The Potential Temperature

In order to able to compare water at different depths, it is necessary to remove the effect of pressure (e.g. compression) on the temperature of water.

We define the potential temperature, , as the temperature that a parcel of water would have if transported to the surface adiabatically (no heat transfer).

(Knauss - figure 2.3)

Potential temperature is a theorectical temperature and is calculated based on the properties of the fluid.

Magnitude of the Compressibility Effect on Temperature

is

Oceanographers use 4 Different Representations for Density

in situ density - 103 kg m-3

calculated using in situ salinity, temperature and pressure

density with contribution from compression removed - calculated by assuming surface pressure

density with contribution from compression removeddensity calculated taking in account the effect of compression on Tprovides the easiest density to use for stability calculations

density with contribution from compression removeddensity calculated taking in account the effect of compression on Tdensity calculated taking in account the effect of pressure on compressibilityprovides the most accurate density to use for stability calculations of deep water that is very nearly neutrally stable

density calculated taking in account the effect of compression on T

calculated using the in situ temperature; hence is in error as a result of the change in temperature due to compression, the adiabatic effect

Stratification

Stability is often expressed in terms of N, the Brünt-Väisälä frequency, also known as the Buoyancy Frequency

N has units of s-1 (i.e., radians per second), it is the (angular) frequency at which a parcel of fluid would oscillate up and down if displaced vertically

Typical values of 2/N, i.e., the period of oscillation:

Seasonal thermocline: 5 minutes

Main thermocline: 20 minutes

Deep water: 20 hours

The Hydrostatic Equation

A motionless fluid has w = 0 always, hence Dw/Dt = 0 and

This is the Hydrostatic EquationIt is an equation with which you should all become familar But it’s an approximation; the correct equation is:

i.e., the error we make using the hydrostatic equation is:

But, even when w ≠ 0 this equation will be an excellent approximation if:

What does Hydrostatic mean?

Pressure variation with depth is approximately hydrostatic

At a depth z, pressure is equal to the weight of the overlying water plus the atmospheric pressure

Simplificationsgravitational acceleration g is essentially constant varies by only a few percent with depth

Thus,

where avg is the average density of the water column between the surface and z (Recall z<0)

Conservation Laws

Conservation laws play a critical role in physics

Loosely speaking, conservation laws state that what goes into a system must come out of it, if there are no sinks or sources of the property within the system

Fundamental conservation equations that we will deal with are:

Conservation of mass, volume, salt, and other substances

Conservation of energy (e.g. heat)

Conservation of linear momentum (mass x velocity)

Conservation of angular momentum (vorticity, like angular momentum)

The Substantial Derivative

This is the acceleration of a parcel of fluid, Lagrangian accelerationLagrangian accel. = Eulerian accel. + Advective accel.

More, generally

The Lagrangian D/Dt term is the rate of change experienced by a given tagged water parcel

The Eulerian term term is the local rate of change at a fixed point. is what you get from a current meter at a fixed point in space

The advective term is which contributes to the change in the property due to the displacement of the parcel. It converts between Eulerian and Lagrangian rates of change

Steady Flow in a Pipe

Example

Consider the steady ( everywhere) flow of an incompressible fluid in a narrowing pipe

A water parcel enters the pipe with velocity u1

And leaves it with velocity u2

u2 > u1 since the pipe narrows

The parcel clearly accelerates as it moves into the narrower region, but the local acceleration is zero, so:

Temperature in a Channel After some small time t, the float has moved x. How much has the float temperature of the float changed?

If nothing heats or cools the water as it is advected to the right, the drifter will not experience any temperature change.

t=0

t=t

That is,

Temperature in a Channel

If nothing heats or cools the water as it is advected to the right, the drifter will not experience any temperature change. That is,

The local rate of change in T will equal the negative of the advective flux. An observer at a fixed location would see an apparent cooling.

