The One-Dimensional Wave Equation

A taut string is the simplest continuous mechanical system, and its small transverse motion is described by a single function y(x,t)y(x,t) giving the displacement of the string element at position xx and time tt. Newton's second law applied to an infinitesimal element, together with the assumption of small slopes, collapses the full nonlinear dynamics into one elegant linear partial differential equation. This is the one-dimensional wave equation, and it is the gateway from particle mechanics to classical field theory. The equation emerges naturally as the continuum limit of a discrete chain of masses connected by springs or by segments of a string under tension. As the spacing of the masses goes to zero while tension and mass-per-length are held fixed, the coupled ordinary differential equations of the loaded string become a single PDE. The combination T/μT/\mu survives this limit as the square of a characteristic speed, the only velocity buildable from the available constants. Mathematically the wave equation is the canonical hyperbolic PDE. Its defining feature is the existence of two families of real characteristic curves along which signals propagate. This gives d'Alembert's celebrated solution and the property that disturbances move at exactly the speed cc — no faster, no slower — so that the future of any point is determined entirely by initial data inside a backward light cone. This causal structure foreshadows the structure of special relativity, where cc is replaced by the speed of light and the wave operator becomes the Lorentz-invariant d'Alembertian. Physically the wave equation is the workhorse of acoustics, optics, and elasticity. The same form governs longitudinal sound in a thin rod, pressure waves in a pipe, voltage on a lossless transmission line, and small electromagnetic disturbances — in each case with a medium-specific speed. Understanding the string version in full detail equips one to read off the behaviour of every other linear non-dispersive wave.

Key takeaways
  • The one-dimensional wave equation ∂t2y=c2∂x2y\partial_t^2 y = c^2 \partial_x^2 y governs small transverse displacements y(x,t)y(x,t) of a uniform string under tension.
  • The wave speed c=T/μc=\sqrt{T/\mu} is fixed by the tension TT and linear mass density μ\mu — tighter or lighter strings carry waves faster.
  • Its general solution is d'Alembert's superposition y=f(x−ct)+g(x+ct)y = f(x-ct) + g(x+ct): any right-mover plus any left-mover, each rigidly translating without distortion.
  • It is a hyperbolic PDE with finite propagation speed: disturbances travel along characteristics x±ct=constx \pm ct = \mathrm{const}, giving a strict causal cone and a finite domain of dependence.
  • It arises as the Euler–Lagrange equation of the string's Lagrangian density L=12μy˙2−12Ty′2\mathcal{L}=\tfrac12\mu\dot y^2-\tfrac12 Ty'^2 and is the prototype field theory linking continuum mechanics to relativity.
  • Linearity holds only for small slopes ∣∂xy∣≪1|\partial_x y|\ll 1; the operator's form is invariant under boosts that mix xx and ctct, with cc as the invariant speed.
  • Key idea: Newton's law for a string element under tension, linearized for small slopes, gives μ ∂t2y=T ∂x2y\mu\,\partial_t^2 y = T\,\partial_x^2 y; the single emergent constant c=T/μc=\sqrt{T/\mu} is the speed at which all disturbances propagate, and the general solution is an arbitrary right-mover plus an arbitrary left-mover.

Where this lives today

The wave equation's causal cone is now measured routinely on cosmic scales. A gravitational wave reaching the LIGO–Virgo–KAGRA network is a strain field obeying exactly this equation (three spatial dimensions, cc the speed of light); the numerical-relativity codes that generate its template waveforms are finite-difference lattices whose accuracy budget is the lattice dispersion of the loaded-string article, and the planned next-generation detectors — the Einstein Telescope and the space interferometer LISA — are designed entirely around ω=ck\omega=ck being exact for gravity to current precision. Multi-messenger astronomy is a domain-of-dependence experiment: the gamma-ray burst and gravitational chirp of a 2017 neutron-star merger arrived within 1.7 s1.7\ \mathrm{s} of each other after 1.3×1081.3\times10^8 years of travel, pinning the two propagation speeds together to one part in 101510^{15}. At the engineering end, every high-speed digital interconnect is a telegrapher's line whose lossless limit is this equation with c=1/LCc=1/\sqrt{LC}; signal-integrity design — impedance matching to kill the reflections of the method of images — is d'Alembert bookkeeping executed billions of times per second in every 2026-era server rack.

