The Earth’s atmosphere is a leaky bucket, with four big holes (and a lot of little ones).
Whole libraries have been filled with talk of a single characteristic emissions layer — a simplistic idea that there is one effective “surface” that radiates to space. It exists in an abstract sense, after sufficient averaging, but it’s a paradigm that doesn’t help us think clearly. In any case, it’s too simple for our purposes in this series. In reality there are many layers that radiate to space, different for each type of molecule that can emit longwave radiation (which means infrared). Then there are the surface and cloud-tops too.
Energy comes in one way but leaves through at least four different paths.
To follow this series you’ll need to understand the concept of four pipes through which energy flows to space. It’s a powerful idea and big advance on the simpler notion of one-pipe-in and one-pipe-out. For those not as familiar with photons and excited molecules, you may want to read the “Background” section at the end of the post first.
For a photon there are a lot of paths to space
Some photons at Earths surface will be at the right wavelength to head straight for Jupiter and stopped by nothing much in the sky. But others will bump into a CO2 molecule which will eat them up (for a while) and get “excited”. Eventually either that excited molecule will spit the photon’s energy right back out in a random direction, or it will run smack into a molecule like Oxygen or Nitrogen (O2 and N2). If a collision happens the excitement (the energy) shifts to another molecule, and so the air generally warms. But O2 and N2 are not greenhouse gas molecules — they can’t release the photon’s energy, so the heat sticks around for a while. Sooner or later the energy from the photon will have been shared and smacked until it hits the jackpot, and ends up in a molecule that is a greenhouse gas, and also happens to be high enough, in thin enough air, to have a sufficiently direct line to space, where it might get ejected in the right direction and leave the Earth forever. This is the top of the emissions layer. It’s different for each type of molecule, and even at different wavelengths for the same molecule. The ones packed in near the surface can’t eject anything to space; there are too many other molecules in the way. The height of the emissions layer turns out to matter a lot — if we thicken the blanket, the layer at the top that can emit is raised, and is also colder. A colder body can’t emit as much.
I know the greenhouse effect is the source of much debate among skeptics; I’ll diplomatically refer people to past discussions of thermodynamics on this site. None of that has changed. We don’t want to rehash that. I think skeptics who have worked doggedly to test the basic theory of greenhouse deserve some credit for intuitively knowing that “something is wrong” with the theory — they are right, CO2 has only a minor effect. But the details matter, and we have to get them correct. Let’s not get stuck in a semantic debate with badly defined words. This is about the net flow of energy — a river that runs inexorably from the Sun to Earth and on to space. Comments that imply that photon “knows” what direction it’s headed in or the temperature of its destination don’t belong here. Let’s talk about the four pipes instead. These diagrams are important.
— Jo
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6. How the Greenhouse Effect Works
Dr David Evans, 28 September 2015, David Evans’ Basic Climate Models Home, Intro, Previous, Next, Nomenclature.
Before proposing a feedback in response to increased CO2 that is omitted from the conventional basic climate model (next post), or the third error in the conventional model, we need to review some aspects of the radiation of heat to space.
All climate modelling at the basic level focuses primarily on how the outgoing longwave radiation (OLR) is affected by changes in the various climate variables. We limit this review to some background and the climate physics of OLR — this is not a complete explanation of the greenhouse effect, just enough for the modelling in this series.
This is all conventional climate science, except that we introduce the terminology of “pipes” for brevity.
The CO2 Blanket
Carbon dioxide absorbs and emits photons with wavelengths around 15 μm (microns). There are many absorption lines (distinct wavelengths, or energies, at which it absorbs) of CO2 around 15 μm, and each line is blurred by factors such as the Doppler effect so that the CO2 can effectively absorb and emit in a narrow but continuous range of wavelengths around each line. The end result is that, in the current atmosphere, CO2 absorb photons from ~13 μm to ~18 μm, with various probabilities. Absorption is less likely at wavelengths further from an absorption line — a photon on such a wavelength can on average travel further through a cloud of CO2. CO2 also has other absorption lines, but these are not so relevant to its effect on climate. These issues are well explained at greater depth on the Barrett-Bellamy website (see the last diagram!).
