1. AOSS 401, Fall 2007 Lecture 2 September 7, 2007 Richard B. Rood (Room 2525, SRB) rbrood@umich.edu 734-647-3530 Derek Posselt (Room 2517D, SRB) dposselt@umich.edu 734-936-0502

2. Class News Ctools site (AOSS 401 001 F07)AOSS 401 001 F07 –Calendar (completed for whole semester) –Syllabus –Lectures Posted on day of –Homework (and solutions) Homework has been posted –Under “resources” in homework folder Due next Wednesday (September 12, 2007)

3. Class news: Schedule issues Currently 4.5 hours are scheduled for a 4.0 hour course. (So we have some flexibility; we can “cancel” 4 classes) –There will be no class on September 14 –There will be no class on October 12 –There will be no class on November 21 –When to schedule final exam?

4. Weather National Weather Service –http://www.nws.noaa.gov/http://www.nws.noaa.gov/ –Model forecasts: http://www.hpc.ncep.noaa.gov/basicwx/day0- 7loop.html http://www.hpc.ncep.noaa.gov/basicwx/day0- 7loop.html Weather Underground –http://www.wunderground.com/cgi- bin/findweather/getForecast?query=ann+arborhttp://www.wunderground.com/cgi- bin/findweather/getForecast?query=ann+arbor –Model forecasts: http://www.wunderground.com/modelmaps/maps.asp ?model=NAM&domain=US http://www.wunderground.com/modelmaps/maps.asp ?model=NAM&domain=US

5. Outline Pressure gradient force Gravitational Force Viscous force Centrifugal Force Coriolis Force Should be review. So we are going fast. You have the power to slow us down.

6. Some basics of the atmosphere Earth: radius ≡ a = 6.37 x 10 6 m atmosphere: depth ~ 1.0 x 10 5 m Mountain: height ~ 5.0 x 10 3 m OceanLandBiosphere

7. Some basics of the atmosphere Troposphere: depth ~ 1.0 x 10 4 m Troposphere ------------------ ~ 2 Mountain Troposphere ------------------ ~ 1.6 x 10 -3 Earth radius This scale analysis tells us that the troposphere is thin relative to the size of the Earth and that mountains extend half way through the troposphere.

8. Newton’s Law of Motion F = ma Force = mass x acceleration In general we will work with force per unit mass; hence, a = F/m And with the definition of acceleration Bold will represent vectors.

9. Newton’s Law of Motion Which is the vector form of the momentum equation. (Conservation of momentum)

10. What are the forces? Pressure gradient force Gravitational force Viscous force Apparent forces Can you think of other classical forces and would they be important in the Earth’s atmosphere? Total Force is the sum of all of these forces.

11. Newton’s Law of Motion Where i represents the different types of forces.

12. How do we express the forces? In general, we assume the existence of an idealized parcel or “particle” of fluid. We calculate the forces on this idealized parcel. We take the limit of this parcel being infinitesimally small. –This yields a continuous, as opposed to discrete, expression of the force. Use the concept of the continuum to extend this notion to the entire fluid domain.

13. An intrinsic assumption There is an equation of state that describes the thermodynamic properties of the fluid, the air.

14. A particle of atmosphere xx yy zz  ≡ density = mass per unit volume (  V)  V =  x  y  z m =  x  y  z ------------------------------------- p ≡ pressure = force per unit area acting on the particle of atmosphere x y z ii jj kk

15. Pressure gradient force (1) xx yy zz p 0 = pressure at (x 0, y 0, z 0 ). (x 0, y 0, z 0 ) x axis p = p 0 + (∂p/∂x)  x/2 + higher order terms

16. Pressure gradient force (2) xx yy zz. x axis p = p 0 + (∂p/∂x)  x/2 + higher order terms p = p 0 - (∂p/∂x)  x/2 + higher order terms

17. Pressure gradient force (3) (ignore higher order terms) xx yy zz. x axis F Bx = (p 0 - (∂p/∂x)  x/2) (  y  z) F Ax = - (p 0 + (∂p/∂x)  x/2) (  y  z) A B

18. Pressure gradient force (4) Total x force = (p 0 - (∂p/∂x)  x/2) (  y  z) - (p 0 + (∂p/∂x)  x/2) (  y  z) F x = F Bx + F Ax = - (∂p/∂x)(  x  y  z) We want force per unit mass F x /m = - 1/  (∂p/∂x)

19. Vector pressure gradient force x y z ii jj kk

20. Viscous force (1) There is in a fluid friction, which resists the flow. It is dissipative, and if the fluid is not otherwise forced, It will slow the fluid and bring it to rest. Away from boundaries in the atmosphere this frictional force is often small, and it is often ignored. (We will revisit this as we learn more.) Close to the boundaries, however, we have to consider friction. Derivation is at end of lecture or in the text.

21. Viscous force (2) OceanLandBiosphere velocity ≡ u m/sec velocity must be 0 at surface

22. Viscous force (3) OceanLandBiosphere velocity ≡ u m/sec Velocity is zero at the surface; hence, there is some velocity profile.

