This is essentially a re-posting of an article of 2 November 2017 which has been edited with additions several times since the original version appeared. The essential physics has not been changed, but I have tried to help the reader understand it more readily along with its implications for the fate of the catastrophic man-made global warming hypothesis.
I will present the Consensus, Settled Science solution to the
parallel plane black body radiator problem and demonstrate that it is
wrong. I will show that it exaggerates the energy of electromagnetic
radiation between the two planes by as much as a factor of two as their
temperatures approach one another. As a result, the calculations of the
so-called Consensus, Settled Science dealing with thermal radiation very often
result in violations of the Law of Conservation of Energy.
Their calculations of thermal radiation greatly exaggerate the density of the infrared photon radiation in the atmosphere and the extent of the absorption of infrared radiation by infrared-active molecules, commonly called greenhouse gases, such as water vapor and carbon dioxide. Their theory of the transport of heat energy between the surface of the Earth and the atmosphere and through the atmosphere is very wrong. It exaggerates the role of thermal radiation greatly and minimizes the role of the water evaporation and condensation cycle and the role of thermal convection. It cannot be emphasized enough how harmful their mishandling of thermal radiation calculations is to their understanding of the critical issues pertaining to the Earth's climate and to man's role in changing the climate through the use of carbon-based (fossil) fuels.
In the case in which the two black body parallel plane radiators have the same temperature, the volume between them becomes that of a black body radiator. The fundamental characteristic of a black body radiator is the constant energy density in the cavity. I will show that the so-called settled science treatment, which wrongly takes the primary characteristic of a black body radiator to be that the power of emission of radiant energy is given by the Stefan-Boltzmann Law, clearly violates the real principal characteristic of a black body cavity, namely that its constant energy density, e, is given by Stefan’s Law as
e = aT4,
where T is the temperature in
Kelvin, a is Stefan’s Constant of 7.57 x 10-16 J/m3K4 , and e
is in Joules per cubic meter.
I have already demonstrated the failure of the settled science
treatment of thermal radiation from black body radiators in the form of
concentric spherical shells in a paper posted on 23 October 2017,
entitled Thermal Radiation Basics and Their Violation by the
Settled Science of the Catastrophic Man-Made Global Warming Hypothesis, but I want to post this
solution with its simpler geometry so that the reason the consensus treatment
is wrong will be even more apparent to thinking readers.
Numerous critics of the consensus science on catastrophic man-made
global warming have argued that the Second Law of Thermodynamics claims that
energy only flows from the warmer body to the colder body, but the consensus
scientists have argued that thermodynamics only applies to the net flow of
energy. I have long argued that the reason that radiant energy only flows
from the warmer to the cooler body is because the flow is controlled by an
electromagnetic field and an energy gradient in that field. I will offer
that proof in this paper with some qualification on the energy flow from the cooler body to the warmer body within a tight constraint. The Second Law of Thermodynamics is not invoked
as the basis of the proof in this paper, but the minimum energy of a system
consistent with the Second Law of Thermodynamics does turn out to be a
consistent solution to the problem of thermal radiation, while the Consensus,
Settled Science theory of thermal radiation does not minimize the system total
energy, does not produce the correct energy density of a black body cavity, and
is not consistent with the Conservation of Energy. I have pointed out its
failure to conserve energy in many prior posts.
In a black body cavity, the electromagnetic radiation is in
equilibrium with the walls of the cavity at a temperature T. The energy
density e is the mean value of
½ E·D + ½ H·B,
where E is the impressed electric field, D is
the displacement, which differs from E when the medium is
polarized (i.e., has dipoles), H is the impressed magnetic
field and B is the magnetic polarization of a medium. If
the cavity is under vacuum, then D = E and B = H in
the cavity volume and |E| = |H|, so e equals |E|2.
The mean value of the energy density of the electromagnetic field in the cavity
depends on the temperature and is created by the oscillating dipoles and higher
order electric poles in the cavity walls. The energy density is independent
of the volume of the cavity. The radiation pressure on the cavity walls
is proportional to the energy density.
This physics may be reviewed by the reader in an excellent
textbook called Thermal Physics by Philip M. Morse,
Professor of Physics at MIT, published in 1965 by W.A. Benjamin, Inc., New
York. Prof. Morse wrote it as a challenging text for seniors and
first-year graduate students. I was fortunate to use it in a
Thermodynamics course at Brown University in my Junior year.
Alternatively, see my post The
Greenhouse Gas Hypothesis and Thermal Radiation -- A Critical Review.
