Guest Post by Willis Eschenbach
Dr. Judith Curry notes in a posting at her excellent blog Climate Etc. that there are folks out there that claim the poorly named planetary “greenhouse effect” doesn’t exist. And she is right, some folks do think that. I took a shot at explaining that the “greenhouse effect” is a real phenomenon, with my “Steel Greenhouse” post. I’d like to take another shot at clarifying how a planetary “greenhouse effect” works. This is another thought experiment.
Imagine a planet in space with no atmosphere. Surround it with a transparent shell a few kilometres above the surface, as shown in Figure 1.
Figure 1. An imaginary planet surrounded by a thin transparent shell a few kilometres above the surface (vertical scale exaggerated). The top of the transparent shell has been temporarily removed to clarify the physical layout. For our thought experiment, the transparent shell completely encloses the planet, with no holes. There is a vacuum both inside and outside the transparent shell.
To further the thought experiment, imagine that near the planet there is a sun, as bright and as distant from that planet as the Sun is from the Earth.
Next, we have a couple of simplifying assumptions. The first is that the surface areas of the planet and the shell (either the outside surface or the inside surface) are about equal. If the planet is the size of the earth and the transparent shell is say 1 kilometre above the surface, the difference in area is about a tenth of a percent. You can get the same answer by using the exact areas and watts rather than watts per square meter, but the difference is trivial. Assume that the shell is a meter above the surface, or a centimeter. The math is the same. So the simplification is warranted.
The second simplifying assumption is that the planet is a blackbody for longwave (infra-red or “greenhouse”) radiation. In fact the longwave emissivity/absorptivity of the Earth’s surface is generally over 0.95, so the assumption is fine for a first-order understanding. You can include the two factors yourselves if you wish, it makes little difference.
Let’s look at several possibilities using different kinds of shells. First, Fig. 2 shows a section through the planet with a perfectly transparent shell. This shell passes both long and shortwave radiation straight through without absorbing anything:
Figure 2. Section of a planet with a shell which is perfectly transparent to shortwave (solar) and longwave (“greenhouse”) radiation. Note that the distance from the shell to the planet is greatly exaggerated.
With the transparent shell, the planet is at -18°C. Since the shell is transparent and absorbs no energy at all, it is at the temperature of outer space (actually slightly above 0K, usually taken as 0K for ease of calculation). The planet absorbs 240 W/m2 and emits 240 W/m2. The shell emits and absorbs zero W/m2. Thus both the shell and the planet are in equilibrium, with the energy absorbed equal to the energy radiated.
Next, Figure 3 shows what happens when the shell is perfectly opaque to both short and longwave radiation. In this case all radiation is absorbed by the shell.
Figure 3. Planet with a shell which is perfectly opaque to shortwave (solar) and longwave (“greenhouse”) radiation.
The planet stays at the same temperature in Figs. 2 and 3. In Fig. 3, this is because the planet is heated by the radiation from the shell. With the opaque shell in Fig. 3, the shell takes up the same temperature as the planet. Again, energy balance is maintained, with both shell and planet showing 240 W/m2 in and out. The important thing to note here is that the shell radiates both outward and inward.
Finally, Fig. 4 shows the energy balance when the shell is transparent to shortwave (solar) and is opaque to longwave (“greenhouse”) radiation. This, of course, is what the Earth’s atmosphere does.
Here we see a curious thing. At equilibrium, the planetary temperature is much higher than before:
Figure 4. Planet with a shell that is transparent to shortwave (solar) radiation, but is opaque to longwave (“greenhouse”) radiation.
In the situation shown in Fig. 4, the sun directly warms the planet. In addition, the planet is warmed (just as in Fig. 3) by the radiation from the inner surface of the shell. As a result, the planetary surface ends up absorbing (and radiating) 480 W/m2. As a result the temperature of the surface of the planet is much higher than in the previous Figures.
Note that all parts of the system are still in equilibrium. The surface both receives and emits 480 W/m2. The shell receives and emits 240 W/m2. The entire planetary system also emits the amount that it receives. So the system is in balance.
And that’s it. That’s how the “greenhouse effect” works. It doesn’t require CO2. It doesn’t need an atmosphere. It works because a shell has two sides, and it radiates energy from both the inside and the outside.
The “greenhouse effect” does not violate any known laws of physics. Energy is neither created nor destroyed. All that happens is that a bit of the outgoing energy is returned to the surface of the planet. This leaves the surface warmer than it would be without that extra energy.
