newton’s second law for a parcel of air in an inertial coordinate system (a coordinate system in...

Newton’s second law for a parcel of air in an inertial coordinate system

(a coordinate system in which the coordinate axes do not change direction and are not accelerated)

objectanonactingForcesdt

dVm i

Fma

In the case of more than one force:

y

y

y

x

x

xz

(toward you)

z(toward you)

z(toward you)

Earth notrotating


Vdm i

An inertial coordinate system and the meaning of the vector and dt

Vd i

iV

Inertial coordinate system:

iV

is with respect to a distant star

dt

Vd i

is with respect to a distant star

Let’s first take a look at the term in the yellow box…

Earth-based (non-inertial) “spherical” coordinate system(favored by meteorologists)

x = positive toward easty = positive toward northz = positive toward zenith

udt

dx wind component toward east

vdt

dy wind component toward north

wdt

dz wind component toward zenith

= longitude

= latitude

dadx cos

addy

Earth-based (non-inertial) “spherical” coordinate system(favored by meteorologists)

x = positive toward easty = positive toward northz = positive toward zenith

wdt

dz

dt

da

dt

dx cos

dt

da

dt

dy

kwjviuV

Where unit vectors are a function of position on the earth (not Cartesian)

kwjviuVi

dt

kdw

dt

jdv

dt

idu

dt

dwk

dt

dvj

dt

dui

dt

Vd i

We need to determine what these are

Since the unit vectors are not constant…

Let’s start withdt

id

z

iw

y

iv

x

iu

t

i

dt

id

What is the relationship between and dt

Vd i

dt

Vd

z

iw

y

iv

x

iu

t

i

dt

id

0t

i

At a point (constant x,y,z) none of the unit vectors change with time so

As one moves north or south the i direction experiences no change so

0y

i

As one moves up or down the i direction experiences no change so 0

z

i

x

iu

dt

id

So:

x

iu

dt

id

From figure on right looking down at north pole at latitude :

ii

cosax

cos

1

ax

i

This gives us the magnitude, but not the direction. Note thatis pointed toward the center of the earth at the original point

i

Looking at the figure at the right:

We see that has components in the and directions

j

k

i

The unit vector describing the direction of has two components:i

cossin kji

So:a

uk

a

uj

a

ukuj

x

iu

dt

idu

tan

cos

cossin

kwjviuVi

dt

kdw

dt

jdv

a

uk

a

uju

dt

dwk

dt

dvj

dt

dui

dt

Vd i

tan

We still need to determine what these are


Let’s do nextdt

jd

z

jw

y

jv

x

ju

t

j

dt

jd

REMEMBER THIS SLIDE?

z

jw

y

jv

x

ju

t

j

dt

jd

does not change with time or elevation, but does change in the x and y directions

j

y

jv

x

ju

dt

jd

Lets first figure out x

j

Look at light gray triangle in (a):

sincos a

Dark gray triangle is shown in a different view in (b)

tan

aor

tan

ax

jj

So:

ax

j tan

Direction ofis the -x direction

j

Now lets figure out y

j

ay jj

Direction of is the directionj k

kay

j 1

a

vk

a

ui

dt

jd

tan

So:

aay

j 1

kwjviuVi

dt

kdw

a

vk

a

uiv

a

uk

a

uju

dt

dwk

dt

dvj

dt

dui

dt

Vd i

tantan

We still need to determine what this is


Let’s do nextdt

kd

z

kw

y

kv

x

ku

t

k

dt

kd


z

kw

y

kv

x

ku

t

k

dt

kd

does not change with time or elevation, but does change in the x and y directions

k

y

kv

x

ku

dt

kd

y

kv

x

ku

dt

kd

x

k

Let’s do first

ax

kk

Direction of is the positive directionk i

iax

k

1

y

kv

x

ku

dt

kd

y

k

Let’s do next

ay

kk

Direction of is the positive directionk i

jay

k

1

k

k

ja

vi

a

u

dt

kd

kwjviuVi

j

a

vi

a

uw

a

vk

a

uiv

a

uk

a

uju

dt

dwk

dt

dvj

dt

dui

dt

Vd i

tantan



a

vu

dt

dwk

a

vw

a

u

dt

dvj

a

uw

a

uv

dt

dui

dt

Vd 222 tantan

or

a

vuk

a

vw

a

uj

a

uw

a

uvi

dt

dV

dt

Vd 222 tantan

What are the correction terms and what do they mean?

