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Home , Centrifugal force, Equations of motion

The of in a Rotating Coordinate System

Chapter 3 Since the is rotating about its axis and since it is convenient to adopt a fixed in the earth, we need to study the in a rotating coordinate system.

Before proceeding to the formal derivation, we consider briefly two concepts which arise therein:

Effective and Coriolis Effective Gravity

g is everywhere normal to the earth’s surface

Ω 2 R 2 R R Ω R g* g g* g

effective gravity g effective gravity on an earth on a with a slight Effective Gravity

If the earth were a perfect and not rotating, the only gravitational component g* would be radial.

Because the earth has a bulge and is rotating, the effective gravitational force g is the vector sum of the normal gravity to the distribution g*, together with a Ω2R, and this has no tangential component at the earth’s surface. gg*R= + Ω 2 The

A line that rotates with the roundabout

Ω A line at rest in an inertial system

Apparent of the ball in a rotating coordinate system Mathematical derivation of the Coriolis force

Representation of an arbitrary vector A(t) Ω A(t) = A (t)i + A (t)j + A (t)k k 1 2 3 k ´ A(t) = A1´(t)i ´ + A2´ (t)j ´ + A3´ (t)k´ inertial j reference system

rotating j ´ reference system i ´ i The of an arbitrary vector A(t)

The derivative of A(t) with respect to time

d A dA dA dA a =+ijk12 + 3 dt dt dt dt the subscript “a” denotes the derivative in an inertial reference frame

ddAdAi′ ′ In the rotating frame a1=++i′′A ... of reference dt dt1 dt dA′ = ii′′′1 +∧+A...()Ω dt 1

⎡⎤d ′ =+∧Ω ()A...1i+′ ⎣⎦⎢⎥dt Let a be any vector rotating with angular Ω

Ω a + da a

θ b da = ||||sinabΩ θ  dt =∧Ω a perpendicular to Ω and a vector r(t)

dar O Want to calculate ua = dt the absolute velocity of an air parcel

The in a rotating frame of reference is dr dr dr dr uijk==12′′′ + ′ + 3′ ′ dt dt dt dt

and uua = + Ω ∧ r Example Ω

uua =+Ω ∧ r

V

ΩRe

Earth’s radius

This air parcel starts relative to the earth with a poleward velocity V.

It begins relative to with an additional eastwards velocity component ΩRe . We need to calculate the absolute if we wish to apply Newton's second law

daua dt Measurements on the earth give only the relative velocity u and therefore the relative acceleration du dt

uua = + Ω ∧ r With A = u a d uud aa=+∧ a Ω u dt dt a d uud aa=+∧+∧∧2ΩΩΩur() dt dt absolute d uud aa=+∧+∧∧2ΩΩΩur() acceleration dt dt

relative Coriolis- Centripetal acceleration acceleration acceleration Centripetal acceleration

Ω Ω Position vector r is split up |Ω| RR rr=+(.ΩΩ )  R

ΩΩ∧∧=()rR Ω∧ (Ω ∧ )r Ω r 2 r =−||Ω R Ω

The Centripetal acceleration is directed inwards towards the axis of and has magnitude |Ω|2R. Newton’s second law in a rotating frame of reference

d u In an inertial frame: ρ aa= F dt

density force per unit mass In a rotating frame:

du ρ[()]+∧+∧∧=2ΩΩΩurF dt Alternative form

du ρρΩρΩ=−Fu2() ∧− ∧Ω ∧ r dt

Coriolis force Centrifugal force Let the total force F = g* + F´ be split up du ρρΩρΩ= Fg*′ +−2() ∧−∧∧ uΩ r dt

Ω ∧ ()||Ω ∧ rR= − Ω 2 With gg*R= + Ω 2

du ρρρΩ= Fg′ +−2 ∧ u dt

The centrifugal force associated with the earth’s rotation no longer appears explicitly in the ; it is contained in the effective gravity. When frictional can be neglected, F’ is the gradient force total pressure

F ′ = −∇ p T per unit volume

du ρρρΩ= Fg′ +−2 ∧ u dt du 1 =− ∇p +gu −2Ω ∧ per dt ρ T unit mass

This is the Euler equation in a rotating frame of reference. The Coriolis force does no

the Coriolis force acts normal to the Ω rotation vector and normal to the velocity. u is directly proportional to the magnitude of u and Ω.

