8. Internal waves modified by rotation – unbounded fluid
The equations of motion, linearized, now are:
(1)
1 ∂p
∂u
− fv = −
ρ o ∂x
∂t
ρo = ρo(z), po = po(z)
(2)
1 ∂p
∂v
+ fu = −
ρ o ∂y
∂t
f = constant – f-plane
(3)
1 ∂p ρg
∂w
=−
−
ρo ∂z ρo
∂t
Basic state:
(4)
∂u ∂v ∂w
+
+
=0
∂x ∂y ∂z
∂p o
= −ρ og
∂z
(5)
∂ρ
dρ
+w o =0
∂z
∂t
∂
of (1) and f x (2)
∂t
∂2u
∂v
1 ∂2
−
f
=
−
∂t
ρ o ∂x∂t
∂t 2
fx (1) and
=>
∂2u
−1 ∂2p
f 2p
2
+
f
u
=
−
ρ o ∂x∂t ρ o ∂y
∂t 2
∂
of (2)
∂t
f ∂p ∂2u 1 ∂2p
∂2 v 2
1 ∂ 2p
f ∂p
∂v
2
= −f u −
= 2+
=> 2 + f v = −
+
f
ρ o ∂y ∂t
ρ o ∂x∂t
ρ o ∂y∂t ρ o ∂x
∂t
∂t
∂2 u
1 ∂2p
f ∂p
2
+
f
u
=
−
−
2
ρ o ∂x∂t ρ o ∂y
∂t
these give (u,v) if we know p
∂2 v
1 ∂ 2p
f ∂p
2
+
f
v
=
−
+
2
ρ o ∂y∂t ρ o ∂x
∂t
Adopt the procedure followed with f = 0
Take
∂
of incompressibility equation (4)
∂t
1
∂ ∂u ∂ ∂v ∂2w
+
+
=0
∂t ∂x ∂t ∂y ∂z∂t
using the horizontal momentum eqs.
1 ∂p ⎤ ∂ ⎡
1 ∂p ⎤ ∂2w
∂⎡
fv
−
−fu
−
=0
+
⎢
⎥
⎢
⎥+
ρ o ∂x ⎦ ∂y ⎣
ρ o ∂y ⎦ ∂z∂t
∂x ⎣
or
∂2w
1 2
+ f(v x − u y ) =
∇ Hp
∂z∂t
ρo
(I)
(like we did for f = 0)
ζ = vx-uy
The equation for ζ is simply obtained forming the vorticity eq. from the horizontal
momentum eqns.
∂
∂w
(v x − u y ) − f
=0
∂t
∂z
Notice that if f = 0
∂
(v x − u y ) = 0
∂t
However, neither ζ nor
different approach:
(II)
Three functions (p, ζ, w) and 2 eqs.
ζ = 0 for all times
∂w
can be eliminated from (I) and (II). So we must follow a
∂z
Evaluate the horizontal divergence from horizontal momentum eqs.
