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?
To all those who think that the Figure 4 is wrong:
If we say we only absorb long-wave radiation in this example, then we get only 240 W/m^2 to the earth. In the first time step, we emitt 240 W/m^2 to the shielding shell (because Heat Capacity is 0). This shell then emits 120 W/m^2 into space and the same back to the planet. In the next step. The planet emits 360 W/m^2 and half of that gets remitted to the planet: 180 W/m2. In the next time step we emit 420 W/m^2 to the shell, the shell divides it and reemitts 210 W/m^2 back to earth. Then we get 450 W/m^2 that is emitted by the planet and thus 225 W/m2 back to earth. This continues until we reach the equilibrium.
E(t) = Solar rate + S'(t)
S'(t) = E(t) /2 (half of it into space)
S(0) = 0
Solar rate = 240 W/m^2
Shell can’t emitt in the initial condition, because it has no energy.
You can easily reiterate this in Excel and you will see that E(t) converges to 480 W/m^2.
Oxbridge Prat says:
November 28, 2010 at 2:27 am (Edit)
Willis, see
Of course. Have two.
Max says:
November 28, 2010 at 2:28 am
Max, as I said earlier, there cannot be an equilibrium, defined as per Willis’s Figure 4 where the surface receives 480 w/m2, until infinity:
cohenite says:
November 28, 2010 at 12:27 am
Hi Willis; these types of thought experiments are interesting and serve a purpose. If I may point out an aspect to do with your figure 4; in the first instance only 240 w/m2 reaches the surface in the form of SW so only 240 w/m2 is reemitted to the disc in the form of LW; of that 240 w/m2 isotropic effect causes 120 w/m2 to be reemitted from the disc back to the surface and 120 w/m2 outwards to space; so in the next instance the surface has 240 w/m2 SW + 120 w/m2 LW for a total of 360 w/m2 and so on; in short this is, of course, a limiting sum geometric series:
Sn=a/1-r where a=1 and r=0.5; or 1/n^2 for n=1 to infinity is 2.
In otherwords the surface will only receive 480 w/m2 at infinity.
I assume that the amount of heat humans create is known to some degree and accounted for in climate modelling (unless it is considered trivial)?
cohenite says:
November 28, 2010 at 12:27 am
Yes. However, remember that radiation happens at the speed of light, so for all practical purposes the system will equilibrate in a finite time.
This reminds me of a story my high school math teacher used to tell about the meaning of “for all practical purposes”. You may be familiar with “Zeno’s paradox”. This paradox states that if we move half the distance to our goal, and then half the remaining distance to the goal, and half again and so on, we will never reach our goal.
Mr. Hegji, my math teacher, gave us a thought experiment. He said “Suppose we lined up all of the boys along one wall of the classroom, and all of the girls along the other. Then suppose each time I clap my hands, both groups move forward until they are half as far apart as before.”
“Now you know”, he said, “that according to Zeno the two groups will never reach each other. However, before I have clapped my hands too many times, the boys and girls will be close enough for all practical purposes …”
So yes, you are correct. We will only reach 480 at infinity. But starting at 240, and halving the distance to 480 each time as in your math, we get to within a tenth of a percent(0.1%) of 480 in just nine steps, and within a hundredth of a percent (0.01%) in 13 steps … close enough for all practical purposes.
TimM says:
November 28, 2010 at 3:24 am
Globally the heat humans create (mostly through burning fuels of all types) is not large. From memory it’s in the hundredths of a watt per square metre.
However, in some densely populated areas like say the Netherlands, again from memory, it can be a couple of Watts per square metre. It is also makes more of a difference the further north you go, and in wintertime.
Whether this is accounted for in the climate models I don’t know. My guess is no for most models.
Max,
Then how do you answer Baa Humbug who says:
November 27, 2010 at 2:56 pm Take a basic strip heater of say 1000W, the above fallacy suggests all we have to do is add a well conducting bar across the front of the element, say an inch away (unwired and not receiving any power, i.e. benign) and voila’ we now have a strip heater putting out twice the energy. Can someone help me with patents please?
