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.
Reed,
You certainly are welcome to do your computations and to see the effects of your model. I was perhaps a little harsh to say “it all falls apart”. But when you make an assumption that is almost completely the opposite of reality (assuming emissivity of snow ~0 when indeed is it ~1) then your conclusions will almost certainly be poor.
To answer a few specific questions:
How do you justify using cavity radiation into a medium containing matter (i.e.,
an atmosphere)?
While BB radiation was originally derived using a cavity, it has been shown that the same curve also works very well for other circumstances — ie some surfaces emit practically the same curve from a flat surface ie their emissivity is ~ 1 over the frequencies of interest. Experimentally, much of the surface acts much like a black body for IR wavelengths.
The material into which the BB radiation travels does not matter — only the temperature of he emitting surface, so the properties of the atmosphere are not important here.
This would seem to justify treating the surface as (approximately) a BB wrt IR.
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?
This is a little more subtle.
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?
As long as the emissivity is independent of the frequencies where (a significant amount of) energy is emitted, the T^4 law is (approximately) hold.
For example, this table http://www.monarchserver.com/TableofEmissivity.pdf suggest that type 301 Stainless has an emissivity of 0.27 @ur momisugly 24 C and 0.57 @ur momisugly232 C. This would mean that the IR power emitted by this surface would go up slightly FASTER than T^4 as the temperature rose from 297 K to 505 K.
On the other hand, cuprous oxide decreases in emissivity as temperature increase, so it would be a little less than T^4
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?
Yes. I’ve even seen it. I used to work with furnaces operating at > 1000 C. The uniformly heated walls and the small window in the door very closely approximate a cavity radiator. If you slide a sample in, everything – metal, ceramic, quartz – soon glows the same color.
(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?
No. If the object is hotter than the chamber, is will cool; if the object is cooler it will warm up. Always.
By any reasonable way of thinking, the atmospheric effect on the surface temperature is the difference between T3 and T1.
I agree that there are many ways to interpret the effect of the atmosphere. And I agree with your calculations:
* a=0; e=1 –> T1 = 278.7 (ideal BB)
* a=0.3; e=1 perfect BB but clouds increase albedo to 0.3 –> T2 = 254.9
* as above, but add GHG –> observed temperature T3 = 288
So ONE EFFECT of the atmosphere (cloud cover) is to lower the temperature ~ – 23 K
A DIFFERENT EFFECT of the atmosphere (known as the greenhouse effect) is to raise the temp ~ +33K
The NET effect of the atmosphere is +10 K
(And actually, the clouds are not the only thing reflecting visible light; this image from Wikipedia http://en.wikipedia.org/wiki/File:Albedo-e_hg.svg suggests that the earth as a whole reflects ~10-15 % already, so the clouds themselves only are responsible for PART of the cooling. This would leave maybe -10 K from the clouds and + 20 from the atmosphere as a whole.)
So by a slightly different reasonable way of thinking, the atmospheric effect [specifically known as the green house effect that wikipedia is discussing] on the surface temperature is the difference between T2 and T1.
Maybe we should go edit that page. 🙂
Bangs head against wall.
Please listen to my question, as follows:
Will you give me proof that Visible light from the Sun heats the Earth?
Because until you do, you can keep on claiming it does and you can keep on describing the mechanism you say is doing this, but it remains an imaginary, unsubstantiated claim until you prove it. Just do it.
And stop giving me daft meaningless ‘examples’ which mean nuttin. And which I’m now tired of explaining mean nuttin.
You may well think that the knowledge you have about this is ‘standard’, but it is different from that which was previously taught.
Since you are now teaching something different it is incumbent on you to provide proof that previous knowledge was wrong.
Stop avoiding actually doing this.
I realise, I’ve been looking at these arguments for some time now, that some might have difficulty believing there ever was any other understanding about this, because it’s very difficult to find any who admit it now with AGWScience takeover.
