Superposition & Linear Media

Superposition is not a separate physical postulate; it is the algebraic shape of the wave equation. Collect the equation into an operator, L[y] ≡ ∂²y/∂x² − (1/v²) ∂²y/∂t², so that a wave is anything with L[y] = 0. Every term in L is first degree in y or in one of its derivatives, and differentiation is itself linear, so L[a y₁ + b y₂] = a L[y₁] + b L[y₂] for any constants a and b. If both waves solve the equation the right-hand side is zero, and the combination solves it too. Change one term and the property evaporates: add even a modest λy² and the combination picks up a cross term 2λ y₁ y₂ belonging to neither wave, so the sum is no longer a solution. Notice also what linearity does not demand. It says nothing about the shapes, frequencies or directions involved — a 40 Hz pulse and a 4 kHz sinusoid superpose exactly as readily as two identical crests — and it does not require the waves to be small, only that the equation governing them stay linear across the range they actually use.

Take a short segment of a taut string, of length Δx along x, carrying tension F and mass per unit length μ. The tension pulls along the string at each end, so the net transverse force is F(sin θ₂ − sin θ₁), with θ the local angle to the x axis. Exactly, sin θ = (∂y/∂x)/√(1 + (∂y/∂x)²); exactly, the segment's mass is μ√(1 + (∂y/∂x)²) Δx, because a displaced segment is longer than Δx; and exactly, that extra length raises F. Three separate nonlinearities, all three governed by the same dimensionless number, the slope ∂y/∂x. Impose (∂y/∂x)² ≪ 1 and all three corrections vanish together: the force becomes F (∂²y/∂x²) Δx, the mass becomes μ Δx, and Newton's second law gives F ∂²y/∂x² = μ ∂²y/∂t², with v = √(F/μ). The condition has a size. A wave whose steepest slope is 0.05 makes sin θ differ from ∂y/∂x by 0.12%; at a slope of 0.10 the error is 0.50%. Linearity on a string is not a yes-or-no property — it is an accuracy, and it degrades as the square of the slope.

Operationally the rule is narrow and precise: at a chosen point and a chosen instant, add the displacements the individual waves would produce there, carrying their signs. If one pulse would put a rope element at +12 mm and another at −4 mm, the element is at +8 mm. Every quantity depending linearly on y inherits the rule, because a linear operation applied to a sum returns the sum of the results: transverse velocity ∂y/∂t, slope ∂y/∂x, transverse force, and — for sound and light — acoustic pressure and electric field. Quantities built from squares inherit nothing. Kinetic energy per unit length is ½μ(∂y/∂t)², potential energy per unit length is ½F(∂y/∂x)², and intensity goes as amplitude squared, so none of them is the sum of the separate values. The same linearity licenses decomposition as well as addition. Because any solution can be written as a sum of sinusoids and each sinusoid then propagates independently at v, one column of air carries an orchestra without the cello smearing the piccolo, and a bounded medium can be analysed one normal mode at a time.

Send a pulse f(x − vt) rightward and an inverted copy −f(x + vt) leftward, timed to coincide at t = 0. Then y = f(x) − f(x) = 0 everywhere: the string is flat. It is not at rest. The wave equation is second order in time, so the state of the string is its displacement together with its transverse velocity, and ∂y/∂t = −v f′(x − vt) − v f′(x + vt) equals −2v f′(x) at that instant — largest exactly where the pulse was steepest. The whole energy is kinetic, and the flat string immediately rebuilds both pulses. Run the experiment with two identical upright pulses and the complement appears: at coincidence y = 2f(x) and ∂y/∂t = 0, a doubled crest momentarily at rest with all its energy potential. The accounting closes exactly. A travelling pulse splits its energy equally between kinetic and potential, so each carries 2U with U = ∫ ½F f′² dx. At full overlap the doubled crest holds ∫ ½F (2f′)² dx = 4U, and the flat moving string holds ∫ ½μ (2v f′)² dx = 4U as well, because μv² = F. Two pulses in, 4U throughout, two pulses out.

Sound compresses air far too fast for heat to flow, so the state follows P = P₀(1 + s)^γ, where s = δρ/ρ₀ is the condensation and γ = 1.40 for air. Expanding, P = P₀[1 + γs + ½γ(γ − 1)s² + …]. Keeping only the term linear in s gives the acoustic pressure p = γP₀ s and, with the continuity and momentum relations, a linear wave equation whose speed is c = √(γP₀/ρ₀). That deserves a correspondence check: √(1.40 × 1.013 × 10⁵ Pa ÷ 1.20 kg m⁻³) = 344 m s⁻¹, which is the measured speed of sound. Now size the discarded term. Ordinary speech at 60 dB carries an intensity of 1.0 × 10⁻⁶ W m⁻², a pressure amplitude of 0.029 Pa and a condensation of only 2.0 × 10⁻⁷. Even at 120 dB, the threshold of pain, the pressure amplitude is 28.7 Pa and s ≈ 2.0 × 10⁻⁴. The first discarded term is down on the linear one by ½(γ − 1)s ≈ 4 × 10⁻⁵, so linear acoustics holds to a few parts in a hundred thousand across the whole range of ordinary hearing.

Push the amplitude up and the speed itself begins to depend on the wave. For sound of finite amplitude the local propagation speed is c + βu, with u the signed particle velocity and β = (γ + 1)/2 = 1.20 in air. Compressions, which carry u forward, outrun rarefactions; the waveform leans forward and steepens, and the steepening breeds harmonics at 2f, 3f, … that were never launched. The distance over which a sinusoid steepens into a shock is x ≈ c²/(βωu₀). At 1.0 kHz and 120 dB, u₀ = 0.070 m s⁻¹ and x ≈ 2.2 × 10² m — far longer than any concert hall, which is why room acoustics is linear, and far shorter than the path of a sonic boom, which is why that is not. The test is sharp. A linear, time-independent medium can change a wave amplitude and its phase, but it can never place energy at a frequency absent from the input. Drive two tones f₁ and f₂ through a nonlinear element and new spectral lines at f₁ + f₂ and 2f₁ − f₂ appear, frequencies neither source carried. Measure one of those and you have measured the failure of superposition.