Radiative Transfer

Planetary Equilibrium Temperature:

For the earth, the albedo is approximately 0.3, but of course is larger during ice ages ...

Let's start with consider a zero layer model for the earth (e.g. no atmosphere).

Solar Constant (So)is 1370 Watts per sq meter. Stefan-Boltzmann constant is 5.67x10-8 Watts per sq meter per degree:

T4 = (1/4*1370/a)(1-A)

T = 278*(1-A)1/4

For A = 0.3 one gets T = 254K

One Layer Model (e.g. our atmosphere)

We assume the following:

Let TA be the temperature of the atmosphere and TS be the temperature of the surface of the planet. Since all the longwave radiation emitted by the planet is absorbed by the atmosphere, none of it leaves the top of the atmosphere. Thus, the energy balance at the top of the atmosphere is given by the same equations as the zero layer model and


The energy emitted upwards but the ground scale as (TS)4 and the energy emitted upwards and downwards by the atmosphere will scale as (TA )4


2(TA )4 = (TS)4

TS = 21/4TA = 21/4TE

on the earth TE = 255 K, then TS = 303 K, which is a higher than the usually assumed value of 288K this is because of some detailed complications associated with the last two assumptions above.

In particular, not all the thermal IR emitted by the planet is absorbed by the atmosphere. The oceans are important here and have been ignored.

Now finally let's consider a two layer model :

The top layer has temperature T1 and the bottom layer has temperature of T2.

Again, energy conservation requires that T1 = TTe

For the top layer:

Energy out goes as T14(emitted up) + T14(emitted down)

The Energy input comes from energy emitted upwards from the second layer so that

Energy in goes as T24


2(T1)4 = 2(TE)4 = (T2)4

For the second layer:

Energy out goes as T24(emitted up) + T24(emitted down)

Energy in = (T1)4 (emitted down by layer 1) + (TS)4 (emitted up by surface)

Using the above relationships get,

2(T2)4 = 4(TE)4 = (TE)4 + (TS)4,

which finally gives

TS = 31/4 TE.

So, a two layer model of the earth would give a temperature of about 336 K, which is much too high.

For an N-layered atmosphere (relevant to Venus), one can derive the general expression:

TS = (N+1)1/4 TE

The key here, is that atmospheric gases are weak absorbers in the visible but strong absorbers in the infrared. So incoming radiation easily reaches the ground, but outgoing radiation has a more difficult time escaping back to space, and some fraction of it is re-radiated back to the ground.

Problem 6.8 in Wallace and Hobbs

Atmosphere is thin layer with absorptivity = 0.1 for visible radiation but 0.8 for infrared radiation.

Let x = radiation emitted by the earth's surface and y = radiation emitted (upwards and downwards) by the atmosphere. E = the net solar radiation absorbed by the earth-atmosphere system.

At the earth's surface, radiative equilibrium demands (since 10% of E is absorbed by the atmosphere)

0.9E + y = x

The absorptivity of the atmosphere in the IR is 0.8 therefore 0.2x escapes back to space. This means

0.2x + y = E

or substituting for x

0.2(0.9E + y) + y = E 1.2y = .82E y =0.68E

And clearly x = 1.58E

Where E = So/4(1-A) or 241 watts per sq meter for A= 0.3.

So for the surface temperature one gets

aT4 = 1.58 x 241 277 K

for the atmosphere 0.8aT4 = 0.68 x 241 245 K


Let's consider the impact of radiation on surfaces. If a surface is opaque then the radiation is either absorbed or reflected. The relative amounts of absorption and reflection for any material are highly wavelength dependent. But at any time:

Eabsorbed + Ereflected = Eincident


aw + rw = 1

where w is wavelength and a and r are absorptivity and reflectivity coefficients for that wavelength.

In the case where the surface is non-opaque, then indicident radiation can be transmitted through the surface (e.g. glass). In that case we have the more general expression:

aw + rw +tw = 1

where t is the transmission coefficient.

Back to real physics in next lecture when we talk about scattering!