However, after spending some considerable hours trying to find for myself the proof I’m asking you to provide, and unable to find it, I found this:
“Many physics teachers traditionally attribute all the heat from the Sun to infrared light. This is inexact – visible light from the Sun accounts for 50 percent of the heating, and electromagnetic waves of any frequency will have a detectable heating effect if they are intense enough.” http://www.newworldencyclopedia.org/entry/infrared
So let’s look at that together. Traditionally still holds good until you prove differently.
The NASA page that I linked to is this same Traditional physics teaching.
Please re-read what it says, I posted it above. It says that the heat we feel from the Sun is Thermal IR.
Prove it is wrong, or shut the hell up about claiming you operate in the discipline of Science.
Because until you actually show that the heat we feel from the Sun comes from Visible Light and is not the Thermal IR as by Standard Science Tradition, you are talking a load of bllcks.
And therefore, your claim that 50% of heating comes from Visible Light is without foundation of proven fact.
And, still looking at the extract I posted, this is followed by a statement which I’ve said often enough here, “electromagnetic waves of any frequency will have a detectable heating effect if they are intense enough.”
For example, when I’ve explained that Visible Light is not thermal, but if concentrated in intensity it can burn.
However, this scientific actual fact does not relate back to the claim in the previous firspart of the sentence. I hope you can see that.
It is not confirming that “visible light from the Sun accounts for 50 percent of the heating.” I hope you can see that.
Now, give me what I’ve asked for. This is a science blog.
Bangs head against wall even harder.
Perhaps you can tell me what would count as “proof” to you???
** I have given an example whereby sunlight raises the temperature of PART of the earth (a piece of cloth, but it holds as well for rocks of different colors or wood of different colors or cars of different colors). By extension, light could heat other parts of the earth (unless you can specifically show either that 1) dark objects are not warmer than light objects or 2) the earth absorbs lights in a way that is fundamentally different than the way cloth and rock and cars and wood absorb light)
** I have quoted the simplest possible equation for the energy of light: E=hf. Visible light has a frequency. Visible light has energy. The earth absorbs that energy (other than a bit which is reflected) Absorbed energy raise the temperature of objects. I even have mentioned several times that ~40% of the energy is visible and ~50 is IR.
** AND NOW — (don’t ask me why I even bothered, but) — I took a concave mirror ~ 0.4 m in diameter. The mirror is covered with glass, so little IR should get thru. The light from the mirror was focused to a fairly small area. I put my hand there and
… PAUSE FOR A DRAMATIC DRUM ROLL ….
my hand got warm (quite hot in fact if I got too close to the focal point)!
I even put an extra pane of glass there to be doubly sure the IR would get absorbed. The light was passing thru 4 layers of glass. My hand STILL got warm.
(And for that matter, since glass blocks most IR, then even a magnifying glass would have to be mostly focusing just the visible light, not the IR to char wood or burn ants).
“You may well think that the knowledge you have about this is ‘standard’, but it is different from that which was previously taught. ”
Yes, What I am discussing IS different from what was previously taught — IF YOU LEARNED PHYSICS before Einstein developed the idea of photons!
“Traditionally still holds good until you prove differently. ”
1) No — tradition is not what determines correct science — especially traditions among non-experts. Simple explanations that are traditionally presented by non-experts are often wrong. I’ve heard lots of poor explanations of the green house effect. And tides. And wings. And friction. And the 2nd Law of thermodynamics.
2) Even so, NOT A SINGLE physics professor I have talked to has or would attribute all the sun’s heating to IR. Find the name of one physics professor who will go on record supporting your view.
3) Even the source you quote says the tradition is wrong!
Tim Folkerts says:
April 1, 2011 at 1:42 pm
Reed,
You certainly are welcome to do your computations and to see the effects of your model. I was perhaps a little harsh to say “it all falls apart”. But when you make an assumption that is almost completely the opposite of reality (assuming emissivity of snow ~0 when indeed is it ~1) then your conclusions will almost certainly be poor.
To answer a few specific questions:
How do you justify using cavity radiation into a medium containing matter (i.e.,
an atmosphere)?
