San José State University

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Thayer Watkins
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 A Coupled Short and Long Wavelength Analysis of the Vertical Temperature Profile of the Atmosphere Established by the Greenhouse Effect

Short wavelength radiation, visible and ultraviolet light, enters the Earth's atmosphere where some of it is absorbed by the ozone and some reflected back out by clouds. The rest passes through to the surface where part is absorbed and part reflected. That which is reflected and passes back up through the atmosphere where again some is absorbed by ozone. The greenhouse effect occurs when the heated surface of the Earth radiates infrared (long wavelength) radiation which greenhouse gases such as water vapor and carbon dioxide absorb and heat the air which in turn radiates infrared radiation upward and downward where it is reabsorbed and reradiated. The analysis is carried out first in terms of the simplest version of the model. That simplest version involves an atmosphere in horizon planes. A spherical model is considered elsewhere.

Radiation of intensity I traveling through a medium experiences diminishment according to the Beer-Lambert Law

#### dI/dz = −λρI

where dz is distance through the medium, ρ is molecular volume density and λ is an absorption coefficient.

The level of density depends upon height and the value of λ depends upon the composition of the air; i.e., the proportional amounts of greenhouse gases such as water vapor, carbon dioxide and ozone in the air and the wavelength of the radiation.

The level of the dimension of the product λρ is inverse length. The values λ and ρ may vary with position z.

The Beer-Lambert Law implies that

#### I(z) = I(0)exp(−∫0zλρds)

The quanty exp(−∫0zλρds) is called the transmission coefficient and the quantity ∫0zλρds is called the optical path length. The analysis is improved, conceptually and analytically, if the optical path length is used as the length variable. Let y be the optical length (or optical height or optical depth) where

#### y = ∫0zλρds

The optical path length depends on the wavelength of the radiation.

#### Let dx = λρdz for long wavelength radiationand dy = κρdz for short wavelength radiation.

The crucial thing is that there are monotonic functions relating x and z and y and z.

The atmosphere does not end at some height hmax but trails off with lower and lower density. Thus z may not have have an upper limit but x and y may nevertheless have finite levels.

Since λρ and κρ have dimension of inverse length, the optical path lengths x and y are dimensionless variables.

## The Case for Short Wavelength Radiation

Let J(z) be the intensity of the short wavelength radiation at height z. The intensity may be measured in various forms but the most convenient is in terms of energy per unit area, watts per square meter. In terms of optical path length the Beer-Lambert Law takes the form

#### dJ/dy = −J

The optical path here is measured from the surface upward. Since there is directionality to the radiation flow, let J- be the downward flowing radiation and J+ the upwardward flowing radiation. The two equations for the intensities are

#### dJ+/dy = −J+and dJ-/dy = J-

The sign is right for J- because the intensity decreases as the optical height y decreases from ymax. Therefore as y increases J- also increases; i.e., dJ-/dy is positive.

At the surface, y=0, there is reflectance so

#### J+(0) = αJ-(0)

where the reflectance coefficient α is called the albedo.

The solutions for the intensities is simple.

#### J-(y) = J-(ymax)exp(y−ymax) and J+(y) = J+(0)exp(−y) thus J-(0) = J-(ymax)exp(−ymax) and J+(0) = αJ-(ymax)exp(−ymax) so J+(y) = αJ-(ymax)exp(−ymax)exp(−y) which reduces to J+(y) = αJ-(ymax)exp(−(y+ymax))

Thus the short wavelength intensities, upward and downward, as functions of optical height y are fully determined.

## The Case of an Absorbing-Emitting Medium

In a medium which absorbs radiant energy the temperature increases and it radiates energy according to the Stefan-Boltzmann formula σT4, which is in the long wavelength band. The Beer-Lambert Law is replaced by the Schwarzchild Equation

#### dI/dz = −λρI + ½λρR(z)

where R(z)=σT4(z). This formula presumes Kirchhoff's Law that the emissivity of a substance is equal to its absorptivity. It also presumes that half of the thermal radiation is in the same direction as the radiation flow.

In terms of the optical path length x the Schwarzchild equation takes the form

#### dI/dx = −I + ½R

The directionality must be taken into account. Let I+ be the radiation flux intensity in the direction of increasing optical height x and I- the flux intensity in the opposite direction.

