Energy Density, Energy Flux and Momentum Flux of Surface Waves
[math]\displaystyle{ \varepsilon(t) = \ \mbox{Energy in control volume} \ \gamma(t) }[/math] :
[math]\displaystyle{ \varepsilon (t) = \rho \iiint_V \left( \frac{1}{2} V^2 + gZ \right) dV }[/math]
Mean energy over unit horizongtal surface area [math]\displaystyle{ S \, }[/math] :
[math]\displaystyle{ \overline{\varepsilon} = \overline{\frac{\varepsilon(t)}{S}} = \rho \overline{ \int_{-H}^{\zeta(t)} \left( \frac{1}{2} V^2 + gZ \right) dZ} = \frac{1}{2} \rho \overline{ \int_{-H}^{\zeta(t)} V^2 dZ} + \overline{ \frac{1}{2} \rho g ( \zeta^2 - H^2 ) } }[/math]
where [math]\displaystyle{ \zeta(t) \, }[/math] is free surface elevation.
Ignore term [math]\displaystyle{ -\frac{1}{2} \rho g H^2 \, }[/math] which represents the potential energy of the ocean at rest.
The remaining perturbation component is the sum of the kinetic and potential energy components
[math]\displaystyle{ \overline{\varepsilon} = \overline{\varepsilon_{kin}} + \overline{\varepsilon_{pot}} }[/math]
[math]\displaystyle{ \overline{\varepsilon_{kin}} = \frac{1}{2} \rho \overline{\int_{-H}^{\zeta(t)} V^2 dZ}, \qquad V^2 = \nabla\Phi \cdot \nabla \Phi = \Phi_X^2 + \Phi_Z^2 }[/math]
[math]\displaystyle{ \overline{\varepsilon_{pot}} = \overline{\frac{1}{2} \rho g \zeta^2 (t)} }[/math]
Consider now as a special case plane progressive waves defined by the velocity potential in deep water (for simplicity):
[math]\displaystyle{ \Phi = \mathbf{Re} \{ \frac{igA}{\omega} e^{KZ-iKX+i\omega t} \} }[/math]
[math]\displaystyle{ \Phi_X = \mathbf{Re} \{ \frac{igA}{\omega} (-iK) e^{KZ-iKX+i\omega t} \} }[/math]
[math]\displaystyle{ = A \mathbf{Re} \{ \omega e^{KZ-iKX+i\omega t} \} }[/math]
[math]\displaystyle{ \Phi_Z = \mathbf{Re} \{ \frac{iSA}{\omega} K e^{KZ-iKX+i\omega t} \} }[/math]
[math]\displaystyle{ = A \mathbf{Re} \{ i \omega e^{KZ-iKX+i\omega t} \} }[/math]
Lemma
Let:
[math]\displaystyle{ \mathbf{Re} \{ A e^{i\omega t} \} = A(t) }[/math]
[math]\displaystyle{ \mathbf{Re} \{ B e^{i\omega t} \} = B(t) }[/math]
[math]\displaystyle{ \overline{A(t)B(t)} = \frac{1}{2} \mathbf{Re} \{ A B^* \} }[/math]
[math]\displaystyle{ \overline{\epsilon_{kin}} = \frac{1}{2} \rho \overline{ ( \int_{-\infty}^0 + \int_0^\zeta ) \left( \Phi_X^2 + \Phi_Z^2 \right) } dZ }[/math]
[math]\displaystyle{ = \frac{1}{2} \rho \int_{-\infty}^0 \left( \Phi_X^2 + \Phi_Z^2 \right) dZ + O (A^3) }[/math]
[math]\displaystyle{ = \rho \frac{\omega^2 A^2}{4K} = \frac{1}{4} \rho g A^2 , \qquad \mbox{for} \ K=\omega^2/g }[/math]
[math]\displaystyle{ \overline{\varepsilon_{pot}} = \frac{1}{2} \rho g {\overline{\zeta(t)}}^2 = \frac{1}{4} \rho g A^2 }[/math]
Hence:
[math]\displaystyle{ \overline{\varepsilon} = \overline{\varepsilon_{kin}} + \overline{\varepsilon_{pot}} = \frac{1}{2} \rho g A^2 }[/math]
Energy flux = rate of change of energy density [math]\displaystyle{ \varepsilon(t) }[/math]
[math]\displaystyle{ P(t) \equiv \frac{d\varepsilon(t)}{dt} , \ \varepsilon = \iiint_V(t) (\frac{1}{2} \rho V^2 +gZ ) d }[/math]
[math]\displaystyle{ P(t) = \frac{d \varepsilon(t)}{dt} = \frac{d}{dt} \iiint_V(t) \epsilon(t) dV = \iint_S(t) \frac{\partial \epsilon(t)}{\partial t} dV + \iint_S(t) \epsilon(t) U_n dS }[/math]
Transport theorem where [math]\displaystyle{ U_n }[/math] is normal velocity of surface [math]\displaystyle{ S(t) }[/math] outwards of the enclosed volume [math]\displaystyle{ V }[/math].