1 ∂2p
∂
u x − fv x = −
ρ o ∂x 2
∂t
1 ∂ 2p
∂
v y + fu y = −
ρ o ∂y 2
∂t
gives
∂
1
(u x + v y ) − f(v x − u y ) = − ∇ 2HP
∂t
ρo
∂2
2
∂ t
(u x + v y ) − f
∂
1 ∂ 2
(v x − u y ) = −
∇ Hp
∂t
ρ o ∂t
2
But from (II)
∂
∂w
(v x − u y ) = f
∂t
∂z
ux+vy = -wz
from (4)
⎡ ∂2
⎤
1 ∂ 2
2 ∂w
⎢
⎥
+
f
=+
∇ Hp (III)
2
ρ o ∂t
⎢⎣∂t
⎥⎦ ∂z
From (3) and (5)
ρo
∂ 2w
∂ 2p
∂ρ ∂2p
dρ
=
−
−
g
=
+g ow
2
∂z∂t
∂t ∂z∂t
dz
∂t
or
⎛ −g dρ o ⎞
1 ∂ 2p
+
w
=
−
⎜
⎟
ρ o ∂z∂t
∂t 2 ⎝ ρ o dz ⎠
∂ 2w
∂2 w
1 ∂2 p
2
+
N
w
=
−
ρ o ∂z∂t
∂t 2
(IV)
Rewrite (III) and (IV)
⎛ ∂2
⎞ ∂w ∂
ρ o ⎜⎜ 2 + f 2 ⎟⎟
= ∇ 2Hp
⎝ ∂t
⎠ ∂z ∂t
⎛ ∂ 2w
⎞
∂2 p
2
ρ o ⎜⎜ 2 + N w ⎟⎟ = −
∂z∂t
⎝ ∂t
⎠
Take ∇ 2H of the second one:
⎡ ⎛ ∂ 2w
⎞⎤
⎤ −∂
∂2
∂w
∂ ⎡∂
]
∇ 2H ⎢ρ o ⎜⎜ 2 + N 2w ⎟⎟⎥ = − ⎢ ∇ 2Hp⎥ = ρ o ( 2 + f 2 )
⎣
⎦
∂t
∂z
∂z
∂z
⎢⎣ ⎝ ∂t
⎥
∂t
⎠⎦
or:
⎡ ∂ 2w
⎤ ∂ ⎡ ∂2 ∂w
∂w ⎤
⎥=0
ρ o∇ 2H ⎢ 2 + N 2w⎥ + ⎢ρ o 2
+ ρ of 2
∂z ⎥⎦
⎢⎣ ∂t
⎥⎦ ∂z ⎢⎣ ∂t ∂z
3
∂2
∂t
∇ 2 w + N 2∇ 2Hw +
2 H
∂2 ⎡ 1 ∂ ⎛ ∂w ⎞⎤ f 2 ∂ ⎛ ∂w ⎞
⎜ρo
⎟⎥ +
⎜ρo
⎟=0
⎢
∂t 2 ⎣ ρ o ∂z ⎝ ∂z ⎠⎦ ρ o ∂z ⎝ ∂z ⎠
or:
2 ⎞
⎛ 2
∂2 ⎡∂2w ∂2w 1 ∂ ⎛ ∂w ⎞⎤ f 2 ∂ ⎛ ∂w ⎞
2⎜ ∂ w ∂ w ⎟
⎢
⎥+
+
N
+
+
ρ
ρ
+
⎟
⎟
⎜
⎜
o
o
⎜ 2
⎟=0
∂t 2 ⎢⎣ ∂x 2 ∂y 2 ρ o ∂z ⎝ ∂z ⎠⎥⎦ ρ o ∂z ⎝ ∂z ⎠
∂y 2 ⎠
⎝ ∂x
Similarly to what we did for the case of no rotation
1 ∂ ⎛ ∂w ⎞ 1 dρ o ∂w ∂2w
+
⎜ρ o
⎟=
ρ o ∂z ⎝ ∂z ⎠ ρ o dz ∂z ∂z 2
⎛ 1 dρ o ∂w ⎞ ∂2w
and ⎜
<< 1 which is also the Boussinesq approximation as in the non⎟/
⎝ ρ o dz ∂z ⎠ ∂z 2
rotating case
Then the master equation simplifies to:
2 ⎞
⎛ 2
∂ 2 ⎛ ∂ 2 w ∂ 2w ∂ 2w ⎞
∂ 2w
2⎜ ∂ w ∂ w ⎟
⎜⎜
⎟⎟ + f
+
+
+
N
+
⎜ 2
⎟=0
∂t 2 ⎝ ∂x 2 ∂y 2 ∂z 2 ⎠
∂z 2
∂y 2 ⎠
⎝ ∂x
becoming the old one if N = 0
Look for w = wo ei(kx+ly+mz- ωt) we get the dispersion relationship
ω2 =
f 2m2 + N 2 (k 2 + l2 )
2
2
k +l +m
m
2
w
l
k
Figure by MIT OpenCourseWare.
