Introduction to Atmospheric Dynamics Chapter 1 Paul A. Ullrich [email protected] How to Read These Slides Defnition: A defnition is an explanation or outline for relevant jargon or terms. Concept: An idea that draws a connection between subjects or provides an answer for a question. Question: What is something that motivates delving into this topic? Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Part 1: Forces in the Atmosphere Radius of the Earth 6371.22 km Atmosphere Depth 100 km Troposphere Depth 10 km Mountain Height 8.8 km Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Question: How do we understand the dynamics of the atmosphere? Answer: The principles of atmospheric dynamics are drawn from basic physical principles. Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Question: What are the basic physical principles that govern the atmosphere? Newton’s Second Law: The change in momentum of an object is equal to the sum of forces acting on that object. d(mv) = F dt i allXi Conservation of Momentum: With no external forces momentum must be conserved. Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Basic Principles of Physics dx Defnition: Velocity is the change of u = position with respect to time dt du Defnition: Acceleration is the change a = of velocity with respect to time dt Hence, for an object of constant mass: d(mu) du 1 = m = ma Newton’s a = F dt dt Second Law m i allXi Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Basic Principles of Physics l How do these forces induce acceleration? l We assume the existence of an idealized “parcel” of fuid. l Forces are calculated on the idealized parcel. l Then take the limit of the parcel being infnitely small. Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Parcel Properties ⇢ Density (kg/m3) z V = ∆x∆y∆z Volume (m3) Δz Mass (kg) m = ⇢∆x∆y∆z y Δx Δy Temperature (K) T x p Pressure (Pa) Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Spherical Coordinates j Spherical coordinates: λ Longitude k φ Latitude j r Radius i Basis vectors: i Eastward basis vector j Northward basis vector k Vertical basis vector Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Defnition: The Material Derivative (expressed with a capital D) denotes the change in a quantity following a fuid parcel. 1 Du 1 a = F = F m i Dt m i allXi allXi u =(u, v, w) 3D velocity vector u Eastward velocity (zonal velocity) v Northward velocity (meridional velocity) w Upward velocity (vertical velocity) Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Question: What forces are important for understanding atmospheric dynamics? 1 Du 1 a = F = F m i Dt m i allXi allXi l Pressure gradient force l Gravitational force l Viscous force l Coriolis and centrifugal force l Total force is the sum of all these forces Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Defnition: A Surface Force acts on the surface of a parcel of fuid, typically due to interactions with neighboring parcels. The magnitude of a surface force is typically proportional to the surface area of the parcel. Examples: Pressure Force, Viscous Force. Defnition: A Body Force acts on the center of mass of a parcel of fuid. The magnitude of the body force is typically proportional to the mass of the parcel. Example: Gravity. Defnition: When a coordinate system (for instance coordinates on the sphere) varies with respect to time and/or space, there is an Apparent Force due to the fact that coordinate vectors are changing following the fuid parcel. Examples: Coriolis, Centrifugal Force. Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Pressure Gradient Force (x0,y0,z0) Pressure at parcel center: p0 = p(x0,y0,z0) Δz z Approximate pressure here via Taylor expansion Δx Δy y @p ∆x p = p + + (∆x2) 0 @x 2 O x ✓ ◆ Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Pressure Gradient Force Δz @p ∆x Δy @p ∆x pL p0 Δx p p + ⇡ − @x 2 R ⇡ 0 @x 2 ✓ ◆ ✓ ◆ F = p A = p ∆y∆z F = p A = p ∆y∆z L L L L R − R R − R Ftot = FL + FR @p ∆x @p ∆x = p ∆y∆z p + ∆y∆z 0 − @x 2 − 0 @x 2 ✓ ◆ ✓ ◆ @p = ∆x∆y∆z Total force acting on fuid parcel − @x ✓ ◆ in the x direction Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Pressure Gradient Force Repeat in all coordinate directions: @p @p @p F = i + j + k ∆x∆y∆z = ( p)∆x∆y∆z tot − @x @y @z − r ✓ ◆ Then computing the force per unit mass (recall this determines the acceleration): F 1 tot = p m −⇢r Total force acting on fuid parcel in all directions Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Viscous Force The viscosity of air is responsible for resisting motion of the fuid. It is a dissipative force, which results slowing a fuid which is not otherwise forced. δA For example, the blue fuid ut parcel experiences a shear stress in the x direction due to motion of the red fuid Ftop~i δy u parcel and green fuid parcel. Fbot~i F (u u)δA ~j ub top ⇠ t − F (u u)δA bot ⇠ b − ~i Observe that if all fuid parcels are traveling at the same velocity, no force will be conferred. Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Viscous Force F (u u)δA top ⇠ t − F (u u)δA bot ⇠ b − δA The resistance of the fuid parcel to shearing is determined by the u t proportionality coefficient Ftop~i µ/δy δy u Where μ is the dynamic F ~i bot viscosity. Observe that this ~j ub quantity is inversely proportional to the thickness ~i of the fuid parcel δy. Why might this be the case? µδA Hence: F + F = (u 2u + u ) top bot δy t − b Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Viscous Force The force per unit mass gives the acceleration: F F µ (u 2u + u ) tot = tot = t − b m ⇢yδA ⇢ δy2 And so in the limit δy 0 δA ! 2 u Ftot µ @ u t = m ⇢ @y2 ✓ ◆ Ftop~i δy u Fbot~i ~j ub ~i Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Viscous Force Extending this derivation to each fow direction then yields the total acceleration due to viscosity. µ Kinematic ⌫ = Viscosity ⇢ F visc = ⌫ 2u m r Note Laplacian (second derivative). What consequences does this have for linearly sheared fow? Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Gravitational Force Mm r Recall Newton’s law of gravity: F = G g r 2 r | | | | Gravitational constant: 11 3 1 2 G =6.67384 10− m kg− s− ⇥ Mass of the Earth: r M =5.972 1024 kg ⇥ But since the atmosphere is essentially a thin shell, we can make the approximation 6 r a a =6.37122 10 m F | | ⇡ ⇥ g = gk GM m − Defne gravity at surface: g = a2 Total gravitational force Question: Can you calculate g from the above information? acting on fuid parcel Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Dynamical Equations of Motion (Cartesian, non-rotang fluid) Du 1 @p = + ⌫ 2u Dt −⇢ @x r Dv 1 @p = + ⌫ 2v Dt −⇢ @y r Dw 1 @p = –g + ⌫ 2w Dt −⇢ @z r Pressure Gradient Gravity Viscosity Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Coriolis / Centrifugal Force The Earth revolves around its axis at ⌦ a certain rate ⌦ . j Coriolis and Centrifugal forces are k known as apparent forces, because they only exist because the reference frame is in motion. The Coriolis force defects fuid parcels as a consequence of the Earth’s rotation. The Centrifugal force attempts to push fuid parcels away from the axis of rotation. Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum Angular velocity is defned as the angular frequency (in units of radians ω per second) multiplied by the radius: v = !r | | r (radius) v Δv Δθ Velocity vector v Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum Ω On the surface of the sphere the radius of rotation is equal to the perpendicular distance from the axis R of rotation: a R = a cos φ | | φ ( φ = latitude) Earth Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum Ω k j Defne a local coordinate system: i = Points towards the East (Zonal) R Longitudinal direction a (at red circle this is directed into the page) φ j = Points towards the North (Meridional) Latitudinal direction (at red circle directed to top-left) k = Points in the vertical Earth Local vertical coordinate (at red circle directed to top-right) Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum Ω k j Given a vector B in the same φ direction as R, whare the R B components of this vector in the a local basis? B = Byj + Bzk φ Verify: B = B sin φ y −| | B = B cos φ Earth z | | Observe: B2 + B2 = B 2 y z | | Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum Ω k j φ R B a What is theVerify: velocity φ of this point only due to the rotation of the Earth? Earth Answer: Velocity is purely zonal u = ⌦ R i | | u = uii Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum Velocity due to rotation of the Ω Earth: k u = ⌦ R j i | | φ Angular momentum vector L is equal to the cross product of the R radial vector and the linear a momentum L = R mu φ ⇥ But R and u are perpendicular to one another, so L = R mui Earth | | | | L = m⌦ R 2 | | | | The angular momentum vector points in the same direction as Ω.