Transcript 02. Conservation of Momentum

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)
ma  F
In the case of more than one force:
m
dV i
dt

 Forces
acting on an object
m

d Vi

dt
 Forces
acting on an object
Let’s first take a look at the term in the yellow box…


d Vi
An inertial coordinate system and the meaning of the vector V i and
dt
y
Earth not
rotating
x
z
(toward you)
Inertial coordinate system:
y
x
z
(toward you)
y
x
z
(toward you)

V i is with respect to a distant star

d V i is with respect to a distant star
dt
Earth-based (non-inertial) “spherical” coordinate system
(favored by meteorologists)
x = positive toward east
y = positive toward north
z = positive toward zenith
dx
u
wind component toward east
v
wind component toward north
dt
dy
dt
dz
 w wind component toward zenith
dt
= longitude
= latitude
dx  a cos  d 
dy  ad 
Earth-based (non-inertial) “spherical” coordinate system
(favored by meteorologists)
x = positive toward east
y = positive toward north
z = positive toward zenith
dx
 a cos 
dt
dy
dt
a
dt
dz
d
d
dt
 w
dt




V  u i  vj  w k
Where unit vectors are a function of position on the earth (not Cartesian)
What is the relationship between




Vi  u i  vj  w k

d Vi
and
dt

dV
dt
Since the unit vectors are not constant…

d Vi

di

dj

dk
 dw
 du
 dv
i
 j
k
u
v
w
dt
dt
dt
dt
dt
dt
dt
Let’s start with

di
We need to determine what these are
dt





di
i
i
i
i

u
v
w
dt
t
x
y
z

di

dt

i
t
u

i
x
v

i
y
w

i
z
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
As one moves up or down the i
direction experiences no change so
So:

di
dt
u

i
x

i
t

i
y

i
z
0
0
0

di
dt
u

i
x
From figure on right looking
down at north pole at latitude :


 i  i 
 x  a cos 

i
1

x
a cos 

This gives us the magnitude, but not the direction. Note that  i
is pointed toward the center of the earth at the original point
Looking at the figure at the right:

that  i
We see
has


components in the j and  k
directions

The unit vector describing the direction of  i has two components:

 
 i  j sin   k cos 




u
 u tan 
di
i
j u sin   k u cos 
u

 j
k
So: u
dt
x
a cos 
a
a
REMEMBER THIS SLIDE?




Vi  u i  vj  w k
Since the unit vectors are not constant…

d Vi

dj

dk
 dw
u
 du
 dv
  u tan 
i
 j
k
 u j
k v
w
dt
dt
dt
dt
a
a
dt
dt

Let’s do

dj
dt


j
t

dj
dt
u
We still need to determine what these are
next

j
x
v

j
y
w

j
z

dj
dt


j
t
u

j
x
v

j
y
w

j
z

j does not change with time or
elevation, but does change in the
x and y directions

dj
dt
u

j
x
v

j
y
Lets first figure out

j
x
Look at light gray triangle in (a):
a cos    sin 
or
 
a
tan 
Dark gray triangle is shown
in a different view in (b)
x 
a
tan 
 
 j  j 


tan 
So:  j

x
a

Direction of  j
is the -x direction

j
Now lets figure out

j
y
y
 y  a 
 
 j  j 



a 

Direction of  j is the

j
1

y
a
So:
1
a

 k direction

k

dj
 v
 u tan 
 i
k
dt
a
a
REMEMBER THIS SLIDE?




Vi  u i  vj  w k
Since the unit vectors are not constant…

dVi




 du
 dv
dw
u
v
dk
  u tan 
  u tan 
i
 j
k
 u j
 k   v  i
k w
dt
dt
dt
dt
a
a
a
a
dt


We still need to determine what this is

Let’s do d k next
dt





dk
k
k
k
k

u
v
w
dt
t
x
y
z

dk
dt


k
t
u

k
x
v

k
y
w

k
z

k does not change with time or
elevation, but does change in the
x and y directions

dk
dt
u

k
x
v

k
y

dk
u
dt
Let’s do

k
x

k
x
v

k
y
first
 x  a 


 k  k   


Direction of  k is the positive i direction

k
1
 i
x
a

dk
u
dt
Let’s do

k
x

k
y
v

k
y
next
 y  a 


 k  k   


Direction of  k is the positive i direction


k
1 

j
dk
u  v 
y
a
 i 
j
dt
a
a

k

k
REMEMBER THIS SLIDE?