[math]\displaystyle{ \frac{\partial \epsilon}{\partial t} = \frac{\partial}{\partial t} \{ \frac{1}{2} \rho V^2 + \rho g Z \} = \frac{1}{2} \rho \frac{\partial}{\partial t} ( \nabla\Phi \cdot \nabla\Phi) }[/math]
[math]\displaystyle{ = \rho \nabla \cdot \left( \frac{\partial\Phi}{\partial t} \nabla\Phi \right) - \rho \frac{\partial\Phi}{\partial t} \nabla^2 \Phi }[/math]
[math]\displaystyle{ P(t) = \frac{d \varepsilon(t)}{dt} = \rho \iiint_V(t) \nabla \cdot \left( \frac{\partial \Phi}{\partial t} \nabla \Phi \right) dV + \rho \iint_S(t) \left( \frac{1}{2} V^2 + gZ \right) U_n dS }[/math]
Invoking the scalar form of Gauss's theorem in the frist term, we obtain:
[math]\displaystyle{ P(t) = \rho \iint \frac{\partial\Phi}{\partial t} \nabla \Phi \dot \vec n ds + \rho \iint \left( \frac{1}{2} V^2 + gZ \right) U_n ds }[/math]
An alternative form for the energy flux [math]\displaystyle{ P(t) \, }[/math] crossing the closed control surface [math]\displaystyle{ S(t) \, }[/math] is obtained by invoking Bernoulli's equation in the second term. Recall that:
[math]\displaystyle{ \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \dot \nabla\Phi + gZ = 0 }[/math]
At any point in the fluid domain and on boundarie.
Here we did allow [math]\displaystyle{ \ P_a \equiv \mbox{Atmospheric pressure} \ }[/math] to be non-zero for the sake of physical clarity. Upon substitution in [math]\displaystyle{ P(t) }[/math] we obtain the alternate form:
[math]\displaystyle{ P(t) = \rho \iint \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} ds - \rho \iint \left( \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} \right) U_n ds }[/math]
So the energy flux across [math]\displaystyle{ S(t)\, }[/math] is given by the terms under the interral sign. They can be collected in the more compact form:
[math]\displaystyle{ P(t) = \iint \left\{ \rho \frac{\partial\Phi}{\partial t} \left( \frac{\partial\Phi}{\partial n} - U_n \right) - ( P - P_a) U_n \right\} ds }[/math]
Note that [math]\displaystyle{ P(t) \, }[/math] measures the energy flux into the volume [math]\displaystyle{ V(t) \, }[/math] or the rate of growth of the energy density [math]\displaystyle{ \varepsilon(t)\, }[/math].
We are ready now to apply the above formulae to the surface wave propagation problem.
Break [math]\displaystyle{ S(t) \, }[/math] into tis components and derive specialized forms of [math]\displaystyle{ P(t) \, }[/math] pertinent to each.
[math]\displaystyle{ S_F : \ \mbox{nonlinear position of the free surface} }[/math]
[math]\displaystyle{ \frac{\partial\Phi}{\partial n} = U_n; \ \mbox{normal flow velocity} \equiv \mbox{normal velocity of free surface boundary; kinematic condition}. }[/math]
[math]\displaystyle{ P = P_a; \ \mbox{fluid pressure} \equiv \mbox{atmospheric} }[/math]
Therefore over [math]\displaystyle{ S_F; \ P(t) \equiv 0 }[/math] as expected. No energy can flow into the atmosphere!
[math]\displaystyle{ S_B: \ \mbox{non-moving solid boundary} }[/math]
[math]\displaystyle{ U_n = 0, \ \frac{\partial\Phi}{\partial n} = U_n; \ \mbox{no-normal flux condition} }[/math]
[math]\displaystyle{ S^\pm: \ \mbox{fluid boundaries fixed in space relative to an earth frame} }[/math]
[math]\displaystyle{ U_n = 0, \ \frac{\partial\Phi}{\partial n} \ne 0 }[/math]
[math]\displaystyle{ S_U: \ \mbox{fluid boundaries moving w. velocity} \ \vec{U} \ \mbox{relative to an earth frame} }[/math]
[math]\displaystyle{ U_n = \vec U \cdot \vec n, \quad \frac{\partial\Phi}{\partial n} \ne 0 }[/math]
This case will be of interest later in the course when we consider ships moving with constant velocity [math]\displaystyle{ U }[/math].
The formulae derived above are very general for potential flows with a free surface and solid boundaries. We are now ready to apply them to plane progressive waves.
Energy flux across a vertical fluid boundary fixed in space.
