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
Argg! I don’t understand some of you people.
In the second ‘experiment’: a warmer body radiates more than a cooler body, period. Doesn’t matter where the warmer body’s temps come from; doesn’t matter what it’s atmosphere is; doesn’t matter if it’s receiving the same amount of energy as the cooler body; if it’s warmer, it radiates more energy (THAT’S THE ENERGY YOU ARE DETECTING WHEN YOU DETERMINE THAT IT’S WARMER THAN THE OTHER BODY!!!).
Lazy Teenager: that would depend of the optical properties of the greenhouse glass. What’s it’s optical depth, at what wavelengths does it emit? Even the best of glass will have different properties (emission, transmission, reflectivity) than a parcel of air. That being said though, I would have to say that this case assumes the plant can be imaged through the glass, else the question is moot.
(Pardon me if I seem hesitant…I am a new poster here, and it has been several decades since my Geography coursework, so I may sound somewhat foolish here and there)
I just want to add that I don’t believe 2X CO2 will actually add 1 degree. Most here probably already know my positions on CO2 which are – I don’t know. The 1 degree was used for the example only.
This fun post turned out to be more confusing than I thought it would so I’ll give another clue. The correct answers can be found through energy balance equations.
jeff: The correct answers can be found through energy balance equations.
Wrong. You have stated that the 2x planet is 1 C warmer at the surface. The thermal or blackbody radiation is therefore by definition higher or brighter.
Over time, steady state conditions may be assumed to exist and total energy in and out may balance but that does not change the fact that the spectrum will be different for these two examples with the hotter surface planet having greater thermal emissions than the cooler planet.
Kev-in-UK asks why the effective height of radiation is higher when the concentration of greenhouse gas is greater. The effective height of radiation from an atmosphere is conventionally taken to be that altitude at which the optical depth becomes unity, as integrated over height from outside the atmosphere. If the concentration of greenhouse gases are doubled, their opacities are also doubled at each level, so the integral becomes unity over a shorter distance down in the atmosphere, i.e., at a greater height. Of course the opacity = (1 – transmissivity) varies with wavelength, so a lot of averaging is involved, but the use of an effective height in this manner is the way that the radius of the solar photosphere is defined. Outward radiation from lower altitudes tends to be absorbed and reradiated at higher altitudes.
Jeff ID:
Then to take a stand on it – noticing you had helpfully posted both the Wiki energy balance graphic and said “it’s not the subtleties but about the bulk differences” – then I’d say the plant in the greenhouse will be imaged as thermally brighter than the plant outdoors.
The greenhouse emits more in the thermal, and that plant is sitting inside it, as opposed to a plant sitting outside in the open, unconstrained. Convection for the plant outside will draw away heat energy because there are no constraints (just parcels of air), whereas convection is unimportant inside the greenhouse, because all it does is just transport heat around inside of a sealed box. It will still emit it heavily in the thermal.
As I’d said it’s been many years since my coursework, but am I viewing this in the proper way?
This is the problem with talking about “global mean temperature”. Because there is day and night, each location has outgoing radiation that continuously varies. If the temperature in the greenhouse is warmer, it will radiate more. In the planet example, the high CO2 planet receives less radiation at the surface, as the sun’s radiation is about 45% in the infrared. In this example, it starts at dawn more than 1 degree warmer, and heats slower than the low CO2 planet, but after sunset, also cools slower than the low CO2 planet. On Earth, due to convection, peak surface temperature occurs around 2pm.
Net-net, the 2pm radiation from orbit should be about the same, as some of the higher surface temperature gets absorbed by the CO2; otherwise, the surface temperature wouldn’t be higher. At some times of day, radiation out observed will differ, unless the high CO2 planet keeps always 1 degree warmer than the low CO2 planet, in which case the radiation out would also always be the same. Lower radiation from the high CO2 planet from 2pm to 6am (cools slower). Higher radiation from 6am to 2pm (warms slower). Early afternoon, about the same.
The planets should maintain the same brightness, but I can’t help thinking there is something I don’t know about involving the rate of emission.
