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.
Ira, David, Tim et al – do you have some kind of block here that’s preventing you from actually taking in my question? I’m asking about the specifics, how and how much does Solar actually heat the Earth.
I’ll try again. AGWScience says that Solar energy, UV, Visible and Near IR, the Shortwave energies, actually heat the Earth which then radiates out Thermal IR (which then gets radiated out/trapped etc.). That’s the basic premise of the AGWScience Greenhouse. This should be bog standard understanding and the calculations at hand for any promoting this hypothesis.
How and how much does UV heat the Earth? How and how much does Visible light heat the Earth? How and how much does Near IR heat the Earth?
Prove it.
I’m telling you, that these do not heat the Earth. These energies are incapable of heating the Earth to the extent that it gets hot enough to radiate out that amount of Thermal IR. These shortwave energies are not hot in themselves, that is, they are not Thermal Energies. Heat is Thermal Energy in transit from one location to another. Therefore, since they are not themselves thermal energy, they do not heat by their heat, because they have none.
Show me the method and calculations, the detail, of how much these particular energies actually heat the Earth which is the BASIC PREMISE of AGW Greenhouse Hypothesis.
If you can’t answer this – you have no scientific credibility in promoting this as real Science.
What is it you don’t understand in my questions here? How and how much? Let’s limit this, for the sake of brevity, to Visible Light, which you, generic, keep stressing is the peak energy radiating out from the Sun heating the Earth (which then radiates out Thermal IR, etc.).
Let’s make it even more simple, not only for the sake of brevity but also because you generic stress how deep it goes into the ocean compared with the other Visible wavelengths, how much does Blue Light from the Sun heat the Earth?
Begin, by showing me how much blue light heats organic matter, solid and liquid, and take it from there.
Myrrh, you said, “et al – do you have some kind of block here that’s preventing you from actually taking in my question? ”
No, not at all in my case, in fact, you have really brought up a very curious question and experiment problem. How the heck could someone ever by experiment prove whether visible light warms and how much compared to unfiltered solar radiation.
Well, it’s not summer yet but when it gets warm enough to play with some water I’m going to do such an experiment. I’m going to take a large black box with something like 1″ of water inside. Record the temperature, expose for a given time and get the delta T to compute the rate. Next is to cover this with a large enough plate of glass with a very thin layer of water flowing across it’s top to filter all but the visible frequencies. Repeat the sequence above.
That is the only way I can come up without some expensive equipment that I no longer have. Does that sound like that should answer your specifications? (Would you believe it either way the results came out? I will. ) I’m going to pre-guess there will be some warming in the visible-only case but tiny compared to the unfiltered case and I’m trying to figure out some way to also filter out all but the lower infrared and repeat it there too.
Myrrh, one more thing.
I just figured out to do that very thing, fiter all but the IR, but I’m also going to need a small pump. If the water flowing over the glass to filter out the IR was not garden hose water but recycled water I can measure the rise of temperature of that water also, the water absorbing the IR. Sound about right?
Myrrh;
Did you read the articles?
No?
Then STFU until you do.
Hans says:
No,we do not know that, for the reasons I have stated in my previous posts if you actually bothered to read and comprehend them. I feel like I am talking to a brick wall. As for your question, I won’t even waste time starting to think about it until you make it clear whether you are asking about (1), (2), or (3) in this post: http://wattsupwiththat.com/2011/03/29/visualizing-the-greenhouse-effect-molecules-and-photons/#comment-637072
Statements like this make me wonder if you are comprehending anything that we tell you. Yes, it can stop there because I have described to you the final steady-state state that the system converges to. This is the solution you actually find if you solve the equations rather than just talk nonsense off the top of your head about what you think happens based on whatever misconceptions of reality you have.
What davidmhoffer described in his post was how the system converges to this final steady-state. It is as if we wrote down the geometric sum 1 + 1/2 + 1/4 + 1/8 + … and David described to you how by successively adding each term you could see what it is converging to. And, I just noted the properties of the final result, which is 2 in this case.