Consider third equation: forcesa

vu

dt

dw

22

Air moving in a straightline at a constant speedinitially southward.u = 0, v = -10 m/s, w = 0

Air will accelerate upward!

The earth curves away from the pathof the air parcel.


Vdm i

Forces

a

vuk

a

vw

a

uj

a

uw

a

uvi

dt

Vdm

222 tantan

Newton’s second law in an inertial coordinate system

Newton’s second law in a spherical coordinate system

Now let’s look at the right side of the equation!

Forces and the governing equations

Three Fundamental Forces in the lower atmosphere

Pressure Gradient ForceGravityFriction

Two Apparent Forces in the lower atmosphere due to the rotation of the earth

Centrifugal ForceCoriolis Force

1. Pressure gradient force: Directed from high pressure toward low pressure.

A wall with higher pressure on its left side (indicated by greater molecular density on the left side)

Simple illustration of a pressure gradient acting on a wall (left) and a molecule (right)

An air molecule with higher air pressure on its left (indicated by greater molecular density on the left side)

Volume of the fluid element

zyxV Mass of the fluid element

zyxm = density (g m-3)

Pressure at centerof the fluid element

0000 ,, zyxpp

Definition of pressure: Pressure is equal to the force applied per unit area

termsorderhigherx

x

pppA

20

Find Pressure at A and B

Use Taylor expansion:

termsorderhigherx

x

pppB

20

zyx

x

ppAreapF AAA

20

Find Force at B and A

zyx

x

ppAreapF BBB

20

Force = Pressure Area

Find net x-direction force per unit mass

zyx

x

ppzy

x

x

ppFFF BAx

22 00

zyxx

pFx

x

p

m

Fx

1

x

p

m

Fx

1

In a similar way:

y

p

m

Fy

1

z

p

m

Fz

1

The vector form of the Pressure Gradient Force can therefore be expressed as

z

pk

y

pj

x

pi

m

F

111

pm

F

1

2. Gravity: Directed toward center of earth

r

r

r

GMmFg

2

Newton’s law of universal gravitation G = 6.67310-11 N m2 kg-2

M = mass of Earthm = mass of fluid elementr = vector from Earth centerr = magnitude of r vector

r

r

r

GMmFg

2

r

r

r

GM

m

Fg

2

Newton’s law of universal gravitation…

…expressed as a force per unit mass

Over the depth of the tropospherethe change in the force of gravity is insignificant

and we can approximate r as the earth radius, a

r

r

a

GM

m

Fg

2

3. Friction: Directed opposite the flow

Friction is manifested as:

A drag force in a very thin layer (a few mm) near the surface

Turbulent mixing of blobs of faster and slower air at altitudes above the surface.

Although largely irrelevant to the atmosphere, we will consider the drag force first and then make a simple

analogy to formulate friction for the rest of the atmosphere

Not moving

moving

Expect that drag force on upper plate will be proportional to:

Speed of the plate (u0)Area of the plate (A)

And inversely proportional to the depth of the fluid (l)

l

AuF 0

Where is the proportionality constant called the dynamic viscosity coefficient

Next let’s define a “Shearing Stress” , as the Force/Unit Area on the plate

l

AuF 0

Let’s also assume the lower plate is moving at an arbitrary speed so that the speed difference is u

Let’s also assume that the distance between the plates is arbitrary and that that difference is z

Then:z

uzx

andz

uzx

as 0z

z

uzx

Force per unit area acting in x direction due to shear in z direction

Now let’s consider how that force acts on a fluid element of volume xyz

Illustration of x component of shearing stress (F/A) on a fluid element

Stress acting across the upper boundary on the fluid below it

2

z

zzx

zx

Use Taylor expansion

Stress acting across the lower boundary on the fluid below it

2

z

zzx

zx

Using Newton’s third law: Stress across the lower boundary on the fluid above it must be:

2

z

zzx

zx

To find the net stress, we want to sum the forces that act on fluid within the fluid element

zyxz

yxz

zyx

z

zzxzx

zxzx

zx

22

Dividing by the mass of the fluid element we have the viscous force per unit mass arising from the vertical shear in the x direction on the fluid element

z

u

zzm

F zxzx

11

z

u

zm

Fzx 1

Assuming is constant, this can be written

2

2

2

2

z

u

z

u

m

Fzx

where is called the kinematic viscosity coefficient

We have figured out one component of the frictional force per unit mass: m

Fzx

Considering how shear can act in each direction to create a frictional force in each direction, there are nine components

m

F

m

F

m

F

m

F

m

F

m

F

m

F

m

F

m

F zzzyzxyzyyyxxzxyxx ,,,,,,,,

2

2

2

2

2

2

z

u

y

u

x

u

m

Frx

2

2

2

2

2

2

z

v

y

v

x

v

m

Fry

2

2

2

2

2

2

z

w

y

w

x

w

m

Frz

Unfortunately, all of the preceding theory applies to the air in a layer a few mm thick above the earth’s surface

2

2

2

2

2

2

z

u

y

u

x

uK

m

Frx

For blobs of air mixing, an analogous “eddy viscosity coefficient”, K, is defined that is calculated based on the average length that an eddy can travel before mixing out its momentum.

In general, vertical wind shear is much stronger than horizontal shear for synoptic scale flows so this reduces to:

2

2

z

uK

m

Frx

2

2

z

vK

m

Fry

0m

Frz

2

2

2

222 1tantan

z

VK

r

r

a

GMp

m

F

a

vuk

a

vw

a

uj

a

uw

a

uvi

dt

Vd

Acceleration

pressure gradientforce

gravity

friction

Force balance in the atmosphere for a non-rotating earth expressed in spherical coordinates

Correction for sphericalCoordinate system

Same statement expressed as 3 orthogonal scalar equations

2

21tan

z

uK

x

p

a

uw

a

uv

dt

du

2

22 1tan

z

vK

y

p

a

vw

a

u

dt

dv

AccelerationCorrection terms for earthbased coordinate system

Pressure gradientForce

Friction

(East-West equation)

(North-South equation)

(Up-Down equation) 2

22 1

a

GM

z

p

a

vu

dt

dw

Gravity

Apparent forces on a rotating earth – the Centrifugal Force

We will derive the force terms twice

First from a more conceptual approach

The second time from a more formal vector approach

Apparent forces on a rotating earth – the Centrifugal Force

Consider an object at rest on the rotating earth

Equator Speed = 465 m/s

40 N: Speed = 356 m/s

60 N: Speed = 232 m/s

Actual speed due to rotation of earth:

A ball whirling on a string experiences a CENTRIPETAL acceleration toward the axis of rotation

The acceleration is equal to where is the angular rotation rate r2

A person on the ball whirling on a string experiences a CENTRIFUGAL acceleration toward the axis of rotation

The acceleration is equal to where is the angular rotation rate r2

Standing still on the rotating earth

Outward acceleration away from Earth’s axis is where is the Earth’s angular rotation rate

r2

s09.86156

2

Where 86156.09 s is the time it takes for the Earth to rotate once with respect to a fixed star (a sidereal day)

281.9 msg

22

12 034.0637810009.86156

2

msmsR

Vectors not to true size!

Earth has distorted its shape into an oblate spheroid in response to this force. The distortion is such that the combined gravitational force and centrifugal force (g*) act exactly perpendicular to the Earth’s (flat) surface.

Rr

rgR

r

r

a

GMkg

2*2

2

We combine the centrifugal force with gravity and forget about it!!

Apparent forces on a rotating earth – the Coriolis Force

In the absence of a twisting force called a torque, air in motion across the earth mustconserve its angular momentum (MVR)where M is the parcel mass, V is velocity aboutthe axis of rotation and R is the distance fromthe axis of rotation.

