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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Vince Causey says:
November 29, 2010 at 9:12 am
old construction worker,
the problem with trying to apply accounting analogies is that they depict an instantaneous position whereas the situation is one of flows. I prefer the following analogy:
Imagine a line of people filing into a building through the in-door, at the rate of 240 people per hour, and filing out the other end through the out-door. This represents the energy flow arriving and leaving the earth. Now suppose that half of the outgoing people pair up with those on the incoming line. You now have 360 people per hour entering the building. These extra people haven’t materialised out of thin air – they came in the original line.
I agree with you 100%. But if you look it the “flow charts”, they do not indicate a “content flow”. They indicate one unit of 240 W/m2 going from point a to point b to point c to d (Fig 4#). To correct the misinterpretation, not Willis’s explanation, I would add “content flow rate of ” above 240 W/m2 where it enters the system.
“”””” bessokeks says:
November 29, 2010 at 6:24 am
Vince Causey says:
“But transfering heat energy from “cold” to “warm” violates the second law of thermodynamics.
There is no way around that.”
“There is a lot of misunderstanding surrounding the second law of thermodynamics.”
This is right 😉 “””””
Well I like the Clausius form of the Second Law.
No cyclic machine can have no other effect, than to transport heat from a source at one Temperature to a sink at a higher Temperature.
Note the restriction to “cyclic” machines.
Machines of any kind involve MATTER. Without matter there is NO HEAT; it is simply the mechancial kinetic energy of matter.
Electromagnetic radiation exists even in the total absence of matter. Whether you consider it a photon stream or an electromagnetic field, it can and does transport energy from anything to anything.
We get too wrapped up in the quantum nature of EM radiation; so it raises questions like; if CO2 molecules are only one in 2500 of the total atmospheric molecules; how do they ever get hit by a photon ?
But if you think of thermal radiation as being an electromagnetic field; then it can propagate throughout space, and encounter any CO2 molecule in its path.
It helps if you understand some radio-physics; specifically antenna theory. Antennas can radiate EM energy because Voltages, and currents in their elements generate fields, and those fields cannot appear instantaneously. An ordinary dipole antenna consisting of two wires pointing in opposite directions and driven at their close ends by an AC Voltage can have a current flow along those wires, because the distant ends are still connected together by a small capacitance, which is capable of carrying an AC current; and because of the flow of that AC current, there also is a magnetic field set up around those wires. The direction of the flux lines of the magnetic field are always perpendicular to the lines of the electric that is due to the varying Voltage along the wires. Because the resultant EMf ield takes time to set up and start to fill all of space, it is possible for the direction of the current and or Voltage to reverse, before the filed gets enough time to collapse; there’s that propagation delay thing again. as a result of those propagating EM fields, free space; behaves somewhat like a transmission line with a characteristic impedance that is known exactly; in fact it is 120 Pi Ohms or 377 Ohms. If you know how to make 377 Ohm paint then you can make yourself invisible by just dunking yourself in it. Any EM wave that strikes you will be totally absorbed with NO reflection, and it will apear to a viewer that the radiation just went right on through you to the end of the universe.
Well don’t get too self confident; because if by definition you do absorb all radiation that hits you (you’re a BB) then you likely will undergo a Temperature rise; and ultimately I guess you might have to emit thermal radiation; but don’t quote me on that.
I like to think of it as something like a kid’s wire loop that she sticks in the soap solution and blows bubbles. The air pressure bellows out the stable flat water film , and surface tension tries to pull it into a sphere attached to the wire loop. Somehow the bubble necks down near the loop and pinches off the bubble so it is free to propagate to the edges of the universe..
The antenna EM waves do the same thing; you get a close in (near field) behavior within about a wavelength or so of the antenna, and then there develops disconnected closed loop fields of mutually perpendicular magnetic, and electric fields that are no longer constrained to the antenna, and they propagate at a velocity c (in free space) and in a direction that is also perpendicular to both the electric and magnetic fields.
Well atoms or molecules that emit “resonance” types of EM radiation are just bloody small antennas; well they are if they have a dipole moment.
Some have argued here that a molecule requires three or more atoms to emit Infrared radiation or absorb it; diatomic molecules do not absorb or emit in the infrared.
Well your should try telling that to an HF molecule which has almost as big a dipole moment as does H2O; and it most certain can and does absorb and emit infrared radiation.
But it is generally true of things like N2 and O2 that they are NOT infrared active, as far as resonance radiation goes; because they are symmetrical.
But it simply is NOT TRUE, that these gases or gases in general DO NOT emit THERMAL RADIATION.
The distinction is a fourth grade science question for “Are You Smarter than a Fifth Grader”.
