Guest post by Willis Eschenbach
There is a lot of misinformation floating around the web about how the greenhouse effect works. It is variously described as a “blanket” that keeps the Earth warm, a “mirror” that reflects part of the heat back to Earth, or “a pane of glass” that somehow keeps energy from escaping. It is none of these things.
A planetary “greenhouse” is a curiosity, a trick of nature. It works solely because although a sphere only has one side, a shell has two sides. The trick has nothing to do with greenhouse gases. It does not require an atmosphere. In fact, a planetary greenhouse can be built entirely of steel. A thought experiment shows how a steel greenhouse would work.
Before we start, however, a digression regarding temperature. The radiation emitted by a blackbody varies with the fourth power of the temperature. As a result, for a blackbody, we can measure the temperature in units of radiation, which are watts per square meter (W/m2). For objects with temperatures found on the Earth, this radiation is in the range called “longwave” or “infrared” radiation. See the Appendix for the formula that relates temperature to radiation.
This means that we can denote the temperature of a blackbody using W/m2 as well as the traditional measures (Fahrenheit, Celsius, Kelvin). The advantage is that while temperature (degrees) is not conserved, energy (W/m2) is conserved. So we can check to see if energy lost is equal to energy gained, since energy is neither being created nor destroyed by the climate.
For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square meter of surface area. How is it heated from the interior? Doesn’t matter, we’ll say “radioactive elements”, that sounds scientific.
The planet is in interstellar space, with no atmosphere and no nearby stars. The equilibrium surface temperature of this planet is, of course, 235 W/m2. To maintain the equilibrium temperature, it constantly radiates this amount of energy out to space. Coincidentally, this is the amount of solar radiation that makes it past the clouds to warm the Earth. If we convert 235 W/m2 to one of our normal temperature scales, it is -19 Celsius (C), or -3° Fahrenheit (F), or 254 Kelvins (K). It’s a possible temperature that the Earth might have if there were no greenhouse effect. That is to say … cold.
Now imagine that the planet gets completely surrounded by a thin black steel shell, located a few thousand meters above the surface, as shown in a cutaway view in the picture above, and in Figure 1 below. What happens to the surface temperature of the planet? (To simplify calculations, we can assume the shell has the same outer surface area as the surface. For an earth-sized planet with a shell two kilometers above the surface everywhere, the difference in area is only six-hundredths of one percent. This assumption makes no difference to the argument presented.)
In order to maintain its thermal equilibrium, including the new shell, the whole system must still radiate 235 W/m2 out to space. To do this, the steel shell must warm until it is radiating at 235 watts per square meter. Of course, since a shell has an inside and an outside, it will also radiate 235 watts inward to the planet. The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell. This will warm the planetary surface until it reaches a temperature where it radiates at 470 watts per square meter. In vacuum conditions as described, this would be a perfect greenhouse, with no losses of any kind. Figure 1 shows how it works.
Figure 1. Building a steel greenhouse. (A) Planet without greenhouse. Surface temperature is 235 W/m2 heated from the interior. (B) Planet surrounded by steel greenhouse shell. Shell radiates the same amount to space as the planet without the shell, 235 W/m2. It radiates the same inward, which warms the planet to 470 W/m2 (29 C, 83°F, 302 K). [Clarification added] Note that the distance from the shell to the planet is greatly exaggerated. In reality, it is close enough to the planet that the difference in the areas can be neglected in practice.
The trick can be repeated again, by surrounding the planet and the shell with a second outer shell. In this two shell case, the planetary surface temperature (in W/m2) will be three times the initial surface temperature.
In nature, planets have atmospheric shells, of course, rather than steel shells. The trick works the same way, however. Rather than being warmed from the inside, the Earth receives 235 W/m2 from the sun. Solar radiation passes through the atmosphere. Outgoing longwave radiation is absorbed by the atmosphere, just as it is absorbed by the steel shell shown in Fig. 1.
So that’s the trick of the greenhouse. It has nothing to do with blankets, mirrors, or greenhouse gases. It works even when it is built out of steel. It depends on the fact that a shell radiates in two directions, inwards and outwards. This radiation keeps the planet warmer than it would be without the shell.