While BB radiation was originally derived using a cavity, it has been shown that the same curve also works very well for other circumstances — ie some surfaces emit practically the same curve from a flat surface ie their emissivity is ~ 1 over the frequencies of interest. Experimentally, much of the surface acts much like a black body for IR wavelengths.
The material into which the BB radiation travels does not matter — only the temperature of he emitting surface, so the properties of the atmosphere are not important here.
This would seem to justify treating the surface as (approximately) a BB wrt IR.
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?
This is a little more subtle.
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?
As long as the emissivity is independent of the frequencies where (a significant amount of) energy is emitted, the T^4 law is (approximately) hold.
For example, this table http://www.monarchserver.com/TableofEmissivity.pdf suggest that type 301 Stainless has an emissivity of 0.27 @ur momisugly 24 C and 0.57 @ur momisugly232 C. This would mean that the IR power emitted by this surface would go up slightly FASTER than T^4 as the temperature rose from 297 K to 505 K.
On the other hand, cuprous oxide decreases in emissivity as temperature increase, so it would be a little less than T^4
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?
Yes. I’ve even seen it. I used to work with furnaces operating at > 1000 C. The uniformly heated walls and the small window in the door very closely approximate a cavity radiator. If you slide a sample in, everything – metal, ceramic, quartz – soon glows the same color.
(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?
No. If the object is hotter than the chamber, is will cool; if the object is cooler it will warm up. Always.
By any reasonable way of thinking, the atmospheric effect on the surface temperature is the difference between T3 and T1.
I agree that there are many ways to interpret the effect of the atmosphere. And I agree with your calculations:
* a=0; e=1 –> T1 = 278.7 (ideal BB)
* a=0.3; e=1 perfect BB but clouds increase albedo to 0.3 –> T2 = 254.9
* as above, but add GHG –> observed temperature T3 = 288
So ONE EFFECT of the atmosphere (cloud cover) is to lower the temperature ~ – 23 K
A DIFFERENT EFFECT of the atmosphere (known as the greenhouse effect) is to raise the temp ~ +33K
The NET effect of the atmosphere is +10 K
(And actually, the clouds are not the only thing reflecting visible light; this image from Wikipedia http://en.wikipedia.org/wiki/File:Albedo-e_hg.svg suggests that the earth as a whole reflects ~10-15 % already, so the clouds themselves only are responsible for PART of the cooling. This would leave maybe -10 K from the clouds and + 20 from the atmosphere as a whole.)
So by a slightly different reasonable way of thinking, the atmospheric effect [specifically known as the green house effect that wikipedia is discussing] on the surface temperature is the difference between T2 and T1.
Maybe we should go edit that page. 🙂
Tim, thank you for your response. I agree with almost everything you said. My using an albedo that is one minus the emissivity is very likely a poor representation of reality, and therefore is likely to give poor conclusions. I wouldn’t quite say it that way. Rather, I’d say it this way: “My conclusions are valid, they just have little resemblance to the real world.” I’ll concede the point–my conclusions may have little relevance to the real world. It’s just that over time I became frustrated and irritated with what I considered to an illogical argument to make the claim that the atmosphere has a 33 degree warming effect on the temperature of the surface of the Earth.
Regarding your point (3) above. I think there is some confusion. I also believe the temperature of the surface will trend to and eventually become the temperature of the cavity walls. If the inserted object’s temperature is higher (lower) than the wall temperature, the object’s temperature will decrease (increase). What I also believe to be true that when applying a graybody version of Planck’s blackbody law, the only temperature that will produce radiation rate equilibrium is the temperature of the cavity walls. Am I missing something?
Now a new question, when integrated over all frequencies, the frequency-dependent portion of Planck’s blackbody radiation law, (f^3) / {e^[h*f/(k*T)] – 1}, not only produces a T^4 dependence, it also produces a multiplicative factor equal to (pi^4) * (k^4) / [15 * (h^4)], where k is Boltzmann’s constant and h is Planck’s constant. Planck’s blackbody radiation law in its integral form contains the multiplicative factors 2 * h / c^2, where c is the velocity of light in a vacuum. Combining these factors, one gets a compositve factor of
2 * (pi^4) * (k^4) / [ 15 * (c^2) * (h^3)].
pi times the value of this composite factor is called sigma, the Stefan-Boltzmann constant. The question I have, is to what degree does integration using a frequency-dependent emissivity affect the value of the composite factor, so that if the T^4 rule applies, can sigma still be used?