Temperature in a Channel

If the water parcel is heated, then the Lagrangian change iswhere H is the heating rate. Imagine the spatial structure doesn’t change with time:

Moving Parcels Veer to the Right (Left) in the Northern (Southern) Hemisphere

varies with latitude Parcels veer to the right in the Northern hemisphere, veering increases with latitude

In the Southern hemisphere, things are reverse, f < 0, parcels veer to the left

f = 0 at the equator and at the polesTo simplify things

if the meridional displacement is very small, make the f-plane approximation: f ≈ constant or if the meridional displacement is moderately small, make the beta-plane approximation: f = f0 + y. At 45oN,

Momentum / Continuity Equations

Scaling the Momentum Equation Consider the x-momentum equation for a small-scale flow (i.e., neglecting

Coriolis and turbulent flow terms):

We want to know the relative importance of the advective terms to the friction terms.We do this by scaling the equation

The velocity scale is UThe spatial scale is LThe velocity varies by U over the spatial scale LWith these scales, this means the time scale is L/UDefining dimensionless velocity, scale and time variables by dividing the numerical scales by these scales:

Scaling the Rest of the Momentum Equation Scaling the other terms the same way yields

So the ratio of the advective terms to the viscous term is

is called the Reynolds number, Re

Without solving for the complete flow field, we can use the Reynolds number to get a qualitative picture:

small Re friction-dominated flow (if flow is steady, so )

large Re (say Re > 104) turbulent flow (for a non-rotating fluid)

Characteristics of the Rossby Number

and

small Ro Coriolis-dominated flow (e.g., Coriolis and pressure gradient forces balance) rotation is important

large Ro rotation is not important

For typical values of large scale circulation

U ~ 0.1 m s-1, f = 10-4 s-1, L = 105 m

Geostrophic Flow

Next, we assumed that Flow is steady Friction terms are small Rossby number is small

Then

The balance between the Coriolis terms and the pressure gradient terms is called the Geostrophic Balance

In the northern hemisphere (f > 0)

Geostrophic flow is parallel to isobars with high pressure to the right

Winds (and currents) circle clockwise around a high-pressure areaCounter-clockwise around a low-pressure area

Geostrophic Flow in a Two-Layer Ocean: Pressure in the Upper Layer

Now let’s look at geostrophic flow in a two-layer ocean

First, look at the upper layer (subscript 1)

At location A, the pressure at depth H (= -z), where (H < H1 - h2),

in the upper layer is

At location B, the pressure at this depth is

Geostrophic Flow in a Two-Layer Ocean: Velocity in the Upper Layer

The velocity in the upper layer is:

Geostrophic Flow in a Two-Layer Ocean: Pressure in the Lower Layer

Pressure at two points in the lower layer:

In the lower layer at z = -(H1 + H2), the pressure at A where is

At the same level, at location B, the pressure is

Geostrophic Flow in a Two-Layer Ocean: Velocity in the Lower Layer

So with and

Thus, the velocity in the lower layer is

Thermal Wind: Differentiating the Horizontal Momentum Equations in the Vertical Now, we will show that for geostrophic flow, vertical gradients of velocity are related to horizontal gradients of density Differentiating the geostrophic balance equations with respect to z:

But, the flow is hydrostatic

and f is not a function of depth, so

Simplifications to The Thermal Wind Equations

These eqns can be simplified using the Boussinesq Approximation Expand the left-hand side of the x-equation:

Now consider the ratio of the two terms on the left-hand side:

Thus is generally negligible compared with

So, with this Boussinesq approximation (that density variations are unimportant except in pressure and buoyancy terms), we have

The Boussinesq Approximation Taking over to the right-hand side, we get:

and

These equations are called the Thermal Wind Equations

To neglect the term is to make the Boussinesq approximations - density variations are not important except in the pressure and buoyancy terms

Note that if is small compared with , then cannot also be small, otherwise there is nothing to balance

The Geopotential Height

A geopotential (horizontal or level) surface is everywhere

We define the change in geopotential thus,

We can relate the change in geopotential (geopotential height) to pressure via the hydrostatic equation:

Hence changes in geopotential are related to changes in pressure

So the geopotential at pb relative to pa is

Determining the Geostrophic Velocity from the Geopotential Height

Consider the simple case of a sloping isobar. Imagine at some depth, the velocity is zero. At that depth, the isobar is parallel to the geopotential surface.

Determining the Geostrophic Velocity (continued)

At point A, let the pressure at a depth located zA above this

surface be pA=p0

The geostrophic velocity at this depth is

At the same depth for location B, the pressure is pB=p0+gz

At point A, the distance between the two isobars is zA, so A=gzA. At B, the distance between the same isobars is zB. Thus, B=gzB. The change in geopotential height between B and A is B- A=g(zB- zA)=gz. Thus, the geostrophic velocity at this pressure is

We can determine the geostrophic velocity from the along-pressure gradients of geopotential height.

In the Course Notes, Section 4.4.3 provides another example where the lower layer is not at rest.

