Introduction to Atmospheric Dynamics Chapter 1

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Defnition: A defnition is an explanation or outline for relevant jargon or terms.

Concept: An idea that draws a connection between subjects or provides an answer for a question.

Question: What is something that motivates delving into this topic?

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Part 1: Forces in the Atmosphere Radius of the Earth 6371.22 km

Atmosphere Depth 100 km

Troposphere Depth 10 km

Mountain Height 8.8 km

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Question: How do we understand the dynamics of the atmosphere?

Answer: The principles of atmospheric dynamics are drawn from basic physical principles.

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Question: What are the basic physical principles that govern the atmosphere?

Newton’s Second Law: The change in momentum of an object is equal to the sum of forces acting on that object. d(mv) = F dt i allXi Conservation of Momentum: With no external forces momentum must be conserved.

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Basic Principles of Physics dx Defnition: Velocity is the change of u = position with respect to time dt

du Defnition: Acceleration is the change a = of velocity with respect to time dt

Hence, for an object of constant mass: d(mu) du 1 = m = ma Newton’s a = F dt dt Second Law m i allXi

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Basic Principles of Physics

l How do these forces induce acceleration?

l We assume the existence of an idealized “parcel” of fuid.

l Forces are calculated on the idealized parcel.

l Then take the limit of the parcel being infnitely small.

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Parcel Properties

⇢ Density (kg/m3)

z V = xyz Volume (m3) Δz Mass (kg) m = ⇢xyz y Δx Δy Temperature (K) T x

p Pressure (Pa)

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Spherical Coordinates

j Spherical coordinates:

Longitude k Latitude j r Radius

i Basis vectors:

i Eastward basis vector j Northward basis vector k Vertical basis vector

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Defnition: The Material Derivative (expressed with a capital D) denotes the change in a quantity following a fuid parcel.

1 Du 1 a = F = F m i Dt m i allXi allXi u =(u, v, w) 3D velocity vector u Eastward velocity (zonal velocity)

v Northward velocity (meridional velocity)

w Upward velocity (vertical velocity)

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Question: What forces are important for understanding atmospheric dynamics?

1 Du 1 a = F = F m i Dt m i allXi allXi l Pressure gradient force l Gravitational force l Viscous force l Coriolis and centrifugal force

l Total force is the sum of all these forces

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Defnition: A Surface Force acts on the surface of a parcel of fuid, typically due to interactions with neighboring parcels. The magnitude of a surface force is typically proportional to the surface area of the parcel. Examples: Pressure Force, Viscous Force.

Defnition: A Body Force acts on the center of mass of a parcel of fuid. The magnitude of the body force is typically proportional to the mass of the parcel. Example: Gravity.

Defnition: When a coordinate system (for instance coordinates on the sphere) varies with respect to time and/or space, there is an Apparent Force due to the fact that coordinate vectors are changing following the fuid parcel. Examples: Coriolis, Centrifugal Force.

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Pressure Gradient Force

(x0,y0,z0) Pressure at parcel center:

p0 = p(x0,y0,z0) Δz z Approximate pressure here via Taylor expansion Δx Δy y @p x p = p + + (x2) 0 @x 2 O x ✓ ◆

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Pressure Gradient Force

Δz

@p x Δy @p x pL p0 Δx p p + ⇡ @x 2 R ⇡ 0 @x 2 ✓ ◆ ✓ ◆ F = p A = p yz F = p A = p yz L L L L R R R R

Ftot = FL + FR @p x @p x = p yz p + yz 0 @x 2 0 @x 2  ✓ ◆  ✓ ◆ @p = xyz Total force acting on fuid parcel @x ✓ ◆ in the x direction

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Pressure Gradient Force

Repeat in all coordinate directions:

@p @p @p F = i + j + k xyz = ( p)xyz tot @x @y @z r ✓ ◆

Then computing the force per unit mass (recall this determines the acceleration):

F 1 tot = p m ⇢r

Total force acting on fuid parcel in all directions

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Viscous Force

The viscosity of air is responsible for resisting motion of the fuid. It is a dissipative force, which results slowing a fuid which is not otherwise forced.

