The Wave Equation as a Continuum Limit

Derivation

The Wave Equation as a Continuum Limit

D-064 Home PU-103 Threads waves · force · symmetry Depends on Normal Coordinates for N Coupled Oscillators, Coupled Oscillators and Normal Modes

Statement

A one-dimensional chain of identical beads of mass \(m\), spaced by \(a\) on a light string of constant tension \(T\) and displaced transversely by small amounts \(y_n(t)\), obeys the discrete equations of motion \( m\,\ddot{y}_n = \dfrac{T}{a}\left(y_{n+1}-2y_n+y_{n-1}\right)\). In the limit \(a\to 0\) taken with the linear mass density \(\mu = m/a\) held fixed, the discrete second difference becomes a second spatial derivative and the chain obeys the one-dimensional wave equation \(\partial_t^2 y = c^2\,\partial_x^2 y\) with a single, dispersionless propagation speed \(c=\sqrt{T/\mu}\).

Why it matters

This is the birth of a classical field. A finite mechanical system with \(N\) coupled coordinates — the very object the normal-mode machinery diagonalises — becomes, in the continuum limit, a smooth medium whose disturbance obeys a partial differential equation. Every string, acoustic column, elastic rod and transmission line ultimately reduces to this passage from "many masses on springs" to "a field \(y(x,t)\)".

It also fixes, from mechanics alone, what "wave speed" means: not a property of the source or the launched pulse, but the ratio of a restoring stiffness (tension) to an inertia (mass per length). The same skeleton \(c^2=\text{restoring}/\text{inertia}\) recurs for sound (bulk modulus over density), for light (\(1/\mu_0\varepsilon_0\)), and for relativistic fields, so this simplest case is the template for all of them.

Assumptions

Transverse displacements are small, slopes \(\partial_x y\ll 1\).The vertical force is linearised as \(T\sin\theta\approx T\tan\theta = T\,\partial_x y\). Drop it and the restoring force is no longer linear in \(y\); the tension itself varies with stretch, giving a nonlinear wave equation with amplitude-dependent speed and shock formation.

The tension \(T\) is uniform along the string and unchanged by the motion.If the string stretches appreciably, \(T\) becomes a function of \(x\) and of the local strain; the coefficient of \(\partial_x^2 y\) is no longer constant and \(c\) is not a single number.

The chain is uniform: all masses \(m\) and spacings \(a\) are equal.A two-mass (diatomic) chain splits the spectrum into acoustic and optical branches; a graded chain gives a position-dependent \(\mu(x)\). Either way there is no single \(c=\sqrt{T/\mu}\).

Only tension restores the beads — no bending stiffness, gravity sag, or friction.Restore bending rigidity and a term \(\propto\partial_x^4 y\) appears (a stiff bar or real piano string), adding dispersion and inharmonic overtones; gravity adds a static sag the wave rides on; friction adds a \(\partial_t y\) damping term.

The limit is taken with \(\mu=m/a\) fixed as \(a\to0\), and the field is smooth on the scale of \(a\) (wavelength \(\lambda\gg a\)).If instead \(m\) is held fixed the mass per length diverges and \(c\to0\); and if \(\lambda\sim a\) the Taylor expansion of \(y_{n\pm1}\) is not dominated by its first retained term — the discrete lattice dispersion survives and the wave equation is the wrong description.

Derivation

\[ F_n^{\perp} = T\sin\theta_{\mathrm{R}} - T\sin\theta_{\mathrm{L}} \approx \frac{T}{a}\left(y_{n+1}-y_n\right) - \frac{T}{a}\left(y_n-y_{n-1}\right) \]

Transverse force on bead \(n\): the string segments to right and left each pull with tension \(T\) along their length, and the vertical component of each is \(T\times(\text{slope})\). The small-slope assumption replaces \(\sin\theta\) by the discrete slope \(\Delta y/a\). A

\[ m\,\ddot{y}_n = \frac{T}{a}\left(y_{n+1}-2y_n+y_{n-1}\right) \]

Newton's second law \(m\ddot{y}_n = F_n^{\perp}\), collecting the three terms. This is exactly the uniform coupled-oscillator chain (prior result: coupled oscillators / normal modes) with effective spring constant \(T/a\) between neighbours; the right side is the discrete second difference of \(y\). A

\[ y_n(t)\ \longrightarrow\ y(x,t)\big|_{x=na},\qquad \mu \equiv \frac{m}{a} \]