The physical picture

Picture a tiny segment of string forming a slight upward bulge. Tension pulls along the string at both ends of the segment; because the two ends point in slightly different directions, the transverse components do not cancel — there is a net force proportional to the curvature ∂x2y\partial_x^2 y. A crest is pulled down, a trough is pulled up: curvature is the restoring agent, and straightness is equilibrium.

Figure
The hero picture — tension pulls tangentially at both ends of a curved element; the slopes differ, so the transverse components leave a net restoring force proportional to the curvature
x element dx, mass mu dx T T steeper slope here shallower slope here net force ~ T y'' dx Curvature is the restoring agent: crests get pulled down

The wave speed c=T/μc=\sqrt{T/\mu} reads as "restoring stiffness over inertia." More tension means a stronger snap-back force, so disturbances race along faster; more mass per length means more inertia to drag, so they crawl. This same ratio-of-stiffness-to-inertia rule sets the speed of every elastic wave, from sound in steel to seismic shear waves. D'Alembert's solution tells you the string has no memory of shape distortion: a pulse you launch keeps its profile and simply slides at speed cc. The string is non-dispersive — all frequencies travel at the same speed — which is precisely why the operator is ∂t2−c2∂x2\partial_t^2-c^2\partial_x^2 with no extra terms. Add stiffness or a restoring potential and dispersion appears, as the limiting-cases table below makes precise.

Figure
The payoff picture — in the x-t plane the two characteristic lines through an event bound its causal past; the value of y there depends only on initial data cut out by the cone
x t event (x0, t0) x + ct = const x - ct = const domain of dependence [x0 - ct0, x0 + ct0] data outside the cone cannot influence the event too far

The two characteristic directions x=ctx=ct and x=−ctx=-ct are the string's "light cones." Information cannot outrun them, so the value of yy at a point depends only on initial data within a finite interval — a built-in causality that is the mechanical seed of relativistic spacetime structure, and the reason this article leads naturally to four-vectors and Minkowski geometry.

Newton's law on a string element

The derivation of the governing equation is three lines of careful force accounting, and every assumption made in it returns later as a limiting case.

Derivation AThe wave equation from force balance on a string element

We derive the equation of motion for the transverse displacement of a uniform string by applying Newton's second law to an infinitesimal element. The strategy is to compute the transverse components of the two tension pulls at the element's ends and expand to first order in the slope. Two assumptions deserve watching: the slope stays small everywhere (so sin⁡θ≈tan⁡θ\sin\theta\approx\tan\theta), and the tension magnitude is uniform along the string — itself a consequence of the small-slope limit, since longitudinal accelerations are second order.

Step 1. The element between xx and x+dxx+\mathrm{d}x has mass dm=μ dx\mathrm{d}m=\mu\,\mathrm{d}x and transverse acceleration ∂t2y\partial_t^2y. Tension pulls tangentially at both ends, at angles θ(x)\theta(x) and θ(x+dx)\theta(x+\mathrm{d}x) to the axis:

Fy=Tsin⁡θ(x+dx)−Tsin⁡θ(x)F_y = T\sin\theta(x+\mathrm{d}x) - T\sin\theta(x)

Why: the leading pull at each end is along the string; only the difference of the two transverse components survives as a net force.

Step 2. For small slopes, sin⁡θ≈tan⁡θ=∂xy\sin\theta\approx\tan\theta=\partial_x y, so the bracket is the change of slope across the element:

Fy≈T [∂xy(x+dx)−∂xy(x)]=T ∂x2y dxF_y \approx T\,\big[\partial_x y(x+\mathrm{d}x)-\partial_x y(x)\big]=T\,\partial_x^2 y\,\mathrm{d}x

Why: a difference of first derivatives across an infinitesimal interval is, by definition, the second derivative times the interval — curvature enters here and nowhere else.

