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.
Ad hominem the last resort of losers. Do not pass go.
So, until you can prove, explain, show, how shortwave Solar Light Visible, UV and Nr Ir not thermal in themselves, and only in a one in three possible reactions with other matter in the real world produce heat in exchanges, and such as Visible and Nr IR only capable in artifical intensity, then the AGWScience KT97 Energy Budget which says these energies alone directly heat up the Earth to produce the amount of Thermal IR claimed, IS PHYSICALLY IMPOSSIBLE and therefore JUNK PASSING ITSELF OFF AS SCIENCE.
And those promoting it? Well, you’ll have to decide for yourselves on that score. The deliberately misdirected, educated to believe impossible junk as if real science, those with a smattering of real science in some field or other taking on trust junk memes from another discipline, those knowing full well that the junk science they are peddling is junk science from knowing real science and are deliberately with malice aforethought working to confuse the unwary, whichever, having crossed over to the nasty sociopaths who deliberately created the meme for nefarious reasons of their own, they’re in the same AGWGa-Ga Land of the damaged, a flat world of others seen as cardboard cutouts, out of touch with real life and therefore, ultimately, themselves.
Woe unto you, scribes and Pharisees, hypocrites! for ye compass sea and land to make one proselyte, and when he is made, ye make him twofold more the child of hell than yourselves. Matt 23:15
Oh, two more things. If you will notice the tau (IR optical thickness) that popped out doesn’t seem to agree with either KT97, TFK09, or FM08. That was the first indication that something may be amiss and off a bit. The transmittance from the surface in this is about 9% higher than FM08. (That figure uses the surface flux per the surface temperature to calculate the window flux)
Also, the evapo-transpiration figures required a s.w.a.g., since evaporation is by logic much greater during the day than nighttime, I had to just pick a best guess, I made the daytime evaporation twice than during the night, that simple. I could find no references on that one. That may be causing some of the small divergence, ~5 W/m2.
And on the mean diurnal range used, I found some references but all were very rough, even on NASA’s planetary data sheet so just set it at the mean, ±6.48 ºC. However, upon changing that figure it made little change overall so that’s probably close.
Myrrh: We can’t “prove” anything to you if we don’t have a common basis to start from. The common basis that Ira, David Hoffer, Tim, Phil. and I share is modern science. So, even though our views on AGW run quite a gamut, we are still able to carry on a conversation.
You, however, reject modern science, labeling anything that doesn’t strike your fancy as “AGWScience” which you apparently think is cute rather than childish and immature. The fact that this science goes back more than a century doesn’t matter to you. Such a person cannot be rationally reasoned with. We ridicule you because any scientific substance just bounces off of you.
Joel – you’re talking absolute nonsense, if “modern science” is teaching you that these Solar energies as depicted in AGWScience’s KT97 heat the Earth, you’re not talking about real world science. If you really think they can do this, then for the last time of asking you here, prove it.
If you can’t tell the difference between Light and Heat energies and how heat is exchanged on a microscopic level then you’re not in the real science world, so stop bullsh*tt*ng about how clever you are or that you understand entropy, it’s not convinced anyone here who actually knows and works with real science fact, and, it didn’t take me long to suss out that the AGW claim is physically impossible on a property level about energy claims as I did about AGW claims about Carbon Dioxide, that’s junior/secondary level, in the real world. Impossible. Physically imposssible.
If you’re not a stooge and you really don’t know this, then think about it.
That fact that you and Tim consistently refuse to give any actual proof, any actual real world explanation shows you don’t have it. If you think it exists, go fetch.
And if you’re both such brilliant scientists as you claim and so far above my level, why aren’t you engaging with Wayne?
Oh, and likewise Ira, David and Phil are of the same ilk as you and Tim here, unable to produce real science because they don’t have it.
RJ says:
As I mentioned, we’ve debunked Postma’s silly 2nd Law arguments to death in this thread. However, we haven’t so much his lapse rate stuff, so let’s take a look at that:
Postma’s basic claim about the lapse rate rather than the greenhouse effect making the surface hotter than radiative equilibrium predicts is wrong because he never explains what determines the effective height in the atmosphere that is at the temperature that radiative equilibrium predicts. And, the answer is that this is set by radiative properties of the atmosphere, that is, by the greenhouse gas concentrations (and distributions for those that are not well-mixed), after which, yes, the stability limit imposed by the appropriate mixture of dry and saturated lapse rates determines the surface temperature in a troposphere heated from below in which convection essentially does not allow the lapse rate to exceed the stability limit.
