The governing equations and the dominant balances of flow in the atmosphere and ocean USPC Summerschool, Utrecht 2016 H. E. de Swart

Institute for Marine and Atmospheric research, Utrecht University, P.O. Box 80.005, 3508 TA Utrecht, the Netherlands Email: [email protected] , website: www.science.uu.nl/∼swart104

The earth’s climate is the result of complex feedbacks between radiation processes, thermody- namic processes (e.g., condensation, freezing), motion of fluids and chemical processes. Thor- ough insight into the dynamics of the climate dynamics requires knowledge of fundamental principles to develop and analyse models. The focus of this lecture is on two important laws that underly all models governing the motion of air and water on a rotating planet. The first is conservation of mass, which results in the continuity equation. The second is Newton’s second law, which yields the momentum equations. The fact that the earth rotates has profound implications for the way that flows in the atmo- sphere and ocean behave. This is mainly due to the manifestation of two apparent forces, the centrifugal force and Coriolis force. The former is a ’static’ force and causes deformation of the earth’s surface. The Coriolis force however is dynamic, as it acts on moving particles, causing them to deflect into a direction perpendicular to their motion. A key aspect of this lecture is that large scale flow in the atmosphere and ocean obeys, to a good approximation, a balance between the horizontal pressure gradient force and the Coriolis force. This so-called geostrophic balance explains why e.g. the mean wind in the atmosphere is directed to the east and why flow around low pressure cells is counterclockwise on the North- ern Hemisphere. The vertical momentum equation of large scale flow reduces to hydrostatic balance, i.e., the vertical pressure gradient force balances the gravity force. The slides of this lecture are added in the annex of this document. Also, references to further literature are given. As a preparation for this lecture I recommend you to read J. Holton (2004), An introduction to dynamic meteorology, chapters 1 and 2. Contents of the lecture:

1. Introduction page 1 2. Basic concepts to describe fluid motion 3 Eulerian/Lagrangian framework, total time derivative, advection 3. Derivation of the continuity equation 8 4. Momentum balance 11 formulation in absolute frame, transfomation to rotating frame manifestatation of centrifugal force and Coriolis force other forces (pressure gradient, tides, friction) 5. Other equations satisfied by moving fluids (brief) 25 6. Scale analysis of large scale flow 27 geostrophic balance hydrostatic balance thermal wind 7. Final remarks 40 Further literature and appendices 41

1 Governing equations and dominant balances for flow in atmosphere and ocean

H.E. de Swart

Introduction

In lectures so far:

Climate system results from feedbacks between

radiation processes
thermodynamic processes
chemical processes
motion of air, water, ice
transport of substances (e.g., moisture, salt, ozon, sediment)

1 Introduction

Here, focus on motion of air and water on a rotating earth

This topic is called geophysical fluid dynamics

Important aspect of fluids on rotating planets (here: Earth): Their dynamics is significantly affected by the Coriolis force

=> deflection of fluid to the right (on Northern Hemisphere)

Explains for example why flow around low pressure is counter-clockwise… ; flow is approximately geostrophic

L L

pressure gradient force Coriolis force

2 Remark:

Not all phenomena in atmosphere and ocean are strongly affected by the Coriolis force, e.g.

This is a matter of scales: if time scale feature <<1 day then effects of earth rotation are not important

Topics:

1. Introduction

2. BasicBasic conceptsconcepts

3. Equations of motion

3.1 Continuity equation 3.2 Momentum equations 3.3 Remarks about other equations

4. GeostrophicScaling and geostrophicflow flow

3 2. Basic concepts

2.1 The continuum hypothesis Motions of fluids occur on spatial scales of mm-10000 km, 23-08-2014 15.00 GMT

www.weather underground

Scales are large compared to distance between molecules Hence, discrete molecular nature of fluid can be ignored; medium is considered as a continuum

‘ ’ Thus, point (x,y,z) in continuum is P small volume element (small compared to scale of motion) P still contains large number of molecules

z (x,y,z) y Δz x Δy Δx Field variables density ρ velocity vector u pressure p averaged quantities, represent net effect temperature T of many molecules moisture/salinity q/s are continuous functions of space and time

4 Note: state of a fluid is determined by 7 variables:

density ρ velocity vector u pressure p temperature T humidity/salinity q/s

=> Seven equations required that govern the fluid motion

These equations follow from physical laws:

P conservation of mass P Newton’s second law P first + second law of thermodynamics

To derive them, some concepts must be introduced

Control volumes

In deriving equations of motion for fluids two types of infinitesimal control volumes in the fluid are used

1. Eulerian frame of reference The control volume is fixed in space sizes are Δx, Δy, Δz

z (x,y,z) y Δz x Δy Δx Particles move through this control volume

5 Control volumes

2. Lagrangian frame of reference The control volume consists of infinitesimal mass of ‘tagged’ fluid particles.

z y Δz x Δy Δx

This control volume moves with the local fluid velocity

Local versus total time derivative

Consider a state variable (e.g., temperature T) in an Eulerian frame of reference.