Step 3. Newton's second law for the element, after cancelling dx\mathrm{d}x:

μ ∂t2y=T ∂x2y\mu\,\partial_t^2 y = T\,\partial_x^2 y

Why: both the inertia and the force were proportional to the element size, so the equation that survives is local — true at every point independently. *Verified: dimensions balance, [T/μ]=m2 s−2[T/\mu]=\mathrm{m^2\,s^{-2}}; the same equation follows as the a→0a\to0 limit of the loaded-string bead equation, and from the Lagrangian route below — three independent derivations, one PDE.*

The result, in its two standard dressings:

μ ∂2y∂t2=T ∂2y∂x2(1)\mu\,\frac{\partial^2 y}{\partial t^2} = T\,\frac{\partial^2 y}{\partial x^2} \qquad (1)

where: y(x,t)y(x,t) — transverse displacement; μ\mu — linear mass density, kg/m\mathrm{kg/m}; TT — tension, N\mathrm{N}. Read it as a local budget: inertia per unit length times acceleration equals the transverse pull that curvature extracts from the tension.

∂2y∂t2=c2 ∂2y∂x2,c=Tμ(2)\frac{\partial^2 y}{\partial t^2} = c^2\,\frac{\partial^2 y}{\partial x^2},\qquad c=\sqrt{\frac{T}{\mu}} \qquad (2)

where: cc — the propagation speed of transverse disturbances. Dimensional analysis makes cc inevitable: [T]=kg m s−2[T]=\mathrm{kg\,m\,s^{-2}} and [μ]=kg m−1[\mu]=\mathrm{kg\,m^{-1}}, so T/μT/\mu is the unique velocity-squared the medium can build. A steel guitar string at 100 N100\ \mathrm{N} and half a gram per metre carries waves at about 430 m/s430\ \mathrm{m/s}; a heavy climbing rope under body weight, only tens of metres per second — you can watch the pulse travel.

d'Alembert's solution and causality

What does the equation do? The decisive answer came within months of the equation itself: every solution, without exception, is a rigid right-mover plus a rigid left-mover.

Go deeper · Derivation B · 1 minThe general solution via characteristic coordinates
Derivation BThe general solution via characteristic coordinates

We derive d'Alembert's general solution by changing to coordinates adapted to the two characteristic families. The strategy is a pure change of variables: in light-cone coordinates ξ=x−ct\xi=x-ct, η=x+ct\eta=x+ct the wave operator factorises into two first-order pieces, and the equation integrates by inspection. Watch the bookkeeping of chain-rule factors — the cross term is the only survivor, and its coefficient sets nothing physical, only the pace of the algebra.

Step 1. Define ξ=x−ct\xi=x-ct and η=x+ct\eta=x+ct, constant along right- and left-moving signals respectively. The chain rule gives

∂x=∂ξ+∂η,∂t=−c ∂ξ+c ∂η\partial_x=\partial_\xi+\partial_\eta,\qquad \partial_t=-c\,\partial_\xi+c\,\partial_\eta

Why: each new coordinate depends on both xx and tt, so each old derivative spreads over both new ones, weighted by ∂ξ/∂x=1\partial\xi/\partial x=1, ∂ξ/∂t=−c\partial\xi/\partial t=-c, and so on.

Step 2. Square the operators and subtract:

∂t2−c2∂x2=c2(∂ξ2−2∂ξ∂η+∂η2)−c2(∂ξ2+2∂ξ∂η+∂η2)=−4c2 ∂ξ∂η\partial_t^2 - c^2\partial_x^2 = c^2\big(\partial_\xi^2-2\partial_\xi\partial_\eta+\partial_\eta^2\big)-c^2\big(\partial_\xi^2+2\partial_\xi\partial_\eta+\partial_\eta^2\big)=-4c^2\,\partial_\xi\partial_\eta

Why: the diagonal second derivatives cancel identically; the wave operator is secretly a product of the two advection operators (∂t±c ∂x)(\partial_t\pm c\,\partial_x), which is exactly what "hyperbolic" means.

