One of the
keystone claims of the catastrophic man-made global warming (CAGW) hypothesis is that
infra-red thermal radiation emitted from a cooler body and incident upon a
warmer body is entirely absorbed by the warmer body. They claim that both
the warmer and the colder body emit photons with a power density proportional
to T4. What is more, the flux
of photons emitted from a body due to its temperature is said to be unchanged by the
nearby presence of a body at a different temperature. As long as each body is at a given
temperature, each is claimed to send out a flux of photons proportional to T4
just as though it were isolated in vacuum and surrounded by a space at a
temperature of absolute zero, 0K.
At temperatures near or less than 300K, they claim that both the warmer and the cooler bodies absorb all of these infra-red photons when they are incident upon a body. Thus, the surface of the Earth absorbs all of the photons emitted by infra-red active gases (commonly called greenhouse gases) in the atmosphere, even though the emitting gas molecules are usually cooler than the surface of the Earth is. The energy of these absorbed photons heats the Earth’s surface and makes it warmer than it otherwise would be. In addition, the surface of the Earth emits photons as though it were interfaced with vacuum and as though it were in radiative equilibrium only, with no other energy transport mechanisms acting on it. This claim is essential to the claim that an increasing concentration of CO2 molecules in the atmosphere will cause a catastrophic increase in the temperature of the Earth’s surface.
These thermal radiation assumptions are built into such Earth Energy Budget Schematics as the 1997 Kiehl-Trenberth diagram featured in the 4th UN IPCC climate report of 2007. The Earth energy budget below produced by NASA in 2013 retains such concepts of physics in its energy fluxes as well. Note that while solar insolation (in yellow) incident upon the Earth's surface is partially reflected, no part of the very large back radiation (in red) is reflected. Note that the surface-emitted radiation is of 117% of the top of the atmosphere solar radiation and back radiation is 100%! Given that reflected solar radiation totals 29%, so atmosphere and surface in total could not possibly absorb more than 71% of the solar radiation incident upon the Earth.
In fact, some of the 23% of solar radiation absorbed in the atmosphere before it reaches the surface must be re-emitted to space before it ever has an opportunity to be incident upon the surface. This fraction should be upwards of half of the 23% of the solar radiation absorbed directly by the atmosphere. Consequently, it is clearly unreasonable to think that more than 48% + 0.5 (23%) = 59.5% of the solar insolation could ever be emitted from the surface by the sum of all surface cooling mechanisms. These mechanisms include convection and evaporation, in addition to thermal radiation. Consequently, the maximum thermal radiation that might be emitted from the surface is 59.5% - 5% - 25% = 29.5%. This is a simple matter of energy conservation.
Nonetheless, a fantastic theory of energy generation due to thermal radiation is put forth by the "settled", "consensus" theory of catastrophic man-made global warming based upon a wild generation of photons ungoverned by any actual consideration of the real many-body electric field conditions. The catastrophic man-made global warming greenhouse hypothesis teeters on a concept of free photons. They require no energy to form, but they deliver energy in spades.
The Earth's surface radiates energy in this scheme as though its temperature were 289.4 K with a black-body surface emissivity of 1.00 and it is radiating into vacuum at a temperature of 0K. Numerous other falsehoods include the claim that only 12/117 = 0.103 of the radiation from the surface of the Earth is not absorbed by the atmosphere and the claim that the atmosphere radiates much more energy to the surface than it does into space.
As I have shown in other posts (such as here, here, here, and here), there are numerous violations of physics in these Earth Energy Budget diagrams that form the essential basis of the catastrophic man-made global warming hypothesis. This post is going to take a very fundamental look at the thermal radiation in a cavity in thermal equilibrium. This is commonly called black body radiation. We will note the conditions actually required and that they are not those commonly stated. Then we will ask which of the conditions and which of the results might be carried over to the behavior of thermal radiation from surfaces not in an idealized black-body cavity. We will then set up some very simple models of bodies in thermal steady state conditions to learn further lessons about thermal radiation which are expected to apply to real-world phenomena and which are critically important to the evaluation of the catastrophic man-made global warming hypothesis as exemplified in the above Earth Energy Budget or any of the many similar energy budgets.
