AOSS 321, Winter 2009
Earth System Dynamics
Lecture 11
2/12/2009
Christiane Jablonowski
cjablono@umich.edu
734-763-6238
Eric Hetland
ehetland@umich.edu
734-615-3177
Today’s lecture
• Derivation of the potential temperature equation
(Poisson equation)
• Dry adiabatic lapse rate
• Static stability, buoyancy oscillations
• Derivation of the Brunt-Väisälä frequency
Thermodynamic equation
(Divide by T)
DT
Dp
cp

J
Dt
Dt
c p DT  Dp J


T Dt T Dt T
Use equation of state
(ideal
gas law)

Rd

T
p
Thermodynamic equation
For conservative motions
(no heating, dry adiabatic: J = 0):
c p DT Rd Dp

0
T Dt
p Dt
D(ln T)
D(ln p)
cp
 Rd
0
Dt
Dt

Derivation of Poisson’s Equation (1)
(integrate
over Dt)
D(ln T)
D(ln p)
cp
 Rd
Dt
Dt
 c p D(ln T)  Rd D(ln p)
p
T
(integrate)
cp
 D(ln T)  R  D(ln p)
d
T0


p0
T 
 p 
c p ln   Rd ln  
T0 
p0 
p0 
T0 
 c p ln   Rd ln  
T 
 p 
Derivation of Poisson’s Equation (2)
p0 
T0 
c p ln   Rd ln  
T 
 p 
p0 
T0 
ln    ln  
T 
 p 
Rd
cp
Rd
cp
p0 
T0  T    
 p 
Poisson’s Equation
 is called potential temperature!
Definition of the
potential temperature 
Rd
cp
p0 
  T  
 p 
with p0 usually taken to be constant with
p0 = 1000 hPa
The potential temperature is the temperature a
parcel would have if it was moved from some
pressure level and temperature down to the surface.

Definition of potential temperature
Rd
cp
p0 
  T  
 p 
Does it makes sense that the temperature T
would change in this problem? We did it
adiabatically. There was no source and sink
of energy.

Annual mean zonal mean temperature T
(hPa)
P 1
r
e
s
s 10
u
r
e
100
1000
Kelvin
260
230
220
260
North Pole
200
300
Equator
210
260
South Pole
Source: ECMWF, ERA40
Annual mean zonal mean
potential temperature 
(hPa)
Kelvin
100
P
r
e
s
s
u
r
e
350
330
285
285
300
1000
North Pole
Equator
South Pole
How does the temperature field look?
Source: ECMWF, ERA40
Dry adiabatic lapse rate
• For a dry adiabatic, hydrostatic atmosphere the
potential temperature  does not vary in the
vertical direction:

0
z
• In a dry adiabatic, hydrostatic atmosphere the
temperature T must therefore decrease with
height.

Class exercise
500 hPa
T?, ?
850 hPa
T = 17 ºC, ?
• An air parcel that has a temperature of 17 ºC at
the 850 hPa pressure level is lifted dry
adiabatically.
What is the temperature and density of the parcel
when it reaches the 500 hPa level?
Dry adiabatic lapse rate: Derivation
Start with Poisson equation:
Rd
cp
p0 
  T  
 p 
Take the logarithm of :
Rd
ln   ln T +
ln p0  ln p
cp
Differentiate with respect
to height

 ln   ln T R d  ln p0  ln p 





z
z
c p  z
z 

1  1 T R d p



 z T z c p p z
= 0 (p0 constant)
Dry adiabatic lapse rate
Use hydrostatic equation
1  1 T R d g


 z T z
cp p
Plug in ideal gas law for p, then multiply by T:

T  T g


 z z c p

0
For dry adiabatic, hydrostatic atmosphere with
z

T g


 d
z c p
d: dry adiabatic lapse rate (approx.9.8 K/km)
Static stability
• We will now assess the static stability characteristics
of the atmosphere.
• Static stability of the environment can be measured
with the buoyancy frequency N.
• N is also called Brunt-Väisälä frequency.
The square of this buoyancy frequency is defined as
g 
N 
 z
2
We will derive this equation momentarily, but first let’s
discuss some static stability/instability conditions.
Static stability
We will now assess the static stability characteristics
of the atmosphere.

