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

A_I (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

A_I = 2πR² ∫₀^π/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

A_I = πR²

which helps to confirm that we have A_I (θ) right. The total incident planar wave power on the sphere (all on one hemisphere) is then

I_T = P₀ A_I = πR² P₀

The total area of the sphere, A_S, is 4πR², so we can rewrite this as

I_T = (1/4) A_SP₀

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 A_I (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 A_I (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 1̊  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

A_I (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 A_I (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 A_I (θ) / πR² over this range is 1.  What this means is that A_I (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.

Extortion of Businesses Using a Safety Regulation Scam

This is a particular examination of a regulatory case which applies to relatively few people.  This makes it a more viable scam which does not generate voter ire during elections.  Some voters are actually deceived into thinking that big government is doing its job of protecting them.  This case is instructive in terms of understanding what motivates most regulatory agencies and about their competency as a regulator.  In that way, it should of interest to everyone who wants rational, ethical government and cares about the rights of the individual.

The state of Maryland under the Dept. of the Environment has a Radiological Health Program which regulates all x-ray generating equipment except the cathode ray tubes (CRT) used in the older TV and computer monitor technology.  Yes, CRTs generate x-rays.  The electron beam that causes the light to appear generates x-rays at moderately high energies such as 25 to 40 KeV energies.

My laboratory operates a scanning electron microscope whose electron beam energy is commonly in the 15 to 25 KeV range and so can generate x-rays with energies up to the electron beam energy.  But, unlike a CRT computer monitor or TV, the SEM has very thick stainless steel walls.  Just as is the case with a CRT, the x-rays are contained within a vacuum system into which no human body part can be inserted while under vacuum.  When the sample insertion door is opened, no x-rays can be generated because if the electron beam-producing filament was on, it would instantaneously burn out and the electron beam would be eliminated.

This SEM instrument and the entire class of them was ruled a couple of years ago in Maryland to be safe enough that it does not have to be listed and no x-ray tube tax is assessed on it by the state of Maryland.  I thought this meant that the only reason my laboratory still had to register as a certified radiation facility was because of our x-ray photoelectron spectrometer (XPS) system.  Apparently not quite so.  Under this belief, and knowing that there is no conceivable safety hazard in operating the XPS system and that it is actually clearly much safer even than either a CRT or an SEM, I attempted to get it exempted from regulation.

As a regulated radiation facility, I have to hire a state-licensed inspector to come to my laboratory every two years and run around with a Geiger counter looking for radiation near the SEM and the XPS system.  He sometimes thinks he can see something, but never enough radiation to warrant any health issue.  What he is actually seeing is the radiation emitted due to very energetic cosmic rays hitting the massive steel structure and generating x-rays at higher levels than does the air when he backs up a couple of feet from the instrument.  He is not seeing any radiation from the inside of the instrument due to its operation.  But, he does not know enough radiation physics to know this.

I pay the state Radiological Health Program a $215 x-ray tube charge every second year for each x-ray tube they require to be listed.  I also have to pay for radiation badges, which in 18 years of operation have never had a quarterly reading other than 0 mRem.  This is despite the fact that I leave a badge on the instrument at the most vulnerable point 100% of the time, so it collects at least 100 times more radiation than any operator possibly could.  It is a hassle to mail these back to the radiation measurement company to be read and recorded due to the unusual size of the envelope and the bulk of the badges.  The post office requires a different amount of postage almost every time.  In addition, I am supposed to have my instruments maintained by a state-licensed company, none of whom know anything about an XPS instrument.  Presumably, this is to assure that safety issues are addressed, but if there are no safety issues, I am baffled as to how they are to address them.