……………………………
I try to make the same point, that the shell has to have some rather special properties to do anything. Also, there is a big mix-up (as usual) between static analysis and dynamic analysis. This shows with what cohenite says:
November 28, 2010 at 12:27 am – In otherwords the surface will only receive 480 w/m2 at infinity.
It’s actually worse than that because less than 50 % of isotropic emission will return to Earth from the shell because of geometry.
Similarly where Michael D Smith says:
November 27, 2010 at 10:29 pm – Thickness will only affect the response time for a change in input. Near 0 = fast response, thick = slow response. As small at 1 atom thickness would still be in equilibrium.
That’s dynamic analysis, not static. You have to wonder how a thickness of 1 atom interacts with a wavelenth of IR light of say 10 microns. It would take a rather long time to show. Logically, a tickness of 1 atom would imply a separation between atoms, so the sphere is not merely transparent to IR, it has holes in it. Ad absurdum, if the mechanism works with holes in the sphere , it will work with one atom. (Don’t think so).
The issue is surely not the existence of a “greenhouse effect”, but its magnitude. Errors abound in calculations for the “non-greenhouse” earth, for example the usual assumption that the surface acts as a perfect “black body”. If it did, radiation scientists wouldn’t need to struggle to devise perfect”black body” reference sources for measurements, they’d just need a handful of dirt or a bucket of sea-water.
The relationship between temperature and radiation is non-linear. Averaging a wide range of one element to calculate the other introduces an error – the wider the range, the larger the error. Kiehl & Trenberth show that for small increments in either radiation or temperature, the relationship can be considered to be linear. However, they do this AFTER averaging wide ranges in radiation to calculate average temperature and vice-versa, and assuming that surface and atmosphere behave as perfect “black bodies”. Also water-vapour is NOT uniformly distributed over the earth’s surface, therefore neither are clouds, and the heat transfer involved in evaporation and condensation is very large, and not properly included in the internal atmospheric “budget” These factors mean that the averaging process used invalidates the “energy-balance” diagrams by as much as several degrees K. For example, averaging the radiation deriving from temperatures of 0°C and 40°C results in a figure which gives 22.1°C rather than 20°C. That’s not insignificant when temperatures are calculated to a tenth of a degree.
Willis Eschenbach says:
November 27, 2010 at 7:50 pm
“Why The Thought Experiment Can’t Represent Earth
It is not powerful enough.
What I have illustrated above is the energy balance in a theoretically perfect single shell planetary “greenhouse”. It shows that the maximum amount that such a system can produce is a doubling of the input.”
This is silly.
1) Plenty of people upthread have pointed out that there isn’t just a single shell; you poo-pooed them for adding complexity.
2) Doubling the input is ample for a simple model of Earth’s climate; you have only made it apparently inadequate by adding additional shells or absorption mechanisms that cut down the energy reaching the surface in places beyond what you assumed, which is an unnecessary complication.
3) Even a single shell can in principle produce any degree of greenhouse warming required. Off hand, I can think of at least three mechanisms:
(i) if the ratio of the thermal emissivities of the lower and upper surfaces is E then the temperature of the shell is enhanced by a factor E^1/4, and the flux re-radiated to the planet is increased by the same factor of E. One could easily engineer a factor of up to ~100 (using, eg, aluminium film), which would push the Earth’s temperature to almost 1000K! In the real atmosphere, with its thick shell, those emissivities do not differ by anything like so much, but they are not identical.
(ii) if the shell is reflective (as you have implicitly assumed by setting the albedo greater than zero) then the amount of energy radiated up and down from the shell can be much less than the amount of thermal energy simply reflected or scattered back down. Again, an aluminium film could boost the heat reaching the surface a hundredfold!
(iii) if the shell is not perfectly conductive, there is a temperature difference between top and bottom, and there is no fundamental limit on how large that temperature difference can be. In the real atmosphere, there is indeed such a temperature gradient.
You could argue, I suppose, that (i) and (iii), in using a shell with different properties for the top and bottom surface, are really two shells; but that pedantic objection would not apply to (ii).
