Guest Post by Willis Eschenbach
Dr. Judith Curry notes in a posting at her excellent blog Climate Etc. that there are folks out there that claim the poorly named planetary “greenhouse effect” doesn’t exist. And she is right, some folks do think that. I took a shot at explaining that the “greenhouse effect” is a real phenomenon, with my “Steel Greenhouse” post. I’d like to take another shot at clarifying how a planetary “greenhouse effect” works. This is another thought experiment.
Imagine a planet in space with no atmosphere. Surround it with a transparent shell a few kilometres above the surface, as shown in Figure 1.
Figure 1. An imaginary planet surrounded by a thin transparent shell a few kilometres above the surface (vertical scale exaggerated). The top of the transparent shell has been temporarily removed to clarify the physical layout. For our thought experiment, the transparent shell completely encloses the planet, with no holes. There is a vacuum both inside and outside the transparent shell.
To further the thought experiment, imagine that near the planet there is a sun, as bright and as distant from that planet as the Sun is from the Earth.
Next, we have a couple of simplifying assumptions. The first is that the surface areas of the planet and the shell (either the outside surface or the inside surface) are about equal. If the planet is the size of the earth and the transparent shell is say 1 kilometre above the surface, the difference in area is about a tenth of a percent. You can get the same answer by using the exact areas and watts rather than watts per square meter, but the difference is trivial. Assume that the shell is a meter above the surface, or a centimeter. The math is the same. So the simplification is warranted.
The second simplifying assumption is that the planet is a blackbody for longwave (infra-red or “greenhouse”) radiation. In fact the longwave emissivity/absorptivity of the Earth’s surface is generally over 0.95, so the assumption is fine for a first-order understanding. You can include the two factors yourselves if you wish, it makes little difference.
Let’s look at several possibilities using different kinds of shells. First, Fig. 2 shows a section through the planet with a perfectly transparent shell. This shell passes both long and shortwave radiation straight through without absorbing anything:
Figure 2. Section of a planet with a shell which is perfectly transparent to shortwave (solar) and longwave (“greenhouse”) radiation. Note that the distance from the shell to the planet is greatly exaggerated.
With the transparent shell, the planet is at -18°C. Since the shell is transparent and absorbs no energy at all, it is at the temperature of outer space (actually slightly above 0K, usually taken as 0K for ease of calculation). The planet absorbs 240 W/m2 and emits 240 W/m2. The shell emits and absorbs zero W/m2. Thus both the shell and the planet are in equilibrium, with the energy absorbed equal to the energy radiated.
Next, Figure 3 shows what happens when the shell is perfectly opaque to both short and longwave radiation. In this case all radiation is absorbed by the shell.
Figure 3. Planet with a shell which is perfectly opaque to shortwave (solar) and longwave (“greenhouse”) radiation.
The planet stays at the same temperature in Figs. 2 and 3. In Fig. 3, this is because the planet is heated by the radiation from the shell. With the opaque shell in Fig. 3, the shell takes up the same temperature as the planet. Again, energy balance is maintained, with both shell and planet showing 240 W/m2 in and out. The important thing to note here is that the shell radiates both outward and inward.
Finally, Fig. 4 shows the energy balance when the shell is transparent to shortwave (solar) and is opaque to longwave (“greenhouse”) radiation. This, of course, is what the Earth’s atmosphere does.
Here we see a curious thing. At equilibrium, the planetary temperature is much higher than before:
Figure 4. Planet with a shell that is transparent to shortwave (solar) radiation, but is opaque to longwave (“greenhouse”) radiation.
In the situation shown in Fig. 4, the sun directly warms the planet. In addition, the planet is warmed (just as in Fig. 3) by the radiation from the inner surface of the shell. As a result, the planetary surface ends up absorbing (and radiating) 480 W/m2. As a result the temperature of the surface of the planet is much higher than in the previous Figures.
Note that all parts of the system are still in equilibrium. The surface both receives and emits 480 W/m2. The shell receives and emits 240 W/m2. The entire planetary system also emits the amount that it receives. So the system is in balance.
And that’s it. That’s how the “greenhouse effect” works. It doesn’t require CO2. It doesn’t need an atmosphere. It works because a shell has two sides, and it radiates energy from both the inside and the outside.
