1

Lecture 1

In the first lecture, we will start with introducing the governing laws for the motion in the atmosphere or the ocean for that matter. The motions of the atmosphere are governed by the fundamental physical laws of conservation of mass, momentum and energy. What is fluid in a mathematician’s eyes---a continuous medium, or continuum. Air parcel or air particle is often referred to as a “point” in the atmospheric continuum. This lecture will be about the mathematical expression (in terms of partial differential equations) of the forces and laws---the building blocks of the atmospheric dynamics.

forcebody force − − − gravitational forcesurface force − − − pressure, friction

forcetrue force

apparent force

⎧⎨⎪

⎩⎪(non-inertial frame of reference)

Newton’s Laws of Motion

First law: In an inertial frame of reference, a mass remains at rest or in uniform motion

unless compelled to change its velocity by an external force.

(inertial frame of reference is coordinate fixed in space)

Second law: ma = F

∑

(vector sum of external forces = product of mass and acceleration)

Third law: force of action is equal and opposite to the force of reaction.

In a non-inertial frame of reference, i.e., not fixed in space, acceleration of a mass may be

thought of as the result of the sum of the true (inertial) forces plus the sum of the

“apparent” forces. Forces acting to accelerate a mass may be (i) body forces whose

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magnitude is proportional to the mass (such as gravitational force) that may be thought of

as acting on a single point, namely the center of mass, or (ii) surface forces that act on the

surface of a mass (or across the boundary that separates of mass of fluid from its

environment) whose magnitudes are independent of the mass.

Note: If we define momentum as p = mv and acceleration is

a ≡ dv / dt , the second law

may be written as

dpdt

=F∑ and the first law becomes

dpdt

= 0 . The latter is just the

statement of conservation of momentum when external forces are zero.

Fundamental forces

1) Gravitation

2) Pressure gradient

3) Friction

Apparent forces

1) Centrifugal

2) Coriolis

3) Gravity

3

FUNDAMETNAL FORCES

Gravitational force

FMm = −

GMmr2

rr

⎛⎝⎜

⎞⎠⎟= mg*

with

G = 6.673×10−11Nm2kg−2 measured in laboratory

ME = 5.988 ×1024 kg mass of Earth

a = 6.37 ×106m mean Earth radius

For typical mass of air on Earth, distance from center of Earth is a + z ( z = altitude

above surface) where 0 ≤ z < 105m .

For z = 0 , g0* =

GMa2

= 9.81ms−2 .

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For z ≠ 0 , g* = GM(a + z)2

=GM

a2 (1+ z / a)2=

g0*

(1+ z / a)2≈ g0

*(1− 2z / a) .

So g* is reduced by 1% for each 32km above the surface and we may treat it as a

constant for the purposes of this course.

Exercise: Can we figure out the mass of the Earth, assuming the universal constant G is

known?

Pressure gradient force

Consider an infinitesimal volume of air, δV = δxδyδz , centered at the point ( x0 , y0 , z0 ) as

illustrated in Figure 1.1 below. (The atmosphere is made of tiny cubes---Cartesian

coordinates)

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δV = δxδyδzp(x0 , y0 , z0 ) = p0

pA = p0 +∂p∂x

δx2

+

pB = p0 −∂p∂x

δx2

+

FAx = −(p0 +∂p∂x

δx2)δyδz

FBx = (p0 −∂p∂x

δx2)δyδz

Fx = FAx + FBx = −∂p∂x

δxδyδz

Fy = −∂p∂y

δxδyδz

Fx = −∂p∂z

δxδyδz

acceleration = force / mass =Fm

= (Fxm,Fym,Fzm)

= −1ρ∂p∂x, − 1

ρ∂p∂y, − 1

ρ∂p∂z

⎛⎝⎜

⎞⎠⎟= −

1ρ∇p

Note that this force is proportional to the gradient of the pressure field, not the pressure

itself.

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Viscous force

Empirical fact: the force tangential to the upper plate required to keep it in uniform

(steady) motion is proportional to the area A, the velocity u0 , and the inverse of the

distance separating the plates l.

F = µAu0 / l

where µ is a constant called dynamic viscosity.

Since the motion is steady ( dudt

= 0 ), for every thin horizontal layer of fluid of depth δz ,

the force imposed on it from above must be exactly the same as the force from below,

and which can be all the way traced back to F = µAu0 / l . This may be expressed in the

form F = µAδu /δz where δu = u0δz / l is the velocity shear across the layer δz .