As far as editing Wikipedia, I’ll leave that to you. 😉 When it comes to anything to do with global warming, from what I’ve been told most Wikipedia editions that contradict the idea that global warming is real, manmade, and catastrophic don’t last long.
Joel Shore says:
April 1, 2011 at 12:15 pm
Reed Coray says:
I don’t know how you interpret those words, but I interpret them as follows: …
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.
You elect to interpret scientific statements in a “charitable way.” That’s fine. I don’t. I don’t mind, in fact I believe in treating the statement’s author in a charitable manner. However, I choose to interpret scientific statements as written using my best understanding of the English language. I decline to defend/support/promulgate a logically-flawed scientific or mathematical argument because there is a “charitable” but illogical way to interpret what it claims.
Joel Shore asks:
When you open a freezer door, can you feel all of that the 270°K (or so) warmth beaming at you? How long do you have to stand there before you get too hot?
“When you open a freezer door, can you feel all of that the 270°K (or so) warmth beaming at you? How long do you have to stand there before you get too hot?”
Compared to what, a freezer full of liquid nitrogen (77 °K)? Neither warms you up, but one will cool you off faster, and to a lower temperature, than the other. And if it’s a freezer door that opens a magic portal into the cold of space, you’re going to cool off REALLY fast.
Joel says;
….. 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)”..
My reply;
Its refreshingly honest for you to admit that you did not understand the work of G&T.
As I’ve said to you in a previous post I dont think you even read the G&T paper.
Perhaps you were asked to sign up in a political sense and your name appears as a of gesture of solidarity with Halpern.
Anyone who has been anywhere near a thermodynamics book knows G&T are correct.
I have made a point of asking people with a background in physics to find ANY mistake in the G&T paper.
So far no mistakes have been found.
Quite eminent physicists have agreed that there are no mistakes.
A number have indicated in particular that their analysis of models and the difficulty of a climate solution for Navier Stokes equation is well founded.
I have collected here a list of the relevant material.
I am going on a short holiday so I will not be able to respond to any question you might raise.
However after a week or so I expect you to have reviewed the material presented here.
You seem to frequent WUWT and I will I hope you will be able to discuss in depth any faults you can find in the G&T paper.
1] “Falsification Of the atmospheric CO2 greenhouse effects within the frame Of Physics” by Gerhard Gerlich and Ralf D. Tscheuschner; International Journal of Modern Physics B, Vol. 23, No. 3 (2009) pages 275-364.
http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf
[2] “Proof of the atmospheric greenhouse effect” by Arthur P. Smith; arXiv:0802.4324v1 [physics.ao-ph]
http://arxiv.org/PS_cache/arxiv/pdf/0802/0802.4324v1.pdf
In this paper Arthur Smith defends the current IPCC position and has the merit of taking issue with G&T for something that they did say.
[3] “Comments on the “Proof of the atmospheric greenhouse effect” by Arthur P. Smith” by Gerhard Kramm, Ralph Dlugi, and Michael Zelger; arXiv:0904.2767v3 [physics.ao-ph]
http://arxiv.org/ftp/arxiv/papers/0904/0904.2767.pdf
Takes issue with Arthur Smith
[4] Comment on ‘Falsification Of the atmospheric CO2 greenhouse effects within the frame Of Physics’ by Joshua B. Halpern, Chistopher M. Colose, Chris Ho-Stuart, Joel D. Shore, Arthur P. Smith, Jorg Zimmermann.
This must be the most embarrassing paper in history as it attacks G&T for things they didn’t say.
