1. WHAT ARE THE FORCES BEHIND ATMOSPHERIC CIRCULATION? 1.Global Circulation as a Giant Sea Breeze. Concepts: Pressure Gradient Force; visualizing pressure with isobars 2.Introduction to the Coriolis Force (with a supporting role played by angular momentum). We want to explain circulation patterns like these, which take place over large enough scales that the rotation of the earth has an effect on moving air parcels: CHAPTER 4: ATMOSPHERIC MOTIONS and TRANSPORT

2. CHAPTER 4: ATMOSPHERIC TRANSPORT Forces in the atmosphere: Gravity Pressure-gradient Coriolis Friction to R of direction of motion (NH) or L (SH) Equilibrium of forces: In vertical: barometric law In horizontal: geostrophic flow parallel to isobars P P +  P pp cc v In horizontal, near surface: flow tilted to region of low pressure P P +  P cc v ff pp

3. Illustration of the Coriolis force. (Left panel). An observer sitting on the axis of rotation (North Pole) launches a projectile at the target. The curved arrow indicates the direction of rotation of the earth. (Right panel) The projectile follows a straight-line trajectory, when viewed by an observer in space, directed towards the original position of the target. However, observers and target are rotating together with the earth, and the target moves to a new position as the projectile travels from launch to target. Since observers on earth are not conscious of the fact that they and the target are rotating with the planet; they see the projectile initially heading for the target, then veering to the right. The Coriolis force is a fictitious force introduced to the equations of motion for objects on a rotating planet, sufficient to account for the apparent pull to the right in the Northern hemisphere or to the left in the southern hemisphere.

4. The geometry of the earth, showing the distance from the axis of rotation as a function of the latitude. r = R cos (the distance from the axis of rotation)  An object on the earth’s surface at a high latitude has less angular momentum than an object on the surface at a low latitude. v = 2  r cos( ) / t where t = 1 day (86400 seconds). The latitude of Boston is 42  ; plugging in numbers, you will find that you are traveling at a constant speed v = 1250 km/h (800 mph!). 1667 km/hr at the equator. Note: sound speed ~ 1440 km/hr

5. Coriolis Force (Northern Hemisphere): An air parcel (mass) begins to move from the Equator toward North Pole along the surface of the earth. The parcel moves closer to the axis of rotation: r decreases The parcel’s angular velocity is GREATER THAN the angular velocity of the earth’s surface at the higher latitude. It deflects to the right of it’s original trajectory relative to the earth’s surface. In the Southern Hemisphere, the parcel would appear to deflect to the left. The angular momentum of an object on the earth due to the planet’s rotation: L= mr 2 . The requirement that L be conserved implies that, if r changes,  must change so as to counteract the change in r, i.e.  = L/(mr2). For example, if r were to decrease by factor 2,  would increase by factor 4 so that L would stay unchanged. Example: skater "spinning up" – note that the skater really does spin up, by doing work (adding energy to the spinning motion !

6. The air parcel is deflected to the right.

7. Coriolis acceleration = F/m = 2  v sin( ). Coriolis acceleration increases as (latitude) increases, is zero at the equator. We thus find in all cases that the Coriolis force is exerted perpendicular to the direction of motion, to the RIGHT in the Northern Hemisphere and to the LEFT in the Southern Hemisphere.

8. A sample calculation:  = 7.5  10 -5 s -1 ; v = 10 m/s (36 km/hr, 21.6 mph); is 42 N (Boston), sin( )= 0.67 Coriolis acceleration = 1  10 -3 ms -2 The change in velocity is: 3.6 m s -1 in 1 hour (3600 s), during which the parcel travels 36 km in its original direction. The change in velocity would be 86 m s -1 in 24 hours if the Coriolis acceleration stayed the same over the whole period. Obviously this will not be the case. Coriolis acceleration = F/m = 2  v sin( ).

9. Deflection of an object by the Coriolis force.  y = [  (  x) 2 / v ] sin( ) (a) A snowball traveling 10 m at 20 km/h in Boston (42N): 20 km/hr = 5.5 m/s;  =7.5  10 -5 s -1 ; sin ( )=.67;  x=10  y = 9.1  10 -4 m (b) A missile traveling 1000 km at 2000 km/h at 42 N. v = 555 m/s,  x=1  10 6 m;  y = 9.05  10 5 m. At Boston ( = 42  N), we find that a snowball traveling 10 m at 20 km/h is displaced by  y = 1 mm (negligible), but a missile traveling 1000 km at 2000 km/h is shifted 100 km (important!). Note the importance of (  x) 2  c = 2  v sin ( ) ; t =  x/v   y = ½  c t 2

10. low pressure high pressure Pressure gradient force N S Motion of an air subjected to a north/south pressure gradient. Pt. A 1, initially at rest; Pt. A 3, geostrophic flow. The oscillatory motion depicted in the previous slide is usually not observed in the real atmosphere, because atmospheric mass will be redistributed to establish a pressure force balanced by the Coriolis force, and motion parallel to the isobars.