Now, it is tempting to think that we could model the Earth in this manner, as a sphere surrounded by a single shell. This is called a “two-layer” model, with the two layers being the surface and the atmospheric shell. In fact, a number of simplified climate models have been built in this way. Unnoticed by their programmers, however, is that is not physically possible to model the Earth as a two-layer system. Figure 2 shows why. Note that in all cases, the system has to remain in thermal equilibrium. This means that the surface must radiate as much as it absorbs, and the shell must also radiate as much as it absorbs. In addition, radiation from the shell to space (upwelling longwave radiation or ULR) must equal radiation from the shell to the Earth (downwelling longwave radiation, or DLR)
Figure 2. Single-shell (“two-layer”) greenhouse system, including various losses. S is the sun, E is the Earth, and G is the atmospheric greenhouse shell around the Earth. The height of the shell is greatly exaggerated; in reality the shell is so close to the Earth that they have about the same area, and thus the small difference in area can be neglected. Fig. 2(a) shows a perfect greenhouse. W is the total watts/m2 available to the greenhouse system after albedo. Fig. 2(b) is the same as Fig. 2(a) plus radiation losses Lr which pass through the atmosphere. Fig. 2(c) is the same as Fig. 2(b), plus the effect of absorption losses La. Fig. 2(d) is the same as Fig. 2(c), plus the effect of thermal losses Lt.
Figure 2(a) shows the same situation as Figure 1(B), which is a perfect planetary greenhouse. In this case, however, it is heated by an amount of energy “W”, which is coming from the sun. The planet receives solar radiation in the amount of “W” from the sun, and longwave radiation “W” from the atmospheric shell. The surface temperature is thus 2W. All energy flows are in Watts/square metre (W/m2).
Figure 2(b) adds two losses. The first is the reflection from the Earth’s albedo (Wo – W). This is energy which never enters the system and is reflected back into space. We are still getting the same energy entering the system (W). The second loss Lr is from radiation which goes from the surface to space through the “atmospheric window”. Because of the second loss, the surface of the Earth does not get as warm as in a perfect system. In a perfect system, the temperature of the surface is 2W. But including the radiation loss Lr, the surface temperature drops to 2W – 2Lr.
Figure 2(c) adds another loss La, which is the solar radiation which is absorbed by the shell. This cuts the radiation hitting the surface down to W – La. Including this loss, the surface temperature is 2W – 2Lr – La.
Figure 2(d) adds the final loss. This is Lt, the thermal loss. It is the sensible energy (energy you can feel) and latent energy (evaporation) which is transported from the surface to the shell by convection. Including all of these losses, the surface temperature of the Earth is 2W – 2Lr – La – Lt.
Now, why can’t the Earth be modeled in this manner? A look at the size of the various losses shows why. Here is the canonical global energy budget, called the “Kiehl/Trenberth” budget, or the K/T budget.
Figure 3. The Kiehl/Trenberth Global Energy Budget. This is a “two layer” representation, with the surface and the atmosphere being the two layers. Lr, the radiation loss, is the 40 W/m2 labeled “Atmospheric Window”. La, the absorption loss, is the 67 W/m2 labelled “Absorbed by Atmosphere”. Lt, the thermal loss, is 102 W/m2. This is the sum of the 24 W/m2 labelled “Thermals” and the 78 W/m2 labelled “Evapo-transpiration”. W, the energy from the sun, is the incoming solar radiation of 342 W/m2 less the 107 W/m2 that is reflected by the surface and the clouds. This means that W is 235 W/m2. SOURCE
What’s wrong with this picture? Note that the temperature of the Earth is 390 W/m2, labeled as “Surface Radiation”. This is 15 C, or 59°F. But from Fig. 2(d), we know that the surface temperature of a greenhouse system with a single shell, as shown in the drawing, is 2W (470) – 2Lr (80) – La (67) – Lt (102) = 221 W/m2. This is far below the known surface temperature of the Earth. In other words, a single shell greenhouse system simply isn’t efficient enough to give a surface temperature which is warm enough to allow for the known losses.
So where is the problem with the K/T budget diagram? The hidden fault is that the upward radiation from the atmospheric layer does not equal the downward radiation. There is 195 W/m2 going to space from the atmospheric shell, and 324 W/m2 going down to the surface.