Vi  u i  vj  w k
Since the unit vectors are not constant…

dVi
 dw
u
 v
 du
 dv
  u tan 
  u tan 
u  v 
i
 j
k
 u j
 k   v  i
 k   w i 
j
dt
dt
dt
dt
a
a
a
a
a 


a
or

dV
  du uv tan  uw    dv u 2 tan  vw    dw u 2  v 2
 k
 i





  j 


dt
a
a 
a
a 
a
 dt
 dt
 dt

dV
  uv tan  uw    u 2 tan  vw    u 2  v 2
 k

 i 


  j 

dt
dt
a
a 
a
a 
a



dV








What are the correction terms and what do they mean?
Consider third equation:
dw
dt
u v
2

2
 forces
a
Air moving in a straight
line at a constant speed
initially southward.
u = 0, v = -10 m/s, w = 0
Air will accelerate upward!
The earth curves away from the path
of the air parcel.
Newton’s second law in an inertial coordinate system
m

d Vi
dt

 Forces
acting on an object
Newton’s second law in a spherical coordinate system

 d V   uv tan  uw 
m
 i 


a
a 

 dt
  u 2 tan  vw    u 2  v 2
 k
j 


a
a 
a



 


 Forces
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 Force
Gravity
Friction
Two Apparent Forces in the lower atmosphere
due to the rotation of the earth
Centrifugal Force
Coriolis Force
1. Pressure gradient force: Directed from high pressure
toward low pressure.
Simple illustration of a pressure gradient acting on a wall (left) and a molecule (right)
A wall with higher pressure on its left
side (indicated by greater molecular
density on the left side)
An air molecule with higher air pressure
on its left (indicated by greater
molecular density on the left side)
Volume of the fluid element
V  xyz
Mass of the fluid element
m   x  y  z
 = density (g m-3)
Pressure at center
of the fluid element
p0  p  x0 , y0 , z 0 
Definition of pressure:
Pressure is equal to the force applied per unit area
Find Pressure at A and B
Use Taylor expansion:
p  x 
p A  p0 

  higher order terms
x  2 
p  x 
p B  p0 

  higher order terms
x  2 
Find Force at B and A
Force = Pressure  Area
F A  p A  Area
F B  p B  Area
A
B

p  x  
   p 0 

   y  z
x  2  


p  x  
   p 0 

   y  z
x  2  

Find net x-direction
force per unit mass


p  x  
p  x  
F x  F A  F B    p 0 


y

z


p




   y  z
 0
x  2  
x  2  


Fx  
Fx
m
p
x

xyz
1 p
 x
Fx
m

1 p
 x
In a similar way:
Fy

m
1 p
Fz
 y
m

1 p
 z
The vector form of the Pressure Gradient Force
can therefore be expressed as

F
  1 p 
 
 i  
m
  x 

F
m
  1 p 
 
j  
  y 

1

p
  1 p 

k  
  z 
2. Gravity: Directed toward center of earth
Newton’s law of universal gravitation


GMm  r 
Fg  
 
2
r
r
G = 6.67310-11 N m2 kg-2
M = mass of Earth
m = mass of fluid element
r = vector from Earth center
r = magnitude of r vector
Newton’s law of universal gravitation…


GMm  r 
Fg  
 
2
r
r
…expressed as a force per unit mass

Fg

GM  r 
 2  
m
r r
Over the depth of the troposphere
the change in the force of gravity is insignificant
and we can approximate r as the earth radius, a