Joel you must have a copy of the published comment
[5] “Reply to ‘Comment on ‘Falsification Of the atmospheric CO2 greenhouse effects within the frame Of Physics’ by Joshua B. Halpern, Chistopher M. Colose, Chris Ho-Stuart, Joel D. Shore, Arthur P. Smith, Jorg Zimmermann” by Gerhard Gerlich and Ralf D. Tscheuschner, International Journal of Modern Physics B, Vol. 24, No. 10 (2010) pages 1333–1359.
http://www.skyfall.fr/wp-content/Gerlich-reply-to-Halpern.pdf
G&Ts reply to the absurd [4]
(6.)
Gerhard Kramm and others with a broader look at current climate science including the “greenhouse effect”
http://www.benthamscience.com/open/toascj/articles/V004/137TOASCJ.pdf
“Will you give me proof that Visible light from the Sun heats the Earth? Because until you do, you can keep on claiming it does and you can keep on describing the mechanism you say is doing this, but it remains an imaginary, unsubstantiated claim until you prove it. Just do it.”
You mean AT ALL? OK, look up on a clear day at noon, and then look down. Is the ground beneath you as bright as the sun? No? OK, so it isn’t reflecting all of the incoming light. So where did the energy of the absorbed light go?
That’s just the ground, which admittedly reflects so much visible light (unless you’re standing on asphalt) that I bet it is primarily heated by infrared wavelengths, which it reflects very little of. (I’m just guessing on that – ask a physicist) In terms of visible light significantly impacting the earth’s energy budget, repeat the experiment over an ocean, and keep in mind that this is the situation over most of the earth’s surface. Most of the visible light is absorbed by the upper layers of the oceans. Water has a high heat capacity, and the absorbed energy is distributed over a huge depth, so a diver isn’t going to feel a temperature increase at noon on a cloudy day just because the clouds break. The oceans are the primary heat reservoir of the energy of absorbed visible light from the sun – not the ground, and not the atmosphere.
@ur momisugly Steve
You might find this intriguing. I located a paper related to the confusion of radiation of a pure argon atmosphere but guess what, article not available. Arggg.
http://adsabs.harvard.edu/abs/2009APS..TSF.C3015B
Under what conditions do accelerating charges radiate? An examination of recent literature
Butterworth, Edward; Cox, Paul
American Physical Society, Joint Fall 2009 Meeting of the Texas Sections of the APS, AAPT, and SPS Postdeadline, October 22-24, 2009, abstract #C3.015
The process by which accelerated charges emit electromagnetic radiation remains surprisingly obscure: even at the advanced level, most textbooks do not treat it in detail, and published reports show a wide variety of descriptions of the process, some of which have led to paradoxes.
…
Against published claims that uniformly accelerated charges do not radiate, Boulware (1980) and de Almeida & Suu (2006) propose that they do, but into a region of spacetime inaccessible to a comoving observer.
…
Piazzese (2003) obtains the result that charges in uniform circular motion do not radiate, subject to particular constraints that limit orbital size; with the result that electrons in Bohr orbits do not radiate, while synchrotron radiation is allowed.
…
No wonder I am confused !!! The radiation specialist are confused too!
And this abstract once again does not even mention thermal collisions. There might be no hope right now in answering that question. But as I mentioned above, that is one unanswered question that AGW and CO2 promoters are hiding under, shaky ground. On one side CO2 is a cooler, not a warmer and on the other side CO2 is not special at all.
@ur momisugly Myrrh
Noticed your note on 67 ºC, and I generally agree (that is unless an pure argon atmosphere really does radiate). That is an interesting question to follow in both astrophysics and spectrometry.
BigWaveDave:
Steve has answered your question. The alternative for the earth is to have an IR-transparent atmosphere, in which case it is basically a direct portal to the 3 deg background radiation of space. So, yes, being exposed to a relative cold atmosphere, even if we take the temperature at the tropopause of ~190 K, will feel like a freakin’ heat wave compared to being exposed to space!
That is what you fail to understand: When we talk about the greenhouse effect, the comparison case isn’t having some nice warm surroundings radiating at the earth instead. The alternative is having all of the terrestrial radiation escape to space.
