Dispersion of Surface Gravity Waves

Derivation

Dispersion of Surface Gravity Waves

D-330 Home PU-307 Threads waves · energy · fields Depends on Potential Flow and Laplace's Equation, Bernoulli's Theorem, Steady and Unsteady

Statement

For small-amplitude, irrotational waves on the surface of an inviscid, incompressible fluid layer of uniform depth \(h\) under gravity \(g\), a sinusoidal free-surface disturbance of wavenumber \(k\) and angular frequency \(\omega\) obeys the dispersion relation \(\omega^2 = g\,k\,\tanh(kh)\), which reduces to \(\omega^2 = gk\) for deep water (\(kh \gg 1\)) and to \(\omega^2 = gh\,k^2\), i.e. a non-dispersive speed \(c = \sqrt{gh}\), for shallow water (\(kh \ll 1\)).

Why it matters

This single relation governs almost every gravity-driven wave on a free liquid surface: ocean swell arriving at a beach, ship wakes, tsunamis crossing an ocean basin, and ripples in a laboratory tank. Because \(\omega\) depends non-linearly on \(k\), the medium is dispersive: components of different wavelength travel at different phase speeds, so a localised disturbance spreads and sorts itself into a spectrum of wave trains.

It is also the cleanest non-trivial example of solving Laplace's equation with a free boundary whose position is itself an unknown. Linearising the free-surface conditions turns an intractable moving-boundary problem into a solvable eigenvalue problem, and the same technique underlies capillary waves, internal waves, and the theory of ship resistance.

Assumptions

Incompressible flow (\(\nabla\cdot\vec{u}=0\)).If dropped, density perturbations couple in acoustic modes and Laplace's equation for the potential no longer holds; one must solve the full compressible-flow equations.

Irrotational flow (\(\nabla\times\vec{u}=0\)), so \(\vec{u}=\nabla\phi\).If dropped, no single velocity potential exists; vorticity (e.g. from a sheared mean current or a viscous boundary layer) modifies the surface conditions and the clean dispersion relation is lost.

Inviscid fluid.If dropped, viscosity damps the waves (amplitude \(\propto e^{-2\nu k^2 t}\)) and adds a tangential-stress condition at the surface; the real part of \(\omega\) is only weakly shifted but the wave now decays.

Small amplitude, \(ak \ll 1\) (\(a\) = wave amplitude).If dropped, the neglected quadratic terms in the free-surface conditions matter: the dispersion relation gains amplitude-dependent (Stokes) corrections and harmonics are generated.

Uniform depth \(h\) and constant gravity; surface tension negligible.If dropped: a sloping bottom makes \(k\) vary along the wave (shoaling/refraction); restoring the capillary term adds \(+\,(\sigma/\rho)k^3\) inside the bracket, giving gravity–capillary waves with a minimum phase speed.

Atmosphere of negligible density and pressure constant at the surface.If dropped, a dense overlying fluid (two-layer system) replaces \(g\) by a reduced gravity \(g' = g\,\Delta\rho/\rho\) and yields the internal-wave dispersion relation.

Derivation

\[ \vec{u} = \nabla\phi, \qquad \nabla^2\phi = 0 \quad \text{for } -h \le z \le \eta(x,t) \]

Irrotationality gives a velocity potential; incompressibility then makes it harmonic (assumed prior result: potential-flow Laplace equation). Take a plane wave travelling in \(x\), with \(z\) measured upward from the undisturbed surface. A

\[ \left.\frac{\partial \phi}{\partial z}\right|_{z=-h} = 0 \]

No flow through the rigid, impermeable, flat bottom: the normal velocity vanishes there. This is the lower boundary condition. A

\[ \frac{\partial \eta}{\partial t} + u\frac{\partial\eta}{\partial x} = \left.\frac{\partial\phi}{\partial z}\right|_{z=\eta} \;\;\xrightarrow{\text{linearise}}\;\; \left.\frac{\partial \eta}{\partial t}\right. = \left.\frac{\partial\phi}{\partial z}\right|_{z=0} \]

Kinematic condition: a fluid particle on the surface stays on the surface, so the surface's vertical velocity equals the fluid's. Dropping the quadratic term \(u\,\partial_x\eta\) and evaluating at \(z=0\) instead of \(z=\eta\) are both first-order in the small amplitude. B

\[ \frac{\partial\phi}{\partial t} + \tfrac{1}{2}|\nabla\phi|^2 + \frac{p}{\rho} + gz = C(t) \;\;\xrightarrow[\;p=p_{\text{atm}}\;]{\text{linearise at }z=0}\;\; \left.\frac{\partial\phi}{\partial t}\right|_{z=0} + g\,\eta = 0 \]