Joel Shore says:
Richard E Smith says:
” ‘These emissions are at the wavelength determined by the blackbody’s temperature. When they return they are either at the same or a lower energy level. Therefore they cannot excite the chicken’s molecules to a greater level of energy than they were before.’ ”
‘Ah…Can you tell me what law of nature actually says this? It sounds like the “Magical 2nd Law of Thermodynamics” to me. While you are at it, can you tell me the experimental evidence in support of this law?’
You can test it experimentally by pointing a spotlight at a wall in your house and holding a mirror so that as the light reflects off the wall it is reflected back to the wall by the mirror. You will find that the reflection from the mirror will illuminate the wall that is not already illuminated by the spotlight. If you direct it at the part already illuminated you will find that it does not brighten that part any more. The ‘extra’ light has no effect. Infra-red is a form of light and so an object cannot heat up itself by its own radiant emissions. Energy cannot multiply itself.
I think a lot of confusion exists over the so-called greenhouse effect because people confuse light and heat. They transfer their experience of achieving greater warmth by the use of blankets and saucepan lids to so-called greenhouse gases. In fact a lid simply confines the area that is being warmed. This both restricts convection and reduces the volume of the area that is heated. It does not add heat to the source of the heat.
Wayne and Myrrh: You are welcome to do all the experimentation that you like. However, I would also suggest you get a physics textbook and actually read it. Comprehend ideas like “conservation of energy”, “the photon model of light”, “blackbody radiation” that have come from centuries of experimentation and hypothesis testing by scientists. Your gaps in knowledge and misconceptions are way too large to be cured by us talking to you in this forum, particularly since you seem to have a mental block to absorbing anything that we say. Maybe you will have less of a block if you read it in a textbook. (The book that we use in the course that I teach is “College Physics” by Knight, Jones, and Field, which I would recommend except that it actually has a page on the greenhouse effect and global warming, so no doubt Myrrh would conclude it is part of the grand AGW conspiracy. Better to find an old textbook from before the time when you think this grand conspiracy overtook all of the physics textbooks.)
Myrrh says: “I’m telling you, that these do not heat the Earth. These energies are incapable of heating the Earth to the extent that it gets hot enough to radiate out that amount of Thermal IR. ”
As long as you are contradicting yourself within the course of 2 sentences, there is little change for any one else to explain anything to you! You are saying:
1) ‘these wavelengths DO NOT heat the earth [at all]’
2) ‘these wavelengths DO heat the earth [just not enough to do the job]’
1) has been explained over and over.
2) is true and is indeed the basis of the Greenhouse effect. Additional energy (thermal IR from the atmosphere) is needed to warm the earth.
Exactly which wavelengths provide how much power would vary from location to location and from season to season. That is a question beyond the scope of a discussion here. If you are really keen to know, dig into the literature!
Richard E Smith says:
April 7, 2011 at 4:24 am
You can test it experimentally by pointing a spotlight at a wall in your house and holding a mirror so that as the light reflects off the wall it is reflected back to the wall by the mirror. You will find that the reflection from the mirror will illuminate the wall that is not already illuminated by the spotlight. If you direct it at the part already illuminated you will find that it does not brighten that part any more. The ‘extra’ light has no effect. Infra-red is a form of light and so an object cannot heat up itself by its own radiant emissions. Energy cannot multiply itself.
Well I’d suggest you conduct that experiment with a well controlled detector rather than the human eye (a square law detector with automatic brightness control)! In any case your experiment suggests that all of the reflected light from the mirror is absorbed when it hits the bright spot. Think about it.
Myrrh says:
April 7, 2011 at 12:09 am
Ira, David, Tim et al – do you have some kind of block here that’s preventing you from actually taking in my question? I’m asking about the specifics, how and how much does Solar actually heat the Earth.
It gets absorbed and heats up whatever absorbs it. A watt of UV heats up a black body the same as vis, ir or electric current for that matter. Evidence given above which you ignored!
I’ll try again. AGWScience says that Solar energy, UV, Visible and Near IR, the Shortwave energies, actually heat the Earth which then radiates out Thermal IR (which then gets radiated out/trapped etc.). That’s the basic premise of the AGWScience Greenhouse. This should be bog standard understanding and the calculations at hand for any promoting this hypothesis.