Inside a black body cavity radiator at a temperature T, the energy
density, or the energy per unit volume of the vacuum in the cavity is constant
in accordance with Stefan’s Law. If one opens a small peephole in the
wall of the black body cavity, the energy density just inside that peephole is
the energy density of the black body cavity and that energy density is
proportional to the square of the electric field magnitude there. The
Stefan-Boltzmann Law states that the flow rate of energy out of the peephole
when the black body cavity is surrounded by vacuum and an environment at T = 0
K, is given as the power P per unit area of the peephole as
P = σT4
Note that P = (σ/a) e and that e in the T=0K sink is equal to
zero. A change of energy density in the vacuum volume immediately inside
the peephole into the black body cavity as given by Stefan's Law to a value of
zero in the T=0K outside environment causes a power of thermal radiation
emission out of the peephole to the outside environment as given by the
Stefan-Boltzmann Law.
Why is it that a surface which is not a peephole into a black body
cavity might act like a black body radiator? It has to be that the energy
density very, very close to that surface has the characteristic of the energy
density in a black body cavity radiator, namely that
e = aT4.
Any flow of energy out of the surface due to its temperature T
must be caused by this electromagnetic field energy density at the surface
generated by the vibrational motion of electric charges in the material of the
surface.
Here is the problem of the parallel plane black body radiators
diagrammed, where TC is the cooler temperature and TH is
the warmer temperature:
Let us first consider the case that each plane is alone and
surrounded by an environment of space at T=0. Each plane has a power
input that causes the plane to have its given temperature. Each plane
radiates electromagnetic energy at a rate per unit surface area of
P = σT4.
Consequently, if neither plane were in the presence of the other
and each plane has a surface area of A on each side and PCO , PCI , PHO , and PHI are all radiation
powers per unit area, we have
PC = APCO + APCI =
2AσTC4
And
PH = APHO + APHI =
2AσTH4 ,
since in equilibrium the power input is equal to the power output
by radiation.
In the consensus viewpoint, shared by many physicists and by
almost every climate scientist, the parallel plane black body radiators above
are believed to emit photons from every surface of each plane even in the
presence of the other plane with a power per unit area of
PH = σTH4 and PC = σTC4,
just as they would if they were not near one another and they only
cast off photons into a sink at T = 0 K. This viewpoint takes the emitted
radiation as a primary property of the surfaces rather than an electromagnetic
field with a known energy density as the primary property of the surfaces.
Thus, when these planes are in one another's presence this consensus viewpoint says that
PC = APCO + APCI -
APHI = 2AσTC4 - AσTH4
PH = APHO + APHI -
APCI = 2AσTH4 - AσTC4
Note that PC becomes zero at TC =
0.8409 TH and below that temperature PC is
negative or a cooling power in addition to radiative cooling. If TH =
288K, then TC = 242.2K, the effective radiative temperature of
the cooler plane to space if the cooler plane is thought of as the atmosphere
and the warmer plane is the surface of the Earth, both the atmosphere and the
surface act like black body radiators, and the atmosphere receives only radiant
energy from the surface. This is in agreement with calculations I have
presented in the past and is a result which I believe to be correct under the
assumptions, even though in some critical respects this consensus viewpoint is
wrong.
If these planes were isolated from one another and each plane
faced only that T = 0 K vacuum, then one would have
eH = aTH4 and eC =
aTC4,
because these are black body radiator surfaces.
eH = aTH4 in this case.
Unlike the case of concentric spherical shells, which I considered
in Thermal
Radiation Basics and Their Violation by the Settled Science of the Catastrophic
Man-Made Global Warming Hypothesis, there is no divergence or convergence
of the photons emitted from either surface. The relationship of the
radiative power P to the energy density due to that electromagnetic radiation
is always e = (a/σ) P as one traverses the distance between the
planes.
Consequently, the energy density between the planes is
e = (a/σ) PHI + (a/σ) PCI = aTH4 +
aTC4
anywhere between the two planes, because photons have energy no
matter which direction they are traveling and they do not annihilate one
another based on their direction of travel. The total energy between the
planes is that of the electric field or it is the sum of the energy of all the
photons in the space between the planes. The energy density e would then
be the total energy of all the photons divided by the volume of vacuum between
the planes.