Unsteady Bernoulli (assumed prior result) along the whole flow, evaluated at the surface where \(p\) equals the constant atmospheric pressure. Absorb constants into \(\phi\); drop the quadratic kinetic term. This dynamic condition supplies the restoring force. B

\[ \left.\frac{\partial^2\phi}{\partial t^2}\right|_{z=0} + g\,\left.\frac{\partial\phi}{\partial z}\right|_{z=0} = 0 \]

Differentiate the dynamic condition (step 4) in time and substitute the kinematic condition (step 3, \(\partial_t\eta = \partial_z\phi\)) to eliminate \(\eta\). One combined free-surface condition on \(\phi\) alone remains. B

\[ \phi(x,z,t) = f(z)\,\sin(kx-\omega t), \qquad \eta(x,t) = a\cos(kx-\omega t) \]

Seek a separable travelling-wave mode; the linear problem has constant coefficients in \(x\) and \(t\), so Fourier modes decouple and it suffices to solve one. A

\[ f''(z) - k^2 f(z) = 0 \;\;\Rightarrow\;\; f(z) = A\cosh\!\big(k(z+h)\big) \]

Insert the ansatz into \(\nabla^2\phi=0\): the \(\sin(kx-\omega t)\) factor gives \(-k^2 f + f'' = 0\). The general solution is \(A\cosh k(z+h) + B\sinh k(z+h)\); the bottom condition (step 2) \(f'(-h)=0\) forces \(B=0\), leaving the \(\cosh\) that automatically satisfies it. A

\[ \big(-\omega^2\big)A\cosh(kh) \;+\; g\,\big(kA\sinh(kh)\big) = 0 \]

Substitute \(f(z)=A\cosh k(z+h)\) into the combined surface condition (step 5) at \(z=0\): \(\partial_t^2\phi \to -\omega^2 f\), and \(\partial_z\phi \to f'(0)=kA\sinh(kh)\). The common factor \(A\sin(kx-\omega t)\) cancels. B

\[ \omega^2 = g\,k\,\frac{\sinh(kh)}{\cosh(kh)} = g\,k\,\tanh(kh) \]

Divide the eigenvalue condition (step 8) by \(A\cosh(kh)\) and solve for \(\omega^2\). A non-trivial mode (\(A\neq0\)) exists only when this holds — the dispersion relation. A

Result

\[ \boxed{\;\omega^2 = g\,k\,\tanh(kh)\;}\qquad c=\frac{\omega}{k}=\sqrt{\frac{g}{k}\tanh(kh)} \]

Reading. The squared frequency is the product of a gravitational restoring rate \(gk\) and a depth factor \(\tanh(kh)\) that runs smoothly from \(kh\) (shallow) to \(1\) (deep). Long waves feel the bottom (small \(kh\), factor \(\to kh\)); short waves do not (large \(kh\), factor \(\to 1\)). Because \(c\) depends on \(k\), the surface is a dispersive medium.

Units check. \(g\) is \(\mathrm{m\,s^{-2}}\) and \(k\) is \(\mathrm{m^{-1}}\), so \(gk\) is \(\mathrm{s^{-2}}\); \(\tanh(kh)\) is dimensionless (its argument \(kh\) is \(\mathrm{m^{-1}\cdot m}\), pure number). Hence \([\omega^2]=\mathrm{s^{-2}}\) and \([c]=\sqrt{\mathrm{s^{-2}}/\mathrm{m^{-2}}}=\mathrm{m\,s^{-1}}\), as required.

Limiting cases

Deep water, \(kh\gg1\) (\(h\gtrsim \lambda/2\)): \(\tanh(kh)\to1\), so \(\omega^2=gk\), \(c=\sqrt{g/k}=\sqrt{g\lambda/2\pi}\). Longer waves are faster; strongly dispersive.
Shallow water, \(kh\ll1\) (\(h\lesssim \lambda/20\)): \(\tanh(kh)\to kh\), so \(\omega^2=gh\,k^2\), \(c=\sqrt{gh}\) independent of \(k\) — non-dispersive; all long waves travel at one speed set by depth.
Intermediate depth: full \(\tanh(kh)\) must be kept; \(c\) lies between the two limits. The transition is centred near \(kh\sim1\), i.e. \(h\sim\lambda/6\).
Group velocity: \(c_g=\dfrac{d\omega}{dk}=\dfrac{c}{2}\!\left(1+\dfrac{2kh}{\sinh 2kh}\right)\), giving \(c_g=c/2\) (deep) and \(c_g=c=\sqrt{gh}\) (shallow).