Choose cartesian coordinates x,y,z

Then, T=T(x,y,z,t) is a field variable Change in time of T in a fixed control volume is

∂T partial derivative of T(x,y,z,t) ∂t for fixed x,y,z

Likewise, define in Eulerian frame of reference the partial derivatives

∂T ∂T ∂T ∂x ∂y ∂z

6 Local versus total time derivative

Now assume that x,y,z change in time

So T=T(x(t), y(t), z(t), t)

Now, the change ΔT during a small time increment Δt is

∂T ∂T ∂T ∂T ΔT = Δt + Δx + Δy + Δz+ ... ∂t ∂x ∂y ∂z

Divide by Δt and take the limit Δt→0 : dT ∂T dx ∂T dy ∂T dz ∂T = + + + dt ∂t dt ∂x dt ∂y dt ∂z

T function T function of of time only x,y,z and t

Local versus total time derivative In particular, when following the motion dx dy dz u= , v = , w= dt dt dt are the velocity components in x,y,z-direction

Use this to define right-hand side of previous expression as DT ∂T ∂T ∂T ∂T ≡ + u +v +w Dt ∂t ∂x ∂y ∂z

advective terms being the total time derivative of field variable T(x,y,z,t) Its meaning: the rate of change of T following the motion

Remark: Some authors write d/dt instead of D/Dt

7 Total time derivative in vector notation

Introduce

∂ ∂ ∂ u= ex u + ey v +ez w ∇= e + e +e x ∂x y ∂y z ∂z velocity vector nabla ‘vector’ with ex, ey, ez unit vectors in the x,y,z direction z

y x Then

DT ∂T ∂T ∂T ∂T DT ∂T = + u +v +w = + (u ⋅ ∇)T Dt ∂t ∂x ∂y ∂z Dt ∂t

Topics:

1. Introduction

2. Basic concepts

3. Equations of motion

3.1 Continuity equation 3.2 Momentum equations 3.3 Remarks about other equations

4. Scaling and geostrophic flow

8 Conservation of mass

Consider a small fixed volume element around point (x,y,z); Sizes of the element are Δx, Δy, Δz

z (x,y,z) y Δz x

Δy Δx Definitions:

ρ : density of fluid -3 -3 (~1.25 kgm for air; ~1030 kgm for sea water) u,v,w : velocity components in x,y,z direction

Conservation of mass

Mass inside the volume element: Δm = ρ Δx Δy Δz

Δm changes if total mass inflow through all sides ≠ 0

z (x,y,z) y ρ u (x− 1 Δx,y,z) ρ u + 1 Δ 2 (x 2 x,y,z) x

∂ ‘ ’ 1 Mass flux right : ρ u (x+ 1 Δx,y,z) ≈ ρ u (x,y,z) + (ρ u) (x,y,z) Δx 2 ∂x 2 ∂ ‘ ’ 1 Mass flux left : ρ u (x− 1 Δx,y,z) ≈ ρ u (x,y,z) − (ρ u) (x,y,z) Δx 2 ∂x 2 Total mass/time entering through left+right side: (flux left - flux right) ΔyΔz

9 Total mass per time entering the volume element through left+right side: ⎡ ∂ 1 ⎛ ∂ 1 ⎞⎤ ⎢ρ u (x,y,z) − (ρ u) (x,y,z) Δx −⎜ ρ u (x,y,z) + (ρ u) (x,y,z) Δx⎟⎥ Δ yΔz ⎣ ∂x 2 ⎝ ∂x 2 ⎠⎦ ∂ = − (ρ u)Δx ΔyΔz ∂x

z (x,y,z) y x

Likewise, ∂ = − (ρv)Δx ΔyΔz through front + back side: ∂y ∂ through top + bottom side: = − (ρ w)ΔxΔyΔz ∂z

Conservation of mass

So, ∂ ⎧ ∂ ∂ ∂ ⎫ (ρ Δx ΔyΔz) =⎨ − (ρ u) − (ρv) − (ρw)⎬ Δx ΔyΔz ∂t ⎩ ∂x ∂y ∂z ⎭ or ∂ρ ∂ ∂ ∂ + (ρ u) + (ρv) + (ρw) = 0 ∂t ∂x ∂y ∂z

This is the continuity equation

In vector notation this equation reads ∂ρ + ∇ ⋅ (ρu) = 0 ∂t

1 0 The continuity equation: alternative expression

Recall,

∂ρ ∂ ∂ ∂ + (ρ u) + (ρv) + (ρw) = 0 ∂t ∂x ∂y ∂z or ∂ρ ∂ρ ∂u ∂ρ ∂v ∂ρ ∂w + u + ρ +v + ρ + w + ρ = 0 ∂t ∂x ∂x ∂y ∂y ∂z ∂z

Use the definition of the total time derivative =>

Dρ ⎛ ∂u ∂v ∂w⎞ Dρ +ρ ⎜ + + ⎟ = 0 + ρ ∇ ⋅ u = 0 Dt ⎝ ∂x ∂y ∂z ⎠ Dt

(for alternative derivation see annex 1)

Topics:

1. Introduction

2. Basic concepts

3. Equations of motion

3.1 Continuity equation 3.2 Momentum equations 3.3 Remarks about other equations

4. Scaling and geostrophic flow

1 1 Objective: To derive the momentum equations in a frame that rotates with the earth

‘Apparent’ forces to be expected (centrifugal, Coriolis)

Strategy: P first equations in an absolute (or inertial) frame P after that transformation results to the rotating frame

Momentum balance; absolute frame

Newton’s second law in an absolute frame

D u a a = ∑ Fi Fi : forces (per mass unit) Dt i

Operator Da/Dt is total time derivative in absolute frame; is velocity in absolute frame ua Transform this equation to momentum balance in a frame on the rotating earth (angular speed of rotation: Ω ≈7.3.10-5 s-1) So D u /Dt has to be converted into Du/Dt a a where D/Dt: total time derivative in rotating frame u : velocity vector in rotating frame

1 2 Momentum balance; rotating frame

Z,z’ r

Ω Ω: rotation vector y’ X,Y,Z: axes in absolute frame x’,y’,z’: axes in rotating frame r : position vector Y X x’

D r Dr By definition: u = a u = a Dt Dt

If u=0, then r is in uniform circular motion, velocity Ω×r

Momentum balance; rotating frame

So

= + Ω × Dar Dr ua u r => = + Ω × r Dt Dt

Note: in the boxed relation r can be replaced by any vector, in particular by => ua D u Du a a = a + Ω × u Dt Dt a D = ()u + Ω × r + Ω × ()u + Ω × r Dt Du = + Ω × u + Ω × ()u + Ω × r Dt

Substitute in Newton’s second law (see 2 slides back) =>

1 3 Momentum balance; rotating frame

The momentum balance in a rotating frame

Du + 2Ω × u + Ω ×() Ω × r = ∑ Fi Dt i

Here, two apparent forces emerge: −2Ω × u Coriolis force/mass −Ω×() Ω×r =Ω2 R centrifugal force/mass Ω

R r |R| : distance to rotation axis

Momentum balance; rotating frame

Manifestation of centrifugal force Fce

To capture the meaning of the centrifugal force the gravitational force is introduced.

Remark: a fluid is subject to the following (non-apparent) forces:

1. Gravitational force 2. Tidal forces discussed 3. Pressure gradient force later 4. Frictional force

1 4 The gravitational force

Its expression is

F = − ∇Φ = g Φ : gravitational potential g* g* * g* and vector is directed towards the centre of the earth. g* Ω

g *

Its length | |≈9.8 ms-2 is approximately constant g* in atmosphere and ocean.

Momentum balance; rotating frame

Manifestation of centrifugal force Fce

If earth would be a sphere, then Fce would have a component along the earth’s surface

However, Fce causes shape of the earth to adjust from sphere → spheroid, such that gravity = + Ω2 g g * R is perpendicular to the surface of the earth.

1 5 Momentum balance; rotating frame

Gravity g has only component -g along z-axis (z-axis vertically upward from earth’s surface)

z=0: the geoid (mean sea level)

Momentum balance; rotating frame

Thus, momentum equations on a rotating earth become

Du + 2Ω × u = g + F + F + F Dt p tide fric with

Fp : pressure force/mass

Ftide : tidal force/mass

Ffric : frictional force/mass

Before deriving expressions for Fp, Ftide, Ffric investigate the effect of the Coriolis force on geophysical flows.

1 6 Momentum balance; rotating frame

Manifestation of Coriolis force Fco

By definition, this apparent force/mass is

Fco = − 2Ω × u

The Coriolis force is perpendicular to velocity => it does no work!

Ω z’

’ u y x’ Fco

Fco causes moving fluid to deflect at right angles.

Momentum balance; rotating frame

Manifestation of Coriolis force Fco

1 7 Manifestation of Coriolis force Fco Introduce coordinates λ : longitude , φ : latitude , r : distance Ω

ey ez ex φ r

λ

At each point three unity vectors: ex : west→east , ey : south→north, ez : upward The velocity vector dλ dφ dr = + + u=rcosφ , v =r , w= u ex u ey v ez w dt dt dt

Manifestation of Coriolis force Fco Ω

R Furthermore,

φ Ω = ey Ω cosφ + ez Ω sinφ

R = − ey R sinφ + ez R cosφ

Thus, in this coordinate system, the Coriolis force

ex ey ez Coriolis parameter f −2 Ω × u = 0 −2Ω cosφ −2Ω sinφ (2x vertical component of Ω) u v w

= ex (−2Ω cosφ w+2Ω sinφ v) − ey 2Ω sinφ u+ez 2Ω cosφ u

1 8 Manifestation of Coriolis force Fco

Recall,

Fco = ex (−2Ω cosφ w+2Ω sinφ v) − ey 2Ω sinφ u+ez 2Ω cosφ u

This result reveals:

If u>0 (eastward) at Northern Hemisphere: Fco deflects particle southward + upward.

If v>0 (northward) at Northern Hemisphere: Fco deflects particle eastward.

