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The Vertical Temperature Profile
Created by Absorption and Emission
of Radiant Energy in a Spherical Atmosphere

This is an analysis of the greenhouse effect in a spherical atmosphere. The analysis in an atmosphere of parallel planes is given elsewhere. That analysis found that the radiation intensity, incoming and outgoing, is a linear function of the optical path lengths (this concept is explained below). The spherical geometry of the atmosphere modifies the analysis.

Consider an infinitesimal element of thickness dr centered at r which subtends a solid angle of dΩ. For background information on the concept of solid angles click here. The area of the cross-section of the cone subtended by the solid angle dΩ at a distance r from the center of the sphere is r²dOmega;, thus the volume is approximately r²dOmega;dr. (In what follows r is the distance from the center of the Earth.) The infinitesmal element is in the nature of a truncated cone. The cone with base at r+½dr has a volume of (1/3)(r+½dr)3Ω whereas the one with a base at r-½dr has a volume of (1/3)(r-½dr)3Ω. The difference is then

dV = (1/3)[3r²dr]dΩ = r²dΩdr

Consider a flux of intensity I which enters the lower face of the infinitesimal element and exits the upper face. The energy entering is then I(r-½dr)(r-½dr)²dΩ and that leaving is I(r+½dr)(r+½dr)²dΩ. Thus the net inflow, to the first approximation, is


This net energy inflow would give a rate of temperature increase dT/dt such that

cpρr²(dT/dt)dωdr = (d(Ir²)/dr)dΩdr

where ρ is the mass density of the substance and cp is the specific heat per unit mass of the substance.

Cancelling the dΩdr term gives

cpρr²(dT/dt) = d(Ir²)/dr

At temperature equilibrium dT/dt=0 so Ir² has to be constant.

When there is an outflow I+ and an inflow I- this condition takes the form that (I+-I-)r² is constant.

The Gradients of Radiations Fluxes

Consider now the flows of radiation in and out of the infinitesimal element. The change in radiation intensity for the outgoing radiation I+ between the upper and lower faces of the element is given by

(r+½dr)²ΩI+(r+½dr) − (r-½dr)²ΩI+(r-½dr)
which reduces to

The absorption of radiation within the element is given by the product of the absorption per unit length (κρI+(r)) the thickness of the element dr and the cross section area (r²dΩ); i.e.,


where ρ is the density and κ is the absorption coefficient.

If the element is at a temperature T and generating radiation at a rate of κρσT4 per unit volume then the total radiation generated in the element is


where the emissivity coefficient is the same (Kirchhoff's Law) as the absorption coefficient and σ is the Stefan-Boltzmann coefficient.

With spherical symmetry there is no net flow through the sides of the element so the half of the thermal radiation flow upward through the upper face and half downward through the lower face.

Thus the balance of flows for the outgoing radiation gives

(d(r²I+(r))/dr)dΩdr = − κρr²I+(r)dΩdr + ½κρσT4r²dΩdr
which upon cancellation of dΩ reduces to
d(r²I+(r)) = [− κρr²I+(r) + ½σT4r²]κρdr

The quantity κρdr may defined as the infinitesimal increment of optical path length dx. For the it is convenient to take the optical path to be the optical height above the surface of the Earth. It is thus zero at the surfac of the Earth and reaches some finite value xmax even if the atmosphere does not have a definite top.

Note that the product κρ has dimension of inverse length so path length is a dimensionless variable.

It is convenient to represent σt4 as R(T). Thus the gradient of outgoing radiation intensity with respect to optical path length is given by

d(r²I+(r))/dx = −r²I+ + ½r²R

Rather than converting r² to path length x let us denote r²I+(r) as J+(x) and. r²R as Q. Thus the gradient equation for the outgoing radiation is

dJ+/dx = −J+ + ½Q

The corresponding equation for the inward flowing radiation is

dJ-/dx = J- − ½Q

Now consider the quantities φ=J+−J- and μ=J++J-. From the above gradient equations it follows that

dφ/dx = dJ+/dx − dJ-/dx = −μ + Q
dμ = dJ+/dx + dJ-/dx = −φ

The net inflow of radiation to the element is given by


If this is always zero it means that J+(r)−J-(r) is constant. But J+−J- is what has been denoted as φ(r). Thus the equation dμ=−φ has the solution

μ = −φx + constant

The integration constant can be evaluated from the boundary conditions. At the top of the atmosphere, x=xmax, the incoming radiation is zero and hence μ at that level is equal to φ. Thus

φ = −φxmax + constant
constant = φ(xmax+1)
and hence the complete solution for μ is
μ = φ(xmax−x +1)

The quantity φ is a derived quantity. The constraint on the system is that the outgoing long wavelength radiation at the top of the atmosphere be equal to the net incoming short wavelength radiation from the Sun. Let S be the intensity of the Sun's radiation at top of the atmosphere multiplied by r². Thus the constraint is

J+(xmax) = S


μ = J+ + J- = J+ + (J+ − φ) = 2J+ − φ
J+ = (μ + φ)/2

At x=xmax μ=φ. This means that at x=xmax J+(xmax)=φ. Therefore φ=S. Thus the full solution to the system is

μ(x) = S(xmax−x + 1)
φ(x) = S
J+(x) = S(xmax−x + 2)/2
J-(x) =S(xmax−x)

Let the relation between optical height x and altitude h be given by

x = f(h)
and hence
h = g(x)

Therefore r = r0 + h, where r0 is the radius of the Earth. Thus

I+(h) = J+(f(h))/(r0 + h)² = S(xmax−f(h) + 1)/(r0 + h)²
I+(h) = J-(f(h))/(r0 + h)² = S(xmax−f(h))/(r0 + h)²

The constant S was defined as (r0+hmax)²α where α is the solar constant, which in watt/m² is 1374.

This means that

I+(h) = α(xmax−f(h) + 1)/((r0 + hmax)/(r0 + h))²
I-(h) = α(xmax−f(h))/((r0 + hmax)/(r0 + h))²

The radius of the Earth is about 6366 km and the maximum height of the atmosphere is about 300 km. Thus the fraction ((r0 + hmax)/(r0 + h))² varies from 1.00 up to about 1.1. Since the variation in the heights in the atmosphere are so small compared to the radius of the Earth the dependence of radiation intensity is largely determined by the term (xmax−f(h)) in the above equations.

In the above diagram the light colored lines are what the radiation intensity would be without the effect of the spherical geometry. The increase indicated is due to the compression of the area due to being closer to the center of the sphere.

(To be continued.)

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