Our initial examination of cavity radiation properties in a cavity under thermal equilibrium in a vacuum will call upon significant knowledge of mathematics, but the explanations are complete with little left as exercises for the reader. We will use thermodynamics concepts mathematically and refer to qualitative concepts of electromagnetic radiation.
Let us examine an enclosure whose walls are at a constant temperature. The radiation within this enclosure is in thermal equilibrium with the walls of the enclosure. In such a case, the energy density of the radiation, e(T), is dependent on the temperature, but independent of the volume enclosed. If the volume enclosed is V, then the total radiation energy, U, is Ve(T). For our enclosure in which the thermal properties are described only in terms of the intensive variables T (temperature) and P, the radiation pressure on the walls, and the extensive variables V and S, the entropy, we have dU = T dS – P dV. Or, T dS = dU + P dV.
The radiation pressure on the walls of the uniform temperature enclosure is equal to twice the momentum, p, component perpendicular to the wall of photons reflected off the wall times half the perpendicular velocity component, where the perpendicular velocity component tells us how many photons strike the wall per second and only half have a perpendicular velocity component toward the wall. The pressure P for a photon gas exerted in the x-direction on area A of the wall will be summed over all i = 1 to N photons:
P = ½ ∑ 2 pix vix
/ V = ⅓ U/V = ⅓ e,
U is just ∑
pix vix + piy viy + piz
viz for photons, so for random motion, the summation over only one
component is one-third of this sum. Note
that the radiation pressure on the wall does not require that one photon
incident upon the wall be absorbed and another emitted. One gets the same result with a photon
reflecting off the wall without absorption just as one would calculate the
pressure on a wall for a perfect gas under the assumption of elastic collisions
with the wall. Such a gas of point
particles has a pressure of ⅔ e however, since the average kinetic energy of
the perfect gas particles is ½ the dot product of the momentum and the velocity
and not the dot product of the momentum and the velocity as is the case for
photons.
Thus,
T dS = dU + P dV = d(Ve) + P dV
= V de + e dV + ⅓ e dV
= V (de/dT) dT + (4/3) e dV
We also have
T dS = T (∂S/∂T)V dT + T
(∂S/∂V)T dV,
So we match
up the terms for dT and for dV:
(∂S/∂T)V = (V/T) de/dT
(∂S/∂V)T = 4e/3T
Since the
order of differentiation does not matter, it is also the case that
{∂[(∂S/∂T)V]/ ∂V}T
= {∂[(∂S/∂V)T]/ ∂T}V
{∂[(V/T) de/dT]/ ∂V}T = {∂[4e/3T]/
∂T}V
(1/T) de/dT = (4/3T) de/dT – 4e/3T2
de/dT
= 4e/T
e = aT4
The equation
for the radiation energy density is Stefan’s Law and a is Stefan’s constant. If we open a peep hole into this cavity, the
flux of energy emitted has this energy density of aT4, though an
equal amount of energy has to then be supplied to the enclosure to maintain it
at temperature T as it loses energy through the peep hole. Within the cavity emitting this energy, we do
not know if the photons incident upon the walls are absorbed and replaced with
the emission of another of the same energy or whether those photons are simply
reflected from the wall. The above
results are the same in either case.
When the cavity above is at a constant temperature Tf with the walls everywhere at that temperature, we know that the emission of photon power from the interior walls equals the absorption of photon power by the walls. If the cavity was at an initial lower temperature Ti and then was heated until it came to a new higher equilibrium temperature of Tf, the emissivity of the walls had to be greater than the absorptivity of the walls during that heating process in order to increase the energy density in the cavity from aTi4 to aTf4. Because the energy density depends on the fourth power of the temperature, a doubling of the temperature requires a factor of 16 times greater energy density. The emissivity of the walls during such a heating process must be much greater than the absorptivity to allow that 16-fold photon energy density increase. The emissivity and the absorptivity of the walls of the cavity are not just a function of the material. They are a function of whether the material is in radiative thermal equilibrium or not as well. It is very important to realize this. When the cavity was at Ti in a radiative equilibrium condition, the emissivity and the absorptivity were equal, but when the cavity was further heated to the new equilibrium temperature of Tf, they were not equal when Ti < T < Tf. Conversely, during a cooling from a higher equilibrium temperature to a new cooler equilibrium temperature, the absorptivity of the walls would have to be greater than the emissivity of the walls.