In order to get enough energy to allow for the known losses, the simplest model requires two atmospheric shells. A perfect greenhouse with two shells would give a surface temperature of 3W, or 705 W/m2. This is enough to allow for the known losses, and still give a surface temperature which matches that of the Earth. Figure 4 shows one such possible system.
Figure 4. Simplest greenhouse global energy budget capable of representing the Earth. Note that all major flows in the K/T energy budget have been maintained. There are two shells, which must be physically separate. These are the lower troposphere and the lowest stratosphere. They are separated by the tropopause.
This budget fulfills all of the requirements for thermal equilibrium. The same amount is radiated upwards and downwards by each layer. The amount absorbed by each part of the system equals the amount radiated. Further verification of this energy budget is the 147 W/m2 emission from just above the tropopause. This converts to a Celsius temperature of about -50 C, which is a typical temperature for the lowest part of the stratosphere.
I have written a simplified Excel radiation/evaporation/convection model which encapsulates the above system. It is available here. Click on the “File” menu on the webpage and select “Download”.
I invite people to play with the model. It is what I term a “Tinkertoy” model, which is what I call the simplest possible model that can represent a particular reality. One of the interesting results from the model is that there can be very little thermal loss between the inner and the outer atmospheric layers. If there is more than a trivial amount of leakage, the surface cools significantly. Among the other insights yielded by the model is that a change equivalent to a doubling of CO2 (an increase of 3.7 W/m2 downwelling radiation at the top of the atmosphere) can be canceled by a 1% increase in the upper and lower cloud reflections.
Experiment with the values, it is an interesting insight into the energy flows in the simplest possible climate model that can represent the Earth’s greenhouse system.
APPENDIX
The formula that relates the temperature to radiation is called the “Stefan-Bolzmann” equation:
R = sigma * epsilon * T^4
where r = radiation (W/m2)
sigma = the Stefan-Boltzmann constant, 5.67 * 10^-8
epsilon = the emissivity of the body, which for a blackbody = 1
T^4 = absolute temperature in Kelvins (which is Celsius plus 273.15) raised to the fourth power
Willis Eschenbach (16:30:19) :
If a chair is brough in from the cold into a room at 25C then placed in the centre, at least 5 2 metres from any other object, that chair will reach the equilibrium of the atmosphere, not the radiation from other objects, as those objects are already thermalised with theh atmosphere and aren’t giving off much radiation. Even air overpowers the effect of material radiation. According to the SB, different objects should be different temperatures at the same atmospheric temperature and should be radiating and absorbing from each other. In truth, they all thermalise with the atmospheric temperature so don’t give off or receive radiation from each other – air is a very poor conductor. I hope you’re not arguing that a human can radiate to the value of 36C to surrounding objects..
However, in more simple terms, if 100w/m2 is the average human radiate flow, then how can it be argued that at 15C, 235w/m2 is emitted from earth to the atmosphere? Take underflor central heating as a more common value and argue the same.
I’m not criticising your model at all, just maintaing that the SB figures have to be divided by 10 to give a realistic energy result. Heat is not the physical constant that physicist seem to think it is. Cooling can be autonomous. What for instance happens when you pull a re hot poker from a kiln? the air surrounding it heats, the poker cools until they are at atmospheric temperture. Heat disippates, yet we’re arguing that heat is a fixed constant, as though it has to migrate elsewhere
P Wilson (16:22:16) : “If the two planets were at the same temperature they would be in equilibrium and wouldn’t radiate to each other.”
This is wrong. There can be situations where there is no NET radiative transfer of energy between the two bodies, but they are always going to each be radiating per the Stefan-Bolzmann equation.
Even the assumption that two planets at the same temperature will have not NET radiative transfer is wrong. Even if they are the same size, just having different albedos would cause a net energy transfer.
Even if they are the same temperature and the same albedo, there will be net energy transfer if the temperature of the planets is different than the background and their sizes are different. You can envision this by imagining the limit case where a larger planet fills nearly one hemisphere of the view looking outward from a tiny speck of a planet. The larger planet sees incoming radiation from what is essentially all outer space (which we will assume is 4K for this thought experiment). Meanwhile, on the tiny speck of a planet, half of view looking out is 4K outer space, but the other half is the much warmer planet.