11. The geostrophic approximation is a simplification of very complicated atmospheric motions. This approximation is applied to synoptic scale systems and circulations, roughly 1000 km. ( It is easiest to think about measuring the pressure gradient at a constant altitude, although other definitions are more rigorous. ) Geostrophy For air in motion, not on the equator, Coriolis Force  Pressure gradient force Air motion is parallel to isobars V geostrophic = V g geostrophic wind (m/s)  7.29 10 -5 radian/s latitude  x distance (m)  P pressure diff. (N/m 2 )

12. Circulation of air around regions of high and low pressures in the Northern Hemisphere. Upper panel: A region of high pressure produces a pressure force directed away from the high. Air starting to move in response to this force is deflected to the right (in the Northern Hemisphere), giving a clockwise circulation pattern. Lower panel: A region of low pressure produces a pressure force directed from the outside towards the low. Air starting to move in response to this force is also deflected to the right, rotating counter-clockwise. Directions of rotation of the wind about high or low centers are reversed in the Southern Hemisphere, as explained earlier in this chapter.

13. The effect of friction around a high pressure region is to slow the wind relative to its geostrophic velocity. This causes the pressure force to slightly exceed the Coriolis force. The three forces add together as shown in the figure. Air parcels gradually drift from higher to lower pressure, in the case shown here, from the center of a high pressure region outward. An analogous flow (inward) occurs in a low- pressure region.

14. Air converges near the surface in low pressure centers, due to the modification of geostrophic flow under the influence of friction. Air diverges from high pressure centers. At altitude, the flows are reversed: divergence and convergence are associated with lows and highs respectively, closing the circulation through analogous processes noted in the sea breeze example

15. Near surface circulation around a low pressure area—March 7, 2006. Jet Stream

16. THE HADLEY CIRCULATION (1735): global sea breeze HOT COLD Explains: Intertropical Convergence Zone (ITCZ) Wet tropics, dry poles General direction of winds, easterly in the tropics and westerly at higher latitudes Hadley thought that air parcels would tend to keep a constant angular velocity. Meridional transport of air between Equator and poles results in strong winds in the longitudinal direction.

17. Reminder of the sea breeze: Distribution of pressure with altitude. The atmosphere expands as it is heated over the land, generating buoyancy and increasing the scale height H. The rate of pressure decline with altitude is reduced, therefore at altitude, the pressure is higher over land than over the adjacent sea, which causes mass to be transferred to the air column over the sea. Surface pressure over the ocean is therefore increased, giving rise to the distribution of pressure shown in the figure.

18. THE HADLEY CIRCULATION (1735): global sea breeze HOT COLD Explains: Intertropical Convergence Zone (ITCZ) Wet tropics, dry poles General direction of winds, easterly in the tropics and westerly at higher latitudes Hadley thought that air parcels would tend to keep a constant angular velocity. Meridional transport of air between Equator and poles results in strong winds in the longitudinal direction. Problems: 1. does not account for Coriolis force correctly; 2. circulation does not extend to the poles.

19. GLOBAL CLOUD AND PRECIPITATION MAP (intellicast.com) Today 11 Oct 2005 Images (3) show colder temperatures as brighter colors

20. Global Circulation and Precipitation as indicated by satellite images ITCZ: location, strength Wet and Dry season in Amazônia Strength and location of polar and subtropical jet streams Images show colder temperatures as brighter colors 11 Oct 200518 Feb 2007

21. TROPICAL HADLEY CELL Easterly “trade winds” in the tropics at low altitudes Subtropical anticyclones at about 30 o latitude

22. Z ln(P) Pressure anomaly scale (mb) land sea warm cold Global winds and pressures, July

23. CLIMATOLOGICAL SURFACE WINDS AND PRESSURES (July)

24. CLIMATOLOGICAL SURFACE WINDS AND PRESSURES (January)

25. cold warm Pressure anomaly scale (mb)

26. TIME SCALES FOR HORIZONTAL TRANSPORT (TROPOSPHERE) 2 weeks 1-2 months 1 year

27. 1) It's a distinct 'block' of air in an environment of … air; we often assume it has volume of 1 m 3. It has to be small enough so that it has uniform properties (T, P, etc). It’s a fictional entity that helps us to think through a physical process. 2) We can follow it (as if it were colored with dye) and it stays together (the same molecules are inside at the end of a process as there originally). 3) At the beginning of any of thought exercise, it has the same characteristics as its surrounding environment. 4) The parcel can change with time, by moving, emitting or absorbing heat radiation, etc --usually in a way we can describe with equations. 5) The environment of the parcel can change too. The parcel changes as a parcel NOT necessarily with the environment. Buoyancy and Lapse Rate The concept of an air parcel