Among the properties of the cavity with volume V in radiative thermal equilibrium at temperature T is that:
U = a V T4
P = ⅓ a T4
S = (4/3) a V T3
We can
calculate the chemical potential, µ, which measures the ease with which the
number n of moles of photons adjusts to keep the energy density constant in the
cavity in radiative thermal equilibrium:
µn = U – ST + PV
µ = 0
Photons are
a very special kind of boson. When they
are in radiative thermal equilibrium in a volume V at a constant temperature T,
their chemical potential is zero. The
number of photons in the cavity is strictly determined by the temperature of
the walls.
Applying Bose-Einstein statistics for a boson gas with a chemical potential of zero and integrating over the photon standing wave states available in the cavity in radiative thermal equilibrium, one finds that
a = π2 k4 / 15 ɦ3
c3 = 7.57 x 10--16 J / m3 K4,
where ɦ is
Planck’s constant h/2π, k is the Boltzmann constant, and c is the speed of
light. Note that a, Stefan's Constant, is not the same as the Stefan-Boltzmann Constant.
The cavity I
have discussed here is almost always called a black-body cavity, with the
implication that every photon incident upon an interior wall is absorbed and
replaced by another photon of equal energy.
This is usually said to be a condition in which the emissivity and the
absorptivity of the walls of the cavity are both 1. Yet, at no point was such a condition ever
used in the derivation of any of the properties of a cavity of volume V under a
condition of radiative equilibrium at temperature T. The only requirement was that a photon
incident upon a wall was either reflected or if it were absorbed, then it had
to be replaced by another photon emitted from the wall at the same energy,
which is easily done because the photons in the cavity at radiative thermal
equilibrium have a chemical potential of zero.
When the thermal radiative properties of a cavity in radiative equilibrium are applied to a surface which is not in thermal radiative equilibrium in a closed cavity condition, we must be very aware of the differences. When such a surface is surrounded by vacuum and all of its thermal conditions are described by its radiative properties, there will be an infinitesimal volume in the vacuum arbitrarily close to the surface which has an energy density e as described above from the equilibrium cavity condition. Let us imagine two infinite planar surfaces which are very close compared to their dimensions, with only vacuum between them and beyond them. If these two surfaces are at the same temperature T and in equilibrium, the energy density e is clearly aT4 everywhere in the volume between the two planes.
Now let us suppose that one of the two planes is radiatively heated from the back side and reaches a new higher temperature Th, at which the temperature becomes stable. As this hotter plane radiates energy at its higher temperature, it will increase the temperature of the nearby plane. If the cooler plane has a heat capacity, it will take a while to heat up to its own new warmer equilibrium temperature. Assuming this initially cooler plane only loses heat by radiating it toward the vacuum at absolute zero, it will achieve that equilibrium new temperature such that it is equal to that of the plane heated on its backside. Once again
the energy density e is clearly aT4 everywhere in the volume between the two planes, where T is the new higher temperature.
