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.
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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
The answer to the plant question is simple. With the “perfect greenhouse” (lets through all radiation but doesn’t allow flow of air), the only way for energy to get in and out is via radiation (+ much less conduction which I’m ignoring here).
With the lack of the greenhouse, heat escapes via convection (hot air is taken away by the wind – assuming the wind is cooler!!!!)
The difference in apparent IR temperature is due to the heat flow via convection (hot air) and therefore the greenhouse plant must be warmer.
However, when you talk about two planets, they both have 100% perfect greenhouse enclosures, since no gas escapes via convection beyond the atmosphere. Discounting short term variations due to night/day, the average thermal IR in must equal the average therm IR out for both worlds, and therefore the apparent IR temperature must be the same.
From which we should draw the very useful insight that the world is a 100% perfect greenhouse irrespective of the amount of CO2 and therefore changing CO2 has absolutely no impact on the greenhouse function of the atmosphere!!!
Only two things can effect the energy content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system.” The longer the “residence time” the greater the energy sink capacity. The greater the energy capacity, the longer it takes for any change to manifest. Convection shortens the residence time of energy within a system.
The observed radiation from both planets will be the same (effectively 255 K black-body radiation). However, the question is if the high-concentration planet has a 1 C (as you suggest ‘arbitrarily’) or 3 C (as climate scientists suggest).
Add: surface temperature change.
I dislike gedanken experiments.
Take your car in the sun, and an equivalent volume next to it. Leave a bottle of water in the shade on the seat and a bottle of water on the pavement in the shade next to it. Which bottle of water will be warmer? As cars easily reach 50C when the outside temperature is 30C the answer is obvious. The one inside the car will be reaching 50C and the one in the shade outside 30C. This is real facts.
Unless you mean to stress the fact that plants control their temperature. 😉 .
The answer would still be that the plant inside will be on the higher side of the controlled range than the one outside.
And for those who use the argument “energy in should be energy out”, note that radiant energy changes frequencies interacting with matter and the camera measures infrared only.
For the plant, the un-boxed plant will appear brighter, while the greenhoused plant will appear duller (cooler) through a hazy (indicating warmth) box.
For the planets, both will appear the same.
Jeff Condon. Sorry to say but as the Greenhouse effect is IMPOSSIBLE under the laws of physics- particularly thermodynamics. It is like expecting water to run uphill. Yes, with a pump (work put into the system) but not possible in the real world. Heat ALWAYS moves from a hot body to a cooler body and this also applies when radiation is the transmission medium as well. Those who try to claim otherwise either never studied a physical science degree or have forgotten the relevant lectures in their chemistry or physics degree.
Therefore your post and questions are totally irrelevant to the question of climate and climate change.
A very interesting thought experiment that is bound to create confusion. It creates confusion due to the confusing way that it is expressed.
I think that almost everyone agrees that the ‘energy in’ must equal the ‘energy out’. This is the radiation balance. It applies to all systems in equilibrium [Conservation of energy, 1st Law of Thermodynamics].
But is this the question being asked?
We are told that this magic camera measures something called “radiative emission temperature “. (Ironically, this confusion creator
is preceded by the words “To be clear”).
Now the plant within the greenhouse will be warmer (that’s what greenhouses do) – almost everyone agrees with that. So the plant’s temperature will be higher.
But the first question is “On our camera, will the greenhouse plant image be Brighter..” What does ‘brighter’ mean? Well we are told this means warmer. So, yes – it will be warmer. But then, to this option, is added the additional explanation – ‘More escaping radiation’. Well, no – the radiation out will still equal the radiation in. So the question is impossible to answer correctly! It depends on what the mythical camera is actually claimed to be measuring. Does it measure temperature or radiation energy? The temperature of the plant is higher but the energy escaping the greenhouse is the same as in the no-greenhouse case.
In these thought experiments, the language needs to be precise and unambiguous. This one, like most, is badly phrased.
