AOSS 321, Winter 2009 Earth System Dynamics

Lecture 112/12/2009

Christiane Jablonowski Eric Hetland

cjablono@umich.edu ehetland@umich.edu

734-763-6238 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

€

c p

DT

Dt−α

Dp

Dt= J

c p

T

DT

Dt−

α

T

Dp

Dt=

J

T(Divide by T)

€

αT

=Rd

pUse equation of state(ideal gas law)

Thermodynamic equation

€

c p

T

DT

Dt−

Rd

p

Dp

Dt= 0

c p

D(lnT)

Dt− Rd

D(ln p)

Dt= 0

For conservative motions (no heating, dry adiabatic: J = 0):

Derivation of Poisson’s Equation (1)

€

c p

D(lnT)

Dt= Rd

D(ln p)

Dt⇔ c pD(lnT) = Rd D(ln p)

c p D(lnT) = Rd D(ln p)p0

p

∫T0

T

∫

⇔ c p lnT

T0

⎛

⎝ ⎜

⎞

⎠ ⎟= Rd ln

p

p0

⎛

⎝ ⎜

⎞

⎠ ⎟

⇔ − c p lnT0

T

⎛

⎝ ⎜

⎞

⎠ ⎟= −Rd ln

p0

p

⎛

⎝ ⎜

⎞

⎠ ⎟

(integrateover Dt)

(integrate)

€

c p lnT0

T

⎛

⎝ ⎜

⎞

⎠ ⎟= Rd ln

p0

p

⎛

⎝ ⎜

⎞

⎠ ⎟

lnT0

T

⎛

⎝ ⎜

⎞

⎠ ⎟c p

= lnp0

p

⎛

⎝ ⎜

⎞

⎠ ⎟

Rd

T0 = T p0

p

⎛

⎝ ⎜

⎞

⎠ ⎟

Rd

c p

≡ Θ

is called potential temperature!

Derivation of Poisson’s Equation (2)

Poisson’s Equation

Definition of the potential temperature

€

=T p0

p

⎛

⎝ ⎜

⎞

⎠ ⎟

Rd

c p

with p0 usually taken to be constant withp0 = 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

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.€

=T p0

p

⎛

⎝ ⎜

⎞

⎠ ⎟

Rd

c p

Annual mean zonal mean temperature T

300

200

230

210220

Source: ECMWF, ERA40

260

260 260

Equator South PoleNorth Pole

Pressure

(hPa)

100

1000

1

10

Kelvin

Annual mean zonal mean potential temperature

300

330

350

285285

Source: ECMWF, ERA40

Equator South PoleNorth Pole

Pressure

(hPa)

100

1000

Kelvin

How does the temperature field look?

• For a dry adiabatic, hydrostatic atmosphere the potential temperature does not vary in the vertical direction:

• In a dry adiabatic, hydrostatic atmosphere the temperature T must therefore decrease with height.

Dry adiabatic lapse rate

€

∂∂z

= 0

• 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?

Class exercise

850 hPa

500 hPa

T = 17 ºC, ?

T?, ?

Dry adiabatic lapse rate: Derivation

Start with Poisson equation:

Take the logarithm of :

Differentiate with respect to height€

=T p0

p

⎛

⎝ ⎜

⎞

⎠ ⎟

Rd

c p

€

lnθ = lnT +Rd

cp

ln p0 − ln p( )

€

∂ lnθ

∂z=

∂ lnT

∂z+

Rd

cp

∂ ln p0

∂z−

∂ ln p

∂z

⎛ ⎝ ⎜

⎞ ⎠ ⎟

⇔1

θ

∂θ

∂z=

1

T

∂T

∂z−

Rd

cp p

∂p

∂z = 0 (p0 constant)

Dry adiabatic lapse rateUse hydrostatic equation

Plug in ideal gas law for p, then multiply by T:

For dry adiabatic, hydrostatic atmosphere with

€

T

θ

∂θ

∂z=

∂T

∂z+

g

c p

€

−∂T

∂z=

g

c p

≡ Γd

d: dry adiabatic lapse rate (approx. 9.8 K/km)

€

1

θ

∂θ

∂z=

1

T

∂T

∂z+

Rdρg

cp p

€

∂∂z

= 0

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

€

N 2 =g

Θ

∂Θ

∂z

We will derive this equation momentarily, but first let’sdiscuss some static stability/instability conditions.

Static stabilityWe will now assess the static stability characteristics of the atmosphere.

€

∂Θ∂z

> 0 statically stable

∂Θ

∂z= 0 statically neutral

∂Θ

∂z< 0 statically unstable

Stable and unstable situations

Check out this marble in a bowl:http://eo.ucar.edu/webweather/stablebowl.html

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see 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

Warmer

Coolerz

T

- ∂T/∂z is defined as lapse rate

Let’s take a closer look: Temperature as function of height

z

Warmer

Coolerz

T

- ∂T/∂z is defined as lapse rate

Let’s take a closer look: Temperature as function of height

z

Warmer

Coolerz

T

- ∂T/∂z is defined as lapse rate

Let’s take a closer look: Temperature as function of height

z

Warmer

Coolerz

T

- ∂T/∂z is defined as lapse rate

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 environmentz

Warmer

Cooler

If the parcel moves up and finds itself cooler than the environment then it will sink. (What is its density? larger or smaller?)

Parcel cooler than environmentz

Warmer

Cooler

If the parcel moves up and finds itself cooler than the environment then it will sink. (What is its density? larger or smaller?)

Parcel warmer than environmentz

Warmer

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?)

Parcel warmer than environmentz

Warmer

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?)

This is our first example of “instability” – a perturbation that grows.