Which there is except for those who willfully ignore it.
How and how much does UV heat the Earth? How and how much does Visible light heat the Earth? How and how much does Near IR heat the Earth?
Prove it.
Done, you are the one who makes the extraordinary statement that a watt of energy at a blue wavelength is incapable of heating an object which absorbs it. What do you think happens to that energy if it is not reflected, where does it go?
I’m telling you, that these do not heat the Earth. These energies are incapable of heating the Earth to the extent that it gets hot enough to radiate out that amount of Thermal IR. These shortwave energies are not hot in themselves, that is, they are not Thermal Energies. Heat is Thermal Energy in transit from one location to another. Therefore, since they are not themselves thermal energy, they do not heat by their heat, because they have none.
Phil. says:
Correction to previous post.
April 7, 2011 at 7:27 am
Myrrh says:
April 7, 2011 at 12:09 am
Ira, David, Tim et al – do you have some kind of block here that’s preventing you from actually taking in my question? I’m asking about the specifics, how and how much does Solar actually heat the Earth.
It gets absorbed and heats up whatever absorbs it. A watt of UV heats up a black body the same as vis, ir or electric current for that matter. Evidence given above which you ignored!
I’ll try again. AGWScience says that Solar energy, UV, Visible and Near IR, the Shortwave energies, actually heat the Earth which then radiates out Thermal IR (which then gets radiated out/trapped etc.). That’s the basic premise of the AGWScience Greenhouse. This should be bog standard understanding and the calculations at hand for any promoting this hypothesis.
Which there is except for those who willfully ignore it.
How and how much does UV heat the Earth? How and how much does Visible light heat the Earth? How and how much does Near IR heat the Earth?
Prove it.
Done, you are the one who makes the extraordinary statement that a watt of energy at a blue wavelength is incapable of heating an object which absorbs it. What do you think happens to that energy if it is not reflected, where does it go?
I’m telling you, that these do not heat the Earth. These energies are incapable of heating the Earth to the extent that it gets hot enough to radiate out that amount of Thermal IR. These shortwave energies are not hot in themselves, that is, they are not Thermal Energies. Heat is Thermal Energy in transit from one location to another. Therefore, since they are not themselves thermal energy, they do not heat by their heat, because they have none.
False as can be proved by focussing light on any light meter (preferably a bolometer, the very existence of which proves you wrong, they’re traditional physics invented in 1878)!
Joel Shore says:
April 5, 2011 at 6:51 pm
Claes is spouting nonsense, as seems to be his custom. This is my favorite thing that Claes has written – Look at Equation (10.3) here: http://www.csc.kth.se/~cgjoh/ambsblack.pdf He admits this equation is correct but then claims the problem is with the interpretation and spouts the standard talking points about no back-radiation, one-way transmission from hot to cold, etc.
——————————————————————————————————————;
I think Johnson does an excellent job of presenting modern physics to the pseudo-sciences.
There are no contradictions. It’s a physics problem.
Johnson is simply reminding the readers that the Laws of Thermodynanics constrain the solutions of the differential equation to physically accessible states.
The radiation portion of the Stefan-Boltzmann equation was provide by Stefan and the thermodynamic foundation was provided by Boltzmann.
If you’re looking for detailed statistical mechanics justification of the constraint T_2 > T_
1, then you should be reading Boltzmann (or Gibbs) instead of Johnson.
Richard E Smith says:
And, you’ve actually tried this experiment before? Three things makes me pretty sure you haven’t:
(1) If it were true, why would people design reflective luminaires for lights?
(2) What happens to the energy in the reflected light? Does it just disappear and how is this consistent with conservation of energy?
(3) You will presumably admit that two different spotlights shined on a wall will produce a brighter spot than one. So, the claim seems to be related to the light coming from one source. So, I would ask you, what makes the light from one source special? Do the photons carry little labels that say what source they came from? And, if so, what about a fluorescent lightbulb that is a meter long…Does light from the two ends constitute light from the same source or a different source? How far apart do the parts of the source have to be before it is no longer considered the same source?