If this second plane has some other uniform means of losing heat than radiation over its surface area, its new equilibrium temperature will not be as high as Th. Let the lower temperature plane have the temperature Tc. Now the energy density in the vacuum arbitrarily close to either side of the hotter plane will be aTh4 and that arbitrarily close to the cooler plane on either side will be aTc4. This is a condition for a plane at a temperature T with a vacuum interface that the energy density e = a T4. Said in a different fashion, the perfectly thermally conducting plane would not be at a temperature T if the energy density of photons in the space immediately at the surface of the plane was not aT4. In the vacuum space in between the two planes the energy density will be a continuously decreasing function from a high value of aTh4 at the hotter plane surface to a lower value of aTc4 at the cooler plane surface. The hotter plane emits and absorbs photons as a surface at an energy density of aTh4. The lower temperature plane absorbs and emits photons as a surface would with an energy density of aTc4 in the vacuum immediately adjacent to the surface. Its lower temperature compared to that of the plane radiantly heated from the backside alerts us that this second plane has some non-radiative means of heat loss.
The energy density between the two planes is not a( Th4 + Tc4), which is what it would have to be if it were true that the photons incident upon the warmer plane from the cooler plane was that due to an energy density of aTc4 even as the emission energy density of photons from that warmer surface was that due to an energy density of aTh4. There has to be a photon energy density gradient between these two planes and this is not consistent with all of the emitted photons from the cooler plane surface being incident upon the warmer plane surface as is usually hypothesized by the advocates of catastrophic man-made global warming. They deny the existence of an electromagnetic field gradient and a gradient in the energy density with such rhetorical whimsy as “The photons emitted by the cooler surface must be absorbed by the warmer surface because the warmer surface cannot poll incident photons to see if they came from a warmer or a colder surface.” The electromagnetic field in effect does just that by controlling what photons are created by the field. We shall see that this has critically important implications for the CAGW hypothesis.
We also have
to take note that there is a flow of photons from the warmer surface to the
cooler surface. This case is not a true
equilibrium condition of the kind found in the cavity where all surfaces were
at the same temperature. The chemical
potential of photons in this case is not zero.
A photon incident upon the cooler surface can now be absorbed by that
surface without the emission of a photon of the same energy. Indeed, the other side of the cooler plane
surface is radiating energy into space, so there is a net flow of energy across
this plane. More photons have to be
absorbed on the side facing the hotter plane than are emitted from the side
facing only vacuum as the cooler plane is initially warmed by the warmer plane. When the temperature becomes constant, then the energy flux of absorbed photons on the warmer side must equal the energy flux of photons out of the cooler plane outer surface facing the cold vacuum plus that of any other cooling mechanism. More photons have to
be emitted from the cooler plane side facing only space than are absorbed on that same surface. Indeed, when space is at absolute
zero, no photons would be absorbed on that side at all. The absorptivity of the surface of the plane
facing the warmer plane is greater than the emissivity of that surface, which
is at variance with the conditions of the cavity in radiative thermal
equilibrium.
The photon radiation pressure P on the plane is proportional to aT4 of the plane. This photon pressure is proportional to the number of photons incident upon the plane, as is the amount of energy absorbed or emitted from the plane. Therefore, we know that the power density of absorbed or emitted photons will be proportional to the energy density e and therefore to aT4. That new proportionality with T4 is given by the Stefan-Boltzmanm constant,
The photon radiation pressure P on the plane is proportional to aT4 of the plane. This photon pressure is proportional to the number of photons incident upon the plane, as is the amount of energy absorbed or emitted from the plane. Therefore, we know that the power density of absorbed or emitted photons will be proportional to the energy density e and therefore to aT4. That new proportionality with T4 is given by the Stefan-Boltzmanm constant,
σ =
5.6697 x 10—8 W / m2K4,
when every incident
photon is absorbed, as it is for a black body.
If the fraction of photons absorbed is less than one, we call that an
absorptivity of the fraction α and if the fraction of emitted photons is less
than the maximum, we call that the emissivity fraction ԑ.