A simple interpretation of the Coriolis effect:

Air moving across the earth’s surface willtry to come to equilibrium at a latitude/altitudewhere its angular momentum equals that of the earth so that the parcel has no relative motion.

Motion on a rotating earth – the Coriolis Force

Consider air moving:

Eastward: The air has greater angular momentumthan the earth beneath it. It will experience an“outward” centrifugal acceleration equatorward

Westward: The air has less angular momentumThan the earth beneath it. It will experience an“inward” centripetal acceleration poleward.

Poleward: The air will progressively move overpoints on the earth with less angular momentum. Air will accelerate eastward relative to the earthbelow it.

Equatorward: The air will progressively move overpoints on the earth with greater angular momentum. Air will accelerate westward relative to the earthbelow it.

Suppose an object is moving eastward on the rotating earth at speed u

The total centrifugal force it will experience will be its angular velocity squared distance to axis of rotation

2

22

2

2R

Ru

R

RuRR

R

uCEN

Centrifugal force(combined with

gravity)

231 101086156

22 msms

s

25222

106378100

10 msm

sm

Magnitude of 2nd term on RHS Magnitude of 3rd term on RHS

Suppose an object is moving eastward on the rotating earth at speed u

The total centrifugal force it will experience will be its angular velocity squared distance to axis of rotation

2

22

2

2R

Ru

R

RuRR

R

uCEN

Centrifugal force(combined with

gravity)

ignore

Coriolis Force

R

RuCOR

2

cos2sin2 ukujcirclelatitudetoparallelmotiononacceleratiCoriolis

sin2 udt

dv

cos2 udt

dw

22 RRRR

uR

Suppose an object is moving equatorward on the rotating earth at speed -v

Apply principle of conservation of momentum

Expand this expression

222 2 RRRRRR

uR

Since and are small, neglect terms with their productu R Ru

RR

uRRRRR

2

22 2

RR

uRRRRR

2

22 2

(From previous slide)

RR

uRRR

2

2

If we assume that << this expression can be approximated as:

uRRR 2

R R

or

Ru 2

Ru 2

For a displacement in the –y direction, from Figure:

sin2sin2 yyu

sin2sin2 vdt

dy

dt

du

Dividing by and taking the limit ast 0t

Ru 2

For a displacement in the z direction, from Figure:

cos2cos2 zzu

cos2cos2 wdt

dz

dt

du

Dividing by and taking the limit ast 0t

Combining the results for east-west and north-south movement of air

fuudt

dv sin2

cos2 udt

dw

cos2cos2sin2 wfvwvdt

du

Where: f sin2

and f is the Coriolis parameter

In a reference frame on the earththe Coriolis effect appears as a force acting on an air parcel.

Motion on a rotating earth – the Coriolis Force

The Coriolis Force

Causes air to deviate to the right of its direction of motion in the Northern Hemisphere (and to the left in the Southern Hemisphere);

Affects the direction an object will move across the earth’s surface, but has no effect on its speed;

Is strongest for fast-moving objects and zero for Stationary objects; and

Has no horizontal component at the equator and has a maximum horizontalcomponent at the poles

In our previous discussion of spherical coordinates, the coordinates were fixed in position.

AAt Point A, for example, the x, y, and z axes always point toward distant stars. At point B, the axes point at different stars, but always in the same direction.

B

A formal way of deriving the Coriolis and Centrifugal Forces

When the earth rotates, the axes at point A find themselves at a later time at point B. The direction the axes at a specific location point is a function of time

To account for the local movement of the coordinate system mathematically, we must consider the behavior of a vector in a rotating coordinate system