If w<0 (downward) Fco deflects particle eastward.

Manifestation of Coriolis force Fco

If w<0 (downward), Fco deflects particle eastward. Example: object falls from a tower Ω north

west Fco east south u

Argument: use again that angular momentum L is conserved, but now ΔR=-cosφ Δr

In case of object that falls from Dom tower (100 m) the deflection is ~1.5 cm.

1 9 Governing equations and dominant balances for flow in atmosphere and ocean

H.E. de Swart

Topics:

1. Introduction

2. Basic concepts

3. Equations of motion

3.1 Continuity equation 3.2 Momentum equations 3.3 Remarks about other equations

4. Scaling and geostrophic flow

2 0 Important results:

1. Definition of total time derivative DT ∂T ∂T ∂T ∂T ≡ + u +v +w Dt ∂t ∂x ∂y ∂z

Its meaning: the rate of change of T following the motion

2. Continuity equation ∂ρ Dρ + ∇ ⋅ (ρu) = 0 or + ρ ∇ ⋅ u = 0 ∂t Dt

It describes conservation of mass

3. Momentum equations on a rotating earth Du + 2Ω × u = g + F + F + F Dt p tide fric with

Fco = − 2Ω × u Coriolis force/mass unit

g : gravity (includes centrifugal force)

Fp : pressure force/mass

Ftide : tidal force/mass

Ffric : frictional force/mass

We will now consider expression for Fp (Ftide, Ffric in notes)

2 1 Tidal forces

.. can cause large response..

Pressure gradient force

Note: pressure itself is a force per surface area. It are the spatial variations in pressure that result in a force/mass.

Derivation (fixed control volume):

z (x,y,z) y * * F Fp p,left ,right x

Pressure force ‘right ’: * ⎛ ∂p 1 ⎞ Fp,right =− p (x+ 1 Δx,y,z)ΔyΔz ≈ −⎜ p (x,y,z) + (x,y,z) Δx⎟ ΔyΔz 2 ⎝ ∂x 2 ⎠ Pressure force ‘left’:

* ⎛ ∂p 1 ⎞ Fp,left = p (x− 1 Δx,y,z)ΔyΔz ≈ ⎜ p (x,y,z) − (x,y,z) Δx⎟ ΔyΔz 2 ⎝ ∂x 2 ⎠

2 2 Pressure gradient force

Total force/mass in x-direction:

* * Fp,left + Fp,right 1 ∂p F = = − p,x ρΔxΔyΔz ρ ∂x

Likewise, total force/mass in y- and z-direction:

* * Fp,fr ont + Fp,back 1 ∂p F = = − z p,y ρΔ Δ Δ ρ ∂ x y z y (x,y,z) y * * x Fp,down + Fp,up 1 ∂p F = = − p,z ρΔxΔyΔz ρ ∂z

Final result in vector notation: 1 F =− ∇p directed from high→low pressure p ρ

Frictional forces

All fluids are subject to friction.

Complete discussion is beyond this lecture Here, only the overall concept.

Consider a typical velocity profile near the bottom:

distance to bottom

velocity

Fluid elements above each other induce shear stresses at contact surfaces (i.e., forces/surface)
Due to friction particles at the bottom do not move (no slip)

2 3 Final momentum equations and continuity equation for geophysical flows

Final momentum equations in rotating frame on earth:

Du 1 + 2Ω×u= −∇p ++ gF + F Dt ρ tide fric

Remark: if Ω=0 and Ffric represents only molecular friction these are called the Navier Stokes equations In general, geophysical flows obey generalisations of the Navier Stokes equations.

Recall, the continuity equation

Dρ + ρ ∇ ⋅ u = 0 Dt

Topics:

1. Introduction

2. Basic concepts

3. Equations of motion

3.1 Continuity equation 3.2 Momentum equations 3.3 Remarks about other equations

4. Scaling and geostrophic flow

2 4 Other equations of motion

Continuity equation and momentum equations do not constitute a closed system:1+3 equations, 5 unknowns (3 velocity components + p + ρ).

Simple closures are established in cases

1. ρ=constant (homogeneous fluid) → lectures of Van de Berg and von der Heydt

2. ρ=ρ (p) (barotropic fluid)

In general, ρ depends on p, T and q/s (T: temperature, q: specific humidity, s: salinity),

So 3 more equations are needed.

Other equations of motion

These three additional equations are

1. Equation of state 2. Thermodynamic (heat) equation 3. Mass conservation for moisture/salt

Ad 1: Equation of state

ρ = ρ(p, T, q) air → ~ ideal gas law

ρ = ρ(p, T, s) sea water → ~ linear in T and s

See Gill (1982) for explicit expressions.

2 5 Other equations of motion

Ad 2: Thermodynamic heat equation

From first+second law of thermodynamics:

DT Dp ρc − α T = ∇ ⋅ ()k ∇T −F + ρQ p Dt Dt T rad

-1 -1 with: cp (Jkg K ) : specific heat at constant pressure -1 α (Κ ) : thermal expansion coefficient -1 -1 kT (Wm K ) : thermal diffusion coefficient -2 Frad (Wm ) : radiative fluxes Q : heating rate per mass (e.g. latent heat release) See Gill (1982) for further details.

