UPDATE: Jeff provides his answer below
Guest post by Jeff Condon (reposted by request from The Air vent)
Derek has been in a war with ScienceofDoom over the what appears to be Planck radiation. I’m actually not sure of his position because it doesn’t make sense yet to me but he left a thought experiment on the thread which could make for some interesting discussion. Some will find it pretty easy, while I bet others will get all tied in knots over it. As a suggestion, taking a thought experiment to an extreme is often a good way to identify a preferred design path or to understand differences in similar situations. I will give the answers in the coming days, they are already written so I can’t back out but as you consider them I’ll warn that this post is not about the subtleties but rather about the bulk differences.
I’m going to paraphrase Derek’s experiment below and then add another of my own. If it’s not the exact same as his it doesn’t matter the idea is still interesting.
For our experiment assume we have a bolometric camera for measuring emitted thermal radiation as an image. IOW a cool toy which in this case happens to detect all EM wavelengths with perfect sensitivity. To be clear, the camera integrates to measure the radiative emission temperature of the object.
We have two plants, one is contained in a transparent box (greenhouse) the other in open air, both thermally stable (temperature isn’t changing). We take an image of the two plants in the early afternoon on our fancy camera, what do you find in the image?
Derek asserts that the greenhouse plant will be warmer and therefore brighter, but lets continue this experiment further.
For our second experiment we have two thermally stabilized earths, one which has today’s CO2 and one which has 2X today’s level. All other conditions are identical and for some quirk of Id-ian physics, they orbit one right behind the other around the sun such that we can observe them simultaneously on our fancy camera. Now the CO2 of the higher concentration planet will block some of the emitted radiation creating the AGW greenhouse effect so the planet has a 1C warmer surface temperature. (For this thought experiment basic physics are required, planets are stabilized and I’m going with a 1C estimate chosen at random). Since it’s my universe, I’m staying on the warm Earth with a functional economy (right side) and sending all the vegetable eating enviros over to the cold economically devastated one on the left. haha- too fun, I probably should stick to the science for this though.
Now from a distant point we observe our otherwise identical worlds using our amazingly fancy camera. What would our camera reveal?
I’ll give the answers to these with supporting explanations tomorrow or the next day depending on how much fun people are having.
============================================================
1/2/2011 Jeff’s answer:
Ok, so the point of this thought experiment was to engage the public in a consideration of the differences between the greenhouse effect and an actual greenhouse. Most here already know that the name itself is a misnomer, but by considering the physics of what is going on we can better understand the argument and better present our opinions on the subject. The majority of the answer below was emailed to Anthony yesterday before he ran the post at WUWT with the note- just to make sure I can’t back out!
I know everyone is wondering what my answers will be to the two greenhouse situations, we’ll see how many will be convinced to change their opinions – or insist that I change mine 😉 . It turns out that both problems are fairly straightforward when considered from an engineering standpoint. In the thermally stabilized systems of the example where temperature is not changing, energy into the system is equal to energy out. We’ll cover the greenhouse vs free air plant situation first. Both plants receive the same energy but the ability to remove heat from the system is limited in the greenhouse plant through convection and evaporation. So in the case of the free air plant, although it is receiving the same energy it has 3 methods of cooling: convection, evaporation and radiation. In the case of the greenhouse, we can consider evaporation and convection negligible so the only option to release the energy is through radiation. Since our camera is only measuring radiation, and since both plants must emit the same energy they receive, the free air plant radiation will sum like this:
Measured EM radiation = Energy in – convection energy – evaporation energy
the greenhouse will sum like this
Measured EM radiation = Energy in – zero convection energy – zero evaporation energy
Therefore the camera will show the greenhouse plant as warmer (brighter) than the free air plant. This holds true even if we include non-zero convection and evaporation for the greenhouse because they are still reduced values requiring a higher EM emission to balance the energy equations.
So now we have the situation where we have two planets, one with more CO2 than the other. We know that CO2 absorbs certain outgoing wavelengths of light. We also know energy in is equal to energy out for both planets. Although this is called the greenhouse effect, it is actually quite different. For both the high and low CO2 planets, the only available cooling mechanism is EM radiation. All the energy coming in has to escape by EM radiation to space, so the equations balance like this.
Measured EM radiation = Energy in
That’s it really. Both planets will measure exactly the same to our camera yet one has a higher surface temperature. The reason this works is that the average energy emission altitude has gone up, allowing a warmer surface yet the net flow is the same.