Note that in principle one is not even limited to the 6000K temperature of the Sun; if one were to allow through only the blue or “hot” end of the spectrum, the planetary temperature could theoretically exceed the source temperature. This does not contravene the second law of thermodynamics, because there is still a net entropy gain (more photons leave the planet+shell than reach it).
Willis Eschenbach says:
November 28, 2010 at 2:26 am
in my Excel model referred to above. I still find a net warming effect. Here is a more complete view of a two-shell system, with the other factors included. The two shells are shown as horizontal gray bands:
Willis, that’s an interesting reworking of the Kiehl-Trenberth energy budget sketch. Have you been able to produce an estimated sensitivity using it?
The so called “greenhouse-effect” does not exist.
Radiation of heat through a body only depends on the temperature of that body and its coefficient for its emissivity. Back radiation cannot influence this. It only could transfer heat back to that body.
But transfering heat energy from “cold” to “warm” violates the second law of thermodynamics.
There is no way around that.
Well, the sensitivity cannot be estimated from that budget. This is because of the great unknown, the power of the feedbacks and the size of the losses.
For example, you can say that 390 W/m2 is emitted by the surface as radiation, and 100 W/m2 is lost from the surface through convection and evapotranspiration. So somewhere around a fifth of the energy is not going into warming the surface.
But that is in the aggregate. It does not tell us what the net effect of a change in the forcing might be. The Clausius-Clapeyron relationship, along with the linear dependence of evaporation on wind speed, would indicate that for an additional forcing above the current temperature, much more than a fifth of the incoming energy would go into evaporation and convection, and correspondingly less would go into heating the surface. This fits with our experience — we know that as heat engines run hotter, parasitic losses increase rapidly.
So while we can’t calculate the sensitivity, it certainly appears just from an examination of the losses that there is not much chance that the net feedback is positive. The losses are too large for the net feedback (which of course includes the parasitic losses) to be a net gain.
Geoff Sherrington says:
November 28, 2010 at 3:41 am
No. Take a basic strip heater. Put it in a vacuum. Measure the temperature. Then surround it with a metal shell. Measure the temperature. The temperature will go up, because of the so-called “greenhouse effect”. The shell radiates some of its heat back to the strip heater, making it warmer.
See my answer to this above.
You are correct. Again, this is being ignored as a simplifying assumption because the effect is small. With the radius of the earth being over 6,000 km, when you are a km or two above the earth the difference between a flat plane and a sphere is not large.
That I don’t know about, I have no idea if it would work if the shell were one atom thick … not sure why or if that is relevant.
Will the mechanism work with holes in the sphere? Yes, they just introduce losses where some amount of the radiation goes out to space. This makes the surface less warm, but the mechanism is still working.
w.
MostlyHarmless says:
November 28, 2010 at 3:44 am
AAARGHH! This is a thought experiment. I used a simplifying assumption, that the earth was a black body for IR. This introduces an error in the conversion of radiation to temperature on the order of about 5%, acceptable for a first-order understanding of what is happening. I said if you want to include it, do so …
However, it makes no difference to the radiation calculations themselves. If 240 watts/m2 is absorbed by a planet, 240 watts/m2 will need to be radiated if the planet is to stay at equilibrium, regardless of whether the planet is a black body or not.
The thermodynamics of this model neglects entropy.
The Earth is not at equilibrium and it while the energy balance may be that the net energy flux integrated over a surface surrounding the Earth may be zero, the system is not at equilibrium, in which the entropy production is zero, but is in an approximate steady state. There are gradients of pressure, temperature, velocities of air and water, i.e.: decreased entropy, which are driven by an energy flux through the system. These would decay if there was an opaque shell surrounding the earth. A simple radiative transfer model striles me as somewhat simplistic.
Here’s the problem that I see. In any of your “fully opaque” scenarios, the 240 units of energy from the sun should be REDUCED because the energy is the sum total of all wavelengths. So if none of the longwave energy can pass from the sun to the earth, then it’s not 240 that hits the earth, it’s 170 (or whatever… just an example.)
Willis — I don’t follow that. If anything, it seems like making more CO2 should decrease the amount of O2 in the atmosphere.
