Change of Coordinates (non-orthogonal)
Different vertical coordinates
Suppose we have a property S(x, y, z, t) and want to express it as S(x, y,  , t) in terms
of a different vertical coordinate  =  (x, y, z, t)  e.g., pressure, so that we look at the
temperature vs. latitude and longitude on the 500mb surface or the 750mb surface. What
is the relationship between derivatives like the rate of change with x along a horizontal line
 S 




 x  z
and the rate of change with horizontal distance along a constant  surface
 S

 x

Let us look at this graphically:
The derivatives in question are
S  S1
 S 

  3
x 2  x1
 x  z
,
S  S1
 S 

  2
x 2  x1
 x 
We can relate these two by using the vertical changes
 S 
  z 2  z1  
 z  x
S2  S3  
 S

 

  2   1 
x
Using this to eliminate S 3 from the rate of change along a horizontal surface gives
S  S1  S   2  1
 S 



  2
x 2  x 1   x x 2  x 1
 x z
 S

 x
 S

 
 


 z 2  z1  2  1

 x x z z
x
2
1 2
1
 S

 x
 S

 
 


  z 

  x 

x 
1
 z

 




x




?
Likewise
 S 
 S 


  
 z x
  x
Thus, to change coordinates we replace
 S

 x
with similar forms for
S
y
and
S
x

 S
 
z
 x
S
t
 z

 


x
by
 S

  
  
 z

 x





x  z
 



x
; the vertical replacement is
 S

 
 S 

 
 z
 z x

 


x


x
General coordinate change
There is a fairly straightforward mathematical procedure for changing coordinates
from one system to another, even if the second is nor orthogonal. Suppose we have a
function S(x) and wish to express it and its derivatives as functions of the new coordinates
ξ. We could use the chain rule to find
j S
S

xi
x i j
(1)
But this may not be adequate, for the following reason. We wish to have coefficients in the
final equations expressed as functions of the new coordinates; however, quantities such as
 1
x 3
are more likely to be known as functions of x.
To accomplish the goal of having all terms expressed in the new coordinates, we begin
with the opposite form
x j S
S

or  x S   S
(2)
i
 i x j
and assume that the
x i
i
terms are functions of ξ. We can express derivatives in the old
coordinate system in terms of derivates in the new system by inverting the transformation
matrix:
 x
S
 i
x i
  j



1
S
 j
or  S  
2
1
xS
(3)
In terms of the Jacobian matrix
 A

 1

 B
( A , B ,C )
 det 1

( 1 , 2 , 3 )

 C
 
 1

A 

 3 

B 

 3 

C 
 3 


A
 2
B
 2
C
 2
We have
 (S, x 2 , x 3 )
 (S, x 2 , x 3 )
S


x1
 ( x1 , x 2 , x 3 )
 (1 ,  2 ,  3 )
 ( x1 , x 2 , x 3 )
(1 ,  2 ,  3 )
etc.
Example
If we take polar coordinates as a specific case, we have the relationship between the
old and new coordinates
x  r cos 
y  r sin 
z  z
x j
So that the transformation matrix matrix Tij 
i
 cos 

T    r sin 

0

in (2) is
sin 
r cos 
0
0

0
1 
1
sin 
r
1
cos 
r
0

0

0

1


The inverse is
T 1

 cos 

  sin 

 0



So that
1
sin   
r
1
 sin   r  cos   
r
  z
 x  cos   r 
y
z
using subscript notation for derivatives.
3
Change in vertical coordinate
If we switch from x, y, z to x  , y  ,ξ , the transformation matrix is

1 0



T  0 1


0 0


z 

x  

z 
y  

z 
 


1


 0

0


z 

 
z 

 
z

1
 
and its inverse is
T 1
z
x 
z

y

0
1
0
Thus we can replace horizontal gradients

z 
z 
vertical derivatives

1 

z
z 
And time derivatives
z 



 t
t
t
z 
in our original equations.
First, we note that the material derivative becomes
D