So yes, dear friends, the “greenhouse effect” is real, whether it is created by a transparent shell or an atmosphere.
And now, for those that have followed the story this far, a bonus question:
Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?
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DocMartyn says: The Earth rotates. At least 50% of the time the Earth is radiating into space and is not absorbing light.
Have I missed a response to this? The graphics appear to assume a Flat Earth, which reinforces what the RealClimate folks have been saying about climate skeptics… 😉
I guess that a couple of relevant factors would be the extinction coefficient or saturation of the long wave absorption capacity of the shell and the permanence or otherwise of the shell (residence time of GHG in our case).
Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?
Too hot?
Third picture shows equilibrium of ~30 C at the surface. Average for Earth is lower than that.
Here is my short explanation of why it is wrong (this could be expanded into several books).
At the real Earth surface, the atmosphere has a molecule every 3.7e-26 square metres. The longwave photons, therefore, have to travel by/through 5.4e+30 (54 with 29 Zeros behind it) layers of atmospheric molecules (opaque shells) on its way to space.
About 20% of shortwave photons from the Sun hits one of those 5.4e+30 molecules on its way to the surface so a longwave photon only needs to hit a few more than 20% to cause just enough lag in the time between the two types of photons so that the Earth’s surface is warmer than it should be. Then factor in the fact that the Earth’s land and ocean surface then absorb 85% of the 85% of the shortwave photons that reach the surface and those land molecules hold onto them for a little while and the surface is a nice comfortable 288K (for some of us).
All physics equations should have “Time” included in them. Given that 1 watt/m2 = 1 joule/m2/second, the shortwave photons actually come in 960 joules/m2/second at the height of the day (which is 3e+21 individual sunlight photons per square metre per second) and 0 joules/m2/second at night (but don’t forget the Cosmic Background Radiation).
The surface temperature at 2 metres, however, only changes by +0.001 joules/m2/second during the day and loses 0.0017 joules/m2/second through the night despite the large difference in shortwave energy coming in during that 24 hour period. Given the longwave photons are at lower energy levels, there are still almost as many thermal photons emitted per second at night as the sunlight coming in at the height of the day. Lots of layers and molecules have to be navigated by that unbelievable number of photons even though they travel at the speed of light and could escape the atmopshere in 0.0003 seconds if they weren’t intercepted so often.
The greenhouse effect description needs to move to the level it occurs at, the quantum level. Technically, each shortwave photon from the Sun ends up in about 8 billion different molecules before it escapes from the Earth system – that is a lot of random walking for a lot of individual photons.
Vorlath says:
November 27, 2010 at 1:30 pm
Vorlath, the shell (shield) absorbs shortwave and emits longwave … as does almost everything on the earth. We are warmed by the shortwave radiation of the sun, and re-radiate that energy as longwave (infra-red) radiation.
Energy can change from one form to another (chemical to mechanical, mechanical to heat, etc.). It cannot be created. So the Figures above show that at equilibrium the total energy entering something equals the energy leaving it, although the energy may leave in a different form from the way it entered the system.
Here’s a detailed explanation including the very important effect of gravity:
http://realplanet.eu/atmoseffect.htm
Malaga View says:
November 27, 2010 at 1:30 pm
In Fig. 4, the surface receives 240 W/m2 from the sun, and 240 W/m2 from the shell … how much should it radiate at equilibrium, if not 480W/m2?
Layer upon layer upon layer
Why is the above diagram of a single-shell planetary “greenhouse” inadequate for explaining the climate system of the earth?
Earth is a sphere?
Your picture illustrates one point on the surface. But if you “zoom out” to show whole earth, you realize it’s receiving radiation on one hemisphere but radiating to space from a full sphere.
Kev-in-UK says:
November 27, 2010 at 1:37 pm
The equilibrium condition is an average condition, not an instantaneous condition. It merely means that over some longish time period the temperature of the planet doesn’t change. Since the temperature is not changing we know that on average the planet is in equilibrium.
And while instantaneous equilibrium certainly doesn’t exist, as you say, long-term equilibrium exists for any planet which doesn’t change average temperature over some longer period.
Since the average temperature of the Earth hasn’t changed even a single degree in a century, we can see that although there are always spatial and temporal imbalances, overall, the earth is at something very near to an equilibrium state.
In my thought experiment, of course, such daily and hourly temporal and spatial imbalances don’t concern us.
Mike Haseler says:
November 27, 2010 at 1:39 pm
Mike, it is merely called the “greenhouse effect”. I put it in quotes and referred to it as the “so-called greenhouse effect” so that foolish people wouldn’t assume that it had anything to do with a real greenhouse … guess I’ll have to re-think that plan.