What I am trying to get you to do here is to think critically about your hypotheses.
Richard E Smith;
You can test it experimentally by pointing a spotlight at a wall in your house and holding a mirror so that as the light reflects off the wall it is reflected back to the wall by the mirror. You will find that the reflection from the mirror will illuminate the wall that is not already illuminated by the spotlight. If you direct it at the part already illuminated you will find that it does not brighten that part any more. The ‘extra’ light has no effect.>>>
Don’t be daft. If you bother to pay close attention instead of just seeing what you want to see, you’ll SEE that you are wrong. Move the reflected spot from the dark part of the wall toward the illuminated part and you will see the dimmer edges of illuminated part show the outline of the reflection as it crosses into them. Just because at the centre the contrast is below what your eye is capable of detecting doesn’t mean it doesn’t exist.
You can’t see air either. Does that mean it doesn’t exist?
Stop contriving experiments based on the limits of human perception and attempting to present them as some sort of evidence. At least stop for a moment and ponder the fact that millions of engineers use these equations every day to design millions of things that WORK. If the principles being expained to you didn’t work exactly like they are being explained to you, then how is it that millions of engineers produce everything from refrigerators to turbines to nuclear submarines, all over the world, in dozens of languages, every single day, based on those exact same principles and manage to build things that WORK? Better still, work AS DESIGNED?
Albert Einstein published a whole set of theoretical equations to describe photons, how they work, how they behave in 1905. If you think your spot light and mirror means anything consider Robert Millikan who was so convinced that Einstein was wrong that he built an experimental environment that took TEN YEARS to put together. His results? AN EXACT MATCH TO EINSTEIN’S EQUATIONS. A ten year experiment with the goal of disproving Einstein’s theories on light and photons, and all he accomplished was to prove Einstein right. In fact, he spent decades doing other experiments trying to find something wrong with Einstein’s equations and didn’t fully admit that his own work confirmed Einstein’s until 1958 when he said it “scarcely permits of any other interpretation than that which Einstein had originally suggested”.
Millikan also conducted rigorous experimentation that confirmed Planck’s Constant.
So if you want to put your silly flashlight and and mirror experiment up as evidence go right ahead. Or you can put aside your misguided notions about what you think you can conclude based on what you can see with the naked eye. Your flash light experiment is up against a ten year, meticulously documented experiment which confirmed Einstein despite being designed to refute him, by a man who won the 1923 Nobel Prize in Physics for that work (amongst others), who also confirmed Planck’s work which in term confirms Stefan-Boltzman.
You want to refute the work of half a dozen Nobel Laureates all confirming each other’s work, all providing the basis for which millions of design engineers do their jobs every single day, repeatedly get the results they designed for, and expect to discredit them all because your personal pair of eyes can’t measure the brightness difference of reflected light from a flashlight and a mirror?
I’ll go one better. If you are right, then laser pointers that you can buy for a few bucks don’t work either. Except they do. By bouncing light back and forth between two mirrors until the photons emerge in a very narrow beam all travelling the exact same direction, creating a bright point that you can see. Take the bulb out, look right at it, and you know what you will see staring straight at the bulb? Nothing, not even a glow, its below what your eye can detect. Put it back in the laser and presto, laser beam.
The evidence that you are dead wrong is all around you, every day. The experiments have been done in mind numbing detail and they all confirm each other. All you have to do is read their work and understand it. Or shine flash lights and mirrors at walls and smugly tell yourself you must be right.
Richard E Smith;
In fact a lid simply confines the area that is being warmed. This both restricts convection and reduces the volume of the area that is heated. It does not add heat to the source of the heat.>>>
Then could you please explain to me why a thick ceramic lid works better than a thin aluminum one? They both have the exact same effect in terms of restricting convections and reducing the volume heated. In fact, why does an aluminum pot heat up faster than a thick ceramic pot? But the ceramic pot eventually gets its contents to a higher temperature? Why is that? Where did the extra heat in the ceramic pot come from to make the contents hotter? Why on the exact same burner do the contents in the ceramic pot heat up slower, but eventually reach a higher temperature? Where did the extra heat come from? And why does it transfer heat to the contents slower, but still over time transfers more?
