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02 November 2017

Solving the Parallel Plane Black Body Radiator Problem and Why the Consensus Science Is Wrong

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.  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.  Such flow of energy from the surface only occurs to regions with an energy density that is lower.  There is no flow of energy from the inside wall of a black body radiator because the energy density everywhere inside the cavity at equilibrium is equal.  P from the interior walls is everywhere zero.  A non-zero P is the result of a non-zero Δe.

In fact, while it is commonly claimed that photons inside the cavity are being 100% absorbed on the walls and an equal amount of radiant energy is emitted from the absorbing wall, the actual case is that the radiant energy incident upon the walls can be entirely reflected from the walls.  Planck had derived the frequency spectrum of a black body cavity from an assumption of complete reflection from the walls.

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



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.  PHI when the hotter plane is surrounded by T=0 environment provides the photon flow near the emitting surface which causes the local energy density to be

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.

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 TH4σ 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.  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 zero 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.  These radiation flows are hugely exaggerated by NASA and in 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 and does not generate a superfluous stream of photons from the colder body to the warmer body and does not have any more photons flowing to the colder body from the warmer body than necessary.  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.

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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 TH4εC a TC4

PHI = (σ/a) Δe = εH σ TH4εC 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 and boundary condition that the energy density is the fundamental driver of the thermal radiation of materials.

Update:  Considerable additions were made on 8 November 2017.
Update:  Additions made on 12 November 2017.
Update:  The extension to gray bodies and other bodies was made on 14 November 2017.
Update:  Explanatory additions on 27 November 2017.
Update:  Further explanatory text added on 29 November 2017.
Update:  Further efforts to help the reader to understand this post and to provide the reader a reason to read it in the introductory paragraphs on 24 March 2018.  I am amazed how difficult many readers find it to alter their viewpoint of thermal radiation even when they have no counter argument to my argument presented here.

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