)^2$ . Since $\mathbf v' = \nabla\phi$ , this is $\tfrac12\rho_0|\mathbf v'|^2$ — the kinetic energy density of the fluid’s motion, exactly the $\tfrac12 mv^2$ of a moving parcel, per unit volume.
The time-derivative term $\dfrac{\rho_0}{2c^2}(\partial_t\phi)^2$ . Since $p' = -\rho_0\partial_t\phi$ , this equals $p'^2/(2\rho_0 c^2)$ — the potential energy density stored in compressing the air, the acoustic analogue of $\tfrac12 k x^2$ in a spring. (The compressibility $1/\rho_0 c^2$ plays the role of the spring’s compliance; 4.4 is where $c^2 = (\partial p/\partial\rho)_s$ came from.)

Two things are worth pausing on, because they are exactly where this route is usually mis-stated.

First, the units check — and they must. A Lagrangian density has to be an energy per unit volume, or the action $S = \int \mathcal L\,dt\,d^3r$ would not carry the units of action (energy $\times$ time). Both terms above are honestly $\text{J/m}^3$ : $\rho_0|\nabla\phi|^2$ is $(\text{kg/m}^3)(\text{m/s})^2 = \text{J/m}^3$ , and $\tfrac{\rho_0}{c^2}(\partial_t\phi)^2$ is $(\text{kg/m}^3)(\text{s}^2/\text{m}^2)(\text{m}^2/\text{s}^2)^2 = \text{J/m}^3$ as well. If you ever see this Lagrangian written with the two coefficients as $\tfrac12\rho_0$ and $\tfrac12\rho_0 c^2$ , the ratio still yields the wave equation but neither term is an energy density — and then calling them “kinetic and potential energy,” or reading the conserved quantity below as the acoustic energy, is no longer literally true.

Second, the roles look swapped, and that is correct. In particle mechanics the time-derivative term ( $\tfrac12 m\dot x^2$ ) is the kinetic energy and the coordinate term is the potential. Here it is the other way round: the term built from $\partial_t\phi$ is the potential (compression) energy, and the term built from the spatial gradient $\nabla\phi$ is the kinetic energy. The reason is concrete: $\phi$ is a velocity potential, not a displacement. Its time derivative is (minus) the pressure, so squaring $\partial_t\phi$ gives a pressure energy; its gradient is the velocity, so squaring $\nabla\phi$ gives a flow energy. What actually matters is what the principle produces, and to that we now turn.

▶ Why this Lagrangian is the right one Derivation

We can check that $\mathcal L$ reproduces the linear acoustics of routes 1–3 rather than just asserting it. Linear acoustics is two equations: linearised Euler and linearised continuity (4.3, 4.2).

Substituting the dictionary $\mathbf v' = \nabla\phi$ , $p' = -\rho_0\partial_t\phi$ into Euler ( $\rho_0\partial_t\mathbf v' = -\nabla p'$ ) makes it an identity — both sides become $-\rho_0\nabla\partial_t\phi$ — so Euler’s equation is satisfied automatically by any $\phi$ . It is continuity that carries the physics. Linearised continuity is $\partial_t\rho' = -\rho_0\nabla\cdot\mathbf v' = -\rho_0\nabla^2\phi$ . Using the equation of state $p' = c^2\rho'$ to trade $\rho'$ for $p' = -\rho_0\partial_t\phi$ gives $\partial_t\rho' = c^{-2}\partial_t p' = -c^{-2}\rho_0\partial_t^2\phi$ . Equating the two expressions for $\partial_t\rho'$ :

-c^{-2}\rho_0\,\partial_t^2\phi \;=\; -\rho_0\,\nabla^2\phi \quad\Longrightarrow\quad \partial_t^2\phi = c^2\nabla^2\phi.

So the physical content we need $\mathcal L$ to reproduce is exactly the wave equation for $\phi$ . The next section shows the Euler–Lagrange machinery grinds $\mathcal L$ into precisely this — which is what “the right Lagrangian” means. (A Lagrangian is never unique: adding a total divergence changes $\mathcal L$ but not the equations, which is why textbooks differ by such terms.)

Turning the crank: Euler–Lagrange

For a field $\phi$ whose Lagrangian density depends on $\phi$ , its time derivative $\partial_t\phi$ , and its gradient $\nabla\phi$ , the stationarity condition $\delta S = 0$ becomes the Euler–Lagrange equation

\partial_t\!\left(\frac{\partial \mathcal L}{\partial(\partial_t\phi)}\right) \;+\; \nabla\cdot\!\left(\frac{\partial \mathcal L}{\partial(\nabla\phi)}\right) \;-\; \frac{\partial \mathcal L}{\partial\phi} \;=\; 0.

