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8/27/10
Physical Oceanography, MSCI 3001
Oceanographic Processes, MSCI 5004
The Equations of Motion

du 1 
= ΣF
dt ρ
Dr. Katrin Meissner
[email protected]
Ocean Dynamics
€
du 1
= ΣF
dt ρ x
dv 1
= ΣF
dt ρ y
dw 1
= ΣF
dt ρ z
Horizontal Equations:
€
Acceleration = Pressure Gradient Force + Coriolis
du
1 dp
=−
+ fv
dt
ρ dx
dv
1 dp
=−
− fu
dt
ρ dy
Or, for a Barotropic Ocean:
du
dη
= −g
+ fv
dt
dx
dv
dη
= −g
− fu
dt
dy
Vertical Equation:
€
€
Pressure Gradient force = Gravitational Force
If the only force acting on a water parcel is the Coriolis force,
the Navier Stokes equations can be simplified to:
If the only force acting on a water parcel is the Coriolis force,
the Navier Stokes equations can be simplified to:
du
= fv
dt
dv
= − fu
dt
du
= fv
dt
dv
= − fu
dt
Northern Hemisphere:
f>0
What does this mean?
u<0
u>0
€
Northern Hemisphere:
f>0
What does this mean?
u<0
u>0
€
y
v>0
x
v<0
y
v>0
v<0
x
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8/27/10
If the only force acting on a water parcel is the Coriolis force,
the Navier Stokes equations can be simplified to:
du
= fv
dt
dv
= − fu
dt
Horizontal Equations:
Acceleration = Pressure Gradient Force + Coriolis
du
1 dp
=−
+ fv
dt
ρ dx
Southern Hemisphere:
f<0
dv
1 dp
=−
− fu
dt
ρ dy
What does this mean?
u<0
u>0
€
If pressure gradients are small:
Inertia currents
€
y
v>0
du
= fv
dt
dv
= − fu
dt
v<0
the water flows
around in a circle
with frequency |f|.
T=2π/f
x
T(Sydney) = 21 hours 27 minutes
€
Scaling arguments:
1. If there are no surface slopes or horizontal density differences then there will
be no pressure gradient force (i.e. left with Coriolis, inertia currents)
Scaling arguments:
What forces are important in a bathtub?
What kind of speeds will the water get up to? What kind of accelerations? What
surface slopes?
What forces are important in a bathtub?
Magnitude of the pressure gradient force:
What kind of speeds will the water get up to? What kind of accelerations?
What surface slopes?
dη
= 0.1m /1m = 0.1
dx
dη
g
= 10ms−2 (0.1) = 1ms−2
dx
du
dη
= −g
+ fv
dt
dx
dv
dη
= −g
− fu
dt
dy
Magnitude of the Coriolis force:
fu = 7 × 10 −5 × 1ms−2 = 7 × 10 −5 ms−2
du
dη
= −g
+ fv
dt
dx
dv
dη
= −g
− fu
dt
dy
Barotropic!
€
Coriolis << Pressure Gradient,
so we can neglect rotation effects
€
Acceleration in the bathtub is driven by pressure
differences (due to changes in surface slopes)
€
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But the ocean is not a bathtub….
Scaling arguments:
1.  If there are no surface slopes or horizontal density differences then there will be
no pressure gradient force (i.e. left with du/dt=fv, dv/dt=-fu, inertia currents):
du
= fv
dt
dv
= − fu
dt
2.  If you are sitting in your bathtub, you are in a barotropic environment AND the
Coriolis force can be €
neglected:
du
dη
= −g
dt
dx
dv
dη
= −g
dt
dy
We will conduct a scaling analysis on our equations
of motion ...
to find further simplifications for motions with a
period greater than ~10 days
du
1
=−
dt
ρ
Scaling Analysis:
dv
1
T~10 days = 8.64 x 105 s ~ 106 s
=−
dt
ρ
u,v ~ U ~ 1cms-1 - 1ms-1
dp
+ fv
dx
dp
− fu
dy
f~ 10-4 s-1
€
Acceleration << Coriolis
€
Geostrophic Balance
Geostrophic Balance
Acceleration is much smaller than Coriolis and Pressure
Gradient Force (this is true almost everywhere in the ocean)
The ocean is in Geostrophic Balance (= balance between
Pressure Gradient and Coriolis Forces)
Ocean is in
“Steady State”
(no acceleration)
du/dt is negligible
€
du
1 dp
=−
+ fv
dt
ρ dx
1 dp
= fv
ρ dx
dv
1 dp
=−
− fu
dt
ρ dy
1 dp
= − fu
ρ dy
•  What does the geostrophic
balance mean physically?
•  Suppose we have a difference
in sea-level height.
•  Water will want to move from
the region of high pressure
towards the region of low
pressure.
