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Ilker Fer

Lecture Notes for GEOF110 Chapter 8 to Geopotential = 2 hours Geostrophy = 4 hours Thermal Wind, barotropic/baroclinic = 2 hours. Ilker Fer. Guiding for blackboard presentation, following Pond & Pickard, Introductory Dynamical Oceanography. Hydrostatic Equilibrium. Stationary : u = v = w = 0

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Ilker Fer

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  1. Lecture Notes for GEOF110Chapter 8to Geopotential = 2 hoursGeostrophy = 4 hoursThermal Wind, barotropic/baroclinic = 2 hours Ilker Fer Guiding for blackboard presentation, following Pond & Pickard, Introductory Dynamical Oceanography

  2. Hydrostatic Equilibrium • Stationary : u = v = w = 0 • Remains stationary • Diverse forces (friction, tides etc.) F = 0 Isobaric (p=constant) surfaces are horizontal. No pressure force to cause horizontal motions. Hydrostatic equation in differential form. GEOF110 Guidelines / 5

  3. Assume • Horizontal isobaric surfaces • (i.e., there are no pressure gradients in the horizontal direction– no slope of sea surface and no slope on pressure surfaces in the interior ocean) • Vertical velocity w = 0 everywhere • No friction (F = 0) • 4) Horizontal velocity do not vary in space u=f(t); v=f(t) Inertial Motion GEOF110 Guidelines / 5

  4. Note: • Horizontal velocity vector has constant magnitude. Integration constant is the horizontal speed. • In the Northern hemisphere (NH), f>0, velocity vector rotates clockwise (CW) in time and turns 360o after one inertial period: In other words: Eq. of motion for a body in NH travelling CW in a horizontal circle at constant linear speed vH and angular speed f = 2sin. For example, wind blows steady in one directions, then stops after water has a speed VH. If there is NO friction, due to its inertia, the motion continues (hence inertial motion). In real world, there’s friction  decay due to friction. GEOF110 Guidelines / 5

  5. Northern Hemisphere t=/2f vH vH Coriolis Force vH B=vH/f t=/f t=0 =2/f (x0, y0) vH vH t=3/2f Since the speed is constant Centripetal acc. = Coriolis acc. Particle path Let’s (x,y) be coordinates of an individual fluid particle: Each particle moves in a circle with centre at (x0,y0), with radius B = vH/f B = inertial radius. In the ocean, at mid-latitudes, f  10-4 1/s For vH = 0.1 m/s  B =VH/f= 1 km For vH = 1 m/s  B = 10 km GEOF110 Guidelines / 5

  6. Period of revolution is • (or remember from: • Time = circumference/speed =2B/VH =2VH/fVH =2/f) • inertial period = half a pendulum day  Ti = ½ Tp 90N Ti 12 h 45N Ti  17 h Equator Ti =  Southern Hemisphere Northern Hemisphere Anti-cyclonic = CCW Cyclonic = CW Anti-cyclonic = CW Cyclonic = CCW L H H L Sidereal day  stjernedøgn Pendulum day  pendeldøgn Inertial oscillations  treghets-svingninger inertial motion is anti-cyclonic in both hemispheres GEOF110 Guidelines / 5

  7. V > vH V < vH Particle path If the ocean, in addition to inertial oscillations, also has a uniform velocity V, say in y-direction: Real ocean: Translation speed + decay due to friction. (Fig. from Wells) Particle path Equator GEOF110 Guidelines / 5

  8. Geopotential We raise a mass M a vertical distance dz against gravity. Ignore friction. Amount of work done (= potential energy gained) = dW=Mgdz Define: Change in geopotential (d) is gdz  Md=dW = Mgdz This is potential energy change per unit mass  joules/kg = m2/s2 Using hydrostatic eq. (dp = -gdz) Note, (S,T,P)=(35,0,P) +  (specific volume anomaly << ) Integrate between 2 levels, z1 to z2: Units: Energy/mass  Joules/kg or m2/s2 [not metre!] GEOF110 Guidelines / 5

  9. For g = 9.8 m/s2 dz = 1 m   = 9.8 J/kg For convenience: a common unit for the geopotential is the dynamic metre. In atmosphere: 1 dynamic meter = Z = /10 In the ocean; 1 dynamic meter = D = -/10 So that D2 – D1 is similar to z2 – z1. Geopotential surface: Level surface. The surface to which the force of gravity is perpendicular. Isobaric surface: Equal pressure surface Imagine a stationary (u=v=w=0) lake, with no currents, waves and no wind. Say atmospheric pressure = 0. The lake surface is the isobaric surface for p = 0. it is also level surface. Deeper isobaric surfaces will also be level surfaces. GEOF110 Guidelines / 5

  10. z Pressure force Isobarflate, p = konst. i i Nivåflate,  = konst. g, loddlinje Sats: Ingen bevegelse; isobarflaten må være en nivåflate. Bevis: Balanse i vertikalen Balanse i horizontalt Hor. komp kan skrives: For å få balanse må vi ha en tilsvarende kraft (F/m) mot høyre: F/m gtani GEOF110 Guidelines / 5