The two equations for the optical path gradients of radiation intensities are

#### dI+/dx = −I+ + ½R and dI-/dx = I- − ½R

These relationships are for long wavelength radiation.

For further analysis it is convenient to define two new variables

#### φ = ½[(I+ − I-)] and μ = ½[(I+ + I-]

The reason for these definitions is the equations for the rates of change of these two variables along the optical path take particularly interesting form. First note that from the gradient equations given above

#### dI+/dx + dI- = −(I+−I-) dI+/dx - dI- = −(I++I-) + R

From these conditions it follows that:

#### dμ/dx = −φ dφ/dx = −μ + R

Now the matter of energy balance and temperature equilibrium must be considered.

## Temperature Balance in an Infinitesimal Element

Consider an infinitesimal element of thickness dz and cross-sectional area dA. For the moment consider only the long wavelength radiation intensities, I+ and I-.

The amount of energy absorbed in the element is λρ(I*+I-)dzdA. The amount radiated out is λρRdzdA. This total is from the previous relationships for the optical path gradients

#### λρ(I*+I-)dzdA − λρRdzdA = dxdA[ (I* − ½R) + (I- − ½R)] = dxdA[−(dI*/dx) + (dI-/dx)] = −[dI*/dx − dI-/dx]dxdA = −(d(I*-I-)/dx)dxdA

The net inflow of radiative energy to the element is

Had the short wave length radiation intensities been included the above result would have been

#### (dAdz)d[(I+-I-) + (J+-J-)]/dz = 0

That net inflow of energy to the infinitesimal element must increase its temperature according to

where ρ has to be mass density, cp is specific heat at constant pressure and t is time.

At equilibrium dT/dt=0 so it must be that

#### (d[(I+-I-) + (J+-J-)]/dz) = 0

This means that (I+-I-)+(J+-J-)] has to be constant over the optical path and hence φ=−½(J+-J-), a known quantity. This means that the differential equation for μ can be solved by integration; i.e.,

#### dμ/dx = −φ = ½(J+-J-) and hence μ(x) = ½∫0x(J+-J-)ds

From the knowledge that φ=−½(J+-J-) and that dφ/dx=−μ+R it follows that

#### R(x) = μ + −½d(J+-J-)/dx

Thus a formal solution has been found for the radiation intensities and temperature as a function of height.

Some further insights into the nature of the solution may be obtained by considering the relationships that exist for the gradients of the J's. Let the ratio of the absorptivities of the two wavelengths of radiation (κ/λ)=(dy/dx) be denoted as ε(x). Then note that

#### dJ+/dx + dJ-/dx = [dJ+/dy + dJ-/dy](dy/dx) = −(J+−J-)ε(x) and dJ+/dx - dJ-/dx = −(J++J-)ε thus (J+−J-) = −[d(J++J-)/dx]/ε(x)

If the atmosphere were well mixed at all levels then ε would be constant. However there are much higher levels of water vapor in the troposphere than the stratosphere. On the other hand, there are higher levels of ozone in the stratosphere than in the troposphere. Nevertheless the case of constant ε is an important special case for mathematical analysis. For this case it would follow that

#### μ(x) = −½(J++J-)]/ε

This holds however only for a vertically uniform atmosphere.

## Equilibrium Solution

The determining condition for the equilibrium of the system is that the outflow of long wavelength energy at the top of the atmosphere must be equal to the net inflow of short wavelength radiation at the top of the atmosphere. At the top of the atmosphere I-=0 and thus I+=2φ and μ=φ. Thus if the net energy input of short wavelength radiation is S=[J-(xmax)−J+(xmax)] then

#### 2φ(xmax) = S φ(xmax) = ½S which means that all quantities may be computed from S.

One interesting implication of the solution is that there is a temperature discontinuity at the surface. In other words, the surface is at a higher temperature than the air just above it. This is actually a familiar phenomenon as at the beach where the sand may be too hot to walk upon barefooted while the air temperature is not uncomfortable.

The analysis indicates that the air temperature decreases with altitude as a result of the greenhouse effect of the absorption of thermal radiation by water vapor, carbon dioxide and the other greenhouse gases except ozone. Ozone absorbs visible and ultraviolet radiation and the temperature effect of this absorption increases with height. In the real atmosphere temperature decreases with height up to a particular level called the tropopause and above that level the temperature increases with altitude (in the stratosphere).

(To be continued.)