In a
previous post, Critique
of The Steel Greenhouse by Willis Eschenbach, I wrote about the Willis
Eschenbach thought experiment model of a perfectly conducting sphere
closely surrounded by a concentric perfectly conducting shell in which these
bodies behave like black body radiators and the only energy loss mechanism is
thermal radiation. That black body
thought experiment was not presented as a good model of the greenhouse gas
effect for the Earth. It was presented
to illustrate that there is a warming effect due to thermal radiation absorption
in the atmosphere, which many call the greenhouse gas effect. I pointed out that Eschenbach is right that
the surrounding shell causes the inner sphere to have a higher temperature than
it would if it were only surrounded by vacuum at T = 0 K because it loses
radiant energy more slowly.
But
Eschenbach makes a very serious error in common with almost all scientists
because he believes the surrounding shell radiates the surface of the inner
sphere with the same radiation that the outer shell radiates from its outer
surface toward T = 0 K surroundings. I
have explained why this is an error in my post Solving
the Parallel Plane Black Body Radiator Problem and Why the Consensus Science is
Wrong. By making this error, he
multiplies the photon energy density by a factor of 3 in the space between the
sphere and the shell, making this one of many ways that the catastrophic
man-made global warming advocates greatly strengthen the greenhouse gas effect. I pointed out that the current NASA Earth
Energy Budget amplifies the photon energy density in the atmosphere even more by
a factor of 12.8! This is critical
because the warming effect due to greenhouse gases is proportional to the
photon energy density at a wavelength times the absorption cross section at
that wavelength. Consequently,
increasing the photon energy density by a factor of 12.8 increases the warming
energy by a factor of 12.8 at all wavelengths at which absorption occurs for
the gas molecule.
In this
post, I will provide the radiative equilibrium solution for a black body sphere
with 2 surrounding black body shells. This
might seem important because the mean free path length for absorption of
longwave infrared photons that can be absorbed by water vapor and carbon
dioxide is very short at the primary absorption wavelengths, though it can be
much longer in the wings of those primary absorption wavelengths. The solution will then be generalized to N
surrounding black body shells since the absorption of photons of different
wavelengths by a greenhouse gas in the atmosphere can take different numbers of
spheres in a model. In fact, for a given
infrared-active molecule, the number of shells is a function of the absorption
cross-section as a function of the wavelength because the mean free path length
varies over orders of magnitude from the main absorption wavelengths out into
the tails or wings of those principal absorption peaks.
Having
developed this nested shell model for black body absorbers/radiators, I will
then generalize this model to account for a sphere surface which has an
emissivity different from the emissivity of the shell surfaces. Because the absorption of greenhouse gases is
over a substantially smaller range of frequencies than is that of the black
body material, one might expect a major change in the result for the inner
sphere equilibrium temperature based on the effective emissivity/absorptivity
of the inner sphere surface and of the nested shell surfaces that might be an
analog to greenhouse gases. This result
proves to be very interesting.
The results will
inform us of interesting properties of an energy loss problem dominated by radiation
loss and absorption. However, this model
is not a good model of the Earth’s greenhouse gas effect. Energy loss and transport from the Earth’s
surface is not dominated by radiant energy loss and absorption. It is dominated instead by the effects of the
Earth’s gravitational field moderated by convection and the evaporation-condensation
cycle of water. The temperature of the
Earth’s surface and of each successively higher layer of air in which an
infrared-active molecule will absorb the longwave radiation from the Earth’s
surface or a lower layer of air is not determined primarily by radiation
transport of energy, but by gravity, convection, and the water cycle.
One has to
remember that there are many competing effects in determining the Earth’s
climate, including many cooling effects by both water vapor and carbon dioxide
that are too often underestimated. Radiative
cooling of the surface is much less than NASA and the UN IPCC claim it is. The alarmist rendition of greenhouse gas
warming does everything it can to amplify that effect, to ignore the many
cooling effects, and to over-emphasize the role to carbon dioxide relative to
that of water vapor. Finally, I will
adapt a portion of the shell model to make an estimate of the total greenhouse
gas effect in the real world of the Earth’s climate. I will then proceed to make an estimate for
the size of the greenhouse effect for the first 400 ppm of carbon dioxide in
the atmosphere and discuss briefly what one can expect as one adds higher concentrations
of carbon dioxide to the atmosphere.