And why is it that if you know the properties of the pots in enough detail, you can use all the equations we’ve been trying to explain to you to predict exactly how long each pot will take to reach maximum temperature and exactly what the temperature of the contents will be? If the equations are wrong, how does that work?
davidmhoffer, joel and phil (I think) have all derided my experiment of reflecting the light from a spotlight back on to the wall where the spotlight is shining. They claim that the reflection from the mirror does add light to the bright spot but that it simply cannot be observed by the human eye. In other words they believe that reflecting light adds to the light from the source.
Ok then. Assume that a torch is shining on a purely reflective surface – the opposite of a blackbody. 100% of the light will be reflected. Now get a mirror and hold it over the reflective surface. Let’s assume that the mirror reflects 10% of the light leaving the surface back to the surface. That’s now an extra 10% to add to the light from the torch. So according to david and joel 110% of the original supply is now being emitted from the surface. Add a few more mirrors and you can keep on increasing the brightness. Are you really suggesting that you can create energy like this? What about the first law of thermodynamics? – energy can neither be created nor destroyed.
david uses the example of lasers to try to refute my observations, but a laser is not created by simply bouncing light between mirrors – the light has to pass through a gain medium with a supply of energy from an electric current. I do not doubt that light can be concentrated. You can do this with something as simple as a magnifying glass. I remember as a boy using a thick magnifying glass to focus the sunlight on to paper to set fire to it in the back garden in summer. Happy days!
david also takes me to task about lids. He asks why a ceramic lid works better than an aluminium one. Simply because it is a better insulator. This is another way of saying it is not losing so much heat to the exterior so it needs less heat to reach the same temperature. It is nothing to do with backradiation. There is no extra heat.
Agile Aspect says:
What Claes is doing is showing the derivation of the equation (correct as far as I have checked; at least I know the final result iscorrect) and then launching into a bunch of pseudoscience mumbo-jumbo to try to distract you from the fact that the equation itself shows the greenhouse effect and totally negates his point. In large part, he does this by flailing against strawmen (like the idea that heat can flow from cold to hot), which nobody is arguing about and partly he does this by launching into completely irrelevant (and incorrect) arguments about the interpretation of the terms.
I explained to you in my last post on the subject why Claes’ interpretation is nonsense. To summarize:
(1) There is good evidence supporting the notion that the colder object is actually radiating to the warmer one, not the least being that one can measure that radiation. (Of course, the warmer one is radiating toward the colder one to…and the 2nd Law tells us that the flow of heat, i.e., the net of the radiative flows, will be from the hot object to the cold object, as the equation itself explicitly shows.)
(2) It doesn’t matter if you interpret one term as being “back-radiation” or not. Calling it “fred” instead isn’t going to change the fact that the equation itself shows how the heat flow from T_2 to T_1 is reduced as T_1 is increased while T_2 is kept constant. This is the origin of the greenhouse effect.
The only slightly interesting question regarding Claes’ exposition is whether someone with his knowledge can really get so confused in the end or whether what his interpretation actually amounts to is deliberate deception. In the work I have seen on this subject, whether it is Claes’ or Gerlich & Tscheuschner’s, there seems to be this bizarre habit of mixing a lot of mathematics that is correct (although not really original) with this pseudoscience mumbo-jumbo interpretation. It is hard to imagine how this cannot be part of a deliberate attempt to deceive but maybe it is just testimony to how people can completely deceive themselves when their ideology prevents them from accepting what the equations are telling them.
Richard E Smith says:
You are getting more and more ridiculous. First of all, there is no rule that says that light can’t reflect off of a surface multiple times. If it is an absorptive surface, some will be absorbed each time; however, for your imaginary, perfect reflector it is not. There is no problem with conservation of energy because none of the energy from the light beam is being removed from the beam by the perfect reflector!