[math]\displaystyle{ \frac{P(t)}{\mbox{width}} = - \rho \int_{-\infty}^{\zeta(t)} \frac{\partial\Phi}{\partial t} \Phi_n dZ = - \rho \left( \int_{-\infty}^0 + \int_0^\zeta \right) \frac{\partial\Phi}{\partial t} \Phi_n dZ = - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} dZ + O(A^3) }[/math]
Mean energy flux for a plane progressive wave follows upon substitution of the regular wave velocity potential and taking mean values:
[math]\displaystyle{ \bar P = - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} dZ = \frac{1}{2} \rho g A^2 + \left( \frac{1}{2} \frac{g}{\omega} \right) }[/math]
or
[math]\displaystyle{ \bar P = \bar \varepsilon V_g, \quad V_g = \mbox{group velocity} = \frac{1}{2} V_p \equiv \frac{1}{2} C }[/math]
It follows from this exercise that the mean energy flux of a plane progressive wave is the product of its mean energy density times a velocity which equals [math]\displaystyle{ \frac{1}{2} }[/math]. The phase velocity in deep water we call this the group velocity of deep water waves and it is defined as:
[math]\displaystyle{ V_g = \frac{1}{2} V_P = \frac{1}{2} \frac{g}{\omega} }[/math]
A more formal proof that this is the velocity with which the energy flux of plane progressive waves propagates is to ask the following question:
[math]\displaystyle{ \longrightarrow }[/math] What needs to be the horizontal velocity [math]\displaystyle{ U_n \equiv U }[/math] of a fluid boundary so that the mean energy flux across it vanishes?
This can be found from the solution of the following equation:
[math]\displaystyle{ \overline{P(t)} = 0 = \rho \ {\overline{\int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} dZ}}^t - U \ {\overline{\int_{-\infty}^0 \left(\frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) dZ}}^t = 0 }[/math]
Where terms of [math]\displaystyle{ O(A^3) }[/math] have been neglected. Note that within linear theory, energy density and energy flux are quantities of [math]\displaystyle{ O(A^2) }[/math]. If higher-order terms are kept then we need to consider the treatment of second-order surface wave theory, at least.
Solving the above equation for [math]\displaystyle{ U }[/math] we obtain:
[math]\displaystyle{ U = \frac{\rho \ {\overline{\int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} dZ}}^t}{{\overline{\int_{-\infty}^0 \left( \frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) dZ}}^t} }[/math]
Upon substitution of the plane progressive wave velocity potential and definition of pressure from Bernoulli's equation we obtain:
[math]\displaystyle{ U \equiv V_g = \frac{1}{2} \frac{g}{\omega} = \frac{1}{2} V_P }[/math]
Note that [math]\displaystyle{ U \equiv V_g }[/math] by definition. If the above exercise is repeated in water of finite depth the solution for [math]\displaystyle{ U }[/math] after some algebra is:
[math]\displaystyle{ U = V_g = \left( \frac{1}{2} + \frac{KH}{\sinh 2KH} \right) V_P }[/math]
with
[math]\displaystyle{ \omega^2 = gK \tanh KH \, }[/math]
It may be shown that the group velocity [math]\displaystyle{ V_g }[/math] is given in terms of [math]\displaystyle{ \omega \ne k \, }[/math] by the relation
[math]\displaystyle{ V_g = \frac{d\omega}{d K} }[/math]
This relation follows from the very elegant "device" due to rayleigh which applies to any wave form:
Consider two plane progressive waves of nearly equal frequencies and hence wavenumbers. Their joint wave elevation is given by
[math]\displaystyle{ \zeta(x,t) = A cos ( \omega_1 t - K_1 x) + A cos ( \omega_2 t - K_2 x) \, }[/math]
where the amplitude is assumed to be common and:
[math]\displaystyle{ \omega_2 = \omega_1 + \Delta \omega, \quad | \Delta\omega | \ll \omega_1 , \omega_2 }[/math]
[math]\displaystyle{ K_2 = K_1 + \Delta K, \quad | \Delta K | \ll K_1 , K_2 }[/math]
Converting into complex notation:
[math]\displaystyle{ \zeta(x,t) = A \mathbf{Re} \{ e^{i\omega_1 t - i K_1 x} + e^{i\omega_2 t - i K_2 x} \} = A \mathbf{Re} \{ e^{i\omega_1 t - i K_1 x} + e^{i\omega_1 t - i K_1 x + i \Delta\omega t - i \Delta K x} \} }[/math]
[math]\displaystyle{ = A \mathbf{Re} \{ e^{i\omega_1 t - i \K_1 x} \left( 1 + e{i\Delta\omega t - i \Delta K x \right) \} }[/math]
The combined wave elevation [math]\displaystyle{ \zeta \, }[/math] vanishes identically where [math]\displaystyle{ F \equiv 0 \, }[/math].
[math]\displaystyle{ F = 0 \, \ }[/math] when:
[math]\displaystyle{ e^{i(\Delta t - \Delta K x)} = -1 }[/math]
or when:
[math]\displaystyle{ \Delta \omega t - \Delta K x = ( 2 n + 1 ) \pi, \qquad n = 0, 1, 2, \cdots }[/math]
Solving for [math]\displaystyle{ x \, }[/math] we obtain:
[math]\displaystyle{ x = \frac{1}{\Delta K} \{ (2n+1)\pi + t \Delta\omega \} \equiv X(t) }[/math]
For values of [math]\displaystyle{ X(t)\, }[/math] given above, [math]\displaystyle{ \zeta \equiv 0 \, }[/math]. These are the nodes of the bi-chrohatic wave train where at all times the elevatio vanishes and hence the evergy density [math]\displaystyle{ \equiv 0 }[/math]. The wave group has the form