Now, let us imagine that these planes are very close together and
the ends are far away and nearly closed. Let us have TC → TH,
then
e → 2aTH4,
but this space between the planes is now a black body cavity in
the limit that TC → TH, and we know by Stefan’s
Law that
e = aTH4
in this case. In addition, we have created a black body
cavity radiator here and P for the walls inside the cavity is actually zero
because the interior is in a state of equilibrium and constant energy density.
P is only P = σ T4 just outside the peephole facing
an environment at T = 0 K. In the above consensus viewpoint case, each
plane surface is emitting real photons, but these cannot annihilate those
photons of the opposite plane. There are no negative energy
photons. These respective photon streams simply add to the total energy
density.
If we did insist on claiming that oppositely directed photons had cancelling energies as is commonly done, then the energy density inside a black body cavity would be zero, not the finite energy given by Stefan's Law. Given the thermal equilibrium in a black body cavity, there are just as many photons traveling in one direction as in the opposing direction, yet their respective energies must be added when determining the energy density. Stefan's Law insists that all photons have positive energy.
If we did insist on claiming that oppositely directed photons had cancelling energies as is commonly done, then the energy density inside a black body cavity would be zero, not the finite energy given by Stefan's Law. Given the thermal equilibrium in a black body cavity, there are just as many photons traveling in one direction as in the opposing direction, yet their respective energies must be added when determining the energy density. Stefan's Law insists that all photons have positive energy.
The consensus treatment of black body thermal radiation doubles
the energy density in a black body cavity, in clear violation of the principal
characteristic of a black body cavity upon which their treatment must be
based. Their treatment greatly increases the energy density between the
planes whenever TC is anywhere near TH, such as is
the case of the temperatures in the lower troposphere compared to the Earth’s
surface temperature. Consequently, the sum of PHI and PCI must
be much smaller than they are thought to be in the consensus treatment of this
problem or in the similar concentric spherical shell problem.
Let us now examine the correct solution to this parallel plane
black body radiators problem. It is the electromagnetic field between the
two planes that governs the flow of electromagnetic energy between the
planes. Or one can say it is the energy density at each plane surface
that drives the exchange of energy between the planes due to the energy density
gradient between the two planes. The critical and driving parameter here
is
Δe = eH - eC = aTH4 –
aTC4 ,
where each black body radiator surface maintains its black body
radiator requirement that the energy density at the surface is given by
Stefan’s Law.
Electromagnetic energy flows from the high energy surface to the
low energy surface, as is the case in energy flows generally.
PHI = (σ/a) Δe = (σ/a) (aTH4 –
aTC4 ) = σ TH4 – σ TC4
PCI = 0,
which is consistent with experimental measurements of the rate of
radiant heat flow between two black body radiators. Note that as TC approaches
TH, PHI approaches zero as should be the case inside
a black body cavity in thermal equilibrium. There is no thermal emission
from either of the black body cavity walls then, except with the absorption of a photon of equal energy. Note also that when TC =
0 K, PHI is given by the Stefan-Boltzmann Law
PHI = σ TH4.
Let us recalculate PH and PC in
this correct formulation of the problem:
PC = APCO + APCI -
APHI = AσTC4 + 0 – [AσTH4 -
AσTC4] = 2AσTC4 - AσTH4
PH = APHO + APHI -
APCI = AσTH4 + [AσTH4 -
AσTC4] - 0 = 2AσTH4 - AσTC4
And we see that the power inputs to each plane needed to maintain
their respective temperatures as they cool themselves by thermal radiation are
unchanged in this correct energy density or electromagnetic field centered
viewpoint from the consensus viewpoint. Experimentally, the relationship
between the power inputs to the thermal radiation emitting planes at given
temperatures are exactly the same. This fact causes the proponents of the
consensus viewpoint to believe they are right, but they nonetheless violate the
energy density requirements of electromagnetic fields and of black body
radiation itself.
Because PCI = 0, back radiation
from a cooler atmosphere to the surface is also either zero or accompanied by an offsetting decrease in the emission from the surface and not 100% of the top of
the atmosphere solar insolation as in the current NASA Earth Energy
Budget. Because PHI = σ TH4 – σ TC4 , the Earth's
surface does not radiate 117% of the top of the atmosphere insolation
either. If the net energy flow by radiation between the surface and the atmosphere is 17%, then any actual back radiation from the atmosphere would be offset by a reduction of radiation from the surface below the 17% level. In other words, if there were a 2% back radiation from the atmosphere, then the surface radiation would be limited to 15% in order to preserve the proper energy density. The NASA radiation flows are hugely exaggerated as are the similar Earth Energy Budgets presented in the UN IPCC reports. See
the NASA
Earth Energy Budget below:
This is very important because reducing these two radiant energy
flows of infrared photons reduces the effect of infrared-active gases, the
so-called greenhouse gases, drastically. Many fewer photons are actually
available to be absorbed or emitted by greenhouse gases than they
imagine. This is a principal error that should cause the global climate
computer models to greatly exaggerate the effects of the greenhouse gases, just
as they have.