That this is true is obvious in the limit of no greenhouse gases: One cannot possibly have the radiation coming from some height up in the atmosphere since such an atmosphere can’t absorb or radiate in the IR.
So, in that case the surface must radiate away the energy that returns to space and the earth’s average surface temperature (or more precisely, 4th root of the average of T^4) is the temperature that achieves radiative balance…about 255 K with the earth system’s current albedo.
Now, as we add greenhouse gases, that level slowly rises, since the atmosphere acquires a larger and larger optical thickness and the radiation that escapes to space has to be from the part of the atmosphere where the optical thickness above it is of order 1. All the while, the temperature at the surface is indeed determined by starting from this effective radiating layer height and extrapolating the temperature down to the surface using the lapse rate.
This is all explained in any basic book that discusses radiative transfer in the atmosphere. Ray Pierrehumbert’s “Principles of Planetary Climate” being one of the best resources for all of the details.
Question: Myrrh, do you notice something about Ira Glickstein, Tim Folkerts, David M Hoffer and Joel Shore and others here, who, while we may differ in our assessment of whether global warming poses a significant danger, and how much is due to human activities, and the exact sensitivity to CO2, all agree on a common language and standard of science?
Answer: We each have a real name that may, for example, be Googled to determine our associations with corporations and academic institutions and websites and other things real people have.
So long as you hide behind Myrrh, I will assume you are merely “the dried oleo gum resin of [certain] species of trees. … small or low thorny shrubs that grow in rocky terrain. … produced by the tree as a reaction to a purposeful wound through the bark and into the sapwood. … myrrh is waxy and brittle … The term is derived from the Aramaic ܡܪܝܪܐ(murr), meaning ‘bitter'”.
Or, you may be an Agent Provocateur, sent to infiltrate and discredit our blog by inducing us to rash responses. Nope, probably not. You’ll remain a mere bitter, thorny shrub unless you identify yourself as a real person.
Ira – my analysis of you et al through your responses in this discussion is as a reader and participator here, who I am makes no soddin difference. You’re the ones setting yourselves up as experts, I’m just a passing oik name doesn’t matter member of the jury from the people analysing your own defence of your claim, which you set up to be discussed on this forum where science is discussed and where names don’t matter – and – my judgement is you’re either total shysters, knowingly and with malice aforethought participating in spreading the Anthropogenic Global Warming Fraud, which I’ve shown is junk impossible science re properties of your Solar short wave heating the Earth and so proved to my own satisfaction through interrogating you and your companion defenders here, or you’re such d*ckh**d scientists that you couldn’t calculate your way out of a paper bag.
You, still, have not provided any proof, any logical real science explanation of how Solar can heat the Earth as in the junk AGWScience KT97 claim. You can’t provide it because it doesn’t exist. I know that.
What I’ve been examining is your reactions to me knowing that. Your running away from engaging with me on this so early on and only coming back to make personal attacks is the best you can do in defending this junk impossible science you’re promoting. There’s nothing ‘modern’ about that, it’s an old story.
And in the end it comes down to personal integrity. You’ve set yourself up to promote this under your own name, that was your decision, I didn’t force you to do that.
It wouldn’t take any reasonably intelligent logical thinker long, any teenager examining the properties of Light and Heat and CO2/Oxygen/Nitrogen/Hydrogen molecules and the Gas Air which is our Atmosphere, to see for themselves that what you claim is physically impossible in the real world. So why don’t you admit it?
What you are party to is the criminal and morally indefensible corruption of access to real world science to the public at large and the world’s children.
What sort of scientist does that make you?
Myrrh
As I mentioned, we’ve debunked Postma’s silly 2nd Law arguments to death in this thread. However, we haven’t so much his lapse rate stuff, so let’s take a look at that:
Have you? Hans Schreuder for one does not seem to agree that it is silly.
http://climaterealists.com/index.php?id=7457
To All,
I am delighted to present to you a scientific paper that illustrates how our atmosphere interacts with the Sun and how the temperatures throughout the atmosphere as well as on the surface are in perfect harmony with basic physics and the laws of thermodynamics.