1


 u  
( w  z t  u  z )
Dt
t
z

and we can define the “vertical” velocity  as
 
1
( w  z t  u  z )
z
so that the material derivative becomes
D



 u  
Dt
t

With this definition, we note that  
D
z as we might expect.
Dt
4
Transformed equations
The horizontal momentum equations become
D
1
u  fk̂  u     
Dt

(e.1)
with   gz begin the geopotential; the hydrostatic balance is

1 
 


 
(e.2a)
while the conservation of mass gives
1 D
1
1 
D
  u 
u   z 
(
z) 0
 Dt
z
z  Dt
implying
1 D
1 D


z    u 

 Dt
z Dt

or
1 D

p    u 
 0
p Dt

(e.3)
Finally, the thermodynamic equation becomes
D
1 D
 2
p0
Dt
c s Dt
(e.4a)
in general. The potential vorticity (with  being the entropy) is
q
g
(  3  u  fk̂ )   3
p
with the  3 notation indicating the vertical derivatives are included.
5
(e.5)
Vertical coordinate function of pressure
When the vertical coordinate is a function of pressure    ( p ) or p  p(  ) , we can
define and p   gpc   simplify the equations to
D
u  fk̂  u  
(p.1)
Dt


  g c b


u 
gp
D
   2c  0
Dt
cs
or
(p.2)
1 
( pc )  0
pc 

 c
g 2  c2
D
b    g

Dt

 2 c s2


(p.3)

 0


The last equation can also be written
 g c  
g 2  c2 

b  u  b    

0
t
2
 2 c s2 

or

b  u  b  S  0
t
(p.4)
with the stratification parameter S
S 
 c
 c2 2
b2
N  b  b

c
2
cs2
(p.5a)
defined in terms of the Brunt-Väisälä frequency
N 2  g

1 
g2
g2
g2
g 2  c
g2
  2  g


b 

 z
c
b c
cs
c s2
b2
c s2
(p.5b)
The PV is
q
1
c
3  u  fk̂  3
(p.6)
We shall use eqns. p1-p4 as our basic set.
The boundary conditions are a bit tricky; if the bottom is at z  hx , y  , we get an
implicit equation for the surface pressure  s x , y , t  :
 x , y , s x , y , t , t   ghx , y 
(b.1)
We also have the kinematic condition


 
D 
D
z 
h
Dt 
Dt

D
  gh  0
Dt
Together, these two imply

D
 s at    s
Dt
6
(b.2)
Linearized equations
The wave equations for this system are

u  fk̂  u   
t

   b

1 
    0
u 
 c  c

b   S  0
t
If we make the particular choice of c   , so that ξ
is just the height in the resting
atmosphere, we have b  g , S  N 2 , and the equations look like the Boussinesq form
except for the  factors in the stretching term. We can separate variables
u  u x , t F z 
,
    x , t F z 
b  bx , t 
,
F

,
 
  1 F
t N 2 
The mass conservation equation gives

   1   
F     uF  0

t    N 2  
giving again the vertical structure eigenvalue equation
1   
1
F 
F
gH e
  N 2 
and the horizontal equations

u  fk̂  u   
t
1  
 u  0
gH e t
The lower boundary condition gives the surface pressure

 s   x ,0 , t   0


s  
1
 x , t F 0 
g
and its evolution
 s
  1 F
  x ,0, t   
t
t N 2 
Often, however, the simpler condition 
0

7

F
N2

F

g
F
0

is used.
at ξ = 0
Isothermal atmosphere
One case that can be worked out completely is the isothermal basic state. Using the
gas law gives p   RT ; the hydrostatic equation then gives
p  p0 exp z H s 
  0 exp z H s 
,
H s  RT g
,
,
p0  0 gH s
__ the density decays exponentially with a scale height Hs. We can just choose ξ =
Hs ln(p0 / p) so that it’s the same as height. The associated density c   as before. When
we calculate the Brunt-Väisälä frequency, we get
N2 
c 
g
g2
g 
g R


1 v  

2
Hs
H
c
H
cs

s 
p 
s cp

and discover that it is constant. The vertical structure equation becomes
2F
 2

1 F
1
R

F
H s 
HsHe cp
Therefore F will have exponential solutions
F  expz H s 
2  
,
Hs R
0
He c p
,
 