It looks forbidding, but it is just “differentiate $\mathcal L$ with respect to each way $\phi$ enters, then take the matching outer derivative.” Three pieces, computed one at a time:

$\dfrac{\partial\mathcal L}{\partial(\partial_t\phi)} = \dfrac{\rho_0}{c^2}\,\partial_t\phi$ . The outer $\partial_t$ makes it $\dfrac{\rho_0}{c^2}\,\partial_t^2\phi$ .
$\dfrac{\partial\mathcal L}{\partial(\nabla\phi)} = -\rho_0\,\nabla\phi$ . The outer divergence makes it $-\rho_0\,\nabla^2\phi$ .
$\dfrac{\partial\mathcal L}{\partial\phi} = 0$ , because $\phi$ appears only through its derivatives, never bare. (This absence is itself meaningful — it is the symmetry that gives energy conservation below.)

Add them:

\frac{\rho_0}{c^2}\,\partial_t^2\phi \;-\; \rho_0\,\nabla^2\phi \;=\; 0 \qquad\Longrightarrow\qquad \boxed{\;\partial_t^2\phi \;=\; c^2\,\nabla^2\phi.\;}

The acoustic wave equation, now for the velocity potential. Because $p' = -\rho_0\partial_t\phi$ , applying $-\rho_0\partial_t$ to both sides shows the same equation governs the pressure, $\partial_t^2 p' = c^2\nabla^2 p'$ — the boxed result of routes 1–3, reached here from a principle rather than a force.

What Noether hands back

The reason to have climbed to a variational principle is Noether’s theorem: every continuous symmetry of the action corresponds to a conserved quantity, and the theorem gives you that quantity by an explicit recipe. For the sound field this is not abstract bookkeeping — it produces the very quantities Chapter 5 is about.

Energy, from symmetry under shifting time. Nothing in $\mathcal L$ depends on $t$ explicitly (that was the $\partial\mathcal L/\partial\phi = 0$ observation), so the action is unchanged if we shift the clock. Noether’s recipe then yields a conserved energy density

\mathcal E \;=\; \underbrace{\tfrac12\rho_0\,|\mathbf v'|^2}_{\text{kinetic}} \;+\; \underbrace{\frac{p'^2}{2\rho_0 c^2}}_{\text{potential}},

the sum of the two energies we identified in $\mathcal L$ — kinetic energy of the moving air plus potential energy of its compression. (The recipe is the field version of $H = \dot x\,\partial L/\partial\dot x - L$ ; running it turns the minus in $\mathcal L$ into a plus here, which is why the Lagrangian is a difference of the two energies while the energy is their sum.) This $\mathcal E$ is exactly the acoustic energy density derived independently in 5.2 — first-principles agreement between two routes.

Conservation is local: $\mathcal E$ obeys a continuity equation $\partial_t\mathcal E + \nabla\cdot\mathbf I = 0$ , with energy flux

\mathbf I \;=\; p'\,\mathbf v'.

This product — pressure times velocity — is the acoustic intensity, the power per unit area a sound wave carries. It is the single most important quantity in Chapter 5, and here it arrives as the conserved current partnered to energy.

Momentum, from symmetry under shifting space. The action is likewise unchanged if we slide the whole field over in space, and Noether’s theorem attaches to that symmetry a conserved momentum density and flux. Physically: a sound wave carries momentum, and when it is absorbed or reflected it pushes on the obstacle — the radiation pressure of 5.6.

where

$\mathcal E$: acoustic energy density (kinetic + potential) J/m^3
$\mathbf I = p'\mathbf v'$: acoustic intensity — energy flux, the conserved current for energy W/m^2

We stop at naming these; Chapter 5 computes them in full. The point of route 4 is that their existence and form are dictated by the symmetries of a single scalar $\mathcal L$ , not discovered term by term.

⏳ The history — Emmy Noether and the theorem behind conservation laws

For two centuries, conservation of energy and momentum were empirical facts, patched into mechanics by hand. In 1918 Emmy Noether, working in Göttingen at Hilbert and Klein’s invitation — and barred from a salaried professorship because she was a woman — proved that they are consequences of symmetry: every continuous symmetry of a system’s action implies a conserved quantity, and vice versa. Time-translation symmetry $\Rightarrow$ energy; space-translation symmetry $\Rightarrow$ momentum; rotation symmetry $\Rightarrow$ angular momentum. Her theorem reframed conservation laws as statements about the sameness of the laws of physics from moment to moment and place to place, and it became the organising principle of all modern field theory, from electromagnetism to the Standard Model. The humble acoustic energy density above is one of its smallest and most concrete instances.

Why this route is not a repeat of the others

Routes 1, 2, and 3 begin from different pictures but all ultimately invoke $F = ma$ — route 1 directly, routes 2 and 3 through the oscillator chain and the molecular bath. Route 4 begins from a different kind of statement altogether: that physical evolution extremises an action. Newton’s law is then a theorem, not an axiom. The equivalence of the two formulations is one of the deep facts of classical mechanics — but the variational form is the one that generalises to gauge theories, general relativity, and the path integral of quantum mechanics, and it is the form that makes conservation laws a corollary of symmetry rather than a separate discovery. For acoustics specifically, that is the whole dividend: the energy and momentum of Chapter 5, handed over by Noether the moment the Lagrangian is written down.

Next lesson: all four routes give the same speed of sound — but each route gives that speed a different meaning.