A couple of years ago, I had a phone conversation with one of the radiological health bureaucrats about getting the absolutely safe XPS exempted.  He flat out told me that the radiation physics and the fact of whether there was any possible danger or not was of no concern to him.  He said their purpose was to make people feel as though they were safe from radiation.  Whether there actually was any radiation or not was of no interest to them.  This time around, I thought I would write them a letter and explain why this instrument should be exempted and why my laboratory should not have to register as a radiation facility.  This letter follows:

Roland G. Fletcher

Program Manager IV

Radiological Health Program

Air & Radiation Management Administration

Maryland Dept. Of the Environment

1800 Washington Blvd.

Baltimore, MD 21230

Phone: 410-537-3000

Dear Roland Fletcher,

Anderson Materials Evaluation, Inc. does not now and never has operated any radiation machine with any safety hazard.  The only radiation producing equipment we operate produces low energy radiation completely contained by a vacuum containment vessel, which cannot be breached without the instantaneous destruction of the filament which creates the electron beam and hence the x-rays.  These instruments are:

•                    Cathode Ray Tube (CRT) Computer Monitors

•                    Scanning Electron Microscope (SEM)

•                    X-ray Photoelectron Spectrometer (XPS)

These are listed in the order of the danger posed by each.  Your agency has rightly acknowledged that the first two do not require registration as a radiation facility.  Therefore, the much less dangerous XPS system does not require registration either.

The CRTs have electron beam energies of 25 KeV and higher and therefore can generate Bremsstrahlung radiation inside their vacuum containment vessel up to an energy equal to the beam energy.  Our SEM is operated in the electron beam energy range from 15 to 25 KeV.  These are acknowledged safe instruments.

Our XPS system is operated at only 10 KeV and the radiation is only emitted into the analytical vacuum chamber after passing through an x-ray monochromator which filters out all x-ray energies except those precisely at the 1.487 KeV energy of the main aluminum Kα characteristic x-ray.  This is an incredibly low energy x-ray, which precisely suits the purpose of having as much of the x-ray absorbed in the outer 20 nanometers of a material surface as possible.  The only photoelectrons useful to us are those emitted from this depth or less.  This beam has no penetrating power!  If it were possible to enter air, its mean free path is still less than 1 millimeter.

I have raised the issue of getting this instrument exempted before and I have been ignored.  I am unable to continue spending money and my time on this pointless nonsense.  The Obama economy has sapped our resources sorely and your new requirement that only licensed x-ray maintenance companies provide the maintenance on this highly unusual system with no safety issues is not one it is possible to comply with.  They know nothing about this system and nothing they can do can possibly make the system safer than it already is.  There are no fail-safes which have any possibility of failing to work.

So, there is no reason for Anderson Materials Evaluation, Inc. to renew its Radiation Facility license # 27-0563 and there is no reason we should be taxed for having an x-ray tube.  It is high time that some semblance of rationality is applied here.

Best wishes,

Charles R. Anderson, Ph.D.

President & Principal Scientist

I would have viewed a reasonable response one in which they simply said OK the XPS instrument is exempted or we will send our radiation physics expert out to your facility to make a determination of whether it might be exempted or not, perhaps at my expense.  The response to this letter strongly suggests that it was viewed as an insurrection to be dealt with even more sternly and forcefully than King George III moved to put down the rebellion in the colonies in America.  Here is the response, not from Mr. Fletcher, but from the Queen herself:

Dr. Anderson,

Thank you for your letter dated December 6, 2013 in response to the Notice of Violation that the Department issued.  I appreciate the concerns that you have raised but each of our 5,000 regulated facilities must adhere to the Maryland regulations in order to ensure the the general public and the occupational worker are not exposed to unnecessary radiation.  

The scanning electron microscope does require the registrant to denote possession of this unit on the Radiation Machine Facility Registration Form (RX1).  Also the State licensed private inspector does inspect and perform an area survey of this unit.  

The X-ray Photoelectron Spectrometer (XPS) must undergo a state certification inspection by one of the state licensed private inspectors. This unit is not exempt from complying with the Maryland regulations.  

Failure to comply with these regulations will subject the facility to further enforcement actions.

Thanks,

Eva  

Eva S. Nair
Division Chief, Radiation Machines
Maryland Department of the Environment
Radiological Health Program
(410) 537-3179

It is interesting that all of my letters to my Congressman and my Senators and any bureaucrat of the state of Maryland start with a thank you for my letter and then an assertion of complete disagreement without any attempt to actually address the argument I gave for my viewpoint.  This is the response to be expected from Imperial Government which prefers a minimum of insurrection, but will put any insurrection down with brutal force if that proves expedient.

The violation she claims is because I had not put in for a renewal of my radiation facility registration by 16 November because they demand 45 days to process those registrations.  In reality, I am only in violation if I operate the XPS instrument after 31 December when my current registration and certification runs out.  So, I probably will not be able to operate that instrument in early January or maybe all of January.  If that condition continued longer than that, I and my employees will all be unemployed.