The “greenhouse effect” does not violate any known laws of physics. Energy is neither created nor destroyed. All that happens is that a bit of the outgoing energy is returned to the surface of the planet. This leaves the surface warmer than it would be without that extra energy.
So yes, dear friends, the “greenhouse effect” is real, whether it is created by a transparent shell or an atmosphere.
And now, for those that have followed the story this far, a bonus question:
Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?
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O H Dahlsveen says: December 6, 2010 at 3:11 pm
All the explanations for the “Greenhouse Theory” I have ever seen or read about rely on the atmosphere being able to stop – or restrict convection.
Then with all due respect you have been reading the wrong explanations. The main problem is that there are two very different processes that are BOTH called “the greenhouse effect.”
* For actual greenhouses, “the greenhouse effect” does indeed depend primarily on the restriction of convection.
* For planets, “the greenhouse effect” depends on the IR that is shining down from the atmosphere. Atmospheres without GHG’s are very ineffective at radiating IR, so they cannot provide any significant IR toward the surface.
Anything that warms the atmosphere will increase the downward IR, which depends on a) the temperature of hte atmosphere and b) the concentrations of various GHG’s. Convection warms the higher levels a bit. Evaporation/condensation warms it more. IR from the surface warms it even further. So this warm atmosphere will radiate IR, some of which will radiate toward the surface and hence provide extra energy for the surface.
@ur momisugly From NY
I don’t quite follow what you are objecting to — I don’t think I am contradicting anything jimmi said.
Let me try one more time. Lets suppose as a starting point, two blackbody discs with an area of 1 m^2. The cool one radiates 100 W/m^2 and the warm one radiates three times as much: 300 W/m^2. The discs are placed 10 m apart facing each other. The rest of space is very cold and provides no appreciable radiation.
Only a small bit of the energy from the cool disc actually reaches the warm disc, since most of the radiation “misses” the warm disc. I estimate the actual flux reaching the warm disc as 0.16 W/m^2, but the actual number is not important. What IS important is that the cool disc will receive 3 times as much power per meter (0.48 W/m^2). Thus cool disc receives more power than it loses. There is a net flux from the warm disc to the cool disc
If you put TEN cools discs side-by-side (or a single disc of area 10 m^2), then EACH cool each square meter radiates 0.16 W/m to the warm disc = 1.6 W/m^2. But the warm discs radiates 0.48 W/m^s to each cool disc = 4.8 W/m^2 total. Thus cool disc receives more power than it loses. There is still a net flux from the warm disc to the cool discs.
If you put a lens between the 1 warm disc and the 10 small discs, you could focus more of the power from the cool disc that would missed the warm disc. Say the lens is big enough to focus 5 times more power from the cool discs. There is now 10 x 1.6
W/m^2 = 16 W/m^2 hitting the warm disc. But as I said, the lens works in both directions, it will also focus 5x as much energy from the warm disc = 48 W/m^2. There is still a net flux from the warm disc to the cool discs.
Not matter how you arrange cool discs, lenses or mirrors, I challenge you to find a specific example that either has a) more energy from cool to warm than from warm to cool or b) more than 100 W/m^2 from the cool surfaces to the warm surface.
Back to my paying job. 🙂
“Tim Folkerts says:
December 6, 2010 at 7:47 pm
I don’t quite follow what you are objecting to — I don’t think I am contradicting anything jimmi said.”
OK, let me point you to exact words:
“Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an[d] direct the photons of the slightly colder coin exactly onto the surface of the smaller coin.”
Would the photons from the colder coin be absorbed by the warmer coin or not?
Yes, the photons from the cooler object will be absorbed by the warmer object, just like photons from the warmer object will be absorbed by the cooler object. Indeed, if we are assuming that the objects are blackbodies, then temperature is really immaterial — any blackbody at any temperature will absorb any photons from any source at any temperature.
(Of course, no object is a true blackbody. How well they actually absorb photons will depend on the wavelength of the incoming light. For example, the shell in the original model was assumed to be a blackbody (emissivity = 1) for LWIR, but a “clearbody” (emissivity = 0) for visible wavelengths.)
