The Steel Greenhouse

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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savethesharks
November 17, 2009 8:03 pm

Thank you for this very enlightening essay, Willis.
The fact that the swirling, eddying, jet-streaming, convectively-bubbled water-vapor-filled atmosphere of Earth, that behaves not unlike it were liquid….makes it almost impossible for those greenhouse-alarmists to pin down….because it ain’t steel, it ain’t glass, hell, it ain’t even a blanket.
Much more benign…and life-giving…than that.
Chris
Norfolk, VA, USA

Stephen Goldstein
November 17, 2009 8:05 pm

Great post and great post-post discussion 😉
I am trying to get used to using W/M^2 as equivalent to temperature — I agree it simplifies the explanation and, at the end of the day, any given level of equilibrium energy flux corresponds to an equilibrium temperature but still . . . .
I also want to add my 2 cents to two comments . . . .
morganovich (16:10:54) –
A compressible fluid under pressure is not warm because it is under pressure but because it was compressed, the increased temperature representing the work (energy) done on it by the compression process. Once compressed, the warm fluid will either cool to ambient temperature (consider a SCUBA diver’s air tank or a CO2 fire extinguisher) or, remain warm if there is an external source of energy (the Venusian atmosphere heated by the Sun).
jt (18:15:34)
I definitely do not understand your explanation “that an electron cannot emit a photon unless it can “find” another electron which is in a suitable configuration to receive said electron.”
It is night here and the sky is clear. If I understand you correctly, the reason that I can see Sirius, the Dog Star, is that 8.6 years ago, an electron on the Sirius photosphere “found” an electron in my retina and emitted a photon. Indeed, many electrons emitted many photons in quite a range of energies ’cause I see a very bright blue-white star.
Yes, I know you were thinking of long wave radiation on a terrestrial scale but why would wave length and distance change the fundamentals?

November 17, 2009 8:11 pm

I’m sorry but I thought the article was tremendous nonsense, and will cause immense damage to the reputation of this website.
If there are physicists watching this weblog then I’m sure they’ll fill you in on exactly why your “steel greenhouse” won’t work in the way Willis has described, and why adding “shells” to the planet will not raise the surface temperature one iota.
In a sense, what Willis has done is manage to make the same mistake as the people who think that the Earth is in “radiative balance” and try to work out a tedious budget of everything going in and out.
It’s wrong from first principles.

Editor
November 17, 2009 8:14 pm

Merrick (19:00:01) :
“Mike Lorrey – I believe my comments regarding temperature and surface radiance are correct with respect to your statement. W/m^2 up and @/m^2 down are the same so (slightly) more W up than down in a system with a shell of real thickness.”
Re: inside vs outside of the shell: your analysis is only true if the shell material is thermally saturated. If, instead, the materially is not thermally saturated there will be waves of energy passing through the material. That the atmosphere warms during the day and cools at night, cools under cloud and warms in broad daylight, cools in broad night time and warms under night clouds, says that the picture is far too complex to treat as a solid shell. That different levels of atmosphere travel at different speeds, are different temperatures, and have different gaseous concentrations tells me you need to treat it as a Matrioshka doll where each shell is its own lava lamp.
Re: inside of the shell vs Earths emissions: my point still stands, and its even more complex, see above. Terrain at different altitudes changes the number of atmosphere shells one is dealing with, and don’t get me started on ocean.
Now we can start getting into more real climate models, but ghu, you need a grid cell resolution of a few kilometers, not 250 or 100 km, and you need to analyse each cell as its own navier-stokes thermodynamic/fluid dynamic system, not just a basic heat in vs heat out childs toy.

Willis Eschenbach
November 17, 2009 8:15 pm

Tom in Florida (15:42:34) :

You stipulate that the Earth acts like a perfect blackbody. Didn’t I read here on this blog that it doesn’t?

You are correct, this is a simplified model. However, including the emissivity changes the temperatures but not the radiation.