Willis, I have a question about your figure 4. Consider the 53 units of energy that are radiated from the ground and are not absorbed by the troposphere. Of these 52, 13 (13/52=24%) are absorbed by the stratosphere, the rest going out into space. However, out of the 321 units radiated outward from the troposphere, 271 (271/321=84%) are absorbed by the stratosphere, the rest going out into space. Shouldn’t these percentages be equal? Similarly, 339/392 of the radiation from the ground is absorbed by the troposphere, while 147/147 of the radiation down from the stratosphere is absorbed by the troposphere. Shouldn’t these portions be equal as well? Surely these layers can’t “tell” what the source of the radiation was and discriminate between them.
Willis,
I give you kudos for your model. Really there can be no harm from modeling and thinking about things. You have got a lot of flak I know, but really this area is hard for everyone. It is very hard to get an intuitive feel for how things work. I don’t claim to know or have an intuitive feel either. I’ve struggled with this for quite a while and finally I feel confident in claiming that without a focusing device of some kind that a black body is the limit of absorption. You can not have a body become hotter than a black body without a coherence (which the sun does provide directionally) and a focusing device to exploit this coherence. A magnifying lens and/or gravity.
But given that, here are some of the main issues I don’t believe you have addressed.
1.) The shells have double the surface area of the planet. Let’s assume there is some sort of constant flux outward from the surface. If there is 100 W/m^2 out, the flux in from the shell will only be 50 W/m^2 in and 50 w/m^2 out (not 100).
If you have infinite shells this mean you will asymptotically approach 200 W/m^2 flux, not ‘infinity’ flux. Two times flux means (2)^0.25 higher temperatures or a mere 20% higher, not even double.
2.) The model has a stated constant energy flux outward. This is not like a greenhouse in that a greenhouse needs a way for energy to enter (a hole). If you have a sphere with a hole in it, that is the standard approximation to a blackbody. The stated result for which is given by the Stephan-Boltzmann law. The maximum temperature that can be reached is equivalent to that of a black body. You are saying that it is possible to make a special hole that allows us to achieve temperatures higher than that of a blackbody. I don’t think so. All holes of all types will be equivalent.
As for the solar heater, you might be right about solar irradiance at surface, I couldn’t find a source. I’ll take your word for it. However I suspect that there is a focusing device at work here. i.e. A reflective back plane that adds to the effective ‘area’ of energy absorption. As you can see the result is close to black body and I feel confident that if focusing is taken into account the result is still consistent and we have not created a way to make a temperature higher than that of a blackbody with absorption alone.
cheers,
Ian
Anton Eagle (16:12:49) :
If as you say it makes no difference … then what difference does it make? How is this a “contradiction” as you claim?
However, you have forgotten the units and ignored the geometry. If the shell gets say ten times as large, the shell radiation in W/m2 will get very small. It will still emit a total of 235 X watts if X is the planet surface area, but the radiation per square metre will be smaller. This will reduce the warming of the planet. Instead of intercepting 235 W/m2, the planet will only intercept some small fraction of that. The shell will still have to emit the total radiation of the planet to space, but the planet will not be warmed as much by the radiation of the shell. Instead of being bathed in radiation of 235 W/m2, it will only intercept 23.5 W/m2 of radiation. Thus it will not warm up as much. Most of the shell radiation will simply be reabsorbed and maintain the shell temperature, rather than being intercepted by the planet.
Well, for starters, see the Second Rule at Willis Eschenbach (11:03:15). Putting “Period.” at the end of your post just indicates that you think it needs additional emphasis, but it adds nothing to the strength of your ideas.
Next, the shells are assumed to be vanishingly thin, as befits a thought experiment. This is because if there is a significant thickness to the shell, the inside will be warmer than the outside.
I fail to see any violations of thermodynamics. Two physicists whose opinions I greatly respect (Joel Shore and Luboš Motl) have said that they don’t see any violations. Your claims otherwise are not spelled out in enough detail to see exactly what laws are being violated, you just keep saying laws are violated, laws are violated. But what laws, and where?