1
1

2
2
1 4
R Hs
c p He
If we start with the case when the argument of the square root is positive, we must eliminate
the large root, since it has an energy density u 2 ~ exp2  1 H s  which grows towards
infinity. Therefore we can only accept the negative sign, giving

F  exp 1 


1 4
R  


c p H e   2H s

The lower boundary condition gives (for ω = 0)
 0

1
 0 , F 1
gH e
or for the full condition
 
Hs N 2
R

g
cp

He 
cp
Hs

Hs
R
cv
1
cp
,
 R 
F  exp
 cp Hs





which will be well-behaved as long as c p  2 R (for the atmosphere
cv , c p , R  718, 1005, 287.1 J kg K o
(Tsonis, An Introduction to Atmospheric Thermodynamics) so that this condition is fine.
The equivalent depth is 40% larger than the scale height. Over one scale height, F
Grows by a factor of exp R c p  1.33 while the kinetic energy density decreases by
 R

exp 2
 1   0.65 . This is the equivalent barotropic mode.
 c

Arep there
 any other modes? The derivation above makes it clear that this is the only mode
with H e  4 R c p H s  1.14H s . What about the modes with complex α
which have energies remaining order one at infinity? The lower boundary condition clearly
requires both the α+ and α- modes; however, the latter will have downward energy flux. To
maintain such a mode, we require a reflecting surface or an energy source
high in the atmosphere. This will not happen for a resting atmosphere; therefore, the only mode
available is the equivalent barotropic mode.




8
Thermodynamics
For an ideal gas, we can simply the thermodynamic using   c p ln 
D
 0
Dt
(p.7)
with the potential temperature being
  0
0

1r
 p 


 p0 
Thus, the buoyancy becomes
bg
c

 p 


 p0 
1 r

 G  
0
(p.8)
With a little work, you can substitute (p.8) into (p.4), using cs2  p  to show that (p.7)
holds. The Brunt-Väisälä frequency is
N2  g

 
ln   g
ln 
z
 c 
,
S  g
c 
ln 
 
Quasigeostrophic form
We start with the momentum equations

 


1


u  fkˆ  3  u  u  kˆ    u  u   bkˆ
t
2


take the curl and look at the vertical component
 
 

  f    2  u    2    f
 u  
 
 t

1
c
The QG form of this is


1 
 

 ug    g  f  f
 

 c  c
 t

with
u g  kˆ  
 g   2
,
,
  f
For the thermodynamics, we recognize that
 c2
2
N2 
 c2  
  
2
9
N 2    S  

 c

so that
S
 

 u g        0

f
 t

Combining these gives

Q  J  , Q   0
t
with the QG PV being
Q   2 
1
c
 c f

S
2

  f

For the atmospheric case, we can use the potential temperature equation times G(ξ) to
write
S  G  

  

and show, from the definition of G  g c  , that
S  g c
1


 
 c2

2
N 2  
as before.
Examples
In this chart, P0 and ρ0 are reference values; the scale height is related to these two by
gH  RT0  R 0  p0  0 .
10
Summarizing
In the atmosphere, we would usually use pressure coordinates
    G  kˆ
D
u  fkˆ  u
Dt
u p
D

Dt
q
0
0


  g  3  u  fkˆ   3 ln 
 f2 
Q   2 
  f
 S 
G
R
p0
 p 


 p0 
1 r
S 
,
1 
ln 
 
but will find the log form convenient if we work with an isothermal stratification so that
  c  0 exp  H 
D

u  fkˆ  u     g kˆ
Dt

 
1 
  0
  u  

H 
 
D
 0
Dt
1
q
 3  u  fkˆ   3 ln 



f
Q   2 
   0 e 
H
,
N2 
2
N2
 
1  



  f
H  
 
g  1
H 
,
   0 e  1 H
For the ocean, we usually use  p0  p  p0 g and ignore the difference between S and
N2
D
u  fkˆ  u
Dt
  u  
D
b
Dt
q
   bkˆ
0
0

1
0
3  u  fkˆ  3b
Q   2 
bb
g2
c s2

,
11
 f2 
  f
 N 2 
S  b 
g2
c s2