1. 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.

2. Magnitude of the Compressibility Effect on Temperature is

3. Oceanographers use 4 Different Representations for Density in situ density - 10 3 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 removed density calculated taking in account the effect of compression on T provides the easiest density to use for stability calculations density with contribution from compression removed density calculated taking in account the effect of compression on T density calculated taking in account the effect of pressure on compressibility provides 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

4. Stratification Stability is often expressed in terms of N, the Br ü nt-V ä is ä l ä 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

5. The Hydrostatic Equation A motionless fluid has w = 0 always, hence Dw/Dt = 0 and This is the Hydrostatic Equation It 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:

6. 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 Simplifications gravitational 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)

7. 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)

8. The Substantial Derivative This is the acceleration of a parcel of fluid, Lagrangian acceleration Lagrangian 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

9. 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 u 1 And leaves it with velocity u 2 u 2 > u 1 since the pipe narrows The parcel clearly accelerates as it moves into the narrower region, but the local acceleration is zero, so:

10. 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,

11. 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.

12. Temperature in a Channel If the water parcel is heated, then the Lagrangian change is where H is the heating rate. Imagine the spatial structure doesn ’ t change with time:

13. 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 poles To 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 = f 0 +  y. At 45 o N,

14. Momentum / Continuity Equations

15. 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 U The spatial scale is L The velocity varies by U over the spatial scale L With these scales, this means the time scale is L/U Defining dimensionless velocity, scale and time variables by dividing the numerical scales by these scales:

16. 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 > 10 4 )  turbulent flow (for a non-rotating fluid)

17. Characteristics of the Rossby Number and small R o  Coriolis-dominated flow (e.g., Coriolis and pressure gradient forces balance) rotation is important large R o  rotation is not important For typical values of large scale circulation U ~ 0.1 m s -1, f = 10 -4 s -1, L = 10 5 m

18. Geostrophic Flow Next, we assumed that Flow is steady  Friction terms are smallRossby 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 area Counter-clockwise around a low-pressure area

19. 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 < H 1 - h 2 ), in the upper layer is At location B, the pressure at this depth is

20. Geostrophic Flow in a Two-Layer Ocean: Velocity in the Upper Layer The velocity in the upper layer is:

21. 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 = -(H 1 + H 2 ), the pressure at A where is At the same level, at location B, the pressure is

22. Geostrophic Flow in a Two-Layer Ocean: Velocity in the Lower Layer So with and Thus, the velocity in the lower layer is

23. 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

24. 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

25. 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

26. 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 p b relative to p a is

27. 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.

28. Determining the Geostrophic Velocity (continued) At point A, let the pressure at a depth located  z A above this surface be p A = p 0 The geostrophic velocity at this depth is At the same depth for location B, the pressure is p B = p 0 +  g  z At point A, the distance between the two isobars is  z A, so  A = g  z A. At B, the distance between the same isobars is  z B. Thus,  B = g  z B. The change in geopotential height between B and A is  B -  A = g(  z B -  z A )=g  z. 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.

29. Summary of Geostrophic Balance

30. 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 Several surfaces defined: isobaric - surface of constant pressure isopycnal - surface of constant density geopotential - `level’ surface, pendicular to gravity

31. 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

32. 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 by which we call the relative vorticity The vorticity is a measure of the tendency for a fluid parcel to rotate

33. 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

34. Conservation of Potential Vorticity Assuming that no torques are applied, the conservation of angular momentum states that: We define as the potential vorticity, , of the fluid parcel The 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

35. Rossby Adjustment Process

36. 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:

37. 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 u 0 and v 0 yields, for steady state values and The wind is blowing eastward but the flow is not eastwards The flow is to the right of the wind for If (no friction), Flow is southward, perpendicular to the wind

38. 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

39. 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

40. The Time-Dependent Solution - The Adjustment to Steady State With the initial velocity being zero at t =0, then andat this time The other boundary conditions are that u 1 =0 and v 1 =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

41. 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 state Flow almost at right angles to the wind if the friction is small

42. 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 U 0 and the Ekman depth

43. A Schematic of the Solution Surface flow is 45 o 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.

44. The Ekman Depth - A Laboratory Experiment An experiment was undertaken to visualize the Ekman layer The experimental conditions were: Table rotation rate Glycerine in water solution What is the Ekman depth? Surface stress is from the right to the left

45. 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

46. 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

47. Meridional Dependences of Ekman and Sverdrup Transports for a Sinusoidal Wind Stress  x has a maximum at 40 o N, zero at 30 o N and a minimum at 20 o N Sverdrup transport is largest where is a maximum: 30 o N (southward) Ekman transport is largest where has maxima: at 40 o N (southward) and at 20 o N (northward) Sverdrup transport is zero at 20 o N and 40 o N

48. Ekman and Sverdrup Transports for a Sinusoidal Wind Stress At 40 o N 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

49. 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 equatorward The 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 same In 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 ”

50. 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

51. 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

52. 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

53. 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 change Integrating 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 

54. Need Either Another Torque or a Change in PV This 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, , or By 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

55. Need an Additional Torque Applied to the Fluid Parcel BUT 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?

56. 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

57. 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