These estimates are not precise, but they are of the proper scale and
will inform us that the warmer Earth surface compared to its overall radiative
temperature as seen from space is in very little part due to the absorption of
longwave radiation by carbon dioxide. It
also makes it clear that the infrared-absorbing effect due to water vapor is
also a small fraction of the 33K effect normally attributed to greenhouse gases,
though that effect is many times the effect due to carbon dioxide.
The inner
core sphere section of unit area has a power input of Q, the temperature of the
sphere is TS and the power per unit area of surface it emits is PS. The first closely surrounding concentric
shell has no power input except that radiated by the sphere. Its own temperature is TO1, which
when the power to the sphere is turned on is 0 K and the shell is only then
warmed by radiation from the sphere which travels at the speed of light to
it. It radiates no photons toward the
sphere, but does radiate photons as its temperature rises toward the second shell
with power PO1. The second
concentric shell has the same parameters but with a 2 in the subscript, rather
than a 1. It also starts from T=0K. Only vacuum exists between the sphere and the
planes so that there are no heat losses except by means of thermal radiation.
When the
power Q to the sphere is first turned on, the sphere has an initial temperature
of TSI, given by the Stefan-Boltzmann Law, since the sphere is at
that instant surrounded by T = 0K.
Q = PSI
= σTSI4
Let the
thermal equilibrium values of each parameter be denoted with the addition of an
E in the subscript, then at thermal equilibrium:
Q = PSE
= PO1E = PO2E
Q = σ TSE4 – σ TO1E4 = σ TO1E4 – σ TO2E4 = σ TO2E4
We can see
that
TSI
= TO2E
From the
right side of the equilibrium equation we see that
TO1E4
= 2 TO2E4
Then
plugging this value in the part of the equilibrium equation involving TSE,
we have
TSE4
– TO1E4 = TO2E4
TSE4
– ( 2 TO2E4 ) = TO2E4
TSE4
= 3 TO2E4 or TSE = 30.25 TO2E
= 30.25 TSI =
1.3161 TSI, since TO2E = TSI
With one
surrounding shell, the equilibrium temperature of the enclosed sphere was given
by
TSE
= 20.25 TSI = 1.1892 TSI
Thus the
second shell causes a sufficient reduction in the cooling rate that the
equilibrium temperature of the sphere is 1.067 times higher than it would be
with one surrounding shell. The rise in
the sphere temperature is less with each added shell.
The Nth
surrounding shell results in a radiative equilibrium sphere temperature of
TSE
= (N+1)0.25 TSI
Thus,
For 10
shells: TSE = 1.8212 TSI
For 100
shells: TSE = 3.1702 TSI
For 1000
shells: TSE = 5.6282 TSI
Imagine
modeling the absorption of surface radiation by water vapor using such a
model. Of course, water vapor is not a
black body absorber or radiator of longwave infrared radiation. For most of the radiation that it absorbs, it
has a mean free path for absorption which is short, though in the wings of an
absorption maximum, the absorption cross section can be much lower and the
corresponding mean free path is much longer.
But for a crude model, one might say that the absorption mean free path
is something like 10 meters. One might
then say that to account for water vapor absorption out to 8000 meters
altitude, one needs 800 shells. Of course,
one would think each shell would absorb only a fraction of the power that a
black body would and it would emit only a fraction of that energy also. Carbon dioxide has a longer mean free path
and it absorbs a smaller fraction of the power that a black body would compared
to water vapor, but it also does not have a relatively sharp cut-off in its
density at higher altitudes in the atmosphere.
One might model it with a much smaller fraction of absorption compared
to water with a shell every 40 meters but with 300 shells to get to an altitude
of 12000 meters.
Let us see
what happens in this simple model if we assign an emissivity to the sphere
surface, ɛS, and an emissivity to shells representing absorptions by a
greenhouse gas, ɛG. The equations
for a two-shell model then become:
Q = ɛS
σ TSI4
At
equilibrium,
Q = ɛS
σ TSE4 - ɛG σ TO1E4 = ɛG
σ TO1E4 - ɛG σ TO2E4 = ɛG
σ TO2E4
ɛS σ TSI4 = ɛG σ TO2E4, so TO2E
= (ɛS/ɛG)0.25 TSI
TO1E4
= 2 TO2E4
ɛS TSE4 = 3 (ɛG
TO2E4) = 3 (ɛS TSI4)
TSE
= 30.25 TSI, the same solution for the equilibrium sphere
temperature we had for the black body emitters and absorbers.