Tim Folkerts says:
December 6, 2010 at 6:52 pm
“We are getting far afield, but I don’t think your example invalidates anything I said.
In relativity all the measurements are — well — relative. Suppose the the system has reached equilibrium. The temperature of the two objects as measured in any given reference would be the same.”
Strictly speaking, it is not possible to measure the temperature other than in the rest frame of each object. One can infer the temperature by observing the emitted radiation, but to do so one has to make certain assumptions – such as that the radiation one observes is thermal and has not been doppler shifted – that are not always valid. The temperature difference in my example is a real or absolute one. We can see this by imagining that the two objects are made of a single material that has a melting point above the temperature of the shell but below that of the planet. The shell will remain solid; the planet will melt.
The significance of this thought experiment is that it is unwise to rely upon archaic formulations of the second law – like “heat cannot flow from a cooler object to a hotter object” – based on temperature or heat flux, because they hide assumptions that occasionally bite back. They are very nearly – but not quite – correct. The correct basis of the second law is entropy, and the key question whether entropy is non-decreasing.
“The temperature difference in my example is a real or absolute one. ”
I see where you are coming from, but I disagree with your terminology. There is no “absolute” reference frame in relativity. These might be called the “proper temperature” is suppose.
“The shell will remain solid; the planet will melt. ”
OK. In different frames lengths are different and clocks run differently. And I strongly suspect materials melt at different temperatures. But hashing out relativistic thermodynamics here is beyond my expertise or motivation. 🙂
But I completely agree that “The significance of this thought experiment is that it is unwise to rely upon archaic formulations of the second law – like “heat cannot flow from a cooler object to a hotter object””
“Tim Folkerts says: December 7, 2010 at 10:33 am
Yes, the photons from the cooler object will be absorbed by the warmer object”
OK, so what happens to the warmer object? Does its temperature go up? After “Then put a lens between an[d] direct the photons of the slightly colder coin exactly onto the surface of the smaller coin.”
@ur momisugly From NY
The warmer object gains the energy, but its temperature does not go up. Specifically, the warm object absorbs the energy from the photon, but thermodynamics/blackbody radiation laws tells us that the warm object will necessarily be losing more energy from its own outgoing photons than it is gaining from the incoming photons. Or put another way, the photons from the cool object will make the warm object warmer than it would have been without the photons from the cool object, but the warm object will still be getting cooler.
It’s kind of like I give you $10 every minute and you give me $5. Do I get richer from your $5? Well, I don’t get poor as quickly, but I am still getting poorer.
And the lens still will not be able to focus more from cold to hot than from hot to cold. At this point I fear you will either have to trust me (and jimmi), or do some heavy-duty analysis with principles from BB radiation, optics, thermodynamics, and surface integrals until you can show a specific counterexample that violates the 2nd law. For over 150 years clever people have been trying to come up with ways to violate the 2nd law — but as far as I know, no one has succeeded. So I am quite confident you will not be able to come up with such a violation.
“Tim Folkerts says: December 7, 2010 at 6:06 pm
Specifically, the warm object absorbs the energy from the photon, but thermodynamics/blackbody radiation laws tells us that the warm object will necessarily be losing more energy from its own outgoing photons than it is gaining from the incoming photons.”
How can the warm object absorbing energy stay at the same or lower temperature and radiate more? This would be against the Stefan–Boltzmann law.
Specifically, the warm object absorbs the energy from the photon, but thermodynamics/blackbody radiation laws tells us that the warm object will necessarily be losing more energy from its own outgoing photons than it is gaining from the incoming photons.”
By the way, this statement is factually incorrect. I sure observed warm objects getting warmer when placed in the sunlight.
Tim Folkerts says:
December 7, 2010 at 4:00 pm
PB: The temperature difference in my example is a real or absolute one.
“I see where you are coming from, but I disagree with your terminology. There is no “absolute” reference frame in relativity. These might be called the “proper temperature” is suppose. ”
It is a misconception that there are no absolutes in relativity. Temperature is such an absolute (that is, the absolute temperature of an object, not the apparent brightness temperature of a radiation source), as is the number of sides of a plane figure. All observers will agree on the latter; and all observers will agree the point at which phase changes etcetera take place, and on the position of chemical equilibria. Proper temperature would also be a proper term.