There is sufficient CO2 in our atmosphere to make it quite opaque at the wavelengths at which CO2 absorbs and emits. A photon near 15 μm can only travel, on average, a few meters through the troposphere before being absorbed by a CO2 molecule. CO2 is a slightly heavier molecule than N2 or O2, so it tends to settles to the bottom of the atmosphere. Nonetheless, CO2 concentrations are moderately uniform throughout the troposphere, gradually decreasing with height, and the CO2 blanket persists well into the lower stratosphere.
OLR consists of the infrared photons that escape to space. The crucial observation here is that, at the wavelengths at which CO2 absorbs and emits, the photons that contribute to OLR nearly all come from the very top layer of CO2.
The CO2 emission layer is the optical upper boundary of the CO2 surrounding the Earth. It is at the effective or average height of emission to space on the wavelengths at which CO2 absorbs and emits: photons on those wavelengths emitted well below the layer are usually absorbed before they reach space, photons on those wavelengths that are emitted upwards from above the layer mainly make it to space, and an observer in space at those wavelengths can only “see” into the atmosphere about as deep as the CO2 emission layer (about one optical depth).
Figure 1: The “greenhouse effect” works by displacing the layer from which OLR is emitted, from the warm surface to some colder place high in the atmosphere
Sometimes we extend the concept to be wavelength-specific: the CO2 emission layer can be at different heights at different wavelengths. At wavelengths where CO2 absorbs with lower probability, the emission layer is at a lower altitude, because photons can go through more of the CO2 cloud yet still escape to space. At the edges of the band of wavelengths at which CO2 absorbs, the emission layer can be down on the surface because it takes a whole atmosphere of CO2 to raise the total probability of absorption to that of one optical depth.
Blankets of Other Greenhouse Gases
Likewise, there is a water vapor emissions layer (WVEL) for water vapor, the effective upper optical boundary of the water vapor in the atmosphere. Water vapor is the main greenhouse gas; CO2 is the second most influential. Because water vapor is not well mixed in the atmosphere, the WVEL can change height frequently in a given location. In the modeling here we use the WVEL concept only as a global average.
There is a separate emission layer for each greenhouse gas — so they exist for methane, ozone, etc.
The Four Main Emission Layers
Earth’s OLR is mostly emitted by four disparate emissions layers:
Figure 2.
The (infrared) atmospheric window is the collection of wavelengths at which photons can pass through the atmosphere without being absorbed by any greenhouse gas. Photons in the atmospheric window emitted by the land or sea surface are free to escape to space if the sky is clear of clouds — but clouds are opaque to those photons and absorb them. Clouds emit infrared at a wide range of wavelengths, similar to the surface, so the OLR in the atmospheric window comes from the surface and from cloud tops.
Above each point on the Earth’s surface, OLR at different wavelengths physically originates from several widely-different altitudes. The Earth’s emission spectrum, such as observed by the Nimbus satellites (later in this series), is clear evidence of the disparate nature of the physical emission layers.
Pipes
We introduce the term “pipe” as shorthand for the outgoing longwave radiation (OLR) over a group of electromagnetic wavelengths and from a particular emissions layer — it’s so much easier to say “CO2 pipe” than “OLR emitted by the CO2 emission layer on the wavelengths absorbed and emitted by CO2”.
Figure 3.
There are also minor pipes for the minor GHGs: ozone, methane, nitrous oxide, etc.
How Much OLR is Emitted by an Emission Layer?
The amount of OLR emitted by a physical emission layer (i.e. escaping through its pipe) is determined almost solely by, and increases with, its temperature — by Planck’s law (which underlies the Stefan-Boltzmann law).