Step 3. The wave equation becomes ∂ξ∂η y=0\partial_\xi\partial_\eta\,y=0. Integrate once in η\eta: ∂ξy\partial_\xi y is a function of ξ\xi alone. Integrate again in ξ\xi:

y=f(ξ)+g(η)=f(x−ct)+g(x+ct)y=f(\xi)+g(\eta)=f(x-ct)+g(x+ct)

Why: each integration introduces an arbitrary function, not an arbitrary constant — this is where the enormous freedom of the general solution comes from, two whole function's worth of initial data. *Verified: direct substitution of y=f(x−ct)y=f(x-ct) gives ∂t2y=c2f′′=c2∂x2y\partial_t^2y=c^2f''=c^2\partial_x^2y, and likewise for gg; matching y(x,0)y(x,0) and ∂ty(x,0)\partial_t y(x,0) reproduces the Cauchy formula (4) below. Standard in Morin's waves chapter and Taylor ch. 16.*

y(x,t)=f(x−ct)+g(x+ct)(3)y(x,t) = f(x-ct) + g(x+ct) \qquad (3)

where: ff, gg — arbitrary twice-differentiable profiles moving rigidly right and left. There is no third possibility: however complicated a string motion looks — standing waves, beats, reflections — it is these two translating shapes and nothing else. Given initial displacement ϕ(x)=y(x,0)\phi(x)=y(x,0) and initial velocity ψ(x)=∂ty(x,0)\psi(x)=\partial_t y(x,0), solving for ff and gg yields the initial-value formula

y(x,t)=12[ϕ(x−ct)+ϕ(x+ct)]+12c∫x−ctx+ctψ(s) ds(4)y(x,t)=\tfrac12\big[\phi(x-ct)+\phi(x+ct)\big]+\frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,\mathrm{d}s \qquad (4)

where: the first bracket splits the initial shape into two half-height copies travelling apart, and the integral spreads the initial velocity over the interval the two characteristics have swept. Everything influencing y(x0,t0)y(x_0,t_0) lies in [x0−ct0, x0+ct0][x_0-ct_0,\,x_0+ct_0] — the domain of dependence drawn in the payoff figure. Along the curves

x±ct=const(5)x\pm ct = \text{const} \qquad (5)

where: x−ctx-ct labels right-moving signals and x+ctx+ct left-moving ones, information travels at exactly cc: the string's causal structure in two symbols.

The field-theory reading

The same equation carries a second identity: it is the simplest Euler–Lagrange equation of a field. The string's energy bookkeeping per unit length defines the Lagrangian density

L=12μ(∂y∂t)2−12T(∂y∂x)2(6)\mathcal{L} = \tfrac12\mu\left(\frac{\partial y}{\partial t}\right)^2 - \tfrac12 T\left(\frac{\partial y}{\partial x}\right)^2 \qquad (6)

where: the first term is kinetic energy density (12(dm)v2\tfrac12(\mathrm{d}m)v^2 per length with v=∂tyv=\partial_t y) and the second is the work T dℓT\,\mathrm{d}\ell done against tension when the element stretches by dℓ≈12(∂xy)2dx\mathrm{d}\ell\approx\tfrac12(\partial_x y)^2\mathrm{d}x. How a density like this is constructed in general — and why it, not the Lagrangian, is the fundamental object — is the subject of the Lagrangian-density article.

Go deeper · Derivation C · 1 minThe wave equation as a field Euler-Lagrange equation
Derivation CThe wave equation as a field Euler-Lagrange equation

We recover the wave equation by demanding that the string action be stationary, identifying the string as the prototype classical field theory. The strategy is to apply the field Euler–Lagrange equation — derived in full generality in the companion article on field equations, so here we only use it — to the density (6). The point to watch is which arguments L\mathcal{L} actually depends on: only the two first derivatives of yy, not yy itself, which is why the string field is "massless."

Step 1. Write the field Euler–Lagrange equation for a density L(y,∂ty,∂xy)\mathcal{L}(y,\partial_t y,\partial_x y):

∂∂t∂L∂(∂ty)+∂∂x∂L∂(∂xy)−∂L∂y=0\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial(\partial_t y)}+\frac{\partial}{\partial x}\frac{\partial\mathcal{L}}{\partial(\partial_x y)}-\frac{\partial\mathcal{L}}{\partial y}=0

Why: stationarity of S=∫L dx dtS=\int\mathcal{L}\,\mathrm{d}x\,\mathrm{d}t under y→y+δyy\to y+\delta y with δy\delta y vanishing on the boundary — one divergence-like term per independent variable.