A planet with no IR absorption in its atmosphere comes to an average surface temperature (really average of T^4) which is dictated by the radiative balance between what it gets from the sun and what it emits back into the coldness as space. A planet with an atmosphere that can absorb IR is able to maintain a warmer temperature at its surface. [As viewed from afar, the planet with its surrounding atmosphere still emits the same amount of radiation and acts as a body having the same average temperature (really average of T^4) as it would in the IR-transparent case; however, most of this radiation originates from higher up in the atmosphere where the temperature can be much colder than it is on the surface.] There is nothing particularly mysterious about this. It is basic physics, in perfect agreement with…in fact, dictated by…the Laws of Thermodynamics and radiative physics.
Tim – re-read my post. I am asking specifically about one kind of light.
I have already worked out that you’re not engaging with what I’m actually saying, hence my reference to daft meaningless examples which I’m no longer willing to waste my time on.
The Traditional Physics is Still Being Taught – reference the newworldencylopedia, and NASA page.
You, IRA are you listening?, generic, are teaching something different from AGWScience.
You are saying that Thermal IR is NOT heating the Earth.
This is A RATHER LARGE CLAIM TO BE MAKING, TO OVERTURN BASIC TRADITIONAL PHYSICS AS STILL BEING TAUGHT.
Who’s claiming the Nobel Prize for Physics for this Revelation?
Are you ever going to engage with my questions, IRA?
When you open the door of a freezer, the cold air comes pouring out, cooling you, just as opening the door of an oven gets hot air streaming out, warming you. OK, get a relatively thin cylinder of plastic that passes both shortwave and longwave IR and stand inside of it, blindfolded.
If you are in a room that is at 300 K (about 27ºC or 81ºF) you will be slightly warm, due to the 300 K radiation on all sides, plus the internal heat generated by your body.
Now, your assistant rolls up an oven on your right and a freezer on the left (or vice-versa, she does not tell you which). The doors of the oven and freezer are opened, but, since you are enclosed in the plastic cylinder, you don’t feel the hot air on one side and the cold air on the other. However, you now have 250 K longwave IR radiation coming at you from one side and 350 K longwave IR radiation coming from the other side. I have no doubt you will be able to tell which side has the oven and which the freezer.
Myrrh:
You are presumably being ignored by Ira (and others including me up until now) because you have displayed no evidence of being teachable. It is a thankless task to try to teach someone who doesn’t want to learn.
You have taken one line from one source and blown it totally out of proportion. What you are referring to is not “traditional physics”; rather, your source claims that it is something that “many physics teachers traditionally” have said that the source itself notes is not really correct. I am skeptical of the claim that many physics teachers traditionally say this, but who knows…I guess I was fortunate enough never to run into one of those physics teachers in the course of getting my PhD in physics.
However, that it is not really correct is beyond dispute. If you want to believe otherwise, there is nothing that we can do to help you.
Myrrh, you’re correct. There is a great amount of IR in the solar radiation and most of the albedo and refection off of the surface is strictly visible and above frequencies which must be subtracted from these higher frequency (left) portion.
Use this: http://i56.tinypic.com/5wk13l.gif
Myrrh says:
No…He is not saying that. The intensity of solar radiation is found by integrating over all wavelengths.
However, most of the intensity coming from the sun is in the UV, visible, and near IR parts of the spectrum. The “thermal IR”, which according to Wikipedia http://en.wikipedia.org/wiki/Thermal_infrared#Different_regions_in_the_infrared is used roughly for those IR wavelengths above 3 microns (although I don’t know how standard such a definition is), doesn’t make up that much of the solar spectrum…In fact, most plots of the solar spectrum don’t even bother to go out further than 2 or 3 microns. Furthermore, what little of it there is from the sun will get absorbed in large part before reaching the earth’s surface.
Bryan says:
It is a little different “spin” to say that I don’t understand their work than to say that I cannot find any interpretation of what they are trying to say that makes any sense whatsoever. You, despite your sniping at me, have steadfastly refused to provide any such interpretation of your own. In fact, you here repeat your accusation that we “attack[] G&T for things they didn’t say” but continue to avoid my invitation to enlighten us on what they did say (that makes any coherent sense) regarding the greenhouse effect and the Second Law.