At a given wavelength, the temperature of a physical emission layer of greenhouse gas is the average temperature of the gas molecules that emit to space on that wavelength — which are close together in height because they are near the top of the cloud of such molecules in the atmosphere, and thus have about the same temperature.
The total OLR can be determined from the temperatures of the various physical emission layers: determine the OLR from each emission layer from its average temperature, then sum the OLRs from each of the emission layers.
The Greenhouse Effect
The greenhouse effect is basically a displacement of OLR, from being emitted at the warm surface to being emitted from a colder place high in the atmosphere, which emits less because it is colder. So the presence of a GHG “traps” some heat. See Fig. 1.
Taking a broader view, by the conservation of energy the OLR is equal to the ASR in steady state. Just consider the four main pipes shown in Fig. 3. The total OLR through the four pipes is equal to the ASR, yet the amount of OLR in each pipe is determined by the temperature of its emission layer. Therefore the temperatures of the four emissions layers must be such that they collectively emit OLR equal to the ASR. If the ASR changes, then there is a corresponding adjustment to the OLR — which may involve warming the surface, changing the height of the WVEL or cloud tops (thus changing their temperature), or adjusting the lapse rate (which changes the relative temperatures of the various emission layers).
Increasing the CO2 concentration reduces the amount of OLR through/in the CO2 pipe. If the CO2 pipe is slightly “squeezed” (to use an analogy), so less OLR goes through it, what is going to happen? Obviously more OLR must flow through the other three pipes, collectively. That will involve the surface pipe carrying more OLR, which would require the surface to warm — and the $64 trillion question is, how much warmer?
(Btw, note that energy can effectively change wavelengths via collisions between molecules. For example, a photon at one wavelength might be absorbed by a GHG molecule, which then transfers some of that energy to a molecule of another species of GHG via a collision, which then emits a photon at a different wavelength.)
The Characteristic Emission Layer is Unrealistically Simple
OLR is sometimes modeled as coming from a single layer, called the “characteristic emission layer”. However this layer is only a modeling construct, an average of the physical emission layers. There is no single surface — a layer of molecules at much the same average temperature — from which OLR physically originates. The temperature of the characteristic emissions layer is not directly relevant to OLR, because it is not where outgoing emissions physically originate.
The characteristic emission layer is too simple for our purposes in this blog post series, a simplification too far. As Albert Einstein famously opined, “Everything should be as simple as possible, but no more so.”
Background
– Photons
Energy (or heat) only enters or leaves the Earth’s climate in significant quantities as radiation, which is best viewed for these purposes as consisting of bazillions of photons. Photons can be thought of as tiny massless particles of energy, traveling close to the speed of light. The higher the frequency of a radiation, the smaller its wavelength and the higher the energy in each of its photons (its energy is proportional to its frequency, which is inversely proportional to its wavelength).
In decreasing order of energy: x-ray photons, ultraviolet photons, visible photons, infrared photons, radio-frequency photons.
– Molecules and Atoms Absorb and Emit Photons
A molecule is group of two or more atoms, tightly bound together by chemical bonds so that it acts as a single object for most purposes. Molecules and atoms vibrate internally in various ways, and these vibrations store energy. Crucially, due to the geometry and internal states of the atom or molecule, it can only store certain exact amounts of energy — it cannot store just any amount or a continuum of values, only certain discrete values (hence the name for the study of atomic and sub-atomic particles, “quantum physics”).
When a photon collides with an atom or molecule, it will be absorbed by the atom or molecule if the energy of the photon exactly matches the difference between the amount of energy currently stored by the molecule or atom and an amount that it could store. If the photon is absorbed, the photon ceases to exist and its energy is transferred to the molecule or atom, which is now in a higher energy state (i.e. it vibrates more).
Conversely, a molecule or atom can emit a photon, creating it from scratch and firing it off in a random direction with energy equal to the difference between the stored energy of the atom or molecule in its initial state and its lower energy level after the emission.