Willis Eschenbach
November 17, 2009 8:22 pm

Monckton of Brenchley (15:30:32) :

… but if, as K&T and you both agree, the surface radiation is ~390 W/m2, equivalent to 288 K (today’s mean global surface temperature) by the SB equation assuming emissivity = 1 for longwave radiation, then the first differential – i.e. temperature change per unit change in radiative flux – is just 0.185 K/W/m2, while the UN gives 0.313 K/W/m2, and also assumes temperature feedbacks on top, giving a final climate sensitivity parameter close to 0.9 K/W/m2, around five times too big. Discuss.

Always a pleasure to hear from you. You are correct, the direct SB equation gives a sensitivity of 0.185 K/W/m2. However, this is for a 1 W/m2 change at the surface.
The UN IPCC, on the other hand, figures it for a 1 W/m2 change at the top of the atmosphere. This gives a larger number.
Next, they assume a large (and unverified) water vapor/cloud/whatever feedback on top of that, to give the 0.9 number. I think the UN IPCC feedbacks are far too large, my sense is that net feedback is negative or the temperature swings would be much larger. In addition, I think that there is a governor system which keeps the temperature within a narrow range. See my post here.
Many thanks, keep up the good work,
w.

Willis Eschenbach
November 17, 2009 8:25 pm

Bill Illis (16:20:51) :

Thanks very much Willis,
In your model, the orange cell (long-wave absorption by GHGs) is set to 95%, what levels do we set it at for changing GHG levels.

Good question, don’t think there’s an answer. IIRC, if you set it to 96.3% or so, it increases the downwelling radiation at the top of atmosphere (TOA) by about 3.7 W/m2. This is what is projected from a CO2 doubling … but this is a tinkertoy model with lots of assumptions.

Willis Eschenbach
November 17, 2009 8:32 pm

Bird Stewart Lightfoot (16:46:20) :

This article should be removed. The author obviously knows nothing about radiative heat transfer.
All of the numbers based on constant heat flux are meaningless because the surface area of the emitting surface is smaller than the surface area of the receiving shell. The actual radiative heat transfer is, ideally, proportional to the fourth power of the temperature, and the analysis must be based on this.
While total energy is conserved, energy flux is not.

The main radiation from the troposphere occurs at somewhere around 3000 -6000 ft. Let’s call it a kilometer and a half. The radius of the earth is about 6378 km, and the area is 511,185,933 sq. km. The area of the radiating surface at an elevation of 1.5 km is 511,666,935 sq. km. As this is an error of 0.1%, it is typically ignored in this type of analysis.
Energy flux in W/m2 is indeed conserved. If a planet is absorbing 235 W/m2, it must emit 235 W/m2 to stay in equilibrium.

Willis Eschenbach
November 17, 2009 8:34 pm

michel (16:49:44) :

“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.”
OK, I’ll bite. How much would 10 shells raise the surface temperature by? Or 100 shells?

In a perfect greenhouse, one shell gives a surface temperature of two times the incoming radiation, two shells give us a surface temperature of three times the incoming radiation, three shells gives four times, and so on.

Willis Eschenbach
November 17, 2009 8:36 pm

Steve (17:12:20) :

You can’t say that the earth radiates 235 watts per square meter from the center and that this amount still ends up being 235 watts per square meter at the surface. As you go out in radius, the square meters of surface increases. For the total amount of energy radiated to be the same, the watts per square meter have to go down.
For example, if you measure the watts per square meter radiated by the sun at a distance of 1 AU from the sun’s surface, you aren’t going to get the same measurement of watts per square meter if you back away to 5 AU.
I haven’t done my geometry in a while, but I recall something about the surface area increase being proportional to the square of the increase in radius.

Sorry for the lack of clarity. I meant that the total energy in the earths core gives a surface radiation of 235 W/m2.

