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 Incoming links: Minimum phase speed of a gravity-capillary wave

Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2021 / Section A / 3 / Solution 2026-10-18

Take the outward wavenumber

k > 0

; the absolute value in the source makes the two propagation directions equivalent. The phase velocity of the gravity-capillary wave obeys

c_{p}^{2} = \frac{ω ^{2}}{k ^{2}} = \frac{g}{k} + \frac{σ}{ρ} k .

Differentiate with respect to

k

:

\frac{d ( c _{p}^{2} )}{d k} = - \frac{g}{k ^{2}} + \frac{σ}{ρ} = 0 \Rightarrow k_{*} = \frac{ρ g}{σ} .

The second derivative is positive, and

c_{p} \to \infty

at both ends of the positive-

k

range, so this is the global minimum. The minimum phase speed of a gravity-capillary wave is

c_{p, m i n} = (\frac{4 g σ}{ρ})^{1/4} .

For

g = 9.81 m s^{- 2}

and the supplied surface tension and mass density,

c_{p, m i n} = 0.231 m s^{- 1} .

The corresponding wavenumber is

k_{*} ≃ 369 m^{- 1}

, giving wavelength

λ_{*} = 2 π / k_{*} ≃ 1.70 cm

. At this minimum,

c_{g} = d ω / d k = c_{p} + k d c_{p} / d k = c_{p}

. Thus the group velocity and phase velocity agree for this particular ripple, although they differ for general gravity-capillary waves.

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