And, if you go far away from both reflectors and ask where the extra 10% comes from, it comes from the 10% that you took out of the original beam by the partial mirror. I.e., the partial mirror transmits 90% on the first pass, and the other 10% can then be transmitted on subsequent passes.
At some point, what this all comes down to is this: For ideological reasons, you don’t want to believe in the greenhouse effect and you are willing to come up with pseudo-scientific nonsense to confirm your incorrect belief from now til the cows come home. There is no convincing someone whose has such religious beliefs prevent him from comprehending the science.
Richard E Smith:
And, this differs from the greenhouse effect how exactly? The greenhouse effect says that additional CO2 makes the atmosphere a better insulator. Hence, less heat is needed to reach the same surface temperature.
All your mumbo-jumbo about “back radiation” doesn’t matter. Forget that anybody ever used the phrase “back radiation”. It is just a phrase sometimes used to explain at another level how the extra insulating effect works. If you don’t like it, don’t use it. (This retired meteorologist strongly counsels against it: http://www.ems.psu.edu/~fraser/Bad/BadGreenhouse.html ) However, what you call something doesn’t change the scientific reality coming out of the equations, which is exactly like what you have described for the ceramic lid case!
Richard E Smith says:
“Ok then. Assume that a torch is shining on a purely reflective surface – the opposite of a blackbody. 100% of the light will be reflected. Now get a mirror and hold it over the reflective surface. Let’s assume that the mirror reflects 10% of the light leaving the surface back to the surface. That’s now an extra 10% to add to the light from the torch. So according to david and joel 110% of the original supply is now being emitted from the surface. Add a few more mirrors and you can keep on increasing the brightness. Are you really suggesting that you can create energy like this? What about the first law of thermodynamics? – energy can neither be created nor destroyed. ”
No excess energy is being created in your scenario.
Let me make a slight variation of your situation. A beam of light shines on a mirror, and the beam bounces off. The beam hits a second mirror head-on and bounces straight back to the first mirror, so that 100% of the energy bounces back to the first mirror. This makes the spot 200% as bright!
Let’s make a very special light — 1000W but it is only on for 1 ns, so it emits a 1 uJ pulse of energy. As the pulse reaches the mirror, 1 uJ of energy arrives at the mirror, and 1 uJ of energy leaves the mirror and 0 uJ of energy are absorbed by the mirror. Energy was conserved.
Then a few ns later, the pulse returns to the mirror. 1 uJ arrives and 1 uJ leaves and 0 uJ are absorbed. Energy is again conserved.
Send 1 pulse or 2 pulses or a billion pulses (1 second worth of light). Energy is conserved for each pulse, and hence conserved overall. The fact that one new pulse and one old pulse arrive at the same time in no way contradicts conservation of energy.
(In fact, with a few modifications, I think this might be an effective way to explain the GH effect.)
Apologies, I’m not going to able to concentrate on this until next Tuesday, and I haven’t been ignoring what y’all been saying, but nothing so far, and yes I have seen pages on your favourite laws etc., many, many, pages before, addresses my specific points, which I would like in English. So far, none of your explanation in English make any sense. A brief response to some of the points, I will go back over this from Tuesday on in more depth.
Phil – the Earth is not a black body. Visible light get very easily scattered by the molecules of oxygen and nitrogen, I gave this explanation earlier in a link. If your contraption could be attached to real life organic matter..? And how does this laser, concentrated light, compare with that which reaches Earth from the Sun? Real life.
A light meter measures how much light there is, not what I’m asking for, but could be useful in the package of explanation I’m waiting for.
Please, I am still demanding specific answers to specific questions and you haven’t given them, much as you think you have.
Done, you are the one who makes the extraordinary statement that a watt of energy at a blue wavelength is incapable of heating an object which absorbs it. What do you think happens to that energy if it is not reflect, where does it go?
Good question, ask a plant.
Then ask a rock.
Generally, I’m beginning to tire of irrelevant or gobbledegook maths in responses, and to top it all, even saying that Trenberth didn’t know what he meant, and then being blamed for “not understanding”..
