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12 September 2019

Calculating the Angular Dependence of Planar Radiation Upon a Sphere

The incidence of planar radiation upon a sphere, such as that from the sun upon the Earth, is an important calculation for climate science, so I am going to demonstrate that calculation.  This post repeats the previous post content, but adds graphs of properties as a function of surface normal angles of the sphere.

Let the amplitude of the planar radiation be P0, which is commonly expressed as W/m2.   The incident radiation on the sphere surface normal to the planar radiation will have an intensity of P0.  As one examines a unit surface area of the sphere which is non-normal to the planar wave of radiation, the unit area of the sphere intercepts less and less of the planar radiation the greater the surface normal angle is with respect to the direction of the plane wave.  Let us call the angle from the center of the sphere to the point on the surface upon which the planar wave has normal incidence and the surface annulus of points equidistance from the normal incidence point θ.  The surface of the hemisphere facing the planar radiation then consists of a normal point and a series of annuli, one for each Δθ that we choose.  The annuli have increasing radii perpendicular to the planar wave of radiation.  The angle θ to describe these annuli varies from 0 to π/2 radians or 90̊.  We will use the radian units.  Let the radius of the sphere be R.
Each annulus has a circumference of 2πR sin θ and it intercepts an annulus of the planar radiation wave with the same circumference, but with a narrower width than the annulus on the sphere surface has for θ > 0.  The width of the planar wave annulus that projects onto the sphere surface is given by Δθ R cos θ.  Thus, at angle θ the area of the plane wave of radiation intercepted by the sphere at angle θ, is
AI (θ) = 2πR2 sin θ cos θ Δθ
Integrating over the range from 0 to π/2 for θ, we find the area of the planar wave of radiation which is intercepted by the sphere.  The integral is
AI = 2πR2 ∫0π/2 sin θ cos θ dθ
The value of the integral is ½ sin2θ, which for these limits of integration yields a value of  ½.  Consequently, we have an obvious result that
AI = πR2
which helps to confirm that we have AI (θ) right. The total incident planar wave power on the sphere (all on one hemisphere) is then
IT = P0 AI = πR2 P0
The total area of the sphere, AS, is 4πR2, so we can rewrite this as
IT = (1/4) ASP0
Averaging over the entire area of the sphere, the total incident plane wave intensity average per unit area is then P0/4.

Let us examine the intensity of radiation on equal areas of the sphere surface now.  The first area is a circle on the sphere whose surface normal is perpendicular to the wave front and whose area is given by 

π ( R sin θ0 )2

The irradiance of this area is P0 times this area, so finding the irradiance per unit area means we divide by the area of this circle and arrive at P0.  To normalize this to one, we further divide by P0. to make the result more universal for any radiant power.

The subsequent total irradiance values are those of the annuli, which are given by AI (θ).  To find the irradiance per unit area on the annuli surface of the sphere for comparison with the irradiance per unit area on the normal facing circle, the AI (θ) values are all divided by the circumference of the annuli times the width of the annuli, so the per unit area relative irradiance values to compare with that of the center circle normal to the radiation wavefront are:

(2πR2 sin θ cos θ Δθ) / (2πRsin θ)( R Δθ) = cos θ

If we make the inner circle sized to have an arc of 1̊ or  π/180, then θ0 = 0.5̊.  Using a step size of   each annulus will subsume an arc of π/180, then we have the normalized irradiance per unit surface area on the sphere for θ shown in the plot below:

Let us now plot the total irradiance of the sphere as a function of the angle between the wave front and the sphere surface normal.  This is just

AI (θ) = 2πR2 sin θ cos θ Δθ,

where πR2 is just the intercepted area of the plane wavefront.  So to keep this a generalized function for any sphere radius R, we will divide AI (θ) by πR2.  Recall that the integral of the product sin θ cos θ over the range of angles from 0̊ to 90̊ is ½, so the integral of AI (θ) / πR2 over this range is 1.  What this means is that AI (θ) / πR2 is the fraction of the total incident radiation on the sphere at the surface normal angles θ.  This fraction of the radiation incident on the sphere for θ is plotted below using a 1̊ step size (The sum of the fractions of the incident radiation is 0.999898506, which is pretty close to 1.):

The total radiation incident upon the sphere maximizes at 45̊ even though the radiation density is greatest at 0̊.  This is because the circular area with a 1̊ arc at 0̊ is very small and the area of the annuli for larger angles gets larger and larger as θ increases, even as the radiation density decreases.

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