€
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8/27/10
Which direction is the
Geostrophic wind? (f <0 SH)
Geostrophic Balance
y
•  As the water starts to move,
the Coriolis effect (rotation)
deflects the water to the right
(NH) or left (SH).
x
•  The water keeps getting
deflected until the force due to
the pressure difference
balances the Coriolis force.
PG
•  This balance is called a
geostrophic balance and the
resulting current is referred to
as a geostrophic current.
CF
V
13
Geostrophy Problem 1
Geostrophy Problem 2
In the NH: Which way does the current flow if sea
level height is increasing towards the South?
Which direction does the water flow around this pressure
feature if it is in the Northern Hemisphere?
You can use the equations, or just think about the forces
NH: f>0
dη/dy<0
u=−
g dη
f dy
u=(-)(+)(+)(-)>0 East!
dη
fv = g
dx
dη
fu = −g
dy
C
€P
y
€
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Anticyclone
Geostrophy Problem 3
Cyclone
Which direction does the water flow around this pressure
feature if it is in the Southern Hemisphere?
Northern
Hemisphere
clockwise
counterclockwise
Southern
Hemisphere
Geostrophy Problem 4
Anticyclonic circulation:
A certain ocean current has a height change of 1.1 m (increasing to the
east) over its width of 100 km at 45° N. How fast is the current flowing?
 ALWAYS around a high pressure system.
 clockwise in the Northern Hemisphere
fv = g
 counter-clockwise in the Southern Hemisphere
dη
Δη
≈g
dx
Δx
f = 2Ωsin(φ) = 1x10-4 s-1
g = 10 ms-2
Δη = 1.1m
Δx = 100x1000 m
V=1 m/s
Cyclonic circulation:
 ALWAYS around a low pressure system.
 counter-clockwise in the Northern Hemisphere
 clockwise in the Southern Hemisphere
€
Summary: Geostrophy is the balance between pressure gradient forces and the
Coriolis force. All major current systems in the ocean can essentially be
considered geostrophic. Geostrophy does not work over short periods of time or
small distances (other forces become dominant). Geostrophy also fails in regions
where friction becomes important.
5
8/27/10
Forces on a Parcel of Water



du 1  
= ( Fg + FC + FP + F f + ...)
dt ρ



du 1  
= ( Fg + FC + FP + F f + ...)
dt ρ
• 
• 
• 
• 
Gravity
€
Coriolis
Pressure
Friction
du 1
= ∑ Fx
dt ρ
dv 1
= ∑ Fy
dt ρ
dw 1
= ∑ Fz
dt ρ
€
The last force to consider is friction. This is only important at
continental boundaries, at the bottom of the ocean, and at
the surface (due to wind).
What will the friction term look like? We know that friction
always tries to retard motion.
€
Effects of Friction
A simple model for the frictional force at the sea floor in the x and y
direction is:
Fx = -ru
Fy = -rv
€
Rayleigh frictional dissipation, r is a coefficient (r ~ 10-7s-1)
du
1 dp
=−
+ fv − ru
dt
ρ dx
dv
1 dp
=−
− fu − rv
dt
ρ dy
Hence the equations of motion become
€
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8/27/10
Fx = -ru
Fy = -rv
Henry Stommel
(1948)
To examine the effects of friction, consider the
simple balance:
du
i.e no pressure gradient forces and
Coriolis + pressure gradient forces
dt
€
Harald Sverdrup
(1947)
no coriolis force. Motion is just
decelerated because of friction.
u(t) = uoe −rt
A solution is:
wind forcing
= −ru
1/r represents the time
it takes for the speed to
drop to 1/e (~1/3) of its
initial value.
€
So that at t=0, u(0)=uo
U
€
Uo
Velocity decreases with
time because of friction.
Coriolis + pressure gradient forces
1/3Uo ≈ Uoe-1
wind forcing
06.09.2010 (Monday)
Mid Semester Break (no class)
13.09.2010 (Monday)
Dr. Caroline Ummenhofer
(and Dr. Matthew England)
20.09.2010 (Monday)
Dr. Caroline Ummenhofer
27.09.2010 (Monday)
Dr. Laura Ciasto
04.10.2010 (Monday)
Labour Day (no class)
1/r
26
Time
Thermal Wind Balance
•  So far we have assumed that density ρ is constant (barotropic)
•  Small horizontal changes in ρ can result in large vertical changes in current/
wind – e.g. near fronts and eddies
1 dp
= fv
ρ dx
• Starting with the geostrophic balance
06.10.2010 (Wednesday!) Me again… on a WEDNESDAY!
