Lecture 1.Pdf

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.

body force − − − gravitational force force surface force − − − pressure, friction

⎪⎧ true force force ⎨ (non-inertial frame of reference) apparent force ⎩⎪

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

1 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   dp  dp may be written as = ∑F and the first law becomes = 0 . The latter is just the dt dt 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

2 FUNDAMETNAL FORCES

Gravitational force

 GMm ⎛ r⎞ * F = − = mg Mm 2 ⎝⎜ ⎠⎟ r r with

G = 6.673 × 10−11 Nm2kg−2 measured in laboratory 24 M E = 5.988 × 10 kg mass of Earth a = 6.37 × 106 m 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 < 105 m .

GM For z = 0 , g* = = 9.81 ms−2 . 0 a2

3 GM GM g* For z ≠ 0 , g* = = = 0 ≈ g*(1− 2z / a) . (a + z)2 a2 (1+ z / a)2 (1+ z / a)2 0

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 = δ xδ yδz , centered at the point ( x0 , y0 ,z0 ) as illustrated in Figure 1.1 below. (The atmosphere is made of tiny cubes---Cartesian coordinates)

δV = δ xδ yδz

p(x0 , y0 ,z0 ) = p0 ∂p δ x p = p + + A 0 ∂x 2 ∂p δ x p = p − + B 0 ∂x 2

∂p δ x F = −(p + )δ yδz Ax 0 ∂x 2 ∂p δ x F = (p − )δ yδz Bx 0 ∂x 2

∂p F = F + F = − δ xδ yδz x Ax Bx ∂x ∂p F = − δ xδ yδz y ∂y ∂p Fx = − δ xδ yδz ∂z  F F F F acceleration = force / mass = = ( x , y , z ) m m m m

⎛ 1 ∂p 1 ∂p 1 ∂p⎞ 1 = − , − , − = − ∇p ⎝⎜ ρ ∂x ρ ∂y ρ ∂z ⎠⎟ ρ

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

5 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.

du Since the motion is steady ( = 0 ), for every thin horizontal layer of fluid of depth δz , dt 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)

6 δu ∂u τ zx = lim µ = µ δ z→0 δz ∂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.)

For general conditions, the stress acting across the upper boundary (a z-surface) on the fluid below can be written as

∂τ δz τ + zx zx ∂z 2 while the stress acting across the lower boundary on the fluid above is

⎡ ∂τ zx δz ⎤ − τ zx − ⎣⎢ ∂z 2 ⎦⎥

(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 δz ⎤ ⎡ ∂τ zx δz ⎤ τ zx + δ xδ y − τ zx − δ xδ y ⎣⎢ ∂z 2 ⎦⎥ ⎣⎢ ∂z 2 ⎦⎥

Dividing this expression by the mass δV = δ xδ yδz , we find the viscous force per unit mass due to vertical shear of the x-component of motion is

1 ∂τ zx 1 ∂ ⎛ ∂u ⎞ = ⎜ µ ⎟ ρ ∂z ρ ∂z ⎝ ∂z ⎠

If µ is constant, we can write

8 1 ∂τ ∂2u zx = υ ρ ∂z ∂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.

COLA IGES ⎛ ∂2u ∂2u ∂2u ⎞ F = υ + + Comment: End of lecture 1 rx ⎝⎜ ∂x2 ∂y2 ∂z2 ⎠⎟ ⎛ ∂2v ∂2v ∂2v⎞ F = υ + + ry ⎝⎜ ∂x2 ∂y2 ∂z2 ⎠⎟ ⎛ ∂2w ∂2w ∂2w⎞ F = υ + + rz ⎝⎜ ∂x2 ∂y2 ∂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

10 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:

11 dθ ω = dt

V1 = V2 = V

δV = δθ V δV δθ = V δt δt

δV dθ r acceleration = lim = V in direction towards the center of circle, i.e., − . δt→0 δt dt r

dθ Now substitute ω = and V = ωr , so dt

dV r = (ωr)(ω)(− ) = −ω 2r dt r

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.

dV r Inertial: = (ωr)(ω)(− ) = −ω 2r = f dt r r

dV Non-inertial: r = 0 = f + ω 2r dt r

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.