November 17, 2009 8:41 pm

Monckton of Brenchley (15:30:32) :
“temperature change per unit change in radiative flux – is just 0.185 K/W/m2”
Amongst the many other key points this raises – such as much lower climate sensitivity than proposed by the IPCC as Lord Monckton points out – also consider how small this # is in the over all radiation budget. Assuming we have warmed about 1deg K (generous certainly) over the last 130 years, that would correspond to a net flux change + 0.185 W/m2 over the last 100 years. Right?
Look at the numbers in figures 3 & 4 for various components of the radiation budget. I have a hard time believing each of those components are known well enough that there isn’t at least a 0.185 W/m2 error bar on them. I havent researched this myself, so if that is incorrect, please give me a reference that shows the accuracy of each of the components of the budget.
So, the key implication from the last paragraph is that very minor changes ANY of the components of the budget could be just as easily responsible for the net temp changes observed as CO2. Why or how have these changes been ruled out?
Also, and maybe more significantly – what about the sun??? Here’s a question for both Lord Monckton & Leif (if you are reading). Hasnt the TSI over the last 130 years trended up by 3 W/m2? (1369 to 1372)? I remember some past post where Leif said these might be outdated #’s & the TSI is more constant (but I cant remember the details), so I am not sure, but based on the above calcs, don’t we only need an increase of 0.185 W/m2 to explain the observed temp changes? Seems pretty plausible / pretty easy to achieve.
Looking forward to your comments on this.

Willis Eschenbach
November 17, 2009 8:41 pm

michel (19:35:56) :

I think its an amazing invention, the steel greenhouse. If I understand it properly.
Right now it is looking like I can produce any temperature I want, its just a matter of how many shells my greenhouse has. I am enormously excited about this, and am going to do a small experiment tomorrow. Unfortunately I only have five or six cans which will fit inside each other, but I will buy a few more tomorrow, and put them all together. Then I will put a battery powered torch in the inside one, and wait for the explosion.
It is a sad commentary on human nature that the first application that comes to mind for this brilliant invention is as usual a bomb. But there you go. It depends how many boxes I can buy. if I can find someone to sell me a few hundred cheap, well, you might even hear the explosion, accompanied by a mushroom cloud, where you live. Looking forward to it. I have always wanted to go out with a bang, and this should do it.

Bear in mind thermal losses. Unless you have a vacuum and the various shells don’t touch each other anywhere, you will get thermal losses. These losses will increase rapidly with temperature, and will keep any physical system from getting too hot.
However, you will be able to get to “pretty warm” without much trouble.

Willis Eschenbach
November 17, 2009 8:45 pm

AlexB (19:42:17) :

I know what your talking about Willis.
A very good model. If only some people would realise that working with GROSS values of radiation is acceptable despite what thier high school physics teacher might have told them. Also that the difference in areas is negligable.
There is no provision in your model however for the effect of thermal losses between the troposphere and stratosphere such as is caused by verticle wind shear.
Verticle wind shear correlates well with temperature so as temperature increases you can expect more thermal losses across the tropopause which will reduce the increase in radiation reflected back to the surface which should reduce the expected surface temperature increase.

I include both convection and evaporation losses in the model I cited at the end of the article. One thing I learned from the model is how many assumptions you need to make to get anything that will come close to modeling the earth. For example, how fast does the evaporation increase as the temperature increases? While Clausius-Clapeyron gives a theoretical answer, in the real world we have spray and wind and changing albedo to complicate things. The GCMs claim to model this, but I has my doubts …

Willis Eschenbach
November 17, 2009 8:50 pm

Gary Hladik (20:01:30) :

Monckton of Brenchley (15:30:32) : “…giving a final climate sensitivity parameter close to 0.9 K/W/m2, around five times too big. Discuss.”
This one really has me confused. Up to now I’ve only seen “climate sensitivity” defined as the temperature increase expected from a doubling of atmospheric CO2. I assumed all the temperature rise was attributed to “radiative forcing” directly and indirectly increased by CO2. I thought the magnitude of the radiative increase was in dispute, but the effect of a known change in radiation wasn’t. Now it is? WUWT?
BTW, according to the Stefan-Bolzmann Equation, a sensitivity of 0.9 K/W/m2 implies the average temperature of the earth is about 170 Kelvins, or -103 degrees Celsius.
I’m putting another blanket on the bed tonight. 🙂

The UN IPCC gives climat sensitivity of ~ 3K per doubling of CO2. Doubling of CO2, in turn, gives a UN IPCC value of 3.7 W/m2 of forcing.
Dividing the first by the second gives a climate sensitivity of 0.8K/W/m2. Same number, different incarnation. Monckton is just using a different way of expressing the same thing.
Finally, the sensitivity is not constant, it is only valid at a certain temperature range. This is because the relationship between temperature and radiation is far from linear.