Notice that the energy and thus the wavelength of an emitted photon are exactly the same as that of an absorbed photon that changed the energy level of the atom or molecule previously — it’s a reversible process, as the energy stored in the atom or molecule moves up or down with the absorption or emission of a photon. Thus a molecule or atom absorbs and emits photons on exactly the same wavelengths.
Nearly all the “objects” in the atmosphere are molecules, principally N2 (nitrogen) and O2 (oxygen); there are very few lone atoms.
A greenhouse gas (GHG) is characterized by being able to absorb and emit photons of relatively low energy, the infrared photons. A GHG molecule consists of three or more atoms — the torsional vibration modes that are made possible by having two or more inter-atom bonds can store relatively low amounts of energy. Most (all?) atmospheric molecules with three or more atoms are GHGs: H2O, CO2, O3 (ozone), CH4 (methane), N2O (nitrous oxide), etc. The atmospheric molecules with only two atoms, principally N2 and O2, are not GHGs — their lowest absorption energy is higher than the energy typically carried by the infrared photons in the atmosphere, so they don’t absorb the photons and they are transparent to the radiation.
Thanks for this.. And now hands up those who don’t believe in photons.
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I have both hands up!
If electromagnetic flux, (not photons), has sufficient four-space power density, i.e. sufficient Planck units (h) of ‘action'(one cycle of work), to be higher than some predefined ‘work function’ (say electron emission of nickel) than such action has a finite probability of occurring. This is called the photoelectric effect.
Your ‘photon’ is but an abstract (fantasy,virtual) gauge boson that mediates that power transfer with the probability thereof. Photons never go, they simply are when needed.
All the best! -will-
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Yes I did mean you. Well recognised.
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How much for a whole bag of photons? Perhaps three at each frequency, and audited? 🙂
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Do you want those in red, green, blue, purple, yellow or mixed sir?
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gai I am looking for some high power sub Hz photons. The type that would be radiated from the antenna that forms out of a planets magnetic tail and modulated by alterations to that tail such as the moon passing through it. Any hints on suppliers?
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Try these guys Siliggy link
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You can also try this supplier They even have Free Shipping on Orders $75+
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The hard ones are the 60 Hz puppies,so much used in the US, Three of which end to end is from NYC to LA. Perhaps those pipeline folk have some way of rolling them up! Fortunately not to many Hz left. Here comes Sliggy with the damned sub Hz crap that has not even one Planck unit (h) of action. This is much like the CAGW Clowns that think increased temperature above the tropopause somehow mean more power!
All the best! -will-
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SOMETHING “goes”. Photons don’t just quantum tunnel through everything, like electrons can in narrow circumstances. Else single- (and for that matter double-) slit experiments would not get the results they do.
So the “travel” has some effect. It also takes finite time.
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For those who don’t believe in photons, try Feynman’s lecture:
https://www.youtube.com/watch?v=eLQ2atfqk2c&list=PL8590A6E18255B3F4
36:00 – 40:30.
And then review 24:00 to 25:15 – “If you don’t like [the way nature works], go to another universe”.
Or check out the book written from these lectures: “QED”
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I can’t see any reason to not use the concept of a photon, a quantum of energy that is as small as electromagnetic energy at some wavelength can be subdivided into. If it works, use it. If it doesn’t, use the electromagnetic field. Each has its use as far as I can see.
60 Hz is a long way from the wavelengths being discussed here, a very long way. But what works, works. If it describes the thing usefully, it’s useful.
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I agree if useful. Way different concepts of duality, work function, action per cycle, and resolution of probability. All are smuched together with intent to confuse! When the context is clearly defined the concept is reviled. The word ‘photon’ by itself has no meaning!
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Will,
You are in disagreement with a lot of the physics world — your privilege too as far as I’m concerned. But the photon is a very useful concept since only certain energy levels will interact with, say, a CO2 molecule. Electromagnetic field theory doesn’t appear to me to deal adequately with that fact (or at least I see no reason to not believe it’s fact). At short wavelengths in the visible spectrum, for example, behavior is a lot more like a particle with only specific energy levels possible than it’s like an em field with any arbitrary energy level being possible.