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

12 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

dΦ and since g is parallel to the local vertical, g = , We can integrate this equation w.r.t. dz z , assuming Φ = 0 at mean sea level, so

z Φ = gdz ∫0

13 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

14 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

15 Now consider particle in spherical coordinate system.

Meridional displacement: δ y

δ y = aδφ

R = acosφ

δ R = −asinφ δφ

Vertical displacement: δz

δ R = cosφ δz

16 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 + δ R , i.e,.

Vi Ri = Vf Rf

The velocity viewed from the inertial frame:

Initial: Vi = ΩR + u

Final: Vf = Ω(R + δ R) + (u + δu)

Solve for δu :

u δ R2 δ R δu = −2Ωδ R − δ R − Ω − δu R R R u ≈ −2Ωδ R − δ R R

Thus,

δu δ R u δ R = −2Ω − δt δt R δt

If the displacement only occurs in meridional direction:

δ R δφ δφ δ R = −asinφ ; a = v ;  = −vsinφ δt δt δt δt

⎛ du ⎞ uv then = 2Ωsinφv + tanφ ⎝⎜ dt ⎠⎟ a

If the displacement takes place only in the vertical direction:

δ R δz δz = cosφ ; = w δt δt δt

17 ⎛ du ⎞ uw then = −2Ωcosφw − ⎝⎜ dt ⎠⎟ a

δ R δφ δz For arbitrary displacement (δ y,δz ): = −asinφ + cosφ δt δt δt du uv uw = 2Ωsinφv − 2Ωcosφw + tanφ − (1.10) dt a a

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

2 2 ⎛ u ⎞ 2 2ΩuR u R Ω + R − Ω R = + ⎝⎜ R⎠⎟ R R2

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

18 Taking meridional and vertical component of R, one yields

⎛ dv⎞ u2 = −2Ωusinφ − tanφ ⎝⎜ dt ⎠⎟ a

⎛ dw⎞ u2 = 2Ωu cosφ + ⎝⎜ dt ⎠⎟ a

For synoptic scale motions, u ΩR , w u ,  

⎛ dv⎞ = −2Ωusinφ = − fu ⎝⎜ dt ⎠⎟ Co ⎛ du ⎞ ⎜ ⎟ = 2Ωvcosφ = fv ⎝ dt ⎠ Co where f ≡ 2Ωsinφ is the Coriolis parameter.

⎛ dw⎞  gravitation  change the apparent weight slightly, depending on whether ⎝⎜ dt ⎠⎟ Co the parcel is moving eastward (lighter) or westward (heavier). For large scale synoptic motions, this term is usually negligible.

We now have

⎛ du ⎞ = 2Ωvsinφ − 2Ωwcosφ ⎝⎜ dt ⎠⎟ Co ⎛ dv⎞ ⎜ ⎟ = −2Ωusinφ ⎝ dt ⎠ Co

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

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

2 at rest: FCF = Ω R

2 2 ⎛ u ⎞ 2 u at motion: F = Ω + R = Ω R + 2Ωu + CF ⎝⎜ R⎠⎟ R

The extra CF force:

u2 F = −2Ωu − yR 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

⎛ u ⎞ 2 Initial: M = Ω + R I ⎝⎜ R⎠⎟

21 ⎛ u + δu ⎞ 2 Final: M F = ⎜ Ω + ⎟ (R − δ R) ⎝ R − δ R⎠

Angular Momentum conservation:

M I = M F ⇒ ΩR2 + uR = ΩR2 − 2ΩRδ R + Ωδ R2 + (u + δu)(R − δ R) ⇒ 0 = −2ΩRδ R + Ωδ R2 − uδ R + δuR − δuδ R δ R ⇒ δu ≈ 2Ωδ R + u R

