Guest Post by Ira Glickstein
This series began with a mechanical analogy for the Atmospheric “Greenhouse Effect” and progressed a bit more deeply into Atmospheric Windows and Emission Spectra. In this posting, we consider the interaction between air molecules, including Nitrogen (N2), Oxygen (O2), Water Vapor (H2O) and Carbon Dioxide (CO2), with Photons of various wavelengths. This may help us visualize how energy, in the form of Photons radiated by the Sun and the Surface of the Earth, is absorbed and re-emited by Atmospheric molecules. 
DESCRIPTION OF THE GRAPHIC
The animated graphic has eight frames, as indicated by the counter in the lower right corner. Molecules are symbolized by letter pairs or triplets and Photons by ovals and arrows. The view is of a small portion of the cloud-free Atmosphere. (Thanks to WUWT commenter davidmhoffer for some of the ideas incorporated in this graphic.)
- During the daytime, Solar energy enters the Atmosphere in the form of Photons at wavelengths from about 0.1μ (micron – millionth of a meter) to 4μ, which is called “shortwave” radiation and is represented as ~1/2μ and symbolized as orange ovals. Most of this energy gets a free pass through the cloud-free Atmosphere. It continues down to the Surface of the Earth where some is reflected back by light areas (not shown in the animation) and where most is absorbed and warms the Surface.
- Since Earth’s temperature is well above absolute zero, both day and night, the Surface radiates Photons in all directions with the energy distributed approximately according to a “blackbody” at a given temperature. This energy is in the form of Photons at wavelengths from about 4μ to 50μ, which is called “longwave” radiation and is represented as ~7μ, ~10μ, and ~15μ and symbolized as violet, light blue, and purple ovals, respectively. The primary “greenhouse” gases (GHG) are Water Vapor (H2O) and Carbon Dioxide (CO2). The ~7μ Photon is absorbed by an H2O molecule because Water Vapor has an absorption peak in that region, the ~10μ Photon gets a free pass because neither H2O nor CO2 absorb strongly in that region, and one of the 15μ Photons gets absorbed by an H2O molecule while the other gets absorbed by a CO2 molecule because these gases have absorption peaks in that region.
- The absorbed Photons raise the energy level of their respective molecules (symbolized by red outlines).
- The energized molecules re-emit the Photons in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
- This frame and the next two illustrate another way Photons are emitted, namely due to collisions between energized GHG molecules and other air molecules. As in frame (2) the Surface radiates Photons in all directions and various wavelengths.
- The Photons cause the GHG molecules to become energized and they speed up and collide with other gas molecules, energizing them. NOTE: In a gas, the molecules are in constant motion, moving in random directions at different speeds, colliding and bouncing off one another, etc. Indeed the “temperature” of a gas is something like the average speed of the molecules. In this animation, the gas molecules are fixed in position because it would be too confusing if they were all shown moving and because the speed of the Photons is so much greater than the speed of the molecules that they hardly move in the time indicated.
- The energized air molecules emit radiation at various wavelengths and in random directions, some upwards, some downwards, and some sideways. Some of the re-emitted Photons make their way out to Space and their energy is lost there, others back down to the Surface where their energy is absorbed, further heating the Earth, and others travel through the Atmosphere for a random distance until they encounter another GHG molecule.
- Having emitted the energy, the molecules cool down.
DISCUSSION
As in the other postings in this series, only radiation effects are considered because they are the key to understanding the Atmospheric “Greenhouse Effect”. I recognize that other effects are as important, and perhaps more so, in the overall heat balance of the Earth. These include clouds which reflect much of the Sun’s radiation back out to Space, and which, due to negative feedback, counteract Global Warming. Other effects include convection (wind, thunderstorms, …), precipitation (rain, snow) and conduction that are responsible for transferring energy from the Surface to the Atmosphere. It is also important to note that the Atmospheric “Greenhouse Effect” and a physical greenhouse are similar in that they both limit the rate of thermal energy flowing out of the system, but the mechanisms by which heat is retained are different. A greenhouse works primarily by preventing absorbed heat from leaving the structure through convection, i.e. sensible heat transport. The greenhouse effect heats the earth because greenhouse gases absorb outgoing radiative energy and re-emit some of it back towards earth.
That said, how does this visualization help us understand the issue of “CO2 sensitivity” which is the additional warming of the Earth Surface due to an increase in atmospheric CO2? Well, given a greater density of CO2 (and H2O) molecules in the air, there is a greater chance that a given photon will get absorbed. Stated differently, a given photon will travel a shorter distance, on average, before being absorbed by a GHG molecule and be re-emitted in a random direction, including downwards towards the Surface. That will result in more energy being recycled back to the Surface, increasing average temperatures a bit.