1 dp
= − fu
ρ dy
• We can differentiate the equations with regard to depth (z)
1 dp
ρf dx
1 dp
u=−
ρf dy
dp
= ρg
dz
v=
€
dv d ⎛ 1 dp ⎞ €
= ⎜
⎟
dz dz ⎝ ρf dx ⎠
du d ⎛ 1 dp ⎞
= ⎜ −
⎟
dz dz ⎝ ρf dy ⎠
dp
= ρg
dz
€
7
8/27/10
Thermal Wind Balance
Thermal Wind Balance
•  So far we have assumed that density ρ isdvconstant
g dρ (barotropic)
dv
1 d
=
= in current/
( ρg)
•  Small horizontal changes in ρ can result in large
vertical changes
dz ρf dx
dz ρf dx
wind – e.g. near fronts and eddies
du
g dρ
du
1 d
=−
=−
( ρg)
dz
ρf dy
dz
ρf dy
• Starting with the geostrophic balance
dp
dp
= ρg
= ρg
dρ
du
u
<< ρ
dz
dz
dz
dz
• We can differentiate the equations with regard to depth (z), then substitute for
dp/dz
€
€
€ dv d ⎛ 1 dp ⎞
dv
1 d ⎛ dp ⎞
dv
1 d ⎛ dp ⎞
1 dp
=
⎜
⎟
=
=
⎜
⎟
⎜ ⎟
v=
dz dz ⎝ ρf dx ⎠
dz ρf dz ⎝ dx ⎠
dz ρf dx ⎝ dz ⎠
ρf dx
du d ⎛ 1 dp ⎞
du
1 d ⎛ dp ⎞
du
1 d ⎛ dp ⎞
1 dp
= ⎜ −
=−
⎜ ⎟
=−
⎟
u=−
⎜ ⎟
dz dz ⎝ ρf dy ⎠
dz
ρf dy ⎝ dz ⎠
dz
ρf dz ⎝ dy ⎠
ρf dy
dp
= ρg
dz
€
€
dp
= ρg
dz
dv
g dρ
=
dz ρf dx
du
g dρ
=−
dz
ρf dy
i.e. horizontal density gradients in temperature (T) and
salinity (S) can explain the change in horizontal velocity
€
with depth (vertical profile of horizontal velocity).
dp
= ρg
dz
dp
= ρg
dz
€
For geostrophic conditions:
The vertical structure of u and v is related to the horizontal
density gradients
€
Thermal Wind Balance:
Cold Core Eddy Example
Thermal Wind Balance
dv
g dρ
=
dz ρf dx
dρ
>0
dx
€
NH: f > 0, so,
€
Velocity is into the page and increases with depth
€
dv
>0
dz
i.e. v gets more
positive with
increasing depth
8
8/27/10
Thermal Wind Balance:
Cold Core Eddy Example
Thermal Wind Balance:
Cold Core Eddy Example
dv
g dρ g Δρ
=
≅
dz ρf dx ρf Δx
y
x
How does the geostrophic
current velocity in the eddy
change with depth?
cold/saline core
x
y
Potential Density through an
Eddy near the Gulf Stream.
€
dv
g dρ
=
dz ρf dx
du
g dρ
=−
dz
ρf dy
Estimate the density
€ gradient at x=40km:
ρ changes from 1026.8 to 1027.2 kgm-3
over 25 km.
Δρ 1026.8 −1027.2
=
= −1.6 × 10 −5
Δx
25km
dv
g
=
× −1.6 × 10 −5
dz ρf
= −1.5 × 10 −3 s−1
27.9
€
Thermal Wind Balance:
Cold Core Eddy Example
dv
g dρ g Δρ
=
≅
dz ρf dx ρf Δx
y
Thermal Wind Balance:
Cold Core Eddy Example
This means that at z = 500m,
dv
= −1.5 × 10 −3 s−1
dz
x
y
let’s see how much v varies over 100m of
depth at z=500m (Δz=100m, Δv=?):
€
€
To work out the actual velocities,
we need one more piece of
information: at depth (say 2000m)
the velocity is 0. This is called the
depth of no motion.
Δv = −1.5 × 10 −3 s−1 × 100m
= −0.15ms−1 = −15cms−1
i.e the velocity shear ~ -15 cm/s per
€ 100m depth increase.
27.9
This means that v (velocity into page) is
getting more negative as we get deeper
x
27.9
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8/27/10
•  Going down in depth, the density surfaces flatten out.
•  We can assume a ‘level of no motion’ where there is no longer a
change in density.
•  Hence we can calculate the change in velocity up through the
water column.
•  Note that on the other side of the eddy, the density surfaces
slope the other way, so the circulation must also be in the
opposite sense.
Thermal Wind Balance:
Cold Core Eddy Example
A physical explanation
NH
f>0
Step 1: denser in the middle so
the surface will be depressed
V=+.3+.15
100m
V=+.15+.15
V=0+.15
V=0
100m
100m
V is positive into the page
v=
1 dp
ρf dx
€
Thermal Wind Balance:
Cold Core Eddy Example
P
NH
f>0
Thermal Wind Balance:
Cold Core Eddy Example
NH
f>0
Step 3: Forget the barotropic
component. What happens at
greater depth?