Making a few assumptions. “Greenhouse” implies that the enclosure is differentially transparent to incoming and outgoing radiation. “Transparent” means it does not reflect all incoming radiation. For the plants I assume“Early afternoon” means the phenomenon is cyclical.
For the planets the problem is different, the camera would integrate, not just across all wavelengths but across all daylight hours because it would see all the daylight hours simultaneously, it is not cyclical.
Plants.
The camera is taking an instantaneous measurement during the course of a cycle.
If the phenomenon is cyclical then instantaneous equilibrium is never reached. Over a number of cycles cyclical equilibrium will eventually be reached – once the system has warmed. Part of the incoming radiation will cause a state change within the system and the energy used to achieve state change is latent, not directly apparent to the observer Whilst it’s warming the system emits less radiation than when it’s warmed because its storing some of it in the form of state change. During that part of the cycle it may appear dimmer in two senses – it’s light temperature is “cooler”, and the amount of outgoing radiation is less, but as energy is added to the system it warms up and the light temperature will brighten to a point where it it is the same as the other system,and then become brighter in terms of light temperature, however it will still be emit less less radiation until it reaches it peak brightness at equilibrium. I believe “brightness”here is intended in the quantitative, sense.
Since it is cyclical, of course, it never reaches instantaneous equilibrium. There is a shift in phase and amplitude (time and quantity) applied to the Greenhouse plant However, twice within each cycle it will achieve a faux equilibrium, in the way a stopped clock tells the right time twice a day. At first the faux equilibria will occur at different times for the greenhouse and non-greenhouse plants, but as the non-reversible state changes are completed by storage of incoming radiation the faux equilibria will be resynchronised. Whist they are in disequilibrium, as Derek says, in the early afternoon the greenhouse plant will be warmer and brighter. But necessarily, during the course of the early afternoon it will also have been dimmer and of equal brightness. It all depends on what you mean by “early” and also how long the experiment has been running.
Planets.
The camera is not taking an instantaneous picture, it is recording a total over all daylight hours. It’s given that the planets are “thermally stabilised” – there are no further irreversible state changes taking place – faux equilibria will be synchronous – and the radiative cycle for both will be in phase. The quantity of radiation from both at all times of day will be the same and will be equal to incoming radiation. The camera will see identical quantities of radiation from bodies which have different colour temperatures, they will appear equally bright.
I like thought experiments.
For 1) I’m going with the same. I’m assuming a decent greenhouse, ie double glazed so the plant will be warmer but the camera’s outside so would be sensing the greenhouse’s emissions, not the plant. I’m thinking the thermal images used by double glazing sales types demonstrate this.
For 2) I’m going with the same as well. But less sure about the camera. If it’s measuring all energy, then energy in should equal energy out but the spectrum would vary between planets.
How about this thought experiment for the two planets. Let them be two balloons.
Both balloons are inflated by a constant flow of air. They are leaky. At some point the inflow of air equals the outflow (equilibrium).
Now I put a few bits of tape on one of the balloons and make it slightly more airtight. (i.e. I add GHG to the second planet.) The air flows in at the same rate. The balloon inflates slightly further than before until once again inflow = outflow.
If the air inside each balloon is taken to represent energy then both balloons emit the same quantity at equilbrium. The slightly improved balloon has more energy in store (a greater volume of air) and thus a higher temperature.
There is something wrong with this picture, but I’m not sure what. Or how to extend it to the plant in the greenhouse.
Kev-in-UK seems to claim that the similarity of the two planets with respect to physical properties would preclude their having different optical depths. However, the postulated difference in the CO2 concentration by itself would result in a difference in the optical depth, as at each level there will be twice as many CO2 molecules interacting with the radiation.