“OK. In different frames lengths are different and clocks run differently. And I strongly suspect materials melt at different temperatures. But hashing out relativistic thermodynamics here is beyond my expertise or motivation. ”
No, no, no. Time runs differently. Clocks run exactly the same. The local physics is independent of the local rest frame (it is impossible to determine its location, velocity or potential by purely local measurements).. Materials melt at the same temperature/pressure wherever they are. There is a slight caveat here in that neither the shell nor the planet is in an inertial frame; since pendulum clocks obviously need the same gravity to work correctly, this would have to be supplied (easy enough with a centrifuge).
“By the way, this statement is factually incorrect. I sure observed warm objects getting warmer when placed in the sunlight.”
But the sun is warmer than the “warm object”! So the “warm object” you are thinking of has become the “cool object” in our discussion, and the sun has become the “warm object”. And it is still completely impossible to have a net heat transfer from a cool object on the earth to the warmer surface of the sun.
“It is a misconception that there are no absolutes in relativity. Temperature is such an absolute … ”
First of all, I said there are no absolute reference frames, which I do beleive to be correct.
Secondly, google turns up several papers discussiong effects of relativity on temperature, so I am not convinced that it is “absolute.”
In any case, this seems like it has become a discussion for a physics/relativity blog, not a climate blog. I am sure there are people more learned in this topic than either of us who could shed much more light on the topic.
Tim Folkerts says:
December 8, 2010 at 8:38 am
“First of all, I said there are no absolute reference frames, which I do beleive to be correct.”
I was talking about the absolute temperatures of the shell and planet. This has nothing to do with absolute frames of reference; and even then it needs to be understood that although velocity is relative, acceleration or gravity is not.
“Secondly, google turns up several papers discussiong effects of relativity on temperature, so I am not convinced that it is “absolute.” ”
I don’t know what alleged effects you are referring to, but that local physics is invariant is the fundamental principle of relativity (the constancy of the speed of light in vacuo being just one aspect of that).
“In any case, this seems like it has become a discussion for a physics/relativity blog, not a climate blog. I am sure there are people more learned in this topic than either of us who could shed much more light on the topic.”
Speak for yourself! As for its relevance: quite a few previous threads here have misapplied relativity (and quantum mechanics) to climate issues (I’m thinking of various solar, solar neutrino and cosmic ray threads, greenhouse effect discussions and others). In this thread, the last week’s discussion has been based upon people talking past each other with muddled notions of heat and temperature, which a grasp of this simple relativistic scenario could have prevented, since it demonstrates how an object at one temperature can heat another to a higher temperature; so this cannot be what the second law prohibits.
@Tim Folkerts December 8, 2010 at 3:34 am
Ok, you meant the warmer object, not just warm. Now, will you answer my question posted December 8, 2010 at 1:54 am ?
At this time Tim Folkerts, you really have got into a deep discussion from all sides, haven’t you?
I’ll just let you know that I have, of course come across both of the “greenhouse theories” you mentioned earlier . What I meant was that the only one that makes any sense, to me – is the one that relates to greenhouses. But I was not clear about that, nor was I clear about what puzzles me about this particular one.
Look at fig.4 with a shell which blocks long wave radiation but lets short wave in. My question should have been; HOW CAN LONG WAVE (BLUE) RADIATION RETURN AS ENERGY TO SPACE IF THE SHELL IS OPAQUE TO IT?
If one speculates that the heat is conducted from the shells inside to the shells outside before being radiated back to space, then there will be a delay and one must ask how long does that conductive delay take?
This is an important question indeed as a correct answer will enable us to calculate how many times the long wave radiation wave (BLUE ARROW) will bounce, at the speed of light, between the shell and the planet’s surface. If one can (as indicated in the drawing Fig.4) double the W/m² for every bounce one may be able to make a very reliable estimate of that planet’s greenhouse warming.
Then again: is the first delay the only delay?
Oh and about the answer you gave me, you say; “Atmospheres without GHG’s are very ineffective at radiating IR, so they cannot provide any significant IR toward the surface. “
Could you please give me one, just one example of such an atmosphere? I.E. an atmosphere devoid of “greenhouse gases” as the comparison may/will prove – or disprove everything concerning AGW.