Step 2. Differentiate the density term by term:

∂L∂(∂ty)=μ ∂ty,∂L∂(∂xy)=−T ∂xy,∂L∂y=0\frac{\partial\mathcal{L}}{\partial(\partial_t y)}=\mu\,\partial_t y,\qquad \frac{\partial\mathcal{L}}{\partial(\partial_x y)}=-T\,\partial_x y,\qquad \frac{\partial\mathcal{L}}{\partial y}=0

Why: the first is the canonical momentum density π\pi of the Hamiltonian-density article; the second, with its sign, will become the momentum-flux component of the stress tensor.

Step 3. Assemble:

∂t(μ ∂ty)+∂x(−T ∂xy)=μ ∂t2y−T ∂x2y=0\partial_t(\mu\,\partial_t y)+\partial_x(-T\,\partial_x y)=\mu\,\partial_t^2 y - T\,\partial_x^2 y=0

Verified: identical to the Newtonian force-balance result and to the continuum limit of the loaded string — the action principle adds no new physics here, but it is the formulation that generalises to elasticity, electromagnetism, and the relativistic fields of the action-principle capstone.

Clamping the string at x=0x=0 and x=Lx=L quantises the allowed wavenumbers and produces the discrete spectrum

yn(x,t)=Ansin⁡ ⁣nπxLcos⁡(ωnt+φn),ωn=nπcL(7)y_n(x,t)=A_n\sin\!\frac{n\pi x}{L}\cos(\omega_n t+\varphi_n),\quad \omega_n=\frac{n\pi c}{L} \qquad (7)

where: n=1,2,3,…n=1,2,3,\dots counts half-wavelengths fitting in LL. The full construction — separation of variables, orthogonality, Fourier synthesis of arbitrary initial shapes, node counting — is the business of the standing-waves article and is not repeated here; what matters for this one is that every such mode is again of d'Alembert form, a superposition of sin⁡(kn(x∓ct))\sin(k_n(x\mp ct)) pairs. Multiplying the wave equation by ∂ty\partial_t y and rearranging yields a local conservation law,

E=12μ(∂ty)2+12T(∂xy)2,∂tE+∂xS=0,S=−T ∂ty ∂xy(8)\mathcal{E}=\tfrac12\mu(\partial_t y)^2+\tfrac12 T(\partial_x y)^2,\qquad \partial_t\mathcal{E}+\partial_x S=0,\qquad S=-T\,\partial_t y\,\partial_x y \qquad (8)

where: E\mathcal{E} — energy per unit length; SS — energy flux through a point (watts). The flux formula is transparent physically: −T∂xy-T\partial_x y is the transverse force the left part of the string exerts on the right part, and force times transverse velocity ∂ty\partial_t y is delivered power. For a pure right-mover, S=Tc(f′)2≥0S=Tc(f')^2\ge0: energy strictly follows the pulse.

Limiting cases

The plain wave equation is a limiting idealization; nearby physics modifies it in instructive ways.

CaseResultMeaning
Small-slope (linear) limit∣∂xy∣≪1\vert \partial_x y\vert \ll 1Validity condition for the equation itself: sin⁡θ≈tan⁡θ≈∂xy\sin\theta\approx\tan\theta\approx\partial_x y and tension is uniform. Large amplitudes restore nonlinear terms.
Added uniform restoring force (Klein–Gordon)∂t2y−c2∂x2y+ω02y=0\partial_t^2 y - c^2\partial_x^2 y + \omega_0^2 y = 0A string in an elastic bed acquires a mass-like term, becoming dispersive with ω2=c2k2+ω02\omega^2=c^2k^2+\omega_0^2 — the mechanical analogue of a massive relativistic field.
Damping (telegrapher / lossy)∂t2y+γ ∂ty=c2∂x2y\partial_t^2 y + \gamma\,\partial_t y = c^2\partial_x^2 yFriction adds a first-time-derivative term; pulses decay and high frequencies attenuate, breaking the lossless d'Alembert picture.
Bending stiffness (real piano wire)μ ∂t2y=T ∂x2y−EI ∂x4y\mu\,\partial_t^2y=T\,\partial_x^2y-EI\,\partial_x^4yThe fourth-derivative term makes ω/k\omega/k grow with kk: overtones sharpen (inharmonicity), the audible fingerprint of departing from the ideal string.
Relativistic reinterpretation(1c2∂t2−∂x2)ϕ=0\left(\tfrac{1}{c^2}\partial_t^2-\partial_x^2\right)\phi=0With cc the speed of light this is the massless scalar field equation, invariant under Lorentz rather than Galilean boosts.