Bryan says:
That is not what cal said. What he said is that the radiation that successfully escapes to space is mainly emitted from close to or a little below the tropopause (because radiation emitted from a lower layer will likely be absorbed again before escaping and higher layers have less emission). The greenhouse effect is not confined above a certain layer; it comes about from a holistic understanding of radiative transfer within an atmosphere where temperature decreases with height.
I don’t even know what that sentence means. In one sense it is exactly backwards…It is because CO2 absorbs so strongly at atmospheric temperatures and pressures that radiation emitted from near the surface is unlikely to escape to space without being absorbed again. However, if you mean that the fact that the temperatures in the troposphere decrease with height is vital in understanding the greenhouse effect, then this is correct. And, that is where jumping from Wood’s experiment to any conclusion about the atmosphere turns out to be wrong.
Within the simple spherical shell models (which consider only radiative effects), one naturally finds that the temperatures of the shells decrease as one goes outward…i.e., the radiative equilibrium condition itself (coupled with the solar radiation mainly being absorbed at the surface) produces such a result. In the real atmosphere, convection prevents the lapse rate from being too steep…but you still get the temperature decreasing as you go up in the troposphere.
And still the examples of short-wave Light energy heating oceans..
AGW can never understand the objections re the Laws of Thermodynamics because AGWScience claims there is a “net” word added to the 2nd Law is because they have to pretend that Solar energy in the short-waves are absorbed by the Earth and so heat it, that the only Thermal IR in the energy budget comes from the thus heated Earth radiating into the atmosphere. They exclude the Thermal IR coming from the Sun that actually, really, heats the Earth, and us, that we can feel, to teach instead that non-thermal photons of light are in effect thermal, are actually felt as thermal. AGWScience has created another physics.
I bring to your attention this old NASA page for children of the classic physics of Light and Heat energies, that they are different. This page is soon to disappear and the clear, straightforward descriptions will go the way of all AGWBSSscience mangling, so that AGWScience can continue to teach that cold molecules radiating back from the atmosphere heat the earth and that a cold interior wall of an igloo can cook a lump of meat if left there for a few hours, and the rest. So for the educational record, (the exclamation marks are in the original, this is for children):
http://science.hq.nasa.gov/kids/imagers/ems/infrared.html
“Infrared light lies between the visible and microwave portions of the electromagnetic spectrum. Infrared light has a range of wavelengths, just like visible light has wavelengths that range from red light to violet. “Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.
Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared. The temperature-sensitive nerve endings in our skin can detect the difference between inside body temperature and outside skin temperature.
Infrared light is even used to heat food sometimes – special lamps that emit thermal infrared waves are often used in fast food restaurants!
Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.”
Because AGW doesn’t teach this, is why there are so many here who can’t understand convection or explanations by others here who do understand the real Thermodynamics Laws, for example:
“If it were possible to get energy to flow from cold to hot then we would have all our energy requirements solved. We would have a perpetual motion machine. This is impossible.” (John Marshall March 29 @ur momisugly1:52 am)
Contrast example: “There is always a net energy flow from a warmer body to a cooler body in accordance with the Laws of Thermodynamics but in fact both bodies still radiate towards each other.
A cooler body doesn’t stop radiating just because it is in the presence of a warmer body.
It is the net rate of energy transfer between the two that changes with no need for the cooler body to effect any direct warming of the warmer body.
the suggestions that the greenhouse effect somehow offends the Laws of Thermodynamics is a non starter and a hindrance to scepticism of the theory of AGW.” (Stephen Wilde March 29 @ur momisugly 3:45 am)
The AGWScience energy budget, as Ira et al teach it, says that it is Solar energies which heat the Earth, that is, Visible Light and the two short waves either side of UV and Near Infrared. These are not hot. We cannot feel them. No matter how far Blue Light, for example, penetrates into the ocean, it will not heat it. UV may burn surfaces, as artifically intensified Blue light can also burn, but these do not raise the temperature of matter the way that Thermal IR does.
Thermal IR heating the Earth is excluded from the AGW Energy Budget.
Why?
Where is the proof that Solar Light energies can heat the Earth, raise the temperature of the Earth?
It’s Thermal IR that can cook food, Near Infrared is not hot.
wayne says:
March 31, 2011 at 1:01 pm
“If there was zero radiation from a pure nitrogen atmosphere, heck, lets make it pure argon as dense as ours, no oceans, no clouds, would it be cold or hot? What do you think?”
I’m still trying to understand what you mean. If there is zero radiation across an entire atmosphere, top to bottom, then it is extremely cold – absolute zero. So it wouldn’t even be an atmosphere, but a solid. I don’t think that is what you mean.
If there is zero radiation in a particular wavelength, there are multiple explanations. The most obvious is that your atmospheric molecules don’t emit at the target wavelength. Or the atmosphere might be the wrong temperature, and it would radiate the target wavelength if it was hotter/colder.