Many above are speaking, of the two planet experiment, maybe without knowing it, in blackbody terms (unrealistic) and forgetting the emissivity for the Stefan-Boltzmann realtion in a more realistic example. Blackbody is very theorlogical whereas a gray body is close to reality, especially if viewed as the sum across all frequencies. Black body uses E = σT^4 where as gray body includes the emissivity of the surfaces as E = εσT^4 with “ε” being the emissivity that you can lookup for a rough estimate for many subtances and surfaces. Just the fact that one planet is 1°C warmer does not immediately imply it emits more without taking into account the planets emmissivity as a whole.
However, in the example given above the warmer planet had more co2 which would increase its emmisivity AND equal absorptivity, and vise versa, in comparson to the other planet which would mean it would be radiating much more that the cooler planet that just the one degree difference implies but it equally would be absorbing more and of an equal amount. The 1 deg difference would not occur in reality, that is a great example of the AGW fallacy.
Like Dr. Feynman said of Q.E.D., it twists your mind a bit, sometime a lot, but that is the way physics is IN REALITY and there is nothing we can do but try to understand it properly and not to twist reality into something that isn’t.
Further, the plant inside the greenhouse will not (in general) reach equilibrium temperature at the same time as the outdoors plant. For example, on the moon, equilibrium temperature is not reached until late afternoon. So, due to lack of convection, the greenhouse plant would take a different amount of time to reach equilibrium. The same should hold, to a lesser degree, on a high CO2 planet. The planet that reaches equilibrium earlier (after noon) will have a higher energy in at that moment than the planet that reaches equilibrium later (sun lower in the sky) and so, the planet with a quicker equilibrium will show a higher temperature.
Note that this logic does not work on the moon, as peak temperature on the moon is much higher, but that is due to the two weeks of sunlight. Peak temperature rise per hour on the moon (late morning) is quite similar to that on Earth. If we had an airless moon that rotated every 24 hours, its peak would be late afternoon, at a lower temperature than Earth. Later equilibrium, lower peak temperature.
Oh, the high CO2 planet should hit equilibrium earlier, based on the statement that the temperature at equilibrium is higher.
“In the second experiment, the two planets again have the same thermal emission temperature. In planet 1, with the doubled CO2, the layer that thermal emission temperature occurs is 460 metres higher than planet 2 and hence the surface temperature on planet 1 is 1.0C higher. Effectively, the lagtime in which the energy enters and then exits planet 1 is a few minutes longer than planet 2. 44.06 hours in planet 1 compared to 43.44 hours in planet 2.”
nice bill.
Jeremy, I take it he’s assuming identical planets, and therefore identical emissivities, otherwise the emitted radiation could vary by a factor of 50 (the polished silver planet versus the carbon planet).
Or you could just change the temperature profile of the planet (hot side versus cold side, or hot regions (black) and cool regioins, while leaving the average temperature unchanged. That can generate huge errors in temperature estimation due to the fourth power in the Stefan-Boltzman law. For example, if you make half the planet absolute zero and double the temperature on the other half of the planet, leaving the average temperature exactly the same, the planet emits eight times more radiant energy. Or you could have a planet where the cities on the night side are lit with 6000K color-temperature LED’s, or one where everything’s been painted with US Navy low-IR emissivity paint, and of course you have Skull Island where there’s a super villain with a telescope aiming a laser at your camera.
Something like this formed the basis of the Gaiaia hypothesis, where the flowers bloom to regulate a planet’s reflectivity and emissivity to create the ideal environment for prancing unicorns.
There are so many complexities to accurate radiative temperature measurement that it boggles the mind. The usual solution is to ask what the temperature actually is and then adjust the sensor’s calibration to match.
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.
I say no change. Both will absorb the same level of upwelling LWIR from the surface. The 2X planet will achieve its quota at a slightly LOWER level. Both planets will experience the SAME quantity of energy being conducted into the atmosphere via molecular collisions. Both planets will handle the absorbed energy through the lapse rate up the atmosphere.
Areas of the tropics experience much more than a doubling of H2O without any dramatic rise in temperature. In fact seasonal temperature change over the tropic oceans is far more stable than seasonal temperature changes over desert regions.
