4. Energy equation for surface gravity waves

Equations of Motion

du ρ dt

= −∇ ˆ p − gρk (1) ρ = constant

∇ •u = o (2) D = constant

Multiply (1) by u

1( 2

ρu • u )t + u • ∇p + gρw = o

∂zIn the linearized case, at every level z w =∂t

and

1[ 2

ρu • u + gρz]t + ∇

• (pu ) = 0

or rate of change (kinetic + potential energy) + divergence (energy flux) = 0

If we integrate from z = -D to z = η, we obtain the kinetic and potential energy and

energy flux per unit horizontal area:

∂∂t

12

η 1∫ ρu • u dz +

−D 2

⎡ ⎤ ρgη2⎢ ⎥ + ∇

⎣ ⎦

η • ∫H (u Hp)dz = 0

−D

as p(η) = 0

and w = 0 at z = -D

∇∂

= ˆ H i ∂x

∂+ J

∂y1 ; gρzdz =2

η z2∫ gρ

−D 21|η =−D 2

gρ(η2 − D2) 1

-D

z

Energy flux

Unit surface area

t (kE + pE)

η

Figure by MIT OpenCourseWare.

Figure 1.

∂or ∂t

[KE + PE]+ ∇ •H Eflux = 0

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:

12

ρg(η2 − D2) we have neglected the term proportional to D2 as an irrelevant

∂D2constant and

∂t= 0.

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(≡ a2). 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) ω2 = gk tanh(kD)

aw φ =k sinh(kD)

cosh k(z + D)cos(kx − ωt)

ρω2a p = −ρgz +k sinh(kD)

cosh k(z + D)cos(kx − ωt)

aω u =sinh(kD)

cosh k(z + D)cos(kx − ωt)

aω w =sinh(kD)

sinh k(z + D)sin(kx − ωt)

2

PE =12

ρga2 cos2(kx − ωt)

KE = +ρ(u2 + w2)

2−D

o∫ dz =

ρa2ω2

2−D

o∫

cos2(kx − ωt) cosh2 k(z + D)sinh2(kD)

+

sin2(kx − ωt) sinh2 k(z + D)sinh2(kD)

dz

Let us now average both quantities over a wave period, indicated by < >

< PE >=14

ρg a2

< kE >= ρa2ω2 14−D

o∫

cosh2k(z + D)sinh2(kD)

dz = as ω2 = gktanh(kD)

=ρ a2ω2 18

sinh(2kD)k sinh2(kD)

=

= ρa2gtanh(kD) sinh(kD)cosh(kD)4sin2(kD)

=14

ρga2

Averaged over a wave period

<PE>=<kE> Equipartition of wave energy between potential and kinetic like in the

oscillator problem. η is a linear oscillator!

And < Etotal >=< KE > + < PE >=ρga2

2

If we now calculate the energy flux vector and average it over one wave period we get:

< Eflux >=< (up)dz >=−D

o∫

=12

ρga2(ω2

gKcoth(kD) c 1

2+

kDsinh(2kD)

⎡

⎣ ⎢ ⎤

⎦ ⎥

3

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

But cg =∂ω∂k

= c 12

+kD

sinh(kD)⎡

⎣ ⎢ ⎤

⎦ ⎥

Thus the period average of the energy equation is:

∂∂t

< E > +∇H •[

c g < E >] = 0

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

(|

k |) only and we can write

∂∂t

< E > +

c g • ∇H < E >= 0

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

only. Then σU = U(x) = kU+ω.

4

MIT OpenCourseWare http://ocw.mit.edu 12.802 Wave Motion in the Ocean and the AtmosphereSpring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.