Let’s quantify this:Characteristics of the environment

€

Tenv = Tsfc − Γz with constant Γ = -∂Tenv

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟ lapse rate

So if we go from z to z + Δz,

then the change in Tenv of the environment is

ΔTenv = Tsfc − Γ(z + Δz) − (Tsfc − Γz) = −ΓΔz

We assume that the temperature Tenv of the environment changes with a constant linear slope (or lapse rate ) in the vertical direction.

Tsfc: temperature at the surface

Let’s quantify this:Characteristics of the parcel

€

So if we go from z to z + Δz,

then the temperature change ΔT of the parcel is

ΔTparcel = Tparcel ,z − ΓdΔz − Tparcel ,z = −ΓdΔz

Γd =g

c p

≡ dry adiabatic lapse rate

We assume that the temperature Tparcel of the parcel changes with the dry adiabatic lapse rate d.

Stable:Temperature of parcel cooler than

environment

€

Tparcel < Tenv

⇔

Γ < Γd

environmentparcel

environment parcel

compare the lapse rates

Unstable: Temperature of parcel greater than

environment.

€

Tparcel > Tenv

⇔

Γ > Γd

environmentparcel

environment parcel

compare the lapse rates

Stability criteria from physical argument

€

>d unstable

Γ = Γd neutral

Γ < Γd stable

Lapse rate of the environment

Dry adiabatic lapse rate of the parcel

Hydrostatic balance

€

environment in hydrostatic balance

0 = −1

ρ env

∂penv

∂z− g

But our parcel experiences an acceleration, small displacement z

€

Dw

Dt=

D2δz

Dt 2= −

1

ρ parcel

∂penv

∂z− g

Assumption of immediate adjustment of pressure.

€

∂pparcel

∂z=

∂penv

∂z

Solve for pressure gradient

€

environment in hydrostatic balance

0 = −1

ρ env

∂penv

∂z− g

⇔ ∂penv

∂z= −gρ env

But our parcel experiences an acceleration

€

Dw

Dt=

D2δz

Dt 2=

−1

ρ parcel

∂penv

∂z− g

D2δz

Dt 2=

−1

ρ parcel

−gρ env( ) − g =gρ env

ρ parcel

− g

D2δz

Dt 2= g

ρ env

ρ parcel

−1 ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

D2δz

Dt 2= g

penv

RdTenv

⎛

⎝ ⎜

⎞

⎠ ⎟RdTparcel

pparcel

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟−1

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥= g

Tparcel

Tenv

−1 ⎛

⎝ ⎜

⎞

⎠ ⎟

use ideal gas law:

€

penv = pparcelRecall:

Rearrange:

€

D2δz

Dt 2= g

Tparcel

Tenv

−1 ⎛

⎝ ⎜

⎞

⎠ ⎟

=g

Tenv

Tparcel − Tenv( )

=g

Tenv

δTparcel

δz−

δTenv

δz

⎛

⎝ ⎜

⎞

⎠ ⎟ δz

=g

Tenv

−δTenv

δz

⎛

⎝ ⎜

⎞

⎠ ⎟− −

δTparcel

δz

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥ δz

Back to our definitions of temperature change

€

Recall : Γ = −∂Tenv

∂z and Γd = −

∂Tparcel

∂z

€

D2δz

Dt 2+

g

Tenv

(Γd − Γ)δz = 0Second-order, ordinary differential equation:

€

D2δz

Dt 2=

g

Tenv

−δTenv

δz

⎛

⎝ ⎜

⎞

⎠ ⎟− −

δTparcel

δz

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥ δz

⇔D2δz

Dt 2=

g

Tenv

(Γ − Γd ) δz

Recall: Dry adiabatic lapse rate

Taking the logarithm of , differentiating with respect to height, using the ideal gas law and hydrostatic equation gives:

€

d − Γ =g

c p

+∂Tenv

∂z=

Tenv

θ

∂θ

∂z

€

d =g

c p

= −∂Tparcel

∂zdry adiabatic lapse rate (approx. 9.8 K/km)

Rearrange:

€

D2δz

Dt 2+

g

Tenv

(Γd − Γ) δz = 0

D2δz

Dt 2+

g

Tenv

Tenv

θ

∂θ

∂z δz = 0

D2δz

Dt 2+

g

θ

∂θ

∂z δz = 0

D2δz

Dt 2+ N 2 δz = 0

With the Brunt-Väisälä frequency

€

N =g

θ

∂θ

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟

1

2

Buoyancy oscillations

€

D2δz

Dt 2+ N 2δz = 0

1) stable, the solution to this equation

describes a buoyancy oscillation with period 2/N

2) unstable, corresponds to growing

perturbation, this is an instability

3) neutral

€

with N =g

θ

∂θ

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟

1

2= g

∂lnθ

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟

1

2

€

∂∂z

> 0,N 2 > 0 :

€

∂∂z

< 0,N 2 < 0 :

€

∂∂z

= 0,N 2 = 0 :

Solution to the differential equation

The general solution can be expressed via the exponential function with a complex argument:

€

z = Aexp(±iNt)

with A: amplitude, N: buoyancy frequency,

If N2 > 0 the parcel will oscillate about its initial level with a period = 2/N.Average N in the troposphere N ≈ 0.01 s-1

€

D2δz

Dt 2+ N 2δz = 0

€

i = −1

Remember Euler’s formula

with x: real number

€

z = Re(Aexp(±iNt))

If N2 > 0 (real) the solution is a wave with period = 2/N (more on waves in AOSS 401)

€

exp(ix) = cos(x) + isin(x)

Physical solution:

Stable solution:Parcel cooler than environmentz

Warmer

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.

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 aTIFF (Uncompressed) decompressor

are needed to see this picture.

Temperature soundings

Sounding of the environment:T inversion

Sounding of the parcel

z