 0 statically stable
z

 0 statically neutral
z

 0 statically unstable
z
Stable and unstable situations
Check out this marble in a bowl:
http://eo.ucar.edu/webweather/stablebowl.html
QuickTime™ and a
TIFF (Uncomp resse d) de com press or
are nee ded to s ee this picture.
Stable and unstable air masses
Stable air:
A rising parcel that is cooler than the surrounding
atmosphere will tend to sink back to its original
position (why?).
Unstable air:
A rising parcel that is warmer than the surrounding
atmosphere will continue to rise (why?).
Neutral air:
The parcel remains at the new location after being
displaced, its temperature varies exactly as the
temperature of the surrounding atmosphere.
Unstable air
Unstable air: makes thunderstorms possible.
Here: visible since clouds rise to high elevations!
Stable air
Stable air: makes oscillations (waves) in the
atmosphere possible, visible due to the clouds!
Stable air
Stable air: makes oscillations (waves) in the
atmosphere possible, what is the wave length?
Stable air
Stable air: temperature inversions suppress rising
motions. Here: stratiform clouds have formed.
Let’s take a closer look:
Temperature as function of height
z
Cooler
z
- ∂T/∂z is
defined as
lapse rate
T
Warmer
Let’s take a closer look:
Temperature as function of height
z
Cooler
z
- ∂T/∂z is
defined as
lapse rate
T
Warmer
Let’s take a closer look:
Temperature as function of height
z
Cooler
z
- ∂T/∂z is
defined as
lapse rate
T
Warmer
Let’s take a closer look:
Temperature as function of height
z
Cooler
z
- ∂T/∂z is
defined as
lapse rate
T
Warmer
The parcel method
• We are going displace this parcel – move it up
and down.
– We are going to assume that the pressure
adjusts instantaneously; that is, the parcel
assumes the pressure of altitude to which it is
displaced.
– As the parcel is moved its temperature will
change according to the adiabatic lapse rate.
That is, the motion is without the addition or
subtraction of energy. J is zero in the
thermodynamic equation.
Parcel cooler than environment
z
Cooler
If the parcel moves
up and finds itself
cooler than the
environment then it
will sink. (What is
its density? larger
or smaller?)
Warmer
Parcel cooler than environment
z
Cooler
If the parcel moves
up and finds itself
cooler than the
environment then it
will sink. (What is
its density? larger
or smaller?)
Warmer
Parcel warmer than environment
z
Cooler
If the parcel moves
up and finds itself
warmer than the
environment then it
will go up some
more. (What is its
density? larger or
smaller?)
Warmer
Parcel warmer than environment
z
Cooler
If the parcel moves
up and finds itself
warmer than the
environment then it
will go up some
more. (What is its
density? larger or
smaller?)
Warmer
This is our first example of “instability” – a perturbation that grows.
Let’s quantify this:
Characteristics of the environment
We assume that the temperature Tenv of the
environment changes with a constant linear slope
(or lapse rate ) in the vertical direction.
 Tenv 
Tenv  Tsfc  z with constant    lapse rate
 z 
So if we go from z to z  z,
then the change in Tenv of the environment is
Tenv  Tsfc  (z  z)  (Tsfc  z)  z
Tsfc: temperature at the surface
Let’s quantify this:
Characteristics of the parcel
We assume that the temperature Tparcel of the parcel
changes with the dry adiabatic lapse rate d.
So if we go from z to z  z,
then the temperature change T of the parcel is
Tparcel  Tparcel,z  d z  Tparcel,z  d z
g
d 
 dry adiabatic lapse rate
cp
Stable:
Temperature of parcel cooler than
environment
Tparcel  Tenv
parcel
environment

  d
environment
parcel
compare the lapse rates
Unstable:
Temperature of parcel greater than
environment.
Tparcel  Tenv
parcel
environment