My co-authors and I have discussed the faults in our comment on their paper. Arthur Smith on his comment alone has discussed other faults. The ball is now in your court, my friend. If you think that G&T is so sensible, then why don’t you tell us what it means? I would be happy to hear an interpretation of their work that both makes sense and is not in direct contradiction to things that they clearly state (like “The atmospheric greenhouse e ffect … essentially describes a fictitious mechanism, in which a planetary atmosphere acts as a heat pump driven by an environment that is radiatively interacting with but radiatively equilibrated to the atmospheric system. According to the second law of thermodynamics such a planetary machine can never exist.”). So far, you have been unwilling to provide one.
wayne says:
That shows the emitted radiation in W/m^2 at the surface of the sun and at the surface of earth (on a log-log plot, I believe). However, as you might not be aware, the earth isn’t located at the surface of the sun. To obtain the intensity of the solar radiation at the radius of the earth’s orbit around the sun, you have to multiply the emitted radiation by the ratio of the square of the radius of the sun to the radius of the earth’s orbit around the sun. After you’ve done that (and corrected for the factor of 4 between the area of a disc and the surface area of a sphere) and put it on a linear-linear scale, you’ll get something more like this: http://atoc.colorado.edu/wxlab/radiation/emissionspec.gif
Myrrh says : “You are saying that Thermal IR is NOT heating the Earth.”
No, I am saying that ALL wavelengths of sunlight are part of the energy going to earth, and therefore ALL wavelengths are providing heating to the earth. Please show ONE place where I said thermal IR is not part of the heating. (Thermal IR is a very SMALL part, but it is a part).
On the other hand, you say “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.”
This seem pretty clear — you do not think blue light (or visible light in general) can help raise the temperature of the land or the oceans.
“Thermal IR heating the Earth is excluded from the AGW Energy Budget.”
Not sure how you came to this conclusion.
One the one hand, thermal IR is effectively excluded from the energy budget when looking at sunlight. This is NOT due to some nefarious AGW scheme or poorly understood science. It is simply because thermal IR (specifically, I am referring to 4 um or longer), is less than 1% of of the incoming solar radiation and hence is only a minor player in the overall energy budget.
On the other hand, thermal IR is BOLDLY included in the energy budget, since thermal IR is the 390 W/m^2 upward radiation from the earth and the 324 W/m^2 downward radiation from the atmosphere.
Let me say one more time — the “Traditional Physics” you keep referring to is NOT traditional physics. There are several people in this discussion who have shown a pretty good level of scientific understanding — none of them seem to agree with your conclusions. I teach “traditional physics” and I know lots of other people who teach “traditional physics” and none of us teach that you can’t feel heat from visible light.
Or perhaps I am still mis-understanding you.
What SPECIFICALLY do you think “traditional physics” teaches about IR, visible light, and their heating effects that you agree with?
What SPECIFICALLY do you think “AGW physics” teaches about IR, visible light, and their heating effects that is incorrect?
Myrrh;
You are some piece of work bud. Take a look at one of your own sentences:
“For example, when I’ve explained that Visible Light is not thermal, but if concentrated in intensity it can burn.”
So…. it isn’t thermal, but if concentrated…it is thermal.
Magic.
Is Glickstine’s ‘science’ and attempt to ‘debunk’ the man-made warming that we are experiencing at present? I do not see in here resume where she is even remotely qualified to comment on GHG’s.
another reviewer wrote “Although there clearly is a greenhouse effect it is infinitesimal compared to the energy retaining effect of the oceans.”
friggin DUH! how did the heat get to the oceans in the first place??!!! from the heat trapped IN THE ATMOSPHERE
Certainly feels thermal and can heat up and burn the average body pretty badly. I believe thast UV light is the only part of the spectrum that causes sea temperature to rise.
@ur momisugly Cassandra King (March 29, 2011 at 9:59 am)
Many thanks for your kind words, Cassandra! 🙂
Richard Monror says……
……”friggin DUH! how did the heat get to the oceans in the first place??!!! from the heat trapped IN THE ATMOSPHERE”…..