If you’re willing I’d like to see a more detailed exposition on your position. I’m always willing to learn.
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“But the photon is a very useful concept since only certain energy levels will interact with, say, a CO2 molecule.”
This is a good example of the rampant misuse of ‘photon’.
Just what certain energy level do you mean? A CO2 gas molecule like any gas molecule has a noise power of kT/t. If you speak of emitting a photon, say a 15 micron ‘photon’ .07 ev just how much energy departs from that molecule? Can one CO2 molecule even do that? How much did the molecule’s temperature change? How do you know? What is the size of the ‘photon’ compared to the size of that CO2 molecule? Does your ‘photon’ go through one or two slits? Can it diffract with itself?
Why use the word when most folk will misinterpret what you mean?
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Indeed!
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Will: Scientists may discuss the noise power in solids, but the proper terminology for gases is a Boltzmann distribution of kinetic, rotational, vibrational, and electronic energy levels. Unlike solids (which often exhibit a continuum of energy levels), rotational, vibrational and electronic states are quantized and transitions between these states are caused by collisions OR by absorption or emission of a discrete particle (a photon) with hv equal to the energy difference between these states. The QM rules governing the behavior of these particles creates phenomena resembling that exhibited by waves, but it does not make them waves. They are discrete particles with a definite amount of energy absorbed or emitted by molecules that gain or lose that amount of energy. There is nothing “noisy” about this behavior; random collisions create a Boltzmann distribution of energy among these QUANTIZED states and kinetic energy. Single molecules have kinetic energy, but not a temperature. Temperature is a thermodynamic concept involving the mean kinetic energy of a large group of rapidly colliding molecules and the 2LoT applies only when temperature is defined.
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The diagrams look fantastic
KK
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What percentage of Surface Radiation is at 15microns?
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Zero percent. The Earth’s surface does not radiate at 15 microns as the opposing radiance from the first 2 meters of atmosphere stops that. Over water sometimes the air is at a higher temperature than the surface. 🙂
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Next time you take a bath, fill the tub with cold water and try to warm it with a hot air gun. 😛
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“Next time you take a bath, fill the tub with cold water and try to warm it with a hot air gun”
…..while you are NOT IN THE TUB…..
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RCDs will ensure that the stupid will survive to breed another generation.
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(“Over water sometimes the air is at a higher temperature than the surface.”)
“Next time you take a bath, fill the tub with cold water and try to warm it with a hot air gun. :-P”
Not very effective indeed, but higher radiance above a surface demands no radiative flux ‘from’ that surface.
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The graph below shows the blackbody radiation flux of an object that is 277K (about 14 degrees Celsius). The flux is just the amount of energy that is emitted from each square meter of the surface per second. The total energy emitted is the total area under the curve. The amount of energy emitted between 13 and 18 microns is shown by the shaded area. So the percentage of the radiation emitted between 13 and 18 microns would just be the percetage of the area of the shaded region relative to the total area under the curve.
In other words, not much.
https://nhill.files.wordpress.com/2015/09/blackbodycurve.jpg
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I had a look at your black body emmission graph at 277K.
I noticed that;
1. The 13-18um emmission band is near the middle of the emission curve,
2. The vertical scale of emission flux is logarithmic.
Consequently the large areas in the lower part of the diagram contribute very little to the total emissions.
My eyeball estimate is that about 20% of the total surface emission could be in the 13-18um band. That may be a low estimate.
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The graph is of spectral radiance not any sort of emission. Emission at any wavelength is always limited or even reversed by opposing radiance.
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Nicholas,
Good graph (presumably 287 K, not 277 K). However, it shows emissions from the surface. For the modeling that works towards the ECS, however, we need only focus on the much simpler problem of emissions that actually make it to space, the OLR. Much of the spectrum in your graph is attenuated by various greenhouse gases, and clouds.