Willis Eschenbach
November 17, 2009 8:52 pm

John A (20:11:22) :

I’m sorry but I thought the article was tremendous nonsense, and will cause immense damage to the reputation of this website.
If there are physicists watching this weblog then I’m sure they’ll fill you in on exactly why your “steel greenhouse” won’t work in the way Willis has described, and why adding “shells” to the planet will not raise the surface temperature one iota.
In a sense, what Willis has done is manage to make the same mistake as the people who think that the Earth is in “radiative balance” and try to work out a tedious budget of everything going in and out.
It’s wrong from first principles.

I’m happy to be proven wrong, that’s how science advances. I await the information.
In the meantime, can you explain why my steel greenhouse won’t work?

Leo G
November 17, 2009 8:56 pm

wow, just want to say thanx for letting an average dude get to play with a climate model. It may be simple, but it really does help in understanding some more of the “science”.
I think i froze everyone to the ninth layer of hell…….
:)~

Alan D. McIntire
November 17, 2009 9:13 pm

In reply to michael- for shells, substitute ‘atmospheres’.
0 atmospheres will give you a wattage flux of 1 ( no greenhouse
effect)
1/3 atmosphere will give you a wattage flux of 1+ 1/3 at the surface.
3 atmospheres will give you a wattage flux of 1+ 3 = 4 at the surface,
n atmospheres will give you a wattage flux of 1 + n at the surface. The temperature will be proportional to the 4th root of 1 + n.
Of course greenhouse gases do not act as black bodies,
nor even as “gray’ bodies. They absorb only certain ranges of
radiation. Again take a planet
with no greenhouse atmosphere. Add 1 “standard greenhouse
atmosphere” of a gas that absorbs radiation
from 1/10 of the total outgoing spectrum, and is transparent in the
rest. What will the warming effect be?
You can compute this using the “harmonic mean”.
For the general formula,break the atmosphere into ranges, p0, p1,
p2…pn. p0+ p1 + …pn had better add up to 1. Take say 3 gases, with
effective ranges of atmospheric absorption = p1, p2, p3.
Throw in p0, the value for no absorption.
p1 has an effective “standard greenhouse atmosphere” of a1, p2 has a
“standard greenhouse atmosphere of a2, etc.
The total greenhouse effect will be
1
_______________________________________
p0/1 + p1/(a1+1) + p2/(a2+1) + p3/(a3+1)
For a concrete example, let p0 =0.3 and of course a0=0, let p1 =
0.6 and a1 = 1 ( maybe that’s water vapor)
let p2 = 0.1 and a2 = 10 (maybe that’s CO2).
Then, total watt flux will be
1
_______________________________ = 1/0.609 = 1.642
0.3/1 + 0.6/(1+1) + 0.1/(10+1)
The ground temperature will be (1.642)^1/4 = 1.132 times top of
atmosphere flux, with a top of
atmosphere flux of 255 K, surface temperature will be 255* 1.132 =
288.66 K.
Let’s double the amount of CO2. Assuming everything else stays
constant, the new temperature flux will be
1
___________________ = 1/0.605 = 1.653
0.3 + 0.3 + 0.1/(20+1)
The temperature at Earth’s surface will increase to (1.653)^1/4 * 255
= 289.14 K. Now
increase the level of CO2 by another factor of 10. Again, assuming
everything else stays constant,
the new temperature flux will be 1.665.
255 * (1.665)^0.25 = 289.66 so increasing CO2 by a factor of 10
only increased temperature by 0.52 K.
There is a limit here of 1.666….. No matter how much CO2 is
increased, forcing in watts can’t go past that level.
Caveats: The temperature assumes the earth is a black body. In
actuality, it approximates a gray body
with an average emmissivity of about 0.95. That implies that if the
effective temperature is 288 K,
watts radiated per square meter of surface will not be 390.7 but
0.95*390.7 =371.165 watts
The model is assuming classical theory rather than quantum theory.
Once quantum theory is
taken into consideration, temperature magnification can go higher.
Spectrum Broadening
Some on this site have stated that H20 can only absorb photons at
specific wavelengths. At some level this
may be true, but in practice it’s not, otherwise the argument about