Consider a vector in a stationary coordinate system

kAjAiAA zyx

The components of the vector in the rotating coordinate system are:

kAjAiAA zyx

where , , and are fixed at local pointsi

j

k

The time derivative of the vector kAjAiAA zyx

The time derivative of the vector kAjAiAA zyx

in a stationary coordinate system would be:

kdt

dAj

dt

dAi

dt

dA

dt

Ad zyx

in the rotating coordinate system are:

dt

kdA

dt

jdA

dt

idAk

dt

Adj

dt

Adi

dt

Ad

dt

Adzyx

zyx

dt

kdA

dt

jdA

dt

idAk

dt

Adj

dt

Adi

dt

Ad

dt

Adzyx

zyx

kdt

dAj

dt

dAi

dt

dA

dt

Ad zyx

Stationary

Rotating

We can write the equation for the rotating frame of reference as:

dt

kdA

dt

jdA

dt

idA

dt

Ad

dt

Adzyx

The derivatives , , and , represent the rate of

change of the unit vectors , , and , that arise because the coordinate system is rotating

dt

id

dt

jd

dt

kd

i

j

k

Let’s determine and expression for

dt

id

Rotation vector, , points upward from north pole

iiBy similar triangles

idt

di

dt

id

t

i

t

0

lim

The magnitude of is given by: dt

id

The direction of is given by: dt

id

idt

id

jj

By similar triangles

jdt

dj

dt

jd

t

j

t

0

lim


jd

The direction of is given by:

jdt

jd

dt

jd


dt

jd


kk By similar triangles

kdt

dk

dt

kd

t

k

t

0

lim


kd

The direction of is given by: dt

kd

kdt

kd


dt

kd


dt

kdA

dt

jdA

dt

idA

dt

Ad

dt

Adzyx

Our previous equation was:

dt

kdA

dt

jdA

dt

idA

dt

Ad

dt

Adzyx

and we can write

jAjAdt

jdA yyy

iAiA

dt

idA xxx

kAkAdt

kdA zzz

then

kAjAiAdt

Ad

dt

Adzyx

Adt

Ad

dt

Ad

Adt

Ad

dt

Ad

For a vector A, the coordinate transformation from a non-rotating to a rotating system is given by:

Let’s apply this to two vectors:

Let be a position vector, perpendicular to the axis of rotation of the earth with magnitude equal to the distance from the surface of the Earth to the axis of rotation

rdt

rd

dt

rda

By definition: rVVa

The absolute velocity of an object on a rotating earth is equal to its velocity relative to the rotating earth + the velocity of the Earth’s rotation

Also apply vector transform to aV

r

aaaa V

dt

Vd

dt

Vd

The absolute acceleration of an object on a rotating earth is equal to its acceleration relative to the rotating earth + acceleration due to the Earth’s rotation

aaaa V

dt

Vd

dt

Vd

rVVa(1) (2)

Put (2) into (1)

rVrVdt

d

dt

Vd aa

rVdt

rd

dt

dV

dt

Vd aa

rVVdt

dV

dt

Vd aa

rVdt

dV

dt

Vd aa

22

CoriolisForce

CentrifugalForce

rVdt

dV

dt

Vd aa

22

w

k

vu

ji

V sin2cos202

sin2 udt

dv

dt

vd aa

cos2 udt

dw

dt

wd aa

cos2sin2 wvdt

du

dt

ud aa

Equations of motion on non-rotating earth

2

21tan

z

uK

x

p

a

uw

a

uv

dt

du

2

22 1tan

z

vK

y

p

a

vw

a

u

dt

dv

AccelerationCorrection terms for earthbased coordinate system


Friction



(Up-Down equation) 2

22 1

a

GM

z

p

a

vu

dt

dw

Gravity

Equations of motion on rotating earth

AnswerWe must add terms to account for the accelerationrequired for air to conserve its angular momentum.

In scalar form the equations of motion for each direction become:

cos2sin2

1tan2

2

wvz

uK

x

p

a

uw

a

uv

dt

du

sin2

1tan2

22

uz

vK

y

p

a

vw

a

u

dt

dv

cos2122

ugz

p

a

vu

dt

dw



(Up-Down equation)

Acceleration

Correction terms for spherical coordinate system


Friction

EffectiveGravity

Coriolis Force

The complete momentum equations on a spherical rotating earth

FOR SYNOPTIC SCALE MOTIONS…..

WHICH TERMS ARE LARGE AND IMPORTANT?WHICH TERMS ARE SMALL AND INSIGNIFICANT?