Resonance emission or absorption is a line spectrum that is characteristic of the specific material (molecule species), and is essentially independent of any Temperature. Well experiment shows that the width of these resonance radiation lines is somewhat Temperature dependent, and also pressure dependent ; due to the Doppler effect (Temperature) and the molecular collision frequency (Temperature and pressure related). But the basic frequencies stay the same.
Thermal radiation as its name implies is a consequence of the Temperature of the emitting material and is essentially independent of the nature of the material; and depends ONLY on the Temperature.
Its immediate Physical cause is the simple fact that the MOLECULE contains electric charges; and those charges undergo motion and acceleration as a result of the translational motion of the molecule itself which is the direct result of the Temperature of the material. Classical Physics requires that accelerated electric charges constantly radiate EM radiation and energy.
As evidence that atomic or molecular structure is unrelated to the thermal radiation; we have the obvious case of a free electron or proton; which don’t exactly have a whole lot of orbitals associated with them; but which are well known to lose energy by radiation in the process of being accelerated in particle beam accelerators. Particularly in circular orbital accelerators, electrons eventually radiate more energy during one revolution than the machine can replace next time the packet returns to the kick site.
That is the whole reason for the Stanford, 2- mile Linear Electron accelerator; and also for the whacking great size of the LHC machine that covers most of Europe (underground). Going in circles is constant acceleration (rate of change of momentum) so constant energy loss.
Charged particles at near light speed crashing into materials at greater than the velocity of light in the material undergo rapid deceleration, and during that time they emit radiation (usually a blue light); well the Cerenkov radiation is usually of that nature. I’m not exactly sure of the distinction between Cerenkov radiation, and Bremsstrahlung (which I gather is German for “Braking Radiation.” ) And I don’t do umlauts so if you need one grab one.
Bottom line is that whereas N2 and O2 may not absorb or emit infrared characteristic line spectra, they most certainly DO emit thermal radiation in a BB like spectrum dependent only on the Temperature.
My Quantum mechanics timed out at about the Master’s level, and I haven’t much kept it up; so I don’t have a good grasp of the quantum mechanics of thermal radiation; but I have seen suggestions that it is during the collision times between molecules, when all those electrons get in each other’s hair, that the quantum selection rules like the Pauli principle and the like, and assist in the thermal emission; but I am out on thin ice on that; so you need to ask the PhDs like maybe Phil; if you really want to delve into that. I’m happy to accept that the Planck derivation of BB radiation didn’t really need any deviation from the Classical physics (statistical mechanics ) of the Wien and Raleigh Jeans derivations; other than to insist that the energy levels couldn’t be closer together than kT/2; whereas the classical partition rules had no minimum energy difference.
And if I got this a bit too screwed up; well you should be able to go and look someplace for the fair dinkum scoop on it; but don’t go to wiki; because they are more screwed up than I am.
I have one more quation Willis,
Why the few km height of the glass shell; why not a few cm or a few atto metres for that matter ?
We know that spectrally selective absorbers do work quite well in collecting solar energy; by having high solar spectrum absoptance; but much lower thermal sepctrum emissivity. Well they work well enough to take the chill off your swimming pool water in winter.
“jimmi says:
November 29, 2010 at 2:19 pm
There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”
Can this be demonstrated/verified via an experiment?
George E. Smith says: November 29, 2010 at 4:49 pm
Thermal radiation as its name implies is a consequence of the Temperature of the emitting material and is essentially independent of the nature of the material; and depends ONLY on the Temperature.
This is not even close to right. The material makes a huge difference. Look at the table here: http://www.bacto.com.au/downloads/IR%20&%20Emissivity.pdf
The energy radiated by an object is not simply P = AσT^4.
Rather it is P = AεσT^4 where ε = emissivity.
The IR emissivity can vary greatly depending on the material (and the surface finish). Polished metal is often below 0.05. Soot is up around 0.95. A painted metal object (ie high IR emissivity) will cool off much faster than a similar size/shape/temperature shiny metal (ie low emissivity) object because the low emissivity object simple cannot radiate IR effectively.
Since N2 & O2 don’t absorb IR well (very low emissivity), they also don’t emit well. If you are waiting for N2 to cool simply by radiation, you will presumably be waiting a very long time.
,blockquote>There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”
Can this be demonstrated/verified via an experiment?
To be honest it is self-evident, but if you know someone with night vision glasses (the thermal imaging type, not the light amplification variety) you could try the following:
Turn out all the lights, turn off the power to your fridge (so the internal light does not come on) put on the glasses and open the fridge door. Do you see anything? If so you have proved it – the glasses are at room temperature, the contents of the fridge are at about 5C, and those imaging glasses should work for objects down to about -5C (according to the specs). I haven’t tried this myself (no night vision glasses) so if anyone can try it I am sure some people would be interested.
