The State of the Atmosphere

1. Atmospheric mass and pressure

2. Temperature structure

3. Geopotential

4. Circulation

5. Water in the atmosphere

Atmospheric mass and pressure

• From the equation of continuity, after integration for a polar cap limited by

WallSVV

vdsdsncdvcdivdvt

.)(

In the long-term mean: 0Wall

dsv

Assuming that the atmosphere is a ideal gas (p=RT) in hydrostatic equilibrium

z

dzgzp )(

we haveRT

dzg

p

dp and by integrating: )exp()(

0

0 z

z

dzRT

gpzp

We can then rewrite the first equation as:

Wallsurfh

vdsdsptg

_

0

1(1)

The rapid decrease of density with height distinguishes the atmosphere from the ocean

Distribution of mass in terms of pressure

• If H_surf=2.56x1014 m2

• PNorth_h=983.6mb, PSouth_h=988.0mb, PGlobe=983.6mb• g=9.8 m/s2

Then

m_North_h=2.57x1018 kg

m_Souh_h=2.58x1018 kg

m_Globe=5.15x1018 kg

...Distribution of mass in terms of pressure

The interhemispherictransport

From Eq. (1) we can calculate the transport trough the equator:

0 0

/][][z

dzdzvv

If p=1mb/month

smRmbmonth

mbR

v

/002.021012

1

1

12

][

2

Global distribution of temperature

Vertical and meridional change

Temporalvariabilityof temperature

Geopotential heights

Mean circulation

• Correlation coefficient between northern hemisphere stratospheric geopotential at 50hPa and an index representing the tropospheric 500hPA NAO.

Hadley model

Hadley (1735) wanted to explain trade wind circulationComplements of E. Kant and J. DaltonFerel (1856): Coriolis force and Geostrophic windHelmholtz: the role of friction-deviation of wind includes turbulent viscosity for the first time

Actual vertical circulation

• Kinetic energy of the atmosphere

• K=KTE +KSE +KM

• K=0.5[u2m+v2

m]

• KTE=0.5[(u‘2+v‘2) m]

• KSE=0.5[u m *2+v m *2]

• KM=0.5([u m] 2 +[v m] 2)

Precipitation