Figure 1.
4
or
⎡N 2 − ω 2 ⎤
m2 = ⎢ 2 2 ⎥(k 2 + l2 )
⎢⎣ ω − f ⎥⎦
Remember:
(m,z)
K
m
q
(l,y)
j
(k,x)
KH
Figure by MIT OpenCourseWare.
Figure 2.
Rewrite the dispersion relationship as:
2
ω =f
2m
K
2
2 kH
+N
2
2
2
m = K sin θ
K
kH = K cos θ
ω 2 = f 2 sin 2 θ + N 2 cos 2 θ
which becomes ω=±Ncosθ
In the atmosphere and ocean N>>f and
if f = 0
N
= 0(100) so that the dispersion curves
f
previously shown still holds. Again, from
r
r
r r
r
∇ • u = 0 and u = u oe i(k•x −ωt)
r r
K • uo = 0
r
r
u is in planes perpendicular to K → transverse waves
5
phase lines are lines of constant p:
r
r
r
∇p = Kp o is parallel to K and perpendicular to u .
Other useful forms of the dispersion relation
ω2=f2sin2θ+N2cos2θ
are
a) N2-ω2 = N2-f2sin2θ-N2cos2θ = N2(1-cos2θ) – f2sin2θ = N2sin2θ-f2sin2θ
N2-ω2 = (N2-f2)sin2θ
b) ω2-f2 = f2sin2θ + N2cos2θ – f2 = -f2(1-sin2θ) + N2cos2θ = N2cos2θ – f2cos2θ
ω2-f2 = (N2-f2)cos2θ
Consider the two limiting cases:
a) f = 0
K
u
w
q
q
q
g
Figure by MIT OpenCourseWare.
Figure 3.
ut = −g
ρ
cosθ
ρo
buoyancy along u
acceleration in u = buoyancy force along u
6
ρ t + wρ oz = 0
→
ρ t + u cosθρ oz = 0
u tt = −
g
g
cosθρt = − (− u cosθρ oz )cosθ
ρo
ρo
u tt + (−
g dρ o
)u cos2 θ = 0
ρ o dz
u tt + N 2 cos2 θu = 0
u is a linear oscillator with ω = N cosθ
b) If N = 0
f
K
q
f
u
q
u
Figure by MIT OpenCourseWare.
Figure 4
r
then the momentum eq. in any direction normal to k is:
r
r
f// component// K
r r
r
ut + ( f × u) = 0
or
r r r
u t + f// × u = 0
f// = f sinθ
r
r
u H + (f sinθ)kˆ '×u = 0
r
kˆ '= unit vector in the direction of K.
r
The motion occurs in circles around K at ω = fsinθ
7
f
K
Figure by MIT OpenCourseWare.
Figure 5.
r
Again cg is by definition the gradient of ω in wavenumber space. Like before, it
is perpendicular to the conical surfaces of constant ω.
Now:
r
N2 − f 2
cg =
− cosθ sinθ[sinθ cosφ,sinθ sinφ,− cosθ
ωΚ
r
The magnitude of cg is now
]
(N2 − f 2 )
− cosθ sinθ
ωΚ
Notice that if f = 0 the magnitude becomes
N2 cosθ sinθ N2 cosθ sinθ N
=
= sinθ
ωΚ
N cosθΚ
Κ
with ω = N cosθ
the previous relationship.
8
Like before, upward propagation of phase implies downward propagation of
energy and vice versa. The particle motions with f≠0, N≠0 are a combination of the two
extreme cases.
