Chapter 4: Fundamental Forces Newton’s Second Law: F=ma In atmospheric science it is typical to consider the force per unit mass acting on the atmosphere:

€

Forcemass

= a

In order to understand atmospheric motion (accelerations) we need to know what forces act on the atmosphere. Body and Surface Forces Body (or volume) force – a force that acts on a volume of the atmosphere Surface force – a force that acts locally upon a part of a fluid Forces in an Inertial Reference Frame Inertial reference frame – a non-accelerating frame of reference Gravity – a force that arises from the mutual attraction between two objects

€

g * = −GMr 2

#

$ %

&

' ( r r#

$ % &

' (

G – gravitational constant (= 6.673x10-11 N m2 kg-2) M – mass of the Earth (kg) r – distance between objects For a unit mass of air at the surface of the Earth: M = 5.9742x1024 kg (mass of Earth) r = 6.37x106 m (radius of Earth)

€

g0* = 9.83 m s-2

Geopotential (Φ): work required to raise a unit mass to height z

€

Φ(z) = g* z( )dz0

z

∫

Note that

€

g* varies with z since it depends on the distance from the center of the Earth. Geopotential height (Z):

€

Z =Φ

g0* =

1g0* g*dz0

z

∫

Pressure Gradient Force Example: Real-time weather map

The force exerted on the left face of this air parcel due to pressure is:

€

pA = pδyδz The force exerted on the right face of this air parcel due to pressure is:

€

− p + δp( )δyδz = − p +∂p∂xδx

%

& '

(

) * δyδz

The net force exerted by pressure on this air parcel is:

€

−∂p∂xδxδyδz

By dividing by the mass of the air parcel (

€

ρδxδyδz) we get the force per unit mass due to changes in pressure (i.e. the pressure gradient force):

€

P g • i = −

1ρ∂p∂x

We can write all three components of the pressure gradient force as:

€

P g = −

1ρ∂p∂x i + ∂p

∂y j + ∂p

∂z k

%

& '

(

) *

In what direction does this force act relative to locations with high and low pressure? Example: Calculation of the pressure gradient force from a weather map Viscous Force

Each layer in this fluid is experiencing a shear stress due to differences in the speed of motion of adjacent layers of fluid.

€

τ • i = µ

∂u∂z

µ - coefficient of shear viscosity (depends on fluid being considered) How does the shear stress change between adjacent layers of the fluid? What does this imply about the net force acting on a layer of this fluid? For a net force to arise there must be a gradient in the shear stress. The force that arises due to a gradient in the shear stress is called a viscous force and can be represented as:

€

F r • i =

µρ∂ 2u∂x2

+∂ 2u∂y2

+∂ 2u∂z 2

$

% &

'

( )

Kinematic viscosity coefficient:

€

ν =µρ

(~ 1.5x10-5 m2 s-1 in the atmosphere)

In three dimensions this force is written as:

€

F r = ν

∂ 2u∂x2

+∂ 2u∂y2

+∂ 2u∂z 2

$

% &

'

( ) i +

∂ 2v∂x2

+∂ 2v∂y2

+∂ 2v∂z 2

$

% &

'

( ) j +

∂ 2w∂x2

+∂ 2w∂y2

+∂ 2w∂z 2

$

% &

'

( ) k

*

+

, , , , , , , ,

-

.

/ / / / / / / /

Hydrostatic Balance For an atmosphere that is at rest, what forces act in the vertical direction?

€

−1ρdpdz

= g*

€

⇒dpdz

= −ρg*

This equation is known as the hydrostatic equation. Hydrostatic balance – a balance between the vertical pressure gradient force and the gravitational force Combining the hydrostatic equation with the ideal gas law gives:

€

ln pp0

"

# $

%

& ' = −

g*

RdTdz

0

z

∫

Using a layer average temperature,

€

T , we can write:

€

p ≈ p0e− z H

where

€

H =Rd Tg*

This equation is known as the hypsometric equation and relates pressure and height.

Using the definition of geopotential we can also write:

€

dpp

= −g*

RdTdz

dΦ = −RdTdpp

ΔΦ = −Rd Tp0

p

∫ d ln p

€

⇒ΔZ = −Rdg0* T

p0

p

∫ d ln p

Thickness (ΔZ) – the difference in geopotential height (Z) between two pressure levels What causes the thickness to increase (or decrease)? Example: Thickness and atmospheric soundings Forces in a Rotating Reference Frame Non-inertial frame of reference – a frame of reference which is undergoing an acceleration Why is a frame of reference fixed to the Earth a non-inertial frame of reference? Angular velocity of rotation of the Earth:

€

Ω =2π

1 sidereal day=

2π86164 sec

= 7.292 ×10−5 s−1

Sidereal day – the time required for the earth to make one complete revolution in an absolute (fixed) coordinate system, that is, with respect to the stars.

Centrifugal Force Rotating table experiment:

What happens to the object when it is placed on the rotating table? What force is required to keep the object stationary on the rotating table?

€

mrω 2 Centrifugal force – outward radial force Centripetal force – inward radial force Is this object experiencing an acceleration when viewed in a frame of reference rotating with the table? Is this object experiencing an acceleration when viewed from a fixed frame of reference? Which forces are acting in each frame of reference? The centrifugal force arises only in observations taken in a rotating frame of reference and is due to the acceleration of the frame of reference.

The centrifugal force and the Earth:

What is the direction of the centrifugal force for an object on the Earth? The centrifugal force is a body force and can be combined with the gravity force to give an effective gravity. Effective gravity: g = 9.81 m s-2 at sea level For atmospheric science applications we use effective gravity (g) rather than gravity (g*). Therefore effective gravity should be used in place of gravity in all of the previous equations in this chapter. The addition of the centrifugal force to the gravity force results in the effective gravity force not being directed towards the center of the Earth. In reality the Earth is not a perfect sphere and the effective gravity is exactly normal to the surface of the Earth at all locations (neglecting topography). Therefore, the effective gravity force acts only in the z-direction. This discussion has only considered an object at rest in a rotating frame of reference. What happens when an object is in motion relative to a rotating frame of reference?