Introduce a smooth field \(y(x,t)\) that interpolates the bead displacements, sampled at the lattice sites \(x=na\), and define the linear mass density \(\mu=m/a\), the intensive quantity to be held fixed. Legal because the displacements are assumed to vary slowly from bead to bead. B

\[ y_{n\pm1}=y(x\pm a,t)=y\pm a\,\partial_x y+\frac{a^2}{2}\partial_x^2 y\pm\frac{a^3}{6}\partial_x^3 y+\frac{a^4}{24}\partial_x^4 y+\cdots \]

Taylor-expand the neighbouring values about \(x\). Requires \(y\) to be at least four-times differentiable in \(x\) — this is where "smooth on the scale of \(a\)" is used, and the whole limit rests on it. C

\[ y_{n+1}-2y_n+y_{n-1}=a^2\,\partial_x^2 y+\frac{a^4}{12}\,\partial_x^4 y+O(a^6) \]

Add the two expansions from step 4. The odd-order derivatives cancel by symmetry (the \(\pm\) terms are equal and opposite); the leading survivor is \(a^2\partial_x^2 y\), and the first correction is the \(a^4\) term. This is the standard second-difference-to-second-derivative identity. C

\[ \mu\,\partial_t^2 y = T\,\partial_x^2 y + T\,\frac{a^2}{12}\,\partial_x^4 y + \cdots \]

Insert step 5 into step 2, then divide both sides by \(a\). The prefactor \(m/a\) on the left becomes exactly \(\mu\); the leading term on the right is \(T\,\partial_x^2 y\), with the correction now carrying an explicit factor \(a^2\). B

\[ \lim_{a\to0}\ T\,\frac{a^2}{12}\,\partial_x^4 y = 0 \quad\Longrightarrow\quad \mu\,\partial_t^2 y = T\,\partial_x^2 y \]

Hold \(\mu\) and \(T\) fixed and send \(a\to0\). Because \(\partial_x^4 y\) stays bounded for a smooth field, every correction carries a positive power of \(a\) and vanishes, leaving a closed equation in the field alone. This is the actual continuum limit; it is legitimate precisely when \(\lambda\gg a\) so that \(a^2\partial_x^4 y\sim(a/\lambda)^2\partial_x^2 y\ll\partial_x^2 y\). C

\[ \partial_t^2 y = \frac{T}{\mu}\,\partial_x^2 y \equiv c^2\,\partial_x^2 y,\qquad c\equiv\sqrt{\frac{T}{\mu}} \]

Divide by \(\mu\) and name the necessarily-positive coefficient \(c^2\). Since \(T>0\) and \(\mu>0\), \(c\) is real. Symbolic answer complete; no numbers have entered. A

\[ y_n=A\,e^{i(kna-\omega t)}\ \text{in step 2}\ \Longrightarrow\ \omega^2=\frac{4T}{ma}\sin^2\!\frac{ka}{2};\quad ka\ll1\Rightarrow\omega\approx ck \]

Consistency check via the exact discrete dispersion relation. Substituting a plane wave into the difference equation gives \(-m\omega^2=(T/a)(2\cos ka-2)=-(4T/a)\sin^2(ka/2)\). For long waves \(\sin(ka/2)\to ka/2\), so \(\omega\to\sqrt{Ta/m}\,k=\sqrt{T/\mu}\,k=ck\) — linear and dispersionless, confirming \(c=\sqrt{T/\mu}\). C

Result

\[ \boxed{\ \partial_t^2 y = c^2\,\partial_x^2 y\,,\qquad c=\sqrt{\frac{T}{\mu}}\ } \]

Reading. Every transverse disturbance of a light, uniformly tensioned string travels without change of shape at one speed \(c\), set only by the tension pulling it back and the mass it must accelerate. The general solution is d'Alembert's \(y=f(x-ct)+g(x+ct)\): a right-mover plus a left-mover, each rigid. Tighten the string and waves speed up (\(\propto\sqrt{T}\)); load it more heavily and they slow down (\(\propto 1/\sqrt{\mu}\)). No dispersion means every wavelength travels alike, so a pulse keeps its profile.

Units check. \([T]=\mathrm{N}=\mathrm{kg\,m\,s^{-2}}\) and \([\mu]=\mathrm{kg\,m^{-1}}\), so \([T/\mu]=\mathrm{m^2\,s^{-2}}\) and \([c]=\mathrm{m\,s^{-1}}\), a speed. In the equation, \([\partial_t^2 y]=\mathrm{m\,s^{-2}}\) and \([c^2\partial_x^2 y]=(\mathrm{m^2\,s^{-2}})(\mathrm{m\cdot m^{-2}})=\mathrm{m\,s^{-2}}\). Consistent.