See Scale Analyses: Table 3.1 P. 60, Table 3.2 P. 64

cos2sin2

1tan2

2

wvz

uK

x

p

a

uw

a

uv

dt

du

sin2

1tan2

22

uz

vK

y

p

a

vw

a

u

dt

dv

cos2122

ugz

p

a

vu

dt

dw

Scale Analysis of the horizontal momentum equations

WfUfH

U

L

p

a

UW

a

U

L

U002

22

cos2sin2

1tan2

2

wvz

uK

x

p

a

uw

a

uv

dt

du

sin2

1tan2

22

uz

vK

y

p

a

vw

a

u

dt

dv

6373854 10101010101010

110 msU 101.0 msW mL 610 mH 410

2233

101

101000 sm

mkg

mbPap

121 smK

*numbers in yellow boxes apply to free atmosphere above boundary layer

Scale Analysis of the vertical momentum equations

110 msU 101.0 msW mL 610 mH 410

2253

5

101

100010 sm

mkg

mbPap

201 msg

cos2122

ugz

p

a

vu

dt

dw

ufgH

p

a

U

L

UW0

2

31157 1010101010

cos2sin2

1tan2

2

wvz

uK

x

P

a

uw

a

uv

dt

du

sin2

1tan2

22

uz

vK

y

P

a

vw

a

u

dt

dv

cos20122

ugz

P

a

vu

dt

dw

ABOVE THE BOUNDARY LAYER

cos2sin21tan

2

2

wvz

uK

x

P

a

uw

a

uv

dt

du

sin2

1tan2

22

uz

vK

y

P

a

vw

a

u

dt

dv

cos20122

ugz

P

a

vu

dt

dw

WITHIN THE BOUNDARY LAYER

fvz

uK

x

P

dt

du

2

21

fuz

vK

y

P

dt

dv

2

21

gz

P

1

0

fvx

P

dt

du

1

fuy

P

dt

dv

1

ABOVE THE BOUNDARY LAYER, ALL HORIZONTAL PARCEL ACCELERATIONS CAN BE UNDERSTOOD BY COMPARING THE MAGNITUDE AND DIRECTION OF THE PRESSURE GRADIENT AND CORIOLIS FORCES

WITHIN THE BOUNDARY LAYER, ALL HORIZONTAL PARCEL ACCELERATIONS CAN BE UNDERSTOOD BY COMPARING THE MAGNITUDE AND DIRECTION OF THE PRESSURE GRADIENT, CORIOLIS AND FRICTIONAL FORCES

The Hidden Simplicity of Atmospheric Dynamics:

THE ATMOSPHERE IS IN HYDROSTATICBALANCE – VERTICAL PGF BALANCESGRAVITY – ON SYNOPTIC SCALES

Note the Total Derivative:

zyxtyxtzxtzy tdt

dt

zdt

dz

ydt

dy

xdt

dx

dt

d

,,,,,,,,

zyxtyxtzxtzy tzw

yv

xu

dt

d

,,,,,,,,

dt

dzw

dt

dyv

dt

dxu ,,Note that:

tyxtzxtzyzyx zw

yv

xu

dt

d

t ,,,,,,,,

The rate of changeof a property (temp)at a fixed point(x,y,z)

The rate of changeof the propertyFollowing a parcel asit moves tothe point (x,y,z)

The advection of the propertyfrom upstream to the point(x, y, z)

=

fvx

P

t

u

z

uw

y

uv

x

uu

zyxtyxtzxtzy

1

,,,,,,,,

fvx

P

dt

du

1

z

uw

y

uv

x

uufv

x

P

t

u

1

Simplified u Momentum Equation expressed in terms of local time derivatives

The rate of changeof west-east windat a fixed point(x,y,z)

PGF COR The advection of the west-east windComponent from upstream to the point(x, y, z)

fuy

P

t

v

z

vw

y

vv

x

vu

zyxtyxtzxtzy

1

,,,,,,,,

fuy

P

dt

dv

1

z

vw

y

vv

x

vufu

y

P

t

v

1

Simplified v Momentum Equation expressed in terms of local time derivatives

The rate of changeof south-north windat a fixed point(x,y,z)

PGF COR The advection of the south-north windComponent from upstream to the point(x, y, z)

Pressure as a vertical coordinate

Pressure decreases monotonically with altitude, and therefore can be used as a vertical coordinate instead of altitude

We will find this coordinate system more convenient mathematically and so our task will be to recast the momentum equations in pressure coordinates

fvx

P

dt

du

1

fuy

P

dt

dv

1

The simplified horizontal momentum equations in height coordinates are given by:

or more succinctly:

Vkfpdt

Vd

1

In order to cast these equations in pressure coordinates, we need to convert the PGF term into an equivalent expression in isobaric coordinates……..