Later, when examining the observed spectrum of OLR such as from the Nimbus instruments, we find that about 20% of the OLR is in the CO2 pipe (that is, emitted by CO2, that is, at the wavelengths at which CO2 typically emits in our atmosphere). Doubling CO2 once would reduce that by about 1.5% (of the total OLR), and three doublings might reduce it from 20% of OLR to 15% of OLR.
Take a look at the last diagram on this page (Barrett’s): the change in the spectrum as the concentration of CO2 rises appears fairly trivial, and at first I was left scratching my head about how it could have much effect. But the answer is along the lines above.
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Looks like my estimate, 20%, was almost spot on then.
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The graph is of spectral radiance not any sort of emission. Emission at any wavelength is always limited or even reversed by opposing radiance. How can something not emitted ever be absorbed. More and more deliberate confusion.
“Take a look at the last diagram on this page (Barrett’s): the change in the spectrum as the concentration of CO2 rises appears fairly trivial, and at first I was left scratching my head about how it could have much effect. But the answer is along the lines above.”
Again deliberate misdirection. Radiative CO2 exitance to space originates at the tropopause. The surface does not radiate in the 14-18 micron band. Nothing to absorb past present or future. All scam!
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At last – a new post WITHOUT equations.
I can understand the words in this particular post MUCH better than one filled with equations – sorry but I’m a better wordsmith that mathematician.
At least I am on the correct – shall we say “wavelength” now 🙂
Cheers,
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Me too. Far less brain burn from reading this. I’m pretty sure I understood most of it.
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Presumably higher frequency radiation reflected directly by clouds and particulates form another “pipe” which will be considered in a later post?
Thanks for this series of posts David, I’m really enjoying them, even though much of the calculus is beyond me.
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No, there are just the four pipes mentioned — plus minor ones for ozone, methane, NO2, etc. I’m not sure I understand your question Rollo.
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Rollo,
Dr Evans takes care of that problem in his definitions:
ASR = A = Absorbed solar radiation (energy from the Sun that is not reflected)
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That is so, gai, and a reasonable approach. But at some point there is variability in the amount of ASR due to cloud fluctuations, which will affect all the pipes. And changes in solar output would have a similar effect.
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Indeed! that variability is called climate! 🙂
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IIRC, Dr Evans said he will address Albedo later. Right now he is staying within the confines of the ‘Accepted’ model.
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Those CO2 molecules that are just below their emission layer – some can emit downwards, then their photon can be captured by a lower CO2 molecule, repeat the several times and you have downwards flow that I have not seen in past accounting examples. Is it affecting the maths balance or does it simply report as a delay?
Geoff
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Geoff, the crucial point is just that OLR tends to come from layers in the atmosphere, and the temperature of a layer determines how much OLR is produced by the layer. The timing of individual photons in the OLR, and what happens beneath the relevant layer at a given wavelength, are of no interest to the basic model.
The basic model just cares about how much OLR there is, and the temperature of the emission layers. We are especially interested in the temperature of one emission layer, the surface. If the surface has to emit more OLR because the CO2 pipe carries less OLR (due to extra CO2), then we know the surface warms. Later we are going to quantify this.
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“The basic model just cares about how much OLR there is, and the temperature of the emission layers. We are especially interested in the temperature of one emission layer, the surface. If the surface has to emit more OLR because the CO2 pipe carries less OLR (due to extra CO2), then we know the surface warms. Later we are going to quantify this.”
This is totally invalid. Both Clowns and you claim, but cannot demonstrate ‘any’ decrease in CO2 exitance with increasing CO2 ppmv.
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let me just give you a large balloon weighing several tonnes as an example. there is no downward energy.There is only energy in pursuit of equilibrium (cooling) going upwards , always upwards .