Limiting cases

Long wavelength (\(\lambda\gg a\)). The regime in which the derivation is exact: \(\omega=ck\) to better than \((ka)^2/24\) in fractional speed; the string behaves as a perfect continuum.
Zone boundary (\(ka\to\pi\), i.e. \(\lambda\to2a\)). The exact \(\omega\to2\sqrt{T/(ma)}\) flattens; the group velocity \(\mathrm{d}\omega/\mathrm{d}k\to0\), a standing pattern of alternating beads. The continuum wave equation misses this entirely.
Zero tension (\(T\to0\)). \(c\to0\): with nothing to restore them the beads do not communicate transversely, and disturbances do not propagate.
Fixed ratio. Only \(T/\mu\) matters, so a light taut string and a heavy slack one can carry waves at identical speed.
Three-dimensional generalisation. Repeating the argument on a cubic lattice gives \(\partial_t^2 y = c^2\nabla^2 y\), the same speed with the Laplacian replacing \(\partial_x^2\).

Breaks when

Short wavelengths, \(\lambda\lesssim a\). Near the zone edge \(k\to\pi/a\) the exact dispersion \(\omega=\tfrac{2c}{a}|\sin(ka/2)|\) saturates at \(\omega_{\max}=2c/a\); the group velocity \(\to0\). The wave equation, which predicts unbounded \(\omega=ck\), fails and the discreteness is physical (the acoustic-phonon branch).
Large amplitude, \(|\partial_x y|\sim1\). The step \(\sin\theta\approx\tan\theta\) breaks; the true equation acquires nonlinear terms, \(c\) becomes amplitude-dependent, and harmonic generation, self-steepening, and shocks appear that the linear equation cannot describe.
Non-uniform or displacement-dependent tension. If stretching changes \(T\), or \(T=T(x)\), the coefficient \(c^2=T/\mu\) is no longer constant; waves refract and reflect off gradients and no single speed exists.
Bending stiffness or damping present. A real stiff string adds \(\propto\partial_x^4 y\) (dispersion, inharmonicity); dissipation adds \(\propto\partial_t y\) and free waves decay — the lossless, non-dispersive wave equation no longer holds.

Failure modes

Using \(m\) instead of \(\mu\). Writing \(c=\sqrt{T/m}\) is dimensionally wrong: \(\sqrt{\mathrm{N/kg}}=\sqrt{\mathrm{m/s^2}}\), the square root of an acceleration. The density \(\mu=m/a\) is what survives the limit.
Keeping \(m\) fixed as \(a\to0\). Then \(\mu=m/a\to\infty\) and \(c\to0\); forgetting to hold \(\mu\) fixed gives a frozen, infinitely dense string.
Dropping the \(1/a^2\) before the limit. The bare second difference \((y_{n+1}-2y_n+y_{n-1})\to0\); only \((\cdots)/a^2\to\partial_x^2 y\) survives. Forgetting the \(a^2\) makes the whole right side vanish.
Taylor-expanding to first order only. Keeping just \(y\pm a\,\partial_x y\) cancels in the symmetric difference and gives zero; the \(\tfrac{a^2}{2}\partial_x^2 y\) term is essential and must be retained.
Confusing the two speeds on a string. \(c=\sqrt{T/\mu}\) is the transverse wave speed, distinct from the longitudinal (sound) speed \(\sqrt{Y/\rho}\) set by Young's modulus; they are generally very different.
Assuming \(\omega=ck\) exactly for the chain. That is only the leading behaviour; the true lattice relation \(\omega=\tfrac{2c}{a}|\sin(ka/2)|\) matters whenever \(ka\) is not small.

Discussion

The essential move is coarse-graining: a set of \(N\) coupled ordinary differential equations, one per bead, collapses into a single partial differential equation for a field \(y(x,t)\). The individual masses disappear from the final equation — only the ratio \(\mu=m/a\) and the coupling \(T\) survive. This is the microscopic origin of a classical field, and the same logic underlies elasticity, acoustics and, with a Lorentz-invariant lattice, relativistic field theory. The normal modes of the finite chain, \(y_n\propto e^{i(kna-\omega t)}\) with \(\omega=\tfrac{2c}{a}|\sin(ka/2)|\), pass smoothly into the plane waves \(e^{i(kx-\omega t)}\) of the continuum as \(ka\to0\).