Consequently,
for N greenhouse gas shells we have: TSE
= (N + 1)0.25 TSI just as with N black body shells.
So, the
greenhouse gas hypothesis is looking as though it could indeed cause a
disastrous increase in the Earth’s surface temperature even though greenhouse
gases are far less efficient absorbers and emitters than are black body
absorbers, right?
Wrong. There is a fatal flaw in the model. That fatal flaw is the assumption that the
only way that heat is transported from the surface of the Earth is by means of
thermal radiation. In reality, much more
heat is transported up through the atmosphere by means of the water evaporation
and condensation cycle and by convection currents. Let us look once again at the NASA Earth
Energy Budget:
There is no
back radiation of 100% and the surface radiation given as 117% in terms of the
top of the atmosphere solar insolation is hugely exaggerated by NASA. The real thermal radiation from the surface
is the difference between these values or 17%.
This is perhaps a decent value for the loss from the surface itself and
of this 12% is lost immediately to space.
This leaves only 5% of thermal radiation from the surface that is
absorbed by the atmosphere along with 5% attributed to convection and 25% lost
by water evaporation. So, in this energy
balance only 5% / (5% + 5% + 25%) = 0.143 of the surface energy is absorbed by
the atmosphere as radiation from the surface.
Even this is
a huge overstatement of the fraction of the surface energy carried off by means
of transport by radiation from shell to shell.
This is the fraction that is absorbed by the first shell of the
many-shell model. Once that first
absorption occurs, the absorbing greenhouse gas molecule passes off the
absorbed energy to the nitrogen and oxygen molecules and the argon atoms that
collide with it 6.9 billion times a second at sea level and 2.1 billion times a
second at 10 Km altitude in the U.S. Standard Atmosphere of 1976. Yes, the greenhouse gas molecule very quickly
comes into temperature equilibrium with the molecules in the same layer of air
with it. These greenhouse gas molecules
can radiate thermal energy to the layer of air just above which is slightly
cooler, but the time between such radiation events is extremely long compared
to the gas collision frequency. Water
molecules radiate only about every 0.2 seconds and carbon dioxide molecules
radiate only about once a second.
Consequently,
almost all of the 5% of surface energy that is radiated to the first shell and
absorbed is thereafter transported by convection. The assumption in the model above that all
energy is transported through the atmosphere as radiation and in stages could
not be more wrong. In fact, while the
little bit of further radiation transport from one layer of air to a cooler
layer of air a short distance above it would be handled in the nested shell
model as a further warming of the surface.
But, this should actually be viewed as a cooling effect, because any
energy transported from air layer to air layer is transported to higher
altitude faster compared to the alternative transport mechanism of
convection. These realizations constitute
a good lesson in the need to check your premises!
Based on
almost all of the 5% of surface radiation being absorbed by the first shell and
then thereafter being transported by convection, about what surface temperature
might we expect? Recall that in the one
shell case, the surface equilibrium temperature is given by
TSE
= 20.25 TSI = 1.189 TSI
At a thermal
radiation fraction of 0.143, the greenhouse gas effect temperature rise would
be about 0.143 (0.189) = 0.027 times TSI. If one takes TSI = 255 K, the radiative
temperature of the Earth system as a whole with respect to space, then the
change of temperature attributable to the greenhouse gas effect for our present
atmosphere is
ΔT = 6.9 K
Note that
6.9 K is only 21% of the 33K difference of the surface temperature with respect
to the Earth’s effective radiative temperature as seen from space. Consequently, the claim that this temperature
difference is due to the absorption of longwave infrared radiation emitted from
the Earth’s surface is exaggerated by nearly a factor of 5.
