6. Internal waves
Consider a continuously stratified fluid with ρo(z) the vertical density profile.
z
ρ
p'
ξ
p
ρο(z)
Figure by MIT OpenCourseWare.
Figure 1.
At a point P raise a parcel of water by small amount ξ from its equilibrium position P
adiabatically. The change in pressure experienced by the parcel is
dp = -ρo gξ
dp
if ρo(z) is the medium density, while the change in density is dρ = 2 [Remember the
cs
⎛ ∂p ⎞
definition of adiabatic compressibility ⎜ ⎟ = c2s]. The buoyancy force acting on the
⎝ ∂ρ ⎠
particle induces an acceleration because the density of the particle at the displaced
position is different from the background density of the medium
ρ o | P'= ρ o | P +
dρ o
ξ and
dz
ρ |particle at P' = ρ |particle at P+dρ
ρ
= o
−ρ ogξ
cs2
1
d2ξ
ρ gξ
= g[(ρ o + ρ oz ξ) − (ρ o − o2 )]
dt
cs
ρo
2
d2ξ
dt
2
=
g
ρ g
[(ρ oz + o2 )ξ]
ρo
cs
d2ξ
gρ oz g 2
+
(−
− 2 )ξ = 0
ρo
dt 2
cs
Define N 2 (z) =
d 2ξ
dt
2
2
−g dρ o g 2
the Brünt-Vaisala frequency or buoyancy frequency
−
ρ o dz cs2
+ N 2ξ = 0
is the equation of the harmonic oscillator with solution e±iNt.
Thus the parcel oscillates about its equilibrium position and the Brünt-Vaisala frequency
is the natural frequency of oscillation determined by the local density stratification and
the fluid’s compressibility. As we saw, in the ocean compressibility effects are negligible
and we can assume
N2 =
−g ∂ρ o
ρ o ∂z
The motion is the oscillation of the particle around the equilibrium position with
frequency N. Let us compare the frequency of surface gravity waves with the buoyancy
frequency and use deep water gravity waves
Δρ
N
1 ∂ρ o 1 / 2
1 ∂ρ o 1 / 2
ω int .
) ~ (−
λ)
≈ o( o )1 / 2
=
= (−
ωsurf.
ρo
kρ o ∂z
ρ o ∂z
gk
k~
1
∂ρ
; λ o = Δρ o
∂z
λ
density change over λ
2
Δρ o ~10−3
over total depth:
ωint << ωsurface .
The frequency of internal waves is much less than the frequency of surface
gravity waves. The restoring force for surface gravity waves is g; the restoring force for
internal waves is
Δρ o
g, reduced gravity.
ρo
Internal waves in an unbounded fluid
Consider an incompressible, stratified non-rotating fluid characterized by a basic
state of rest and hydrostatic equilibrium:
u ≡ o, ρ = ρ o (z), p = p o (z) and
dp o
= −gρ o
dz
The fluid experiences small amplitude perturbations around the basic state, so we can use
the linearized equations of motion, assuming also that the motion is frictionless and
adiabatic:
(1) ρ o
∂u
∂p
=−
∂t
∂x
ρtotal = ρo+ρ; ρ<<ρo
(2) ρ o
∂v
∂p
=−
∂y
∂t
ptotal = po+ρ; p<<po
(3) ρ o