Coriolis Force For an object in motion relative to a rotating frame of reference we need to consider:

- an additional centrifugal force - changes in the relative angular momentum of the object

What is the centrifugal force acting on an object moving towards the east?

€

CFtotal = m Ω+u

Rφ

$

% &

'

( )

2 R φ

= mΩ2 R φ +2mΩu

Rφ

R φ + m u2

Rφ2

R φ

The first term on the RHS of this equation is the same as the centrifugal force for an object at rest, and is incorporated into the effective gravity. The last term on the RHS of this equation is assumed to be small since

€

Rφ >> u. The remaining term on the RHS of this equation can be written as:

€

CFmotion = −2mΩusinφ j + 2mΩucosφ

k

and is the centrifugal force due to the motion of the object that we need to consider in our study of the atmosphere. This component of the centrifugal force arises only for zonal (east/west) motion.

Conservation of angular momentum For a rotating object the angular momentum of the object is given by:

€

Iω = 12mRφ

2Ω The angular momentum (Iω) of this object will remain constant. What happens to the terms in this equation if the object is moving towards the North Pole (v>0)?

€

12mRφ

2Ω = 12m Rφ + δR( )2 Ω+

δuRφ + δR

%

& '

(

) *

€

Rφ2Ω = Rφ

2Ω+ 2RφδRΩ+Rφ2δu

Rφ + δR

As the object moves towards the pole its zonal velocity (u) increases. Noting that

€

δR = −vsinφδt we get:

€

δu = 2Ωvsinφδt As

€

δt→ 0 we then have:

€

dudt

= 2Ωvsinφ

Conservation of angular momentum results in a zonal acceleration for objects moving in a meridional (north/south) direction. Similarly, conservation of angular momentum also results in objects moving in the vertical direction experiencing a zonal acceleration equal to:

€

dudt

= −2Ωwcosφ

Conservation of angular momentum does not lead to an acceleration for objects moving in a purely zonal direction. The accelerations due to the additional component of the centrifugal force and due to conservation of angular momentum for an object in motion in a rotating frame of reference can be combined to form a single force known as the Coriolis force.

€

FCoriolis

m= 2Ωvsinφ − 2Ωwcosφ( )

i − 2Ωusinφ

j + 2Ωucosφ

k

Using vector notation, and noting that

€

Ω =Ωcosφ

j +Ωsinφ

k the Coriolis

force can also be expressed as:

€

FCoriolis

m= −2 Ω × u

In what direction does the Coriolis force act relative to the motion of the air? Does this differ in the Northern and Southern hemispheres? Example: The Coriolis force on the roof of Duane Physics The Navier-Stokes Equation The sum of the forces acting in each direction gives the acceleration experienced by an air parcel.

€

DuDt

= −1ρ∂p∂x

+ ν∂ 2u∂x2

+∂ 2u∂y2

+∂ 2u∂z 2

&

' (

)

* + + 2Ωvsinφ − 2Ωwcosφ( )

DvDt

= −1ρ∂p∂y

+ ν∂ 2v∂x2

+∂ 2v∂y2

+∂ 2v∂z 2

&

' (

)

* + − 2Ωusinφ( )

DwDt

= −g − 1ρ∂p∂z

+ ν∂ 2w∂x2

+∂ 2w∂y2

+∂ 2w∂z 2

&

' (

)

* + + 2Ωucosφ( )

What does each term in these equations represent?

These equations are known as the Navier-Stokes equations. We can rewrite the Navier-Stokes equations in an Eulerian framework as:

€

∂u∂t

+ u∂u∂x

+ v∂u∂y

+ w∂u∂z

= −1ρ∂p∂x

+ ν∂ 2u∂x2

+∂ 2u∂y2

+∂ 2u∂z 2

&

' (

)

* + + 2Ωvsinφ − 2Ωwcosφ( )

∂v∂t

+ u∂v∂x

+ v∂v∂y

+ w∂v∂z

= −1ρ∂p∂y

+ ν∂ 2v∂x2

+∂ 2v∂y2

+∂ 2v∂z 2

&

' (

)

* + − 2Ωusinφ( )

∂w∂t

+ u∂w∂x

+ v∂w∂y

+ w∂w∂z

= −g − 1ρ∂p∂z

+ ν∂ 2w∂x2

+∂ 2w∂y2

+∂ 2w∂z 2

&

' (

)

* + + 2Ωucosφ( )

Perturbation Pressure We can rewrite the vertical momentum equation by defining a reference pressure, p0, that is in hydrostatic balance with a reference density, ρ0, as:

€

dp0dz

= −ρ0g

The total pressure is then given by:

€

p = po z( ) + pd Substituting this definition into the vertical component of the Navier-Stokes equation gives:

€

ρDwDt

= −ρg − ∂∂z

p0 z( ) + pd( ) + ρν∂ 2w∂x2

+∂ 2w∂y2

+∂ 2w∂z 2

&

' (

)

* + + ρ 2Ωucosφ( )

= −ρg + ρ0g −∂pd∂z

+ ρν∂ 2w∂x2

+∂ 2w∂y2

+∂ 2w∂z 2

&

' (

)

* + + ρ 2Ωucosφ( )

DwDt

= −ρ − ρ0( )ρ

g − 1ρ∂pd∂z

+ ν∂ 2w∂x2

+∂ 2w∂y2

+∂ 2w∂z 2

&

' (

)

* + + 2Ωucosφ( )