Almost all
of this 6.9K warming effect is due to water vapor, not the 400 ppm of carbon
dioxide in the atmosphere. The
absorption by carbon dioxide compared to that by water vapor is about 8 times
less. Thus the portion of the total infrared
warming effect due to carbon dioxide is one ninth of 6.9 K, which is 0.8 K. My recent post Using
Heat Transport Powers of the NASA Earth Energy Budget to Prove that Carbon
Dioxide has an Insignificant Effect on Surface Temperatures,
estimated that the present 400 ppm of CO2 in the atmosphere causes a
temperature increase of only 0.2 K, but this is the result of a different
approach to the issue and it took into account the increased absorption of solar
insolation caused by carbon dioxide in the atmosphere. The fraction of the 6.9 K net greenhouse gas
temperature increase that can be attributed to carbon dioxide overall is then
about 3%.
Given that
the 0.2 K net warming effect of carbon dioxide is already mostly saturated,
additions of further carbon dioxide to the atmosphere will have very little
effect on the surface temperature of the Earth.
What is more, the feedback effect of water vapor at present common levels
of water vapor is more likely negative than positive and is certainly very
small.
The reason
that the surface of the Earth is about 33 K warmer than the effective radiative
temperature of the Earth system as a whole as seen from space is because most
of the energy dissipated by the Earth’s surface and through the troposphere (the
lower atmosphere) is transported upward by a combination of water evaporation
and convection. It is very important
that transport by thermal radiation is a much more minor actor in the transport
of heat in the troposphere. It is
critically important that most of the Earth’s heat is radiated to space from
the upper part of the troposphere which is at temperatures that are close to
those of the Earth system as observed from space. This actually forces the air at the altitudes
where most of the radiation to space occurs to be near and slightly lower than
this low effective radiative temperature of the Earth system. Once that is a given, then the temperature
gradient in our troposphere due to the Earth’s gravitational field dictates the
surface temperature with adjustments due to such cooling mechanisms as
convection and the effects of the water cycle and clouds.
Your solution for the case where the sphere and shells have emissivities not equal to one is obviously wrong. I can provide you with the full solution for a single shell if you like and we can discuss the differences. However, if you simply consider the case of one shell with a blackbody sphere and make the shell have a near zero emissivity and a transmissivity of close to 1, then the solution for the temperature of the sphere should approach the solution for no shells. I'm pretty sure that you will never be able to solve this problem correctly without using the fact that the shells emit in both directions (out and in) equally and according to their temperature and emissivity as indicated using the S-B law.
ReplyDeleteCharles. Thank you for the clarity of the mathematics – I now understand how you determine that the power-generating sphere will become hotter as each successive shell is added around the sphere. Furthermore, I was particularly surprised to see that the ɛS / ɛG term dropped out – really sweet.
ReplyDeleteHowever, proper consideration of Radiant Energy Density (something that you made me aware of) remains a problem for me in your solution. I do hope that you don’t think me ignorant: I’m sincerely trying to get to the root of my misunderstanding of this issue. As I see it, the radiant energy density immediately above the surface of that blackbody, at temperature TSE , must be = a TSE^4 , and in this particular case that Radiant Energy Density entirely comprises photons that were forced from that surface then;
a) in a parallel plane model, that same Radiant Energy Density should be seen immediately above the inside surface of the first plate and it is this same Radiant Energy Density which will determine the inside surface temperature of the first plate. On the basis that all of the radiant energy arriving at the inside surface of the first plate is all absorbed (and thermalised into kinetic energy) and is all conducted away from the inside surface across to the outside surface of the first plate without developing a kinetic energy density gradient within the material of the plate then (i) the temperature of the outside surface of the first plate will be the same as the inside surface and (ii) the radiant energy density immediately above the outside surface of the first plate will be the same as the radiant energy density above the inside surface. Therefore, I can only imagine that; T01E = TSE
b) in a spherical model, a diminished Radiant Energy Density having been reduced by a factor of (RSPHERE / RSHELL)^2 should be seen immediately above the inside surface of the first shell and it is this diminished Radiant Energy Density which will determine the inside surface temperature of the first shell. On the basis that all of the radiant energy arriving at the inside surface of the first shell is all absorbed (and thermalised into kinetic energy) and is all conducted away from the inside surface across to the outside surface of the first shell without developing a kinetic energy density gradient within the material of the shell then (i) the temperature of the outside surface of the first shell will be the same as the inside surface and (ii) the radiant energy density immediately above the outside surface of the first shell will be the same as the radiant energy density above the inside surface. Therefore, I can only imagine that;
T01E = ((RSPHERE / RSHELL)^2 )^0.25 TSE = (RSPHERE / RSHELL)^0.5 TSE
I agree that, in the parallel plane model, if the surface temperatures of the power-generating body and the plate were identical then there would be no heat transferred from the power-generating body to the plate i.e. Q=0. However, I still imagine that radiant exitance would still be flowing from power-generating body to the plate (to replenish the radiant energy emitted from the outside surface of the plate). So, as I naively see it, radiant exitance will always be emitted from the power-generating body (at a rate corresponding to the power it’s generating). At the receiving surface of the receiving body, this will all be thermalised into kinetic energy but only some of it be considered as heat (if the receiving body becomes warmer). However, if the other emitting surface(s) of the receiving body are already at the same temperature as the receiving surface of the receiving body then none of this incoming power will be considered as ‘heat’ but will instead all be used to maintain the same temperature by replacing the energy being dissipated from the other emitting surfaces of that body.