Summary of Geostrophic Balance

Barotropic Flow and Property Surfaces Barotropic flow is flow that does not depend on depth

This means that the horizontal pressure gradient does not change with depth Which in turn means that the horizontal gradient of the weight of the overlying fluid must be independent of depth; if it changed, then the pressure gradient would haveto change This means that pressure surfaces must parallel density surfaces. Isobars parallel isopycnals

This means that pressure surfaces must parallel density surfaces. Isobars parallel isopycnalsSeveral surfaces defined: isobaric - surface of constant pressure isopycnal - surface of constant density geopotential - `level’ surface, pendicular to gravity

Baroclinic Flow and the Reference Velocity

Baroclinic flow occurs when isobars and isopycnals cross each other;

The geopotential method allows us to estimate the velocity shear, the baroclinic part of the flow

It does not allow us to determine the barotropic component

We must obtain the barotropic component in some other fashion:

Historically, oceanographers assumed that there would be a `level of no motion’ somewhere deep in the ocean where the flow is quiescent

This assumption could be wrong, especially in shallow coastal regions or in areas where deep ocean currents are present

Use actual currents at a particular depth with an ADCP or current meter

The Vertical Component of Vorticity - the Relative Vorticity

In most large-scale flows, we only consider the vertical component of vorticity.

The vertical component of vorticity is given bywhich we call the relative vorticityThe vorticity is a measure of the tendency for a fluid parcel to rotate

Planetary and Total Vorticity

The vorticity due to the rotating earth, (the planetary vorticity) is f (the Coriolis parameter)

The total or absolute vorticity (vorticity viewed from an inertial coordinate system) is f +

For large-scale oceanic flows, the relative vorticity is typically much smaller than the planetary vorticity

The ratio of the relative vorticity to planetary vorticity,

, is the Rossby Number, Ro, which is typically <<1

Conservation of Potential VorticityAssuming that no torques are applied, the conservation of angular momentum states that:

We define as the potential vorticity, , of the fluid parcelThe potential vorticity of a water parcel is conserved unless external torques are applied to it (wind, friction, etc)

If you stretch a water column of water, its radius decreases, its moment of inertia decreases and its angular momentum increases

Rossby Adjustment Process

QuickTime™ and aYUV420 codec decompressor

are needed to see this picture.

QuickTime™ and aYUV420 codec decompressor

are needed to see this picture.

QuickTime™ and aYUV420 codec decompressor

are needed to see this picture.

QuickTime™ and aYUV420 codec decompressor

are needed to see this picture.

The Basic Assumptions for the Adjustment Problem How does the ocean adjust to an applied wind stress?

We want to determine for the mixed layer as a whole, while the fluid is accelerating from to its final steady state value.Assume: A surface mixed layer of thickness H

The wind blows steadily exerting a uniform stress, , independent of x, y and t.

Non-linear terms are small compared to the other terms: Level surface, constant density

Friction is linearly related to the mixed-layer velocity:

The Steady State Solution

To determine as the fluid accelerates we need to solve:

We begin by finding the steady solution, then the transient solution

The steady state solution, , must satisfy the above with

Solving for u0 and v0 yields, for steady state values

and

The wind is blowing eastward but the flow is not eastwardsThe flow is to the right of the wind for If (no friction),

Flow is southward, perpendicular to the wind

Deviation from the Steady State Solution

Any imbalance in these forces will lead to an acceleration, a time dependence in the velocity

Note, the Coriolis force and frictional force depends on the velocity of the water, which is initially zero

The Time-Dependent Solution - The Adjustment to Steady State

Assume the complete solution is the sum of the steady solution already found and a transient part

where is the time-dependent velocity

Substituting for u, v in the original equations

we obtain

since

The Time-Dependent Solution - The Adjustment to Steady State

With the initial velocity being zero at t=0, then and at this time

The other boundary conditions are that u1=0 and v1=0 as t∞ in order to get the steady state solution for the total velocity

The solution to these differential equations with these boundary conditions is

The Physics of the Complete Solution

The solution contains two separate physical behavioral patterns:

Inertial oscillations given by

This is harmonic motion at angular frequency If one arbitrarily disturbs the ocean (gives it a kick), it will tend to oscillate at the local inertial period With friction present, this oscillation decays away

Tendency toward steady stateFlow almost at right angles to the wind if the friction is small

The Solution With these boundary conditions, the solution for u and v to the equations of motion

is

where+ for f positive, - for f negative

The surface velocity U0

and the Ekman depth

A Schematic of the Solution

Surface flow is 45o to the right of the wind

Deflection increases to the right in successively deeper layers

The depth at which is important is on the order of the Ekman depth - our definition is where the velocity is in the opposite direction of the surface velocity and 4% of it.