So, once again, I am forced to apply for a radiation facility certification when the most dangerous radiation sources in my laboratory are the CRTs we still use on a number of our computers.  Logically and rationally, any CRTs in residences and offices throughout the state should be regulated by the Radiological Health Program given that my equivalent or safer instruments are.  Imagine the insurrection if this were done.  That would be a loser at the polls, so it is not done.  But, it is always easy to extort businesses, whose owners are but a small fraction of the vote.

Now in the mind of a big government bureaucrat, no possible instance of misleading the public into believing they are being protected by that big government can be missed.  The more facilities they regulate, the bigger their income from the x-ray tube tax, the bigger their budget from the taxpayers, and the higher their inflated salaries with the least risk of unemployment.

Yes, medical and dental x-ray instruments do have real safety issues.  But, this Maryland agency does not at all care if an instrument has no safety issues by virtue of its function, design, and radiation physics.  It is only interested in setting itself up under the pretense of providing safety.  It reveals itself as having no actual radiation physics expertise, which makes one wonder how it can be doing a credible job of protecting the People from those radiation-producing instruments which can do real harm.

The Emperor has no clothes.  But the nude Emperor is vicious and brutal.  This is nearly an invariable rule with government regulatory agencies.  They are particularly a plague upon businesses, because that is where the money is and because most of them have little political clout.  They care not a fig for the legitimate function of government of protecting the sovereign rights of the individual to life, liberty, property, and the pursuit of happiness.  They get rich by violating the individual right to earn a living providing one is doing no harm to others.  The Emperor lives high on the hog pretending he is protecting the People, few of whom understand that most of those claims are fraudulent.

What if the Atmosphere had no Greenhouse Gases?

According to the "settled science", the Earth's surface temperature would be about 33K or 33C colder on average than it is now if there were no so-called greenhouse gases.  The settled scientists treat the Earth's surface and its atmosphere as a single system which is in radiative equilibrium with the sun and space.  The Earth system is assumed to be a black body radiator and the calculated temperature of such a black body radiator is about 255K or about 33K colder than the approximate observed average temperature of 288K.  They forget to note that the Earth's surface temperature is about 288K, but the surface is not in radiative equilibrium with the sun and space.  Therefore, it is completely wrong to compare the average Earth surface temperature with the temperature of the equivalent effective black body radiator even if the Earth's surface were a black body radiator.

Let us dispense with the infrared active gases, the so-called greenhouse gases, so that the Earth's surface is in radiative equilibrium with the absorbed solar insolation and the infrared emissions into space.  In the greenhouse gas game, clouds are considered to cause part of the effect and this is natural given that without water vapor they would not exist.  Let us look at the Kiehl-Trenberth Earth Energy Budget of 1997, which the IPCC 4th Report of 2007 claimed was the settled science, to find a few numbers we will need to perform a simple calculation.

Fig. 1.  The Kiehl-Trenberth Earth Energy Budget of 1997, representing the settled science according to the IPCC 4th Report of 2007.

Now, I have found any number of faults with this diagram in previous posts.  But, given the claims of the settled science I am at least justified in throwing some of its consequences back at them.  In addition, a few of the numbers on certain factors ought to be about right, or they have spent about $140 billion of U.S. taxpayer money with unbelievable incompetence.

So, let us look at the energy inputs to the Earth's surface.  The average daily power input at the top of the atmosphere is 342 W/m^2.  The 77 W/m^2 that reflects off clouds does not do so, since we have taken the clouds away.  The 67 W/m^2 absorbed by the atmosphere is much reduced.  Ozone and even O2 to a small degree do absorb some of the energy.  Let us be generous in estimating how much is absorbed by the non-greenhouse gases and say they absorb 12% of the incoming sunlight, rather than the 19.6% of the K-T diagram above.  By being generous in this, we are lowing the radiation that will reach the surface to warm it.  So (12%/19.6%) 67 W/m^2 = 41 W/m^2 is still absorbed by the atmosphere and does not reach the surface.