I don’t know if my earlier attempts to contact you have been successful – or if you do not want to respond. If this is the case, please say so just once, and I’ll stop bothering you with my thoughts.
Best Regards,
Steve Titcombe
I have looked for any references on the internet for your theory of an electromagnetic gradient preventing cooler body IR radiation to a hotter body and cannot find any. Is this your original theory or did you get it from somewhere else? The physicists on Physics forum all say that you are wrong. Since this is a key point for the possibility of back radiation, you need proof of your theory.
ReplyDeleteYou said "My recent post Using Heat Transport Powers of the NASA Earth Energy Budget to Prove that Carbon Dioxide has an Insignificant Effect on Surface Temperatures, estimated that the present 400 ppm of CO2 in the atmosphere causes a temperature increase of only 0.2 K, but this is the result of a different approach to the issue and it took into account the increased absorption of solar insolation caused by carbon dioxide in the atmosphere. The fraction of the 6.9 K net greenhouse gas temperature increase that can be attributed to carbon dioxide overall is then about 3%."
ReplyDeleteWhether the CO2 catches the solar energy before it hits the surface or after should not affect the numbers if there is back radiation. Therefore your 2 methods of calculating the amount of K attributed to CO2 dont agree.
Also in one of your 1st blogs you said that you disagreed with the 33 K warming provided by grenhouse gases and you put the figure at 9K. Now you are back to 33K What gives?
In your June article you said "and the loss of surface energy due to the evaporation of water are not very well-established numbers." I did the numbers working backward from the hydrological cycle of total evapotranspiration and NASA is definitely in the ballpark with that figure. The only other one I trust is the incoming solar insolation 340 before reflection. . Are there other ones that have been actually measured? Is the 240 outgoing LWIR accurate?
ReplyDeleteI believe that has been measured.
I see that you still have not corrected your post.
ReplyDeleteThe solution for a core with emissivity ec and reflectivity (1-ec) surrounded by a shell with emissivity es and transmissivity (1-es) is:
Ts^4 = (Q/sigma)/(2-es)
Tc^4 = es Ts^4 + Q/sigma/ec = (Q/sigma)(2-es+es*ec)/ec/(2-es)
Would you like more details on how this solution is derived?
Linked to this article in my new climate historyscope, best on the Net.
ReplyDeletehttp://www.historyscoper.com/climatescope.html
Anonymous: The emissivity for a surface is really just an adjustment for the energy density of the electromagnetic field immediately outside the surface. A gray body will have a surface energy density which is less than that of a black body. The generated photon flux between two surfaces is then driven by the difference in their surface energy densities.
ReplyDeleteSteve Titcombe: When the power generating body with an input power of Q emits radiation toward another body starting at T=0K, it will initially emit power at a rate given by
ReplyDeleteQ = PS = σ TSI^4
Where TSI is its initial temperature, given by the simple application of the Stefan-Boltzmann Law of thermal radiation.
But as the energy density builds up between the power generating body and the first shell plane, the emission from the power generating body decreases continuously until an equilibrium condition is achieved. When an equilibrium condition is achieved, the temperature of the first shell is higher than it was initially (I assumed it was at T=0K to start), so
Q = PSE = σ TSE^4 – σ TO1^4
Now TO1 is greater than zero, so the only way this power of emission can still be equal to Q is if TSE is greater than TSI.