The shear stress is defined as viscous (surface) force per unit area (unit=N/m2)

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τ zx = limδ z→0µ δuδz

= µ ∂u∂z

where subscripts indicate the component of the shearing stress in the x direction due to

the vertical shear (in z direction) of the x velocity component.

From the molecular point of view, the shearing stress results from a net downward

transport of momentum by the random motion of the molecules. And the shearing stress

is equivalent to the net downward momentum flux per unit time per unit area.

For the simple 2-dimensional steady-state motion illustrated in Fig. 1.3, the net viscous

(body) force (in units of N per unit mass) acting on the elements of fluid is zero, because

the shearing stress acting across the top boundary of each fluid element is just equal and

opposite to that acting across the lower boundary.

(Caution: distinguish between force and stress; between viscous surface force and viscous

body force.)

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For general conditions, the stress acting across the upper boundary (a z-surface) on the

fluid below can be written as

τ zx +∂τ zx∂z

δz2

while the stress acting across the lower boundary on the fluid above is

− τ zx −∂τ zx∂z

δz2

⎡⎣⎢

⎤⎦⎥

(This is just the opposite of the stress acting across the lower boundary on the fluid

below.) The net viscous force due to shear stress on the fluid element:

τ zx +∂τ zx∂z

δz2

⎡⎣⎢

⎤⎦⎥δxδy − τ zx −

∂τ zx∂z

δz2

⎡⎣⎢

⎤⎦⎥δxδy

Dividing this expression by the mass δV = δxδyδz , we find the viscous force per unit

mass due to vertical shear of the x-component of motion is

1ρ∂τ zx∂z

=1ρ

∂∂z

µ ∂u∂z

⎛⎝⎜

⎞⎠⎟

If µ is constant, we can write

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1ρ∂τ zx∂z

= υ ∂2u∂z2

υ = µ / ρ is the kinematic viscosity = 1.46 ×10−5 m2s−1 for standard atmosphere at sea

level.

The same derivation applies in y- and z-directions and x- y-surfaces.

Frx = υ ∂2u∂x2

+∂2u∂y2

+∂2u∂z2

⎛⎝⎜

⎞⎠⎟

Fry = υ ∂2v∂x2

+∂2v∂y2

+∂2v∂z2

⎛⎝⎜

⎞⎠⎟

Frz = υ ∂2w∂x2

+∂2w∂y2

+∂2w∂z2

⎛⎝⎜

⎞⎠⎟

For the atmosphere below 100km, υ is so small that molecular viscosity is negligible

except in a thin layer within a few centimeters of the earth’s surface where the vertical

shear is very large. Away from this surface molecular boundary layer, momentum is

transferred primarily by turbulence (as discussed in Chapter 5 of Holton).

Exercise: Derive the viscous forcing in the x, y, z directions using the illustration below

as a guide

COLA IGESComment: End of lecture 1

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11

APPARENT FORCES

1. Centripetal acceleration and centrifugal force

Apparent forces occur to the equation of motion, due to non-inertial frame of reference.

The coordinate system of choice for meteorology is one that is fixed w.r.t. Earth The

coordinate system rotates itself and a parcel is at rest w.r.t this coordinate is under

acceleration. Therefore, the geocentric reference frame is non-inertial.

If we fix the coordinate system on the Earth and assume the rotation rate is uniform, two

apparent forces (centrifugal force and Coriolis force) enter into the inertial picture we

have already considered. To understand this system, we consider a simple system, that is,

a ball of mass m is attached to a string and whirled through a circle of radius r r at a

constant angular velocity ω .

Viewed from a fixed/inertial coordinate:

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ω =dθdt

V1 = V2 = VδV = δθ VδVδt

= V δθδt

acceleration = limδ t→0

δVδt

= V dθdt

in direction towards the center of circle, i.e., − rr

.

Now substitute ω =dθdt

and V =ωr , so

dVdt

= (ωr)(ω )(− rr) = −ω 2r

This well-known acceleration is called centripetal acceleration as perceived from fixed

inertial coordinate. Observed from a system rotating along with it (as you are riding a

merry-go-round), the ball is stationary and the force exerted by the string ( fr ) is balanced

by a centrifugal force (the force you feel on a rotating Mary-Go-Round), which is just

equal and opposite to the centripetal acceleration.