Other equations of motion

Ad 3: Mass conservation for moisture/salt

Definitions:

density of moisture in air: ρw = q ρ

density of salt in sea water : ρs = s ρ Thus, q and s are mass fractions.

The equations read

Dq 2 Ds 2 = k ∇ q = kw ∇ s Dt a Dt

2 -1 with: ka, kw (m s ) : diffusion coefficients. See Gill (1982) for further details.

2 6 Topics:

1. Introduction

2. Basic concepts

3. Equations of motion

3.1 Continuity equation 3.2 Momentum equations 3.3 Remarks about other equations

4. Scaling and geostrophic flow

Scaling

Consider momentum equations (+ continuity equation). They describe all scales of motion:

P Very difficult to obtain solutions of full equations P Not possible to gain insight in this way

2 7 Scaling

Often simplifications are made

Depending on phenomena of interest, certain terms in the equations appear to be dominant with respect to others

Procedure:

1) Present equations in a suitable coordinate system 2) Choose typical scales that characterise the phenomenon of interest 3) Evaluate magnitude of different terms in the equations 4) Retain only the dominant terms 5) Analyse the resulting, simplified system

Scaling

Step 1: Coordinate system Choose oblate spheroidal coordinates λ, φ, r on rotating earth, such that r=r0: geoid Ω

ex : west→east ey ez e : south→north e y x e : upward φ z λ depend on position

The velocity vector dλ dx dφ dy dr dz u = e u+e v+ e w u=rcosφ ≡ , v =r ≡ , w= ≡ x y z dt dt dt dt dt dt

2 8 Scaling

The momentum equations in vector notation:

Du 1 + 2Ω×u= −∇p ++ gF + F Dt ρ tide fric

Now,

Du D = ()ex u + ey v + ez w Dt Dt

Du Dv Dw De Dey De = e + e + e + x u + v + z w x Dt y Dt z Dt Dt Dt Dt and

D ∂ ∂ ∂ ∂ ∂ ≡ + u⋅ ∇ = + u + v + w Dt ∂t ∂t ∂x ∂y ∂z

Scaling

For spheroidal coordinates Ω Dex u = ()ey sinφ −ez cosφ Dt rcosφ ey ez Dey utanφ v ex = − e −e Dt x r z r φ

De u v λ z = e +e Dt x r y r

(see e.g. Holton, 2004)

Next Coriolis force. Recall (from previous slides),

2Ω × u = ex (−2Ω cosφ w+2Ω sinφ v)−ey 2Ω sinφ u+ez 2Ω cosφ u

2 9 Scaling

Pressure gradient force: 1 1 ∂p 1 ∂p 1 ∂p − ∇p = − e −e −e ρ x ρ ∂x y ρ ∂y z ρ ∂z Gravity:

g = −ez g etc.

So momentum equations in components:

Du uv tanφ uw 1 ∂p ∂Φ − + − 2Ωsin φ v + 2Ωcosφ w = − − t + F Dt r r ρ ∂x ∂x x Dv u2 tanφ vw 1 ∂p ∂Φ + + +2Ωsin φ u = − − t + F Dt r r ρ ∂y ∂y y Dw u2 + v 2 1 ∂p ∂Φ − − 2Ωcosφ u = − − g − t + F Dt r ρ ∂z ∂z z metric Coriolis pressure gravity tide friction gradient

Scaling

Step 2: Choice of phenomena of interest

Here, focus on large-scale flow at midlatitudes:
horizontal length scale L >> vertical length scale H
L << r0 (radius of the earth)
velocity scale U such that L/U >> rotation period Two prototype examples:

weather systems Agulhas current and eddies

3 0 Scaling: large-scale flow

Step 3: Evaluate magnitude of different terms

Example: large-scale atmospheric flow Typical scales:

6 4 7 L~10 m H~10 m r0~10 m -1 -1 -4 -1 U~10 ms W=(H/L)U~0.1 ms f0~10 s 3 2 -2 -2 Δp/ρ~10 m s ρ = ρ0(z) + ρ’ [ρ’ ]/[ρ0]~10 Evaluate terms in horizontal momentum equations, e.g. x-equation: Du uv tanφ uw 1 ∂p ∂Φ − + − 2Ωsinφ v+2Ωcosφ w= − − t + F Dt r r ρ ∂x ∂x x 2 U U UW Δp ΔΦt f0 U f0 W [Fx ] L /U r0 r0 ρ0L L magnitude 10-4 10-5 10-6 10-3 10-6 10-3 10-7 10-7 (ms-2)

Scaling: large-scale flow

Likewise, for the y-momentum equation.

Thus, to a first approximation the horizontal momentum balance for large-scale flow reads

1 ∂p 1 ∂p − f 0 v ≈ − , f0 u ≈ − ρ0 ∂x ρ0 ∂y

with f0=2Ω sinφ0 : Coriolis parameter at a fixed latitude.

