Ocean currents: some misconceptions and some dynamics

Transcription

1 Ocean currents: some misconceptions and some dynamics Joe LaCasce Dept. Geosciences October 30, 2012

2 Where is the Gulf Stream? BBC Weather Center

3 Where is the Gulf Stream? Univ. Bergen news website (2011)

4 Where is the Gulf Stream? Dagbladet (2012)

5 Where is the Gulf Stream? Discovery news (Oct, 2012)

6 keeps us warm Forskningsrdet

7 No it doesn t Aftenposten (2012)

8 will shut down!

9 will shut down!

10 No it won t Introduction Dagbladet (2012)

11 But it s good enough for Hollywood... Link

12 Where is the Gulf Stream?

13 Introduction Where is the Gulf Stream? Benjamin Franklin (1770)

14 Sea surface temperature (NOAA)

15 Gulf Stream cross section

16 North Atlantic winds

17 Gulf Stream Introduction A strong current (1 m/s velocities) Driven by the winds Associated with a strong temperature front Significant vertical shear, with strongest velocities at the surface Lies on the western side of the North Atlantic

18 Similar systems Schmitz (1996)

19 Outline Introduction Hydrostatic balance Geostrophic balance Thermal wind relations A simplified model of the Gulf Stream

20 Hydrostatic balance

21 Hydrostatic balance Downward accerleration: mg Upward acceleration: [p(z + dz) p(z)] A If the section has density ρ, then m = ρv = ρa dz The fluid is not moving (static) if: or: [p(z + dz) p(z)] A = mg = ρga dz dp dz = ρg

22 Hydrostatic balance The hydrostatic balance also applies to moving fluid Then write: z p = ρg The balance is used in most weather models

23 Pressure and fluid acceleration

24 Pressure and acceleration Newton s second law: F = ma A pressure difference exerts a net force on a region of fluid

25 Pressure and fluid acceleration δv = δx δy δz

26 Pressure gradient Using a Taylor series, we can write the pressure on the right side of the box: p R = p(x 0, y 0, z 0 ) + p δx x Similarly, the pressure on left side of the box is: p L = p(x 0, y 0, z 0 ) p δx x

27 Pressure gradient The force on the right hand side (directed inwards): On left side: So the net force is: F R = p R A = [p(x 0, y 0, z 0 ) + p δx x 2 ] δyδz F L = p L A = [p(x 0, y 0, z 0 ) p δx x 2 ] δyδz F x = F L + F R = p δx δy δz (1) x

28 Pressure gradient The volume weighs: m = ρ δx δy δz So: a x t u = F x m = 1 p ρ x

29 Pressure gradients So we can write: t u = 1 ρ x p t v = 1 ρ y p t w = 1 ρ z p g

30 Rotation Introduction

31 Rotation Introduction δθ = Ωδt Assume Ω = const. (reasonable for the earth) Change in position r is δr, the arc-length: δ r = r sin(γ)δθ = Ω r sin(γ)δt = ( Ω r) δt

32 Rotation Introduction So: d r dt = Ω r This is the motion of a fixed vector. For a moving vector: ( d r dt ) F = u F = u R + Ω r So the velocity in the fixed frame is equal to that in the rotating frame plus the rotational movement

33 Acceleration Introduction Can apply similar arguments to find: ( d u F dt ) F = ( d u R dt ) R + 2 Ω u R + Ω Ω r We have two additional terms: Coriolis acceleration 2Ω u R Centrifugal acceleration Ω Ω r

34 Centrifugal acceleration

35 Centrifugal acceleration

36 Centrifugal acceleration It doesn t affect the motion The vertical component simply modifies gravity: g = g ( Ω Ω r) ˆk A tiny correction (0.034 m/sec 2 maximum)

37 Rotation vector 2 Ω = 2Ωcosθ ĵ + 2Ωsinθ ˆk

38 Coriolis acceleration The vertical component is the most important. So: 2 Ω u [0, 0, 2Ωsin(θ)] (u, v, w) where = ( fv, fu, 0) f = 2Ωsin(θ)

39 Momentum equations t u fv = 1 ρ x p t v + fu = 1 ρ y p t w = 1 ρ z p g (There are additional terms, due to advection, which we ll ignore)

40 Coriolis acceleration Example: What is the Coriolis acceleration on a car driving east at 50 km/hr? Ignoring the pressure gradient terms: t u = fv t v = fu f = 2Ωsin(60) = 2(2π) sin(60) = sec 1

41 Coriolis acceleration So: fu = ( )(13.9) = m/sec 2 The acceleration is to the south t u = 0 t v = m/sec 2

42 Coriolis acceleration The Coriolis force is to the right of the motion So the force does no work it causes circular motion But note is much smaller than gravity (g = 9.8 m/sec 2 )

43 Coriolis acceleration What if in southern hemisphere? Say we re at 60 S instead. f = 2Ωsin( 60) = sec 1 So: t v = m/sec 2 Acceleration is to the north, to the left of the parcel velocity.

