P=Mv ∑F =Ma =M Dv Dt = Dp Dt

Lesson 5 Boussinesq Approximation Boussinesq approximation Recall for an incompressible fluid: ∇ ⋅ v = 0 Conservation of Momentum Derivation of the Equations of Motion Assume density variations are small and (EOM) therefore can compare local variations of Navier-Stokes equations the density field to uniform background Must be able to identify terms and coordinate density ρ ′ systems << 1 ρ 0 Valid for nearly all circumstances in the ocean and lower atmosphere Conservation of Momentum Conservation of Momentum Newton’s Second Law Use this to derive our equations of motion Relates the rate of change of absolute Navier-Stokes Equations momentum following the motion in an Let’s re-examine the net forces acting on a inertial reference frame to the sum of the fluid parcel forces acting on the fluid = This information helps to construct equations p mv that calculate the net force - and therefore dv dp acceleration - acting on a parcel ∑F = ma = m = Sum the forces acting in each direction and dt dt equate it to the acceleration of the parcel 1 Primary Forces Where are we going… Lesson 5 Vertical Components Vector form of EOM Gravity Centrifugal force Spherical form of EOM PGF Coriolis Shallow Cartesian (rectangular) Horizontal Componets PGF Primitive f-plane Viscous Equations β-plane Coriolis Equations of Motion: Goal EOM: Derivation Now, to deal with the rotating frame of Du 1 ∂P = − + υ∇ 2u + (2Ωv sin φ − 2Ωwcos φ) reference.... Dt ρ ∂x Inertial (absolute) frame Dv 1 ∂P = − + υ∇ 2v − (2Ωusin φ) = ˆ + ˆ + ˆ Dt ρ ∂y A Axi Ay j Azk Non-inertial (rotating) frame ∂ Dw 1 P 2 = ′ˆ ′ + ′ˆ ′ + ′ˆ ′ = − + υ∇ w + (2Ωucos φ) − g A Ax i Ay j Az k Dt ρ ∂z 2 EOM: Derivation EOM: Derivation Relationship between inertial and rotating DaU a = DU r + Ω × − Ω2 frames of reference: 2 U r r Dt Dt Da A Dr A = + Ω × A Acceleration following the motion in an Dt Dt inertial frame is equal to the acceleration following the motion in a rotating frame Relationship between absolute velocity plus the coriolis acceleration and and velocity relative to Earth centripetal acceleration = + Ω × U a U r r EOM: Vector EOM: Spherical Coordinates Rewrite to get N2L relative to a rotating Assume the Earth’s departure from a coordinate frame: sphere is negligible Vector EOM Expand vector form of EOM so that the surface of the Earth corresponds to a coordinate surface λ φ dU r 1 2 Coordinate axes ( , , z): = − Ω × − ∇ + + υ∇ −π ≤ λ ≤ π 2 U r P g U r λ: longitude ( ) ρ π π dt φ (− ≤ φ ≤ ) : latitude 2 2 1 2 3 4 5 z: height above Earth’s surface a: radius of the Earth Unit vectors to describe motion 3 EOM: Spherical Coordinates EOM: Spherical Coordinates Relative velocity vector becomes: Zonal (East-West) distance V = u iˆ + v ˆj + w kˆ = φ λ dx acos d Where: Meridional (North-South) distance dλ u ≡ acos φ dy = ad φ dt dφ Vertical distance v ≡ a dt dz = da dz w ≡ dt EOM: Derivation EOM: Derivation Diˆ ∂iˆ u Dˆj ∂ˆj ∂ˆj utan φ v = u = (sin φˆj − cos φkˆ ) = u + v = − iˆ − kˆ Dt ∂x acos φ Dt ∂x ∂y a a Determine the magnitude and direction Determine the magnitude and direction 4 EOM: Derivation EOM: Derivation Dkˆ ∂kˆ ∂kˆ u v Substitute into relative velocity equation: = u + v = iˆ + ˆj Dt ∂x ∂y a a DV Du uv tan φ uw Dv u2 tan φ vw Dw u2 + v 2 = ( − + )iˆ + ( + + ) ˆj + ( − )kˆ Determine the magnitude and direction Dt Dt a a Dt a a Dt a Describes the spherical expansion of the acceleration following the motion Recall, our vector EOM had our force terms…now we have to expand the force terms to get the complete EOM in spherical coordinates EOM: Spherical Where are we going… Du uv tan φ uw 1 ∂P − + = − + 2Ωv sin φ − 2Ωw cos φ + F Vector form of EOM Dt a a ρ ∂x x Spherical form of EOM 2 φ ∂ Dv + u tan + vw = − 1 P − Ω φ + 2 usin Fy Dt a a ρ ∂y Shallow Cartesian (rectangular) Lesson 6 Dw u2 + v 2 1 ∂P − = − − g + 2Ωucos φ + F Primitive f-plane Dt a ρ ∂z z Equations β-plane 5.

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