Another commenter challenges the concept of effective height of radiation. The analysis can be done on a wavelength-by-wavelength basis. Of course energy will be somewhat differently distributed among wavelengths on the two planets, but for those wavelengths that interact with the greenhouse gases the planet with the greater concentration will tend to appear slightly larger and dimmer than will the planet with the lesser concentration, and if the energy emitted is integrated over all wavelengths, the energy from the planet with the more concentrated greenhouse gas will come from a somewhat larger, but dimmer atmosphere, while the total energy emitted will be the same.
anna v says:
January 2, 2011 at 3:55 am
…the camera measures infrared…
According to the question, the camera is 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.
Seems you have missed the point of the thought experiment, which is to avoid complicating the understanding of the class with real-world complexities. You may be correct in the real world, but you have not helped to clarify that energy balance is the first step necessary for understanding. Almost everyone is jumping in to this question at the wrong point…
For the Two Plant question, the answer is — not enough information. Because Energy is not Temperature, without knowing the energy exchange with the surroundings, you don’t know the Temperature. Example scenarios with different conduction, convection or radiant energy assumptions, including some intentionally absurd to make the point – thinking ‘outside the box’ (chuckle):
* Four plants, two in glass boxes (GB) and two Open. Place all 4 on my kitchen table .. after they come to equilibrium, temperatures are all the same
* Move one each, GB and Open … one to the Oven the other to the Refrigerator … Temperatures will be Different. and it doesn’t matter which one you put into the oven. You only said I had to measure the temperature in the afternoon, but I took some liberties with where.
* Take the remaining two, one Open & one GB, move them to the back yard on a sunny day .. one under a shade tree, the other in open sun. They will probably be different, but we are not sure because we have not specified other environmental conditions. Measuring in the afternoon makes it highly unlikely that the temperatures would be constant over any time period. What if it was cloudy or raining that day? Living plants transpire, so might humidity be a factor? Would it be different if we had 2 rocks?
* Two identical plants at the Florist .. one in the cooler case (meets the glass box requirement) and one displayed on the counter – temperatures different.
* Two aquatic plants at the bottom of a lake or pool … both reach equilibrium with surrounding water – so both same.
* Go to the desert, dig a pit and place the Glass Box plant at the shaded bottom of the pit, place the Open plant on the nearby surface. The Open plant is exposed to a constant breeze and is warmed or cooled to reflect that convective exchange with the air. The GB plant is not exposed to convection with the air because of the GB and the pit, and it is not heated directly by the sun, but it is radiating out to space and loses heat (the water in the pot may even freeze). When I measure the temperatures of the two, they will probably be different because of different energy exchange with their environments.
* Take two identical plants, one in a glass box and one open to the air … run the experiments and measure the temperatures.
“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.”
No other variables are provided or measured, therefore none can be assumed. Both images will look the same. “all EM wavelengths with perfect sensitivity” This is the null hypothesis. However if the images are not the same, then you have a basis for “grants for life”.
It’s always the same missunderstanding of the greenhouse effect:
A greenhouse warms up in the sun, because convection of the heated air is limited to the inside of the greenhouse. Hot air can’t leave the greenhouse. It has nothing to do with the blocking of IR-radiation by the glas of the greenhouse. So during sunshine (“early afternoon”) the greenhouse heats up, and of course you can measure increased thermal radiation with your camera.
The earth is totally different, it can’t be compared with a greenhouse! The heat from the surface is transported to the higher atmosphere through convection. The only thing that could increase the temperature at the surface of the earth is a more dense atmosphere. The composition of the atmosphere has nearly no effect! So the second question only makes sense, if the warmer earth has a more dense atmosphere. Of course in that case the incoming and the outgoing thermal energy will be the same than in the cooler earth (with the less dense atmosphere).
PS: Sorry for the poor english …
For clarity I will post my original simile that Jeff Id refers to.
It can be found in the comments on this thread at Jeff’s blog, the Air Vent.
http://noconsensus.wordpress.com/2010/12/19/fixing-the-basic-agw-calculations/#comment-44043
I wrote there,
” Question, have you ever looked at a thermal camera picture of a greenhouse and it’s surroundings. ?