∂w
∂p
= − − ρg
∂t
∂z
ρ = perturbation density
(4)
∂u ∂v ∂w
=0
+
+
∂x ∂y ∂z
(5)
∂p
∂ρ
+w o =0
∂z
∂t
p = perturbation pressure
The basic reference state cancels out in 3
If u = v = o and p = o, we have the special solution
3
ρo
∂w
∂ρ
∂ρ
= −ρg;
= −w o
∂t
∂z
∂t
∂2w
∂ρ
g∂ρ o
∂2w −g ∂ρo
)w = o
ρ o 2 = −g = +
w → 2 +(
∂t
ρ o ∂z
∂z
∂t
∂t
General equation
Differentiate (4):
∂ ∂u ∂ ∂v ∂ ∂w
+
+
=0
∂t ∂x ∂t ∂y ∂t ∂z
Substitute into it (
∂u ∂v
, ) given by (1) and (2)
∂t ∂t
1 ∂p
1 ∂p
∂
∂
∂ ∂w
(−
) + (−
)+ ( )=0
∂x ρ o ∂x ∂y ρ o ∂y ∂z ∂t
1 2
∂2w
∇ Hp
=
∂z∂t ρ o
(I)
where ∇ 2H =
∂2
∂2
+
∂x 2 ∂y 2
Eliminate ρ between (3) and (5)
ρo
∂2w
∂2p
∂ρ
∂ 2p
dρ
=
−
=
−
−
g
+ gw o
2
∂z∂t
∂z∂t
∂t
∂z
∂t
∂2w
g ∂ρ o
1 ∂2p
+ (−
)w = −
ρ o ∂z
ρ o ∂z∂t
∂t 2
or
∂2w
1 ∂2p
2
w
=
−
+
N
ρ o ∂z∂t
∂t 2
(II)
Eliminate p between (I) and (II) taking ∇ 2H of (II)
∇ 2H
∂2
∂t
∇ 2 w + N 2∇ 2Hw = −
2 H
Take to the LHS and factorize
1 ∂2
1 ∂2 ∂
∂w
2
p
=
−
)
∇
(ρ o
H
2
2
ρ o ∂z∂t
ρ o ∂t ∂z
∂z
∂2
∂t 2
4
1 ∂
∂2 ⎡ 2
∂w ⎤
w
+
∇
(ρ
)⎥ + N 2∇ 2Hw = 0
⎢
o
2⎣ H
ρ
∂z
∂z
⎦
∂t
o
Internal wave equation
Consider the last term in square bracket:
1 ∂ρ o ∂w ∂2w
∂w
1 ∂
)=
(ρ o
+
ρ o ∂z ∂z ∂z 2
∂z
ρ o ∂z
Ratio
∂ρ o ∂w
∂z ∂z = 1 ∂ρ o • 1 d 2 = d ∂ρ o << 1 = d
∂2 w
ρ o ∂z d
D
ρ o ∂z
ρo 2
∂z
d = vertical scale of w ~ thickness of thermocline
and D is the ocean depth or atmospheric height -> valid also in the atmosphere. So we
can ignore the first term and we simplify to:
⎛ ∂2
∂2
∂2 ⎤
∂2 ⎞
∂2 ⎡ ∂2
⎟⎟w = 0
⎢
⎥w + N 2 ⎜⎜
+
+
+
2
∂t 2 ⎣⎢∂x 2 ∂y 2 ∂z 2 ⎦⎥
∂y 2 ⎠
⎝ ∂x
total Laplacian
horizontal Laplacian
∂2
∂t
2
∇ 2w + N 2∇ 2Hw = 0 internal wave equation.
Notice that if N = 0, no stratification, ∇ 2ω = 0 the motion is irrotational -> surface
gravity waves.
Consider now a plane wave solution in three-dimensions
w = wocos(kx+ly+mz-ωt)
(k,l,m) wave number
Substituting into the internal wave equation, we obtain the dispersion relationship
ω2 =
N 2 (k 2 + l2 )
(k 2 + l2 + m2 )
5
or ω 2 =
N 2k 2H
K2
k
ω = ±N H
K
(m,z)
K
m
q
(l,y)
j
(k,x)
KH
Figure 2.
Figure by MIT OpenCourseWare.
ω = ±N cos θ
And the wave numbers are
k = K cosθcoφ, l = K cosθsin φ; m = K sinθ
The dispersion relation for internal waves is of a quite different character from the
dispersion relation for surface waves:
ω of surface waves depends only on the magnitude of K and not on its direction
ω of internal waves is independent of the magnitude of K and depends only on the
orientation of the wave vector, i.e. on the angle θ of the wavenumber K with the
horizontal.