Sorry I was not more clear. Your shell is the atmosphere. The atmosphere is always replenished in that there’s leakage into space. The atmosphere is (re)generated by life, so it’s not a steady state where increase of A necessarily means B decreases.
Specifically I was thinking of photosynthesis. If the number of plants is a steady state, then more CO2 = decrease in O2. On the other hand more CO2 = more plant growth (more plants?), hence more O2 released. More O2 = slightly more atmosphere. All things being equal there ought to be more plant activity with more CO2 meaning there ought to be more O2 emitted as well. In real life the volume of the atmosphere, which is a living system of sorts, ought to change slightly as parts are lost and subsequently replenished, so we know it’s being created. But enough to be measureable? I don’t know.
(Mental model: CO2 concentration goes up, plant volume up. More O2 created. Atmospheric thickness goes up. )
I also don’t know how the ppm # is created. If machinery counts up a million molecules and derives ppm, then OK, I’m dead wrong. OTOH if the machinery is calibrated against a steady state and ppm = derivation from this state, then what’s being measured is an absolute CO2 molecule count vs pctg of atmosphere, hence this count may not mean what it ought to mean.
Is it far off enough to make a difference? No idea. I’m simply wondering if we’re measuring what we think we’re measuring.
No I’m not questioning the greenhouse effect or offering alt theory here. Mostly it’s a nagging question regarding the reality of the hard numbers, and I’d thought of your list of known losses etc in your thought experiment.
Thanks for listening.
Willis
“For example, you can say that 390 W/m2 is emitted by the surface as radiation, and 100 W/m2 is lost from the surface through convection and evapotranspiration. So somewhere around a fifth of the energy is not going into warming the surface.”
“there is not much chance that the net feedback is positive”
So there must be 390+100 watts absorbed by the surface. There is only about 164 watts solar absorbed + lets say 36 watts solar re-emitted from the atmosphere, i.e there is an 490/(164+36) amplification. Why wouldn’t an additional 4 W/m2 (~1% change) see about the same 2.45 amplification?
The greenhouse effect was shown to be bogus by R W Wood.
He showed that;
1. A glasshouse only worked by stopping convection.
2. The remaining real residual radiative effect was so small that it could be almost ignored.
See page 17
http://www.friendsofginandtonic.org/assets/hutton%20-%20climate%20change.pdf#page=17
Here is a paper that more people should read.
Especially as it comes from a source with no “spin” on the AGW debate.
The way I read the paper is it gives massive support for the conclusions of the famous Woods experiment.
Basically the project was to find if it made any sense to add Infra Red absorbers to polyethylene plastic for use in agricultural plastic greenhouses.
Polyethylene is IR transparent like the Rocksalt used in Woods Experiment.
The addition of IR absorbers to the plastic made it equivalent to “glass”
The results of the study show that( Page2 )
…”IR blocking films may occasionally raise night temperatures” (by less than 1.5C) “the trend does not seem to be consistent over time”
http://www.hort.cornell.edu/hightunnel/about/research/general/penn_state_plastic_study.pdf
OMG, I really got desperate while reading some of the comments. Perhaps 30%. Is the general commenter here THAT clueless?
The next time I see a post which explains something very basic and/or proposes readers to solve a problem, I promise not to read any of the comments. It puts my morale down. Twice was I tempted to answer with expletives… It is incredible the authority with which absolutely clueless people can talk about things that they cannot understand.
Willis, really, just give up. Whoever cannot understand the GHE, after the several posts that have appeared here in the last months about the subject, will never do. It’s not worth the effort.
Following on from my post at 1:32 am.
Received average W/m^2 for half a cycle = 240 = Wr
Transmitted average W/m^2 for a full cycle = Wt
For 1 cycle: Wr+2Wt = 0 (steady state). Therefore Wt = Wr/2 = 120
This is true with and without a ‘greenhouse’ effect. So what lifts average temps from -18 deg? The answer is in the semi-transparent nature of the atmosphere to solar radiation, particularly IR, which creates a lag or hysteresis. With hysteresis, the work done on the system is the area under the hysteresis graph. In this instance, the work done is the raising of average temperature from -18 deg, an increase in potential energy. To understand lag, imagine yourself as a photon travelling into the atmosphere along-side a group of others: as it is semi-transparent, some, including yourself, pass through but on the way out some get out but you don’t. Your eventual departure has been delayed: you are lagging behind the others.