With multiple shell plates, the power absorbed by each is the same and the power absorbed by each is also equal to the power that the last shell outer surface emits to the T=0K environment outside it.
Note that because the temperature of the power generating surface is greater when equilibrium is achieved, the emission of photons from that surface is enough to maintain the energy density in the space between it and the first shell plane without any emission of photons from the surface of the shell plane facing the generating surface. The first shell plane only emits photons toward the second shell if there is one or to T=0K if there is no second shell.
Alan Tomalty First Comment: I am not aware of anyone offering a similar proof that the prolific and wasteful generation of photons in the model that every surface emits thermal radiation as though it were surrounded only by a T=0K environment. Of course, some scientists have claimed that this is not the case based on the 2nd Law of Thermodynamics, though that really requires the addition of a minimum energy solution. Such minimum energy solutions are very often correct for physics problems, so it should not surprise us that it turns out to be correct here also. However, my approach is from the direct consideration of the energy density of electromagnetic fields and attends to the principal characteristic of a black body cavity, namely that it has a constant energy density given by Stefan's Law.
ReplyDeleteI provided proof of my theory in the prior post entitled Solving the Parallel Plane Black Body Radiator Problem and Why the Consensus Science Is Wrong, which is referred to in the article above. If your friends at the Physics Forum say I am wrong, then let them address my proof with an argument rather than just an opinion or some claim of consensus. I clearly showed that their consensus misuse of the Stefan-Boltzmann equation causes a doubling of the energy density of a black body cavity in violation of Stefan's Law. One can also show many instances in which their misuse of the Stefan-Boltzmann Law leads to violations of The Conservation of Energy.
Alan Tomalty (Second Comment): First of all, there is no back radiation except in the occasional cases when the air at a higher altitude is warmer than that at a lower altitude. In the normal equilibrium case, this is not so. Second, if there were back radiation, the absorption of solar radiation would still matter, since the absorbing molecules would then supposedly radiate the absorbed energy in all directions equally, including toward space and much more importantly, those molecules would transfer almost all of that energy to infrared-inactive molecules which would then primarily transport that energy upward by convection. The interception of solar energy in the atmosphere keeps it from the surface, from which the important cooling mechanisms are all working to transport the surface energy upward through the entire atmosphere, rather than just a portion of the atmosphere.
ReplyDeleteThere is a 33K difference between the Earth system radiative temperature as seen from space and the Earth's average surface temperature, at least approximately. Some people call that entire difference the greenhouse effect. Others call the greenhouse effect just that warming that occurs directly due to infrared active molecules absorbing long wave infrared radiation emitted from the Earth and its atmosphere. There are actually many warming and cooling mechanisms and one can parse them in many ways to include or exclude them from being part of the greenhouse effect. So, consider the way I have done that in each instance in which I have dealt with this issue. One of the key issues is whether one counts the way the Earth's gravitational field creates a temperature gradient in the atmosphere as separate though dependent upon the role of infrared-active gases or one only talks about energy transported through the atmosphere by thermal radiation between layers of air as the greenhouse effect.
Reply to Ultra: I expect that the top of the atmosphere incoming and outgoing radiation numbers should be about the best numbers we have in terms of direct measurements. I am curious about how you did your calculations on the surface water latent energy of evaporation and what you mean by NASA being in the ballpark.
ReplyDeleteCharles,
ReplyDeleteYour response is yet again incorrect. I provided you with the solution for the case where the surface is reflective and the shell is transmissive. If the shell is perfectly transmissive then the solution for the core temperature must be the same as if the shell is not there. Hence, your solution is obviously wrong. Why don't you post my solution so that your readers can see the difference?
This is a very simple problem where you are claiming that the correct temperature for the core and shell are DIFFERENT from those determined from radiative transfer theory. Furthermore, when something is not a blackbody it does matter if it is transmissive, reflective, or some combination of the two. If there were a single shell with emissivity of zero and REFLECTIVITY of one, then the temperature of the core would go to infinity.
Sorry Charles, but this is a case where you can't just handwave away your error with words. Your solution gives a nonsensical result.