AlexB (16:29:05) and Curt (16:45:40), thank you for understanding the thought experiment. You are right, it is basic thermodynamics.
Charlie (17:15:10) :
it pays to use smaller scale examples, so if a hot frying pan is left to cool, in a room the heat loss is from the pan – if objects are radiating energy they’re losing heat, until at such point the pan is the same temperaure as the air of the room. What happened to that heat? did it disappear? did it cool autonomously until it reached equilibrium to the 2nd law of thermodynamics? did the heat escape into the room and distribute itself around? Did it escape from the room and into the atmosphere? I don’t think there is an argument that heat has a permanent property that the last argument suggests – as it can thermalise to the atmosphere’s temperature when it reaches such a point that the flows between the two are barely measurable. I doubt that the heat of the pan will transfer itself to the next pan, back and forth.
stephan (16:53:18) :
I’m not sure I understand you. Why could we not physically measure the radiative flux? This is done on the Earth all the time, and there are instruments specifically designed to measure it from both the ground and satellites … and I assure you that the upwelling and the downwelling flux are rarely equal. See e.g. here and here.
Incidentally i’m not arguing that heat is “destroyed” in violation of the st law – but that the cooling of an object doesn’t depend on the heat it has received by another object, according to its entropy, as molecules and electrons become less excited during the cooling process.
addendum oops. atoms, not molecules
SteveBrooklineMA (17:15:39) :
What you say is certainly possible. However, the radiation is at a different frequency depending on the temperature at which it is radiating, so there is no reason to assume that the absorption will be equal. The set of numbers I chose were selected to replicate the main flows of the K/T budget. Other numbers will work as well. I encourage you and others to play with my model and see what other numbers work, or what other assumptions work.
Ian Schumacher (17:17:58) :
Um, no. If there is 100 W/m2 in from the sun, then there has to be 100 W/m2 out of the shell or else the planet will heat infinetely rather quickly – at whatever rate is generated by the imbalance. This is a pretty basic concept, so if you don’t understand it, you aren’t in any position to argue further.
This is also the reason, btw, that there is any storage, aka “gain,” in this system in the first place. The instant the shell is “applied” to the model a new transfer function is applied. Through superposition, this step can actually be applied to the input itself, which implies that the result can be obtained by modeling the step response of a simple 1st order feedback system.
Nope. Since you got the first point wrong, which already doubles the energy, this is wrong by extension.
Uh, no, he’s not saying that. The planet is radiating at its temperature, and the surface of the shell is radiating at its temperature. The difference, of course, is that the shell must radiate in all directions equally, but the output is constrained to be equal to the input.
There is a “hole,” btw, in that LWIR is passed without disturbance into the system, but 50% of the SWIR is radiated inward after being absorbed by the shell (in the feedback derivation, that would be a coefficient of 0.5, which results in a “gain” of 1/(1-0.5) = 2). This is an approximation, for a simple model of course, but the point is that most of the input resides in LWIR and most of the output resides in SWIR. Not perfect, but Willis makes that clear: it’s a simplification.
Willis, please note that when I put “Period.” at the end of my response to Anton, I should have also put /sarc. 🙂
Mark
Ian Schumacher (17:17:58) :
Willis,
I show in Figure 1 how it can be done.
Agreed. That is why in Fig. 1 the constant flux outward from the surface is 470 W/m2, and at the shell it is half of that, 235 W/mw in and 235 W/m2 out.
I encourage you to draw up the two shell system which is the expansion of Fig. 1. At equilibrium, the outer shell is at 235 W/m2. The planetary surface is at 705 W/m2 (three times the source). The inner shell gets 235 W/m2 from the outer shell and 705 W/m2 from the surface for a total of 940 W/m2, so it radiates at 470 W/m2 (two times the source) up and down.
In a real planetary greenhouse, you have a “one-way” hole in the form of an atmosphere that allows solar radiation in and blocks longwave radiation from leaving. This is why we can achieve temperatures above blackbody on the Earth’s surface.
While a focusing device would improve performance, it is not necessary.
w.
Re: Willis Eschenbach (17:23:33)
It’s a pleasure. Thank you for the excellent post, the most stimulating I have read in a while. I must say I admire your patients though. I would have torn my hair out long ago and the insides of my computer would now be engaging with radiative heat transfer with the troposphere from their scattered locations around my lawn.