  11. Vi fant at Coriolis-kraften var viktig. En måte å få balanse på er å ha en hastighet V1 inn i papirplanet (in the northern hemisphere) slik at  Geostrofisk likning Balance between the Coriolis force and the pressure force This would permit us to get speed from slope of isobaric surface, i. We cannot measure p accurately, men måler (S,T,p) og beregne p fra dp = - gdz. Forsatt vanskelig å måle “i”, fordi havflaten ikke er en nivåflate. Måler derfor bare relative vinkler. Slopes are small  e.g., f = 10-4 1/s, v = 1 m/s, tani = 10-5  1 m/100 km. We determine the difference between i1 at level z1 and i2 at level z2. We get the velocity at z1 RELATIVE to that at z2 velocity shear dV/dz. Geostrophic wind speed in the atmosphere can be calculated accurately by direct measurements of pressure at known levels. GEOF110 Guidelines / 5

  12. z B x V0 A 0 z1, p1 V1 p=p1 z2, p1 a 1 i1 z3, p2 V2 p=p2 z4, p2 b 2 i2 L Surface p=pA=konst GEOF110 Guidelines / 5

  13. v p/y p/nH VH u y p/x fv x fVH fu Utledening av praktisk geostrofisk likning fra bevegelseslikningene: We do not neglect the Coriolis force, although it is small. It is balanced by the other small term, pressure force. We can only neglect small terms if there are larger terms. Coriolis = - Pressure Force Force • Pressure gradient is initiated • Fluid starts to move downgradient • Coriolis force acts to the right (NH)Alternatively: • 1. Wind stress piles water to one side with Coriolis deflecting the movement • 2. Pressure gradient balances GEOF110 Guidelines / 5

  14. z C Isobarflate, p =p1= konst. dz i Nivåflate,  = 1 konst. A dx B : Pressure gradients on constant z (= constant ) cannot be measured. Forandringen i nivået på en isobarflate Bruk denne likningen i 2 nivåer: GEOF110 Guidelines / 5

  15. Where V1 and V2 are the horizontal velocity components perpendicular to direction of nH at levels 1 and 2. GEOF110 Guidelines / 5

  16. Thermalvind-likningene Derived to show how T variations in horizontal lead to vertical variations in geostrophic wind. • Notes: • We can take f out of /z, as it does not change with z. • In (u)/z: Vertical variation in density is much less than shear • (u)/z   u/z (consistent with Bossinesq) Note: Again we determined the vertical variation of velocity (i.e., shear) For /x; /y : Use e.g. virtual potential temperature in atmosphere, sigma_t in the ocean (upper 1000 m). Light water on the right rule (NH) GEOF110 Guidelines / 5

  17. a) Level of no motion (LNM, klassisk) b) Kontiunitets-prinsipper c) Strømmålinger d) Overflate-topografi fra satellitt =konst. fast slow high Absolutte hastigheter Typical “deep” LNM is assumed due to blief of weak currents in deep ocean. This can be wrong as observations show strong currents- but they are localised and probably transient. On the average and long time scale deep LNM is a good approximation. Dynamisk topografi: Plotter overflatens geopotensial-topografi relativt til en annen flate Vi fant: Velg overflaten som det ene nivå, mange stasjoner A, B, C, …. Gir da muligheten til å plotte et kart av ’er i forhold til et annet nivå. Steep slope  large speed GEOF110 Guidelines / 5

  18. North Atlantic Current Gulf Stream Kuroshio Current ACC GEOF110 Guidelines / 5

  19. Dynamic topography is usually based on deep LNM. Surface currents are generally strong and will not be affecte d much. BUT total volume transport integrated over full depth can be wrong! Say in 4 km water, LNM assumed at 1 km, and upper 1 km mean geostrophic velocity is 10 cm/s. Say the actual mean current below 1 km is 2 cm/s. Error in velocity estimate is than 20%. Geostrophic volume transport is 0.1x1000 = 100 m3/s per m. Actual volume transport is 0.12x1000+0.2x3000 = 180 m3/s per m  80% error. We know speed goes to zero at the sea bed. Why not use bottom as LNM reference layer? The velocity profile approaches zero at the bottom due to FRICTION. But friction terms are neglected in geostrophic equations. We cannot use geostrophy when friction is important. GEOF110 Guidelines / 5

  20. Relasjon mellom isobar-flater og nivå-flater Typical LNM is about 1000 m in Pacific, between 1000-2000 m in Atlantic. d) SLOPE CURRENT  BAROTROPIC. This would give zero geostrophic velovcity (correct) but absoluite velocity is not zero. UNLIKELY because T,S variations are likely to cause variations in isobaric slopes. usannsynlig GEOF110 Guidelines / 5

  21. Relasjon mellom isobar-flater og isopyknaler Isopyknal: konstant tetthets-flater Hvis tettheten bare er en funksjon av trykket (f.eks. S og T konstante), r = r(p)  BAROTROPT massefelt Isobarflater og isopyknaler er parallel. Hvis tettheten også er en funksjon av andre parametere kan tetthets og isbarflatene ha forskjellige helning  BAROKLINT massefelt GEOF110 Guidelines / 5

  22. GEOF110 Guidelines / 5

  23. 1 d 2 z No flow x 0 For the 2-layer flow in panel c of the previous figure, interface slope is about 100 times steeper than the sea-surface slope. Derive: GEOF110 Guidelines / 5

  24. GEOF110 Guidelines / 5

  25. GEOF110 Guidelines / 5

  26. Some disadvantages of the geostrophic method: • Only relative currents • Selection of LNM • LNM reaches bottom as stations get shallower • Only mean values over stations with large distance apart (sometimes this can be a benefit, e.g. smooths short-term variations, such as internal waves) • Friction is ignored • Break-down near Equator where Coriolis force is so small that friction can be important. • Will include any effect of long-period transient currents. GEOF110 Guidelines / 5

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