You also specify “as dense as ours”, and in a previous comment you specified “same pressure at sea level”. Since the atmosphere will approximately follow ideal gas laws, you have defined a priori the temperature of the atmosphere. Instead of asking a question, you have provided the answer via the assumptions.
The question is “Given an atmosphere of equal mass to the current atmosphere, but consisting entirely of nitrogen (or argon), what would the density and pressure be at sea level? (and thus, the temperature)” Note that in this question “at sea level” refers to a height of a meter or two above the ground. The average pressure exerted by the atmosphere directly upon the ground’s surface at sea level will be unchanged, assuming the total mass of the atmosphere is the same.
My hypothesis is that, given an equal mass of just nitrogen (or argon), the atmosphere will be denser at sea level because it will be colder. The atmospheric pressure at sea level (let’s say 2 meters above surface) will be lower, because the percentage mass of the atmosphere located below knee level will have gone up. The height of the top-of-atmosphere will also be lower.
What I can’t hypothesize is how much denser/colder a pure nitrogen or argon gas atmosphere would be (at sea level) compared to the one we’ve got. And note that the radiation experienced by a person, such as us, standing in direct sunlight would skyrocket during the day and plummet during the night. The surface of the moon, with practically no atmosphere, hits over 115 Celsius at midday and drops below -170 Celsius at night (at it’s equator). Our atmosphere both reflects/absorbs harmful radiation during the day and radiates helpful energy down to the surface at night.
Thanks Steve, interesting viewpoint. Seems almost every parameter of that is known to a close estimate. That just might deserve a quick simulation of such a scenario to see what that would show.
You and I agree perfectly on the soaring temperatures during the day and frigid nights but I was including much weight on conduction and convection as the prime heater, so much that night would never be able to catch up with limited convection.
The reason I perked up on that is it’s a perfect example to see what a lapse rate would be in such a scenario. Thanks for your viewpoint again. Very interesting.
The Earth’s atmosphere is greenhouse regulated by the water cycle, it would be 67°C without water.
Not to mention thirsty :^)
Tim
Nasif’s experiments and calculations and those of many others that he references show very much that he is correct. Real life observations would also show that he’s correct in the negligible contribution of CO2. I will accept any other views unreservedly if you would post on that site and show him where he is wrong. You can show I’m wrong easily but that’s different, it doesn’t make him wrong.
I also believe his point that most scientists, even the best assume that the earth and CO2 molecules radiate as black bodies with an emissivity of 1.0
Joel Shore says:
March 30, 2011 at 5:47 pm
Sort of true…although really you are forgetting a couple of things. One is that I believe the albedo of the earth in the absence of clouds is still about 8% or so. The second is that such a colder earth would presumably have more ice on it and would therefore have even a higher albedo than this.
At any rate, the full statement is that the atmosphere (including clouds) produces two effects: One is about 33 K of warming due to the greenhouse effect and the other is some cooling due to the albedo of the clouds. So, yes, the net effect of the atmosphere is not a full 33 K of greenhouse warming. However, again, the sort of calculations you discuss are under the assumption of no change in surface albedo [actually, you wrongly assumed 0 albedo]…In reality, if you removed CO2 from the atmosphere, not only would you remove a lot of the water vapor too, you would also increase the surface albedo of the earth.
Joel Shore says:
March 30, 2011 at 5:55 pm
Reed –
On re-reading what you wrote, I realize that you didn’t assume a zero albedo for the earth as I said in my last post but you did assume, as Tim Folkerts points out, that the emissivity is the same in the visible as it is in the far IR…which is indeed not a very good assumption. Emissivities of the earth’s surface in the visible vary widely with the type of surface, whereas emissivities in the far IR tend to be very close to 1 for nearly all surfaces. (I believe I’ve read that the only place that there can be any significant deviations from 1 is in some desert regions.)
Joel, taking your last statement first. If you’re going to use the rule that the rate radiation leaves a surface is proportional to T^4, then the emissivity must be independent of frequency. A T^4 rule with a frequency dependent emissivity may be a good approximation, but it is NOT theoretically correct. Thus, I contend that the use of a T^4 law implicitly implies a frequency independent emissivity.
If the emissivity is independent of frequency and an inert surface is placed in a cavity whose walls are at a common temperature, the steady-state temperature of the surface will be equal to that temperature. If you assume the surface is at a single temperature and you use Planck’s graybody radiation law with constant emissivity to compute that temperature, I believe you will find that one minus the albedo must be equal to the emissivity. Thus, if you’re going to use a T^4 radiation rule to compute the rate energy leaves a surface, I believe you are implicitly assuming (a) both the emissivity and the albedo will be frequency independent, and (b) the emissivity will be equal to one minus the albedo.