So in this steady-state system, the
outside surface of the cooler plane is emitting photons into vacuum in
accordance with the energy density e = aTc4 at the
rate per unit area of εc σ Tc4, where εc
is the emissivity of the cooler plane outer surface. This plane
is also losing energy by non-radiative means at the rate Qout. The absorption of energy by the cooler plane
on the side facing the warmer plane is given by αc σ Tc4, which is the flow of
energy into the cooler plane. The origin
of the energy was the emission from the warmer surface given by εh
σ Th4. Thus,
we have
εh
σ Th4 = αc
σ Tc4 = Qout + εc σ Tc4
Qout
= εh σ Th4 - εc σ Tc4 and
Qout = (αc – εc) σ Tc4
So the energy outflow by
non-radiative means is equal to the difference of the radiated energy flux of the
planes by the surfaces of each emitting toward the cooler outside environment as given by the first expression for Qout. This can readily be mistaken as
the energy flow between the planes, though it is not actually so. The planes will not have a different steady state
temperature unless Qout is non-zero, so the absorptivity of the
cooler plane surface facing the warmer plane must be greater than the emissivity
of the outside surface of that same cooler plane as given by the second expression of Qout. The absorptivity and the emissivity of this plane will only be equal when this plane has the same temperature as the other plane. More photon energy must be absorbed than is
emitted from the cooler plane. Since the
photon energy emitted from the warmer plane is equal to the photon energy
absorbed by the cooler plane and the photon energy density between the planes
is nowhere greater than that immediately outside the warmer plane, all of the
absorbed photons come from the warmer plane.
No photons emitted from the cooler
plane are to be found in the vacuum between the planes. If they were,
there would be a clear violation of the Law of Energy Conservation, at least
given that there is reason to believe that the one lesson that one takes from
the physics of a cavity in thermal equilibrium is that the internal energy
density is given by e = aT4 and that in general, surfaces at
a temperature of T will have an energy density in vacuum immediately adjacent
to them at the same energy density such a surface would have in the cavity at a
thermal equilibrium condition. In this case, not only do no
thermally-emitted photons from the cooler plane become absorbed by the warmer
plane surface, but none are even incident upon that surface. That this is
the case is extremely contrary to the physics assumed in the catastrophic
man-made global warming hypothesis.
To apply this model to climate physics with the warmer plane the surface of the Earth, many changes have to be made. Among these, Qout for the cooler plane of air near the surface would be a combination of that fraction of infra-red radiation absorbed by water vapor and carbon dioxide molecules which was transferred to nitrogen and oxygen molecules prior to its re-emission as infra-red radiation due to the high rate of molecular collisions and also due to the loss of heat as the air slab rises in convection. Within one mean free path length for the absorption of the long wave infra-red emitted from the surface, the Qout value becomes substantial as radiant energy is converted into kinetic energy in infra-red inactive molecules.
To apply this model to climate physics with the warmer plane the surface of the Earth, many changes have to be made. Among these, Qout for the cooler plane of air near the surface would be a combination of that fraction of infra-red radiation absorbed by water vapor and carbon dioxide molecules which was transferred to nitrogen and oxygen molecules prior to its re-emission as infra-red radiation due to the high rate of molecular collisions and also due to the loss of heat as the air slab rises in convection. Within one mean free path length for the absorption of the long wave infra-red emitted from the surface, the Qout value becomes substantial as radiant energy is converted into kinetic energy in infra-red inactive molecules.
Now let us
assume that we have the same radiative heat applied to the outside of the
warmer plane as before and we introduce a perfect gas between the two planes
which has no molecules in it capable of absorbing infra-red energy or any
electromagnetic wave energy of any sort.
Heat transport between the two planes is now both direct radiative
transport and a conductive transport due to gas molecules colliding with the
warming wall and picking up heat as an increase in molecular kinetic energy and
carrying that energy to the cooler wall.
What is the response of the system?
Clearly, the temperature difference between the walls is reduced. The hotter wall temperature drops and the
cooler wall temperature increases. As
the energy flow due to the gas molecule transport of energy increases, the
radiative heat flow decreases as the temperature difference between the
surfaces decreases.