saturation would be correct, and nobody would give a damn one
way or the other about further changes in greenhouse gases.
Suppose CO2 can absorb energy only at exactly 14 nanometers. In
practice, CO2 molecules are moving around with
speeds of rougly 400 meters/second. Since this is small relative to
light speed, non-relativistic physics can be applied.
Since 400 meters/second is rougly 1/750,000 the speed of light,
molecules moving towards a photon will see
wavelengths 1/750,000 shorter as actually exactly 14 nanometers in
length, and can interact with them. Likewise,
molecules moving directly away will see the wavelengths as
1/750,000
longer, and can interact with them, so
in practice there are a range of molecular speeds and wavelengths
interacting with the photons, which is why
you get a band rather than a single set of lines.
One result of quantum mechanical theory is Heisenberg’s uncertainty
princible
dE* dt> 1/2 h
where E is energy in joules, t is time in seconds, and h is Planck’s
constant, 6.626* 10^(-34) joule seconds.
Energy can’t be measured closer than 1/2(6.626) = 3.313*10^(-34)
joule
seconds.
Molecules are constanlty absorbing photons, then radiating the
photons
out in fractions of a second.
Say the average time between collisions is 1/1,000,000 second. Then
the energy absorbed can
vary by a factor of 3.313* 10^(-26) joules because of the uncertainty
principle. The collision rate
is a function of temperature and pressure. Increase the pressure and
you increase the frequency of collisions. Increase the temperature
and you increase the frequency of
collisions. Because of that uncertainty principle, the
gas can absorb energy at the standard energy + or – that 3.313 * 10^
(-26) joules.
The combinded broadening from a normal speed distribution of
molecules
and the broadening due to the uncertainty
principle is addressed in the Voigt temperature profile.
http://en.wikipedia.org/wiki/Voigt_distribution
Of course there’s not a neat solution for the above equation, you’ve
got to solve it numerically using something like:
http://en.wikipedia.org/wiki/Numerical_integration
That’s essentially what modtran does.
http://geosci.uchicago.edu/~archer/cgimodels/radiation.html
I picked up this information on the Voigt profile from a post by
Michael Hammer from
Jennifer Marohasy’s blog.
Specifically this letter:
“Comment from: michael hammer May 15th, 2009 at 11:22 pm
SJT: sorry but again i have to disagree with you. Line absorption
profiles follow a Voigt profile. I have integrated the area under
this
profile at various concentrations. When i also included the effect of
the analysis described in my first post on Jennifer’s site I then got
an almost perfect agreement to a logarithmic response. Hence the
analysis I did (which converts the loss from 10^-N to 1/N) needs to
be
incuded to get the logarithmic response.”

jt
November 17, 2009 9:14 pm

Stephen Goldstein (20:05:22) why would wave length and distance change the fundamentals?
As I read it it has nothing to do with wave length or distance. The argument is that among electrons energy transfer takes place along the light cone between two electrons and involves both the retarded and advanced Schrodinger waves. The two electrons enter into a common resonance state and while in that condition are not separated either in distance or in proper time because the time dilation and length contraction along the light cone reduce the relative distance along the photon trajectory between the interacting electrons to 0 and the past/future time interval along the photon trajectory to 0 as well. As a result the quantum of energy exchanged is never actually in a state of “free flight” between the two interacting electrons. The proposition is that every photon that is emitted is certain to be absorbed at some location in the universe and that an electron cannot emit except when in resonance with another which can absorb. In other words, electrons do not emit photons “on spec”. I refer you to Mr. Carver’s book for the details: Collective Electrodynamics – Quantum Foundations of Electromagnetism pub. The MIT Press ISBN 0-262-13378-4, 0-262-63260-8 (pb).