Caveats: Now I warned that some will get tied in knots over the nuance of this example. There are all kinds of subtleties of the situation which cause minute differences in the planet example. For instance, increasing CO2 will increase the albedo to incoming light, reducing reflected energy and we get a microscopically higher energy in and therefore were our camera of perfect accuracy we could measure a very slightly higher measured radiation from the warmer planet. If this is your explanation, we are in agreement. There are other details as well, but in bulk the answers are A – greenhouse plant is brighter, and B – both are the same.
I read several comments which got the right answer, Carrick was the first to write the correct answer in the comments at tAV, although he kept the answers subtle enough that people had to read it carefully. If you were one who got them both, congratulations. If you are unconvinced by my explanations, ask away and I’ll do my best.
Jeff
“This was a poorly worded thought experiment ” Sorry, I did my best. Can you explain how your answer differed?
Jeremy says:
January 2, 2011 at 8:11 am
“This was a poorly worded thought experiment ”
Sorry, I did my best. Can you explain how your answer differed and why?
Uhh… Guys?
If done correctly these two experiments would show that denser materials conduct heat better than lighter ones..
Comparable real world experiments were done back in about 1909 by R.W. Wood in England (I think). He found that plants in the closed, CO2 enriched, environment grew better but were not warmer than comparable plants in comparable closed green houses without the additional CO2. Kind of proving the obvious, but nevertheless a real result obtained by doing real work.
Jeff,
In my case it’s not a “thought” experiment because I own an actual greenhouse. The plants are warmer inside than outside, no doubt about it. That’s why I built my greenhouse — to keep plants warm — and it works.
It is stunning to me that the “warmer in the greenhouse” answer in poll question #1 did not get 100% agreement. Maybe I read the question wrong, but I have reread it numerous times.
For your next poll you should ask this crowd whether the Pope is Catholic or not, or where do they think bears defecate.
John of Kent says:
January 2, 2011 at 4:30 am
“Heat ALWAYS moves from a hot body to a cooler body and this also applies when radiation is the transmission medium as well”
This is the ‘simplified’ explanation which works for most things and is taught quite widely.
At the molecular level heat is radiated equally in all directions.
The reason heat moves from hot to cold is that colder objects are radiating less heat in all directions then the warmer objects.
As a simple thought experiment
Put two rotary sprinklers on either end of your yard.
Attach a fire hose to one and a garden hose to the other.
Both rotary sprinklers are pumping out water equally in all directions.
Which way does the water flow?
Observing the flow of mud we’ve just created in our back yard, the water(heat) is flowing from the fire hose side to the garden hose side.
But a close examination of the garden hose sprinkler shows it is spraying water equally in all directions and some of that water is ending up on the fire hose side.
Of course CO2 molecules aren’t connected to the water mains, so to further our thought experiment we replace the sprinkler on the garden hose side with a baby pool.
Eventually the baby pool fills up with water and the water overflows the sides of the pool equally in all directions, some of it flowing back towards the fire hose sprinkler and some flowing out into our neighbors yard.
So called ‘green house gases’ are basically baby pools. They absorb IR energy and then overflow in all directions equally.
Ok, I didn’t consider the experiment correctly, because I answered for both cases that the measured radiation would be the same (not thinking about convection, for example). Of course in a real, practical application, you would adjust the calibration of the bolometer to take into account atmospheric transmission and emissivity, which must be different between the greenhouse and the open plant. I wonder how much of a difference though?
Wayne says:
Jan 1, 2011 at 7:23 pm
“Is this not why black objects in an IR camera show as bright and white objects (of anything with low absorption) show as dark. How can this be if their temperature is the same?”
I don’t know what this refers to since I never made any such statement. If anything, it is the exact opposite to what I said. Also, an infrared camera will only measure a fraction of the e/m spectrum. It is not the perfect device given in the thought experiment and is therefore sensitive to spectral shifts caused by either temperature changes or the wavelength dependence of the emissivity.
“In the second statement you are wrong, you said “The two cases can be at different temperatures for the same radiant intensity, because they will have different emissivities.”. If the have different emissivities they also have different absorptivities and the effects equate and the temperatures MUST BE EQUAL in a given radiative field (in the same room).”
Again, this does not address my statement, which refers to the thought experiment given. In a closed room, at thermal equilibrium, you are correct. But, in the present case the sun is the source of illumination and the objects are in equilibrium with all their surroundings, not just the sun’s surface. However, I thank you for your comments. I could have expressed myself more clearly.
Jeff Id says:
January 2, 2011 at 7:14 am
“We know that CO2 absorbs certain outgoing wavelengths of light. “
Should read, “absorbs and emits“ as do all substances including O2 and N2. So what?
“Both planets will measure exactly the same to our camera yet one has a higher surface temperature. The reason this works is that the average energy emission altitude has gone up, allowing a warmer surface yet the net flow is the same.”