Why does this happen? The electromagnetic field in the volume just outside the warmer plane is determined by the oscillating dipoles in the surface. The amplitude of the oscillations is determined by the temperature. But when a gas molecule collides with this surface and picks up energy from the surface, it does so by decreasing the kinetic energy of one or more of these oscillating dipoles. Those oscillating dipoles are no longer able to pour all of their energy into creating as high an electromagnetic field as they did in the absence of the gas molecule collisions with the surface. The reduced field outside the surface must result in a reduced photon energy density outside the plane surface. Thus, what would have been a photon energy density of aTh4, must now be less than that. We know that this results in a decrease in the number of wave nodes, so the value of Stefan’s constant a is now too large. The energy density of photons is now bTh4, where b is less than a. The emission of photons is accordingly reduced.
Let us now add a film of water to the inside surface of the warmer plane. If the plane is not below the freezing temperature of the water, some of the water will evaporate. This will cool the warmer surface, taking energy out of the oscillation of some of the dipoles in the surface. Thermal radiation emissions will decrease once again, both due to the temperature of the plane being decreased and because the constant in the photon energy density, bT4, just outside the surface will decrease once again. The difference in temperatures between the warmer and the cooler planes will once again decrease. Ah, but you say that water vapor is a greenhouse gas. It is capable of absorbing infra-red radiation. Perhaps that causes a warming effect.
The latent heat of evaporation for water is a large heat which is not easily offset by other effects. If the water vapor concentration is similar to that in air and if the air is near STP conditions, then each water vapor molecule absorbing infra-red radiation is about 4 times (10 times according to Prof. Happer) more likely to lose that infra-red absorbed energy to the surrounding air molecules than it is to re-emit the infra-red energy due to the very high frequency of gas molecule collisions. Half of what it re-emits is on its way toward the cooler plane again in any case. The 80% transferred to heating the non-infra-red active molecules of the air just speeds up their kinetic transport of energy to the cooler plane. So only about 10% of the energy absorbed heads back to the hotter plane, which just equals the 10% of the energy radiated toward the cooler plane. This leaves the net effect of the photon absorption by the water vapor molecule to be a heating of the air further from the surface of the warm plane, which increases the air molecule transport of energy between the planes. It is very unlikely that the 10% of the radiant energy absorbed by water vapor and sent back to the hotter surface is a very big effect compared to the cooling effect caused by the evaporation of water on the warmer surface and the speed up of the conductive heat transfer in the gas.
What is more, the excitable water vapor molecule holds more energy due to its additional modes of internal excitation compared to perfect gas molecules. Thus, it can transport more energy to the cooler plane and transfer that greater energy to the plane upon colliding with it. Those internal modes of excitation cause the water molecule to have a higher heat capacity. The water molecule is also lighter than N2 and O2, so at a given temperature, its velocity is greater and it speeds up the transport of heat to the cooler plane relative to the heavier gas molecules.
Complex molecules have been used to fill the space between double glass windows in experiments. It was thought that doing so would impede heat flow as the complex molecule absorbed infra-red energy emissions from the warmer pane of glass. What is found is that the increased heat capacity of the complex molecules simply allows them to transport more energy and transfer it to the cooler glass pane upon colliding with it. These molecules do not succeed in slowing the heat transport at all. The fill gases that work the best are both heavy and simple. Being heavy, their velocity at a given temperature is less than that of a lighter gas molecule. Being simple, such as the monatomic inert gases argon and xenon are, they have a low heat capacity. Both are heavier than N2 and O2, the principal gases in air. In particular, CO2 performs badly as an insulating fill gas in such a case. Dry air is actually a fairly good insulating gas.
There are other important differences between the Earth’s surface and the walls of a cavity. Water covers about 71% of the Earth’s surface. Solar insolation incident on a water surface is not absorbed almost entirely within a micrometer of the water surface. It is absorbed in the first few tens of meters instead. Once that energy is absorbed and distributed over that considerable depth, heat flows to still deeper depths due to the thermal gradient that usually exists with the deeper water being colder. While the absorption of radiant energy on the land surface does occur much, much closer to the surface, heat also flows through soil, sand, and rock there to sub-surface depths. The heat flow from the near surface to deeper depths on either land or water varies throughout a day as the thermal gradient in these zones changes quite a bit through the daily cycle. So, the surface is subject to still another cooling mechanism during the day and often a warming mechanism at night, causing a constantly varying effect in the daily cycle. Thermal equilibrium conditions locally rarely exist and even steady state conditions rarely exist.