As I have pointed out in the past, one fatal consequence of the
exaggeration of thermal radiation from the surface of the Earth is readily
calculated from the fact that even if the atmosphere acted as a black body
absorber, which it does not, it can only absorb the thermal radiation said to
be emitted from the surface at an absurdly low atmospheric temperature.
Observe that the power of infrared radiation from the surface which is absorbed
by the atmosphere PSA is
PSA = σ ( TS4 – TA4)
PSA = (1.17 – 0.12) (340 W/m2)
The first equation is from the discussion above with TS the
surface temperature and TA the temperature of the atmosphere.
The second is according to the NASA Earth Energy Budget above. If one
takes TS to
be 288K, then the temperature of the atmosphere required to absorb as much
infrared energy as NASA claims is absorbed is 155.4K. There is no such
low temperature in the Earth’s atmosphere. To find so low a temperature,
one has to go far out into the solar system many times the radius of Earth’s
orbit. That being the case, any such thermal radiation absorbed by matter
at such a low temperature is as much lost in the Earth Energy Budget as is the
power equal to 12% of the solar insolation emitted from the Earth’s surface
which NASA allows passes through the atmospheric window into space. It
should be apparent to the reader that the NASA Earth Energy Budget is nonsense.
It is not at all surprising that physics adheres to a minimum
total energy in the system. The total flow of photons from the warmer and the
colder bodies are constrained by the energy density requirement between the
bodies. The presence of two bodies in
equilibrium at non-zero temperatures causes a reduction in the total photon
generation in the system of the two bodies.
The normal radiation of each body compared to its being alone in a T=0K
environment is reduced. By means of the electromagnetic field between these two planes, the photon
emission of the planes is coupled and affected by the presence of the other
plane. This is in no way surprising for an electromagnetic field
problem. One needs to remember that photons are creatures of
electromagnetic fields. Opposing streams of photons do not
annihilate one another to cancel out energy, they simply add their
energies. Treating them as though one stream has a negative energy and
the other a positive energy is just a means to throw the use of the
Conservation of Energy out the window. That is too critical a principle
of physics to be tossed out the window.
Extending the Solution to Gray Body Thermal Radiators
and Other Real Materials:
Many real materials do not behave like black body radiators of
thermal radiation. Those that do not radiate as black body materials
would, radiate less than the black body radiator would. Why would they
radiate less? This is because they do not create as high an
electromagnetic field energy density at their surfaces as does a black body
radiator. From Stefan’s Law for a black body radiator, the energy density
at the surface is
e = aT4
but for a gray body radiator the energy density at each wavelength
λ is
e(λ) = εaT4.
This means the energy
density at any given frequency is a constant fraction ɛ of that of a black body
radiator, with 0 ≤ ɛ ≤ 1.
In general, a material
may not behave like either a black body or a gray body radiator at a particular
wavelength. In that case,
e(λ) = ε(λ)aT4,
where the fraction of the black body output at wavelength λ is
variable.
An isolated material surrounded by vacuum and a T=0 K environment
then has a power per unit area output of
P(λ) = εσT4 for a gray body and ε is seen to be
the emissivity, and
P(λ) = ε(λ)σT4, for a general material, such as carbon
dioxide or water vapor, where the absorption and emission become variable
fractions of that of a black body as a function of wavelength.
For our two parallel plates above, if both are gray bodies, then
between the plates
Δe = eH – eC = εH a TH4 – εC a
TC4
PHI = (σ/a) Δe = εH σ TH4 – εC σ TC4
PCI = 0.
Here we see that the emissivity which determines the electromagnetic field energy density at the surface is also playing the role of the absorptivity at the absorbing colder surface. So of course, Kirchhoff’s Law of thermal radiation that the emissivity equals the absorptivity of a material in a steady state process applies. There is really nothing at all to prove if one starts with the primary fact that the energy density is the fundamental driver of the thermal radiation of materials and we know its boundary conditions.
Update: Clarified that any
emission from a cooler body in equilibrium with a warmer body results in still
less emission from the warmer body on 3 September 2018.
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