Can this be demonstrated/verified via an experiment?>>
Well, how do you suppose an igloo works? Or a quinzey? Do you know how the first settlers insulate their houses? By piling snow up against them. Now how could that cold cold snow POSSIBLY make it warmer inside the house? How could you POSSIBLY warm yourself up by hollowing a hole out of a snowbank and sitting inside it? Ever done any winter camping? Sit on top of a pile of snow at -30 and you will die before dawn. Dig a hole in the side of the exact same pile of snow and crawl in. You’ll still die unless you remember to poke a small hole inthe top to allow for circulation of the air. If you remember, and you’ve dug the hole no larger than necessary, you’ll likely want to take your jacket off to cool down after a couple of hours.
Tim,
Oh Oh. You’re right, but now you get to reprise my argument with Svalgaard. 🙂
I’m still trying to figure out the problem with George’s 377ohm paint. I suspect it has something to do with the characteristic impedance of air not being the same as vacuum, but that’s not entirely satisfying.
George?
Willis,
Thank you for the response to my first post. You did misunderstand my poorly phrased suggestion. I was not saying that you would not find any warming. I tried to say you would not find any warming from GHG’s!!!
BIG difference!!
jimmi and davidmhoffer,
“There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”
Well, it may seem self-evident, which is not always the case with quantum physics, but real experiment is needed to verify that a photon from a colder object is indeed absorbed (meaning energy transfer?) and that many more [photons] are going in the opposite direction. I am not sure your proposed test with fridge will demostrate it, perhaps I will see reflected photons from outside? Igloo example is also not clear regarding the specific statement above.
And speaking of “many more,” if we were to use a focusing lense or parabolic mirror to increase number of photons from a colder object so that there are more going in this direction from colder to warmer, will the warmer object absorb them and increase its temperature?
from NY says:
November 30, 2010 at 3:26 am
“Well, it may seem self-evident, which is not always the case with quantum physics, but real experiment is needed to verify that a photon from a colder object is indeed absorbed”
How about you look outside next time it’s snowing and explain how your eyeball at 98.6F can see a snowflake at 32F.
If you can see it you’ll have proved to yourself that photons from colder objects fall upon warmer objects.
If that doesn’t convince you I really don’t know what will.
@Bryan says:
“Heat always travels from hot to cold never the reverse.”
o.k.
“Thermal energy can be radiated and absorbed by both hot and cold objects.”
of course – but thermal (IR) radiation of a colder body cannot be absorbed by a relatively warmer body. This does not mean, that a colder body cannot radiate to a warmer body. Think of radiation as a flow of energy which can be described with two parameters: quantity (amount) and quality (frequency). To transfer heat by radiation you need beside quantity also a certain quality
@terry Oldberg says:
“The entity which, in a Kiehl-Trenberth diagram, is labelled as the “back radiation” should be relabelled as a “vector irradiance” with specified spectral characteristics. Relabelling it would help to eliminate the appearance of a second law violation.”
The second law gets violated when radiation should transform into heat.
Think of this:
Kiehl-Trenberth shows back radiation with 324 W/qm, radiation from sun with 168 W/qm. Why do we build sun collectors and not back radiation collectors. Back radiation is available round the clock, 365 days a year and is delivering double as much energy as radiation from sun. Has anybody ever tried to look at back radiation as a source of energy? Ask yourself why!
@Tim Folkerts says:
“Put the same rock (in a vacuum container to avoid conduction & convection) in a freezer at 255 K. The rock still radiates 480 W^m2. But the container radiates 240 W/m^2 back to the rock (again assuming the container is a blackbody). The net loss is 240 W/m^2.”
Thermodynamics know three kinds of systems:
1.open (exchange of energy and matter)
2.closed (exchange of energy)
3.self-contained (no exchange)
“The rock will cool half as fast as before (at least initially).”
Really???
the second law applies for 1 and 2
@Tim Folkerts says:
“The real consideration is the solid angle subtended by the radiating source. If the cooler container completely surrounds the hot object, then the rate that energy arrives at the hot object is the same, whether the container is a little larger or much larger.”
please rethink…
@jimmi says:
“There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction. ”
This means a large cold objekt can warm a small, warmer object, because it is only a question of quantity…
“By the way, this can apply to conduction as well – if you think carefully about what happens when you allow two gases at different temperatures to mix, you will realize that some of the flow of both molecules and energy has to be from what was ‘colder’ towards what was ‘hotter’. Again it is just the net flow that is hot to cold.”
think about Brown…
@george E. Smith says:
“Well I like the Clausius form of the Second Law”
Well, I love it… But where is the point?
from NY says:
November 29, 2010 at 6:31 pm
” “jimmi says:
November 29, 2010 at 2:19 pm
There is absolutely nothing to stop a photon from a colder object being absorbed by a warmer object, it is just that there are more going in the opposite direction.”