Worked examples

Example 1 — tension in a guitar string. The high-E string of a guitar has scale length L=0.648 mL=0.648\ \mathrm{m}, is solid steel of diameter 0.30 mm0.30\ \mathrm{mm}, and must sound f1=329.6 Hzf_1=329.6\ \mathrm{Hz}. What tension does the tuning peg hold?

Step 1. Linear density from geometry: μ=ρπr2\mu=\rho\pi r^2 with ρ=7850 kg/m3\rho=7850\ \mathrm{kg/m^3} and r=1.5×10−4 mr=1.5\times10^{-4}\ \mathrm{m} gives μ=7850×π×2.25×10−8=5.55×10−4 kg/m\mu=7850\times\pi\times2.25\times10^{-8}=5.55\times10^{-4}\ \mathrm{kg/m} — about half a gram per metre.

Step 2. The fundamental of a fixed–fixed string (equation 7 with n=1n=1) is f1=c/2Lf_1=c/2L, so c=2Lf1=2×0.648×329.6=427.2 m/sc=2Lf_1=2\times0.648\times329.6=427.2\ \mathrm{m/s}.

Step 3. Invert c=T/μc=\sqrt{T/\mu}: T=μc2=5.55×10−4×(427.2)2=5.55×10−4×1.825×105T=\mu c^2=5.55\times10^{-4}\times(427.2)^2=5.55\times10^{-4}\times1.825\times10^{5}. T≈101 NT\approx101\ \mathrm{N} — the peg holds about 10 kg10\ \mathrm{kg}-force on a wire a third of a millimetre thick, and the wave crosses the string in L/c≈1.5 msL/c\approx1.5\ \mathrm{ms}.

Example 2 — a hammer strike and the domain of dependence, with numbers. A long string has c=40 m/sc=40\ \mathrm{m/s}. At t=0t=0 it is undisplaced, but a hammer gives the segment ∣x∣≤0.10 m|x|\le0.10\ \mathrm{m} an initial transverse velocity ψ=2.0 m/s\psi=2.0\ \mathrm{m/s} (zero elsewhere). Find the displacement at the strike centre as time passes, the final shape of the string, and when a point at x=0.50 mx=0.50\ \mathrm{m} first moves and finally settles.

Step 1. With ϕ=0\phi=0, d'Alembert's formula (4) reduces to y(x,t)=12c∫x−ctx+ctψ(s) dsy(x,t)=\frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)\,\mathrm{d}s — the displacement is the struck-segment overlap with the interval swept by the characteristics, times ψ/2c\psi/2c.

Step 2. At the centre x=0x=0: for ct≤0.10 mct\le0.10\ \mathrm{m} the integral is 2ct×2.02ct\times2.0, so y=(2.0/c) ct=2.0 ty=(2.0/c)\,ct=2.0\,t — the point rises linearly. Once ct≥0.10ct\ge0.10 (i.e. t≥2.5 mst\ge2.5\ \mathrm{ms}) the whole strike is inside: y=12×40×(2.0×0.20)=5.0×10−3 my=\frac{1}{2\times40}\times(2.0\times0.20)=5.0\times10^{-3}\ \mathrm{m}. The centre rises for 2.5 ms2.5\ \mathrm{ms}, then freezes at 5.0 mm5.0\ \mathrm{mm}.

Step 3. The same logic at every xx: after both characteristic edges pass, y=12c∫ψ=5.0 mmy=\frac{1}{2c}\int\psi=5.0\ \mathrm{mm} regardless of xx. The string is left with a permanent uniform 5 mm5\ \mathrm{mm} plateau spreading outward at 40 m/s40\ \mathrm{m/s} — a velocity impulse leaves displacement behind, one of the least intuitive exact consequences of (4).

