1. Horizontal Pressure Gradients Pressure changes provide the push that drive ocean currents Balance between pressure & Coriolis forces gives us geostrophic currents Need to know how to diagnose pressure force Key is the hydrostatic pressure

2. Horizontal Pressure Gradients Two stations separated a distance  x in a homogeneous water column (  = constant) The sea level at Sta. B is higher than at Sta. A by a small distance  z Hydrostatic relationship holds Note,  z/  x is very small (typically ~ 1:10 6 )

3. Horizontal Pressure Force

4. Pressure Gradients @ Sta A seafloor p h (A) =  g z @ Sta B seafloor p h (B) =  g (z +  z)  p = p h (B) - p h (A) =  g (z +  z) -  g z  p =  g  z HPF  p/  x =  g  z/  x =  g tan  or HPF per unit mass = g tan  [m s -2 ]

5. Horizontal Pressure Force

6. Geostrophy What balance HPF? Coriolis!!!!

7. Geostrophy Geostrophy describes balance between horizontal pressure & Coriolis forces Relationship is used to diagnose currents Holds for most large scale motions in sea

8. Geostrophic Relationship Balance:Coriolis force = fu HPF = g tan  Geostrophic relationship: u = (g/f) tan  Know f (= 2  sin  ) & tan , calculate u

9. Estimating tan  Need to slope of sea surface to get at surface currents New technology - satellite altimeters - can do this with high accuracy Altimeter estimates of sea level can be used to get at  z/  x (or tan  ) & u geo Later, we’ll talk about traditional method

10. TOPEX/Poseidon

11. Satellite Altimetry

12. Satellite measures distance between it and ocean surface Knowing where it is, sea surface height WRT a reference ellipsoid is determined SSH elli made up three important parts SSH elli = SSH circ + SSH tides + Geoid We want SSH circ

13. Modeling Tides Tides are now well modeled in deep water SSH tide = f(time,location,tidal component) Diurnal lunar O1 tide

14. The Geoid The geoid is the surface of constant gravitational acceleration Varies in ocean by 100’s m due to differences in rock & ocean depth Biggest uncertainty in determining SSH circ

15. The Geoid

16. Groundtracks 10 day repeat orbit Alongtrack 1 km resolution Cross-track 300 km resolution

17. Validation Two sites – Corsica – Harvest RMS ~ 2.5 cm

18. Mapped SSH SSH is optimally interpolated Cross-shelf SSH  SSH ~20 cm over ~500 km tan  =  z/  x ~ 0.2 / 5x10 5 or ~ 4 x 10 -7

19. Geostrophic Relationship Balance:Coriolis force = fu HPF = g tan  Geostrophic relationship: u = (g/f) tan  Know f (= 2  sin  ) & tan , calculate u

20. Calculating Currents Know tan  = 4x10 -7 Need f (= 2  sin  ) –  = ~37 o N – f = 2 (7.29x10 -5 s -1 ) sin(37 o ) = 8.8x10 -5 s -1 u= (g/f) tan  = (9.8 m s -2 / 8.8x10 -5 s -1 ) (4x10 -7 ) = 0.045 m/s = 4.5 cm/s !!

21. Mapped SSH u = 4.5 cm/s Direction is along  ’s in SSH The California Current

22. Geostrophy Geostrophy describes balance between horizontal pressure & Coriolis forces Geostrophic relationship can be used to diagnose currents - u = (g/f) tan  Showed how satellite altimeters can be used to estimate surface currents Need to do the old-fashion way next

23. Geostrophy Geostrophy describes balance between horizontal pressure & Coriolis forces Geostrophic relationship can be used to diagnose currents - u = (g/f) tan  Showed how satellite altimeters can be used to estimate surface currents What if density changes??

24. Our Simple Case Here,  tan  & u are = constant WRT depth

25. Barotropic Conditions A current where u  f(z) is referred to as a barotropic current Holds for  = constant or when isobars & isopycnals coincide Thought to contribute some, but not much, large scale kinetic energy

26. Barotropic Conditions

27. Isobars & Isopycnals Isobars are surfaces of constant pressure Isopycnals are surfaces of constant density Hydrostatic pressure is the weight (m*g) of the water above it per unit area Isobars have the same mass above them

28. Isobars & Isopycnals Remember the hydrostatic relationship p h =  g D If isopycnals & isobars coincide then D, the dynamic height, will be the same If isopycnals & isobars diverge, values of D will vary (baroclinic conditions)