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“Those CO2 molecules that are just below their emission layer – some can emit downwards, then their photon can be captured by a lower CO2 molecule, repeat the several times and you have downwards flow that I have not seen in past accounting examples. Is it affecting the maths balance or does it simply report as a delay? Geoff”
The emission you claim (toward higher radiance), cannot emit as per Maxwell’s equations. In any case, such emission is claimed in Climate Clown Models (CCM) but has never ever been demonstrated or observed in this physical! 🙂
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Will.
I have previously observed that back radiation cannot be real.
KK
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Geoff: As radiation of a given wavelength passes an infinitesimal distance (ds) through molecules that can absorb and emit that wavelength, the intensity of the entering radiation (I_0) changes (dI) according to the Schwarzschild eqn:
dI = emission – absorption
dI = n*o*B(lambda,T)*ds – n*o*I_0*ds
dI = n*o*[B(lambda,T) – I_0]*ds
dI/ds = n*o*[B(lambda,T) – I_0]
where n is the density of absorbing molecules, o is their absorption cross-section at this wavelength, B(lambda,T) is the Planck function used for blackbody radiation and T is the local temperature. This is the equation used in all radiative transfer calculations.
If radiation has passed far enough through these molecules that absorption and emission have come into equilibrium, dI/ds = 0 and I_0 = B(lambda,T) – the radiation has blackbody intensity at that wavelength. The derivation of Planck’s Law assumes radiation at equilibrium with its surroundings.
When the temperature is low enough or I_0 is high enough that emission is negligible, the equation simplifies to Beer’s Law.
For radiation traveling upward, I_0 comes from below where T is usually warmer and B(lambda,T) is bigger. So the term in brackets is usually negative. In that case, increasing n (the density of GHG) makes dI/ds more negative. This is the GHE. The GHE won’t exist in an isothermal atmosphere.
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“Geoff: As radiation of a given wavelength passes an infinitesimal distance (ds) through molecules that can absorb and emit that wavelength, the intensity of the entering radiation (I_0) changes (dI) according to the Schwarzschild eqn:”
The Schuster Schwarzschild two stream approximation is never applicable to non luminous (continuum) atmospheres!
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Will: Last time I checked, the earth’s atmosphere was luminous in the infrared. The equation I cited is used in radiation transfer calculation for non-scattering atmospheres in LTE. LTE exists to about 100 km and there is. Little scattering of infrared. You can confirm this in Grant Petty’s “A first course in Atmospheric Radiation” or at http://www.barrettbellamyclimate.com/page47.htm
I believe that MODTRAN and other software do radiative transfer calculation by inumerically integrating the Schwarzschild equation along a vertical path from the surface to space (OLR) and space to the surface (DLR) and over all thermal infrared wavelengths. The 3.7 W/m2 forcing for 2XCO2 was calculated this way. The needed absorption cross sections are calculated using data in the HITRAN database and change with altitude (pressure and temperature broadening. See:
http://scienceofdoom.com/2011/03/12/understanding-atmospheric-radiation-and-the-greenhouse-effect-part-nine/
Perhaps I’m misinformed about some of these basics of atmospheric radiation despite my owning a copy of Petty. If so, I’d appreciate a reliable reference to help me learn rather than disdainful dismissal.
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David,
I posted this on 5 but belongs here!
At CO2 100ppmv, 12km, 10kpa Optical depth @ 14.6 microns is 40 meters.
At CO2 200ppmv, 12km, 10kpa Optical depth @ 14.6 microns is 20 meters.
At CO2 400ppmv, 12km, 10kpa Optical depth @ 14.6 microns is 10 meters.
At CO2 800ppmv, 12km, 10kpa Optical depth @ 14.6 microns is- 5 meters.
Using a 14.2-15.2 micron bandwidth, how much did EMR exit flux @14.6 microns to space change from a 100 to 800 ppmv CO2? How much did broadband EMR exit flux to space change going from a 100 to 800 ppmv CO2?
How much did temperature anywhere change? Is this amount even measurable? The basic Climate Clown Model (CCM) represents fantasy, not anything connected with this Earth’s atmosphere.