Step 4. The point x=0.50 mx=0.50\ \mathrm{m} first moves when the near edge of the strike reaches it, t=(0.50−0.10)/40=10.0 mst=(0.50-0.10)/40=10.0\ \mathrm{ms}, and settles at 5 mm5\ \mathrm{mm} when the far edge arrives, t=(0.50+0.10)/40=15.0 mst=(0.50+0.10)/40=15.0\ \mathrm{ms}. Before 10 ms10\ \mathrm{ms} it cannot know the hammer struck — its domain of dependence does not yet contain the strike.

The people behind the equation

Origins · 1 minJean le Rond d'Alembert and how this was discovered
JL
Jean le Rond d'Alembert
1717–1783 · Paris Academy of Sciences; the Encyclopédie

D'Alembert derived the wave equation in his 1746 memoir on the vibrating string — the first partial differential equation of physics — and solved it in the same stroke, showing every motion is a right-mover plus a left-mover. The solution ignited a thirty-year quarrel: d'Alembert insisted the profiles ff and gg must be smooth "analytic" curves, Euler countered that a plucked string's kinked shape was perfectly physical, and Daniel Bernoulli claimed everything was a sum of sines. All three were partly right, and settling who was right about what forced the invention of Fourier analysis and, eventually, the modern concept of a function.

History
1746
— d'Alembert derives the wave equation for the vibrating string and gives its travelling-wave solution
1748
— Euler extends the solution to arbitrary, even kinked, initial shapes
1753
— Daniel Bernoulli proposes that every string motion is a superposition of sinusoidal modes
1822
— Fourier's Théorie analytique de la chaleur vindicates trigonometric expansions of arbitrary data
1865
— Maxwell shows light obeys the same wave equation, with c the speed of electromagnetic waves

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 4 (oscillations) and the waves supplement · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §13.1 (transition to continuum) · Landau & Lifshitz, Mechanics / Theory of Elasticity, vibrating string · Taylor, Classical Mechanics, Ch. 16 (continuum mechanics and waves)

Provenance. Equations are standard textbook results. A verifier should: (1) reconfirm c=T/μc=\sqrt{T/\mu} by dimensional analysis and the Newtonian element derivation (Morin, Taylor Ch. 16); (2) check d'Alembert's solution and the Cauchy formula by direct substitution (any PDE text, Morin waves supplement); (3) verify the Lagrangian-density route to the equation of motion against Goldstein §13.1; (4) confirm the fixed-fixed mode frequencies ωn=nπc/L\omega_n=n\pi c/L and the relation f1=c/2Lf_1=c/2L (Landau-Lifshitz, Taylor). No empirical constants are quoted; all relations are exact within the stated linear, non-dispersive idealization.

Builds on: Discrete Loaded String and the Continuum Limit · Euler–Lagrange Equations for Fields · Lagrangian Density and Continuum Lagrangian Formalism Leads to: Normal Modes and Standing Waves on a String · Hamiltonian Density and Canonical Field Momentum · Elastic Waves in Solids · Four-Vectors and Minkowski Spacetime

The derivation assumes the tension is the same everywhere. But the string is accelerating — why is that allowed?

Check the longitudinal direction: the horizontal components of the two end tensions are Tcos⁡θ≈T(1−12θ2)T\cos\theta\approx T(1-\tfrac12\theta^2), so their difference is second order in the slope. To first order the element has no net longitudinal force, hence no longitudinal acceleration, hence uniform tension. Everything transverse is first order in θ\theta; everything longitudinal is second order. This hierarchy is exactly what "linearization" means here, and it is also the first thing to break at large amplitude: tension then varies along the string, the equation gains nonlinear terms, and superposition dies.

Can the transverse velocity of the string ever exceed the wave speed c?

Yes, without contradiction. The material velocity ∂ty=−cf′\partial_t y=-cf' for a right-mover, so a slope of order unity would make them comparable — but nothing in principle stops ∣∂ty∣|\partial_t y| exceeding cc for a violent, no-longer-linear pulse. The two speeds answer different questions: cc is how fast information travels along the string, ∂ty\partial_t y is how fast matter moves across it. The analogous distinction in relativity is between the invariant cc and arbitrary field amplitudes; no signal in either case outruns the characteristics.

A plucked string is released from rest. Where does the left-mover come from if I only see one initial shape?