Addressing anonymous again: Of course if you have only one shell and it does not absorb infrared radiation emitted by the power generating surface, then it is as though that shell is not there for this problem. However, this model specifically makes the shells either black bodies or gray bodies. For the model to apply, the shell does have to have a temperature which can respond to the radiation from the powered surface through some absorption of energy. Otherwise, as in the case you are claiming, the shells have no interaction at all with the power generating surface and this model in the equations does not address that case.
ReplyDeleteTo Anonymous who says: "I see that you still have not corrected your post.
ReplyDeleteThe solution for a core with emissivity ec and reflectivity (1-ec) surrounded by a shell with emissivity es and transmissivity (1-es) is:
Ts^4 = (Q/sigma)/(2-es)
Tc^4 = es Ts^4 + Q/sigma/ec = (Q/sigma)(2-es+es*ec)/ec/(2-es)"
No photons are incident upon the core, so its reflectivity is irrelevant. You are thinking in terms of the willy nilly emission of photons model as though every surface emits photons as though it is surrounded only by T=0K environment. That does not apply in the model I am presenting. The only parameter that matters for the gray body is the energy density that it sets up at its surface when it is at temperature T, which is es a Ts^4 in where a is Stefan's Constant.
Yes Charles, I know that the shells can be gray bodies. A shell with an emissivity APPROACHING zero is gray body. In that limit the solution to the problem for the temperature of the core must approach the case where there are no shells. Your solution does not agree with this limit. Furthermore, your solution makes no distinction between a gray reflecting shell and a gray transmissive shell.
ReplyDelete"No photons are incident upon the core, so its reflectivity is irrelevant."
Come again? Matter at finite temperature emits radiation in the form of photons in a manner that is only dependent on the temperature of the emitting matter. What are you claiming is happening to the photons that are emitted by the inner surface of the shell?
"You are thinking in terms of the willy nilly emission of photons model as though every surface emits photons as though it is surrounded only by T=0K environment."
So, does that mean that you think that the following description is incorrect?
"Although when I say two things at the same temperature, which means you have balance, it doesn't mean they have the same energy in them. It just means it's just as easy to pick energy off of one as to pick it off the other, so as you put them next to each other, nothing apparently happens. They pass energy back and forth equally. The net result is nothing."
Are you trying to claim that when two objects are at equal temperature they are NOT passing energy back and forth equally?
"That does not apply in the model I am presenting."
Right. You have made up a model that does not agree with established physics. The reason that matter emits photons is due to the "vibration" of charge at fintite temperature. How that matter vibrates and emits does not depend on the temperature of the matter that the object is emitting to.
Again, this is a diversion. The simple fact is that your formula produces an absurd result for the case where the emissivity of the shell is 10^-10.
Reply to Anonymous: I was not clear in the sentence you quoted stating that no photons were incident upon the core. I was intending to say that no photons generated by the inner surface of the shell were incident upon the core surface, since the inner surface of the shell does not generate photons because the energy density at the surface of the inner shell is lower than that at the surface of the core. But even this qualification on my part is not adequate. I will further correct myself below.
ReplyDeleteAs for the case in which a shell has an emissivity that approaches zero, then if the energy Q is to flow through it and beyond it, the temperature of the shell becomes very high. One can introduce a direct transmission of energy radiated from the core by making the shell perforated, but that was not built into the model I was developing.
Neither did the model I was developing assume any core absorption and re-emission of photons from the core to compensate for any absorption, but one could easily add that as well. It does not even matter whether photons are emitted from the shell actually, as long as the energy density at the core surface is es aT^4 and that at the inner and outer walls of the shell are eg aT^4. In other words, one can have a certain amount of photon generation by the shell, but any photon generation by the shell is offset by a decrease of photon generation from the core. The overall energy density between core and shell is fixed by the electromagnetic field gradient between them and we know the energy density of that field at the surfaces of both the core and the shell surfaces.
"since the inner surface of the shell does not generate photons because the energy density at the surface of the inner shell is lower than that at the surface of the core"
ReplyDeleteSorry Charles, but this is nonsense. As I explained, the reason that matter emits photons is due to the "vibration" of charge at finite temperature. How that matter vibrates and emits does not depend on the temperature of the matter that the object is emitting to. This is basic physics. What you are claiming has no basis in physics. It's made up nonsense.
"As for the case in which a shell has an emissivity that approaches zero, then if the energy Q is to flow through it and beyond it, the temperature of the shell becomes very high."