This result is known as the geostrophic balance

3 1 Geostrophic flow

The flow that exactly satisfies the geostrophic balance is called the geostrophic flow (components ug and vg).

The corresponding vertical velocity wg =0 (see later).

Hence, in vector notation, the geostrophic velocity is

1 ug =ez × ∇p ρ0 f0

Note: the geostrophic relationship is diagnostic.

Geostrophic flow

Properties of geostrophic flow:

1. Vector ug is in horizontal plane and perpendicular to ∇p, so, the velocity is along lines p=constant

2. Magnitude of geostrophic velocity is proportional to |∇p|

Fp

ug y p=p1-Δp p=p 1 x Fco p=p1+Δp

3 2 Geostrophic flow

Properties of geostrophic flow:

3. Vector ug is divergence free, i.e., ∂∂uv gg+=0 ∂∂xy so the velocity field is governed by a stream function ψ, such that ∂ψ ∂ψ u = − , v = g ∂y g ∂x

In fact, ψ= p/(ρ0 f0)

Geostrophic flow

Necessary condition for observed flows to be geostrophic is

2 inertia << Coriolis, i.e., U /L << f0 U , or

U Ro ≡ << 1 Ro is the Rossby number f 0 L

Remark: not all phenomena satisfy this condition..

3 3 Geostrophic flow

Geostrophic flow relation explains why overall flow in the atmosphere is from west to east:

Geostrophic flow

The geostrophic relationship also explains why flow around Agulhas flow around low-pressure eddies is anticlockwise cells is anticlockwise (NH)

L

pressure gradient force Coriolis force

3 4 Next, the vertical momentum balance

Example: large-scale atmospheric flow Recall, the typical scales:

6 4 7 L~10 m H~10 m r0~10 m -1 -1 -4 -1 U~10 ms W=(H/L)U~0.1 ms f0~10 s 3 2 -2 -2 Δp/ρ~10 m s ρ = ρ0(z) + ρ’ [ρ’ ]/[ρ0]~10

Vertical momentum balance

Dw u2 + v 2 1 ∂p ∂Φ − − 2Ωcosφ u = − − g − t + F Dt r ρ ∂z ∂z z 2 W U p0 ΔΦ f U g t []F 0 ρ z L /U r0 0 H H magnitude 10-7 10-5 10-3 10 10 10-6 10-7 (ms-2)

Hydrostatic balance

Thus, to a good approximation, the vertical momentum equation can be written as

1 ∂ p − = g ρ∂z

which is the hydrostatic balance (pressure equals weight of fluid above it)

3 5 Hydrostatic balance

Remark: care is needed here, because

p = p0(z) + p'(x,y,z,t)

with horizontal mean p0(z) determined by horizontal mean density ρ , i.e., 0 1 ∂ p −−0 g = 0 ρ∂0 z

Now, [p0] >> deviations p’

but deviations p’ determine the horizontal pressure gradient

Hydrostatic balance

Question: is p’ also obeying the hydrostatic balance? Answer: yes, since

1 ∂p 1 ∂ − − g = − ()p0 + p' − g ρ ∂z ρ0 + ρ' ∂z 1'⎡⎤ρ∂⎡pp ' 1 ⎤∂ ' ⎡⎤ ≈−⎢⎥1' −⎢ −ρρ0 gg+ − ≈− ⎥ g+ ⎢⎥ ρρ00⎣⎦⎣ ∂zz ρ0 ⎦ ∂ ⎣⎦

buoyancy

Magnitude of the latter two terms ⎡ ⎤ ⎡ ⎤ ρ' −1 −2 1 ∂p' −1 −2 ⎢ g⎥ ~ 10 ms , ⎢ ⎥ ~ 10 ms ⎣ ρ0 ⎦ ⎣ ρ0 ∂z ⎦ are still large compared to other terms in the full vertical momentum equation

3 6 Geostrophic + hydrostatic balance

Consider geostrophic relationship, divide by ∂p/ ∂z

=> ∂ p ugy−11∂ ⎛⎞∂ z ∂∂pp==⎜⎟ ∂∂zzρ00f ρ∂00 fy⎝⎠p

In last step used that on a surface of equal pressure ⎛ ∂p⎞ ⎛ ∂p⎞ ⎛ ∂p⎞ dp= ⎜ ⎟ dx +⎜ ⎟ dy +⎜ ⎟ dz = 0 ⎝ ∂x ⎠ ⎝ ∂y⎠ ⎝ ∂z ⎠ Now use hydrostatic balance =>

1 ⎛ ∂Φ⎞ 1 ⎛ ∂Φ⎞ u = − ⎜ ⎟ v = ⎜ ⎟ g and likewise g ⎝ ⎠ f0 ⎝ ∂y ⎠ p f0 ∂x p with Φ=gz the geopotential height of a pressure surface. These relationships are often used in dynamic meteorology