Willis,
There is no such thing as a one-way hole. This is the key issue. If we state a theoretical one-way hole then yes you can achieve your result. But there is no such thing as a one-way hole in the real physical world. This would be Maxwell’s Demon.
Anyway we are at an impasse.
Cheers and goodnight.
Yes, I misinterpreted the way Ian was describing this, but your response puts it in the proper context, i.e., the surface of the planet is doubled. This means Ian’s subsequent statements regarding how the shells stack are still incorrect.
Mark
AlexB (14:19:13) :
“I am utterly astounded by the number of people claiming to be physicists who don’t understand fundamental physics. Which fundamental law does this model contradict? ”
Exactly my thought as I’ve read through these comments. There seems to be no end of people who can’t tell the difference between the radiation emitted by a surface, and the net exchange between two surfaces. Anything with a temperature greater than 0 K (meaning, everything) will emit radiation, regardless of what is around it. Why is that so difficult?
Joel Shore: I was going to say the same thing about saturation. It’s a tinkertoy model, but from the direction it’s going, one can see where the saturation arguments go wrong. Yes, the original radiation from the earth at a given wavelength might be absorbed, but that isn’t the end of the story.
Really, I had no idea so many people around here actually disputed the basic mechanism itself, given that it’s based on pretty simple physics. I thought the main dispute was over the magnitude of the feedbacks. Instead, we’ve got people arguing about the magnitude of the SB constant? Oh well.
Mark T,
Really this whole argument is begging for an experimental isn’t it? I would like to, but admittedly, I’m too lazy.
Do you think that you can have a sphere with a hole in it to accept external radiation that can become hotter inside then a black body? Just wondering. That is the key. I’m trying to think what would such a thing look like. It really does require a one-way mirror, but no such device exists (see above). Ah well. Cheers to you also and good night (I’ll still check replies if you care to reply about ‘super’-black bodies, etc.
Willis Eschenbach (16:30:19) :
. . . Both radiation flows warm the object that they strike.
Small point, I’m sure, but this explanation could be contributing to some of the confusion exhibited hereabouts . . . .
IMO, the verb “to warm” refers to something exhibiting an increase in temperature.
Your explanation clearly describes the two energy flows between the two and that the net energy flow is from the warmer body to the cooler. But this notion of the cooler body “warming” the warmer body also shows up here and elsewhere in the thread. (In at least on post you explain something like “warmer than it would be, otherwise.”)
Surely you agree that, in terms of temperature, until they reach an equilibrium temperature, since the net energy flow is from the warmer to the cooler, the cooler object only warms and the warmer object only cools.
Hope that helps.
Very good article. I get the feeling from a lot of the posts that some people have a real problem with science or are being obtuse to confuse the issue on purpose (personally, I think the latter). This was intended to be a very simplified description of what is going on in the atmosphere with many assumptions to make it simple. We do this in physics and other disciplines all of the time. For example, a hockey puck slides across a frictionless surface…. Some of the posters here would then argue that there is no such thing as a frictionless surface. While that is true, for this problem, including friction does nothing to help us demonstrate an elastic collision. That is the point. The earth is not a perfect black body. The principle is the same and a black body is easier to work with. There may be some reflection from the steel sphere. OK, assume it has a perfectly non-reflective surface. And so on for the demonstration. Adding all of the variables back in results in a very complicated model and one that we don’t even really know if we have all of the factors included. It just brings all of the climate models into more question not less. Thank you Mr. Eschenback.
Ian Schumacher (18:26:48) :
The atmosphere allows incoming solar radiation through. It blocks outgoing longwave radiation. That’s basic physics, I haven’t a clue how you can continue to argue with that.
If I remember my high school physics correctly, there are three ways for energy to propogate: radiation, convection and conduction.
Anton Eagle, the difference between your thought experiment of ten thin shells vs one shell ten times thicker is that you are changing the energy transfer from radiation to conduction. The energy propogation in a thicker shell is via conduction which exhibits different characteristics to energy propogation via radiation. The thought experiment only considers energy radiation.