The greenhouse will be warmer than it’s surroundings, so it will be a brighter image than it’s surroundings.
AGW “theory” says the greenhouse is warmer because it traps radiation,
YET the thermal image will clearly show beyond doubt that the greenhouse IS radiating MORE than it’s surroundings
(with the same solar input).
What does this show. ?
It shows whatever is cooling the surroundings is far more powerful than radiation, AND
that it is not radiation doing most of the cooling (otherwise the higher radiating greenhouse would be cooler than it’s surroundings).
If one opened the doors and windows in the greenhouse, especially on a breezy day,
the greenhouse would soon reduce it’s temperature to that of it’s surroundings.
Logically the temperature difference was removed from the greenhouse to the surroundings or aloft by air.
Air transporting sensible, and latent heat (of water vapourisation), I would suggest.
When the doors and windows are closed it logically follows that the increase in temperature inside the greenhouse is due to
the reduced transport of sensible and latent heat (of water vapourisation) from inside the greenhouse to the surroundings.
Obviously even with the doors and windows of the greenhouse are shut,
some conduction and convection at the greenhouse glass surfaces occurs, and
this explains why a greenhouse remains warmer than it’s surroundings for some time after sunset.
Hopefully this has illustrated that a greenhouse works by reducing conduction and convection,
NOT by “trapping” radiation, as AGW pseudo climate science, “physics”, and “theory” suggest. ”
The pdf that my simile came from can also be found at the GWS forum on this thread.
http://www.globalwarmingskeptics.info/forums/thread-1028.html
yours,
Derek.
Is your camera inside the box or outside the box?
I don’t know how Anthony wants to handle the answer. I can be added to the top post, added as a new post or just left here. I’ll leave it to the moderator of the morning to figure out what is best.
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 wrote the reply in terms of energy Joules, it would be more accurate to say power Joules/Sec or Watts but since the system is stabilized the answer works fine.
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
[Left in queue for Anthony’s review/editing/comments. Robt]
Sean Houlihane says:
January 2, 2011 at 5:35 am
According to the question, the camera is a cool toy which in this case happens to detect all EM wavelengths with perfect sensitivity.
Please note the exact description
of the camera just before introducing the thought experiment:
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.
Thermal radiation is infrared by definition. So is the “radiative emission” in outside temperatures (0C to 60C).
So saying “energy in equals energy out” ignores the reflected non infrared energies (and energy taken by convection too, which btw is the major contributor in the temperature difference )and cannot be used as an argument.
Since the greenhouse effect clearly works for greenhouses ( nothing to do with C02 of course) the plant in the greenhouse will be hotter so if ( big if) this clever camera is measuring plant temperature it will measure it as hotter by detecting a different spectrum of energy and more energy. However I doubt this is the real thought experiment proposed.. please clarify Jeff.
Back to the question. The two planet problem is a different case since we are now dealing with, I presume a “top of the atmosphere question” where heat in must equal heat out at equilibrium so the fact that the surface( I presume) is 1c warmer is irrelevant.
In the greenhouse case the greenhouse will have a warmer surface than the air at the same level in the open air so even if the clever instrument is measuring this it will still be measuring more radiation and a shorter wavelength spectrum. Of course if the instrument measures the radiation from the greenhouse/ plant for the top of the atmosphere then the scattering effect of the atmosphere will all but dissipate this plant/greenhouse surface difference.
Jeff,
This was a poorly worded thought experiment – you describe your bolometric camera as measuring “thermal” radiation rather than the entire spectrum of energy.
Those who took your words at face value and assumed you were talking about measurement of emitted “thermal energy” will have a different answer from your final answer above.
These parameters are so vague there can be no answer. “thermally stable” is not a measurement but the description of a condition. It says nothing about the comparative conditions of the two plants.
One plant could be in an environment of thermally stability at 100C the other at 30C. Why even mention what time of day it is since apparently both plants live in a permanently stable environment?
You are posing a junk question and so it does not require an answer.