In this context, the evaporation of water occurs at a rate which increases exponentially with temperature. Therefore, the water evaporation rate is commonly much higher in the afternoon period of a day and is varying considerably through the course of the day. As is the case with the conduction of heat through water, soil, sand, and rock; the conduction, convection (thermals), and advection (wind) of heat through the air away from the surface also depends upon the commonly varying temperature differentials between the surface and the lower atmosphere through the daily cycle.
Consequently, the Earth’s surface is subject to not just varying thermal radiation absorption and emission in the daily cycle, but it is also subject to varying water evaporation cooling, varying air conduction, convection, and advection cooling, and varying rates of heat flow to deeper depths below the surface or even from deeper depths toward the surface. The kind of radiative thermal equilibrium in the cavity which leads to a photon chemical potential of zero, to emissivity and absorptivity of the cavity wall being equal, and to a surface emitting a number of photons and a total photon energy dependent only on the temperature of the surface generally does not exist during the daily cycle of the Earth’s surface. It is very important that this be borne in mind when discussing the radiant energy transport based on cavities with walls at a constant temperature and only in radiative equilibrium.
Advocates of the catastrophic man-made global warming hypothesis tend to make many ill-considered assumptions about radiative heat flow. They do not understand cavity radiation in thermal equilibrium, they do not understand radiative heat flow between surfaces of differing temperatures, and they do not understand how radiative transport of heat works when other mechanisms are present that also transport heat. I have taken pains to point out a number of these misconceptions in my post Why Greenhouse Gas Theory is Wrong -- An Examination of the Theoretical Basis.
Among the consequences specifically of the competing energy transport mechanisms is the fact that the Earth’s surface as a whole has an effective emissivity of a bit less than 0.5. This commonly drives advocates of the common theory for catastrophic man-made carbon dioxide emissions global warming bananas. They keep pointing at measurements that claim that water, soil, and organic plant materials have emissivities near 0.90 to 0.95, meaning that the value of ε in ε σ T4 is 0.90a to 0.95a. When other heat transport mechanisms are prevented in an experimental measurement, the surface emissivity due to radiation alone will increase relative to its value when other heat transport mechanisms are available. In addition, as discussed above, a surface emissivity or absorptivity is not just a function of its material, but is also a function of whether the surface is changing its temperature. The surface of the Earth is locally changing its temperature all the time.
If the Earth's surface temperature were only a function of absorbed solar insolation and its thermal emission, evaporation, and convection cooling, then the emissivity of the Earth's surface would average a mere 0.375 using the values of the NASA Earth Energy Budget above for these values. This is because the radiation emitted is equal to the absorbed solar insolation, 48%, minus the convection loss, 5%, minus the evaporation energy loss of 25%. Then only 18% would be left for thermal radiation emissions from the surface and 18%/48% = 0.375. That it is actually higher than this, is evidence that the Earth's surface temperature is not due only to the absorption of solar insolation. Its temperature is raised by the action of gravity on the molecules of the atmosphere and that allows a higher effective thermal emissivity in the range of 0.43 to 0.48.
I hope this post will help those interested to gain a better understanding of these critical thermal radiation issues and the implications for the failed CAGW hypothesis.
21 May 2015 Update: Important additional deductions were made showing that photons from a cooler body are not absorbed by a warmer body, at least in one simple system.
26 May 2015 Update: I added an Earth Energy Budget diagram to provide more context for how the discussion to follow applies to the usual explanation of the physics of the catastrophic man-made global warming hypothesis. I also added bold and italics to emphasize several important deductions about thermal radiation. Further comments about the effective emissivity of the Earth's surface were also added near the end of this post.
27 and 30 May Update: I changed the discussion of the two parallel radiating planes in vacuum.
19 June: Further editing.