23. Viscous force (4) (How do we think about this?) The drag on the moving plate is the same as the force required to keep the plate moving. It is proportional to the area (A), proportional to the velocity of the plate, and inversely proportional to the distance between the plates; hence, Proportional usually means we assume linear relationship. This is a model based on observation, and it is an approximation. This is often said to be “Newtonian.” The constant of proportionality assumes some physical units. What are they? u(0) = 0 u(z) u(h) = u 0 h F = μAu 0 /h

24. Viscous force (5) Where Laplacian is operating on velocity vector ≡ u = (u, v, w)  ≡  /  kinematic viscosity coefficient

25. Surface forces Pressure gradient force and the viscous force are examples of a surface force. Surface forces are proportional to the area of the surface of our particle of atmosphere. Surface forces are independent of the mass of the particle of atmosphere. They depend on characteristics of the particle of atmosphere; characteristics of the flow.

26. Highs and Lows Motion initiated by pressure gradient Opposed by viscosity

27. Where’s the low pressure?

28. Geostrophic and observed wind 1000 mb (ocean)

29. Body forces Body forces act on the center of mass of the parcel of fluid. Magnitude of the force is proportional to the mass of the parcel. The body force of interest to dynamic meteorology is gravity.

30. Newton’s Law of Gravitation Newton’s Law of Gravitation: The force between any two particles having masses m1 and m2 separated by distance r is an attraction acting along the line joining the particles and has the magnitude proportional to G, the universal gravitation constant.

31. Gravitational Force

32. Gravitational force for dynamic meteorology Newton’s Law of Gravitation: M = mass of Earth m = mass of air parcel r = distance from center (of mass) of Earth directed down, towards Earth, hence - sign

33. Adaptation to dynamical meteorology a is radius of the Earth z is height above the Earth’s surface Can we ignore z, the height above the surface? How would you make that argument?

34. Gravity for Earth a2a2 =g 0 a mg 0 a

35. Gravitational force per unit mass

36. Our momentum equation + other forces Now using the text’s convention that the velocity is u = (u, v, w).

37. Apparent forces

38. Back to Basics: Newton’s Laws of Motion Law 1: Bodies in motion remain in motion with the same velocity, and bodies at rest remain at rest, unless acted upon by unbalanced forces. Law 2: The rate of change of momentum of a body with time is equal to the vector sum of all forces acting upon the body and is the same direction. Law 3: For every action (force) there is and equal and opposite reaction.

39. Back to basics: A couple of definitions Newton’s laws assume we have an “inertial” coordinate system; that is, and absolute frame of reference – fixed, absolutely, in space. Velocity is the change in position of a particle (or parcel). It is a vector and can vary either by a change in magnitude (speed) or direction.

40. Apparent forces: A mathematical approach Non-inertial, non-absolute coordinate system

41. Two coordinate systems x y z x’ y’ z’ Can describe the velocity and forces (acceleration) in either coordinate system.

42. One coordinate system related to another by:

43. Velocity (x’ direction) So we have the velocity relative to the coordinate system and the velocity of one coordinate system relative to the other. This velocity of one coordinate system relative to the other leads to apparent forces. They are real, observable forces to the observer in the moving coordinate system.

44. Two coordinate systems y z z’ axis is the same as z, and there is rotation of the x’ and y’ axis z’ y’ x’ x

45. One coordinate system related to another by: T is time needed to complete rotation.

46. Acceleration (force) in rotating coordinate system The apparent forces that are proportional to rotation and the velocities in the inertial system (x,y,z) are called the Coriolis forces. The apparent forces that are proportional to the square of the rotation and position are called centrifugal forces.

47. Apparent forces: A physical approach

48. Circle Basics ω θ s = rθ r (radius) Arc length ≡ s = r θ Magnitude

49. Centrifugal force: Treatment from Holton ω Δθ ΔvΔv Magnitude r (radius)

50. Centrifugal force: for our purposes

51. Now we are going to think about the Earth The preceding was a schematic to think about the centrifugal acceleration problem. Note that the r vector above and below are not the same!

52. What direction does the Earth’s centrifugal force point? Ω Ω2RΩ2R R Earth

53. What direction does gravity point? Ω R Earth a

54. What direction does the Earth’s centrifugal force point? Ω Ω2RΩ2R R Earth So there is a component that is in the same coordinate direction as gravity (and local vertical). And there is a component pointing towards the equator We are now explicitly considering a coordinate system tangent to the Earth’s surface.

55. What direction does the Earth’s centrifugal force point? Ω Ω2RΩ2R R Earth So there is a component that is in the same coordinate direction as gravity: ~ aΩ 2 cos 2 (  ) And there is a component pointing towards the equator ~ - aΩ 2 cos(  )sin(  ) Φ = latitude

56. So we re-define gravity as

57. What direction does the Earth’s centrifugal force point? Ω Ω2RΩ2R R Earth And there is a component pointing towards the equator. The Earth has bulged to compensate for the equatorward component. Hence we don’t have to consider the horizontal component explicitly.

