Capillary wave

ρ {\displaystyle \rho }

ρ ′ {\displaystyle \rho '}

∇ ϕ {\displaystyle \nabla \phi }

∇ ϕ ′ {\displaystyle \nabla \phi '}

ϕ ( x , y , z , t ) {\displaystyle \phi (x,y,z,t)}

ϕ ′ ( x , y , z , t ) {\displaystyle \phi '(x,y,z,t)}

Three contributions to the energy are involved: the potential energy V g {\displaystyle V_{g}} due to gravity, the potential energy V s t {\displaystyle V_{st}} due to the surface tension and the kinetic energy T {\displaystyle T} of the flow. The part V g {\displaystyle V_{g}} due to gravity is the simplest: integrating the potential energy density due to gravity, ρ g z {\displaystyle \rho gz} (or ρ ′ g z {\displaystyle \rho 'gz} ) from a reference height to the position of the surface, z = η ( x , y , t ) {\displaystyle z=\eta (x,y,t)} :^[6]

V g = ∬ d x d y ∫ 0 η d z ( ρ − ρ ′ ) g z = 1 2 ( ρ − ρ ′ ) g ∬ d x d y η 2 , {\displaystyle V_{\mathrm {g} }=\iint dx\,dy\;\int _{0}^{\eta }dz\;(\rho -\rho ')gz={\frac {1}{2}}(\rho -\rho ')g\iint dx\,dy\;\eta ^{2},}

assuming the mean interface position is at z = 0 {\displaystyle z=0} .

An increase in area of the surface causes a proportional increase of energy due to surface tension:^[7]

V s t = σ ∬ d x d y [ 1 + ( ∂ η ∂ x ) 2 + ( ∂ η ∂ y ) 2 − 1 ] ≈ 1 2 σ ∬ d x d y [ ( ∂ η ∂ x ) 2 + ( ∂ η ∂ y ) 2 ] , {\displaystyle V_{\mathrm {st} }=\sigma \iint dx\,dy\;\left[{\sqrt {1+\left({\frac {\partial \eta }{\partial x}}\right)^{2}+\left({\frac {\partial \eta }{\partial y}}\right)^{2}}}-1\right]\approx {\frac {1}{2}}\sigma \iint dx\,dy\;\left[\left({\frac {\partial \eta }{\partial x}}\right)^{2}+\left({\frac {\partial \eta }{\partial y}}\right)^{2}\right],}

where the first equality is the area in this (Monge's) representation, and the second applies for small values of the derivatives (surfaces not too rough).

The last contribution involves the kinetic energy of the fluid:^[8]

T = 1 2 ∬ d x d y [ ∫ − ∞ η d z ρ | ∇ Φ | 2 + ∫ η + ∞ d z ρ ′ | ∇ Φ ′ | 2 ] . {\displaystyle T={\frac {1}{2}}\iint dx\,dy\;\left[\int _{-\infty }^{\eta }dz\;\rho \,\left|\mathbf {\nabla } \Phi \right|^{2}+\int _{\eta }^{+\infty }dz\;\rho '\,\left|\mathbf {\nabla } \Phi '\right|^{2}\right].}

Use is made of the fluid being incompressible and its flow is irrotational (often, sensible approximations). As a result, both ϕ ( x , y , z , t ) {\displaystyle \phi (x,y,z,t)} and ϕ ′ ( x , y , z , t ) {\displaystyle \phi '(x,y,z,t)} must satisfy the Laplace equation:^[9]

∇ 2 Φ = 0 {\displaystyle \nabla ^{2}\Phi =0}   and   ∇ 2 Φ ′ = 0. {\displaystyle \nabla ^{2}\Phi '=0.}

These equations can be solved with the proper boundary conditions: ϕ {\displaystyle \phi } and ϕ ′ {\displaystyle \phi '} must vanish well away from the surface (in the "deep water" case, which is the one we consider).

Using Green's identity, and assuming the deviations of the surface elevation to be small (so the z–integrations may be approximated by integrating up to z = 0 {\displaystyle z=0} instead of z = η {\displaystyle z=\eta } ), the kinetic energy can be written as:^[8]

T ≈ 1 2 ∬ d x d y [ ρ Φ ∂ Φ ∂ z − ρ ′ Φ ′ ∂ Φ ′ ∂ z ] at  z = 0 . {\displaystyle T\approx {\frac {1}{2}}\iint dx\,dy\;\left[\rho \,\Phi \,{\frac {\partial \Phi }{\partial z}}\;-\;\rho '\,\Phi '\,{\frac {\partial \Phi '}{\partial z}}\right]_{{\text{at }}z=0}.}