The 3 options for the planet experiment are also all wrong.
The term brightness is too precise here.
The energy balance for both planets must be the same. The wavelength distribution of energy emitted by both planets is different. In other words the IR “colour” is different.
Kev-in-UK says:
January 1, 2011 at 4:44 pm
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alcuin says:
January 1, 2011 at 4:27 pm
The effective height from which thermal radiation leaves the higher concentration planet will a bit greater than that of the lower concentration planet….
sorry, but why?
the concentration of CO2 in the atmosphere is described as doubled – but there is no mention of any increase in atmospheric depth (all other conditions are described as identical)?
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The atmospheric depth has not increased but the optical depth at CO2 absorption wavelengths has.
Warning! I am using optical depth as absorption depth. Some specialities use a reversed definition of optical depth
O H Dahlsveen challenges
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I feel certain that if a “Greenhouse effect” natural, forced or otherwise exists then there MUST be somebody out there who can explain it, calmly and rationally.
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There aren’t many out there and you would not be able to recognize one if you saw it, because there are too many junk explanations out there as well.
Jeffs is easy, but I voted on the first experiment on the basis of convection still being in play. Maybe this needs clarifying?
The term brightness is too precise here.
That should read IMprecise.
alcuin says:
January 1, 2011 at 9:47 pm
Sorry, – I meant, why in this case? – the planets are supposed to be identical in all aspects apart from the Co2 conc. This is presented as an ‘ideal’ situation with all other equal properties, so I am assuming the atmospheric ‘envelope’ is the same (density, mass, depth, etc).
40 years of climate science and there is a debate on this simple question?! Even amongst skeptics? I wonder how they would fare over at RC.
I hope we get more of this – we need to finally get the basics right.
Here is my shot at the planet question:
The image of the “warmer” (right) planet appears larger but dimmer. The overall received radiation (brightness) from the two planets nevertheless is identical.
My line of thought: The image of the “warmer” planet gets blurred, because more radiation comes from the absorption bands of CO2 in the (higher) atmosphere and less directly from the surface. If you would use a filter for just the CO2-bandwith, the “cooler” planet would look smaller but brighter while the “warmer” planet would look larger but dimmer.
Firstly Jeff, I strongly disagree wirh the diagram you show from Wikipedia. This shows 195 watts of emission to space from the atmosphere with about 40 watts/sqM from the surface. Have a look at some of the IRIS data from the Nimbus satellite, this shows the emission versus wavenumber (reciprocal of wavelength) as seen from space. These plots are normally shown with Plank black body curve overlays corresponding to different temperatures. The apparent black body temperature of the emission shows where the emission is coming from since it specifies the apparent temperature of the source. What these curves show exceptionally clearly is that the emission to space in the atmospheric window is far higher than 40 watts. Further, the atmospheric emission to space (at the GHG wavelengths) occurs from a region at about 220K (ie: the tropopause) and this is far to cold to emit 195 watts/sqM even if it was an iddeal black body emitting at all wavelengths. In fact the surface emission is more like 170 watts/sqM with the tropopause only emitting about 75 watts/sqM most of which comes from absorption of solar near infra red absorption by water vapour high up in the atmopshere (ie: near the tropopause).
With regard to your second thought experiment the answer is quite clear. You specified the two planets to both have stable temperatures which means they are both in thermal equilibrium. Thermal equilibrium means that energy in equals energy out so both will radiate exactly the same amount of energy HOWEVER, there is a difference. The warmer planet will be radiating slightly more energy per micron over a slightly smaller range of wavelengths. Put another way, the higher green house gas concentrations slightly widens the GHG absorption bands leaving a slightly smaller range of wavelengths over which the planet surface can radiate without impediment. If it radiates the same total amount of energy the energy density at those wavelegnths where it can radiate increases slightly.
In fact even this is a slight simplification. At the GHG wavelengths radiation is not entirely blocked. Surface radiation at those wavelengths is absorbed and replaced by radiation from the tropopause. Since the tropopause is much colder than the surface (except at the poles) the energy radiated at the GHG wavelegnths is considerably reduced. Non the less the overall impact is the same, slightly higher radiation at the non GHG wavelengths coupled with slightly wider GHG absorption bands.