A For example, the blue fuid ut parcel experiences a shear stress in the x direction due to motion of the red fuid Ftop~i y u parcel and green fuid parcel. Fbot~i F (u u)A ~j ub top ⇠ t F (u u)A bot ⇠ b ~i Observe that if all fuid parcels are traveling at the same velocity, no force will be conferred.

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Viscous Force F (u u)A top ⇠ t F (u u)A bot ⇠ b A The resistance of the fuid parcel to shearing is determined by the u t proportionality coeﬃcient

Ftop~i µ/y y u Where μ is the dynamic F ~i bot viscosity. Observe that this ~j ub quantity is inversely proportional to the thickness ~i of the fuid parcel δy. Why might this be the case? µA Hence: F + F = (u 2u + u ) top bot y t b

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Viscous Force

The force per unit mass gives the acceleration: F F µ (u 2u + u ) tot = tot = t b m ⇢yA ⇢ y2

And so in the limit y 0 A ! 2 u Ftot µ @ u t = m ⇢ @y2 ✓ ◆ Ftop~i y u Fbot~i ~j ub

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Viscous Force

Extending this derivation to each fow direction then yields the total acceleration due to viscosity.

µ Kinematic ⌫ = Viscosity ⇢

F visc = ⌫ 2u m r

Note Laplacian (second derivative). What consequences does this have for linearly sheared fow?

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Gravitational Force Mm r Recall Newton’s law of gravity: F = G g r 2 r | | | | Gravitational constant: 11 3 1 2 G =6.67384 10 m kg s ⇥ Mass of the Earth: r M =5.972 1024 kg ⇥ But since the atmosphere is essentially a thin shell, we can make the approximation

6 r a a =6.37122 10 m F | | ⇡ ⇥ g = gk GM m Defne gravity at surface: g = a2 Total gravitational force Question: Can you calculate g from the above information? acting on fuid parcel

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Dynamical Equations of Motion (Cartesian, non-rotang ﬂuid)

Du 1 @p = + ⌫ 2u Dt ⇢ @x r Dv 1 @p = + ⌫ 2v Dt ⇢ @y r Dw 1 @p = –g + ⌫ 2w Dt ⇢ @z r

Pressure Gradient Gravity Viscosity

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Coriolis / Centrifugal Force

The Earth revolves around its axis at ⌦ a certain rate ⌦ .

j Coriolis and Centrifugal forces are k known as apparent forces, because they only exist because the reference frame is in motion.

The Coriolis force defects fuid parcels as a consequence of the Earth’s rotation.

The Centrifugal force attempts to push fuid parcels away from the axis of rotation.

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum

Angular velocity is defned as the angular frequency (in units of radians ω per second) multiplied by the radius: v = !r | | r (radius) v Δv

Δθ Velocity vector

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum Ω

On the surface of the sphere the radius of rotation is equal to the perpendicular distance from the axis R of rotation: a R = a cos | |

( = latitude) Earth

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum Ω k j Defne a local coordinate system:

i = Points towards the East (Zonal) R Longitudinal direction a (at red circle this is directed into the page)

j = Points towards the North (Meridional) Latitudinal direction (at red circle directed to top-left)

k = Points in the vertical Earth Local vertical coordinate (at red circle directed to top-right)

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum Ω k j Given a vector B in the same direction as R, whare the R B components of this vector in the a local basis? B = Byj + Bzk Verify: B = B sin y | | B = B cos Earth z | | Observe: B2 + B2 = B 2 y z | |

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum Ω k j R B a What is theVerify: velocity of this point only due to the rotation of the Earth?