Not only is our Earth’s atmosphere and its ground temperature explained, planet Venus is also shown to comply with the same laws of physics and thermodynamics.
Best regards,
Hans Schreuder
I just got through extending the same spreadsheet from
Kiehl & Trenberth 1997 energy budget here:
http://i56.tinypic.com/avc5g.jpg
to
Trenberth, Fasullo, Kiehl 2009 energy budget:
http://i53.tinypic.com/ir6lie.jpg
Ferenc Miskolczi 2008 will be next but it will take much more time. Seems you must satisfy about twelve simultaneous equations along with meeting the budget constraints
Look closely at these two, especially at the total downwelling solar radiation around noon on both. It took at big jump up between these papers. Seems albedo took a drop.
Also notice the effective tau (IR optical thickness) rose much closer to Miskolczi’s found constant of 1.8675 over the last sixty years but caused very little change in the surface to TOA flux.
This is getting interesting after getting a per hour, cosine weighted, view. The numbers looks a whole lot different when the sun doesn’t shine during the night, doesn’t it.
RJ says:
What the heck is that supposed to prove? That Hans Schreuder is ignorant? Point taken.
Now are you going to engage the science or just point out that there are lots of ignorant people on the web?
RJ says:
April 24, 2011 at 3:51 am
As I mentioned, we’ve debunked Postma’s silly 2nd Law arguments to death in this thread. However, we haven’t so much his lapse rate stuff, so let’s take a look at that:
Have you? Hans Schreuder for one does not seem to agree that it is silly.
Really, and Hans Schreuder is some sort of authority is he?
In looking at the site you referenced I noticed this:
“Lastly, there is one fundamental law of physics that relates to blackbody emission of thermal energy: it is absolutely fundamentally impossible for a blackbody to further warm itself up by its own radiation. This is actually true for all objects, but we‟ll just keep referring to blackbodies here since that‟s what the subject is about. For example, imagine a blackbody which is absorbing energy from some hot source of light like a light-bulb, and it has warmed up as much as it can and has reached radiative thermal equilibrium. The blackbody will then be re-emitting just as much thermal infrared energy as the light energy it is absorbing. However, because the blackbody doesn‟t warm up to a temperature as hot as the source of light, its re-emitted infrared light is from a lower temperature and thus of a lower energy compared to the incoming light that it is absorbing. Now here‟s the clincher: imagine that you take a mirror which reflects infrared light, and you reflect some of the infrared light the blackbody is emitting back onto itself. What then happens to the temperature of the blackbody? One might think that, because the blackbody is now absorbing more light, even if it is its own infrared light, then it should warm up. But in fact it does not warm up; it‟s temperature remains exactly the same. ”
Note that he asserts “But in fact it does not warm up”, no evidence given, no equations! In fact it will get hotter, he conveniently forgets the (T^4-To^4) term.
I’d like to see him explain why a thermocouple when surrounded with a radiation shield gets hotter using his perverted physics?
Joel again claims to have demolished Postma’s paper.
Joel is big on empty claims as the G&T ‘comment ‘ paper proves.
The adiabatic lapse rate depends on the thermodynamic consequences of the kinetic theory of gases in a gravitational field.( 9.8K/km)
Nothing whatsoever to do with radiation as even scienceofdoom agrees.
The dry adiabatic lapse rate is modified through the content of phase change of water in the atmosphere as this narrative shows.
http://science.hq.nasa.gov/kids/imagers/ems/infrared.html
So on occasions when the modified lapse rate is say 7K/km then if the cloud level is say 5km and the temperature there is -30C then by applying the current lapse rate the ground surface temperature would be 5×7 =35 degrees C higher that the clouds that is 278K or 5C.
Just back from a short holiday so have not managed to read all other posts in detail.
Bryan says:
Thanks, Bryan, for the little lecture which does not contradict anything that I said in my post. (The one note that I should make on my original post is that when I say “greenhouse gases”, that should really also include the condensed greenhouse gas of water vapor, i.e., clouds.)