Thermal wind relations

Calculate vertical shear of the geostrophic flow, using

g ⎛ ∂z ⎞ g ⎛ ∂z ⎞ u = − ⎜ ⎟ vg = ⎜ ⎟ g ⎝ ∂ ⎠ f0 ⎝ ∂y ⎠ p f0 x p For example,

∂ug ∂ug ∂p ∂ug = ≈ − ρ g ∂z ∂p ∂z 0 ∂p and ⎛ ⎛ ⎞⎞ ⎛ ⎞ ⎛ ⎞ ∂ug g ∂ ∂z g ∂ −1 1 ∂ρ = − ⎜ ⎜ ⎟⎟ = − ⎜ ⎟ ≈ − ⎜ ⎟ ∂ ∂ ∂ ∂ ρ ρ 2 ∂ p f0 ⎝ y ⎝ p⎠⎠ p f0 y ⎝ g⎠ p 0 f0 ⎝ y ⎠ p So, ⎛ ⎞ ∂ug g ∂ρ ∂vg −g ⎛ ∂ρ⎞ = ⎜ ⎟ = ⎜ ⎟ ⎝ ⎠ ∂z ρ0 f0 ⎝ ∂y ⎠ p ∂z ρ0 f 0 ∂x p

which are the thermal wind relations

3 7 Thermal wind relations

Thermal wind is difference in geostrophic winds at different vertical levels; follows from

⎛ ⎞ ∂ ⎛ ⎞ ∂ug g ∂ρ vg −g ∂ρ = = ⎜ ⎟ ⎜ ⎟ ∂ ρ ⎝ ∂ ⎠ ∂z ρ0 f0 ⎝ ∂y ⎠ p z 0 f 0 x p

Background of the name: dynamic meteorology. For air the ideal gas law yields p=ρRT,

-1 -1 so (∇ρ)p /ρ=-(∇Τ)p /Τ gas constant (287 J kg K )

If density and pressure surfaces coincide (barotropic fluid), the geostrophic flow is independent with height

Thermal wind relations

In atmosphere: due to meridional temperature gradient at midlatitudes the western wind increases with height

p=p1-Δp

ug p=p1 z y x

Sketch is for Northern Hemisphere

3 8 Finally, the continuity equation

1 Dρ + ∇ ⋅ u = 0 ρ Dt

Use ρ = ρ (z) + ρ’ and write it in earth’s coordinates => 0

1Dρ ' w∂ρ ∂∂∂ uvwtan φ 2 w +0 +++ − v +=0 ρDt ρ∂ z ∂ x ∂ y ∂ z r r

[ρ'/ρ0 ] W U U W U W

L /U H L L H r0 r0 magnitude 10-7 10-5 10-5 10-5 10-5 10-6 10-8 (s-1)

Estimates for large-scale atmospheric flow (see e.g. 6 slides back)

So, to lowest order the continuity equation for large-scale flow reads

∂u ∂v 1 ∂ + + ()ρ0w ≈ 0 ∂x ∂y ρ0 ∂z

Now, recall that the horizontal flow is to lowest order geostrophic, and obeys

∂u ∂v g + g = 0 ∂x ∂y

As vertical velocity at earth surface is zero

(to lowest order), it follows that

w ≈ 0

3 9 Final remarks

Thus, if the flow obeys the condition U Ro ≡ << 1 Ro is the Rossby number f 0 L Then, to lowest order it obeys

1 1 ∂p u ≈ ug ≡ez × ∇p − = − g ρ0 f0 ρ ∂z geostrophic balance hydrostatic balance

Powerful results, but P degenerated system (any pressure field can be chosen)

P no time evolution of variables P vertical motion is known to be essential

Final remarks

To obtain equations that govern the evolution of large-scale flow in atmosphere/ocean departures from geostrophic flow should be considered, i.e.

u ≈ ug + Ro u1 + .... u1 is ageostrophic

Dynamics of small ageostrophic corrections: evolution equation for the vorticity

∂v ∂u ω = g − g = ∇ 2ψ ~ rotation of fluid columns ∂x ∂y and ψ the stream function that was introduced earlier

Main effect of earth’s curvature: Coriolis parameter varies with latitude (see lectures of Van de Berg/Von der Heydt)

4 0 Final remarks

For motions with do not satisfy Ro <<1, but still obey L>>H (e.g., polar lows, storm surges...) =>

P main terms in horizontal momentum balance: inertia+Coriolis+pressure gradient => no degeneration (!)

P curvature of earth can be neglected: f-plane approximation

P vertical momentum balance is still hydrostatic

Books on ocean and atmospheric dynamics

Apel, J.R. 1985. Principles of ocean physics. Academic Press. Extensive discussion of basic equations of motion, incl. tidal forces. Focus on ocean.

Batchelor, G.K. 1967. An introduction to fluid dynamics. Cambridge Univ. Press. Standard work on basic fluid dynamics (inertial frame). Extensive discussion on continuum hypothesis and derivation of basic equations.

Cushman-Roisin, B. and J.M. Beckers, 2010. Introduction to geophysical fluid dynamics. Academic Press. No derivation of momentum equation in inertial frame. Thorough discussion of equation in rotating frame + geostrophic/hydrostatic balance.