Thank you, Willis, for an interesting article. I’ll admit that I didn’t immediately understand how the two sides of the steel sphere affected the surface temperature, but I’m pleased that I worked it out before I read the comment about what would happen if the steel sphere was replaced with a perfect insulator.
Ian Schumacher (18:34:13) :
Maybe. I already mentioned one. I have another, but it is conservation of mass, not energy, and the mechanism for actually implementing the mass idea would be rather complex (though doable).
This is the point you keep making that doesn’t make sense. There is no set temperature of a black body. If there is sufficient energy storage, the temperature will necessarily be higher.
Keep in mind, Willis is making a few ideal assumptions that cannot absolutely hold in the real world, which necessarily implies his results are upper limits.
A dichroic mirror, actually, may get you close. It lets different wavelengths pass in different directions. For Willis’ example, the shell is transparent to long wave IR, but reflects 50% of short wave IR. Since the earth absorbs LWIR, but then radiates SWIR, you have feedback, allowing energy to increase within the system, though only during the transient state. Once equilibrium is reached, there is no more warming and things continue as Willis has shown (well, in the ideal).
Mark
Anton Eagle says:
There is nothing wrong with this. The assumption in the model is that each shell is a blackbody, meaning it absorbs all the radiation impinging on it and emits according to the Stefan-Boltzmann Law with epsilon = 1. So, indeed, having 10 blackbody shells leads to a higher temperature than having one blackbody shell.
If you wanted to have something that converged to a specific result in the limit that you take the number of shells to infinity, you would want to scale epsilon for the shells so that the atmosphere has a constant absorption as you increase the number of shells.
John A says:
As Willis explains, these are statements about net flows. It is not as if a cooler object detects a warmer object approaching it and magically stops emitting radiation to it. They both emit radiation to each other but the warmer body emits more to the cooler body than the cooler does to the warmer.
The Second Law of Thermodynamics is not magic. It has its basis in statistical physics. Consider another example with molecular flow instead of heat flow: Say, you have two chambers with a wall between them, one at a pressure of 1 atmosphere and another evacuated to 0.01 atmosphere. Now, let’s say you punch a hole in the wall between them. What will happen is that there will be a net flow of molecules from the high pressure to the low pressure chamber. However, this does not mean that magically no molecules from the low pressure chamber move into the high pressure chamber. In fact, some will. It is just that many more will go the other way. And, the reason, from a statistical point of view, that the net flow is from the high pressure chamber to the low pressure chamber is simply that there are many more molecules impinging on the hole from the high pressure side than from the low pressure side. And, once you are dealing with situations involving any sort of macroscopic amount of material, the statistics become overwhelming, so that the chance of the pressure in the high pressure chamber increasing while the pressure in the low pressure chamber decreases becomes so astronomically small that it ain’t going to happen even if you carry out the experiment once a second for the entire lifetime of the universe. (If you did the same experiment with, say, only 5 molecules in one chamber and 1 in the other, then while most of the times [on average 5/6] that you tried the experiment you would see the number of molecules in the high pressure chamber decrease and the number in the other increase, if you ran the experiment enough times you would in fact see a case where the 1 molecule in the low pressure chamber went into the high-pressure chamber so that the flow went the “wrong way”. The Laws of Thermodynamics are an emergent property of macroscopic systems that involve enough molecules that the probabilities of net flows going the wrong way become astronomically small.)
although the cooler planet emits radiation toward the warmer planet, supposing that one were nominally 100K and the other 150K, at some point, supposing they were fixed to receive one another’s thermal influuence they would thermalise, if there were not the presence of another radiating body, although radiation goes in all directions, not just towards other hypothetical planets, and may lead to Kelvin’s heat death hypothesis where there was no thermal energy left. Kelvin extrapolated that heat loss is endemic to nature, based on this 2nd law of thermodynamics, but for the purposes of our climate, we still have the sun.
carrot Eater wrote that some are apt to forget that anything above absolute zero emits radiation – well yes it does, because its losing heat, but depends on the temperature surrouding it. If an area of space is 10K and an object suddenly enters it at 4K, eventually the 4k object will reach 10K, supposing the absence of other radiative sources. Its improbable that the 10K region of space will lose heat to a 4K object