Malaga View says:
November 27, 2010 at 1:46 pm
You are missing a couple of things. One is that as you move away from a fire, you absorb less energy because you intercept less of the radiation. But in the figures above, all inward radiation from the shell is intercepted by the earth.
The second thing is that air absorbs some of the heat from the fire, just like the atmosphere absorbs some of the heat from the earth … but my thought experiment is in a vacuum.
Willis Eschenbach says:
November 27, 2010 at 2:36 pm
Vorlath, the shell (shield) absorbs shortwave and emits longwave … as does almost everything on the earth. We are warmed by the shortwave radiation of the sun, and re-radiate that energy as longwave (infra-red) radiation.
———————-
I’m talking about Figure 4, the third graph.
Your model is inadequate because the earth rotates and so the incoming radiation will oscillate. In other words, your model is static, while the real world i dynamic.
My question was:
A number of people have proposed that the problem with the model is that it is missing something – clouds, atmosphere, water, feedbacks, the list is long. But that’s not the main problem with the model. Even if those were fixed, this model wouldn’t work
There is a reason that this model cannot physically represent the Earth, even in a simplified manner. And unless I’ve missed it, no one has mentioned it yet.
I think it’s erroneous to add the Wm2 from the sun to the Wm2 from the shell (atmosphere) which effectively doubles the energy i,e, the original 240Wm2 becomes 480 Wm2.
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?
The greenhouse effect is the most abused non-existing item on this planet.
Let me explain:
CO2 does have an absorption in the far IR. However, H2O does also. CO2 sits in a window of H2O vapour.
Both of them absorb the black body radiation of the earth. The fact is, earth is NOT a perfect black body. It is a reflecting, colourful body, which means that it reflects a lot of its incoming energy. When emitting, it will only emit radiation with a max temp of 300K. Which means that it does nothing to heat the atmosphere. All atmospheric gasses emit the same amount of radiation at their black body temperatures. Since there is no boundary layer, emissions cancel out.
Ther is no green house effect.
The shortcoming in the model is that each molecule absorbs heat emitted from the earth, the re-radiates it out spherically, meaning that half of the energy is returned to the earth, and that this process repeats itself until vacuum is reached. So the surface will be warmer than the top, though the earth will always find equilibrium and emit its 240 W/m2. Or did I not get the right answer?
My use of the greater than symbol has messed up the html! Again:
Day/night cycles are not considered. Day: outgoing is less than incoming. Night: outgoing is greater than incoming. The change (raising) of surface temperature is the difference between these two budgets.
Dear Folks,
There is a simple way to understand this effect. It is a case of: Radiant Power in = Radiant Power out
The radiant power in is simply (S alpha piR^2), where S is the solar constant (watts/meter^2), alpha is the absorption coefficient, and piR^2 is the area of the Earth’s disk in terms of its radius R.
The radiant power out is given by the grey-body version of Stefan’s Law, equal to epsilon signma T^4 (4piR^2), where epsilon is the emission coefficient, sigma is the Stefan-Boltzmann constant, T is the surface temperature of the Earth, and the expression in parentheses is the surface area of the sphere of the Earth.
If we equate them, we get
S alpha piR^2 = epsilon sigma T^4 (4piR^2)
By cancellation and rearrangement of terms, the temperature is solved as
T = (S alpha / 4 epsilon sigma)^1/4
Now we are in a position to understand the influence of the various parameters. S and sigma are physical constants and are not subject to variation (in this analysis). The absorption coefficient, alpha, may vary from zero (perfect reflection) to unity (perfect absorbtion) and characterizes the Earth’s absorption of radiant power from the sun. If alpha = 0, the Earth cannot absorb any power and will cool to zero temperature. Likewise, the emission coefficient, epsilon, may vary from zero to unity and characterizes the Earth’s emission of thermal power from Stefan’s Law. If epsilon = 0, the Earth cannot emit any power and the temperature will increase infinitely.
The crux is this: if alpha = epsilon (both close to unity), then a certain equilibrium temperature will be reached that will be close to the “airless planet” temperature (for S = 1400 w/m2 and sigma = 5.67 x 10–8 w/m2K4 this will be T = 280.3 K or 7.15 C).
However, if alpha > epsilon, as is the case when there is imperfect emission (grey-body conditions, epsilon < 1), then the equilibrium temperature will be higher (if alpha/epsilon = 1.3 then T = 299.3 K or 26.15 C). This is what happens with a re-radiative atmosphere; it is simply equivalent to the Earth’s emission coefficient being reduced by the thermal radiation backscatter from the atmosphere.