Consequently, the solar radiation that does reach the surface is (342 - 41)W/m^2 = 301 W/m^2.  But now we note that not all of the radiation from the sun that is incident upon the surface is absorbed.  The fraction absorbed is (168/198) = 0.848 in the diagram.  If we assume the same fraction of the radiation incident on the surface is absorbed, then we have an absorbed power density in the surface of 0.848 (301) W/m^2 = 255 W/m^2.

Let us, after the style of the greenhouse gas catastrophe believers, apply the Stefan-Boltzmann Law for radiation, which is

P/A = ε σ T⁴,

in which P is the power in Watts, A is the absorbing or radiating area in square meters, σ = 5.6697 x 10^-8 W/m²K⁴ the Stefan-Boltzmann constant, T is the temperature in Kelvin, and ε = 1 (the emissivity) for the case of the black body radiator.

In this case, the Earth's surface is in radiative equilibrium with space because there are no gases in the greenhouse gas free atmosphere to absorb the thermal infrared radiation emitted by the Earth's surface.  Consequently, the black body surface absorbing 255 W/m^2 would be at a temperature of 259K, which is 29K colder than the approximate average temperature of 288K.

However, when pushed a bit, even the catastrophic man-made global warming alarmists concede that the Earth's surface is not a true black body radiator.  They say that the water and the organic materials that cover much of the Earth's surface all have an emissivity near 0.95.  They claim the Earth's surface emissivity is 0.95, nearly as high as that of a black body radiator.  Using that emissivity instead of that of the black body radiator of 1, the calculated temperature becomes 262.3K.  This is 25.35K colder than the best estimate for the Earth's average temperature, which is about 14.5C or 287.65K.

Physicists do not think of water, plants, and minerals as near black body radiators.   A black body radiator is supposed to have the property that it readily absorbs radiation of all wavelengths and emits radiation in a continuous spectrum.  It has to have a harmonic oscillator response at each and every frequency, both to absorb and to re-emit radiation energy.  In a black body cavity, the emissivity of a surface and the absorptivity must be equal.  Those who back the greenhouse gas hypothesis of strong back radiation and warming of the Earth's surface, claim that not only is the surface emissivity nearly that of a black body radiator, but that 100% of any back-radiation from greenhouse gases is absorbed by the surface.  In emission mode, the Earth's surface is either a black body emitter or nearly so, they say, and it is a black body absorber of the thermal infrared radiation emitted from the Earth's surface.

Real materials do not have excitable harmonic oscillator responses at all frequencies.  In fact, infrared spectroscopy has long taken advantage of the discrete and characteristic absorptions that different materials have to identify those materials.  We can use an infrared spectrometer to determine whether a material has water in it.  We use it to distinguish polypropylene from polyethylene, polystyrene, and cellulose.  It can be used to identify various sugars, starches, and proteins.  Yet, the climate alarmists often claim that organic materials all behave very much like black body radiators and absorbers, emitting and absorbing radiation at all frequencies, at least all thermal infrared frequencies.  Well, no they do not.

That the Earth's surface materials are not black body radiators and absorbers matters.  For instance, if the emissivity is calculated which would cause the Earth with no greenhouse gases to have an average temperature of 287.64K or 14.5C with the absorbed solar insolation average power density we calculated above of 255 W/m^2, then the emissivity would be 0.657.  Is this a plausible effective emissivity for the real Earth surface?  And note that if the surface emissivity is less than 0.657, then there have to be other cooling mechanisms to bring down the equilibrium radiative temperature which then would be higher than 14.5C.

Let us look at a portion of the Earth's radiative output according to the IRIS satellite looking down on the dry Sahara Desert in the figure below.  Note that both at the low wave number end and the high wave number end this spectrum is truncated.  It is truncated at the low wave number end because most infrared spectrometers use window and other materials which absorb too much infrared radiation at lower wave numbers.  There are specialized near infra-red instruments that can make measurements in that lower number range by using different materials.  I do not know why this spectrum is cut-off at the larger wave number end.  My infrared spectrometer acquires spectra from 400 to 7800 wave numbers.  Customarily, we cut our spectra off at 4000 wave numbers, because we almost never see any absorptions in materials between 4000 and 7800 wave numbers.  This figure also presents a model that calculates infrared absorption by greenhouse gases, which curve is shown in red.  At the moment, we are not interested in this.