I agree, real world materials have emissivities/albedoes that are a function of frequency. And if you want to incorporate them into your temperature analysis, that’s fine. But don’t use the T^4 radiation law without clearly explaining why it is a good approximation. In addition, the rate of energy transfer via conduction is not proportional to T^4, but more nearly proportional to the difference in temperature between the two end surfaces of the conductive object. Since energy is transfered from the Earth surface to the Earth atmosphere via conduction and the Earth’s atmosphere radiates, don’t forget to include these effects as well. Then there’s convection, and evaporation, and … so on. Finally, I don’t know how to apply the T^4 law to radiation from a gas. The T^4 law is used in conjunction with a surface area, and a gas doesn’t have a definable surface. Bottom line, the complete computation of the atmosphere’s effect on Earth surface temperature is waaaaaay beyond my capabilities. I’m not against using simple models, but I want the use of a model to be internally consistent. In particular, I don’t want someone to use both a T^4 emission dependence and a frequency dependent emissivity; and I definitely don’t want someone to disregard atmospheric cooling effecs when estimating the effect an atmosphere has on surface temperature .
Now to your first point. OK, I forgot to include the fact that a colder Earth would have significantly more ice–thereby increasing the Earth’s albedo. However, (a) I believe my omission is insignificant compared to Wikipedia’s omission that when computing the effect of the atmosphere on Earth surface temperature, Wikipedia completely ignores the fact that it’s the atmosphere that reflects most incoming solar radiation. Wikipedia is eager to give “credit” to atmospheric effects that warm the Earth’s surface, while completely ignoring atmospheric effects that cool the surface of the Earth.
But even more important than that, (b) the increase in albedo that would accrue with more ice implies a corresponding decrease in emissivity. If the ice doesn’t exist everywhere, it will predominately exist in the polar regions (and I have to believe even Wikipedia would concede that point). As such, the effect of more ice on the surface temperature of a conductive Earth is to raise the Earth’s temperature not lower it. The reason for this is as follows. A one square meter of Earth surface area absorbs energy at a rate that is proportional to the size of the area (one square meter) and the COSINE of the angle between the surface normal and the direction of incoming radiation; but one square meter of Earth surface radiates energy at a rate that is proportional to the area (one square meter) without being modified (attenuated) by a COSINE function. At the poles, the angle between the incoming radiation and the normal to the Earth surface is large. As such, a given increase in albedo will have a smaller effect on absorbed energy than it will have on radiated energy. In particular, the conversion of one square meter of polar sea water to one square meter of ice will increase the albedo of that square meter. However, if that albedo increase causes one Watt less of energy to be absorbed, then for a steady-state Earth the reduction in the rate at which energy is emitted from that one square meter will be greater than one Watt. To counter this net negative change in the energy rate, the surface temperature will have to increase, not decrease.
I have generated a PDF file with equations that show this to be the case. Specifically, I segmented the Earth into three regions (two polar regions of equal size, and the remainder of the Earth surface, which I call the equatorial region). I assume the albedo of the polar regions is higher than the albedo of the equatorial region–as would be the case for ice in the polar regions and sea water in the equatorial region. For a conductive Earth, the Earth surface temperature will be uniform. This means that the Earth surface will be all ice or all sea water. However, we can hypothesize an Earth whose polar and equatorial regions are comprised of differenct material where the Polar regions have higher albedos than the equatorial region. Since I want to use a T^4 law for radiation, I require (a) that both albedos be frequency independent, and (b) the both emissivities be one minus their corresponding albedo. For this Earth model we ask: “What will be the surface temperature of such an Earth compared to the surface temperature of an Earth with (a) the polar albedo everywhere, and (b) the equatorial albedo everywhere?” What’s interesting is that as long as the equatorial region albedo is not unity, independent of the equatorial region albedo, provided the polar region albedo is greater than the equatorial region albedo, the Earth’s surface temperature with mixed albedos will be higher than a uniform temperature Earth with a single albedo (either the polar region albedo, the equatorial region albedo or some other albedo). If you’re interested, I’ll Email the PDF file to you. You probably don’t have my Email address, and I’m reluctant to post it in an open forum. However, you can either post your Email address in a comment; or like me, if that’s not acceptable to you , you can tell Anthony that I give him permission to forward my E-mail address to you. You can send me an E-mail, and I’d be happy to respond.
Ira – 🙂
There’s an English expression, don’t know if it’s both sides of the pond, to get something ar*se about t*t. AGW has this wonderfully absurd comic book science which does that at every point of argument for its survival, it can best be seen as going through the looking glass with Alice as it creates impossible scenarios for our side of the mirror physical properties. Because, for this example, “greenhouse” means only warming in AGW the key role of the Water Cycle in our real greenhouse atmosphere of cooling has to be ditched and there are two outcomes of this.