Assuming
w = wo ei (kx +ly +mz−ωt )
p = po ei (k×+ly+mz−ωt ) ; ρ = ρ o ei (k×+ly+mz−ut )
r
u = u o ei (kx+ly+z−ωt )
the relationship between w and p follows from
∂2w
1 ∂2p
2
+
N
w
=
−
ρ o ∂z∂t
∂t 2
w=
which gives
p
− mω po
Κω
=−
N2 − ω 2 ρ
(N2 − f 2 )sinθ ρ o
where we have used
N2-ω2 = (N2-f2)sin2θ
m = K sinθ
From
∂2u
1 ∂ 2p
f ∂p
+f u= −
−
2
ρ o ∂× ∂t ρ o ∂y
∂t
2
∂2v
1 ∂2p
f ∂p
2
+
f
v
=
−
+
2
ρ o ∂y∂t ρ o ∂x
∂t
The relations follow between (u, v) and p:
u=
kω + ilf p
lω + ikf p
; v=
ω 2 − f 2 ρo
ω 2 − f 2 ρo
9
If the x-axis is chosen to be in the direction of the horizontal component of the wave
vector K ≡ KH; l = 0
u=
− ik H f p
p
; v=
ω 2 − f 2 ρo
ω 2 − f 2 ρo
k Hω
using ω 2 − f 2 = (N 2 − f 2 )cos2 θ
and
KH = Κ cosθ
we finally get
u=
v=
Κω
p
= − tanθw
(N − f )cosθ ρ o
2
2
− iΚf
if
p − ifu
=
=
+
tanθw
ω
ω
(N2 − f 2 )cosθ ρ o
The perturbation density ρ is obtained from
∂ρ
∂ρ
+w o = 0
∂t
dz
iN2
ρ
=
W
⇒
ρo
gω
g dρ o
N2 = −
ρ o dz
Take now the real parts
ω = wo – cos (kx+my-ωt)
k≡kH; l = 0
⎧ u = − tanθ Re(w) = − tanθw o cos(kx + mz − ωt)
⎪
f
if
⎨
⎪⎩ v = Re( ω tanθw) = − ω tanθ sin(kx + mz − ωt)
10
⎧
iN 2
N2
−N 2
⎪⎪ρ = Re(
w) = −
wρ o sin(kx + mz − ωt) →
ρ ow o sin(kx + mz − ωt)
gω
gω
gω
⎨
ρ o (N 2 − f 2 )sinθ
ρ o (N 2 − f 2 )sinθ
⎪
Re(w) = −
w o cos(kx + mz − ωt)
⎪⎩p = −
kω
kω
Total energy <E> averaged again over one wavelength will obey the same energy
equation as the Coriolis force does no work and hence does not contribute to the energy
equation.
< kE >=
1
ρ o < u 2 + v2 + w2 >
2
⇒ PE = ρg
PE = ρwg
dz
dt
From the adiabatic equation
∂ρ
dρ
+w o = 0
dz
∂t
From N 2 = −
w=
and
PE =
g2
⇒
g dρ o
ρ o dz
w= −
⇒
1
∂ρ
dρo / dz ∂t
−
1
dρ o / dz
∂ρ
ρ o N 2 ∂t
g
=+
g
ρo N2
hence
∂ρ
g 2 ∂ρ2
=
ρo N2 ∂t
2ρ o N2 ∂t
ρ
The total energy averaged over one wavelength is the same as in the non-rotating case
< E >=
K2 1
1
w
ρ o w2o
= ρ o ( o )2
2
cosθ
K2H 2
However in the rotating case energy is not equi-partitioned
<KE> is increased by rotation
11
< KE > ω 2 + f 2 sin 2 θ
=
< PE > ω 2 − f 2 sin 2 θ
<PE> is decreased
aga
tio
n
r
The energy flux is again <F> = <E> cg
Gro
rop
ligh
up
wa
Ph
as
ep
t, to
hea
vy,
pro
pag
rd
atio
n
Hig
aw
h
ay
Low
(a)
Hig
K
h
Resultant
Coriolis
Gravity
y
d
u
0
(b)
Figure by MIT OpenCourseWare.
Figure 6.
12
MIT OpenCourseWare
http://ocw.mit.edu
12.802 Wave Motion in the Ocean and the Atmosphere
Spring 2008
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