Tim says:
April 7, 2011 at 5:15 am
2) is true and indeed the basis of the Greenhouse effect. Additional energy (thermal IR from the atmosphere) is needed to warm the earth.
That’s not true, for all practically purposes their ability to heat anything is insignificant. That’s not at all saying just not enough to do the job. You said these Solar energies are 99% responsible for heating the Earth. HOw? And then you say all the thermal IR floating around is heating it. So UV drills a centimetre of heat into something, how much is that contributing to warming the Earth for it to radiate out the claimed Thermal IR? How intense does Blue light have to be to burn? Is that what we get from the Sun? A reference I gave earlier, I think in one of Ira’s discussions, says Blue light is Benign unless intensified, and then it can burn. But this is artificial.
As AGW claim in the PREMISE of the AGW Energy Budget, it’s 100% Solar, so forget your extra IR needed. Quite specifically it says that it’s the SOLAR energies which Heat the Earth and the Heated Earth gives off x Thermal IR from this source. Whatever the figure that is, what is it?
This, and only this, is what I’m talking about. The BASIC premise of the AGWScience Greenhouse Hypothesis. Remember the AGW greenhouse? It doesn’t give a damn about anything thermal IR heating the Earth, it’s only these Solar UV,Visible,NearIR which are said to heat the Earth which then radiates out Thermal IR. I’m not interesting in what happens next. Just the Solar in, Thermal IR out of the basic premise.
Exactly which wavelengths provide how much power would vary from location to location and from season to season. That is a question beyond the scope of a discussion here. If you are really keen to know, dig into the literature!
Find me some… It shouldn’t be beyond the scope of AGWScience explanation and I haven’t seen anything on this. That’s why I’m asking here of those who are promoting this idea…
How much does Blue Visible Light heat the oceans? How much does Blue light heat a metre depth of water? I’m looking for actual proof that these heat the actual Earth as it is claimed. That this is not readily available is absurd.
I’ve already given a reference to Blue Light being used in raising plant life; it is not hot, it can therefore be placed close to the plants. It doesn’t raise the temperature of the plants, it doesn’t cook them. They are doing something else with that energy, they’re not using it to keep themselves toasty warm.
..stored chemical energy until used by the plant for photosynthesis perhaps.
Ira and Hans – The Trenberth Energy Budget is the AGWEnergy Budget, Ira, you use it. So how does Solar convert to Thermal IR? Miraculously? One second it’s Solar and then hey presto it’s Thermal IR?
Wayne, I’m not sure what you meant in the first about running water filtering out all but the Visible frequencies. I would imagine that it is then highly reflective to short waves, scattering. Thermal IR is bigger, but I suppose it depends on how fast the water is flowing, but water is a very good absorber of it.
Oh, as an aside to everyone here, I’ve recently read someone say that the Sun doesn’t emit infrared! The AGWEnergy Budget in all its glory of education.
Richard E Smith;
In other words they believe that reflecting light adds to the light from the source.>>>
We only believe this because the math says it should be so and when we use instruments that measure these things it turns out that it matches the math. What do you think causes that? Pure coincidence?
Add a few more mirrors and you can keep on increasing the brightness. Are you really suggesting that you can create energy like this?>>>
What I’m suggesting is that you understand neither the explanation given to you, nor your own example.
david uses the example of lasers to try to refute my observations>>>
I’m not trying to refute your observations. I’m trying to point out that by carefull observation of your own experiment you will find out your observations are wrong.
but a laser is not created by simply bouncing light between mirrors – the light has to pass through a gain medium with a supply of energy from an electric current.>>>
Does it really? Have you ever built one? Do you know how? Do you even know what a gain medium is? I for one have no idea what you think that is. On the other hand I’ve actually built a low power laser powered by a flashlight battery with two mirrors and a cardboard tube lined with tinfoil and a 3 watt bulb.
He asks why a ceramic lid works better than an aluminium one. Simply because it is a better insulator.>>>>
GASP! You mean just like CO2?
This is another way of saying it is not losing so much heat to the exterior so it needs less heat to reach the same temperature.>>>>
GASP! You mean just like CO2?
