Atmospheric Motion

Dynamics

Newton’s second law F m

x dx2 Mass × Acceleration = Force mF= dt2 Thermodynamics

Concerned with changes in the internal energy and state of moist air. TMD Lecture 2

Newton’s second law The Coriolis force

Newton’s second law A line that rotates with F dx2 the roundabout m mF= dt2 x Ω A line at rest in an inertial system Mass × Acceleration = Force applies in an inertial frame of reference.

But we like to make measurements relative to the Earth, which is rotating! Apparent trajectory of the ball in a rotating To do this we must add correction terms in the equation, coordinate system the centrifugal and Coriolis accelerations. The centripetal acceleration/centrifugal force Effective Gravity

g is everywhere normal to the earth’s surface

2 outward v 2 =Ω r Ω 2 R 2 force r R R Ω R g* Ω v g g* g

r v2 Inward −=−Ω2r acceleration r gg= *+Ω2R effective gravity g on a spherical earth effective gravity on an earth with a slight equatorial bulge

When frictional forces can be neglected, F is the pressure Effective Gravity gradient force total pressure

F = −∇p T force per unit volume If the earth were a perfect sphere and not rotating, the only gravitational component g* would be radial. ⎛⎞du ρ−Ω∧=+ρ⎜⎟2 uFg dt Because the earth has a bulge and is rotating, the effective ⎝⎠ gravitational force g is the vector sum of the normal gravity to du 1 per the mass distribution g*, together with a centrifugal force =− ∇p T +gu −2Ω ∧ Ω2R, and this has no tangential component at the earth’s dt ρ unit mass surface. gg*R=+Ω2

This is Euler’s equation of motion in a rotating reference frame. The Coriolis force does no work Perturbation pressure, buoyancy force

dp0 Define ppzpT = 0 ()+ where =−g ρ dz 0 the Coriolis force acts normal to the Ω rotation vector and normal to the velocity. p0(z) and ρ0(z) are reference pressure and density fields u is directly proportional to p is the perturbation pressure the magnitude of u and Ω. Euler’s equation becomes g = (0, 0, −g) −2Ω ∧ u

Du 1 ⎡ρρ− 0 ⎤ +∧=−∇+2Ω ugp ⎢ ⎥ Note: the Coriolis force does no work because u(⋅ 2Ω ∧ u)≡ 0 Dt ρ ⎣ ρ ⎦ the buoyancy force

Mathematical formulation of the continuity 1 equation for an incompressible fluid Important: the perturbation pressure gradient −∇p ρ ⎛⎞ρ−ρ0 and buoyancy forceg ⎜ ⎟ are not uniquely defined. vv+ δ ⎝⎠ρ

1 ⎛⎞ρ−ρ ww+ δ But the total force −∇+ p g 0 is uniquely defined. ρρ⎜⎟ δy ⎝⎠ u δx uu+ δ

11⎛⎞ρ−ρ0 δz Indeed − ∇+ppgg⎜⎟ =−∇T + ρρρ⎝⎠ w v The mass continuity equation Rigid body dynamics

Incompressible fluid ∇⋅u =0 Mass m Force F

∂ρ + ∇⋅()0 ρu = x Compressible fluid ∂t

2 ∇⋅((z))0 ρu = dx Anelastic approximation o Newton’s equation of motion is: mF= dt2

Problem is to calculate x(t) given the force F

Fluid dynamics problems Fluid dynamics problems

Ø The force field is determined by the overall constraints Ø The aim of any fluid dynamics calculation is to calculate provided by the flow field U(x,y,z,t) in a given region subject to appropriate boundary conditions and the constraint of – the requirement of continuity continuity. – the boundary conditions Ø The calculation of the force field (i.e. the pressure field) Ø In particular, the pressure field at any instant is may not be necessary, depending on the solution method. determined by the flow configuration

– I will now illustrate this with an example! – Let us forget about density differences and rotation for this example A mathematical demonstration LO HI U isobars D1u = −∇p ' ∇⋅u= 0 Dt ρ

HI Momentum equation Continuity equation

The divergence of the momentum equation gives:

U 2 ∇∇⋅ρ⋅∇p' =− ( uu) HI LO pump streamlines This is a diagnostic equation!

But what about the effects of rotation? Assumptions: inviscid, irrotational, incompressible flow

buoyancy form Newton’s 2nd law vertical component

mass × acceleration = force p =+p(z)p′ To dpo Put where =−g ρ o Dw∂ p ρ=ρ(z) +ρ′ dz ρ =−T −g ρ o Dt∂ z Dw 1∂ p′ ⎛⎞ρ−ρo Then =− +b where bg=− ⎜⎟ Dtρ∂ z ⎝⎠ρ buoyancy force is NOT unique Buoyancy force in a hurricane

⎛⎞ρ− ρo ρ ()z bg=− ⎜⎟ o ⎝ ρ ⎠

it depends on choice of reference density ρo(z)

1p∂∂ 1p' but −−=−+T gbis unique ρ∂zz ρ∂ ρ ()z o

Initiation of a thunderstorm negative buoyancy z tropopause

outflow

θ = constant original heated air θ = constant positive buoyancy LFC LCL

Τ+ΔΤ T inflow U(z) negative buoyancy Some questions positive buoyancy HI outflow p' Ø How does the flow evolve after the original thermal has reached the upper troposphere?