The point is that the (stability limit) on the lapse rate is set by convection but the effective level from which the radiation escapes to space is set by those gases and condensed gases in the atmosphere that can absorb and emit IR radiation.
Wayne,
Here’s a little feedback.
* I am impressed so far. You have a great start on modeling heat flow throughout the day.
* You have the hottest time right before sundown (5-6 PM time slot). Every place I have been, the max temperature is well before sundown. For a more accurate model, you might find a way to more accurately estimate the temperature profile during the 24 hr cycle.
* Near dawn & dusk, I would expect the absorption by the atmosphere to be enhanced relative to the absorption by the surface. At dawn & dusk, the light has to travel a long way thru the atmosphere (which, among other things, makes the sun look dim and red and makes the UV much less severe — even for a surface facing directly toward the sun).
* I assume “TOA” is “Top of Atmosphere”. Since the average value is ~235 W/m^2, I assume this must be the total IR emitted as measured from above the atmosphere. How do you estimate this value? I am surprised that this number varies so much, since I don’t imagine the surface, clouds, or atmosphere to vary enough form night to day to cause such a huge swing in IR radiation.
Phil., I read Hans Schreuder’s exerpt you included four times and I find nothing wrong in his statements concerning physics, I don’t even know who he is but will look up his work if I can find it later. You then said:
“I’d like to see him explain why a thermocouple when surrounded with a radiation shield gets hotter using his perverted physics?”
Well, I can answer that and I can guarantee you it is not further warmed by radiation from the thermopile itself. Matter cannot add any warmth to itself with it’s own radiation. Surely you know that. For a moment I thought you knew this basic science. As radiation leaves a surface, the surface cools, if any of that radiation returns to that surface it cancels the initial cooling, so, in that limited view and ignoring geometry and distances, nothing has happened at all. Much as the chicken in an unpowered mirror coated oven. Nothing happens. Are you simply forgetting the very valid wave/particle duality?
So, it must come from the fact that the shield, due to its proximity and applying the inverse square projections, the added warmth the thermometer (or thermopile) registers is from the externally warmed shield itself, maybe plus the original energy source too (can’t quite imagine the exact placement of these parts). Warmth always has to come from something else than itself, even in the case with a perfect mirror.
Myrrh says:
April 22, 2011 at 4:57 pm
Phil. says:
April 22, 2011 at 10:08 am
Re my: “BUT, what kind of energy? What are these energies capable of doing?
“Heating of course!”
Yeah right, that’s why all the green plant life in the world says no thanks to Green Visible Light and bounces it away…
So, take that amount out of the AGWScience Energy Budget KT97, for a start.
See my post to Wayne April 17, 2011 at 6:49 am http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/#comment-643521
and in that is the link to possible outcomes: http://www.pol-us.net/lllt/phototherapy.html
“When a photon is absorbed by a etc.”, which extract I posted from it and I’d assumed you’d read as you posted immediately after mine and corrected my slip of sodium for incandescant…
..so why are you ignoring it now as if it was never there? Because it doesn’t matter as long as you can keep repeating the meme, all propaganda is good propaganda..? That’s Tim’s way, are you a stooge in training?
Well as far as the physics is concerned it’s only partially right, since electronic spectra refer to UV/visible wavelengths not IR which are more usually vibrational spectra. As pointed out the ocean absorbs much of the light incident on the planet and doesn’t have difficulty absorbing green.
Phil says re my “What effect do they have because of their particular characteristics in reacting with other matter? What are each actually doing when they give up their energies?
“Mostly heating,”
See immediately above, so weak Green Light that it bounces of plants, and link to explanation of possible reactions, which you would do well to contemplate.
When it bounces of plants it isn’t giving up its energy!
“I notice that you chose to ignore the point that the Earth’s surface is 70% water which absorbs over 90% of the light (of all solar wavelengths) incident upon it..”
Didn’t we go through this before..? So? Look up reflection, refraction, scattering etc. and remember, from now on, that Blue light gets to deeper depths because it isn’t absorbed, it gets scattered all the way down just as it does in the still cold sky high above, bouncing off the water through the cold depths of the ocean, pinball knocking itself out finally senseless in the black depths of the deep. Where’s the heat from it in that?
Because it is absorbed ultimately and every photon that is gives up its energy as heat.