Fg

GM  r 
 2  
m
a r
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
moving
Not 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)
F 
 Au 0
l
Where  is the proportionality constant called the dynamic viscosity coefficient
F 
 Au 0
l
Next let’s define a “Shearing Stress” , as the Force/Unit Area on the plate
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:
 zx  
u
z
and
 zx  
u
z
as
z  0
 zx  
u
z
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 xyz
Illustration of x component of shearing stress (F/A) on a fluid element
Use Taylor expansion
Stress acting across the upper
boundary on the fluid below it
 zx 
  zx  z
z
2
Stress acting across the lower
boundary on the fluid below it
 zx 
  zx  z
z
2
Using Newton’s third law: Stress across the lower boundary on the
fluid above it must be:
 z 

zx
   zx 

z 2 

To find the net stress, we want to sum the forces that act on
fluid within the fluid element
  zx  z 
  zx  z 
  zx





x

y




x

y

xyz
 zx

 zx

z 2 
z 2 
z


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
1   zx
1   u 




m
 z
 z  z 
F zx
F zx
m

1   u 


 z  z 
Assuming  is constant, this can be written
F zx
m
  u
2

 z
2
 u
2

z
2
where  is called the kinematic viscosity coefficient
F zx
We have figured out one component
of the frictional force per unit mass:
m
Considering how shear can act in each direction to create
a frictional force in each direction, there are nine
components
F zx F zy F zz
,
,
,
,
,
,
,
,
m m m m
m
m m m m
F xx
F xy
F xz
F yx
F yy
F yz
  2u  2u  2u
   2 

2
2
m

x

y

z

F rx
  2v  2v  2v
   2 

2
2
m

x

y

z

F ry








  2w  2w  2w
  


2
2
2
m

x

y

z

F rz




Unfortunately, all of the preceding theory applies to the
air in a layer a few mm thick above the earth’s surface
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.
  2u  2u  2u
 K 


2
2
2
m

x

y

z

F rx




In general, vertical wind shear is much stronger
than horizontal shear for synoptic scale flows so
this reduces to:
Frx
m
 u
2
K
z
2
Fry
m
 v
F rz
z
m
2
K
2
0
Force balance in the atmosphere for a non-rotating earth
expressed in spherical coordinates

dV
  uv tan  uw 
 i 


dt
a
a


Acceleration
  u 2 tan  vw    u 2  v 2
 k
j 


a
a 
a



F


2

1
GM  r 
 V




p



K


2
2
 m

a r
z

Correction for spherical
Coordinate system
gravity
friction
pressure gradient
force
Same statement expressed as 3 orthogonal scalar equations
(East-West equation)
(North-South equation)
(Up-Down equation)
du
uv tan 


uw

1 p
dt
a
a
 x
dv
u tan 
vw
1 p
2

dt
a
dw
u v
dt
2

a


a
2

 y
1 p
 z
 u
2
K
z
2
 v
2
K
z
2

GM
a
2
Pressure gradient
Gravity
Acceleration
Force
Correction terms for earth
Friction
based coordinate system
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
Actual speed due to rotation of earth:
60 N: Speed = 232 m/s
40 N: Speed = 356 m/s
Equator Speed = 465 m/s
A ball whirling on a string experiences a
CENTRIPETAL acceleration toward the axis of
rotation

The acceleration is equal to   r where  is the angular rotation rate
2
A person on the ball whirling on a string
experiences a CENTRIFUGAL acceleration
toward the axis of rotation

The acceleration is equal to  r where  is the angular rotation rate
2
Vectors not to true size!
Standing still on the rotating earth
Outward acceleration
away from Earth’s axis is
2
 r where  is the
Earth’s angular rotation
rate
 
2
86156 . 09 s
Where 86156.09 s is the time it takes for the Earth to rotate
once with respect to a fixed star (a sidereal day)
g  9 . 81 ms
2
2
2

2
1 
2
 R 
s   6378100 m  0 . 034 ms
 86156 . 09

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.





GM  r 
2
* r 
2
gk   2     R  g     R
a r
r
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 must
conserve its angular momentum (MVR)
where M is the parcel mass, V is velocity about
the axis of rotation and R is the distance from
the axis of rotation.
A simple interpretation of the Coriolis effect:
Air moving across the earth’s surface will
try to come to equilibrium at a latitude/altitude
where 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 momentum
than the earth beneath it. It will experience an
“outward” centrifugal acceleration equatorward
Westward: The air has less angular momentum
Than the earth beneath it. It will experience an
“inward” centripetal acceleration poleward.
Poleward: The air will progressively move over
points on the earth with less angular momentum.
Air will accelerate eastward relative to the earth
below it.
Equatorward: The air will progressively move over
points on the earth with greater angular momentum.
Air will accelerate westward relative to the earth
below 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