Inertial: dVdt

= (ωr)(ω )(− rr) = −ω 2r = fr

Non-inertial: dVr

dt= 0 = fr +ω

2r

2. Gravity force

To simplify what comes next, we group together the fundamental gravitational force and

the centrifugal force and refer to the resultant as the gravity force.

Fundamental gravitational: g* = GM(a + z)2

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Centrifugal force: Ω2R

Gravity force: g ≡ −gk ≡ g* +Ω2R

Note that now g is no longer pointing toward the center of the Earth (but perpendicular

to the local vertical). If the material that forms the earth has no resistance to flow, the

earth should adjust its surface till it ⊥ g . In particular, mass will deform to push the

surface away from the center near the equator and towards the center near the poles

oblate spheroid. The oblateness is about 21 km/6370 km, or ~1/300.

One can define a potential function Φ in such a way that

g = −∇Φ

and since g is parallel to the local vertical, g = dΦdz

, We can integrate this equation w.r.t.

z , assuming Φ = 0 at mean sea level, so

Φ = gdz0

z

∫

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equals work required to lift a unit mass to height z from the mean sea level. Φ is called

geopotential and is a function of z only.

3. Coriolis force and the curvature effect

Fig. Merry-go-round

Movie – Merry-go-round

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Returning to merry-go-round example—disk rotating at uniform rate ω with Alice sitting

at point A and Bob at point B. In the rotating frame of reference, Alice and Bob perceive

each other to be at rest. Suppose Alice throws a ball at Bob. She will throw the ball

straightly towards Bob, but while the ball is in the air, the merry-go-round turns

underneath it. In an inertial frame of reference, the velocity vector of the ball is resultant

of the velocity of the point A at the moment of the throw (Vr ) plus the velocity of the

throw itself, Vt . Thus the ball’s velocity in the inertial frame is to the right of the line A-

B. In addition, during the flight of the ball, the merry-go-round turns so that B is no

longer in the same direction w.r.t. an inertial frame. The ball in fact follows a curved

path---turning to the right in the rotating frame. In the rotating frame, we must posit a

force that apparently deflects the motion by acting at a right angle to the velocity vector.

Since the force is ⊥ to the motion, FCOR ⋅V = 0 so Coriolis force can only change the

direction of motion, not its speed (so it does not do any work).

Mathematical form of Coriolis force

Zonal component of Coriolis force---Conservation of angular momentum

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Now consider particle in spherical coordinate system.

Meridional displacement: δy

δy = aδφ

R = acosφ

δR = −asinφ δφ

Vertical displacement: δz

δR = cosφ δz

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Angular momentum conservation states that: Initial zonal velocity Vi times the initial

distance from axis of rotation R equals final zonal velocity Vf times final distance from

axis R + δR , i.e,.

ViRi = Vf Rf

The velocity viewed from the inertial frame:

Initial: Vi = ΩR + u

Final: Vf = Ω(R + δR) + (u + δu)

Solve for δu :

δu = −2ΩδR −uRδR − Ω

δR2

R− δu δR

R

≈ −2ΩδR −uRδR

Thus,

δuδt

= −2ΩδRδt

−uRδRδt

If the displacement only occurs in meridional direction:

δRδt

= −asinφ δφδt

; aδφδt

= v ; δRδt

= −vsinφ

then dudt

⎛⎝⎜

⎞⎠⎟= 2Ωsinφv + uv

atanφ

If the displacement takes place only in the vertical direction:

δRδt

= cosφ δzδt

; δzδt

= w

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then dudt

⎛⎝⎜

⎞⎠⎟= −2Ωcosφw −

uwa

For arbitrary displacement (δy,δz ): δRδt

= −asinφ δφδt

+ cosφ δzδt

dudt

= 2Ωsinφv − 2Ωcosφw +uvatanφ −

uwa

(1.10)

Meridional and vertical component of Coriolis force

Now consider an object that is set in motion in the eastward direction. Because the object

is rotating faster than the earth, the centrifugal force on the object will be increased. The

excess of the centrifugal force over that for an object at rest is

Ω +uR

⎛⎝⎜

⎞⎠⎟2

R − Ω2R =2ΩuRR

+u2RR2

The terms on the r.h.s. represent deflecting forces, which act outward along the vector R.