You see I believe in most that is so far been discovered about radiation, but I do not believe that radiation from an average temperature of 15 °C, or less is powerful enough to have any impact, even if the basic “Greenhouse Effect Theory” is correct.
I am looking forward to an answer from you because, yes – some of your answers out there are pretty good.
Lastly, think about this one: In nature everyting increases or decreases one step or one unit , whatever that unit is, at the time. Now think about temperature (T) and an increase to the power of 4.
Or as it is often portrayed: “The temperature goes up with 2 °C which means that radiation increase by 2 x 2 x 2 x 2 = 16 times. “Oh yes somebody’s constant is involved as well, – I know -. But it hould be 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1…….. Do you follow? No, well think about it some more.
But when all is said and done I still cannot accept we know all there is to know about radiation. Of course everybody’s working outs concur. they will every time if everybody are using the same flawed theories!
“At this time Tim Folkerts, you really have got into a deep discussion from all sides, haven’t you?”
You can say that again! I seem to have hit the trifecta of heavy-duty physics!
* quantum mechanics (IR absorption by polyatomic molecules)
* relativity (redshifted photons)
* thermodynamics (second law/energy transfer from warmer to cooler surfaces)
“HOW CAN LONG WAVE (BLUE) RADIATION RETURN AS ENERGY TO SPACE IF THE SHELL IS OPAQUE TO IT? ”
I think the key idea is that the energy does not directly “return to space”. If the shell is at 255 K, it will radiate 240 W/m^2 into space in the form of long wavelength infrared (LWIR). This is calculated with the the Stefan-Boltzmann law. This LWIR is “created” by the surface independent of how the surface got to that temperature.
Of course, if there was no other energy source, then the shell would start to cool off since it is losing energy. So we need a heater. We could heat the shell with an electric heater. We could heat it with a star radiating 240 W/m^2 of LWIR onto the surface. Or in Fig 4, we could heat the shell indirectly by having the planet absorb 240 W/m^2 of SWIR/visible light. The planet then heats the shell from below with its own LWIR. This LWIR from the planet is absorbed and ceases to exist. Instead it simply becomes thermal energy of the shell. Then the shell can create its own LWIR as in the first paragraph.
“… will enable us to calculate how many times the long wave radiation wave (BLUE ARROW) will bounce, at the speed of light, between the shell and the planet’s surface. ”
Again, I don’t think this is the right picture of what is happening. The LWIR doesn’t “bounce back and forth” If we are assuming the surfaces are blackbodies, then the LWIR is completely absorbed when it reaches the surface.
If the shells are not perfect blackbodies, then some of the light will indeed reflect at each surface. But objects that are not BBs will also emit less light. So there would not be lots of extra IR bouncing around anyway. Changes in emissivity could adjust the equilibrium temperatures a bit, as Paul Birch said November 28, 2010 at 3:56 am
“If one speculates that the heat is conducted from the shells inside to the shells outside ”
Yes, the original model does indeed postulate that the shell is a perfect thermal conductor and that it will have the same temperature everywhere. The LWIR heating the inner surface spreads out evenly thru the shell (inside and out, sunny side and dark side). If this is not the case, then there is some “delay” in the energy getting thru the shell and the analysis will get more complicated, which Paul also addressed in the post i just mentioned.
“Could you please give me one, just one example of such an atmosphere [without GHG’s]?”
I can’t think of any real planet or moon that has such an atmosphere. GHG’s like CO2, H2O and CH4 are common around the universe. But we can certainly measure the IR absorption of various gases and extrapolate what an entire atmosphere of that gas woudl be like as it relates to IR absorption. I would suggest looking here: http://scienceofdoom.com/2010/07/24/the-amazing-case-of-back-radiation-part-two/
“Or as it is often portrayed: “The temperature goes up with 2 °C which means that radiation increase by 2 x 2 x 2 x 2 = 16 times. “Oh yes somebody’s constant is involved as well, – I know -. But it should be 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1…….. Do you follow? No, well think about it some more.”
The RATIO of the power increases with the RATIO of the temperature.
