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
<PE>=<kE> 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
MIT OpenCourseWare
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12.802 Wave Motion in the Ocean and the Atmosphere
Spring 2008
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