Only in your thought experiment!
Jeff Id, the tail does not wag the dog.
If the emission altitude increases it will be because the incoming energy has increased. Not because there is more CO2 in the atmosphere. See my earlier post which is a proper thought experiment because it contains no ambiguous, arbitrary or false statements.
http://wattsupwiththat.com/2011/01/01/greenhouse-thought-experiment/#comment-564125
Will,
“If the emission altitude increases it will be because the incoming energy has increased.”
Increasing/decreasing incoming energy by itself shouldn’t change the emission altitude. The resident time in the atmosphere for the incoming energy won’t change so the altitude won’t change. If you change the absorptivity of the atmosphere (particularly in the long wave regions of the spectrum) the energy stays resident in the lower atmosphere for a longer time on average before re-emission. This results in the average emission altitude increasing yet the same amount of energy being emitted from the body.
Jeff, wish you would have mentioned that evaporation would also play into your thought experiment. My explanation above no longer apply per se.
Might be fair to rewind YOUR explanation excluding evaporation from the experiments and see if you then find many answers above are correct in that light.
Jeff, you can’t throw excluded constraints in at the explanation phase and have a consistent response. Unknowns that impact the results should be in the list of constraints of the experiment in the beginning, not in the answer you just happen to select.
Pretty lame if you ask me (and one of the reasons I didn’t bother with this “thought experiment–I got to the end and had a sense that all was not complete.)
I think Jeff ID’s answer regarding the planets has an inconsistencey, possibly just a typo. He says:
“Both planets will measure exactly the same to our camera yet one has a higher surface temperature. The reason this works is that the average energy emission altitude has gone up, allowing a warmer surface yet the net flow is the same.”
If the average energy emission altitude increases, the radiating spherical surface increases in area, and if the energy radiated is to be kept the same, the surface must have a slightly lower radiation temperature. The temperature difference would not be large, but the difference would be necessary if the energy is to be balanced.
Wayne,
Certainly if the greenhouse didn’t stop evaporation or convection you would have the same situation as with free air. If it stops convection but not evaporation, the answer is unchanged. I specified the windows as transparent, although you could get into all kinds of creative thinking about which wavelengths it was transparent to.
The problem was simply intended to illustrate that the greenhouse effect doesn’t have a lot in common to do with an actual greenhouse.
Will:
Here’s another illustration of people with basic problems with the correct physical quantities, and the results of that semantic confusion….
Solar irradiance is measured in units of Joules/second (power not energy). It need not change in order to have the surface temperature rise:
Increasing the CO2 in the atmosphere results in more stored thermal energy via an increased surface temperature.
More CO2 causes more thermal energy to be stored. So what you claimed was flat wrong.
RockyRoad:
It’s always great to see people respond to fun little exercises in the spirit of comity in which they were posed. 😉
One correction to my comment to Will, irradiance is measured in power per unit area. No edit button and no preview makes it hard to fix these little warts once you notice them.
alcuin,
DeWitt estimated the effect you describe at tAV. It is one of those minor things I mentioned above.
“Nope. The luminosity is the same. For 2x CO2 the total power will be spread out over a slightly larger surface area so it will be slightly dimmer, not brighter. And an increase in effective altitude of 150 m (Tsurf + 1 C at a lapse rate of ~6.5 K/km) changes the surface area by 0.005%. If planet 1 radiates 240 W/m2 then planet 2 radiates 239.9889235 W/m2 and Teff is lower by 0.003 K. Good luck measuring that.”
RockyRoad
“Jeff, you can’t throw excluded constraints in at the explanation phase and have a consistent response. ”
No idea what you are saying. Quite a few people got the point of the thought experiment as it was, several didn’t. I’ll take full responsibility for anyone’s error who can correctly explain why their assumptions led to correct answers for their assumptions but different from my intent.
“Increasing/decreasing incoming energy by itself shouldn’t change the emission altitude.”
Wrong! Tell that to the thermosphere or rather Diurnal atmospheric bulge which tracks the solar point around the Earth bulging up to an altitude of 600 km at noon and disappearing back below the mesosphere at night.
“The resident time in the atmosphere for the incoming energy won’t change so the altitude won’t change.”
So if I boil a kettle for long enough the water will exceed 100º C will it?
Really Jeff that is sloppy stuff.
“If you change the absorptivity of the atmosphere (particularly in the long wave regions of the spectrum)”
You will also increase the emission. For every action there is a reaction. I have already pointed this out but as with all proponents of the “Greenhouse Effect” you appear to suffer from selective reading.