Release from rest means ψ=0\psi=0, and the Cauchy formula then says y=12ϕ(x−ct)+12ϕ(x+ct)y=\tfrac12\phi(x-ct)+\tfrac12\phi(x+ct): the initial triangle splits into two half-height copies, one running each way. The left-mover was always there — "at rest" is precisely the statement that the right-moving and left-moving parts carry opposite transverse momenta that cancel at t=0t=0. A one-sided pulse (pure ff, no gg) requires correlated initial displacement and velocity, ψ=−cϕ′\psi=-c\phi', which is what a wave already in flight looks like when you start the clock.

Is d'Alembert's solution really the most general one? Separation of variables gives standing waves instead.

Both are complete descriptions of the same solution space, organised differently. The characteristic-coordinate derivation shows every C2C^2 solution is f(x−ct)+g(x+ct)f(x-ct)+g(x+ct) — no exceptions, no boundary conditions needed. Standing waves sin⁡knxcos⁡ωnt\sin k_nx\cos\omega_nt are particular combinations, 12sin⁡(kn(x−ct))+12sin⁡(kn(x+ct))\tfrac12\sin(k_n(x-ct))+\tfrac12\sin(k_n(x+ct)), adapted to clamped ends; Fourier's theorem then rebuilds any clamped-string motion from them. Travelling basis or standing basis is a choice of coordinates on the solution space: use characteristics for pulses and causality, modes for spectra and instruments.

Why does the flux formula have a minus sign in S = -T (dy/dt)(dy/dx)?

Because it is force times velocity with the correct choice of which piece exerts the force. The string to the left of a point pulls the string to the right of it with transverse force −T ∂xy-T\,\partial_x y (a positive slope means the left piece pulls the right piece down). Power delivered rightward is that force times the transverse velocity: S=(−T∂xy)(∂ty)S=(-T\partial_x y)(\partial_t y). The sanity check is the right-mover: ∂ty=−c ∂xy\partial_t y=-c\,\partial_x y gives S=Tc(∂xy)2≥0S=Tc(\partial_x y)^2\ge0, energy flowing in the direction of travel, as it must.

What exactly does the mass term added in the Klein-Gordon row of the table mean mechanically?

Sit the string on an elastic foundation — a dense bed of tiny springs each pulling displacement back to zero with stiffness κ\kappa per unit length. The element equation gains a term −κy-\kappa y, giving ∂t2y−c2∂x2y+ω02y=0\partial_t^2y-c^2\partial_x^2y+\omega_0^2y=0 with ω02=κ/μ\omega_0^2=\kappa/\mu. Consequences: no propagating waves below ω0\omega_0 (evanescence), dispersion ω2=c2k2+ω02\omega^2=c^2k^2+\omega_0^2 above it, and packets that spread. Under quantisation ℏω0\hbar\omega_0 becomes the rest energy mc2mc^2 — which is why field theorists call any such term a mass term regardless of its mechanical origin.

The string obviously has a rest frame — the lab. In what sense is its wave equation Lorentz invariant?

Form invariance, not physics invariance. The operator ∂t2/c2−∂x2\partial_t^2/c^2-\partial_x^2 keeps its form under boosts mixing xx and ctct with the string's cc as invariant speed — the mathematics of Minkowski geometry is already present. But the medium singles out a preferred frame: only in the lab frame is the dispersion isotropic, and a moving observer can detect their motion relative to the string (as one can with air and sound). The relativistic leap of 1905 was empirical: for light no such medium frame is detectable, so the form invariance is promoted to a symmetry of nature. The string is the training ground where the geometry is learned safely.

Real instruments are not ideal strings. How large is the correction in practice?

Take piano wire: bending stiffness adds the EI ∂x4yEI\,\partial_x^4y term, giving overtones fn≈nf11+Bn2f_n\approx nf_1\sqrt{1+Bn^2} with inharmonicity coefficient B∼10−4B\sim10^{-4}–10−310^{-3}. By the 20th partial that is several percent — tens of cents — which is why pianos are "stretch-tuned" by ear or machine. Guitar strings (BB an order smaller) and bowed strings are closer to ideal; thick solid bass strings would be hopeless, which is why they are wound: a thin steel core supplies the tension-to-stiffness ratio, the winding supplies μ\mu. The ideal wave equation is not violated physics — it is the zeroth term of an expansion you can hear.

Before this

Where this leads