It is nothing to do with backradiation.>>>
GASP! Call it something else if you want. The process is identical.
There is no extra heat.>>>
GASP! Right again! Just like with CO2!
But per your own admission, an insulator (your word) means you need (your words) “less heat to reach the same temperature”. I presume you will also admit then that the same heat would result in a higher temperature? Or do you want to say no and look completely foolish?
CO2 is just an insulator, albeit a very weak and over hyped insulator, but that is what it is. Since the amount of heat coming in doesn’t shange because of CO2, but it insulates against heat going out, higher temperatures result. You’ve just explained it with your own words!
http://www.natscience.com/Uwe/Forum.aspx/optics/1023/Material-transparent-to-light-but-opaque-to-IR
Found the above discussion, maybe of some use. Interesting last post about Lambert’s Law – not in real life, link I posted somewhere about plant life utilising IR for photosynthesis much deeper than that – according to Lambert they would be dead.
I think this is the same recurring problem with ‘some of the laws’, people quote these as if they’re applicable in real life because they relate to a particular discrete context, and as the oximeter link explained, adjustments have to made if you don’t want gigo.
http://equipmentexplained.com/physics/respi_measurements/oxygen/oximeter/pulse_oximeter.html
“Early on, we discussed how the pulse oximeter uses Beer’s and Lambert’s Law (absorbance depends on concentration and path length) as part of its factors that it uses to compute oxygen saturation. Unfortunately, there is a problem. In physics, the Beer and Lambert law have very strict criteria to be accurate. For an example, the light that goes through the sample should go straight through like the lights rays in the image below. However, in real life this does not happen.”
Anyway, as I’ve complained about before, giving me out of context laws as if they’re the answer to everything is nonsense in so much of AGW, carbon dioxide which can rise up through air even though it’s heavier, might make sense in a topsy turvy world of ideal gases, but not in real life.
So, still asking for real information.
Myrrh;
So, still asking for real information.>>>
No use. It just bounces of your Armour of Density.
But I do note that having been given exactly what you asked for, the exact equations, the exact numbers, wavelength by wavelength, frequency by frequency, examples of the exact experiments, exactly documented, even the names of the people who did them and when, articles explaining in detail the exact answers to every question, you now babble on about something else entirely.
I concluded that you either:
1. Read what I told you to and didn’t understand it.
Or
2. You didn’t bother to read it.
One of these people just doesn’t belong here….
Once again, I give up!
Myrrh, it is unrealistic t expect strangers to teach you all of physics and climate science for free via the internet.
“I’m beginning to tire of irrelevant or gobbledegook maths”
Math *is* the language of science. If calculus and infinite series and differential equations are goggledegook to you, you need to learn some more math before claiming you are right and everyone else is wrong. .
“How much does Blue light heat a metre depth of water? ”
There are plenty of references — go find one. Then phrase your question more precisely, like “how much energy is deposited in a square meter of ocean water by light between 400 nm and 450 nm from natural sunlight.” If you can’t calculate this yourself (or at least get a pretty good estimate), then you need to learn some more science before claiming you are right and everyone else is wrong.
“This, and only this, is what I’m talking about. The BASIC premise of the AGWScience Greenhouse Hypothesis. Remember the AGW greenhouse? It doesn’t give a damn about anything thermal IR”
The AGWScience Greenhouse Hypothesis is EXACTLY about the thermal IR — how thermal IR is emitted and absorbed by the surface and the atmosphere! You need to learn some more science before claiming you are right and everyone else is wrong.
“So how does Solar convert to Thermal IR? Miraculously? One second it’s Solar and then hey presto it’s Thermal IR?”
To paraphrase Clarke’s 3rd Law “Any sufficiently advanced science is indistinguishable from magic.” You need to bring yourself up several notches in scientific understanding until this conversion seem perfectly obvious, rather than a miracle!
“Find me some [literature] ”
We are not here as your personal research division. Find some and read them and discuss them. Do a little more of your own work.
And with that — I am done with Myrrh in this thread.