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Taking meridional and vertical component of R, one yields

dvdt

⎛⎝⎜

⎞⎠⎟= −2Ωu sinφ −

u2

atanφ

dwdt

⎛⎝⎜

⎞⎠⎟= 2Ωu cosφ +

u2

a

For synoptic scale motions, u ΩR , w u ,

dvdt

⎛⎝⎜

⎞⎠⎟ Co

= −2Ωu sinφ = − fu

dudt

⎛⎝⎜

⎞⎠⎟ Co

= 2Ωvcosφ = fv

where f ≡ 2Ωsinφ is the Coriolis parameter.

dwdt

⎛⎝⎜

⎞⎠⎟ Co gravitation change the apparent weight slightly, depending on whether

the parcel is moving eastward (lighter) or westward (heavier). For large scale synoptic

motions, this term is usually negligible.

We now have

dudt

⎛⎝⎜

⎞⎠⎟ Co

= 2Ωvsinφ − 2Ωwcosφ

dvdt

⎛⎝⎜

⎞⎠⎟ Co

= −2Ωu sinφ

In the Northern Hemisphere, deflection is to the right of the motion:

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In the southern hemisphere, deflection is to the left of the motion.

Coriolis force and Curvature effect---A 2-dimensional perspective:

a. Movement in EW-direction u induced centrifugal force: FCF

Fig. 1A Forces due to u-component

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at rest: FCF = Ω2R

at motion: FCF = Ω +uR

⎛⎝⎜

⎞⎠⎟2

R = Ω2R + 2Ωu + u2

R

The extra CF force:

FyR = −2Ωu − u2

R

where the first term is the Coriolis force acting in y direction; the second term is part of

the curvature effect on the forces in y-direction.

b. Movement in radial direction conservation of angular momentum

Fig. 1B Forces due to v-component

Initial: MI = Ω +uR

⎛⎝⎜

⎞⎠⎟R2

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Final: MF = Ω +u + δuR − δR

⎛⎝⎜

⎞⎠⎟(R − δR)2

Angular Momentum conservation:

MI = MF

⇒ΩR2 + uR = ΩR2 − 2ΩRδR +ΩδR2 + (u + δu)(R − δR)⇒ 0 = −2ΩRδR +ΩδR2 − uδR + δuR − δuδR

⇒δu ≈ 2ΩδR + u δRR

Force in u-direction:

Fx = limδ t→0

δuδt

= vr 2Ω +uR

⎛⎝⎜

⎞⎠⎟ ,

where vr =dRdt

.

Together, we have

Fx = vr 2Ω +

uR

⎛⎝⎜

⎞⎠⎟, whichis 0 if vr is 0

FyR = −u 2Ω +uR

⎛⎝⎜

⎞⎠⎟, which is 0 if u is 0

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B. On a sphere-3D:

Fig. 1C Forces due to v-component

Identity: R = acosφvr = vsinφ − wcosφ

For the x-component: simply apply the identity above to Fx

Fx = 2Ω +u

acosφ⎛⎝⎜

⎞⎠⎟vsinφ − wcosφ( )

For the y- and z-components, project Fx onto y- and z-direction.

Fy = FyR sinφ = − 2Ω +u

acosφ⎛⎝⎜

⎞⎠⎟u sinφ

Fz = −FyR cosφ = 2Ω +u

acosφ⎛⎝⎜

⎞⎠⎟u cosφ

This is the same as eqn (1.11) in Holton.

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In vectorial form, FCo = − f

k ×VH

Structure of the Static Atmosphere

Ideal gas law: Thermodynamic variable—pressure, density, temperature—are related to

each other by equation of state:

p = ρRT

or pα = RT (α ≡ ρ−1)

R is gas constant, = 287 J kg-1 K-1 for dry air

Hydrostatic ‘law’: For atmosphere at rest, motion=0 and acceleration=0, so all external

forces must be in balance. Thus, vertical component of pressure gradient

force must be equal and opposite to gravity force:

−1ρdpdz

= g hydrostatic balance

Multiplying by ρ and integrating with altitude:

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p(z) = ρgdzz

∞

∫

Using the definition of geopotential (∇Φ = −gk ), dΦ = −gdz , and noting that pα = RT

dΦ = −gdz = −1ρdpdzdz = −αdp = −

RTpdp = −RTd ln p

Integrating gives

dΦz1

z2∫ = −R T dln pp1

p2∫ = R T dln pp2

p1∫

or

Φ(z2 ) − Φ(z1) = R T dln pp2

p1∫

Define geopotential height Z = Φ / go

Thickness: ΔZT ≡ Z2 − Z1 =Rgo

T dlnpp2

p1∫

Where ΔZT is the thickness of the atmosphere layer between the pressure surfaces p2

and p1 .