Visible light can’t even do that. It penetrates a little further before being reflected or absorbed, not creating any heat in giving up its energy. How then is this blue light creating heat in the KT97? It isn’t.
If it gives up its energy in being absorbed it must end up as heat (1st Law).
Re my: ” A photon of blue light does not give up its energy in heat to a plant in photosynthesis”
“Really, it takes over 10 photons of light to fix one molecule of CO2 in the form of glucose, how efficient a process do you think that is?”
Non sequitur, that’s not producing heat.
It certainly is, all that H2O transpiring from the leaves started off as liquid water in the roots, where do you think the latent heat of vaporization came from?
“The remaining nonsense deleted, it’s the same stuff endlessly repeated. Take a bolometer (as I’ve suggested before but you failed to comprehend) and measure the heat carried by different wavelengths and you’ll find that photon to photon blue is much hotter than thermal IR.”
I comprehended very well indeed. You confused two instruments, bolometer and light meter. Bolometers measure thermal IR, and that with great difficulty, anything else, the lighter less hot such as blue, is nonsense to try and measure with them.
I did not confuse the two, I meant a bolometer. They do not just measure ‘thermal IR’ they measure all wavelengths by absorbing them with high efficiency and measuring the heat, usually with a thermopile. They are not difficult to use, I used one in my lab from the UV through visible to IR with no problems at all.
wayne says:
April 24, 2011 at 5:53 pm
Phil., I read Hans Schreuder’s exerpt you included four times and I find nothing wrong in his statements concerning physics, I don’t even know who he is but will look up his work if I can find it later. You then said:
“I’d like to see him explain why a thermocouple when surrounded with a radiation shield gets hotter using his perverted physics?”
Well, I can answer that and I can guarantee you it is not further warmed by radiation from the thermopile itself. Matter cannot add any warmth to itself with it’s own radiation. Surely you know that. For a moment I thought you knew this basic science. As radiation leaves a surface, the surface cools, if any of that radiation returns to that surface it cancels the initial cooling, so, in that limited view and ignoring geometry and distances, nothing has happened at all. Much as the chicken in an unpowered mirror coated oven. Nothing happens. Are you simply forgetting the very valid wave/particle duality?
So, it must come from the fact that the shield, due to its proximity and applying the inverse square projections, the added warmth the thermometer (or thermopile) registers is from the externally warmed shield itself, maybe plus the original energy source too (can’t quite imagine the exact placement of these parts). Warmth always has to come from something else than itself, even in the case with a perfect mirror.
You’re right it is the shield that is doing the heating, however it is cooler than the thermocouple but warmer than the background!
Let Tg be the temperature of the gas being measured, Tt be the temperature of the ThC, Tb the background temperature and Ts the shield temperature, and Tg is greater than Tt which is greater than Ts which is greater than Tb.
In the absence of the shield the radiational heat loss depends on (Tt^4-Tb^4) whereas in the presence of the shield the heat loss depends on (Tt^4-Ts^4). Thus in the presence of the shield the measured temperature is higher because of the greater amount of ‘back radiation’. Your argument is flawed because it doesn’t take account of the continuous influx of heat. The convective heat transfer to the ThC has to balance the radiative heat loss, by using a shield the ThC temperature gets closer to the true gas temperature.
wayne says:
And, you have discovered this law how exactly? It corresponds to no Law of Physics that I know of.
Look., there are two cases:
(1) The object is receiving no energy from an internal or external source at a higher temperature: In that case, the object can only cool down. However, its cooling can still be slowed by reflecting or otherwise sending back some of the energy that it is radiating away. The reason that this radiation can’t heat it to a higher temperature than it started out at is because there can never be more radiation sent back to it than it gives off. It is simply conservation of energy, not any magical law at work.
(2) The object is receiving energy from an internal or external source at a higher temperature: In that case, the object will equilibrate to a temperature were the radiation emitted (or, more generally if other methods of heat transferred are operating…the heat transferred away) is equal to the amount of energy it is receiving from the internal or external source. Since the heat transfer depends on the surroundings as well as the object’s temperature, the object’s steady-state temperature will depend on the surroundings…and, in particular, in how much of the energy that it radiates is returned to it. So, yes, energy returned to the object by reflection or “back radiation” can indeed warm the object in the sense of yielding a higher steady-state temperature as compared to when this energy is not returned to the object.
