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
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constants are constants. If this planet emits 235w/m2, and eeven if totally blocked to outgoing radiation, its still 235 w/m2 ad infinitum, so it won’t accumulate to 235*some multiplier or else 235 outgoing+235 incoming. Its emitting as much as it blocks, which is efectively the same energy
The basic problem with this post is that it gives a complicated analogy to a complicated system. It just doesn’t simplify things much. Furthermore, it reverses the heat source unnecessarily. And Figure 1 puts people who know how surface scales with radius, off their feed.
Re: “Please read up on the Stefan-Boltzmann equation …” This is a mean spirited statement not intended to illuminate. Like one of the other commenters, I’ve designed heat sinks for commercial production. To convert temperature to W/m^2 requires assuming an emissivity which depends to an incredible degree on the choice of finish. In addition, the emissivity of surfaces typically changes with temperature.
My point is that “temperature” is a simple concept that everyone understands. “W/m^2” is not. And the relationship between them is a mess. An analogy with an insulated bottle is simpler:
Light from the sun gets into a (transparent) bottle and heats it up. The amount it heats it up is proportional to the thickness of the insulation. As an insulator, CO2 provides a thickness that is proportional to the logarithm of the amount of CO2 which is why global warming is not much of a threat.
put another way, 60F will not surpass 60F with the shell number increase, just that that 60F will cover a wider area distribution than 1 shell
Willis & Mark T,
Ok I am dense, I fully admit that, but I am open to learning 😉 And for some reason this problem fascinates me. So I’m thinking through the shells again. Ok step by step least see if I can get this.
Iteration:
1.) 100 W/m2 out … hit shell
2.) shell give 50 back and 50 out to space.
3.) 50 from shell hits earth, warms it up, goes back out as 50 watts, hits shell.
4.) 25W/m2 back, 25W/m^2 out to space.
We can see were this is going, so:
From our special source we have 100W/m2 out .. as stated .. always.
hitting the surface we have 50 + 25 + 12.5 … = 100W/m2
going out from surface we have same … 100W/m2
From shell in we have 50 +25 + 12.5 …. = 100 W/m2 in.
From shell out we have .. same = 100 W out.
We have a source of 100 W/m^2 and we have 100 W/m^2 leaving … great. Everything seems in balance.
Outer shell will always have 100W/m^2 leaving from it.
Ok so next shell 100W/m^2 leaving, means 200W/m^2 from first shell and 200 W/m^2 from surface (plus out source of 100W/m^2).
Ok, you are right. I should have thought through it more carefully the beginning. My mistake. Basically this is like extra layers of insulation I guess.
Now, point #2.
In you thought experiment we have this constant source. This source is the ‘sun’ through the visible window. Ok so to apply this to the steel shell model the sunlight has to come in to hit the surface. We can do that by punching a hole in the steel shell that lets in 100W/m^2 equivalent amount of sunlight energy. Or we can say the shell has a special property. It is a high-pass filter that lets in most of the sunlight, but rejects (acts like a solid shell) for low frequencies. Let’s pick the second one.
This high frequency light comes in, hit the cold surface, warms it up a bit and emits a black body spectrum of frequencies outward, most of these being below our cutoff frequency of our shell.
The radiation that is above the cutoff will easily exit the shell. The remaining radiation will (after bouncing around back and forth a bit such) generate new radiation above the cutoff frequency.
Let’s make the model even simpler and assume that the outside shell doesn’t radiate any energy at all outward. Its a perfect reflector. A perfect reflector is equal to your shells inside shells if you have infinite of them. To show this, imagine your internal energy source again. You have a perfect reflector shell. The flux will keep increasing forever given infinite time to build up. This is the same as with infinite shells, but easier to manage mentally.
So we have a sphere that is a perfect reflector. All energy comes through our frequency hole. All energy leaves through this same hole.
100 W/m^2 comes in through the hole. reflects around. How much comes out? 100W/m^2.
What is the temperature inside?
The distribution of radiation frequency is give by Plank’s law. At first glance it seems that we need to have a temperature large enough that when we integrate the radiation above our cutoff frequency we get 100w/m^2 total.
But this is wrong for the same reason that thinking that the radiation in a spherical cavity with a physical hole in it will need to achieve a temperature so that radiation aimed directly at the hole goes out (integrate over all radiation at the right magic angles to get out). Its wrong because it ignores the constant reflections and ‘repopulating’ of these angles (for the hole example) or repopulating the radiation above the cutoff frequency (for the case of a frequency hole).
Just like the physical hole, the frequency hole is the same.
A hole in frequency is the same as a hole in space. A cavity with a hole in it will achieve temperatures near a black body. Doesn’t matter what kind of hole this is.
At least that’s my theory 😉
Le Sage’s Theory of Gravitation.
I have fixed your Wikipedia link in what I quoted above so people can read your theory at their leisure, and I invite you to discuss it on another thread.
The wikipedia link doesn’t work. Or rather, there is no content. This one looks rather more promising:
Joel Shore:
Unfortunately you are wrong.