For example, if the temperature goes up with 2 °C from 300K to 302K (1.0067 times hotter), the power increases by a factor of (302^4 / 300^4) = 1.0067^4 = 1.027 times more energy. You would have to DOUBLE the temperature (for example, 300K to 600K) to get 16x power. Google “Stefan-Boltzmann Law” for more info.
“But when all is said and done I still cannot accept we know all there is to know about radiation. Of course everybody’s working outs concur. they will every time if everybody are using the same flawed theories!”
Certainly we don’t know all there is to know about radiation (or anything for that matter!). But ultimately the theories have to match observations. We have LOTS of observations of of glowing objects to confirm the basics of Stefan-Boltzmann. (Applying it to a complex system like the atmosphere becomes more of a challenge).
from NY says: December 8, 2010 at 4:38 pm
Ok, you meant the warmer object, not just warm.
Sorry if I was not clear. In discussions like this I assume that “warm” means “warmer than the other object”.
* A “warm” object at -200 C will radiate net energy to a “cool” object at -250 C
* A “warm” object at 20,000 C will radiate net energy to a “cool” object at 15,000 C
“How can the warm object absorbing energy stay at the same or lower temperature and radiate more? “
I think we are facing the same sort of semantics here. There are a couple issues that are interconnected.
* A warm object will radiate more thermal EM energy per m^2 per sec than a cooler object with the same emissivity. This is the stefan-boltzmann law.
* A warm object will radiate more thermal EM energy per sec to a cool object than the cool object radiates to the warm object (independent of size, location, emissivity, or mirrors/lenses). This is the 2nd Law of Thermodynamics (ignoring relativistic effects 🙂 ) .
So the warm object can absorb energy and cool because it is radiating even more energy than it receives. Even if both objects cool, the warm object is “warmer than the cool object” but “cooler than before”. So it radiates “more energy than the even cooler object” but it radiates “less energy than it did before”.
Keeping track of precisely what is meant by “warmer” or “more than” is critical, and I was perhaps making too many assumptions that others were interpreting the word the way I was intending in a particular context.
Paul Birch says: December 8, 2010 at 11:35 am
“In this thread, the last week’s discussion has been based upon people talking past each other with muddled notions of heat and temperature, which a grasp of this simple relativistic scenario could have prevented”
With all due respect, I disagree. From my view, most of the “muddled notions” are much more basic. For example:
* “warm” vs “warmer”
* Multiple claims that Fig 4 is completely wrong in its energy balance.
* muddled analogies of “truckloads of energy” and hot rocks and electric heaters.
* misunderstandings about the nature of GHG’s.
I don’t think relativistic thermodynamics will help clarify those issues!
I have probably spent too much time trying to explain things in this thread. Part of the problem is background — from PhD’s to “soundbite scientists”. (I just coined that phrase! 🙂 ) . Another problem is motivation — some are eager to learn while others already have their minds made up. And finally, what gets resolved here will really not matter to more than a handful of people — the same issues will likely come up again next week in a different entry at a different blog with different people.
But the teacher in me hates to see poor science go by without some discussion …
Tim, you are dodging the question. The Stefan-Boltzmann law states, that “the total energy radiated per unit surface area of a black body per unit time (known variously as the black-body irradiance, energy flux density, radiant flux, or the emissive power), j*, is directly proportional to the fourth power of the black body’s thermodynamic temperature T.” Look up Wiki for example.
A body radiates in a vacuum too. The situation we are discussing is having a coin with certain temperature radiating according to its temperature and surface in a vacuum, and then having photons from a colder coin directed to this coin.
You claim that 1) “Yes, the photons from the cooler object will be absorbed by the warmer object”, 2) “The warmer object gains the energy, but its temperature does not go up.” 3) “Specifically, the warm object absorbs the energy from the photon, but thermodynamics/blackbody radiation laws tells us that the warm object will necessarily be losing more energy from its own outgoing photons than it is gaining from the incoming photons.”
This contradicts the Stefan-Boltzmann law. The temperature does not go up, yet the warmer coin radiates more than before. That is what I am questioning and you are dodging.
The problem starts with your acceptance of absorption of photons from the colder body. You don’t explain how this extra energy is going out at the same temperature.
There is no problem if colder photons are not absorbed.