Physically, the finite speed \(c\) is a competition between stiffness and inertia. Tension \(T\) is the restoring agency that wants to straighten the string; density \(\mu\) is the inertia that resists acceleration. Their ratio is a squared speed by dimensional necessity — no other combination of \(T\) and \(\mu\) has units of velocity. That \(c\) is independent of amplitude and waveform is the defining signature of a linear, non-dispersive medium: shapes translate rigidly, and d'Alembert's solution \(y=f(x-ct)+g(x+ct)\) is exact.

The \(O(a^2)\) term discarded in step 7, \(\tfrac{a^2}{12}\partial_x^4 y\), is not merely an error to be thrown away — retained, it yields the leading lattice dispersion and connects to the Boussinesq and Korteweg–de Vries hierarchies once weak nonlinearity is added. The discrete chain is therefore a regulator: it imposes a shortest wavelength \(2a\) and a maximum frequency \(\omega_{\max}=2c/a\), exactly the role a lattice plays in regularising a quantum field theory. The continuum wave equation is the infrared limit of a theory whose ultraviolet completion is the atomic lattice.

Common misconceptions. The wave equation does not say the string moves at \(c\) — each bead moves transversely and slowly (\(\dot y\ll c\) for small amplitude); it is the pattern of displacement that propagates at \(c\). And \(c\) is set entirely by the medium (\(T,\mu\)), never by how the wave was launched: plucking harder changes amplitude and energy, not speed.

Worked examples

\[ c=\sqrt{\frac{T}{\mu}},\qquad f_1=\frac{c}{2L} \]

A steel guitar string: tension \(T=80\ \mathrm{N}\), linear density \(\mu=3.1\ \mathrm{g\,m^{-1}}=3.1\times10^{-3}\ \mathrm{kg\,m^{-1}}\), vibrating length \(L=0.65\ \mathrm{m}\). Find the wave speed and fundamental frequency. Symbols first. A

\[ c=\sqrt{\frac{80\ \mathrm{N}}{3.1\times10^{-3}\ \mathrm{kg\,m^{-1}}}}=\sqrt{2.58\times10^{4}\ \mathrm{m^2\,s^{-2}}}=1.61\times10^{2}\ \mathrm{m\,s^{-1}} \]

Insert numbers with units; the radicand carries \(\mathrm{m^2/s^2}\). A

\[ f_1=\frac{c}{2L}=\frac{160.6\ \mathrm{m\,s^{-1}}}{2(0.65\ \mathrm{m})}=1.24\times10^{2}\ \mathrm{Hz} \]

The fundamental has a half-wavelength spanning the string, \(\lambda_1=2L\). A

\[ c\approx 1.61\times10^{2}\ \mathrm{m\,s^{-1}},\qquad f_1\approx 124\ \mathrm{Hz} \]

Reading. Near the B\(_2\) of a bass line. Doubling \(T\) would raise \(c\) by \(\sqrt2\approx1.41\) and the pitch by the same factor (about a perfect fifth).

\[ v_{\text{phase}}=\frac{\omega}{k}=c\,\frac{\sin(ka/2)}{ka/2},\qquad c=\sqrt{\frac{T}{\mu}},\ \ \mu=\frac{m}{a} \]

Test the continuum approximation against the exact lattice dispersion. Chain: \(m=0.020\ \mathrm{kg}\), spacing \(a=0.050\ \mathrm{m}\), tension \(T=10\ \mathrm{N}\); examine a wave of wavelength \(\lambda=10a\). Symbols first. C

\[ \mu=\frac{0.020\ \mathrm{kg}}{0.050\ \mathrm{m}}=0.40\ \mathrm{kg\,m^{-1}},\qquad c=\sqrt{\frac{10\ \mathrm{N}}{0.40\ \mathrm{kg\,m^{-1}}}}=5.0\ \mathrm{m\,s^{-1}} \]

Density from \(m/a\); continuum speed from \(\sqrt{T/\mu}\). B

\[ k=\frac{2\pi}{\lambda}=\frac{2\pi}{10a}=\frac{2\pi}{0.50\ \mathrm{m}}=12.57\ \mathrm{m^{-1}},\qquad \frac{ka}{2}=\frac{(12.57)(0.050)}{2}=0.3142 \]

Wavenumber from wavelength; form the argument \(ka/2\) that controls the correction. B

\[ v_{\text{phase}}=c\,\frac{\sin(0.3142)}{0.3142}=5.0\left(\frac{0.3090}{0.3142}\right)\ \mathrm{m\,s^{-1}}=4.92\ \mathrm{m\,s^{-1}} \]