R
2

u R

u  

2
CEN      R   R  2  u

2
R
R
R


2
Centrifugal force
(combined with
gravity)
2
2
 10 ms
1
 10
86156 s
Magnitude of 2nd term on RHS
3
ms
2
2
2
10 m s
2
 10
5
ms
2
6378100 m
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

R
2


u R

u  

2
CEN      R   R  2  u

2
R
R
R


2
ignore
Centrifugal force
(combined with
gravity)
Coriolis Force
COR  2  u

R
R
Coriolis


accelerati on  motion parallel to latitude circle    j 2  u sin   k 2  u cos 
dv
  2  u sin 
dt
dw
dt
 2  u cos 
Suppose an object is moving equatorward on the rotating earth at speed -v
Apply principle of conservation of momentum
2 
u 
2
R    
R  R 
R  R 

Expand this expression
2 

u   2
2
R    
R

2
R

R


R

R  R 



Since  u and  R are small, neglect terms with their product  u  R
2
2
2

R u
 R   R  2 R R 
R  R
2
2

 R   R  2 R R 

2 R R  
If we assume that  R << R
2
R u
R  R
(From previous slide)
2
R u
R  R
this expression can be approximated as:


2 R R   R u
or
u  2 R
u  2 R
For a displacement in the –y direction,
from Figure:
 u   2     y sin     y 2  sin 
Dividing by  t and taking the limit as t  0
du
dt

dy
dt
2  sin   v 2  sin 
u  2 R
For a displacement in the z direction,
from Figure:
 u   2     z cos      z 2  cos 
Dividing by  t and taking the limit as t  0
du
dt

dz
dt
2  cos    w 2  cos 
Combining the results for east-west and north-south movement of air
du
 v 2  sin   w 2  cos   fv  w 2  cos 
dt
dv
  2  u sin    fu
dt
dw
dt
 2  u cos 
Where: 2  sin   f
and f is the Coriolis parameter
Motion on a rotating earth – the Coriolis Force
In a reference frame on the earth
the Coriolis effect appears as a force acting on an air parcel.
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 horizontal
component at the poles
A formal way of deriving the Coriolis and Centrifugal Forces
In our previous discussion of
spherical coordinates, the
coordinates were fixed in
position.
A
B
At 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.
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




A  Ax i  A y j  Az k
 

where i , j , and k are fixed at local points
The components of the vector in the rotating coordinate system are:




A  A x i   A y j   A z k 




The time derivative of the vector A  A x i  A y j  A z k
in a stationary coordinate system would be:

dA
dA x  dA y  dA z 

i 
j
k
dt
dt
dt
dt




The time derivative of the vector A  A x i   A y j   A z k 
in the rotating coordinate system are:




d A y 
d A d A x 
d A z 
di 
dj 
dk 

i
j 
k   A x
 A y
 A z
dt
dt
dt
dt
dt
dt
dt

dA
dA x  dA y  dA z 

i 
j
k
dt
dt
dt
dt
Stationary
Rotating

dA

di 

dj 

dk 
d A y 
d A x 
d A z 

i
j 
k   A x
 A y
 A z
dt
dt
dt
dt
dt
dt
dt
We can write the equation for the rotating frame of reference as:

dA

dt
The derivatives

dA
dt

di 
dt
 A x
,

di 

dj 
dt
dt
 A y

dj 
dt
 A z

dk 
dt

dk 
, and
dt
, represent the rate of



change of the unit vectors i  , j , and k , that arise because
the coordinate system is rotating
Let’s determine and expression for

di 
Rotation vector,  , points
upward from north pole
dt

di 
The magnitude of
dt
 
By similar triangles  i  i 
is given by:


 d
 
lim   i  
di 

 
 i
 i 
t  0  t 
dt
dt
The direction of

di 
dt

di 
dt
is given by:
 
 i
Let’s determine and expression for

dj 
dt
Rotation vector,  , points
upward from north pole


By similar triangles  j   j 

dj 
The magnitude of
dt
is given by:


 d
 
lim   j  
dj 

 
 j
 j 
t  0  t 
dt
dt
The direction of

dj 
is given by:
dt

dj 
dt
 
   j
Let’s determine and expression for

dk 
Rotation vector,  , points
upward from north pole
dt
The magnitude of

dk 
dt
By similar triangles k   k 
is given by:


 d
 
lim   k  
dk 


 k
 k 


t  0  t 
dt
dt
The direction of

dk 
dt

dk 
is given by:
dt
 
 k
Our previous equation was:

dA


dA
dt
dt
 A x

di 
dt

dj 
 A y
dt
 A z

dk 
dt
and we can write
A x

di 
dt
 





 Ax   i     Ax i 


A z
then

dk 
dt

dA
dt

dA
dt


A y

dj 
dt
 





 A y   j    A y j 
 


 A z   k     A z k 


dA
dt

d A
dt

 A x

di 
dt

 A y

dj 
dt
 A z



dk 
dt




   A x i   A y j   A z k 


dA
dt



d A
dt
 
   A

For a vector A, the coordinate transformation from
a non-rotating to a rotating system is given by:

dA
dt


d A
 
   A
dt
Let’s apply this to two vectors:

Let r 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

dar
dt


dr
 
r
dt

 
Va  V    r
By definition:

Also apply vector transform to V a

d aV a
dt


dVa
dt
 
   Va
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
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

d aV a

dt

dVa
dt
 
   Va

 
Va  V    r
(1)
Put (2) into (1)

d aV a

dt

d aV a
dt

d aV a
dt
d
dt
 

 
V r  V r




 

  
dV
dr


  V      r
dt
dt

dV


  
  V   V      r
dt

d aV a
dt

dV

2
 2  V   r
dt
Coriolis
Force
Centrifugal
Force
(2)

d aV a

dt

i

j
 
2   V  0  2  cos 
u
d aua

du
dt
dt
d a va
dv

dt
dt
d a wa
dw
dt

dt
v
dV

2
 2  V   r
dt

k
 2  sin 
w
 v 2  sin   w 2  cos 
 2  u sin 
 2  u cos 
Equations of motion on non-rotating earth
(East-West equation)
(North-South equation)
(Up-Down equation)
du
uv tan 


uw

1 p
dt
a
a
 x
dv
u tan 
vw
1 p
2

dt
a
dw
u v
dt
2

a


a
2

 y
1 p
 z
 u
2
K
z
2
 v
2
K
z
2

GM
a
2
Pressure gradient
Gravity
Acceleration
Force
Correction terms for earth
Friction
based coordinate system
Equations of motion on rotating earth
Answer
We must add terms to account for the acceleration
required for air to conserve its angular momentum.
In scalar form the equations of motion for each direction become:
(East-West equation)
(North-South equation)
(Up-Down equation)
du

uv tan 

uw

1 p
dt
a
a
 x
dv
u tan 
vw
1 p
2

dt
a
dw
u v
2

dt
a


a
2

 y
1 p
 z
 u
2
K
z
2
 v
 2  v sin   2  w cos 
2
K
z
2
 2  u sin 
 g  2  u cos 
Pressure gradient
Effective
Force
Gravity
Correction terms for
Friction
Coriolis Force
spherical coordinate system
Acceleration
The complete momentum equations on a spherical rotating earth
du

uv tan 

uw

1 p
dt
a
a
 x
dv
u tan 
vw
1 p
2

dt
a
dw
u v
dt
2

a


a
2

 y
1 p
 z
 u
2
K
z
2
 v
 2  v sin   2  w cos 
2
K
z
2
 2  u sin 
 g  2  u cos 
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
Scale Analysis of the horizontal momentum equations
du

uv tan 

uw

1 p
dt
a
a
 x
dv
u tan 
vw
1 p
2

dt
U
a
2
10
10
U  10 ms


2
UW
a
4

a
U
L
p

1

a
5
10
8

 y
1 kg m
3
2
K
z
2
K
z
U
L
2
10
H
3
1
10
 2  u sin 
2
f 0U
7
10
L  10 m
6
 10 m s
3
 2  v sin   2  w cos 
2
 v
p
W  0 . 01 ms
1000 Pa 10 mb 
 u
2
2
f 0W
3
10
6
H  10 m
4
K 1m s
2
1
*numbers in yellow boxes apply to free atmosphere above boundary layer
Scale Analysis of the vertical momentum equations
u v
2
dw

dt
a
UW
U
L
a
10
7
p



5
1
1 p
 g  2  u cos 
 z
p
2
10
U  10 ms
2
g
H

1
10
1
L  10 m
10
W  0 . 01 ms
10 Pa 1000 mb 
f 0u
1
10
3
6
H  10 m
2
g  1 0 ms
4
5

