6.2 Spherical waves and the inverse-square law

For a harmonic monopole, the pressure field at distance $r$ from the source has the form

p(r, t) \;=\; \frac{P_0\, a}{r}\, \cos\!\big(\omega t - k(r - a) + \varphi\big),

with $P_0$ the pressure amplitude at the source surface $r = a$ . The two essential properties:

Amplitude decays as $1/r$ . Pressure halves for every doubling of distance.
Phase changes linearly with distance. A kick at the source at $t = 0$ arrives at radius $r$ at time $t = r/c$ .

Intensity and the inverse-square law

Intensity in a plane wave is $\langle I \rangle = P^2 / (2 \rho_0 c)$ . The plane-wave formula is the local relationship between pressure amplitude and intensity, and it applies to spherical waves wherever the wavefront is locally flat (i.e. at distances $r \gg \lambda$ ). Substituting the spherical amplitude:

\langle I \rangle \;=\; \frac{1}{2 \rho_0 c} \cdot \frac{P_0^2 a^2}{r^2} \;\propto\; \frac{1}{r^2}.

Intensity falls as $1/r^2$ — the inverse-square law. Energy conservation, restated: a fixed total power is being spread over a spherical surface whose area grows as $r^2$ , so intensity per unit area drops as $1/r^2$ .

In decibels, the inverse-square law is 6 dB per doubling of distance:

\Delta L_I \;=\; 10 \log_{10}\!\left(\frac{1}{4}\right) \;=\; -6\,\text{dB}.

A point source 1 m away is at, say, 80 dB. The same source 2 m away is at 74 dB; 4 m away, 68 dB; 8 m away, 62 dB. Each successive doubling subtracts 6 dB.

Total radiated power

Integrating the intensity over a sphere of radius $r$ :

P_\text{rad} \;=\; \langle I \rangle \cdot 4\pi r^2 \;=\; \frac{2 \pi P_0^2 a^2}{\rho_0 c}.

The factor of $r^2$ cancels — the total power radiated through any spherical surface enclosing the source is the same, as it must be for an isolated source in a non-absorbing medium. This is one of those tautological-but-useful checks: if you computed a $1/r^3$ falloff from a point source, you’d know something was wrong.

Near field vs. far field

The clean $1/r$ falloff and plane-wave-like local relationships hold for $r \gg \lambda$ — the far field. Close to a small source ( $r \sim a$ or $r \lesssim \lambda$ ) the situation is more complicated. The acoustic impedance becomes complex (it has a reactive component representing energy that swings back and forth between the source and a thin layer of surrounding air without radiating away). For a small spherical source of radius $a$ at angular frequency $\omega$ , the radiation impedance is

Z_\text{rad}(\omega) \;=\; \rho_0 c \cdot \frac{(k a)^2 + i k a}{1 + (k a)^2}.

For $k a \ll 1$ (small source compared to wavelength), the real part is $\sim (ka)^2 \cdot \rho_0 c$ — much less than $\rho_0 c$ — and the source is a poor radiator. The imaginary part is $\sim k a \cdot \rho_0 c$ , much larger than the real part — the source is mostly reactively coupled to the surrounding air, sloshing it back and forth without radiating.

This is why small sources at low frequencies are inefficient radiators: a 5-cm woofer cone trying to radiate 50 Hz ( $\lambda \approx 7$ m) has $k a \approx 0.04$ , so its real radiation resistance is $\sim 10^{-3} \rho_0 c$ . It mostly pushes air back and forth rather than radiating it away. Big subwoofers exist for this exact reason: you need either a bigger $a$ or a horn-loaded structure that effectively enlarges the source for low frequencies.

The far-field rule of thumb

For practical work: a source’s “far field” begins at roughly $r = 2 a^2 / \lambda$ (for a piston of radius $a$ ). Closer than that, the field has reactive structure and direct $1/r$ falloff doesn’t apply. Farther, you can use the simple inverse-square law.

For a 1-m-diameter speaker at 1 kHz ( $\lambda = 0.34$ m), the far field begins at $r \approx 2 (0.5)^2 / 0.34 \approx 1.5$ m. Beyond about 5 m the simple model is excellent.

What follows

Cylindrical sources behave differently — geometric spreading is one-dimensional rather than two-dimensional in their case, and intensity falls as $1/r$ rather than $1/r^2$ . We meet them next.