58. Centrifugal force of Earth Vertical component incorporated into re- definition of gravity. Horizontal component does not need to be considered when we consider a coordinate system tangent to the Earth’s surface, because the Earth has bulged to compensate for this force. Hence, centrifugal force does not appear EXPLICITLY in the equations.

59. Apparent forces: A physical approach Coriolis Force http://climateknowledge.org/figures/AOSS 401_coriolis.movhttp://climateknowledge.org/figures/AOSS 401_coriolis.mov

60. Next time Coriolis Force –Read and re-read the section in the text. Pressure as a vertical coordinate –Geopotential

61. Weather National Weather Service –http://www.nws.noaa.gov/http://www.nws.noaa.gov/ –Model forecasts: http://www.hpc.ncep.noaa.gov/basicwx/day0- 7loop.html http://www.hpc.ncep.noaa.gov/basicwx/day0- 7loop.html Weather Underground –http://www.wunderground.com/cgi- bin/findweather/getForecast?query=ann+arborhttp://www.wunderground.com/cgi- bin/findweather/getForecast?query=ann+arbor –Model forecasts: http://www.wunderground.com/modelmaps/maps.asp ?model=NAM&domain=US http://www.wunderground.com/modelmaps/maps.asp ?model=NAM&domain=US

62. Derivation of viscous force Return to lecture body.

63. Viscous force (1) There is in a fluid friction, which resists the flow. It is dissipative, and if the fluid is not otherwise forced, It will slow the fluid and bring it to rest. Away from boundaries in the atmosphere this frictional force is often small, and it is often ignored. (We will revisit this as we learn more.) Close to the boundaries, however, we have to consider friction. Return to lecture body.

64. Viscous force (2) OceanLandBiosphere velocity ≡ u m/sec velocity must be 0 at surface

65. Viscous force (3) OceanLandBiosphere velocity ≡ u m/sec Velocity is zero at the surface; hence, there is some velocity profile.

66. Viscous force (4) (How do we think about this?) Moving plate with velocity u 0 Linear velocity profile u(0) = 0 u(z) u(h) = u 0 h u(z) = (u 0 -u(0))/h × z

67. Viscous force (5) (How do we think about this?) The drag on the moving plate is the same of as the force required to keep the plate moving. It is proportional to the area (A), proportional to the velocity of the plate, and inversely proportional to the distance between the plates; hence, Proportional usually means we assume linear relationship. This is a model based on observation, and it is an approximation. The constant of proportionality assumes some physical units. What are they? u(0) = 0 u(z) u(h) = u 0 h F = μAu 0 /h

68. Viscous force (6) (How do we think about this?) Recognize the u 0 /h can be represented by ∂u/∂z Force per unit area is F/A is defined as shearing stress (  ). Like pressure the shearing stress is proportional to area. u(0) = 0 u(z) u(h) = u 0 h F = μA(∂u/∂z)

69. Viscous force (7) (How do we think about this?)  zx = μ(∂u/∂z), which is the viscous force per unit area in the x direction, due to the variation of velocity in the z direction Force per unit area; hence, like pressure. u(0) = 0 u(z) u(h) = u 0 h F zx /A = μ(∂u/∂z) ≡  zx

70. Viscous force (8) (Do the same thing we did for pressure) xx yy zz  zx0 = stress at (x 0, y 0, z 0 ). (x 0, y 0, z 0 )  zx =  zx0 + (∂  zx /∂z)  z/2  zx = -(  zx0 - (∂  zx /∂z)  z/2)

71. Viscous force (9) xx yy zz. C D F Czx = (  zx0 + (∂  zx /∂z)  z/2)  y  x F Dzx = -(  zx0 - (∂  zx /∂z)  z/2)  y  x

72. Viscous force (10) = (∂  zx /∂z)  z  y  x F zx = F Czx + F Dzx We want force per unit mass F zx /m = 1/  ∂  zx /∂z

73. Viscous force (11) (using definition of  )

74. Viscous force (12) Assume μ constant  ≡  /  kinematic viscosity coefficient

75. Viscous force (13) Do same for other directions of shear (variation of velocity)

76. Viscous force (14) Do same for other directions of force velocity vector ≡ u = (u,v,w)

77. Viscous force (15) Where Laplacian is operating on velocity vector ≡ u = (u, v, w) Return to lecture body.

78. Summary Pressure gradient force and viscous force are examples of surface forces. They were proportional to the area of the surface of our particle of atmosphere. They are independent of the mass of the particle of atmosphere. They depend on characteristics of the particle of atmosphere; characteristics of the flow. Return to lecture body.

79. Our surface forces other forces Now using the text’s convention that the velocity is u = (u, v, w). Return to lecture body.

80. Highs and Lows Motion initiated by pressure gradient Opposed by viscosity Return to lecture body.