Breaks when

Finite amplitude (\(ak\) not \(\ll1\)). The linearisation of the surface conditions fails; Stokes' expansion gives \(\omega^2=gk\tanh(kh)\,[1+(ak)^2 F(kh)+\dots]\), waves steepen, and near \(ak\approx0.44\) (deep water) the crest reaches a \(120^\circ\) angle and breaks.
Very short wavelength (capillary regime). When \(\lambda \lesssim\) a few cm, surface tension \(\sigma\) dominates: the bracket becomes \(g+\sigma k^2/\rho\), so \(\omega^2=(gk+\sigma k^3/\rho)\tanh(kh)\). Pure gravity-wave dispersion is then wrong, with a phase-speed minimum near \(\lambda\approx1.7\,\mathrm{cm}\).
Strongly sheared or rotational background flow. A depth-varying current or wind-driven shear injects vorticity, invalidating \(\vec u=\nabla\phi\); the result is Doppler-shifted and can become unstable (Kelvin–Helmholtz).
Rapidly varying depth. If \(h\) changes appreciably over one wavelength the uniform-layer normal mode is invalid; refraction, shoaling, and reflection require ray theory or the mild-slope equation.

Failure modes

Dropping the \(\tanh\). Writing \(\omega^2=gk\) for a shallow tank or a tsunami. Always test \(kh\): only \(kh\gtrsim\pi\) justifies \(\tanh\to1\).
Confusing phase and group speed. Using \(c=\omega/k\) where energy transport (hence arrival of a wave packet or a tsunami's leading edge) is set by \(c_g=d\omega/dk\); in deep water these differ by a factor of two.
Measuring \(z\) from the bottom. Then the bottom condition sits at \(z=0\) and the surface at \(z=h\); students often keep the surface conditions at \(z=0\) by habit, producing \(\coth\) instead of \(\tanh\).
Sign/argument slips in \(\eta\) and \(\phi\). Using the same trigonometric phase for \(\eta\) and \(\phi\); the kinematic condition forces a \(90^\circ\) phase offset (\(\cos\) vs \(\sin\)), and forgetting it corrupts the amplitude relation \(A\).
Using \(k=1/\lambda\) instead of \(k=2\pi/\lambda\). A factor \(2\pi\) error that propagates through every number.
Radians vs hertz. Reporting \(\omega\) as a frequency \(f\); remember \(\omega=2\pi f\) and \(T=2\pi/\omega\).

Discussion

The physics is a competition between inertia and a gravitational restoring force, mediated by the geometry of the flow beneath the surface. A crest carries excess weight of fluid; gravity pushes it down, but the fluid displaced must go somewhere, and Laplace's equation dictates that the disturbance decays into the depth as \(\cosh k(z+h)\) — reaching the bottom for long waves, dying out within a depth \(\sim1/k\) for short ones. That penetration depth is exactly what \(\tanh(kh)\) encodes: when the motion reaches the bottom it is throttled (shallow limit), when it does not the bottom is irrelevant (deep limit).

Dispersion has a visible signature. Drop a stone in a pond and the pattern is not a single expanding ring but a train of ripples, because each wavelength travels at its own speed and the components separate. On the ocean, a distant storm radiates a broad spectrum; the long, fast swell arrives days before the short, slow components, and coastal forecasters read the swell period to infer both the storm's distance and its age.

The shallow-water limit is where this connects to the wider physics of waves. There \(\omega=\sqrt{gh}\,k\) is linear in \(k\), so the medium is non-dispersive and the governing equation is the ordinary wave equation with speed \(c=\sqrt{gh}\) — the same mathematical structure as sound and light in vacuum. This is why a tsunami, whose wavelength (\(\sim100\,\mathrm{km}\)) hugely exceeds the ocean depth (\(\sim4\,\mathrm{km}\)), behaves as a shallow-water wave crossing an ocean at jetliner speed while barely rising above the surface, then shoals catastrophically as \(h\) drops near shore and \(c\) and wave height change.