6
Consider the displacement ζ along a line of constant phase by definition perpendicular to
the phase line
z
K
θ
x
θ
Fz
dz θ
θ
Figure by MIT OpenCourseWare.
Figure 3.
Thus particle motion is along the wave crests i.e., along the lines of constant phase
d2ζ
dt
2
+ N 2ζ = 0
We can see this also considering
(u,v,w) = (u o ,v o ,w o )e i(kx + ly+ mz−ωt)
Continuity equation gives
kuo+lvo+mwo = 0
=> K • u = 0
Fluid velocity is perpendicular to the wave vector, i.e. is along the crests of the waves
=> wave motion is transverse
7
K
u
q3
u
high
u
q2
low
q1
high
Figure 4.
phase propagation
Figure by MIT OpenCourseWare.
When the wave vector is horizontal, m = o, the particle motion is purely vertical and
ωmax = N
A displacement ζ along the phase line gives a vertical displacement
dz = ζcosθ
The buoyancy force in the vertical corresponding to the displacement dz is
Fz = -N2dz = -N2ζcosθ
The component of this force along the phase line is
Fζ = -N2ζcos2θ
and the motion of the particle along the phase line is governed by:
d2ζ
dt
2
= Fζ
→
d2ζ
dt
2
+ (N 2 cos 2 θ)ζ = 0
Thus ±Ncosθ is the frequency of oscillation of a particle along the phase line. If θ = 0,
we have vertical oscillations with frequency N.
8
Dispersion effects
For internal waves, the surfaces of constant frequency in wavenumber space are
the cones θ = constant, i.e.
ω = constant
The phase velocity | c |=
on the cone.
ω N
= cosθ is directed along the wave vector and therefore lies
K K
The group velocity c g = i
∂ω ˆ ∂ω ˆ ∂ω
is the gradient of ω in
+j
+k
∂k
∂l
∂m
wavenumber space and therefore by definition is perpendicular to the surfaces
ω = constant. It follows that c g is at right angles with (K,c) .
When the group velocity has an upward component the phase speed has a downward
component.
Let us explicitly evaluate the group velocity
k 2 + l2
N 2k 2H
Write ω 2 = N 2 2 2
=
k + l + m2
K2
∂ω N m2 k
N
=
= sin θ(sin θ cos φ)
2
∂k K K k H K
∂ω N m2 l
N
=
= sin θ(sin θ sin φ)
2
∂k K K k H K
k m
N
∂ω
= −N H 3 = − sinθ(cosθ)
K
∂m
K
Note
ω ∂ω
N2
= − 2 cos2 θ < 0
m ∂m
K
9
The vertical phase velocity is always opposite to the vertical group velocity. Waves
propagating their phase upwards will be propagating their energy downwards and
viceversa,
or
cg =
N
sinθ(sinθcos φ,sinθsin φ,−cosθ) .
K
Explanation of the Figure
For internal waves, wave crest AA’ has moved to 4 wavelengths downward to the
left, the wave group has moved upward parallel to the crests, at right angle to the phase
propagation.
1
For surface gravity waves in deep water, c g = c
2
wavecrest AA’ has again moved 4 wavelengths downward to the left; the wave group has
moved in the same direction but at half the speed, that is two wavelengths: AA’, with
respect to the group, has moved 2 wavelengths.
m
0
0.2
0.4
0.6
0.8
K
cg
1.0
Figure 5.
Figure by MIT OpenCourseWare.
10
z
( K, c )
cg
u
q
x
q
cg
u
(K, c )
Figure 6.
Figures by MIT OpenCourseWare.
A
A'
60o
Figure 7a.
z
11
A
A'
60o
z
Figure 7b.
Figures by MIT OpenCourseWare.
A
A'
60o
z
Figure 7c.
12
MIT OpenCourseWare
http://ocw.mit.edu
12.802 Wave Motion in the Ocean and the Atmosphere
Spring 2008
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