The normal physical well known real science facts are no longer understood because not taught by AGW and the greatly contorted explanations devised in an attempt to justify that greenhouse gases only heat produce people who believe impossible things are real and have no concept of actual physical reality.
So, we end up with zilch AGW understanding of the Law of Thermodynamics and in its place a contorted explanation for the unneccessarily added word “net” in which is proposed non-thermal energies are thermal. As non-thermal Light energies are not radiating heat or creating heat capable of raising temperature (without burning surface, as does UV), then the “net” is irrelevant.
To the extent that you et al end up believing that Solar energy, non-thermal Light, is thermal, you can believe impossible things like the cold insides of igloos radiate back energy capable of cooking lumps of meat, that back radiation from a source heated by the transfer of energy from a hotter source raises the temperature of the hotter source, etc. For you, generic, the atmosphere has no depth or weight or volume but is instead empty space with molecules zipping around in Brownian motion diffusing as if an imaginary ideal gas quite regardless that these molecules have non-ideal gas weight and volume and can’t do this for real. You miss all the funnies like filling one’s attic with CO2 to warm one’s house, because you, generic, have no sense at all of the real physical world around you.
The Water Cycle cools the Earth by convection in taking up heat from earth via water vapour which is lighter than air and condensing out at higher colder levels of the atmosphere releasing heat away from the Earth.
A hot little Carbon Dioxide molecule on the ground swept up in the movement of a hot volume of air, which is wind, will lose its heat as it rises, in this also cooling the Earth heated by the thermal energies from the Sun which are not Solar but Thermal IR. With a heat capacity of even less than Nitrogen and Oxygen, CO2 as quickly gives up its heat as it gets it, practically instantly. A cold CO2 molecule can not heat warmer matter around it because it is itself receiving heat and any heat it has from being in warmer matter, hot air, will be released the moment that air gets cold. As cold air, the volume of fluid gas, is not radiating energy to heat a warmer object (we can tell etc. see NASA description because we have an inbuilt ability to tell the difference between outside and inside body temperature), neither is cold CO2.
So it’s not enough to say that “CO2 re-radiates in every direction, half downwards” and such, if that radiation is non-thermal it can do nothing to raise the temperature of the Earth.
So, show me proof that Visible Light from the Sun is absorbed by the Earth and raises its temperature.
In the real world in real science and not make believe through the looking glass with Alice AGW, we know your claim is nonsense, because we can use real science understanding in this our real physical world: http://www.ehow.com/facts_7868896_fluorescent-light-good-growing.html
Benefits of Fluorescent bulbs – energy efficient and do not produce much heat, can place bulbs close to plants without risking burn damage, why? Because they produce mainly blue light. (Plants also need red light for growing on).
Here: http://growlightuk.com/tabid/58/ProductID/17/Default.aspx?gclid=CJ_034CN-6cCFUEa4QodTmjiqA
Oh what’s up? This lamp produces visible light of blue and red with virtually no heat. Does it somehow extract the heat these lights produce in your claim that Solar light is thermal or in your Solar budget heating the Earth are blue and red lights excluded because these colours do not produce heat? Are these bulbs energy efficient because it takes less energy to produce visible light?
Come on Ira, everyone, let’s nail this once and for all. What is the physical reality here?
Reed Coray says: March 31, 2011 at 10:15 pm
“I require (a) that both albedos be frequency independent, and (b) the both emissivities be one minus their corresponding albedo. “
Your whole set of calculations falls apart with this assumption. Snow has an albedo of close to 1 (for visible light), but also an emissivity of close to 1 (for IR).
Google snow emissivity and the first two hits say:
“Water, ice, and snow generally have a high emissivity, 0.94 to 0.99, across the thermal infrared region. Snow is unusual in that it has a high reflectance in the solar (visible) region where most of the downwelling energy is during the day, and a very high emissivity in the thermal region.”
http://www.comp.glam.ac.uk/pages/staff/pplassma/MedImaging/PROJECTS/IR/CAMTEST/Icewater.htm
“Snow emissivity is in the range 0.96-1.00 for all grain sizes and viewing angles. ”
http://astrogeology.usgs.gov/Projects/PlanetaryIcesWorkshop/abstracts/swarren.html
Reed Coray says:
Not really. The conditions under which the T^4 rule will hold to a good approximation for not-too-large fractional changes in T (remembering that T is an absolute temperature scale!) are much weaker than the conditions under which you can assume that the emissivity in the far IR is equal to one minus the albedo in the visible. In particular, the T^4 rule (for relatively small changes in T) requires just that the emissivity doesn’t vary too rapidly with wavelength over the wavelength regime for which emission in that temperature regime is significant. Assuming that emissivity in the far IR is equal to one minus albedo in the visible assumes that the emissivity doesn’t change much when the wavelength varies by an order of magnitude or more!