To find the dispersion relation, it is sufficient to consider a sinusoidal wave on the interface, propagating in the x–direction:^[7]

η = a cos ( k x − ω t ) = a cos θ , {\displaystyle \eta =a\,\cos \,(kx-\omega t)=a\,\cos \,\theta ,}

with amplitude a {\displaystyle a} and wave phase θ = k x − ω t {\displaystyle \theta =kx-\omega t} . The kinematic boundary condition at the interface, relating the potentials to the interface motion, is that the vertical velocity components must match the motion of the surface:^[7]

∂ Φ ∂ z = ∂ η ∂ t {\displaystyle {\frac {\partial \Phi }{\partial z}}={\frac {\partial \eta }{\partial t}}}   and   ∂ Φ ′ ∂ z = ∂ η ∂ t {\displaystyle {\frac {\partial \Phi '}{\partial z}}={\frac {\partial \eta }{\partial t}}}   at z = 0 {\displaystyle z=0} .

To tackle the problem of finding the potentials, one may try separation of variables, when both fields can be expressed as:^[7]

Φ ( x , y , z , t ) = + 1 | k | e + | k | z ω a sin θ , Φ ′ ( x , y , z , t ) = − 1 | k | e − | k | z ω a sin θ . {\displaystyle {\begin{aligned}\Phi (x,y,z,t)&=+{\frac {1}{|k|}}{\text{e}}^{+|k|z}\,\omega a\,\sin \,\theta ,\\\Phi '(x,y,z,t)&=-{\frac {1}{|k|}}{\text{e}}^{-|k|z}\,\omega a\,\sin \,\theta .\end{aligned}}}

Then the contributions to the wave energy, horizontally integrated over one wavelength λ = 2 π / k {\displaystyle \lambda =2\pi /k} in the x–direction, and over a unit width in the y–direction, become:^[7][10]

V g = 1 4 ( ρ − ρ ′ ) g a 2 λ , V st = 1 4 σ k 2 a 2 λ , T = 1 4 ( ρ + ρ ′ ) ω 2 | k | a 2 λ . {\displaystyle {\begin{aligned}V_{\text{g}}&={\frac {1}{4}}(\rho -\rho ')ga^{2}\lambda ,\\V_{\text{st}}&={\frac {1}{4}}\sigma k^{2}a^{2}\lambda ,\\T&={\frac {1}{4}}(\rho +\rho '){\frac {\omega ^{2}}{|k|}}a^{2}\lambda .\end{aligned}}}

The dispersion relation can now be obtained from the Lagrangian L = T − V {\displaystyle L=T-V} , with V {\displaystyle V} the sum of the potential energies by gravity V g {\displaystyle V_{g}} and surface tension V s t {\displaystyle V_{st}} :^[11]

L = 1 4 [ ( ρ + ρ ′ ) ω 2 | k | − ( ρ − ρ ′ ) g − σ k 2 ] a 2 λ . {\displaystyle L={\frac {1}{4}}\left[(\rho +\rho '){\frac {\omega ^{2}}{|k|}}-(\rho -\rho ')g-\sigma k^{2}\right]a^{2}\lambda .}

For sinusoidal waves and linear wave theory, the phase–averaged Lagrangian is always of the form L = D ( ω , k ) a 2 {\displaystyle L=D(\omega ,k)a^{2}} , so that variation with respect to the only free parameter, a {\displaystyle a} , gives the dispersion relation D ( ω , k ) = 0 {\displaystyle D(\omega ,k)=0} .^[11] In our case D ( ω , k ) {\displaystyle D(\omega ,k)} is just the expression in the square brackets, so that the dispersion relation is:

ω 2 = | k | ( ρ − ρ ′ ρ + ρ ′ g + σ ρ + ρ ′ k 2 ) , {\displaystyle \omega ^{2}=|k|\left({\frac {\rho -\rho '}{\rho +\rho '}}\,g+{\frac {\sigma }{\rho +\rho '}}\,k^{2}\right),}

the same as above.

As a result, the average wave energy per unit horizontal area, ( T + V ) / λ {\displaystyle (T+V)/\lambda } , is:

E ¯ = 1 2 [ ( ρ − ρ ′ ) g + σ k 2 ] a 2 . {\displaystyle {\bar {E}}={\frac {1}{2}}\,\left[(\rho -\rho ')\,g+\sigma k^{2}\right]\,a^{2}.}

As usual for linear wave motions, the potential and kinetic energy are equal (equipartition holds): T = V {\displaystyle T=V} .^[12]