Force in u-direction:

δu ⎛ u ⎞ Fx = lim = vr ⎜ 2Ω + ⎟ , δt→0 δt ⎝ R⎠

dR where v = . r dt

Together, we have

⎛ u ⎞ F = v 2Ω + , whichis 0 if v is 0 x r ⎝⎜ R⎠⎟ r

⎛ u ⎞ F = −u 2Ω + , which is 0 if u is 0 yR ⎝⎜ R⎠⎟

22 B. On a sphere-3D:

Fig. 1C Forces due to v-component

R = acosφ Identity: vr = vsinφ − wcosφ

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

⎛ u ⎞ Fx = 2Ω + (vsinφ − wcosφ) ⎝⎜ acosφ ⎠⎟

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

⎛ u ⎞ F = F sinφ = − 2Ω + usinφ y yR ⎝⎜ acosφ ⎠⎟

⎛ u ⎞ F = −F cosφ = 2Ω + u cosφ z yR ⎝⎜ acosφ ⎠⎟

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

23    F fk V In vectorial form, Co = − × H

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:

1 dp − = g hydrostatic balance ρ dz

Multiplying by ρ and integrating with altitude:

24 ∞ p(z) = ρgdz ∫z

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

1 dp RT dΦ = −gdz = − dz = −αdp = − dp = −RTd ln p ρ dz p

Integrating gives

z2 p2 p1 dΦ = −R T dln p = R T dln p ∫z1 ∫p1 ∫p2 or

p1 Φ(z2 ) − Φ(z1 ) = R T dln p ∫p2

Define geopotential height Z = Φ / go

R p1 Thickness: ΔZT ≡ Z2 − Z1 = T dlnp ∫p go 2

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

and p1 .

(adopted from Marshall and Plumb, 2008)

Defining a layer mean temperature

−1 p1 p1 T = T dln p ⎡ dln p⎤ ∫p2 ⎣⎢∫p2 ⎦⎥

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,

26 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) − p0 ⎤ ⎡(p0 + δ p) − p0 ⎤ ⎛ δz ⎞ ⎢ ⎥ = ⎢ ⎥ ⎜ ⎟ ⎣ δ x ⎦z ⎣ δz ⎦x ⎝ δ x⎠ p

Taking limit, we get

27 ⎛ ∂p⎞ ⎛ ∂p⎞ ⎛ ∂z ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ⎝ ∂x ⎠ z ⎝ ∂z ⎠ x ⎝ ∂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,

1 ⎛ ∂p⎞ ⎛ ∂z ⎞ ⎛ ∂Φ⎞ − ⎜ ⎟ = −g⎜ ⎟ = −⎜ ⎟ ρ ⎝ ∂x ⎠ z ⎝ ∂x⎠ p ⎝ ∂x ⎠ p

Similarly,

1 ⎛ ∂p⎞ ⎛ ∂Φ⎞ − ⎜ ⎟ = −⎜ ⎟ ρ ⎝ ∂y ⎠ z ⎝ ∂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.

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

p − p p − p δz p − p C A = C B + B A δ x δz δ x δ x

∂p ∂p ∂z ∂p In the limit δ x, δz → 0 , = + ∂x s ∂z ∂x s ∂x z

∂p ∂p ∂s ∂p ∂p ∂s ∂z ∂p Now = , so = + ∂z ∂s ∂z ∂x s ∂x z ∂z ∂x s ∂s

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

Note that if s = p ,

∂p ∂p ∂p ∂z = + = 0 ∂x p ∂x z ∂z ∂x p

Using the hydrostatic assumption,

∂p ∂p ∂z ∂z ∂Φ COLA IGES = − = ρg = ρ Comment: End of lecture 2 ∂x z ∂z ∂x p ∂x p ∂x p