  d
environment
parcel
compare the lapse rates
Stability criteria from physical
argument
  d unstable
  d neutral
  d stable
Lapse rate of
the environment
Dry adiabatic lapse rate
of the parcel
Hydrostatic balance
environment in hydrostatic balance
1 penv
0
g
env z
But our parcel experiences an
acceleration, small displacement z
Dw D z
1 penv


g
2
Dt
Dt
 parcel z
2
Assumption of immediate adjustment of pressure.
p parcel penv

z
z
Solve for pressure gradient
environment in hydrostatic balance
1 penv
0
g
 env z
penv

 g env
z
But our parcel experiences an
acceleration
Dw D2z
1 penv


g
2
Dt
Dt
 parcel z
D2z
1
g env

g
genv   g 
2
Dt
 parcel
 parcel
 

D2z
env
use ideal gas law:

g
1

2


Dt
 parcel 
 p R T
  T

D2z
d parcel
parcel
env
 g
1 g
1


2


Dt


Rd Tenv  p parcel  
  Tenv
Recall: penv  p parcel
Rearrange:
Tparcel 
D z
 g
1
2
Dt
 Tenv

g

Tparcel  Tenv 

Tenv
g Tparcel Tenv 



 z
Tenv  z
z 
g  Tenv   Tparcel 

 

z
Tenv  z   z 
2

Back to our definitions of temperature
change
Tparcel
Tenv
Recall :   
and d  
z
z
D2z
g  Tenv   Tparcel 

 

z
2
Dt
Tenv  z   z 
D2z
g


(  d ) z
2
Dt
Tenv
D z
g

(d  )z  0
2
Dt
Tenv
2
Second-order, ordinary
differential equation:

Recall: Dry adiabatic lapse rate
Taking the logarithm of , differentiating with respect
to height, using the ideal gas law and hydrostatic
equation gives:
g Tenv Tenv 
d    

cp
z
 z
Tparcel
g
d 

cp
z
dry adiabatic lapse rate
(approx. 9.8 K/km)
Rearrange:
D2z
g

(d  ) z  0
2
Dt
Tenv
D2z
g Tenv 

z  0
2
Dt
Tenv  z
D2z
g 

z  0
2
Dt
 z
D2z
2

N
z  0
2
Dt
1
2
g  
With the Brunt-Väisälä frequency N  

 z 



Buoyancy oscillations
D z
g    ln  
2
 N z  0 with N  
  g

2
 z   z 
Dt
2
1
2
1
2

1)
 0,N 2  0 : stable, the solution to this equation
z
describes
a buoyancy oscillation with period 2/N

2)
 0,N 2  0 : unstable, corresponds to growing
z
perturbation, this is an instability
3)   0,N 2  0 : neutral
z
Solution to the differential equation
D z
2
 N z  0
2
Dt
2
The general solution can be expressed via
the exponential function with a complex argument:

z  Aexp(iNt)
with A: amplitude, N: buoyancy frequency, i 
If N2 > 0 the parcel will oscillate about its initial
 a period  = 2/N.
level with
Average N in the troposphere N ≈ 0.01 s-1

1
Remember Euler’s formula
exp(ix)  cos(x)  isin( x)
with x: real number
Physical solution:
z  Re(Aexp(iNt))
If N2 > 0 (real) the
solution is a wave with
period  = 2/N (more on
waves in AOSS 401)
Stable solution:
Parcel cooler than environment
z
Cooler
If the parcel moves
up and finds itself
cooler than the
environment then it
will sink (and rise
again). This is a
buoyancy oscillation.
Warmer
Stable and unstable air masses
Picture an invisible box of air (an air parcel). If we
compare the temperature of this air parcel to the
temperature of air surrounding it, we can tell if it is
stable (likely to remain in place) or unstable (likely to
move).
http://eo.ucar.edu/webweather/stable.html
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Temperature soundings
z
Sounding of
the parcel
Sounding of
the
environment:
T inversion