28. Buoyancy force: Forces on a solid body immersed in a tank of water. The solid is assumed less dense than water and to area A (m 2 ) on all sides. P1 is the fluid pressure at level 1, and P1x is the downward pressure exerted by the weight of overlying atmosphere, plus fluid between the top of the tank and level 2, plus the object. The buoyancy force is P1 – P1x (up  ) per unit area of the submerged block. P1x D2D2 D1D1

29. The buoyancy force and Archimedes principle. 1. Force on the top of the block: P2  A =  water D 2 A g (A = area of top) weight of the water in the volume above the block 2. Upward force on the bottom of the block = P1  A =  water D 1 A g 3. Downward force on the bottom of the block = weight of the water in the volume above block + weight of block =  water D 2 A g +  block (D 1 - D 2 ) A g Unbalanced, Upward force on the block ( [2] – [3] ): F b =  water D 1 A g – [  water D 2 A +  block (D 1 - D 2 ) A ] g =  water g V block –  block g V block = (  water –  block ) V g weight of block BUOYANCY FORCE = weight of the water (fluid) displaced by the block Volume of the block = (D 1 – D 2 ) A

30. VERTICAL TRANSPORT: BUOYANCY Object (  z z+  z Fluid (  ’) Balance of forces: Note: Barometric law assumed a neutrally buoyant atmosphere with T = T’ T T’ would produce bouyant acceleration

31. Question: Where does the energy come from for an air parcel to do this work on the atmosphere? Vertical transport: Pressure, work, and Temperature

32. Change of atmospheric temperature with altitude (  pressure ) Atmospheric pressure vs altitude follows the barometric law,  P=-  g  z. Let's think of an ideal case where the buoyancy forces and the weight of an air parcel are perfectly balanced at every altitude, and we neither add or remove heat as the parcel moves. Because an air parcel expands as pressure is lowered, it must do work on the atmosphere as it moves up. The only source of energy is the motion of the molecules, and therefore the air parcel must get colder as it moves up. Two steps are needed to understand how an air parcel that moves up or down changes it temperature. Step 1. Figure out the exchanges of energy between the air parcel and the environment as the parcel changes its pressure, using the definition of heat capacity and Boyle's law. Step 2. Relate this energy balance to the change in altitude, using the barometric law.

33. P V P1V1P1V1 (P 1 +  P 1 )( V 1 +  V 1 ) = P 2 V 2 Boyle's Law: P 1 V 1 = P 2 V 2 How can we use Boyle's Law to determine the change in V when P changes, for a parcel of air (at constant temperature)? Boyles Law: P 2 V 2 = P 1 V 1 + P 1  V +  V 1  P +  P  V = P 1 V 1 P 1  V = —  V 1  P, or  P/  V = — P 1 /V 1 This is an example of how we can understand the relationship between two properties of air (or any gas), when both change together, by dividing the process into very small steps where one changes while the other is held constant, then hold the first constant and change the one initially held fixed. Boyles law  V/V 1 = ─  P/P 1 P 1 +  P 1 = P 2 ; V 1 +  V 1 = V 2

34. How do we get energy out of molecular motion: Heat capacity or Specific heat of a substance The specific heat (C p ) of a substance is defined as the energy needed to raise the temperature of 1 kg by 1 o K (the "p" denotes that the pressure is held constant). This energy goes into the thermal motions of the atoms and molecules (think of a "golf-ball atmosphere"). The specific heat is a quantity we can measure for any gas. It tells us how much energy we extract from the motion of the molecules to lower the temperature of 1 kg by 1 o K. The energy obtained by lowering T is the negative of this amount: [ Energy that must be added to a parcel to change T by  T ] = m c p  T [ Energy obtained (total) by lowering T by  T ] = — m c p  T. Work done against (or by) atmospheric pressure to change the pressure of an air parcel by  P is given by P  V. ( e.g., for the cylinder at the right, Work =  h F = P A  h = P  V ) - m c p  T = P  V (basic energy balance) P P hh Piston with top area A, volume Ah h

35. - mc p  T = P  V (basic energy balance) V  P = - P  V (Boyle’s law) =>> - mc p  T = - V  P  P = - g  Z (Barometric law) =>> - mc p  T = ( V  g  Z  V = m = mass of parcel We see that for an air parcel moving vertically in a hydrostatic atmosphere (barometric law applies), - c p  T = g  Z  T /  z = -g/c p = - 9.8 o K/km This change in temperature with altitude is called the "adiabatic lapse rate". c p = 1005 J/kg/K; g = 9.8 m s -2 =>> - g / c p = — 9.8 x 10 -3 K/m or — 9.8 K/km.