The Ekman Depth - A Laboratory Experiment

An experiment was undertaken to visualize the Ekman layerThe experimental conditions were:Table rotation rate Glycerine in water solution

What is the Ekman depth?

Surface stress is from the right to the left

QuickTime™ and aSorenson Video decompressorare needed to see this picture.

QuickTime™ and aSorenson Video decompressorare needed to see this picture.

Total Transport in the Ekman Layer

We call the Ekman transport

Note that this transport is entirely independent of the eddy viscosity - it doesn’t depend on how the stress is distributed in the mixed layer

Mass Conservation Eliminates a Term

But

from continuity

So, the vertically integrated momentum equations become

where

This is the Sverdrup Relation

which is the fundamental equation of large-scale wind-forced circulation - north-south integrated mass transport is proportional to the curl of the wind stress

Meridional Dependences of Ekman and Sverdrup Transports for a Sinusoidal Wind Stress

x has a maximum at 40oN, zero at 30oN and a minimum at 20oN

Sverdrup transport is largest where is a maximum: 30oN (southward)

Ekman transport is largest where has maxima: at 40oN (southward) and at 20oN (northward)

Sverdrup transport is zero at 20oN and 40oN

Ekman and Sverdrup Transports for a Sinusoidal Wind Stress

At 40oN in our example, the Sverdrup transport is zero and the Ekman transport is to the south, so the deep geostrophic must be to the north and the same magnitude of the Ekman transport

Two ways of Looking at This

There are really two ways of looking at this:The Ekman convergence squashes the lower layer and conservation of PV forces it equatorwardThe negative wind-stress curl applies a negative (clockwise) surface torque and the angular momentum of a parcel consequently dccreases (entire water column)

These are effectively the sameIn PO we tend to view the balance as describe in the first case

The Ekman layer is considered to be VERY thin, hence, does not contribute significantly to the angular momentum balance

The Ekman layer applies no horizontal stress to the lower layer

at the base of the Ekman layer

Instead, the Ekman layer communicates the surface torque to the water below by vertical pumping - “Ekman pumping”

Wind Stress in the Subtropical North Atlantic How does potential vorticity relate to Sverdrup flow?

Suppose that the wind stress has negative curl, as in the subtropical North Atlantic

Changes in Vorticity Due to Convergences in the Subtropical

The Ekman transport is greater at C than at A

u(C) > u(A)

Surface water accumulates between A and C (e.g., at B)

The water columns in the lower layer are squashed

Since is conserved in the lower layer, the reduction in H must be matched either by a reduction in or in f

Parcels Must Move South

(small Rossby number)

So, changes in cannot alter +f significantly

f must decrease the fluid moves South

Negative wind stress curl Ekman convergence squashing water columns beneath the Ekman layer Southward flow

The Physics

The wind stress curl. , applies a torque to cylinders of water to This torque changes the angular momentum, L, of the cylinders

For L to change, I (1/H) and/or must change

Sverdrup says that must changeIntegrating to a fixed depth, H, keeps I from changing

But assuming means that the water column will not spin up relative to a coordinate system fixed to Earth

So the only way to change is to change sin; i.e. change

Need Either Another Torque or a Change in PVThis violates the fundamental conservation law for angular momentum:

To resolve this, Either the PV must change as the parcel goes around the gyre

By decreasing the relative vorticity, , orBy decreasing its moment of inertia, 1/H

Or there must be a positive torque applied to the parcel somewhere along its trajectory that compensates for the wind stress curl

Need an Additional Torque Applied to the Fluid ParcelBUT in the long term, a continual decrease in PV as the parcel goes around and around the gyre is unreasonable:

Either a parcel’s vorticity would have to decrease continually, Or its thickness would have to increase continually,

So we need an additional torque . Where does it come from?

The flow must be zero at a coastal boundary,So the boundary applies a stress to the fluid

If this provides the needed positive torque, on which side of the basin will we find the return flow?

Additional Torque Cannot Come from the Eastern Boundary

Let’s assume that the return flow is on the eastern boundary

The return flow on the eastern boundary is inconsistent with PV = 0

Additional Torque Must Come from the Western Boundary

Now assume that the return flow is on the western boundary

The return flow on the western boundary is consistent with PV = 0