44 Dominant balances But not all terms are the same size. Consider the x-momentum equation: t u fv = 1 ρ x p The forcing is by the pressure gradient. But are the two terms on the left side equally important? To find out, we scale them.

45 Dominant balances The first term has a scale of U T The Coriolis term has scale of fu. So the ratio of the first to second terms is: U TfU = 1 ft

46 Dominant balances Which term dominates depends on relative sizes of T and f 1. f 1 is called the inertial period At mid-latitudes, this is roughly one day

47 Short time scales For T f 1, the first term is most important. So we d have: t u 1 ρ x p At short times, the earth s rotation isn t important This is true for surface waves (at the beach) and for flow in a bathtub

48 Geostrophic balance The time scales for the Gulf Stream are longer days to weeks. Then we have: and: fv 1 ρ x p fu 1 ρ y p These are the geostrophic balances

49 Geostrophic balance

50 Geostrophic balance

51 North Atlantic winds

52 Hurricane Sandy

53 Sea surface temperature (NOAA)

54 Geostrophic balance However the geostrophic balance fails at equator because: f = 2Ωsin(0) = 0 So at the equator the time derivative terms dominate The equatorial flow is as on a non-rotating planet

55 Boussinesq approximation In the ocean, we often make the Boussinesq approximation fv = 1 ρ x p 1 ρ c x p This is because the density of seawater is nearly constant A result of this is that seawater is incompressible

56 Thermal wind The Boussinesq approximation allows us to combine the geostrophic and hydrostatic balances: z (fv) = z 1 ρ c x p = 1 ρ c x z p Using the hydrostatic relation, we get: f z v = g ρ c x ρ

57 Thermal wind Also: f z u = g ρ c y ρ This is the thermal wind balance. If there is a density gradient in x, there is a corresponding shear in the N-S velocity, v

58 Thermal wind

59 Jet Stream Introduction

60 Gulf Stream cross section

61 Gulf Stream cross section Now we understand why there is a temperature front with the Gulf Stream and why the velocities have vertical shear. But why does it lie on the western side of the North Atlantic? To see why, we must start with Nansen...

62 Nansen and the icebergs

63 Ekman s solution Ekman (1905) constructed a model to explain this. He postulated that friction in the surface boundary layer breaks geostrophy. So the flow obeys: fv = 1 ρ c fu = 1 ρ c x p + ν 2 z 2 u y p + ν 2 z 2 v

64 The Ekman spiral

65 The Ekman spiral

66 Sverdrup transport Now the Ekman flow at the surface can drive large scale flow. This was shown by Sverdrup (1947). Take the geostrophic relations (for the interior ocean): fv = 1 ρ c fu = 1 ρ c x p y p We can eliminate the pressure if we take the y-derivative of the first and add to the x-derivative of the second

67 Sverdrup transport Result is: f ( x u + df v) + v y dy = 0 Because the flow is incompressible, we have: So we have: x u + y v + z w = 0 v df dy = f z w

68 Sverdrup transport For simplicity, let s say: df dy = β = const. We can integrate the relation over the depth of the ocean: 0 β v dz = βv = fw 0 fw H H Assuming the flow vanishes at depth, we have: βv = fw 0 This is the Sverdrup relation. Vertical motion at the sea surface can drive N-S transport.

69 Add the Ekman layer

70 The surface circulation in the North Atlantic

71 Westward Ho But this still doesn t explain why the Gulf Stream is in the west. The solution to that problem came in

72 Stommel (1948) Like Ekman (1905), Stommel (1948) added friction but this time to the Sverdrup balance: βv = fw 0 + F Then he specified the surface winds, which determined w 0 via the Ekman relation. The problem can be solved on 1-2 pieces of paper

73 Stommel s solution

74 Stommel s solution Schmitz (1996)

75 Summary Introduction From basic theoretical considerations, we know: is in geostrophic and hydrostatic balance The velocity shear is in thermal wind balance with the density The velocities are forced by the winds at the surface The current is intensified on the western side of the basin

76 Gulf Stream shutdown? Wunsch (2004)

77 The thermally-forced circulation

78 Theoretical solutions Pedlosky (1969), Salmon (1986), LaCasce (2004)

79 The North Atlantic Circulation (WHOI)

80 Atlantic vs. Pacific (WHOI)

81 Stopping the thermohaline circulation Murton et al. (2004)

82 Observations of variability Observations suggest the inflow to the Nordic Seas increases by as much as 25 % in winter, but this may be due to stronger winds 20th century simulations yield 5-10 % variations in the strength of the thermohaline circulation Doubled CO2 leads to a 25 % decrease in the thermohaline circulation in 100 years (but no shut-down) The IPCC concluded that thermohaline shut-down was highly unlikely in the foreseeable future

83 Climate models (MIT GCM)