When the hole is opened, the total entropy of the system increases which takes energy out of the system (because entropy is a form of energy in the form of disorder) even if it is isolated from the rest of the Universe.
If you look up “Maxwell’s Demon” and its resolution you will see why cold does not flow to hot even in a statistical sense.
In Willis’ case he’s arguing that because everything radiates (everything above absolute zero) then the cooler shell radiates back to the planet’s surface. No it doesn’t.
If you took a bowl of hot soup and surrounded it on all sides with ice, then the radiative flux from the ice does not heat the soup at all. Instead the ice melts, taking energy away from the soup and converting that energy into a change of state from ice to water, and increasing entropy of the total system.
People here think its all about radiative fluxes and bouncing backwards and forwards when the reality is that the laws of thermodynamics are violated.
Here’s the link to the story of how Maxwell’s demon works in statistical thermodynamics: http://www.auburn.edu/~smith01/notes/maxdem.htm
Mark T,
Yes i keep mentioning this hole in a sphere thing. Well its not me really. That the example given in text books as an approximation to a blackbody. A sphere with a hole in it is ‘close’ to an ideal blackbody. They say (the texts, Stephan-Boltzmann, etc) that a spherical cavity with a hole in it will achieve temperatures of a blackbody.
So that’s why i keep going back to it. Blackbody temperature for a given external flux is the maximum temperature.
I don’t think it matters what kind of hole. You need a one-way device to achieve a higher temperature. Such a device would be equivalent to Maxwell’s demon and appears (to my demonstratively weak mind) to violate the second law of thermodynamics since it would allow heat from a cold side to move to a hotter place.
Basically many people are saying we ‘can’ create a special hole that allows us to have a temperature hotter than a black body in the same environment. I’m doubtful, but listening, but I admit that for this particular argument we are probably at an end, without an actual experiment to show it quite simply one way or the other.
P Wilson (19:47:09) :
Clearly you misunderstand the thought experiment. The planet emits 235 w/m2 at equilibrium. In other words, at that point it is emitting as much as it absorbs.
Now, suppose we wrap it in a perfectly insulating blanket. The radioactive elements are still producing heat, but it has nowhere to escape. As a result, the planet’s temperature starts to rise.
So while constants are constants, the temperature of the planet in the thought experiment is not constant. That’s why e.g. ovens have insulation. The energy coming from the gas flame is constant … but the oven without insulation is not the same temperature as the the temperature of the same oven with insulation.
I think of conductive heat transfer as being the rather direct transfer of the motion of one atom or molecule to another. Would it be wrong to think that the radiative mode of heat transfer is also going on within a solid or liquid or gas?
Willis- with this reverse greenhouse (heat from within) will there still be changes in the atmosphere ie cool and warm phases? Will the steel green house provide a constant temp worldwide? Will there be polar ice caps?
Willis,
You have the patience of a Saint.
carlbrannen (19:57:41) :
How is a planet with a steel shell around it a “complicated analogy”?
I am sorry you felt it was mean spirited. It was not meant as such. I said it because you claimed that
I assumed you said that because you didn’t understand that for blackbodies we can measure temperature equally well either in W/m2 or in Kelvins. My bad.
Regarding emissivity, in my thought experiment I specified both the planet and the shell as being black bodies. How is this “complicated”? I really don’t know what to say.
As I mentioned before, this cannot be done in units of temperature, because unlike W/m2, Kelvins are not conserved.
Yours is a very poor way to explain the greenhouse effect. Why? Because the real greenhouse effect in the atmosphere does not work like that. I am trying to explain how a greenhouse
, not give an inaccurate analogy to insulation. I am explaining it by positing a thought experiment of a steel atmospheric greenhouse, which works by exactly the same principles of the real atmospheric greenhouse. It is not the same as insulation, as you claim, which is why a Thermos bottle works better than an insulated bottle.
Willis Eschenbach (20:31:54) Re:the thought experiment
Reminds me of the igloo and the Eskimoes (?spelling) being naked inside with a small whale oil lamp for heating in the middle of subzero arctic temperatures.
A real experiment.
Hank Henry (20:42:03) :
Radiation goes on within a gas, but not, as far as I know, within a solid
par5 (21:22:15) :
Since my thought experiment has no atmosphere, none of them are possible.
John A: You are completely and utterly wrong. You would get flunked out of any undergraduate engineering thermodynamics or heat transfer course. If I were your instructor in such a course (and I have taught courses with this subject matter), and you kept maintaining these views, not only would I flunk you, but I would be having a conversation with the dean about how the hell you got into the program in the first place.
Bodies radiate based on their emissivity and their temperature alone, without regard to what they are radiating towards. They have no way of “knowing” what they are radiating towards. This is easily demonstrable in a lab. Put two plates of different temperatures parallel to each other, and you can still measure the radiative energy from the colder plate with a radiometer.
A radiometer that is warmer than the device radiating towards it can still measure the radiation from that object — according to your arguments, that would be impossible.