I don’t think Wikipedia claims to be computing the total effect of the atmosphere on the Earth’s surface temperature. The computation is of the total greenhouse effect due to the atmosphere. I.e., we can divide the atmosphere’s effects into different contributions, rather than just looking at its net effect.
Basically, everything beyond this point in your post relies on an assumption that the emissivity of ice is significantly different from 1 in the far infrared (and, in particular, that the reduction in emissivity is comparable to the albedo of ice in the visible). Here is the data of infrared emissivities showing that this assumption is incorrect: http://www.comp.glam.ac.uk/pages/staff/pplassma/MedImaging/PROJECTS/IR/CAMTEST/Icewater.htm (For comparison, here are some albedos for ice and snow: http://www.arcticice.org/albedos.htm )
“So, show me proof that Visible Light from the Sun is absorbed by the Earth and raises its temperature.”
Since you seem to not trust the most basic theories of light like E = hf, then lets just use a simple experiment.
Dark clothing will make you warmer in sunlight than light clothing.
–> dark clothing absorbs more energy from the sunlight
The emissivity of cloth doesn’t change much with color.
–> any color of clothing will absorb similar amounts of IR
Sunlight consists almost entirely of visible light & IR.
–> The visible light must be what makes the difference.
“Oh what’s up? This lamp produces visible light of blue and red with virtually no heat.”
My gracious! These are LED lights! They use a completely different process to produce light! Since they are not hot glowing objects, there is absolutely no reason to expect anything closer to a black body curve for the light produced!
Joel Shore said:
I and many others have endeavored to explain it to you. If you still cannot understand it, I don’t think it is all of our faults.
Can you give a quantifiable, verifiable explanation of it then, please?
martin mason:
You believe his point based on what exactly? How extensively have you studied the literature or even a basic textbook discussing radiative transfer in the atmosphere?
In fact, let’s test if that belief even makes sense. Can you define for me what it means to assume that CO2 molecules radiate as black bodies with an emissivity of 1.0 because frankly I am a bit confused on the concept being that emissivities is a macroscopically-defined quantity that doesn’t, as far as I know, even make sense to talk about on the scale of individual molecules. I think you have to talk about things like the optical depth of a certain thickness of atmosphere, but hey, if you know differently I’d be glad to hear about it!
Myrrh says:
April 1, 2011 at 3:43 am
Come on Ira, everyone, let’s nail this once and for all. What is the physical reality here?
Certainly not the crap you keep churning out!
Anthony, this is the top Science blog?
And we have to keep reading the nonsense this guy produces!
The loss of temperature by an ascending gas volume is adiabatic i.e. no loss of heat!
CO2 has a higher molar heat capacity than N2 or O2, ~28 vs ~20.
Not to mention all his cockamamie ideas about ‘thermal’ vs ‘non-thermal’ radiation which have been rebutted here ad nauseam and his belief that the gas laws don’t apply!
Joel Shore says:
March 31, 2011 at 1:45 pm
Yes, when you pressurize gas, it warms. However, if you pressurized some gas and then let it sit in equilibrium with its surroundings for billions of years, it will equilibrate with its surroundings. The earth’s atmosphere is not undergoing continual gravitational collapse.
The atmosphere is constantly receiving energy from the surface via conduction, convection and latent heat via the phase changes of water. It hasn’t been in equilibrium with it’s surroundings for billions of years.
To use an analogy the contents of a compressed gas cylinder will stay hot if you light a fire under the cylinder.
wayne says:
March 31, 2011 at 1:01 pm
But I still would ask you on that hypothetical situation, if I could: If there was zero radiation from a pure nitrogen atmosphere, heck, lets make it pure argon as dense as ours, no oceans, no clouds, would it be cold or hot? What do you think?
It would depend only on the reflectivity, absorptivity and emissivity of the surface and the distance from the sun.
Myrrh;
Come on Ira, everyone, let’s nail this once and for all. What is the physical reality here?>>>
Phil;
Anthony, this is the top Science blog?
And we have to keep reading the nonsense this guy produces!>>>
Sad is it not? The price of a forum in which contrary opinions are not suppressed is that people who don’t know what they don’t know get to distract and confuse everyone else with mind boggling mirepresentations of science. Sadder still, it is a mirror unto the real world where the contrary opinions have become the science and the scientists are accused of misrepresentation.
Any sufficiently advanced magic is indistinguishable from science.
Max Hugoson says:
March 29, 2011 at 6:26 pm
There is a distribution of energies of the ensemble of molecules. There is EXCHANGE. Therefore, collisions between O2 and N2 molecules on the “high end” of the distributed energy, LOSE that translational kinetic energy to CO2 and H2O molecules.
WHICH, can…with significant probability, re-emit an IR photon.
