4.7 Route 3 — from kinetic theory and momentum flux

A third derivation, this time from the molecular side. We saw in chapter 1 that pressure is the macroscopic time-average of molecular momentum delivery to surfaces. A pressure perturbation is a local, transient excess or deficit in that momentum flux. By tracking the flux carefully and applying conservation laws, we can derive the wave equation from kinetic theory alone — without ever writing down a Euler equation or an equation of state explicitly. The result is the same wave equation, with $c$ in terms of molecular speeds.

The picture

A column of gas at equilibrium has uniform pressure $p_0$ . Molecules cross any plane in equal numbers in both directions, carrying equal mean momentum per unit area per unit time — so the net momentum flux across the plane is zero.

Now imagine a small region of the gas is compressed. Locally, the number density rises, the temperature rises slightly (adiabatic compression), and the mean molecular speed increases. Molecules leaving this region carry slightly more momentum than those entering it — there is a net momentum flux outward. This momentum-flux gradient is what we identify with pressure gradient in the fluid-mechanics picture, but here it’s derived directly from molecular kinematics.

Sketch of the derivation

A complete derivation needs the Boltzmann equation and the Chapman–Enskog procedure — beyond the scope of a single lesson. We sketch the route in three steps and quote the result.

Step 1: pressure as momentum flux. For a gas with isotropic Maxwell–Boltzmann velocity distribution at temperature $T$ ,

p \;=\; \frac{1}{3}\, n m \langle v^2 \rangle \;=\; n k_B T,

(re-derived in lesson 1.2). The pressure on a surface is the flux of normal momentum into it.

Step 2: conservation of mass and momentum. Counting molecules crossing a small control volume and the momentum they carry, in linearised form,

\partial_t n' \;+\; n_0 \nabla \cdot \langle \mathbf{v}' \rangle \;=\; 0,

m n_0 \partial_t \langle \mathbf{v}' \rangle \;=\; -\nabla p',

with $\langle \mathbf{v}' \rangle$ the local mean flow velocity. These are continuity and Euler, derived from molecular counting rather than from fluid intuition.

Step 3: pressure–density relation from molecular speeds. A small adiabatic compression of an ideal gas heats it by $\delta T / T_0 = (\gamma - 1)\, \delta n / n_0$ , which, combined with $p = n k_B T$ , gives $\delta p / p_0 = \gamma\, \delta n / n_0$ , i.e.

p' \;=\; \frac{\gamma p_0}{n_0}\, n' \;=\; \frac{\gamma p_0}{\rho_0}\, \rho' \;\equiv\; c^2\, \rho'.

Same as route 1’s equation of state, but derived here from molecular kinematics + equipartition.

Step 4: combination. Identical to the combination step in route 1. Out drops

\partial_t^2 p' \;=\; c^2\, \nabla^2 p',

with

c^2 \;=\; \frac{\gamma p_0}{\rho_0} \;=\; \frac{\gamma k_B T}{m}.

⏳ The history — Newton's wrong number, Laplace's fix

In Principia (1687), Newton computed the speed of sound assuming isothermal compression — i.e., that the temperature of the gas stays fixed during a sound wave’s compressions and rarefactions (Newton 1687). His formula $c = \sqrt{p_0 / \rho_0}$ gives about 280 m/s for air, which was already known by then to be about 15% low (Mersenne and others had timed the round-trip of cannon-fire echoes).

The discrepancy stood for 130 years. In 1816 Laplace pointed out that the compressions in a sound wave are too fast for heat to flow between adjacent regions — they are essentially adiabatic. The right formula is then $c = \sqrt{\gamma p_0 / \rho_0}$ , and for diatomic air $\gamma = 7/5$ , recovering $c \approx 343$ m/s (Laplace 1816).

The factor $\gamma$ — the ratio of specific heats $c_p / c_v$ — is the same $\gamma$ that distinguishes adiabatic from isothermal in thermodynamics, and it counts the active molecular degrees of freedom. For a monatomic gas (helium) $\gamma = 5/3$ ; for a diatomic gas like air at room temperature $\gamma = 7/5$ (translation + rotation); for a polyatomic gas with active vibrational modes, $\gamma$ approaches 1 from above. Laplace’s correction connects acoustics to thermodynamics to kinetic theory in a single step.

What kinetic theory adds: $c \sim$ thermal speed

The form $c = \sqrt{\gamma k_B T / m}$ is the kinetic-theory result. Compare to the thermal speed of a molecule:

v_\text{rms} \;=\; \sqrt{\langle v^2 \rangle} \;=\; \sqrt{3 k_B T / m}.

So $c / v_\text{rms} = \sqrt{\gamma/3}$ . For diatomic air with $\gamma = 1.4$ , this is $\sqrt{1.4/3} \approx 0.68$ . The speed of sound is about 70% of the thermal molecular speed. This is not a coincidence: a pressure perturbation propagates by the same molecular motions that constitute the thermal bath. It cannot propagate faster than the molecules carrying it — and indeed it propagates a bit slower because each molecule’s velocity is randomly directed, so only a fraction of the speed projects onto the propagation direction.

For air at 20°C, $v_\text{rms} \approx 500$ m/s and $c \approx 343$ m/s. The numbers fit.

What this route adds

A bridge to chapter 1. The wave equation and Brownian motion both live on top of the same molecular bath. Sound is the coherent motion; thermal jitter is the incoherent motion. Kinetic theory is the framework that holds both.
A correction to “Newton’s value”. Newton’s 1687 calculation gave $c = \sqrt{p_0/\rho_0}$ , which is the isothermal speed — about 280 m/s for air. He was wrong by 20% because he did not realise the compressions are too fast for heat to flow. Laplace’s 1816 fix amounts to inserting $\gamma$ , and from the kinetic-theory perspective $\gamma$ comes from counting how many degrees of freedom can absorb the temperature rise during adiabatic compression. Air’s diatomic structure (with rotational degrees of freedom available at room temperature but not vibrational) gives $\gamma = 7/5$ , just right.

▶ Why $\gamma$ corrects Newton's value

Newton’s reasoning: pressure is proportional to density in isothermal compression, so $\delta p / \delta \rho = p_0 / \rho_0$ and $c_\text{Newton} = \sqrt{p_0 / \rho_0}$ . For air this gives $\sqrt{101325 / 1.2} \approx 290$ m/s — about 15% low.

The fix: compressions in a sound wave are adiabatic, not isothermal. Heat doesn’t have time to flow out of compressed regions in one acoustic period. For an ideal gas, the adiabatic compressibility is $\gamma$ times the isothermal compressibility, so $\delta p / \delta \rho = \gamma\, p_0 / \rho_0$ and $c = \sqrt{\gamma p_0 / \rho_0}$ . For air ( $\gamma = 1.4$ ), this gives 343 m/s.

The factor of $\sqrt{1.4} \approx 1.18$ accounts for the difference. It is the same $\gamma$ that distinguishes $c_p$ from $c_v$ in the heat capacity — and from kinetic theory it counts the active degrees of freedom of the molecule.

Next, and last: a derivation that starts from neither $F = ma$ nor molecular counting, but from a variational principle.

4.7 Route 3 — from kinetic theory and momentum flux

The picture

Sketch of the derivation

What kinetic theory adds: c∼c \simc∼ thermal speed

What this route adds

What kinetic theory adds: $c \sim$ thermal speed