Fig. 2.  A portion of the infrared emission spectrum of the Earth in the Sahara Desert region as seen from space.  Note that the infrared behavior of materials in the Earth's surface from 400 to well beyond 1500 wave numbers is the wavelength of relevant response for these materials.  The satellite spectrometer measures power and does not actually measure the black body temperatures given by the dashed lines.

Since water covers about 70 to 71% of the Earth's surface, let us examine the infra-red absorption spectrum of water.  It is presented in Fig. 3.

Fig. 3.  The FTIR infrared absorption spectrum of tap water.  This is not the absorption spectrum of a black body absorber.

Let us assess this water spectrum.  First, the absorption from 4000 to 7800 wave numbers was zero.  Second, if we generously assume that the maximum absorption here at about 400 wave numbers is a wavelength at which water is sort of a black body absorber, then if at every other wavelength it absorbed the same amount of radiation, we might take this to indicate a material with an emissivity of 1 over these infrared wavelengths which are found in the same range as the emissions from the Earth.  Note that if CO2, whose main absorption is shown in Fig. 2, were causing radiation to be emitted from it back to surface water, the water absorption spectrum does not indicate black body absorption in the vicinity of 675 wave number.

In most other parts of the water spectrum the absorption is even lower.  The integrated area under the absorption curve compared to the area if absorption were everywhere from 400 to 4000 wave numbers equal to the maximum absorption is a fraction of 0.192.  The settled science of the global warming community says that the emissivity is equal to the absorptivity.  This is not actually true for non-black body materials, but if it were true for water, then it would be reasonable to think of this as the average emissivity over this range, and this is much smaller than 0.95.

Could it be that land material emissivities are gigantic and very black body like?  The greenhouse gas proponents claim it is for organic materials, such as plant life.  Let us examine a blade of green grass.

Fig. 4.  The FTIR infrared absorption spectrum of a green blade of grass.  Note that land surface temperatures are supposed to be made 1.5 meters above a field of grass.  This absorption spectrum is not that of a black body absorber.  Grass is full of water and the absorption peaks for water are clearly in this spectrum with some additional peaks due to the organic components in grass blades.

If we take the absorptivity of the green grass blade as one when the absorption across the range is equal to its maximum amount, then the ratio of the actual absorption to that is only 0.207.  A green leaf from a bush is similar, though slightly more absorbing by the maximum absorption criterion without taking into account the absolute absorbance from the spectrum.  The maximum peak absorbance for the bush leaf was less than that of the grass blade.

Fig. 5.  The FTIR infrared spectroscopy absorption spectrum for a green leaf from a bush.  As with a blade of grass, this leaf is full of water and the water peaks are all present.  

Taking the ratio of the actual infrared absorption to that if there were uniform absorption at the maximum value, the absorptivity is then 0.269.  This method should produce an upper bound on the absorptivity.  Then if we say as the greenhouse gas advocates do that the absorptivity equals the emissivity, we have an emissivity of less than 0.269.  Comparing this to the green grass blade, it is actually clear that the maximum absorption was lower for the bush leaf than for the blade of grass.  It is not really significant that the ratio value calculated is larger than that for the blade of grass.  What is important is that there is good reason to believe neither plant is even close to that of a black body radiator.

Is the story different for soil?  The FTIR infrared absorption spectrum follows:

 
Fig. 6. The infrared absorption spectrum of a moist and fairly rich soil.  The maximum absorption just above 400 wave number is actually very high, but the integrated absorption ratio to that of the constant absorption at the maximum rate is very low.

The soil spectrum has less water in it than the grass and bush leaf spectra showed.  If we most generously take the maximum absorption to be that of a black body radiator, the actual absorption is only 0.123 of that.  Again, if the emissivity were equal to this upper bound absorptivity level, than it would also be only 0.123.  Again, this is far short of 0.95.

Of course soil has both inorganic minerals in it and decayed plant materials.  Let us examine the infrared absorption spectrum of a feldspar mineral:

Fig. 7.  The FTIR infrared spectroscopy absorption spectrum of a mineral feldspar material.  This particular feldspar has sodium oxide, potassium oxide, aluminum oxide, and silicon dioxide in three different chemical phases.  Two of the phases are significantly hydrated.  This then is an example of a complex mineral of a type fairly common in the Earth's crust.