1 kg m
3
 10 m s
5
2
2









 

 
ABOVE THE BOUNDARY LAYER
du

uv tan 

uw

1 P
dt
a
a
 x
dv
u tan 
vw
1 P
2

dt

a
dw
a
u v
2

dt

 y
2

a
 u
2
K
z
2
 v
 2  v sin   2  w cos 
2
K
1 P
 z
z
2
 2  u sin 
 0  g  2  u cos 
WITHIN THE BOUNDARY LAYER
du

uv tan 

uw

1 P
dt
a
a
 x
dv
u tan 
vw
1 P
2

dt

a
dw
dt
a
u v
2

a

 y
2

 u
2
K
z
2
 v
 2  v sin   2  w cos 
2
K
1 P
 z
z
2
 2  u sin 
 0  g  2  u cos 
The Hidden Simplicity of Atmospheric Dynamics:
du

dt
 x
dv
1 P

dt
du

dt
dv
dt
1 P

1

1

0
 y
 fv
 fu
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
P
 u
HORIZONTAL PARCEL ACCELERATIONS
K
 fv
2
x
z
CAN BE UNDERSTOOD BY COMPARING
2
THE MAGNITUDE AND DIRECTION OF THE
P
 v
K
 fu PRESSURE GRADIENT, CORIOLIS AND
2
y
z
FRICTIONAL FORCES
2
1 P
 z
g
THE ATMOSPHERE IS IN HYDROSTATIC
BALANCE – VERTICAL PGF BALANCES
GRAVITY – ON SYNOPTIC SCALES
Note the Total Derivative:
d

dt
dx   
dy  




dt   x  y , z , t dt   y
Note that:
u 
dx
dt
,

dz   
dt   





 
 x , z , t dt   z  x , y , t dt   t  x , y , z
v
dy
,
w
dt
dz
dt
  
  
  
  


 u
 v
 w
 



dt
  x  y , z ,t
  z  x , y ,t   t  x , y , z
  y  x , z ,t
d
   

  
d
  
  


 u 
 v 
 w

 


  t  x , y , z dt
  z  x , y , t 
  y  x , z ,t
   x  y , z , t
The rate of change
of a property (temp)
at a fixed point
(x,y,z)
=
The rate of change
of the property
Following a parcel as
it moves to
the point (x,y,z)
The advection of the property
from upstream to the point
(x, y, z)
Simplified u Momentum Equation expressed in terms of local time derivatives
du
dt

1 P
 x
 fv
 u 
1 P
 u 
 u 
 u 

u
 v 
 w


 fv



 x
  x  y , z ,t
  z  x , y ,t   t  x , y , z
  y  x , z ,t
 u
u
u 



 fv   u
v
w
t
 x
y
z 
 x
u
The rate of change
of west-east wind
at a fixed point
(x,y,z)
1 P
PGF
COR
The advection of the west-east wind
Component from upstream to the point
(x, y, z)
Simplified v Momentum Equation expressed in terms of local time derivatives
dv
dt

1 P
 y
 fu
 v 
1 P
 v 
 v 
 v 

u
 v 
 w


 fu



 y
  x  y , z ,t
  z  x , y ,t   t  x , y , z
  y  x , z ,t
 v
v
v 



 fu   u
v
w
t
 y
y
z 
 x
v
The rate of change
of south-north wind
at a fixed point
(x,y,z)
1 P
PGF
COR
The advection of the south-north wind
Component 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
The simplified horizontal momentum equations in height coordinates are given by:
du

dt
dv

dt
or more succinctly:

dV
dt

1

1 P
 x
1 P
 y
 fv
 fu
 
 p  fk  V
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
dp
on a constant pressure surface
Expanding…
 dp  p
 p 
 p 
 p 