This is the problem with so-called thought experiments: they try to substitute complex intertwined non-linear systems with simplistic ideas, loosing physical reality on the path.
In a word (or two), they are meaningless models, not experiments at all.
I noticed some comments referring to equivalent radiation altitude and the question as to whether it gets higher or lower as GHG concentration rises. The reason for the confusion is because the concept of an equivalent radiation altitude is nonsense and leads to paradoxes. The total energy radiated to space is the sum of the radiation at each wavelength. But the radiation at each wavelength can and does vary widely with wavelength. In the atmospheric window the surface radiates directly to space. At the GHG wavelengths surface energy is absorbed very rapidly by the atmopshere. Radiation at these wavelengths comes from the last 1 optical depth of the atmosphere. But one should remember that the total density of the atmospheric column is extremely high at the GHG wavelengths. For example at present CO2 concentrations at 14 microns the total optical depth is about 3000 abs. Thus only the top 1/3000 of the CO2 column can radiate to space. Where is this? For reasons I wont go into now (its a bit lengthy but simply look at the Nimbus experimental data for confirmation) its roughly at the tropopause or lower stratosphere.
There is no wavelength where the middle tropopause (the level of the so called equivalent radiation altitude) can radiate to space. No radiation emanates to space from the so called equivalent radiation altitude. What can change is not the radiation altitude but rather the proportion which comes from the surface and the proportion coming from the tropopause.
Here is another thought experiment:
I have three electric night storage heaters. The power setting is the same on each so the amount of energy consumed during the ON period ( from say 12:30 am to 7:30 am) is exactly the same for each heater. The observation period is 24 hours.
A) Is left as is.
B) I have removed all the thermal bricks. (No atmosphere)
C) I have added 4 extra bricks from B. (Increase in absorption-emmission of atmosphere)
Questions:
1. Does C) become warmer than A) because of the extra 4 bricks?
No, it just takes slightly longer to warm up and cool down in any 24 hour period. To increase the temperature you would need to increase the energy input.
2. Does B) become cooler than A) because of the lack of bricks?
No, it just becomes unbearably hot very quickly during the ON period and loses all of its heat in the first few minutes of the OFF period. (Like the Moon)
Conclusion:
The so called “Greenhouse Effect” is nothing more than a cheep parlour trick which utilises the complexities of atmospherics and fluid dynamics and most peoples inability to simultaneously consider all of the variables.
The TOA is the true heat emitting surface of the Earth, we are subsurface dwellers. Changing the composition of the atmosphere by trace amounts can do nothing to the over all temperature of the atmosphere. The true determining factor of atmospheric temperature is the mean temperature at equilibrium of its main components O2 and N2 @ur momisugly 99%.
Adding a trace amount of a substance which can absorb and emit a narrow band of radiation so weak as to be practically undetectable to our human senses, cannot alter the mean temperature of 99% the atmosphere, O2 and N2 at equilibrium.
CO2 does not heat the atmosphere any more than water vapour does. For all the warming effects of water vapour there are equal numbers of cooling effects. The same is the case for CO2, That which absorbs, equally emits, AKA Kirchhoffs Law.
E in = E out.
The Sun drives the global mean temperature and therefore the climate on Earth.
http://www.spinonthat.com/CO2_files/The_Diurnal_Bulge_and_the_Fallacies_of_the_Greenhouse_Effect.html
Jeremy says:
January 1, 2011 at 5:01 pm
Assuming the glass is totally transparent (to all EM radiation) then the trapped air in the box and by consequence the plant itself will be at a higher temperature due to no cooling from convection.
Therefore the plant in the box must be hotter and therefore it must emit more blackbody thermal radiation which will be seen by the camera.
Ah, but would it actually be “seen by the camera?” Imagine two people sleeping on separate beds in a room. One is lying on top of the bed, the other is under a duvet. The one on top is comfortable and cool, the one under the duvet is sweating from the warmth. But which one do you think would look brightest to a heat-sensitive camera looking down on the two beds?
Answer: the head of the person who is sweating under the duvet.
Think about it. {/joking}