Can this be demonstrated/verified via an experiment?”
nice question…
😉
@davidmhoffer says:
“Well, how do you suppose an igloo works? Or a quinzey? Do you know how the first settlers insulate their houses? By piling snow up against them. Now how could that cold cold snow POSSIBLY make it warmer inside the house? How could you POSSIBLY warm yourself up by hollowing a hole out of a snowbank and sitting inside it? Ever done any winter camping? Sit on top of a pile of snow at -30 and you will die before dawn. Dig a hole in the side of the exact same pile of snow and crawl in. You’ll still die unless you remember to poke a small hole inthe top to allow for circulation of the air. If you remember, and you’ve dug the hole no larger than necessary, you’ll likely want to take your jacket off to cool down after a couple of hours.”
what do you want to prove by that?
@from NY says:
“And speaking of “many more,” if we were to use a focusing lense or parabolic mirror to increase number of photons from a colder object so that there are more going in this direction from colder to warmer, will the warmer object absorb them and increase its temperature?”
Hey, welcome to the club 😉
This “netting”-thing is just bullsh… They call it “modern” physics. Makes me wonder where “fashion” will go in the future…
bessokeks says:
“… thermal (IR) radiation of a colder body cannot be absorbed by a relatively warmer body. This does not mean, that a colder body cannot radiate to a warmer body…”
I’m not sure how you can seriously say those two sentences — they contradict each other! A “colder body radiating to a warmer body” IS thermal radiation of a colder body being absorbed by a warmer body.
Or how about this.
* The sun (5800 K) emits some 15 um IR as part of its thermal radiation.
* The earth (300 K) emits some 15 um IR as part of its thermal radiation.
* The atmosphere (250 K) emits some 15 um IR as part of its thermal radiation.
Can an object that is absorbing the 15 um photons know where those photons came from? Of course not! Whether the object absorbing the light is 200K or 275K or 350K or 3000K, it will absorb the 15 um photons from any temperature source exactly the same.
“Why do we build sun collectors and not back radiation collectors? ”
Because heat engines require a temperature DIFFERENCE to operate. The thermal radiation from the cool atmosphere can slow the cooling of the surface, but could never be used to operate a heat engine where the “cool side” is the warmer earth.
Because photoelectric cells require HIGHER ENERGY visible light photons to operate — focused low energy IR photons (no matter how bright or well focused) will not work.
“And speaking of “many more,” if we were to use a focusing lense or parabolic mirror to increase number of photons from a colder object so that there are more going in this direction from colder to warmer, will the warmer object absorb them and increase its temperature?”
A clever attempt, but ultimately it is wrong.
Consider using a lens to focus sunlight and burn word. Normally the sun provides about 160 W/m^2. If you use a 5x magnifying glass, what you have done is make the sun look 5x bigger from the perspective of the focused spot on the wood surface(=25x more area). That 25x “larger sun” will radiate 25x more light to the focus = 4000 W/m^2. That is a lot – enough to char the wood. But it is still way less than the 60,000,000 W/m^ the sun radiates from its surface.
Even if you focused sunlight from all direction, you can’t focus more photons to an object than you could get by surrounding the object completely by a thermally radiating object of the same temperature.
The same goes for the example above. Focusing photons from a cold object to a second object will never do more than surrounding the object by the colder object — which will cool the second object as expected. So the supposition “so that there are more going in this direction from colder to warmer” is never possible. (Or you WOULD violate the laws of thermodynamics and win a Nobel Prize.)
bessokeks
….”two parameters: quantity (amount) and quality (frequency). To transfer heat by radiation you need beside quantity also a certain quality”…..
I don’t think were in a big disagreement here.
The colder body cannot raise the temperature of a warmer body above its own colder temperature.
Which means the absorption of photons by the warmer has no effect.
The vector dealing of the radiations as outlined by Terry Oldberg above gives the best approach.
bessokeks
I was told in another thread that you can’t “use a focusing lense or parabolic mirror to increase number of photons” of LW because it comes from all directions (unlike the radiation from the Sun).
Bryan says:
November 30, 2010 at 6:27 am
“The colder body cannot raise the temperature of a warmer body above its own colder temperature.”
A room temperature blanket can be used to raise the temperature of a warm bodied person. I understand the main article, but it seems like a lot of extra work over saying “The atmosphere acts like a blanket.”