−2Ω ∧ u

Note: the Coriolis force does no work because u(⋅ 2Ω ∧ u)≡ 0 pressure, force

dp0 Define ppzpT = 0 ()+ where =−g ρ dz 0 p0(z) and ρ0(z) are the reference pressure and density fields p is the perturbation pressure

Important: p0(z) and ρ0(z) are not uniquely defined g = (0, 0, −g) Euler’s equation becomes

Du 1 ⎡ρρ− 0 ⎤ +∧=−∇+2Ω ugp ⎢ ⎥ Dt ρ ⎣ ρ ⎦ the buoyancy force 1 Important: the perturbation pressure gradient −∇p ρ ⎛⎞ρ −ρ0 and buoyancy forceg ⎜⎟ are not uniquely defined. ⎝⎠ρ

1 ⎛⎞ρ −ρ0 But the total force −∇+ p g ⎜⎟ is uniquely defined. ρρ⎝⎠

11⎛⎞ρ−ρ0 Indeed −∇+ppgg⎜⎟ =−∇+T ρρρ⎝⎠ A mathematical demonstration

D1u =− ∇p 'b + kˆ ∇ ⋅u = 0 Dt ρ

Momentum equation Continuity equation

The divergence of the equation gives:

2 ∇p' =−⎡ ∇⋅ρ⋅∇uu −∇⋅ρ b kˆ ⎤ ⎣ ()()⎦

This is a diagnostic equation! Newton’s 2nd law

mass × acceleration = force

ρDw=−∂p − gρ Dt ∂z buoyancy form

p = p ()z + p ' o dp Put where o =−g ρ ρ=+ρρ()z ' o o dz

Dw 1∂ p' ⎛⎞ρ−ρo Then =− +b where bg=− ⎜⎟ Dtρ∂ z ⎝⎠ρ buoyancy force is NOT unique

⎛⎞ρ−ρ bg=− ⎜⎟o ⎝⎠ρ

it depends on choice of reference density ρo(z)

1p∂∂ 1p' but −−=−+gbis unique ρ∂zz ρ∂ Buoyancy force in a hurricane

ρ ()z o

ρ ()z o Initiation of a thunderstorm z

θ = constant

Τ + ΔΤ T U(z) negative buoyancy tropopause

outflow original heated air θ = constant positive buoyancy LFC LCL

inflow

negative buoyancy Some questions

» How does the flow evolve after the original thermal has reached the upper troposphere?

» What drives the updraught at low levels?

– Observation in severe thunderstorms: the updraught at cloud base is negatively buoyant!

– Answer: - the perturbation pressure gradient positive buoyancy

HI outflow p'

original heated air

LFC LO LCL

HI inflow HI HI negative buoyancy Scale analysis

» Assume a homogeneous ρ = constant. » Euler’s equation becomes:

Du 1 +∧=−∇2Ω u p Dt ρ

U 2 ΔP scales: 2ΩU L ρL

DDtu / UL2 / U Then ~ ==Ro 222Ω ∧ u ΩΩU L Rossby number Extratropical cyclone

2Ω = 10−4 s−1

L = 106 m L U = 10 ms−1

U 10 Ro == = 10−1 2Ω L 10−46× 10 Tropical cyclone

U = 50 m/sec

2Ω = 5 X 10−5 L = 5 x 105 m

U 50 Ro == = 2 2 Ω L 510× −55× 510× Dust devil

L

L = 10 - 100 m U = 10 ms−1 2Ω = 10−4s−1 U 10 Ro == =→10−−34 10 2Ω L 10−4 × (, 10 100 ) Waterspout

L = 100 m

U = 50 ms−1

2Ω = 10−4s−1

Ro = 510× 3

L Aeroplane wing

5 L = 10 m U = 200 m s−1 2Ω = 10−4s−1 Ro = 210× The Rossby number

Flow system L U m s−1 Ro

Ocean circulation 103 - 5 × 103 km 1 (or less) 10−2 - 10−3

Extra-tropical cyclone 103 km 1-10 10−2 - 10−1

Tropical cyclone 500 km 50 (or >) 1

Tornado 100 m 100 104

Dust devil 10-100 m 10 103 - 104

Cumulonimbus cloud 1 km 10 102

Aerodynamic 1-10 m 1-100 103 - 106

Bath tub vortex 1 m 10-1 103 Summary

(i) Large scale meteorological and oceanic flows are strongly constrained by rotation (Ro << 1), except possibly in equatorial regions. (ii)Tropical cyclones are always cyclonic and appear to derive their rotation from the background rotation of the earth. They never occur within 5 deg. of the where the normal component of the earth's rotation is small. (iii) Most tornadoes are cyclonic, but why? (iv) Dust devils do not have a preferred sense of rotation as expected. (v) In aerodynamic flows, and in the bath (!), the effect of the earth's rotation may be ignored. Coordinate systems and the earth's sphericity

