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Introduction to Synoptic Scale Dynamics

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Introduction to Synoptic Scale Dynamics Introduction to Synoptic Scale Dynamics J. H. LaCasce, UiO Contents 1 General dynamics 2 1.1 Derivatives ................................. 2 1.2 Continuityequation............................. 3 1.3 Equationsofmotion............................. 5 1.4 Scalingthehorizontalacceleration . ..... 11 1.5 Geostrophicbalance............................. 13 1.6 Thequasi-horizontalmomentumequations. ..... 16 1.7 Othermomentumbalances . 17 1.7.1 Geostrophicflow .......................... 18 1.7.2 Cyclostrophicflow . 18 1.7.3 Inertialflow............................. 19 1.7.4 Gradientwind............................ 20 1.8 Hydrostaticbalance............................. 21 1.9 Theequationsofstate............................ 23 1.10 TheBoussinesqApproximation . 24 1.11 PressureCoordinates . 25 1.12 Thermalwind ................................ 28 1.13 Thevorticityequation. 32 2 Barotropic flows 38 2.1 ShallowWaterEquations . 38 2.2 Conservationofpotentialvorticity . ..... 42 2.3 Linearsystem................................ 43 2.4 Constantf,flatbottom ........................... 44 2.5 Gravitywaves,norotation . 45 2.6 Gravitywaveswithrotation. 48 2.7 Geostrophicadjustment. 50 2.8 Kelvinwaves ................................ 53 2.9 Rossbywaves................................ 55 1 1 General dynamics The motion in the atmosphere and ocean is governed by a set of equa- tions, known as the Navier-Stokes equations. These equations are used to produce our forecasts, for the weather and also for ocean currents. While there are details about these equations which are uncertain (for example, how we parametrize processes smaller than the grid size of the models), they are for the most part accepted as fact. Let’s consider how these equa- tions come about. 1.1 Derivatives A fundamental aspect is how various fields (temperature, wind, density) change in time and space. Thus we must first specify how to take derivat- ives. Consider a scalar, ψ, which varies in both time and space, i.e. ψ = ψ(x,y,z,t). This could be the wind speed in the east-west direction, or the ocean density. By the chain rule, the total change in the ψ is: ∂ ∂ ∂ ∂ dψ = ψ dt + ψdx + ψ dy + ψ dz (1) ∂t ∂x ∂y ∂z so: dψ ∂ ∂ ∂ ∂ ∂ = ψ + u ψ + v ψ + w ψ = ψ + ~u ψ (2) dt ∂t ∂x ∂y ∂z ∂t ·∇ where (u,v,w) the components of the velocity in the (x,y,z) directions. On the left side, the derivative is a total derivative. That implies that ψ on the left side is only a function of time. This the case when ψ is observed following the flow. For instance, if you measure temperature in a balloon, 2 moving with the winds, you only see changes in time. We call this the Lagrangian formulation. The derivatives on the right side though are par- tial derivatives. These are relevant for an observer at a fixed location. This person records temperature as a function of time, but her information also depends on her position. An observer at a different location will generally have a different records (depending on how far away she is). We call the right side the Eulerian formulation. 1.2 Continuity equation δy δz ρ δ ρ δ δ ρ δ δ []u + u y z u y z δx x x + δ x δx Figure 1: A infinitesimal element of fluid, with volume δV . Consider a box fixed in space, with fluid (either wind or water) flowing through it. The flux of density through the left side is: (ρu) δyδz (3) Using a Taylor expansion, we can write the flux through the right side as: ∂ [ρu + (ρu)δx] δyδz (4) ∂x If these density fluxes differ, then the box’s mass will change. The net rate of change in mass is: 3 ∂ ∂ ∂ M = (ρδxδyδz)=(ρu) δyδz [ρu + (ρu)δx] δyδz ∂t ∂t − ∂x ∂ = (ρu)δxδyδz (5) −∂x The volume of the box is constant, so: ∂ ∂ ρ = (ρu) (6) ∂t −∂x Taking into account all the other sides of the box we have: ∂ρ ∂ ∂ ∂ = (ρu) (ρv) (ρw) = (ρ~u) (7) ∂t −∂x − ∂y − ∂z −∇ · We can rewrite the RHS as follows: (ρ~u) = ρ ~u + ~u ρ (8) ∇ · ∇ · ·∇ Thus the continuity equation can also be written: ∂ρ ∂ρ + (ρ~u) = + ~u ρ + ρ( ~u) = ∂t ∇ · ∂t ·∇ ∇ · dρ + ρ( ~u)=0 (9) dt ∇ · The first version of the equation is its Eulerian form. It states that the density at a location changes if there is a divergence in the flux into/out of the region. The last version is the Lagrangian form. This says that the density of a parcel of fluid advected by the flow will change if the flow is divergent, i.e. if: ~u =0 (10) ∇ · 6 4 We can also obtain the continuity equation using a Lagrangian (mov- ing) box. We assume the box contains a fixed amount