Two- and three-dimensional waves

From one dimension to three. The 1D wave equation $\partial^2y/\partial x^2 = v^{-2}\,\partial^2y/\partial t^2$ has only one spatial direction to differentiate. In 2D or 3D there are two or three, and they combine into the Laplacian operator $\nabla^2$, which simply sums the second derivative along every axis:

where: $\psi(\vec r,t)$ is the disturbance at position $\vec r=(x,y,z)$ and time $t$, and $v$ is the wave speed in the medium. Setting $\partial/\partial y = \partial/\partial z = 0$ collapses this back to the familiar 1D equation, so nothing from the string is lost — only added to.

Plane waves. The simplest solution that satisfies this equation everywhere in space is a plane wave,

where: $A$ is the amplitude, $\vec k$ is the wavevector pointing along the direction of propagation with magnitude $|\vec k|=2\pi/\lambda$, $\omega$ is the angular frequency, and $\phi$ is a phase constant. Substituting this form into $\nabla^2\psi = v^{-2}\partial_t^2\psi$ shows it is only a valid solution when $\omega = v|\vec k|$ — the same dispersion relation as in 1D, now written with the vector's magnitude. A wavefront is a surface of constant phase, $\vec k\cdot\vec r-\omega t=\text{const}$; for a plane wave these are flat planes perpendicular to $\vec k$, all sliding along together at speed $v=\omega/|\vec k|$.

Spherical waves. A point source that radiates equally in every direction instead produces a spherical wave,

where: $r=|\vec r|$ is the distance from the source and $k=2\pi/\lambda$. Its wavefronts are concentric spheres centred on the source, expanding outward at speed $v$. The prefactor $A/r$ is not optional decoration — it is forced by energy conservation. Total power crossing any sphere of radius $r$ must be the same regardless of $r$ (no energy is created or destroyed in transit), and intensity is power per unit area, so

since intensity goes as amplitude squared. Contrast the three geometries: a 1D wave on an ideal string has no room to spread, so its amplitude stays constant; a 2D circular ripple spreads its energy around a circle of circumference $2\pi r$, giving amplitude $\propto 1/\sqrt r$; a 3D spherical wave spreads over a sphere of area $4\pi r^2$, giving amplitude $\propto 1/r$. More dimensions to spread into means faster amplitude decay.

There is a second, sharper way to see why plane waves matter beyond just being the simplest solution: far from any source, real waves look locally like plane waves. A small patch of a huge spherical wavefront, viewed up close, is nearly flat — exactly as a small patch of the Earth's surface looks flat even though the Earth is a sphere. That is why so much of optics and acoustics is built entirely on plane-wave mathematics even though most real sources (light bulbs, loudspeakers) are effectively point sources radiating spherical waves.

A loudspeaker acts as a point source radiating sound uniformly in all directions with total acoustic power $P = 0.4\ \text{W}$, travelling at $v = 340\ \text{m/s}$ with frequency $f = 1000\ \text{Hz}$. Find (a) the wavelength and wavenumber, and (b) the intensity at $r_1 = 2\ \text{m}$ and $r_2 = 8\ \text{m}$, checking that the amplitude ratio follows the $1/r$ law.

Step 1. Get the wave's spatial and temporal rates. $\lambda = v/f = 340/1000 = 0.34\ \text{m}$, so $k = 2\pi/\lambda = 2\pi/0.34 \approx 18.5\ \text{rad/m}$, and $\omega = 2\pi f \approx 6283\ \text{rad/s}$. The wave is $\psi(r,t) = (A/r)\cos(kr-\omega t)$ with $k\approx 18.5\ \text{rad/m}$.

Step 2. Apply $I(r) = P/(4\pi r^2)$ at both distances. At $r_1 = 2\ \text{m}$: $I_1 = 0.4/(4\pi\times 2^2) \approx 7.96\times10^{-3}\ \text{W/m}^2$. At $r_2 = 8\ \text{m}$: $I_2 = 0.4/(4\pi\times 8^2) \approx 4.97\times10^{-4}\ \text{W/m}^2$.

Step 3. Check the ratio against the inverse-square prediction. $I_1/I_2$ should equal $(r_2/r_1)^2 = 16$; numerically $7.96\times10^{-3}/4.97\times10^{-4} \approx 16.0$, confirming the law. Since amplitude $A\propto\sqrt I \propto 1/r$, the amplitude at $r_1$ is $4\times$ the amplitude at $r_2$ — exactly what $A/r$ predicts, since $r_2/r_1 = 4$.

Answer: $\lambda = 0.34\ \text{m}$, $k \approx 18.5\ \text{rad/m}$, $I_1 \approx 7.96\times10^{-3}\ \text{W/m}^2$, $I_2 \approx 4.97\times10^{-4}\ \text{W/m}^2$, ratio $16$, consistent with amplitude $\propto 1/r$.

Competitive stretch

A plane wave $\psi(\vec r,t) = A\cos(\vec k\cdot\vec r-\omega t)$ has $\vec k = (3\hat x+4\hat y)\ \text{rad/m}$ and $\omega = 50\ \text{rad/s}$. Then $|\vec k| = 5\ \text{rad/m}$, so $\lambda = 2\pi/5 \approx 1.257\ \text{m}$, direction $\hat k = 0.6\hat x+0.8\hat y$ ($53.1^\circ$ from the $x$-axis), and $v = \omega/|\vec k| = 50/5 = 10\ \text{m/s}$. A wavefront through the origin reaches $(6,8,0)\ \text{m}$ when $\vec k\cdot\vec r=\omega t$: $(3)(6)+(4)(8)=50=50t$, so $t=1\ \text{s}$ — matching $t=10/v=1\ \text{s}$ since $(6,8,0)$ lies along $\hat k$ at distance $10\ \text{m}$.

• Confusing $k$ (scalar wavenumber) with $\vec k$ (wavevector). In 3D the phase uses the dot product $\vec k\cdot\vec r$, not a plain product; $|\vec k|=2\pi/\lambda$ is only the magnitude, and direction information is lost if you drop the vector notation.
• Applying the 1D constant-amplitude rule in higher dimensions. A string wave does not spread out geometrically, so its amplitude is constant. A spherical wave spreads over a growing sphere, so amplitude must fall as $1/r$ — a purely geometric, energy-conservation effect, not damping or absorption.
• Thinking all wavefronts are planes. A plane wave's wavefronts are flat planes; a spherical wave's are concentric spheres. Far from a point source a small patch of a spherical wavefront looks locally flat, which is exactly why the plane-wave approximation is used so widely.
• Forgetting the dispersion relation. Plugging $\psi = A\cos(\vec k\cdot\vec r-\omega t)$ into the wave equation only works if $\omega = v|\vec k|$; this constraint is easy to forget when just quoting the plane-wave formula.

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