There are no “magical” rules about radiation from an object relative to other radiation. The photons do not wear little tags that identify where they came from.
I should add that Hans makes it quite clear that his setup involves the second case that I considered above and thus his statement that the object’s temperature cannot change when some radiation is reflected back falls into the realm of pseudo-scientific nonsense.
wayne says:
Okay, now that I read this more closely, particularly the last sentence, I realize that your misunderstanding may run much deeper than I thought:
(1) Even assuming the proposed wave cancellation mechanism you might think operates is actually operating, your argument doesn’t make sense. If it cancels the initial cooling, then that means you now have a different situation than you had before when the radiation that was emitted and escaped. So, either it will cool slower (if it is not being heated by a source) or it will equilibrate to a higher temperature (if it is). So, even if your argument (about wave interference occurring) was right, your argument (about the thermodynamic result) would not be right. (If such counter-factuals can be thought of as sensible at all.)
(2) Wave interference doesn’t work the way you seem to think it works. First of all, it is rather hard to get radiation coherent enough that you can see interference effects. That is why most interference measurements are done with lasers, which are able to produce coherent radiation over reasonable distances. Second of all, whether you get constructive or destructive interference of waves at a particular point depends on details of the geometry. I don’t understand why you particularly expect destructive interference for your particular geometry.
(3) Basically, interference just redistributes energy spatially. You still have conservation of energy operating. So, for example, if you get destructive interference at a certain point, you will have constructive at another and when you average over the whole region, you still have the same amount of power.
Wayne,
I have never seen “warmth” in any physics equation. Could you define what you mean by “warmth”? Could you write out the equation is used to calculate “warmth” or to calculate how much warmth is added to an object??
Tim
Phil, sorry, in some other environment we could probably converse but not here. This is wasting everyone’s time and I have more science on my stack to finish up.
Btw, you example & equations seem right on. My source was continuous also of course, no flaw. The thermocouple is the measurement point and anything limiting paths to cool will of course raise the temperature above what it would be without it, it just was not radiation from the thermocouple that raised the temperature, it is not an energy source and it cannot increase it’s own temperature with it’s own radiation. None of the radiative energy that resonates (radiative equilibrium between the thermocouple and the shield) that keeps the temperature higher came from the thermocouple itself, it all came from the shield. The shield is warmer than the background. No word of “back radiation” will make it so. Easy to become a paradox isn’t it. I see where so many go astray, so easy to forget everything else going on at the same moment that many times exactly cancel. Your right, SB takes all into account if you use it right.
I’ve decided to stay away from the term “back radiation”, up this thread it really seemed screwed some otherwise eager to learn minds up. Seemed to take on an existence of itself as if it wasn’t just radiation pointed downward. From now on just radiation with a direction vector. Later.
Joel says:
There are no “magical” rules about radiation from an object relative to other radiation.
Of course there are, they’re called properties. That’s why we have a blue sky not a black or white when the Sun shines, because the different colours have different effects when they meet matter.
The photons do not wear little tags that identify where they came from.
Doesn’t matter where they came from re:
So, yes, energy returned to the object by reflection or “back radiation” can indeed warm the object in the sense of yielding a higher steady-state temperature as compared to when this energy is not returned to the object.
Energy from a hotter goes from higher to lower states, the higher energy will always be acting on the lower energy states, energising them, that’s how heat spontaneously travels. The lower energy colder are being bashed by the higher energy hotter, unless, work is done to change that.
What work is being done and being done by what, to send lower energy heat to the higher in your “back radiation” from a colder atmosphere sending this “magical warmth” back to a warmer Earth?
Wayne
“Phil., I read Hans Schreuder’s exerpt you included four times and I find nothing wrong in his statements concerning physics, I don’t even know who he is but will look up his work if I can find it later.”
The exerpt was from the Postma paper. I thought this paper was very good and would be interested in your views on it if you have a chance to read it.
Hans Schreuder. He contributed a number of charters to the climate slayers book. (Ch 13, 14, 15 and 16). And seems to be tied up with the Ilovemyco2 website.