A subtler point is that the boundary itself is the unknown. Formally this is a free-boundary problem: the domain \(-h\le z\le\eta(x,t)\) depends on the solution \(\eta\) we are trying to find. Linearisation resolves this by evaluating the surface conditions on the mean surface \(z=0\) and discarding quadratic terms, which is consistent to first order in \(ak\). Carrying the expansion further (Stokes waves) shows the free surface renormalises the frequency and generates bound harmonics; the deep-water train even suffers the Benjamin–Feir modulational instability, a genuinely non-linear effect entirely absent from the linear dispersion relation. The linear result is thus the leading term of a controlled perturbation series, not an exact law.

Common misconceptions. "Water moves along with the wave" — it does not; to leading order fluid particles trace closed (deep water: circular) orbits and there is no net transport, only the phase pattern propagates. "Deeper water always means faster waves" — true only in the shallow/intermediate regime where \(c=\sqrt{gh}\); once \(kh\gg1\) the speed saturates at \(\sqrt{g/k}\) and further depth is irrelevant.

Worked examples

Ocean swell, deep water. Wavelength \(\lambda=100\,\mathrm{m}\), depth \(h=200\,\mathrm{m}\), \(g=9.81\,\mathrm{m\,s^{-2}}\). Find the period, phase speed and group speed, and check the deep-water assumption.

\[ k=\frac{2\pi}{\lambda}=\frac{2\pi}{100}=6.28\times10^{-2}\,\mathrm{m^{-1}},\qquad kh=6.28\times10^{-2}\times200=12.6 \]

\(kh=12.6\gg1\Rightarrow\tanh(kh)\approx1.000\): deep water is justified (indeed \(h>\lambda/2=50\,\mathrm m\)).

\[ \omega^2=gk=9.81\times6.28\times10^{-2}=0.616\,\mathrm{s^{-2}} \;\Rightarrow\; \omega=0.785\,\mathrm{rad\,s^{-1}} \]

\[ T=\frac{2\pi}{\omega}=\frac{6.283}{0.785}=8.0\,\mathrm{s},\qquad c=\frac{\omega}{k}=\frac{0.785}{6.28\times10^{-2}}=12.5\,\mathrm{m\,s^{-1}},\qquad c_g=\tfrac12 c=6.25\,\mathrm{m\,s^{-1}} \]

\[ T\approx8.0\,\mathrm{s},\quad c\approx12.5\,\mathrm{m\,s^{-1}},\quad c_g\approx6.3\,\mathrm{m\,s^{-1}} \]

Reading. A textbook 8-second swell moves at about 12.5 m/s (45 km/h), but its energy — and the packet you would time — advances at half that.

Tsunami, shallow water. Wavelength \(\lambda=200\,\mathrm{km}\) in an ocean of depth \(h=4000\,\mathrm{m}\). Find the phase speed and the time to cross a \(6000\,\mathrm{km}\) basin; verify the shallow-water limit.

\[ k=\frac{2\pi}{2.00\times10^{5}}=3.14\times10^{-5}\,\mathrm{m^{-1}},\qquad kh=3.14\times10^{-5}\times4000=0.126 \]

\(kh=0.126\ll1\Rightarrow\tanh(kh)\approx kh\) (error \(<1\%\)): shallow water, \(c\approx\sqrt{gh}\).

\[ c=\sqrt{gh}=\sqrt{9.81\times4000}=\sqrt{3.924\times10^{4}}=198\,\mathrm{m\,s^{-1}}\;(\approx713\,\mathrm{km\,h^{-1}}) \]

\[ t=\frac{L}{c}=\frac{6.00\times10^{6}\,\mathrm m}{198\,\mathrm{m\,s^{-1}}}=3.03\times10^{4}\,\mathrm s=8.4\ \mathrm{hours} \]

\[ c\approx198\,\mathrm{m\,s^{-1}},\qquad t\approx8.4\ \mathrm{hours} \]

Reading. Because \(h\ll\lambda\), the tsunami is non-dispersive and crosses the ocean at jetliner speed; in the deep sea its amplitude is only tens of centimetres, growing dangerous only as \(c=\sqrt{gh}\) collapses in shoaling water.