Contrariwise, if alpha is reduced (as by higher albedo from increased clouds), the equilibrium temperature will be lower. As this equation makes clear, albedo has direct leverage over the result, whereas CO2 has very indirect leverage.
The mechanics of heat transfer within the atmosphere and to and from the oceans and land masses are details that affect the values of alpha and epsilon, but are not otherwise involved in the equilibrium radiative thermal balance among the Earth, Sun, and outer space.
The earth is not in a vacuum.
Willis Eschenbach says:
November 27, 2010 at 2:39 pm
In Fig. 4, the surface receives 240 W/m2 from the sun, and 240 W/m2 from the shell … how much should it radiate at equilibrium, if not 480W/m2?
Because a Thermos Flask helps keep things hot… but it doesn’t make them hotter… your only input is 240 W/m2 from the sun!
Willis, you have had the same answer twice; your diagram needs to show
Sun
————–Shield
EEEEEEEEE Earth
————–Shield
The shield radiates at the same temperature all over the sphere, but half the sphere is in darkness. This is why the heat flux budget diagrams are such bollocks, the Earth rotates. We do know that the Earth generates heat from internal radioactive decay; something missing from most heat budget.
Willis, I must be missing something too…
but I do not quite understand your query…the reason, in basic terms that the planetary surface is ‘warmer’ is because the rate of thermal energy (or rather just energy) transfer is reduced or ‘slowed down’ by the shell.
In terms of light transmission through a substance, the light is bent and deviated, resulting in slower (altered) transmission (i.e. refraction – which is different for different wavelengths too!) and this means the light travels a longer path. This could be analagous to the thermal ‘storage’ in that the atmosphere acts as a deviation for the thermal ‘light’ energy (ok – its radiation, but it is an analogy) – thereby lengthening its stay in the atmosphere, which is not a vacuum and its contents can be excited (vibrated) by the passing radiation, thereby allowing it to warm the atmosphere.
Mind you – it’s 11pm here in UK, and after a very nice red wine, with nothing on the TV – this may not be making much sense! LOL
Willis,
Your model does show some essential features that determine planetary “greenhouse effects”. However, it does not tell the full (still simplified) story of what actually happens. The following is a somewhat more complete version. The point is that it is actually the combination of the fact that a lapse rate becomes established in combination with the movement of the outgoing radiation to a high altitude that is required.
Atmospheric greenhouse effect:
Once a sufficient amount of atmospheric long wavelength IR absorbing gases (called atmospheric greenhouse gases by convention) are present, they reduce the direct radiation transmission to space enough so that convection and relayed radiation becomes the dominant modes to transport absorbed energy from the surface to a high altitude where it is radiated to space. The only way the energy can be radiated to space from the high altitude is from the absorbing and radiating gases (or clouds, but that is another story). Once this situation is encountered, adding more absorbing and radiating gases (but not enough to significantly increase the total mass of the atmosphere) do not change the fact that convection is still the dominant heat transport mechanism within the atmosphere. In fact, this reduces the radiation conduction, so that convection carries even more of the total energy to the high altitude. However, adding more absorbing gas does raise the altitude of outgoing radiation somewhat (not because of the tiny added mass, but because the height where the radiating gas concentration is suitable to radiate to space is increased). It is this increase in altitude of the outgoing radiation that results in the slight temperature increase.
Once the dominant mode of heat transport is convection (buoyancy, winds, and turbulent mixing), the atmosphere will form and maintain (on the average) an adiabatic lapse rate. This lapse rate is a temperature gradient due to the cooling effect of rising gas in a dropping pressure (due to gravity). Evaporation from the surface and condensing water vapor in the atmosphere change the level of the lapse rate from the dry air value, and this is called the wet adiabatic lapse rate. The outgoing radiation has to equal the incoming absorbed radiation unless the temperature is changing, but here we consider the case where the temperature has leveled off for simplicity (as it has on Earth for the last decade or so). In that case, the match of radiation out to space, from a particular effective altitude, to absorbed input radiation, determines the effective temperature of the gas at that altitude. This temperature is then added to the lapse rate times the effective altitude of outgoing radiation, and this gives the ground effective temperature. The combination of moving the location of the effective level of the atmosphere, where the radiation to space occurs, to a higher altitude, and adding the effect of the lapse rate times increased altitude is the source of higher surface and low altitude temperatures.