36. The change in temperature with altitude in the atmosphere. The example is from 30 degrees north latitude in summer.

37. ATMOSPHERIC LAPSE RATE AND STABILITY T z  = 9.8 K km -1 Consider an air parcel at z lifted to z+dz and released. It cools upon lifting (expansion). Assuming lifting to be adiabatic, the cooling follows the adiabatic lapse rate  : z “Lapse rate” = -dT/dz ATM (observed) What happens following release depends on the local lapse rate –dT ATM /dz: -dT ATM /dz >   upward buoyancy amplifies initial perturbation: atmosphere is unstable -dT ATM /dz =   zero buoyancy does not alter perturbation: atmosphere is neutral -dT ATM /dz <   downward buoyancy relaxes initial perturbation: atmosphere is stable dT ATM /dz > 0 (“inversion”): very stable unstable inversion unstable stable The stability of the atmosphere against vertical mixing is solely determined by its lapse rate.

38. EFFECT OF STABILITY ON VERTICAL STRUCTURE

39. WHAT DETERMINES THE LAPSE RATE OF THE ATMOSPHERE? An atmosphere left to evolve adiabatically from an initial state would eventually tend to neutral conditions (-dT/dz =  at equilibrium Solar heating of surface and radiative cooling from the atmosphere disrupts that equilibrium and produces an unstable atmosphere: Initial equilibrium state: - dT/dz =  z T z T Solar heating of surface/radiative cooling of air: unstable atmosphere ATM   ATM z T initial final  buoyant motions relax unstable atmosphere back towards –dT/dz =  Fast vertical mixing in an unstable atmosphere maintains the lapse rate to  Observation of -dT/dz =  is sure indicator of an unstable atmosphere.

40. The change in temperature with altitude in the atmosphere. The example is from 30 degrees north latitude in summer.

41. IN CLOUDY AIR PARCEL, HEAT RELEASE FROM H 2 O CONDENSATION MODIFIES  RH > 100%: Cloud forms “Latent” heat release as H 2 O condenses  9.8 K km -1  W  2-7 K km -1 RH 100% T z  WW Wet adiabatic lapse rate  W = 2-7 K km -1

42. Atmospheric temperature and dewpoint for a typical summer day shows the "planetary boundary layer" or "atmospheric mixed layer", that develops as the sun heats the ground in the daytime. This graph is drawn from actual data obtained by Harvard's Forest and Atmosphere Studies group during an experiment (code name "COBRA") over North Dakota in August, 2000.

43. What you see… Puffy little clouds, called fair weather cumulus, occurring over land on a typical afternoon. The lapse rate in the mixed layer is approximately adiabatic, and air parcels heated near the ground are buoyant. Each little cloud represents the top of a buoyant plume. (Photograph courtesy University of Illinois Cloud Catalog).

44. -40040 1000-9.5-6.4-3.0 600 4.2km -9.3-5.4 200 11.8km -8.6 Dry Adb.-9.8 Moist pseudo-adiabatic lapse rate Air is heated by release of latent heat when water condenses: T will decline less rapidly than the dry adiabat Temperature (C) Pressure (Mb) Ambient T 15 (->35) -13 -58  = -g/(c p +  w/  T ) = latent heat of vaporization (J/kg);  w/  T=change in spec humidity/K

47. [Photo: S. Wofsy, Manaus, Brazil, 1987.]

48. VERTICAL PROFILE OF TEMPERATURE Mean values for 30 o N, March Altitude, km Surface heating Latent heat release Radiative cooling (ch.7) - 6.5 K km -1 2 K km -1 - 3 K km -1 Radiative cooling (ch.7) Radiative heating: O 3 + h   O 2 + O O + O 2 + M  O 3 +M heat

49. DIURNAL CYCLE OF SURFACE HEATING/COOLING: ventilation of urban pollution z T 0 1 km MIDDAY NIGHT MORNING Mixing depth Subsidence inversion NIGHTMORNINGAFTERNOON 

50. SUBSIDENCE INVERSION typically 2 km altitude

51. FRONTS WARM FRONT: WARM AIRCOLD AIR WIND Front boundary; inversion COLD FRONT: COLD AIR WARM AIR WIND inversion

52. TYPICAL TIME SCALES FOR VERTICAL MIXING Estimate time  t to travel  z by turbulent diffusion: 0 km 2 km 1 day “planetary boundary layer” tropopause 5 km (10 km) 1 week 1 month 10 years