“the energy stays resident in the lower atmosphere for a longer time on average before re-emission. “
Not if the emission increases with the absorption, which it does according to Kirchhoffs Law. We are talking about radiation at light speed here Jeff.
“This results in the average emission altitude increasing yet the same amount of energy being emitted from the body.”
Does it? How?
Jeff; as I stated before (see my posts timed 2:13 am and 2:30 am) I strongly disagree with your claim the emission altitude increases. If your view is correct, how could one correctly determine from space which was the hotter planet? If it was simply a case of a higher emission altitude then the hotter planet would seem to be slightly larger in diameter and since we both agree the total emission is the same the emission per unit area would appear to be lower leading one to conclude the hotter planet was in fact cooler!
Yes, I agree with DeWitt’s evaluation of the situation. With regard to the practical problem of observing the difference, such considerations did not seem to affect other aspects of the proposed camera; the goal, as I understood it, was to tie down the principles involved. The point is that the radiating surface at the TOA of the 2x CO2 planet would be cooler, not warmer.
Will,
“Wrong! Tell that to the thermosphere or rather Diurnal atmospheric bulge which tracks the solar point around the Earth bulging up to an altitude of 600 km at noon and disappearing back below the mesosphere at night.”
You are correct that warming the air will make it less dense and increase the altitude. This is why emission altitude is sometimes referred to as a pressure rather than distance. We’re talking about an entirely different effect here.
“So if I boil a kettle for long enough the water will exceed 100º C will it?
Really Jeff that is sloppy stuff.”
I can’t follow your point but disparaging remarks won’t help you get answers from me.
“You will also increase the emission. For every action there is a reaction. I have already pointed this out but as with all proponents of the “Greenhouse Effect” you appear to suffer from selective reading.”
Not really. You see we’re discussing the long wave in this case which comes from the planetary surface and atmosphere, very little comes from the sunlight directly. I explained this nuance in my answer above – see caveats. Power in equals power out.
“This results in the average emission altitude increasing yet the same amount of energy being emitted from the body.”
Does it? How?”
What is going on is that the long wave radiation emitted from the atmosphere and surface gets absorbed and re-emitted over and over as the energy travels from ground to space. By adding more absorbing material the probability of striking an absorbing molecule goes up, shortening the mean path length of each photon. This also increases the probability that a photon will be absorbed in the upper atmosphere before escaping, increasing the mean height of emission.
For those interested I would like to give an example of why I get hot and bothered about over simplified hypothetical situations. This is a real actual question from an end of year 12 physics exam (we used to call it matriculation) here in Melbourne (some years ago). (this was the exam that determined entry to university!).
A space craft is in a stable circular orbit around the earth. It fires its rocket for a short time so as to increase its speed. After this firing the space craft is in a new circular orbit. Is the new orbit (1) of greater radius (2) the same radius (3) smaller radius.
The answer marked correct was (3). The reasoning given by the science teachers association (who set the exam) was that the space craft speed had been increased and higher speed corresponds to a smaller orbit.
Alternative point of view, increasing the speed increases the total energy of the space craft (same potential energy but more kinetic energy) therefore after the burn it must be in a higher energy orbit and that means a larger circular orbital radius.
That would seem to be a paradox. The true answer is that the question makes an invalid assumption/claim. After the burn the space craft is at the perigee of an elliptical orbit not a circular one. At the perigee the speed is higher than for a circular orbit of the same radius but the overall energy of the orbit is greater. The answer marked as correct was in fact wrong and worse, based on a fundamental misunderstaning of what is going on.
So did it matter? If you were a student wanting to get into you first choice course at university yes it certainly did. In this case, understanding why “equivalent radiation altitude” is wrong and what the true picture is, is fundamental to an understanding of the action of GHG in the atmosphere. If one wants to understand what is really going on and thus be able to make correct decisions it matters even more now.
michael hammer says:
January 2, 2011 at 12:20 pm
I left an explanation of why it happens for will. The mean path length and probability of photons being absorbed changes. The probability of escape for each emitted photon in a higher CO2 atmosphere goes down causing the mean emission altitude to shift to a higher position in the atmosphere. The whole warming effect is just this delay in the power flow from absorption to emission causing an energy buildup.
I don’t really enjoy talking much about it because people then assume I’m a big supporter of AGW extremism. I am not, but the basics are real.
alcuin,
“The point is that the radiating surface at the TOA of the 2x CO2 planet would be cooler, not warmer.”
I avoided these minuscule effects because there are several. In the caveats above, would a 2X C02 planet absorb enough extra incoming sunlight to offset the change in emission area? The effects are both small but of opposite sign.