dx

dy



 dz p
p
p


 x  y,z
 z  x, y
 y  x,z
(Subscripts indicate differentiation carried out holding subscripted variable constant)
 p 
 p 
 p 
 dy p  
0
 dx p  
 dz p
 x  y,z
 z  x, y
 y  x,z
Now expand dz as a function of x and y
 p 
 z 
 p 
  p     z 


 dy p
0
dy p  
 dx p  


 dx p  

 x  y,z
  z  x , y    x  y , p
 y  x,z
 y  x, p




 p 
 z 
 p 
  p     z 



 dy p
0
dx

dy

dx






p
p
p



 x  y,z
 z  x, y   x  y , p
 y  x,z
 y  x, p




Rearrange:
  p 
  p 
 p   z  
 p   z


0  

dx





 
  p

 


   x  y , z   z  x , y   x  y , p 
   y  x , z   z  x , y   y
 
  dy p
 x , p 
For this statement to be true for all dx p and dy p , the statements in
each square bracket must equal zero
 p 
 p   z 





 

 x  y ,z
 z  x, y  x  y , p
 p

 y

 p   z
   
 
 z  x, y  y
 x,z


 x, p
 p 
 p   z 





 

 x  y ,z
 z  x, y  x  y , p
Use hydrostatic equation
 p 
 z 




g



 x  y ,z
 x  y , p
 p

 y
P
z

 p   z
   
 
 z  x, y  y
 x,z


 x, p
  g
 p
 
 y

 z
    g 
 x,z
 y


 x, p
Divide both sides by 
1  p 
 z 
 


g



  x  y ,z

x

 y, p

1  p

  y

 z 
   g 

 x,z
 y  x, p
or
1  p 
  
 





  x  y ,z

x

 y, p
where
  gz

1  p

  y

  
   

 x,z
 y  x, p
is called the geopotential height
1  p 
  
 





  x  y ,z
 x  y , p


1

1  p

  y

  
   

 x,z
 y  x, p
 p y , z    y , p
We will now drop the use of subscripts and write the
momentum equations in p coordinates
Pressure coordinates
In pressure coordinates, the horizontal momentum equations (above the
boundary layer) become:
du

dt

x
 fv
dv
dt


y
 fu

dV
dt
 
  p   f k  V
where , the geopotential
height, is given by  = gz
and f = 2sin
and the vertical momentum equation becomes

p

RdT
p
(using the ideal gas law)
The Total Derivative in pressure coordinates
d

dt
dx   
dy  




dt   x  y , p , t dt   y
Note that:
u 

dp  



 x , p , t dt   p
dx
,
dt
v

dt   


 
 x , y , t dt   t  x , p , z
dy
dt
Note that the vertical velocity in pressure coordinates has a different form:
 
dp
dt
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:
d

dt
dx   
dy  




dt   x  y , p , t dt   y
Note that:
u 

dp  



 x , p , t dt   p
dx
dt
,
v

dt   


 
 x , y , t dt   t  x , p , z
dy
dt
  
 
  



 u

v





dt
  x  y , p ,t
  y  x , p ,t
 p
d

  

 
 x , y ,t   t  x , y , p
   
  
 
d
  


 u 
 v 
  
 

dt
 t  x, y, p
   x  y , p , t
  y  x , p ,t
 p
The rate of change
of a property (temp)
at a fixed point
(x,y,z)
=
The rate of change
of the property
Following a parcel as
it moves to
the point (x,y,z)




 x , y , t 
The advection of the property
from upstream to the point
(x, y, z)
Simplified u Momentum Equation in pressure coordinates
expressed in terms of local time derivatives
du

dt

x
 fv

 u
u
u 



 fv   u
v

t
x
y
p 
 x
u
The rate of change
of west-east wind
at a fixed point
(x,y,z)
PGF
COR
The advection of the west-east wind
Component from upstream to the point
(x, y, z)
Simplified v Momentum Equation expressed in terms of local time derivatives
dv
dt


y
 fu

 v
v
v 



 fu   u
v

t
y
y
p 
 x
v
The rate of change
of south-north wind
at a fixed point
(x,y,z)
PGF
COR
The advection of the south-north wind
Component from upstream to the point
(x, y, z)