Density is higher at the
centre than further away.
Step 2: Start by figuring out
the pressure force due to the
surface slope: Pressure
increases moving away from
centre
 barotropic component
Remember the
fishtank experiment
Remember the
fishtank experiment
ρ1
<
ρ2
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8/27/10
Thermal Wind Balance:
Cold Core Eddy Example
Thermal Wind Balance:
Cold Core Eddy Example
NH
f>0
Step 3: Forget the barotropic
component. What happens at
greater depth?
Density is higher at the
centre than further away. So
the pressure force, just due
to density would be in the
positive x-direction (on the
right hand side of the gyre).
The density gradient causes
a pressure force that
increases with depth.
Step 4: Add up barotropic
component (black) and
baroclinic component (blue).
Keep repeating this down the
water column, until there is
no density gradient (i.e. at
the bottom of the cold core
eddy)
A smaller pressure gradient
force means a smaller
geostropic velocity with depth
Opposite happens on the left
side of the gyre.
Remember the
fishtank experiment
ρ1
<
NH
f>0
v=
 baroclinic component
ρ2
1 dp
ρf dx
€
Thermal Wind Balance:
Cold Core Eddy Example
Thermal Wind Balance:
Cold Core Eddy Example
NH
f>0
Step 5: If we add up the
forces due to the surface
slope and due to the density
gradient we get a pressure
gradient force that decreases
with depth.
C
P
P
NH
f>0
Finally we need to
use the geostrophic
relationship:
If we have a pressure
gradient in the x
direction, it will create
a geostrophic velocity
in the y direction,
proportional in
strength to the
pressure gradient.
C
v=
1 dp
ρf dx
€
11
8/27/10
Example:
Derive the rotation of these cold and warm water
eddies, in the SH, using the thermal wind
balance.
Given that u & v = 0 at 2000m (depth of no motion) estimate the surface current.
Thermal Wind Balance
For geostrophic flow (i.e. pressure is balanced by Coriolis):
• Geostrophic flow in the presence of
horizontal density gradients
• Horizontal density gradients (T,S) can
explain vertical velocity changes
dv
g dρ
=
dz ρf dx
du
g dρ
=−
dz
ρf dy
• To know absolute velocity we need extra
information (i.e. we need to know
absolute velocity at some depth)
€
-  Can figure out surface velocity from surface heights.
-  Often we assume that at a certain depth (e.g. 2000m) velocities are zero –
this is called “the depth of no motion”.
-  Once we know the velocity at the surface or the depth of no motion we
can calculate velocity at all other depths using the thermal wind equation.
Summary: Ocean Dynamics
Most of the motion in the ocean can be understood in terms of Newton’s Law that
the acceleration of a parcel of water (how fast its velocity changes with time –
du/dt) is related to the sum of forces acting on that parcel of water.
We can split the forces, velocities and accelerations into south-north (y,v), west-east
(x,u) and up-down (z,w) components.
In the vertical direction the acceleration is related to the difference between the
water weight and the bouyancy (or pressure) force. When there is a vertical
density gradient this leads to oscillations (Brunt Väisälä frequency N). The
density gradient tries to inhibit vertical motion (and mixing). These vertical
accelerations are generally very weak, so we get the hydrostatic equation.
If the hydrostatic equation is integrated over depth, it just says that the
pressure at a point just equals the weight of water above that point.
Acceleration in the horizontal can be driven by a number of different forces:
(1)  The pressure gradient force. This exists whenever there is a surface slope and/
or a horizontal density gradient.
(2)  Coriolis (because we live on a rotating planet). It is very weak, so we only feel its
effect over long times (> few days) and large distances (> 10s of km). Coriolis
only affects moving fluids, deflecting to the right in the NH and to the left in SH.
(3)  Friction. Also only acts on moving water. Always acts to slow down motion.
Important at the boundaries of the ocean.
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Summary: Ocean Dynamics
Or for a constant density (barotropic ocean):
du
1 dp
=−
+ fv − ru
dt
ρ dx
dv
1 dp
=−
− fu − rv
dt
ρ dy
€
du
dη
= −g
+ fv − ru
dt
dx
dv
dη
= −g
− fu − rv
dt
dy
Over much of the ocean, the flow is steady (i.e. du/dt=dv/dt=0) and friction is
negligible, so we are left with the geostrophic balance i.e. pressure gradient forces
€ at right angles to the pressure gradient.
balance coriolis. The current moves
P
1 dp
1 dp
= fv ,
= − fu
ρ dx
ρ dy
C
When there is a horizontal density gradient the velocity changes with depth.
This can be calculated using the thermal wind equations
€
dv
g dρ
=
dz ρf dx
,
du
g dρ
=−
dz
ρf dy
€
13