Yes but that probability is lower if the energy is acquired collisionally than if acquired radiatively. When excited radiatively the appropriate vibrational energy level is filled and is then capable of emitting however collisions reduce the lifetime of the state so that emission has a low probability. When excited collisionally that energy is unlikely to solely excite the vibrational level in fact it’s more likely to excite the translational energy of the CO2 molecule which has zero probability of emitting a photon.
martin mason says:
March 31, 2011 at 8:31 pm
Tim
Nasif’s experiments and calculations and those of many others that he references show very much that he is correct. Real life observations would also show that he’s correct in the negligible contribution of CO2. I will accept any other views unreservedly if you would post on that site and show him where he is wrong. You can show I’m wrong easily but that’s different, it doesn’t make him wrong.
You can’t show Nasif he’s wrong, the reason he’s banned on here as I recall is because of his abusive attacks on other posters, Leif Svalgard for example.
I also believe his point that most scientists, even the best assume that the earth and CO2 molecules radiate as black bodies with an emissivity of 1.0
Believe it if you like but any physical scientist who assumes that gaseous CO2 molecules radiate as a black body with an emissivity of 1.o should have their degree rescinded.
bananabender says:
Okay, so what source of energy do you propose is being ignored when it is concluded vi radiative balance that the earth’s surface temperature in the absence of an IR-absorbing atmosphere (with the same total albedo as present) would be approximately 255 K?
Tim Folkerts says:
April 1, 2011 at 3:59 am
Reed Coray says: March 31, 2011 at 10:15 pm
“I require (a) that both albedos be frequency independent, and (b) the both emissivities be one minus their corresponding albedo. “
Your whole set of calculations falls apart with this assumption. Snow has an albedo of close to 1 (for visible light), but also an emissivity of close to 1 (for IR).
Google snow emissivity and the first two hits say:
“Water, ice, and snow generally have a high emissivity, 0.94 to 0.99, across the thermal infrared region. Snow is unusual in that it has a high reflectance in the solar (visible) region where most of the downwelling energy is during the day, and a very high emissivity in the thermal region.”
http://www.comp.glam.ac.uk/pages/staff/pplassma/MedImaging/PROJECTS/IR/CAMTEST/Icewater.htm
“Snow emissivity is in the range 0.96-1.00 for all grain sizes and viewing angles. ”
http://astrogeology.usgs.gov/Projects/PlanetaryIcesWorkshop/abstracts/swarren.html
Tim,
My “whole set of computations” does NOT fall apart with these assumptions. Like Wikipedia, when computing the surface temperature of the Earth without an atmosphere, I assume a model. I admit my model does not match the real world–no model does; but I believe my assumptions are internally consistent and don’t violate any laws of thermodynamics. Do you believe (a) my model is internally inconsistent, and/or (b) violates a law of thermodynamics?
Relative to the real world, I can make the argument that Joel’s and I guess your argument also “falls apart”. Specifically, how do you justify the T^4 rule? How do you justify using cavity radiation into a medium containing matter (i.e., an atmosphere)? How do you justify ignoring conduction and convection? How do you justify ignoring energy transfer via state change (water to vapor), movement of that vapor to various altitudes, radiation from that vapor, etc?
Let me ask you. (1) Does the T^4 radiation law require an emissivity that is independent of frequency? If not, what is the frequency dependence that produces a T^4 law? (2) Will an inert surface, independent of the material that comprises the surface, when placed in the vacuum of a cavity whose walls are at a uniform temperature T, eventually attain the temperature T? (3) If you apply Planck’s graybody radiation law to that surface, will the rate of energy reception equal the rate of energy emission for any temperature other than T?
But all my arguments that question the applicability of Wikipedia’s model pale when compared to my criticism of Wikipedia’s logic. To compute the effects of the atmosphere on Earth surface temperature, Wikipedia chose to assume for emission purposes a blackbody Earth, and for reception purposes assume an albedo that is a composite of atmosphere and surface. Wikipedia then attributes the difference solely to the atmosphere. I can’t buy this. For example, suppose the Earth’s surface was lampblack (not a perfect absorber, but close) with a highly conductive (thermally) interior. For such an Earth, without an atmosphere the surface temperature, T1, would come close to that of a blackbody in space, which I believe is approximately 278 K. To this surface add an atmosphere of clouds. The clouds reflect 30% of the incoming solar radiation. The lampblack surface will now “see” less solar energy. We compute the steady-state temperature, T2, of a lampblack Earth using the reduced radiation input value. We measure the surface temperature T3 in the presence of the clouds. We then make the claim that the atmosphere is responsible for increasing the surface temperature by the difference between of the T3 and T2. This is nonsense. Without the atmosphere the surface temperature was T1. With the atmosphere the surface temperature is T3. By any reasonable way of thinking, the atmospheric effect on the surface temperature is the difference between T3 and T1. How you come up with a model for T1 is open to disucssion. But whatever model you use, it should NOT contain an atmosphere.