The temperature of the shell is given in my solution for the transmissive case. If there is Q being emitted by the core, then (1-e_shell) Q is transmitted through the shell. If e_shell is 10^-10 then the amount of transmitted energy is essentially Q. In order to emit Q, the temperature of the core is (Q/sigma/e_core)^0.25.
For the reflective case, your comment is actually correct about the temperature of the shell. It will approach infinity, and so will the temperature of the core. But this is not what your solution gives.
Seriously Charles, any good scientist would see that their solution gives a nonsense answer in this limit. Do you honestly believe that whether the shell is transmissive or reflective, or with a high emissivity or low emissivity, cannot affect the temperature of the core? Instead of realizing the absurdity of this and abandoning your solution, you double down and try to justify it. It's not a good look. This is basic radiative heat transfer.
Given that your "model" is grossly at odds with temperature predictions, why don't you test it out with a simple experiment?
Charles, take 2 lasers; one with blue light one with red light. Make sure that the red light laser is much more powerful than the blue light one. Point them at each other and turn them on. You will see the blue light laser do damage to the red one even though in the end the blue one will be completely destroyed by the red one.
ReplyDeleteCharles, Connect a bunch of powerful white lights together in a 3 foot circle Shine a weak coloured light at them . You will see the ooloured light on the surface of those other lights. If there was an electromagnetic gradient stopping the weaker photons fromhe coloured light, you would not see the colured light at the other end ,where all the stronger lights are.
ReplyDeleteAnonymous (one of so many) above says:
ReplyDelete"As I explained, the reason that matter emits photons is due to the "vibration" of charge at finite temperature. How that matter vibrates and emits does not depend on the temperature of the matter that the object is emitting to."
The oscillation of electric charge dipoles and higher order electric charge poles in matter creates an electromagnetic field as I have noted many times. Let us consider a black body cavity whose walls we are heating up. The walls will emit photons from the oscillating electric dipoles as the temperature is being increased and the electromagnetic field inside the cavity is increasing in energy density. During this period of increasing wall temperature, the energy density at the wall is greater than that in the cavity elsewhere for an instant until photons from the wall fill the cavity volume. In this stage, photons are generated because there is an instantaneous difference in the energy density immediately at the wall surface and in the rest of the cavity volume. When the temperature of the walls is then no longer increased, the energy density in the cavity will become a constant energy density given by Stefan's law. There is no longer an energy density difference between the wall surface and the remainder of the cavity volume, so no photons need to be generated to provide additional energy to the cavity volume to bring the cavity volume everywhere into equilibrium with the wall surface energy density. Since the energy density in the cavity volume is now in equilibrium with the wall energy density, there is no need to produce additional packets of energy traveling at the speed of light, that is photons.
Now you will likely say that the photons are emitted according to the Stefan-Boltzmann Law, but the energy at the walls is held in equilibrium because for every photon emitted, one is also absorbed at the wall. This works as a solution at the wall boundary condition. The problem, as I have shown in a simple thought experiment with two closely spaced parallel walls is that this requires a violation of Stefan's Law for the energy density uniformly in the volume of the black body cavity. Photons emitted by one wall according to the Stefan-Boltzmann equation as though the temperature of the other wall was T = 0K applied to both walls doubles the energy density in the black body cavity volume that is known from Stefan's Law when you apply your idea of one photon emitted and one absorbed by each wall. The way to preserve Stefan's Law is to note that photons are energy packets formed in response to an electromagnetic field gradient.
Ultra, you have posed an interesting idea and I have not done that experiment, but I expect that it would work as you say. A blue light photon is at a higher energy than is a red light photon. When you say the red laser is more intense, this means there are enough more red frequency photons than blue frequency photons to produce a greater total power flow of red photons than the total power flow of blue photons.
ReplyDeleteThe generation of photons of a particular frequency is a response to an electric field gradient at that frequency. In the context of black body or gray body radiators in equilibrium with one another, the lower temperature body will have a lower energy density at every frequency. In the context of radiation from the surface of the Earth being absorbed by CO2, the radiation from the surface at the frequency absorbed by CO2 is at a higher energy density than is the energy emitted by the CO2 molecule at a lower temperature. These are cases in which photons are generated to flow from the higher energy density emitter to the lower energy density emitter.