Ø What drives the updraught at low levels? original heated air – Observation in severe thunderstorms: the updraught at cloud base is negatively buoyant! LFC LO – Answer: - the perturbation pressure gradient LCL

HI inflow HI HI negative buoyancy

The geostrophic approximation Choose rectangular coordinates: For frictionless motion (D = 0) the momentum equation is k = (0,0,1) z Ω = Ωk Du 1 +∧=−∇2Ω u p Dt ρ Let Ro → 0 u perturbation pressure k h 1 y 2Ω ∧=−∇u p ρ x This is called the geostrophic equation velocity components u = (u,v,w), u = u + wk We expect this equation to hold approximately in synoptic h scale motions in the atmosphere and oceans, except possibly uh = (u,v,0) is the horizontal flow velocity near the equator. 1 Take k ∧ 2Ω ∧=−∇u p The geostrophic wind ρ (k ⋅ u)k = (0, 0 w) 1 1 22()pΩ∧kku() ∧ =Ω[] kuku ⋅ − =− k ∧∇ uk=∧∇p ρ hh2Ωρ

s p = (∂p/∂x, ∂p/∂y, 0) − uh h Ø The geostrophic wind blows parallel to the lines (or more strictly surfaces) of constant pressure - the isobars, with 1 low pressure to the left. uk=∧∇p hh2Ωρ Ø Well known to the layman who tries to interpret the ∂p newspaper "weather map", which is a chart showing and 0 = isobaric lines at mean sea level. ∂z Ø In the southern hemisphere, low pressure is to the right. This is the solution for geostrophic flow.

Choice of coordinates Geostrophic flow

Ø For simplicity, let us orientate the coordinates so that x points in the direction of the geostrophic wind. pressure gradient force Ø Then v = 0, implying that ∂p/∂x = 0 . isobar low p 1 ∂p u =− u 2Ωρ ∂y

high p isobar Ø Note that for fixed Ω , the winds are stronger when the isobars are closer together and, for a given isobar Coriolis force separation, they are stronger for smaller |Ω|. (Northern hemisphere case: > 0) 1 A mean sea level isobaric chart over Australia Note also that the solution uk=∧∇p hh2Ωρ ∂p and 0 = ∂z

L tells us nothing about the vertical velocity w.

Ø For an incompressible fluid, ∇⋅ u = 0 .

Ø Also, for geostrophic flow, ∇h ⋅ uh = 0 . Ø then ∂w/∂z = 0 implying that w is independent of z. H H H

If w = 0 at some particular z, say z = 0, which might be the ground, then w ≡ 0.

The geostrophic equation is degenerate! Vortex flows: the gradient wind equation

Ø The geostrophic equation is degenerate, i.e. time derivatives have been eliminated in the approximation. Ø Strict geostrophic motion requires that the isobars be straight, or, equivalently, that the flow be uni-directional. Ø We cannot use the equation to predict how the flow will evolve. Ø To investigate balanced flows with curved isobars, including vortical flows, it is convenient to express Euler's equation in Ø Such equations are called diagnostic equations. cylindrical coordinates. Ø In the case of the geostrophic equation, for example, a knowledge of the isobar spacing at a given time allows Ø To do this we need an expression for the total horizontal us to calculate, or 'diagnose', the geostrophic wind. acceleration Duh/Dt in cylindrical coordinates. Ø We cannot use the equation to forecast how the wind velocity will change with time. The radial and tangential components of Euler's equation The case of pure circular motion with u = 0 and ∂/∂θ ≡ 0. may be written v2 1 ∂p ∂u ∂u v ∂u ∂u v2 1 ∂p +=fv ++u +w −−=−fv r ρ ∂r ∂t ∂r r ∂θ ∂ρz r ∂r

∂v ∂v v ∂v ∂v uv 1 ∂p ++u +w ++=−fu ∂t ∂r r ∂θ ∂ρz r r ∂θ Ø This is called the gradient wind equation. Ø It is a generalization of the geostrophic equation which takes The axial component is into account centrifugal as well as Coriolis forces. ∂w ∂w v ∂w ∂w 1 ∂p ++u +w =− Ø This is necessary when the curvature of the isobars is large, ∂t ∂r r ∂θ ∂ρz ∂z as in an extra-tropical depression or in a tropical cyclone.

The gradient wind equation Force balances in low and high pressure systems

1 ∂p v2 Write 0 =− + +fv ρ ∂r r Cyclone Anticyclone VV terms interpreted as forces

Ø The equation expresses a balance of the centrifugal force LO HI (v2/r) and Coriolis force (fv) with the radial pressure gradient. CO PG CO Ø This interpretation is appropriate in the coordinate system  CE defined by r and θ , which rotates with angular velocity v/r. HI LO CE PG 1 2 1 1 22 rp∂ 1 ∂p v2 vfrfr=− + + The equation 0 =− + +fv 2 4 ρ ∂r ρ ∂r r is a diagnostic equation for the tangential velocity v in terms Ø In a low pressure system, ∂p/∂r > 0 and there is no of the pressure gradient: theoretical limit to the tangential velocity v.

1 2 1 1 22 rp∂ Ø In a high pressure system, ∂p/∂r < 0 and the local value of vfrfr=− + + 2 2 4 ρ ∂r the pressure gradient cannot be less than −ρrf /4 in a balanced state.

Ø Therefore the tangential wind speed cannot locally exceed Choose the positive sign so that geostrophic balance is rf/2 in magnitude. recovered as r →∞ (for finite v, the centrifugal force Ø This accords with observations in that wind speeds in tends to zero as r →∞). anticyclones are generally light, whereas wind speeds in cyclones may be quite high.

Limited wind speed in anticyclones

In the anticyclone, the Coriolis force increases only in proportion to v: => this explains the upper limit on v predicted by the gradient wind equation.

V

CO= fv HI v2 CE = r CO

CE PG End of L2