Evaluate the sinc factor; it is slightly below 1, so the lattice runs slow. C

\[ v_{\text{phase}}\approx 4.92\ \mathrm{m\,s^{-1}}=0.984\,c\quad(1.6\%\ \text{slow}) \]

Reading. Even at \(\lambda=10a\) — ten beads per wavelength — the continuum wave equation is accurate to under \(2\%\). The discreteness error scales as \((ka)^2/24\), shrinking quadratically as the wavelength grows relative to the spacing, which validates the \(a\to0\) limit.

Problems

A nylon string has \(\mu=1.2\times10^{-3}\ \mathrm{kg\,m^{-1}}\) and carries transverse waves at \(c=140\ \mathrm{m\,s^{-1}}\). Find the tension.
Solution
From \(c=\sqrt{T/\mu}\), \(T=\mu c^2=(1.2\times10^{-3}\ \mathrm{kg\,m^{-1}})(140\ \mathrm{m\,s^{-1}})^2=(1.2\times10^{-3})(1.96\times10^{4})\ \mathrm{N}=23.5\ \mathrm{N}\).
A string sounds a fundamental of \(220\ \mathrm{Hz}\) at tension \(T_0\). By what factor must the tension change to raise the fundamental to \(330\ \mathrm{Hz}\), with length and density fixed?
Solution
\(f_1=\tfrac{1}{2L}\sqrt{T/\mu}\propto\sqrt{T}\), so \(T\propto f_1^2\). Ratio \(T'/T_0=(f_1'/f_1)^2=(330/220)^2=(1.5)^2=2.25\). The tension must increase by a factor \(2.25\).
A chain has beads \(m=5.0\ \mathrm{g}\) at spacing \(a=2.0\ \mathrm{cm}\) under tension \(T=8.0\ \mathrm{N}\). Compute (a) the continuum speed \(c\) and (b) the maximum angular frequency \(\omega_{\max}=2c/a\) the chain can carry.
Solution
(a) \(\mu=m/a=5.0\times10^{-3}/0.020=0.25\ \mathrm{kg\,m^{-1}}\); \(c=\sqrt{T/\mu}=\sqrt{8.0/0.25}=\sqrt{32}=5.66\ \mathrm{m\,s^{-1}}\). (b) \(\omega_{\max}=2c/a=2(5.66)/0.020=566\ \mathrm{rad\,s^{-1}}\), i.e. \(f_{\max}=\omega_{\max}/2\pi\approx90\ \mathrm{Hz}\).
Starting from the exact lattice dispersion \(\omega=\tfrac{2c}{a}\sin(ka/2)\), show that the phase speed falls below \(c\) by a fractional amount \(\approx(ka)^2/24\) at long wavelength, and evaluate it for \(ka=0.40\).
Solution
\(v_{\text{ph}}=\omega/k=c\,\dfrac{\sin(ka/2)}{ka/2}\). Expand \(\sin u=u-u^3/6+\cdots\) with \(u=ka/2\): \(\dfrac{\sin u}{u}=1-\dfrac{u^2}{6}=1-\dfrac{(ka)^2}{24}\). The fractional shortfall is \((ka)^2/24\). For \(ka=0.40\): \((0.40)^2/24=0.16/24=6.7\times10^{-3}\), so \(v_{\text{ph}}\approx0.993\,c\), about \(0.67\%\) slow. (Check: exact \(\sin(0.20)/0.20=0.19867/0.20=0.9933\).)
A steel wire of radius \(r=0.25\ \mathrm{mm}\) and density \(\rho=7.9\times10^{3}\ \mathrm{kg\,m^{-3}}\) is under tension \(T=60\ \mathrm{N}\). Find its linear density \(\mu\) and the transverse wave speed \(c\).
Solution
Cross-section \(A=\pi r^2=\pi(2.5\times10^{-4}\ \mathrm{m})^2=1.963\times10^{-7}\ \mathrm{m^2}\). Linear density \(\mu=\rho A=(7.9\times10^{3})(1.963\times10^{-7})=1.55\times10^{-3}\ \mathrm{kg\,m^{-1}}\). Speed \(c=\sqrt{T/\mu}=\sqrt{60/1.55\times10^{-3}}=\sqrt{3.87\times10^{4}}=1.97\times10^{2}\ \mathrm{m\,s^{-1}}\).