Energy Equation for Surface Gravity Waves

4. Energy equation for surface gravity waves Equations of Motion ρ du = −∇p − gρkˆ dt ∇•u = o ρ = constant (1) (2) D = constant Multiply (1) by u 1 ( ρu • u) t + u • ∇p + gρw = o 2 In the linearized case, at every level z w = ∂z and ∂t 1 [ ρu • u + gρz]t + ∇ • (pu) = 0 2 or rate of change (kinetic + potential energy) + divergence (energy flux) = 0 z Unit surface area η Energy flux -D t (kE + pE) Figure by MIT OpenCourseWare. Figure 1. If we integrate from z = -D to z = η, we obtain the kinetic and potential energy and energy flux per unit horizontal area: η ⎤ ∂⎡η 1 1 2 ⎢ ∫ ρu • udz + ρgη ⎥ + ∇ H • ∫ (u H p)dz = 0 ∂t ⎣−D 2 2 ⎦ −D as p(η) = 0 and w = 0 at z = -D η ∂ ∂ 1 z2 1 2 2 ; ∫ gρzdz = gρ | η ∇ H = ˆi + J −D= gρ(η − D ) 2 2 2 ∂x ∂y −D 1 or ∂ [KE + PE]+ ∇ H • E flux = 0 ∂t Rate of change = horizontal divergence of wave energy flux Bar denotes the quantities per unit horizontal area Notice: 1) In the expression for the integrated potential density: 1 ρg(η2 − D2 ) we have neglected the term proportional to D2 as an irrelevant 2 constant and ∂D 2 = 0. ∂t 2) In the integral for the kinetic energy we can integrate only to z = 0. In fact we are calculating energy to second order in the wave amplitude. To do this, for PE, we must integrate to η to obtain η2 (≡ a 2 ). In the KE, the integral to η would include a correction of 0(u2η)≡o(a3), hence negligible. Let us now consider specifically the surface gravity wave field in one horizontal dimension (x,z,t): η = a cos(kx − ωt) φ= ω 2 = gk tanh(kD) aw cosh k(z + D)cos(kx − ωt) k sinh(kD) p = −ρgz + ρω 2a cosh k(z + D)cos(kx − ωt) k sinh(kD) u= aω cosh k(z + D)cos(kx − ωt) sinh(kD) w= aω sinh k(z + D)sin(kx − ωt) sinh(kD) 2 1 PE = ρga 2 cos 2 (kx − ωt) 2 ⎤ ⎡ 2 ⎢cos 2(kx − ωt) cosh k(z + D)⎥ ⎥ ⎢ sinh 2(kD) ⎥ o ρ(u 2 + w 2 ) o ρa 2ω 2 ⎢ + KE = + ∫ dz = ∫ ⎥ dz ⎢ 2 2 −D −D 2 ⎢ 2 sinh k(z + D)⎥ ⎥ ⎢sin (kx − ωt) 2 (kD) sinh ⎥⎦ ⎢⎣ Let us now average both quantities over a wave period, indicated by < > < PE >= 1 ρg a 2 4 1 cosh2k(z + D) dz = 2 4 sinh (kD) −D o as ω2 = gktanh(kD) < kE >= ρa 2ω 2 ∫ =ρ a 2ω 2 = ρa 2gtanh(kD) 1 sinh(2kD) = 8 k sinh 2(kD) sinh(kD)cosh(kD) 4 sin 2 (kD) = 1 ρga 2 4 Averaged over a wave period = Equipartition of wave energy between potential and kinetic like in the oscillator problem. η is a linear oscillator! And < E total >=< KE > + < PE >= ρga 2 2 If we now calculate the energy flux vector and average it over one wave period we get: o < E flux >=< ∫ (up)dz >= −D 2 ⎡1 kD ⎤ 1 2 ω coth(kD) c ⎢ + = ρga ( ⎥ 2 gK ⎣ 2 sinh(2kD)⎦ 3 But cg = ⎡1 ∂ω kD ⎤ =c⎢ + ⎥ ∂k ⎣ 2 sinh(kD)⎦ Thus the period average of the energy equation is: ∂ < E > +∇ H •[ cg < E >] = 0 ∂t Thus we have the important result that the energy in the wave propagates with the group velocity. If the medium is homogeneous, c g = ∂ω (| k |) only and we can write ∂k ∂ < E > + c g • ∇ H < E >= 0 ∂t For an observer moving horizontally with the group velocity the energy averaged over one phase of the wave is constant. Dispersion relationship for waves moving on a current Suppose I have a wave encountering a current U(x,y), the dispersion relationship is modified by the Doppler shift becoming σ = k(x,y)• U(x,y)+ ω where ω = gk tanh(1k))D is the intrinsic frequency Consider in fact the 1-D example U = U(x) only. Then σ = kU+ ω . 4