How you would show that in a diagram, I don’t know but that diagram will have to show 1/2 cycle input and time lag.
——-
Willis E.,
Your answer is not the most fundamental answer to your own question.
So, let’s play the game with you instead of your game with us!
MY COUNTER QUESTION: “What is the most fundamental reason why the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?”
John
People,
lots of interesting stufff here, but we are all overcomplicating and over critising this simple thought experiment to explain the fundamental mechanics of the green house effect. Not the earths atmosphere, BUT the greenhouse effect.
My GUESS is that the thin shell is described only in its characteristics of being transparent or opaque (as to whether or not it passes or blocks shortwave or long wave radiation.) It does not take into account its mass (in this example not stated so taken as zero) and as such does not account for the heat that it itself would store, how long it would heat up and also cool down….
depending on the temparature differential from one side to the other of the shell.
@Geoff:
Well, Willis posted a good reply for the heater system up there, so I won’t repeat it here a second time.
But when it comes to static between dynamic systems, we have a steady-state system as long as all boundary constraints stay the same, as Willis showed in his figures. He only displayed the steady-state, because (and I only give my opinion here) of time constraints and simplification.
What I added was merely to show how you arrive from the initial condition of only solar radiation influx to the steady-state over a certain amount of time (read: iterations). As you can see the pure mathematical model NEVER actually arrives at 480 W/m^2, which is the convergation limitation of the dynamic model AND not a fault in the thinking. From there on, the model is in a steady-state until the influx from the solar radiation changes.
There is no violation of any kind of Energy law. The Earth in this model is only an intermitting energy source, BUT not the ultimate energy source, that is still the sun. In reality, the Earth heat capacity would further delay the arrival of a steady-state (along with several other factors which would help avoid any steady-state for a longer period). If you want to check the energy balance, you have to cut the whole dynamic system free at the shell. Then you have as much energy going away from the shell as is coming in (480 comes in, 240 goes into space and 240 is reemitted back to earth). The same you can do for the Earth side of the equation: 240 + 240 is coming in (from the sun and from the shell) and 480 is emitted to the shell. There is no “new” energy or energy creation on earth, it is just blind power, which you can also find in hybrid drive trains (as any engineer can tell you).
In those drive trains you have to ways to transmitt power and thus torque from the engine to the wheels. First you might have a mechanic transmission and a hydraulic transmission (for the sake of simplicity). The power is divided at a planetary gear just behind the combustion engine and then part of it is transmitted via the hydraulic transmission and the rest through the mechanic transmission. It is then added to a single shaft by three gears. We there have an addition and/or subtraction of torque and it might be that we actually have power that is only “circulating” between the two transmission and IS NOT leaving it due to their relative rotation speeds. I could go more into detail, but I don’t have the time to design some accompanying drawings that would help illustrating the point.
Also, similar effects can be observed in electrical systems (alternate current): http://en.wikipedia.org/wiki/Reactive_power#Real.2C_reactive.2C_and_apparent_power
It is sometimes a difficult to grasp concept, but nonetheless valid and proven.
quote:
Max,
Then how do you answer Baa Humbug who says:
November 27, 2010 at 2:56 pm Take a basic strip heater of say 1000W, the above fallacy suggests all we have to do is add a well conducting bar across the front of the element, say an inch away (unwired and not receiving any power, i.e. benign) and voila’ we now have a strip heater putting out twice the energy. Can someone help me with patents please?
……………………………
I try to make the same point, that the shell has to have some rather special properties to do anything. Also, there is a big mix-up (as usual) between static analysis and dynamic analysis. This shows with what cohenite says:
Greenhouses work by containing convection and not by preventing the re-transmission of radiant energy.
Whatever the “effect” on the Earth’s climate it has little to do with what happens in greenhouses.
It should more properly be called ” the insulation effect” as the atmosphere limits the transmission of heat energy by its attenuation within a material.