Gill, A.E. 1982. Atmosphere-ocean dynamics. Academic Press. Extensive discussion of basic equations of motion. Discusses both atmosphere and ocean.

Holton, J. 2004. An introduction to dynamic meteorology (4th ed.). Academic Press. Extensive discussion of basic equations of motion. Focus on atmosphere.

Marshall, J. and R. A. Plumb, 2007. Atmosphere, Ocean and Climate Dynamics: An introductory text. Academic Press.

McWilliams, J.C. 2006. Fundamentals of geophysical fluid dynamics. Cambridge Univ. Press. Includes derivation of basic equations and discussion geostrophic balance. Level: advanced.

Pedlosky, J. 1987. Geophysical fluid dynamics. (2th ed.). Springer Verlag. Advanced level. No derivation of momentum equation in inertial frame. Thorough discussion of equation in rotating frame + geostrophic/hydrostatic balance.

Vallis, G. K., 2006. Atmospheric and Oceanic Fluid Dynamics: Fundamentals and Large-Scale Circulation. Cambridge University Press. Includes derivation of basic equations and discussion geostrophic balance. Advanced level.

4 1 Annex 1: Derivation continuity equation without introducing a coordinate system

Consider fixed volume element V, bounded by surface area A, outward pointing normal vector n

V n

u Change of mass in V=flux through surface ∂ ∫∫∫ ρ dV =− ∫∫ ρu⋅ndA = −∫∫∫ ∇⋅(ρu)dV ∂t V A V Gauss theorem Result must be valid for arbitrary V =>

∂ρ + ∇ ⋅ (ρu) = 0 ∂t

Annex 2: Frictional forces Situation in case of flow in x-direction with vertical shear:

distance (x,y,z) z to bottom y x velocity u

Fluid above induces stress τxz acting across upper surface

τxz>0 : downward transfer of positive x-momentum/area/time For a Newtonian fluid (air, water; not ice) ∂ u τ~ ρν xz ∂ z where ν (m2s-1) is kinematic viscosity coefficient

Fluid below induces stress -τxz acting across lower surface

4 2 Annex 2: Frictional forces

Question:

What are typical values of viscosity coefficients?

Answer:

This depends on the medium (air/water), but also on the scales of motion that are explicitly resolved

If smallest scales are resolved (down to ~mm) then ν represents only molecular viscosity :

(ν∼10-5 m2s-1 for air, ν∼10-6 m2s-1 for water)

Annex 2: Frictional forces

Most models: smallest scale that is resolved is >> mm => τxz also represents net transfer of momentum due to unresolved scales of motion

These contributions are very large, because in general a geophysical flow is turbulent (many small-scale eddies)

3 4 => ν related to τxz can be factor 10 -10 larger !

4 3 Annex 2: Frictional forces

So, in case of linear shear flow

distance (x,y,z) z to bottom y x velocity u

Then τ = τ : xz,top xz,bottom volume element experiences no net force

Annex 2: Frictional forces

In general: frictional force due to transfer of x-momentum in z-direction is

(x,y,z) distance z to bottom y x velocity u

* ⎡⎤ Fxz=⎣⎦ττ xz,top− xz ,bottom ΔΔxy

⎡⎤∂τ xz 11⎛⎞∂τ xz ∂τ xz =+⎢⎥ττxz Δz −⎜⎟xz − Δz ΔΔ xy= ΔΔΔxyz ⎣⎦∂∂zz22⎝⎠∂ z

1 ∂τ So, corresponding force/mass = xz ρ∂z

4 4 Annex 2: Frictional forces

Hence, frictional force in x-direction due to gradient in transfer of x-momentum in z-direction is

11∂τ xz ∂∂⎛⎞uw ∂ Fxz =+~ ⎜⎟ρνz ρν x ρ ∂zz ρ∂⎝⎠ ∂ z ∂ x

Second term added because τxz= τzx For geophysical flows viscosity coeff. νx, νz are different

Generalisation:

Define τij as transfer of i-momentum in j-direction (here i,j=x,y,z)

Annex 2: Frictional forces

Thus also frictional forces in x-direction due to 1 ∂τ - gradient transfer x-momentum in x-direction: xx ρ ∂x 1 ∂τ xy - gradient transfer x-momentum in y-direction: ρ∂y

z y x So, total frictional force/mass in x-direction

1 ⎡⎤∂τxx ∂τ xy ∂τ xz Ffric,x =⎢⎥ ++ ρ∂⎣⎦xyz ∂ ∂ Likewise, components in y- and z-direction:

1 ⎡⎤∂τyx ∂τ yy ∂τ yz 1 ⎡⎤∂τ zx ∂τ zy ∂τ zz Ffric, y =⎢⎥ ++ , Ffric,z =⎢⎥ ++ ρ∂⎣⎦xyz ∂ ∂ ρ∂⎣⎦xyz ∂ ∂

4 5 Annex 2: Frictional forces

Frictional force/mass in x-direction in tensor notation

1 F = ∇⋅τ fric ρ % where ~τ is the transpose of tensor τ

τ is the shear stress tensor, with components τij, defined with respect to a chosen coordinate system

4 6