Problems

Deep-water ripple. Find the period and phase speed of a deep-water gravity wave of wavelength \(\lambda=2.0\,\mathrm{m}\) (\(g=9.81\,\mathrm{m\,s^{-2}}\)).
Solution
\(k=2\pi/2.0=3.14\,\mathrm{m^{-1}}\); deep-water \(\omega^2=gk=9.81\times3.14=30.8\,\mathrm{s^{-2}}\Rightarrow\omega=5.55\,\mathrm{rad\,s^{-1}}\). \(T=2\pi/\omega=1.13\,\mathrm{s}\); \(c=\omega/k=5.55/3.14=1.77\,\mathrm{m\,s^{-1}}\) (or \(c=\sqrt{g\lambda/2\pi}=\sqrt{9.81\times2/6.283}=1.77\,\mathrm{m\,s^{-1}}\)).
Regime test. Water of depth \(h=3.0\,\mathrm{m}\) carries a wave of wavelength \(\lambda=30\,\mathrm{m}\). Classify the regime and compute \(c\) exactly, then compare with the shallow-water estimate.
Solution
\(k=2\pi/30=0.209\,\mathrm{m^{-1}}\), \(kh=0.209\times3.0=0.628\): intermediate depth (neither \(\ll1\) nor \(\gg\pi\)). \(\tanh(0.628)=0.556\). \(\omega^2=gk\tanh(kh)=9.81\times0.209\times0.556=1.14\,\mathrm{s^{-2}}\Rightarrow\omega=1.07\,\mathrm{rad\,s^{-1}}\); \(c=\omega/k=1.07/0.209=5.12\,\mathrm{m\,s^{-1}}\). Shallow estimate \(\sqrt{gh}=\sqrt{29.4}=5.42\,\mathrm{m\,s^{-1}}\), about 6% high — the depth factor is starting to matter.
Group velocity in deep water. Starting from \(\omega=\sqrt{gk}\), show that \(c_g=\tfrac12 c\).
Solution
\(c_g=\dfrac{d\omega}{dk}=\dfrac{d}{dk}\big(g^{1/2}k^{1/2}\big)=\tfrac12 g^{1/2}k^{-1/2}=\tfrac12\sqrt{g/k}\). Since \(c=\omega/k=\sqrt{gk}/k=\sqrt{g/k}\), we get \(c_g=\tfrac12 c\). Deep-water energy travels at half the phase speed, so crests appear to run through a wave group from behind and vanish at its front.
Basin crossing. A shallow-water wave crosses an ocean \(L=9000\,\mathrm{km}\) wide of uniform depth \(h=5000\,\mathrm{m}\). Estimate the travel time.
Solution
\(c=\sqrt{gh}=\sqrt{9.81\times5000}=\sqrt{4.905\times10^{4}}=221\,\mathrm{m\,s^{-1}}\). \(t=L/c=9.00\times10^{6}/221=4.07\times10^{4}\,\mathrm{s}=11.3\,\mathrm{hours}\). (Assumes uniform depth; real bathymetry refracts and slows the wave.)
General group velocity. Differentiate \(\omega^2=gk\tanh(kh)\) implicitly to derive \(c_g=\dfrac{c}{2}\left(1+\dfrac{2kh}{\sinh 2kh}\right)\), and confirm the deep- and shallow-water limits.
Solution
Write \(\omega^2=gk\tanh(kh)\). Differentiate: \(2\omega\,d\omega=g\big[\tanh(kh)+kh\,\mathrm{sech}^2(kh)\big]dk\), so \(c_g=\dfrac{d\omega}{dk}=\dfrac{g\big[\tanh(kh)+kh\,\mathrm{sech}^2(kh)\big]}{2\omega}\). Using \(\omega^2=gk\tanh(kh)\) to replace \(g\tanh(kh)=\omega^2/k\) and \(c=\omega/k\): \(c_g=\dfrac{c}{2}\Big[1+\dfrac{kh\,\mathrm{sech}^2(kh)}{\tanh(kh)}\Big]=\dfrac{c}{2}\Big[1+\dfrac{2kh}{\sinh 2kh}\Big]\), where \(\dfrac{\mathrm{sech}^2(kh)}{\tanh(kh)}=\dfrac{2}{\sinh 2kh}\) (since \(\sinh 2kh=2\sinh kh\cosh kh\)). Deep water \(kh\to\infty\): \(2kh/\sinh 2kh\to0\), \(c_g\to c/2\). Shallow water \(kh\to0\): \(\sinh 2kh\to2kh\), so \(2kh/\sinh 2kh\to1\) and \(c_g\to c=\sqrt{gh}\), consistent with the non-dispersive limit.