This is an unusual viewpoint.
So we have no further use for the Sun!
By the way, if you want to figure out the fraction of power in a certain wavelength range for a blackbody of a given temperature, here is an online Excel file that will do it for you: faculty.virginia.edu/ribando/modules/xls/HTTplnkslaw.xls
Using it shows that for a blackbody at 6000 K (which approximates the sun quite well), less than 2% of the energy is emitted at wavelengths above 3 microns and less than 0.3% of the energy is emitted at wavelengths above 6 microns.
By contrast, for a blackbody at 290 K (approximating the earth’s surface), ~99.4% of the energy is emitted at wavelengths above 3 microns and over 96% of the energy is emitted at wavelengths above 6 microns.
This shows that there is very little overlap between the spectra of solar and terrestrial radiation.
Backradiation acts as a perpetuum mobile.
Thermodynamic theory already incorporates all directions of radiation in the surroundings in equilibrium with a radiative surface.
Ira Glickstein, PhD says:
March 30, 2011 at 10:10 pm
“…..Thus, if the longwave IR source was completely surrounded by the shell, the temperature could potentially increase to correspond to double energy, but, in the real world, it would reach equilibrium well before that……
I am not familiar with the details of a blackbody cavity, but I can imagine something like an electric soldering iron. With a given power input, the tip, in free air, would reach a temperature of X. If you put the whole soldering iron in a metal box, and suspended the metal box in free air, the tip would reach a higher temperature, all else being equal………You could calculate the new temperature using the Stefan Boltzmann (fourth power) Law. The new tip temperature would correspond to nearly twice the power.
Note however that no extra energy is being created. At equilibrium, the metal box will radiate (and convect) away the exact amount of power as is being input to the soldering iron, energy in = energy out. The only difference will be that the tip of the soldering iron will be warmer when enclosed in the metal box than when it was in free air. ”
This box would be a perpetuum mobile.
Now lets look at the radiation energy density I mentioned earlier. Radiation energy density (RED) is the photon density (J/m^3) in the space surrounding a radiating surface in equilibrium with the Stefan-Boltzmann temperature, it’s simply related to the temperature as S-B * 4/c. And this RED will spread out over distance!!!
So, we switch on the soldering iron that warms up to say 400 K, and it creates a RED corresponding to 400 K in the space around it. Then we put the box over it and the RED hits the box which is say 300 K, and the radiation from the iron will heat it up to 400 K, then the RED from these walls will also have increased from a density corresponding to 300 K initially to a level of 400 K. But that means the whole box now has a uniform RED, so what happens now? Nothing, because everything would be in equilibrium.
In reality the walls won’t even become 400 K but 399,9…K at the most as long as the box radiates on the outside (and the RED inside spreads out), so the box will never get as warm as the energy source!
Ira, thinks the RED will get equal and then magicly increase near the walls and backfire to the source, and that is impossible. The RED in the box depends on the source. The RED is the force acting on the walls and this results in the temperature of the wall which in return results in a RED from the walls and that will correspond to 400 K at the most.
So many seem to forget that you need a gradient(delta) in temperature, pressure, concentration or density to get a flow. And if the gradient becomes zero the flow stops. This is the Second Law working, always trying to spread things out at the lowest energy level.
So the die hards will say: because the walls also radiate I still believe in backradiation and this must do something.
Well, look at the link where does the factor 4 come
from?. Read and weep, radiation energy density already incorporates all directions of radiation. For the simple perpendicular radiation, half of the energy density in the waves is going toward the walls and half is coming out. But of course RED incorporates all possible angels of radiation because it describes a volume and you get an average factor of 4.
Imagine a second wall on the right and one will see that nothing changes at all for the RED.
You can also look at it as a pressure.Radiation
pressure is RED/3. So you will have a pressure coming from the iron, what pressure will the walls receive? It is clear that they will never receive a higher pressure than the source and so never be able to establish a higher temperature.
So heating by backradiation does not exist, it is radiative energy density pulled out of context.