Consider the differential on a constant pressure surfacedp

Expanding…

pyx

p

zx

pzy

p dzz

pdy

y

pdx

x

pdp

,,,

(Subscripts indicate differentiation carried out holding subscripted variable constant)

pyx

p

zx

pzy

dzz

pdy

y

pdx

x

p

,,,

0

Now expand as a function of x and ydz

p

px

ppyyx

p

zx

pzy

dyy

zdx

x

z

z

pdy

y

pdx

x

p

,,,,,

0

p

px

ppyyx

p

zx

pzy

dyy

zdx

x

z

z

pdy

y

pdx

x

p

,,,,,

0

Rearrange:

p

pxyxzx

ppyyxzy

dyy

z

z

p

y

pdx

x

z

z

p

x

p

,,,,,,

0

For this statement to be true for all and , the statements in each square bracket must equal zero

pdx pdy

pyyxzy x

z

z

p

x

p

,,,

pxyxzxy

z

z

p

y

p

,,,

pyyxzy x

z

z

p

x

p

,,,

pxyxzxy

z

z

p

y

p

,,,

Use hydrostatic equation gz

P

pyzy x

zg

x

p

,,

pxzx

y

zg

y

p

,,

Divide both sides by

pyzy x

zg

x

p

,,

1

pxzx

y

zg

y

p

,,

1

or

pyzy xx

p

,,

1

pxzxyy

p

,,

1

where is called the geopotential heightgz

pyzy xx

p

,,

1

pxzxyy

p

,,

1

pyzyp ,,

1

We will now drop the use of subscripts and write the momentum equations in p coordinates

In pressure coordinates, the horizontal momentum equations (above the boundary layer) become:

Pressure coordinates

fvxdt

du

fuydt

dv

where , the geopotential height, is given by = gzand f = 2sin

p

TR

pd

and the vertical momentum equation becomes

Vkfdt

Vdp

(using the ideal gas law)

The Total Derivative in pressure coordinates

zpxtyxtpxtpy tdt

dt

pdt

dp

ydt

dy

xdt

dx

dt

d

,,,,,,,,

dt

dyv

dt

dxu ,Note that:

Note that the vertical velocity in pressure coordinates has a different form:

dt

dp

The term, , represents the rate of change of pressure following parcel motion as a parcel rises or descends in the atmosphere

Note the Total Derivative:

zpxtyxtpxtpy tdt

dt

pdt

dp

ydt

dy

xdt

dx

dt

d

,,,,,,,,

pyxtyxtpxtpy tpyv

xu

dt

d

,,,,,,,,

dt

dyv

dt

dxu ,Note that:

tyxtpxtpypyx pyv

xu

dt

d

t,,,,,,,,

The rate of changeof a property (temp)at a fixed point(x,y,z)

The rate of changeof the propertyFollowing a parcel asit moves tothe point (x,y,z)

The advection of the propertyfrom upstream to the point(x, y, z)

=

p

u

y

uv

x

uufv

xt

u

Simplified u Momentum Equation in pressure coordinatesexpressed in terms of local time derivatives

The rate of changeof west-east windat a fixed point(x,y,z)

PGF COR The advection of the west-east windComponent from upstream to the point(x, y, z)

fvxdt

du

p

v

y

vv

x

vufu

yt

v

Simplified v Momentum Equation expressed in terms of local time derivatives

The rate of changeof south-north windat a fixed point(x,y,z)

PGF COR The advection of the south-north windComponent from upstream to the point(x, y, z)

fuydt

dv

newton’s second law for a parcel of air in an inertial coordinate system (a coordinate system in...

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