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(adopted from Marshall and Plumb, 2008)

Defining a layer mean temperature

T = T dln pp2

p1∫ dln pp2

p1∫⎡⎣⎢⎤⎦⎥−1

and a layer mean scale height H ≡ R T / g0 , we have

ΔZT = H ln(p1 / p2 )

Thus the thickness of a layer bounded by isobaric surfaces is proportional to the mean

temperature of the layer. Pressure decreases more rapidly with height in a cold layer than

in a warm layer. It also follows immediately that in an isothermal atmosphere, the

geopotential height is proportional to the natural logarithm of pressure normalized by the

surface pressure,

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Z = −H ln(p / p0 )

where p0 is the pressure at Z = 0 . Thus in an isothermal atmosphere the pressure

decreases exponentially with geopotential height by a factor of e−1 per scale height.

Q: what is the structure of density in an isothermal atmosphere?

Pressure as a Vertical Coordinate

Since monotonic relationship exists between pressure and height, one could use pressure

as the independent vertical coordinate and height as a dependent variable. With the aid of

Fig. 1.11, we evaluate partial differentiation holding pressure constant.

Considering only the x, z plane, we see from Fig.1.11 that

(p0 + δ p) − p0δx

⎡⎣⎢

⎤⎦⎥z

=(p0 + δ p) − p0

δz⎡⎣⎢

⎤⎦⎥x

δzδx

⎛⎝⎜

⎞⎠⎟ p

Taking limit, we get

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∂p∂x

⎛⎝⎜

⎞⎠⎟ z

= −∂p∂z

⎛⎝⎜

⎞⎠⎟ x

∂z∂x

⎛⎝⎜

⎞⎠⎟ p

Note the minus sign is included because δz < 0 for δ p > 0 .

The pressure gradient term in z -coordinate can be expressed as, after substitution from

the hydrostatic equation,

−1ρ

∂p∂x

⎛⎝⎜

⎞⎠⎟ z

= −g ∂z∂x

⎛⎝⎜

⎞⎠⎟ p

= −∂Φ∂x

⎛⎝⎜

⎞⎠⎟ p

Similarly,

−1ρ

∂p∂y

⎛⎝⎜

⎞⎠⎟ z

= −∂Φ∂y

⎛⎝⎜

⎞⎠⎟ p

Density no longer appears explicitly in the pressure gradient force.

Generalized Vertical Coordinate

Any function of height s(z) may be used to mark off the vertical dimension if and only if

it is single-valued ∀z, s(z1) = s(z2 )⇒ z1 = z2{ } and

monotonic EITHER z1 > z2 ⇒ s(z1) > s(z2 ) OR z1 > z2 ⇒ s(z1) < s(z2 ){ } . This is also

true for any such function of pressure due to the fact that the hydrostatic law ensures that

p is single-valued, monotonic w.r.t. z .

Consider a candidate vertical coordinate s(x, y, z,t) with the above properties.

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Suppose pressure at points A , B and C is pA , pB and pC . Making use of the trick

pC − pA = pC − pB + pB − pA , then

pC − pAδx

=pC − pBδz

δzδx

+pB − pAδx

In the limit δx, δz→ 0 , ∂p∂x s

=∂p∂z

∂z∂x s

+∂p∂x z

Now ∂p∂z

=∂p∂s

∂s∂z

, so ∂p∂x s

=∂p∂x z

+∂s∂z

∂z∂x s

∂p∂s

These identities will be used to derive the equation of motion in sigma vertical

coordinate.

Note that if s = p ,

∂p∂x p

=∂p∂x z

+∂p∂z

∂z∂x p

= 0

Using the hydrostatic assumption,

∂p∂x z

= −∂p∂z

∂z∂x p

= ρg ∂z∂x p

= ρ ∂Φ∂x p

COLA IGESComment: End of lecture 2