of fluid, so that it conserves it mass, M. Then the relative change of mass is also conserved: 1 d M =0 (11) M dt The mass is the density times the volume of the box, so: 1 d 1 dρ 1 dV (ρV ) = + =0 (12) ρV dt ρ dt V dt Expanding the volume term by using the chain rule: 1 dV 1 ∂δx 1 ∂δy 1 ∂δz = + + = V dt δx dt δy dt δz dt 1 ∂x 1 ∂y 1 ∂z ∂u ∂v ∂w δ + δ + δ + + (13) δx dt δy dt δz dt → ∂x ∂y ∂z as δ 0. So: → 1 dρ + ~u =0 (14) ρ dt ∇ · which is the same as (9). Again, the density changes in proportion to the velocity divergence; the divergence determines whether the box shrinks or grows. If the box expands/shrinks, the density decreases/increases, to preserve the box’s mass. 1.3 Equations of motion The continuity equation pertains to mass. Now we consider the fluid velo- cities. We can derive expressions for these from Newton’s second law: ~a = F~ /m (15) The forces acting on a fluid parcel (a vanishingly small box) are: 5 pressure gradients: 1 p • ρ∇ gravity: ~g • friction: ~ • F For a parcel with density ρ, we can write: d 1 ~u = p + ~g + ~ (16) dt −ρ∇ F This is the momentum equation, written in its Lagrangian form. Under the influence of the forcing terms, on the RHS, the air parcel will accelerate. Actually, this is the momentum equation for a non-rotating earth. There are additional acceleration terms which come about due to rotation. As op- posed to the real forces shown in (16), rotation introduces apparent forces. A stationary parcel on the earth will rotate with the planet. From the per- spective of an observer in space, that parcel is traveling in circles, complet- ing a circuit once a day. Since circular motion represents an acceleration (the velocity is changing direction), there is a corresponding force. Ω δΑ Α δΘ γ γ Figure 2: The effect of rotation on a vector, A, which is otherwise stationary. The vector rotates through an angle, δΘ, in a time δt. 6 Consider such a stationary parcel, on a rotating sphere, with its position represented by a vector, A~ (Fig. 2). During the time, δt, the vector rotates through an angle: δΘ=Ωδt (17) where Ω is the sphere’s rotation rate. We will assume Ω = const., which is reasonable for the earth on weather time scales. The change in A is δA, the arc-length: δA~ = A~ sin(γ)δΘ=Ω A~ sin(γ)δt =(Ω~ A~) δt (18) | | | | × So we can write: δA~ dA~ limδ 0 = = Ω~ A~ (19) → δt dt × If the vector is not stationary but moving in the rotating frame, one can show that: dA~ dA~ ( ) =( ) + Ω~ A~ (20) dt F dt R × The F here refers to the fixed frame and R to the rotating one. If A~ = ~r, the position vector, then: d~r ( ) ~u = ~u + Ω~ ~r (21) dt F ≡ F R × So the velocity in the fixed frame is just that in the rotating frame plus the velocity associated with the rotation. Now consider that A~ is velocity in the fixed frame, ~uF . Then: d~u d~u ( F ) =( F ) + Ω~ ~u (22) dt F dt R × F 7 Substituting in the previous expression for uF , we get: d~u d ( F ) =( [~u + Ω~ ~r]) + Ω~ [~u + Ω~ ~r] (23) dt F dt R × R × R × Collecting terms, we get: d~u d~u ( F ) =( R ) +2Ω~ ~u + Ω~ Ω~ ~r (24) dt F dt R × R × × We now have two additional terms: the Coriolis and centrifugal accelera- tions. Plugging these into the momentum equation, we obtain: d~u d~u 1 ( F ) =( R ) +2Ω~ ~u + Ω~ Ω~ ~r = p + ~g + F~ (25) dt F dt R × R × × −ρ∇ Consider the centrifugal acceleration. This is the negative of the cent- ripetal acceleration and acts perpendicular to the axis of rotation (Fig. 3). The force projects onto both the radial and the N-S directions. This sug- gests that a parcel in the Northern Hemisphere would accelerate upward and southward. But these accelerations are balanced by gravity, which acts to pull the parcel toward the center and northward. The latter occurs because rotation changes the shape of the earth itself, making it ellipsoidal rather than spherical. The change in shape results in an exact cancellation of the N-S component of the centrifugal force. The radial component on the other hand is overcome by gravity. If this weren’t true, the atmosphere would fly off the earth. So the centrifugal force modifies gravity, reducing it over what it would be if the earth were stationary. Thus we can absorb the centrifugal force into gravity: g′ = g Ω~ Ω~ ~r (26) − × × 8 Ω R 2 Ω R g* g Figure 3: The centrifugal force and the deformed earth. Here is g is the gravitational vector for a spherical earth, and g∗ is that for the actual earth. The latter is an oblate spheroid. How big a modification is this? One can show that the centrifugal force at the equator is about 0.034m/sec2—or roughly 1/300th as large as g.
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