Joel Shore says:
April 1, 2011 at 4:07 am
I don’t think Wikipedia claims to be computing the total effect of the atmosphere on the Earth’s surface temperature. The computation is of the total greenhouse effect due to the atmosphere. I.e., we can divide the atmosphere’s effects into different contributions, rather than just looking at its net effect.
I insert here a quote from Dr. Glickstein who quoted Wikipedia:
Ira Glickstein, PhD says:
March 30, 2011 at 8:13 am
from Wikipedia
If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C. However, since the Earth reflects about 30% (or 28%) of the incoming sunlight, the planet’s effective temperature (the temperature of a blackbody that would emit the same amount of radiation) is about −18 or −19 °C, about 33°C below the actual surface temperature of about 14 °C or 15 °C. The mechanism that produces this difference between the actual surface temperature and the effective temperature is due to the atmosphere and is known as the greenhouse effect.
I don’t know how you interpret those words, but I interpret them as follows: (1) the atmosphere is responsible for the difference between the actual surface temperature and the “effective” surface temperature. (2) The “effective” surface temperature is based on the temperature of a ideally thermally conductive blackbody radiator in radiation rate equilibrium with the solar power density at the the top of the Earth’s atmosphere attenuated by 30% (28%). And (3) the name given to this effect (temperature difference) is the “greenhouse effect.” Based on Wikipedia’s words, I don’t see how you can argue that “…we can divide the atmosphere’s effects into different contributions, rather than just looking at its net effect“, and claim that Wikipedia meant to assign the name the “greenhouse effect” to one of those “contributions”.
BigWaveDave says:
Well, I don’t know what you are asking for that we haven’t already given or isn’t available in the literature. The atmospheric greenhouse effect can be thought of at a variety of different levels of detail, from simple toy “shell” models to detailed radiative-convective calculations. Ira Glickstein, Willis Eschenbach, Roy Spencer, and others have spent whole posts here explaining the effect, and even though they and I agree on rather little in regards to AGW, we are all in agreement on this aspect of the science.
What about it exactly is tripping you up? Or maybe, to put it a different way, I said:
Your response was that this violates the 2nd Law of Thermodynamics. So, explain to me exactly why you think it does. Do you believe it impossible for a colder object to even emit radiation toward a warmer object? Or, do you believe that it can but the warmer object won’t absorb any of the radiation? Or, do you believe that the warmer object can absorb the radiation but that it won’t cause it to have any additional thermal energy than it would have had otherwise?
I am very confused about exactly what aspect of this process you believe violates the 2nd Law.
Reed Coray says:
Sorry…This is a false equivalence for a couple of reasons. First, all models or theoretical constructs of any kind involve making assumptions and approximations. However, not all such assumptions and approximations are equally good. What Tim and I have shown is that you made an assumption that is clearly not any good at all and your conclusions will quite clearly fail to apply to the real world because of it.
By contrast, I gave you reason to believe that in many cases, it is in fact quite good approximation to assume the T^4 dependence holds. Heck, the dependence of radiative emission on temperature is so strong that at ANY frequency, the emission a that frequency is a monotonically increasing function of temperature. (People often have the picture of a shifting curve of emission power because you often look at normalized emission curves…but, in fact, it is a curve that grows rapidly with temperature that all of it lies above the corresponding curve for any lower temperature.)
The second reason is that many of the issues that you mention are actually addressed by more complete models. So, yes, it is true that very simple models of the greenhouse effect neglect convection, assume simple emissive properties, etc. However, the justification for these assumptions not mattering to the final answer is that the more complex models give the same results at least for the things that we are using the simpler models to illustrate.
That’s why I emphasize the importance of a hierarchy of models: A very complex model without simple models is simply a “black box” that gives very little insight to what the basic physics is. A very simple model without more complex models to back it up causes one to be unsure if the results shown by the model apply to the real world or are just an artifact of the approximations. However, if you take them all together, you can use the more complex models to back up what you find with the simple ones and then use the simple ones to get a simple picture of what is going on.
When I read someone else’s work, I try interpret it in the most charitable way…i.e., the way that actually makes sense (assuming I can figure out a way in which it does make sense…which isn’t true, for example, for the work of Gerlich and Tscheuschner). If one chooses an interpretation that doesn’t make sense and then complains bitterly about it, I don’t really see what that accomplishes. Sure, maybe one could argue that the wording in that Wikipedia article could have been a little clearer. But, I don’t really see the need to make a federal case out of it. One just clarifies what they meant…that they were specifically talking about the magnitude of the atmospheric greenhouse effect, i.e., the magnitude of the effect due to the absorbance of the atmosphere in the far infrared part of the spectrum…and moves on.