Relative to the strongest absorption peak, the absorptivity overall in this range of the infrared spectrum is about 0.117.  If the emissivity were equal, it would not be greater than 0.117 also.

Finally, let us consider a common, but simple mineral, sand:

 Fig. 8.  The infrared absorption spectrum of dry sand is shown.  The strongest absorption peak is actually very weak compared to those of the other materials considered.

Treating the strongest peak as locally the equivalent of a black body radiator is surely wrong.  Give this peak all the spinach you want and it will not become black body radiator like!  This is a weakly absorbing peak.  But, again being otherworldly generous, if it were the level of absorption of a black body radiator, this sand would have an absorptivity of no more than 0.098.  This is about an order of magnitude lower than the Earth surface emissivity claimed by greenhouse gas supporters to be 0.95.

In another post, I found the Earth's effective surface emissivity to be between about 0.45 and 0.50.  Because the absorptivity values in the infrared spectra I have given above are so low, either the emissivity values are significantly higher than the absorption values or there would appear to be a problem.  Now while the emissivity values can be higher for a non-black body material, much of the explanation is found in another simple matter which has been overlooked by the incredibly well-funded settled scientists of the greenhouse gas theory.

The power density side of the Stefan-Boltzmann equation P/A = ε σ T⁴, uses a very simple and gross idea of area.  The square meter is defined by a square with sides of 1 meter each.  But when the Earth is radiating from a grassy field, the radiating surface area is not 1 square meter inside that square with sides of 1 meter length.  The relevant emitting surfaces are the area of every blade of grass and any exposed soil between the blades of grass.  The emitting surface area may be 2 or 2.5 times the nominal area of 1 square meter of ground.  Even the sandy beach will have the surface area given by following the surface area of every grain of sand.  Surely there is emission from the ground under the tree, the trunk, the branches, and the leaves in many land areas.  Even the oceans will have maybe 1.2 times the emitting surface area of the nominal square which is used on the power side of the radiation equation due to waves and ripples.  

The energy the surface absorbs from the sun is distributed to some depth in the oceans or into the soil or rock surface on land.  The heat absorbed on average each day is far more reasonably provided by the simple 1 meter by 1 meter average power density provided by the sun.  But that energy is then also supplied to each bit of infrared radiation emitting surface area as the emission depletes the stored energy in the surface and the near surface.  Thus, effectively, the area on the right side of this Stefan-Boltzmann equation representing surface emission is almost always larger than the area on the left side taken to represent the area of solar insolation absorption.  Most surfaces on the Earth surface will definitely have a larger emitting surface area.  This has the effect of multiplying the emissivity of the emitting material by the ratio of the greater surface area emitting.  The ratio of the emitting surface area to the solar radiation area is multiplied times the normal laboratory measured emissivity of the emitting material.  The increase in the effective emissivity of the Earth surface is substantial.

Now note that the greenhouse gas claim that infrared absorbing gases cause a very large increase in the Earth's surface temperature is based on a claim that the emissivity of water and plants is 0.95 or even 1.0.  If this were so, then this observation of the need to account for the greater emitting surface area would create a whole different problem for their theory.  What they claim is a measured 0.95 emissivity for water would become a super black body emitter with an effective emissivity of say (1.2)(0.95) = 1.14 due to the waves of the ocean increasing its emitting surface area.  The situation over the prescribed grassy field for land surface temperature measurements would be even worse.  One might easily have an emitting area two times the nominal square area, so the grassy surface emissivity would be 2 (0.95) = 1.9.  Imagine all of that radiation converging on their temperature station only 1.5 meters above this super, super black body radiator!

This conclusively demonstrates that it is very plausible based on laboratory spectra on the materials from the surface of the Earth that the effective emissivity of the surface of the Earth is less than 0.5.  Any effective Earth surface emissivity less than 0.657 as I showed means that there is no warming effect due to the so-called greenhouse gases. Indeed, an effective Earth surface emissivity of less than 0.5 is consistent with a net cooling effect due to the infrared active gases of water vapor, carbon dioxide, methane, and nitrous oxide.  I have stated elsewhere why this is what we should expect.  I will write more about this in the near future.