29. Baroclinic Conditions

30. Baroclinic vs. Barotropic Barotropic conditions – Isobar depths are parallel to sea surface – tan  = constant WRT depth – By necessity, changes will be small Baroclinic conditions – Isobars & isopycnals can diverge – Density can vary enabling u = f(z)

31. Baroclinic vs. Barotropic

33. Baroclinic Flow Density differences drive HPF’s -> u(z) Hydrostatics says p h =  g D Changes in the mean  above an isobaric surface will drive changes in D (=  z) Changes in D (over distance  x) give tan  to predict currents Density can be used to map currents following the Geostrophic Method

34. Baroclinic Flow Flow is along isopycnal surfaces not across “Light on the right” u(z) decreases with depth

35. Geostrophic Relationship Balance:Coriolis force = fu HPF = g tan  Will hold for each depth Geostrophic relationship: u(z) = (g/f) (tan  (z))

36. Example as a f(z)  A   B Goal:  1 or  z 1

37. Example as a f(z) Define p ref - “level of no motion” = p o Know p 1 @A = p 1 @B ->  A g h A =  B g h B  z = h B - h A = = h B -  B h B /  A = h B ( 1 -  B /  A )

38. Example as a f(z) u = (g/f) (  z/  x) = (g/f) h B ( 1 -  B /  A ) / L If  A >  B (1 -  B /  A ) (& u) > 0 If  A <  B (1 -  B /  A ) (& u) < 0 Density  ’s drive u

39. Example as a f(z) Two stations 50 km apart along 45 o N  A (500/1000 db) = 1028.20 kg m -3  B (500/1000 db) = 1028.10 kg m -3 What is  z, tan  & u at 500 m??

40. Example as a f(z)  z = h B - h A = h B ( 1 -  B /  A ) Assume average distance (h A ) ~ 500 m  z = (500 m) (1 - 1028.10/1028.20) = 0.0486 m = 4.86 cm tan  =  z / L = (0.0486 m)/(50x10 3 m) = 9.73x10 -7

41. Example as a f(z) u(z) = (g/f) (tan  (z)) f = 2  sin  = 2 (7.29x10 -5 s -1 ) sin(45 o ) = 1.03x10 -4 s -1 u = (9.8 m s -2 /1.03x10 -4 s -1 ) (9.73x10 -7 ) = 0.093 m s -1 = 9.3 cm s -1

42. Geostrophy as a f(z) u = (g/f) h B ( 1 -  B /  A ) / L This can be repeated for each level Assumes level of no motion Calculates only the portion of flow perpendicular to density section Calculates only baroclinic portion of flow

43. Level of No Motion Level of no motion assumption misses the barotropic part of u(z)

44. Example of an Eddy in Southern Ocean http://gyre.umeoce.maine.edu/physicalocean/Tomczak/IntExerc/advanced4/index.html 30 km

45. So Ocean Example

46. Remember p h =  g h  z = h A - h B = h B (1 -  A /  B ) Start @ 2500 db & work upwards in layers Often specific volume, , or its anomaly, , are used

47. So Ocean Example u(z) = (g/f) tan  (z) Adjust level of known motion

48. Surface Currents from Hydrography Only the baroclinic portion of the current is sampled Need a level of no/known motion Need many, many observations Can get vertical structure of currents

49. Dynamic Height Hydrostatics give us p h =  g D Given isobars & average , D represents the dynamic height Let  (0/1000 db) = 1028.30 kg m -3 D(0/1000 db) = p h / (g  (0/1000 db)) = 1000 db (10 4 Pa/db)/(9.8 m s -2 * 1028.30 kg m -3 ) = 992.32 dyn meters

50. Dynamic Height California Cooperative Fisheries Investigations (CalCoFI) Understand ocean processes in pelagic fisheries Started in 1947

51. Dynamic Height January 2000 - CalCoFI Cruise 0001

52. Dynamic Height  D(0/500db) Shows CA Current Recirculation in the SoCal Bight

53. Dynamic Height Dynamic height anomaly,  D(0/1500db)

54. Dynamic Height

55. Surface Currents from Hydrography

56. Surface Currents from Altimetry Satellite altimeters can estimate the slope of the sea surface Both barotropic & baroclinic portions of current are determined Only surface currents are determined

57. Surface Currents from Altimetry

58. Geostrophy Barotropic vs. baroclinic flow Flow is along lines of constant dynamic height (light on the right) Baroclinic portion can be diagnosed from CTD surveys Quantifies the circulation of upper layers of the ocean