In any introductory engineering heat transfer text, you will see that the net radiative heat transfer between two objects (1 and 2) is given as:
K * (T1^4 – T2^4)
where K is a system constant incorporating the emissivities and geometric properties of the two object. This expression automatically incorporates the fact that the cooler body is radiating towards the warmer body, albeit less than the warmer body is radiating towards the cooler body.
By your argument, this equation is completely wrong. Your claim is that the temperature of the cooler body is irrelevant, and that the net heat transfer from the warmer body is the same whether the cooler body is 0.001C colder or at absolute zero.
If you really believe your argument, you should not be spending your time on blog wars like these. You should be doing whatever it takes to shut down any system that was designed using the expression I gave above (which is ever power plant, every power electronic system, among many others), because according to you, they were designed under completely fallacious assumptions, and therefore could be totally unsafe.
DeWitt Payne (21:24:17) :
I believe that the only stupid questions are the ones that you don’t ask. And I think that if I put up a controversial post, it is my responsibility to at least try to answer the questions and objections.
However, I have been known to lose it …
anna v (22:20:27)
Snow is a very efficient insulator, and is what keeps the ground from getting so cold in the winter that many plants and organisms wintering over underground would otherwise be killed. It has nothing to do with radiation. The body heat and whatever other heat sources they have warm the air which is trapped, much like in a greenhouse.
@jt:
I had that same idea of non-flying photons some years ago. The consequences wouldn’t be obvious here on earth because there are always plenty of ‘receivers’ around. But thinking about a test of this hypothesis leads to very nasty unsymmetries on the space time relationship of the two objects regarding the direction of a transfer. Ignoring those I came up with the following: When, say in the ISS, the sky is scanned with a laser (or other light source) there would be measurable differences in the energy consumption depending on the nuber of receivers available in the respective direction. Showing that amount of consumed energy on a screen would show the number of receivers in the scanned area. Taking as target our sun what should we expect to see at time ‘now (here)’ at the place where the sun was 8 min’s before now, it is now, will be in 8 min’s? When the transfer takes no time from the viewpoint of light the ‘negotiation’ will. And this is the reason why we will not have a look into the future and seeing now if our sun exploded within the next 16 minutes.
I am layman when it comes to physics (I am a computer programmer by profession), so I am not going to comment on the physics itself. It is quite interesting, though, to see how the concept of thermodynamics gets tossed around and how the commenters accuse each other of not understanding the first thing about it. It is almost like hearing programmers debate the pros and cons of Java.
However, I am a little at loss trying to figure out the purpose of this entire thought experiment. As far as I can understand it simply says that anything that blocks the outgoing radiation will make the planet warmer. A layer of steel will do it. Probably a layer of cotton as well, or a layer of old newspapers. Or – as in the case of the Earth – an atmosphere. And – all things equal – as long as nothing changes, everything will remain the same, once the system has reached thermal equilibrium.
But the interesting thing, really, is what happens when things do change. What happens when the system is NOT in thermal equilibrium, as is the case with our planet right now? According to Hansen et al 2004, the Earth is receiving 0.85 +- .15 W/m2 more than it is emitting into space. How fast will the surface temperature of the planet rise or fall as a response to shifting properties of the “shell” (for example, when the mix of greenhouse gases in the atmosphere is changed)?
Interesting model.
Is there a way to re-design this shell system using an external radiation flow ?
The analogy may be accurate, but it does not feel so at first glance, and the number of comments confirm this.
Well all I can say is that if the comments here are a reflection of the confusion currently existing amongst professional scientists then there is no surprise about the media panics arising from virtually all new observations of natural real world changes, not just climate shifts.
I see that many contributors here are far more scientifically qualified and experienced than I am yet seem unable to see the wood for the trees where multiple factors need to be considered simultaneously and their respective significances unravelled and weighted appropriately.
Willis has given us a limited but useful scenario that brings to the fore certain aspects of AGW theory that are quite frankly untenable.
In doing so he has had to ignore certain aspects which his simplified model cannot deal with but they are matters it does not need to deal with to make the point.
Hell, that’s the whole reason for thought experiments and simplified modelling. They are not designed for complete and total accuracy. They are designed to emphasise and more clearly illustrate limited but important issues.
In fact all those ‘wonderful’ climate models that excite us so much are no more accurate in their description of reality than is Willis’s model but at least he is using his for a specific and limited illustrative purpose and not pretending that his model is adequate to justify political policy decisions which have the potential to redraw human societies worldwide and forever.
I hate it when people give me the anal ogies ( or any other sort of ogies!). I also hate it when people choose to move indescriminately between the statistical world of quantum theory and the classical model. As far as I know, no-one has yet got(ten) the Nobel Physics Prize for a unified theory.
I too admire the patience of Willis and Joel. This is a simple radiation example which does illustrate the greenhouse effect well. Textbook stuff. It is described in, for example, this set of Oxford University undergrad lecture notes (top page 9).