carlbrannen
November 17, 2009 9:17 pm

This post is wrong for a rather large number of reasons. It doesn’t take into account the radius and W/m^2 correctly, but most importantly, W/m^2 is not, nor ever has been, equivalent to temperature. It’s just a bad idea.
A better analogy to the greenhouse effect is “insulation”. You can get that by steel spheres and air gaps, but it’s better to analyze it as insulation. With different thickness of insulation, a fixed amount of heat flow results in a temperature difference proportional to the thickness.

yonason
November 17, 2009 9:37 pm

Willis Eschenbach (20:36:37) :
“In a perfect greenhouse, one shell gives a surface temperature of two times the incoming radiation, two shells give us a surface temperature of three times the incoming radiation, three shells gives four times, and so on.”
So, how many shells will it take to get the core to 2,000,000 degrees?

par5
November 17, 2009 10:09 pm

I think this is a great thought experiment- look at how many are getting involved, bringing in their expertise. Thanks Willis. Yes, the earth is capable of receiving more radiation than it emits- and visa versa. The first cartoon about the radiative budget (K/T) has always bothered me for this reason. Science at work on Anthonys’ blog…

Anton Eagle
November 17, 2009 10:12 pm

Logic. Science and logic should be inseperable.
This article is utter nonsense, and a simple logical analysis reveals this without even resorting to physics (and I am a physicist).
Using the author’s argument… the mythical planet radiates 235 W/m2. The steel shell absorbs this and re-radiates it all in both directions. The planet obligingly absorbs the re-radiated 235 coming off the shell, and re-radiates it, for a total radiation “budget” of 470 W/m2. The author conveniently stops at this point.
But wait… there’s more. If we are to follow the author’s logic, then the shell should absorb this 470 W/m2 and re-radiate it again… 470 out and 470 in. The planet should then absorb this 470… add it to the orignal 235 coming from the core, and now be radiating 705 W/m2… this would continue till everything anihilated at a gazillion degrees. Utter nonsense.
As requested by others above… please remove this article.

Thomas
November 17, 2009 10:14 pm

Wait what? You add a shell, it obtains thermal equilibrium with the earth such that it radiates the same magnitude flux. Right: so you’ve got:
F(earth) = Flux(shell) (1)
but then you say:
Flux(shell in) = Flux(earth) and also Flux(shell out) = Flux(earth)
This is obviously wrong because
Flux(shell) = Flux(shell in) + Flux(shell out)
which is simply in violation of (1)
So really you should have
Flux(shell in) + Flux(shell out) = Flux(shell) = Flux(earth)
If you want, you can assume that the shell radiates equally in and out and that would make each 1/2 of the earth’s flux.
….

November 17, 2009 10:29 pm

Stephen Goldstein (20:05:22) :
I am trying to get used to using W/M^2 as equivalent to temperature”

It isn’t. Firstly, radiation depends on a body’s resemblance to a black body. There is no reason a shell as postulated could not have different reflectance on the two sides (or even on parts of one side, as with the Earth and ocean/land/forest/bitumen etc.). Secondly, there is not a linear relation between WM^-2 and temperature, since radiation is proportional to T^4.
And I have my doubts that there is any reason to equate the reflectances of the inner and outer sides, since the rising temp on the inside will certainly give a different apparent reflectance.
As for the article, it starts out determined to be misunderstood by showing equal flux density at two different radii, which is just plain false. Sorry, Willis, but you will be misunderstood if you write fallacies and expect the reader to pick up on all the approximations. And there is, of course, no approximation with the Earth’s core and the surface, these radii are wildly different. If you want to promote this further, it should be rewritten in terms of total radiation and the approximation introduced only where appropriate.

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