» Many of the flows we shall consider have horizontal dimensions which are small compared with the earth's radius. » In studying these, it is both legitimate and a great simplification to assume that the earth is locally flat and to use a rectangular coordinate system with z pointing vertically upwards. » Holton (§2.3, pp31-35) shows the precise circumstances under which such an approximation is valid. » In general, the use of spherical coordinates merely refines the theory, but does not lead to a deeper understanding of the phenomena. Take rectangular coordinates fixed relative to the earth and centred at a point on the surface at .

Ω

k k j j i i

Φφ

equatorÄquator φ Ω Ω

||sinΩ φ k ||cosΩ φ j f-plane or β-plane φ

ΩΩ=φ+φ||cos||sinj Ω k

2||cos||sinΩ ∧=u Ω φ∧+j u Ω φ∧ku In component form

⎡−+22ΩΩvwsinφφ cos ⎤ ⎢ ⎥ 2Ω ∧=u ⎢ 2 Ω u sin φ ⎥ ⎢ ⎥ ⎣⎢ − 2 Ω u cosφ ⎦⎥

I will show that for middle latitude, synoptic-scale weather systems such as extra-tropical cyclones, the terms involving cos φ may be neglected. 2Ω ∧=u 2||cosΩ φ∧+ju 2||sinΩ φ∧ku

The important term for large-scale To a good approximation

22Ω ∧ ukufu= ||sinΩ φ ∧ = ∧

f = 2 | Ω | sin φ fk= f

Coriolis parameter Scale analysis of the equations for middle latitude synoptic systems

» Much of the significant weather in middle is associated with extra-tropical cyclones, or depressions. » We shall base our scaling on such systems. » Let L, H, T, U, W, P and R be scales for the horizontal size,

vertical extent, time, |uh|, w, perturbation pressure, and density in an extra-tropical cyclone, say at 45° latitude, where f (= 2Ω sin φ ) and 2Ω cos φ are both of order 10−4.

UmsW==10−121;; 10− ms− LmkmHmkm==1063();(); 10 10 4 10 and TLU==/ ~1053 s (~ 1 dayP );δ 10 Pamb ( 10 ) Rkgm= 1 −3 . horizontal momentum equations

Du 1 ∂p −+22ΩΩvwsinφφ cos =− Dt ρ ∂x Dv 1 ∂p +=−2Ωu sin φ Dt ρ ∂y scales U2/L 2 ΩU sin φ 2ΩW cos φ δP/ρL orders 10−43 10− 10− 6 10− 3

Du 1 h +∧fpku =−∇ Dt hhρ vertical momentum equations

Dw 1 ∂p −=−−2Ω u cosφ T g Dt ρ ∂z

scales UW/L 2ΩU cosφδPT/ρΗ g

orders 10−−73 10 10 10

negligible

the atmosphere is strongly hydrostatic on the synoptic scale.

But are the disturbances themselves hydrostatic? Question: when we subtract the reference pressure p0 from pT, is it still legitimate to neglect Dw/Dt etc.?

vertical momentum equations Dw 1∂ p − 2ucosΩφ=−+ b Dtρ∂ z

scales UW/L 2ΩU cosφδP/ρΗ∗ gδT/T0

−−73 − 1 − 1 orders 10 10≥ 10 10

still negligible H* = height scale for a assume that H* ≤ H perturbation pressure difference δp of 10 mb assume that δT ≈ 3oK/300 km In summary Dw 1∂ p − 2ucosΩφ=−+ b Dtρ∂ z

orders 10−−−−7311 10≥ 10 10

1p∂ 0b= −+ ρ∂z

In synoptic-scale disturbances, the perturbations are in close hydrostatic balance

Remember: it is small departures from this equation which drive the weak vertical motion in systems of this scale. The hydrostatic approximation

The hydrostatic approximation permits enormous simplifications in dynamical studies of large-scale motions in the atmosphere and oceans. End of Chapter 3