From NY,
Are you postulating that the warm object and/or the cold objects have some sort of other heaters? I was assuming the objects were simply warm and slowly cooling as they lost heat. That might be the source of our miscommunication.
Just to make it more concrete, lets use some numbers. Suppose we have a blackbody sphere with a surface area of 1 m^2 @ur momisugly T = 303K radiating 480 W/m^2 (the “warm object”) We have available a thin shell blackbody with a surface area of 4 m^2 (twice the radius of the warm object) at a temp of 255 K which would radiate 240 W/m^2 = 960 W total (the “cool object”). (The inner surface will also radiate 960 W, but all of this would be absorbed by other parts of the inner surface.)
CASE 1: Neither is heated.
1A) The warm object is in the middle of nowhere. It loses -480 W to space, absorbs 0 W from other sources. It starts at 303 K but cools as it loses -480 W.
1B) The warm object is surround by the cool object. The warm object still radiates -480 W, all of which goes to the cool object. The warm object absorbs 240 W from the outer shell (the inner surface of the shell radiates 960 W, but 720 W would go back to the shell, while 240 W would go to the warm object. The warm object now has a net loss of only -240 W so it will cool, but not a quickly as before.
For the cool object, the outer surface radiates away -960 W. The inner surface radiates -960 W as well, but 720W goes back to other parts of the inner surface. The warm object radiates 480 W to the inner surface of the cool object. The cool object has a net flow of -960 W (to space), -240 W to the warm object, and +480 W from the warm object = -720 W net loss. (so it cools more slowly than if it was by itself radiating -960W to space.)
Over time, all the temperatures and radiation will drop, but there will always be more power warm –> cool than cool –> warm.
CASE 2: Both have thermostats.
1A) The warm object is in the middle of nowhere. It loses -480 W to space, so the thermostat causes the heater to provide 480 W. It stays at 303K
1B) The warm object is surround by the cool object. The heater will provide 240W to the warm object to keep it from cooling. The heater provides 720 W to the cool object. The temperatures of all objects stay the same and the analysis of CASE 1 applies.
CASE 3: Both have heaters (480W for the hot object; 960 W for the cool object, which would independently maintain the current temperatures).
1A) The warm object is in the middle of nowhere. It loses -480 W to space, but the heater to provide 480 W. It stays at 303K
1B) The warm object is surround by the cool object. The heater will provide 480 W to the warm object to keep it from cooling. The net flow for the warm object is -480 W from radiation + 480 W from the heater and + 240 W from the cool object. The warm object starts to warm up.
The net flow from the cool object is -960 W to space + 960 W from the heater – 240 W to the warm object +480 from the warm object. The net power is +240 W. It also warms up.
Eventually, the cool object will warm up until it radiates away all the power from the heaters = 960 W + 480 W = 1440 W = 360 W/m^2. The warm object will be getting 360 W from the cool object, and 480 W from the heater, so it will warm until it radiates 840 W/m^2.
CASE 1B specifically matches my claims that you quoted. Stefan-Boltzmann is satisfied for all cases. 2nd Law of Thermodynamics is satisfied for all cases. There would be a HUGE problem if the photons from the cooler object were not absorbed by the warmer object.
And if you use just part of the outer shell (eg cut out a coin-shaped section), the results will be somewhere between (A) and (B) but we will still have the appropriate laws satisfied. You could also change the shape of the center object to a coin — the calculations would just be a little more complicated.
I *think* that covers pretty much all the options and hopefully this no longer “dodges” which ever specific circumstances you were thinking of.
Tim, this is not what we discussed. The case was suggested by bessokeks December 4, 2010 at 9:14 am
“Imagine two coins opposite to each other. one small and warm, one 10 times bigger and slightly colder. Then put a lens between an direct the photons of the slightly colder coin exactly onto the surface of the smaller coin. You would transfer roughtly 10 times the energy the warm coin radiates onto the warm coin…”
This is what we were talking about all the time, coins, not spheres. Just read my last post again, please:
“The situation we are discussing is having a coin with certain temperature radiating according to its temperature and surface in a vacuum, and then having photons from a colder coin directed to this coin.”
And read the bessokeks’ description for details. Thanks.