Earth Answer: Velocity is purely zonal u = ⌦ R i | | u = uii

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum

Velocity due to rotation of the Ω Earth: k u = ⌦ R j i | | Angular momentum vector L is equal to the cross product of the R radial vector and the linear a momentum L = R mu ⇥ But R and u are perpendicular to one another, so

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum

Angular momentum due to rotation Ω of the Earth: k j L = m⌦ R 2 Rotation only | | | | R Write the radial vector in terms of a: a R = a cos | | So for a fuid parcel at rest relative to the surface: L = m⌦a2 cos2 | | Earth Angular momentum is relevant since it is a conserved quantity. That is, D L | | =0 Dt

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum

Assume the fuid parcel now Ω possesses some zonal velocity u. k j The angular momentum due to its zonal velocity is L = mu R Zonal velocity R | | | | only a Angular momentum including both rotation of the Earth and relative velocity u: L | | = ⌦ R 2 + u R Earth m | | | | u = R 2 ⌦ + | | R ✓ | |◆

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Meridional Displacement Ω k j R

Imagine the air parcel moves R south (or north). What happens? We get some change ΔR in the magnitude of R.

Earth

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Meridional Displacement

Ω Imagine the air parcel moves k south (or north). What happens? j R But if angular momentum is conserved, then u must change: R L u | | = R 2 ⌦ + m | | R ✓ | |◆ u + u =(R + R)2 ⌦ + | | R + R ✓ | | ◆ Earth Solve (assuming small increments): u u 2⌦R R ⇡ R | |

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Meridional Displacement

Ω Imagine the air parcel moves k south (or north). What happens? j R u u 2⌦R R ⇡ R | | R For southward displacement R = sin y (y = a) (y<0) Total change in velocity: u u 2⌦ sin y + sin y Earth ⇡ a cos

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Meridional Displacement

Ω Imagine the air parcel moves k south (or north). What happens? j R u u 2⌦ sin y + sin y ⇡ a cos R Divide through by Δt, take limit: Du Dy u Dy = 2⌦ sin + tan Dt Dt a Dt

Du uv =2⌦v sin + tan Earth Dt a

Coriolis Term Geometric Term

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Angular Momentum

Displacements of a fuid parcel in the meridional direction lead to the apparent forces:

Du uv =2⌦v sin + tan Dt a

What about if the fuid parcel is displaced zonally? i.e. if it experiences an acceleration

What about if the fuid parcel is displaced vertically?

Left as an exercise for the reader… Essentially requires repeating a similar analysis.

Then: Combine all three possibilities to obtain all Coriolis and geometric terms

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Curvature / Coriolis in 3D

Du uv uw =2⌦v sin + tan 2⌦w cos Dt a a Dv u2 vw = 2⌦u sin tan Dt a a Dw u2 + v2 =2⌦u cos + Dt a

Curvature Coriolis

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Coriolis Parameter Du uv uw =2⌦v sin + tan 2⌦w cos Dt a a Dv u2 vw = 2⌦u sin tan Dt a a Dw u2 + v2 =2⌦u cos + Dt a

Defnition: The Coriolis Parameter f is defned as f =2⌦ sin

For w small, the Coriolis force is then given by: Du Dv = fv = fu Dt Dt

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Coriolis Force

Defection is to the right in the Northern hemisphere and to the left in the Southern hemisphere.

Du Dv = fv = fu Dt Dt

Tendency is for fuid parcels to move in circles.

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Centrifugal Force ⌦

j k

Fcent

Centrifugal force always works perpendicular to the axis of rotation: mv2 F = R cent r Typically small, so vertical component absorbed into the gravitational term and horizontal component absorbed into curvature.

Paul Ullrich The Equations of Atmospheric Dynamics March 2014 Dynamic Equations of Motion (Spherical geometry, rotang ﬂuid)

Du uv tan uw 1 @p + = +2⌦v sin 2⌦w cos + ⌫ 2u Dt r r ⇢r cos @ r Dv u2 tan vw 1 @p + + = 2⌦u sin + ⌫ 2v Dt r r ⇢r @ r Dw u2 + v2 1 @p = g +2⌦u cos + ⌫ 2w Dt r ⇢ @r r

Curvature Pressure Gradient Gravity

Coriolis Viscosity

Paul Ullrich The Equations of Atmospheric Dynamics March 2014