All the best! -will-
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Will,
By “EMR exit flux” I assume you mean “OLR”.
Changing the CO2 concentration makes very little difference close to 15 microns, because the CO2 EL is in the stratosphere at those wavelengths — moving the CO2 EL up or down makes almost no difference, because in the stratosphere the temperature gradient is very small. Increasing the CO2 concentration from 100 ppm to 800 ppm, at 14.6 microns I am pretty sure the CO2 EL would be in the stratosphere at around 220 K at and between both concentrations — so the answer to your question is “an insignificant amount”.
However, out in the wings of the CO2 blockage (the indention in the spectrum of the OLR, such as a Nimbus diagram), there would be substantial differences, because at some wavelengths (maybe nearer 13 and 18 microns) the CO2 EL is in the troposphere so changing the CO2 concentration from 100 ppm to 800 ppm would raise the CO2 EL at these wavelengths considerably — making it colder and emit less. Because there are three doublings from 100 to 800 popm, the total OLR in the CO2 pipe would decrease by about 3 times D-sub-R,2X, or about 3 * 3.7 W/m2, or about 11 W/m2.
Please see the fourth (last) diagram at this page on Barrett’s website. It shows how the spectrum of OLR changes in the wavelengths at which CO2 absorbs, as the concentration changes from 380 ppm to 1,000 ppm. For more context, see the rest of the page there, and perhaps the pages before.
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Dr Evans, perhaps you would like to include this illustration in the text.
http://www2.sunysuffolk.edu/mandias/global_warming/images/stratospheric_cooling.jpg
04
” However, out in the wings of the CO2 blockage (the indention in the spectrum of the OLR, such as a Nimbus diagram), there would be substantial differences, because at some wavelengths (maybe nearer 13 and 18 microns) the CO2 EL is in the troposphere so changing the CO2 concentration from 100 ppm to 800 ppm would raise the CO2 EL at these wavelengths considerably — making it colder and emit less. Because there are three doublings from 100 to 800 popm, the total OLR in the CO2 pipe would decrease by about 3 times D-sub-R,2X, or about 3 * 3.7 W/m2, or about 11 W/m2.”
There is almost no pressure or velocity broadening of CO2 lines at pressures below 40kPa! What you are seeing is the continuum emissivity of H2O in whatever phase. This again is where the HiTran models go completely haywire! You would find none of your claimed 11 W/m^2 decrease in flux.
The Barrett stuff is again the misuse of the HiTran data base computation for political gain. His 330 ppm 100 m path transmission plot appears to be for 101kPa not 10kPa. Such also looks no different from 100 to 800 ppmv CO2. There certainly is no such thing as an CO2 emission level that changes temperature or optical depth in any significant way.
I do not mean to criticize your approach, whatever works for you is fine. However, if I can poke holes, ‘they’ can poke bigger holes. Please do not underestimate money!
All the best! -will-
716
Hi Will. So when are you going to publish your model? You certainly must have one, right?
122
“Hi Will. So when are you going to publish your model? You certainly must have one, right?”
No, I do not even want a model, such is always useless!! I just measure what ever is!
Today’s scientists have substituted mathematics for experiments, and they wander off through equation after equation and eventually build a structure which has no relation to reality.”— Nikola Tesla
1315
Will: Doubling CO2 causes about a 1.5% change in OLR. You don’t expect to see a big difference between the spectrum of OLR when CO2 doubles.
70
Dr. Evans
Congratulations on your superb work.
I have often seen it claimed that the effect of CO2 is logarithmic rather than arithmetic, and therefore has little influence above 200 ppm.
Does this correspond with your explanation here?
20
Rod, yes the effect of CO2 is logarithmic, and while it has decreasing influence for each 10 ppm increment as the concentration increases, nonetheless increasing the CO2 concentration always has some influence on the climate.
In both the conventional model and the alternative model we will develop, the CO2 influence (green box) portrays the logarithmic effect by using the variable L, the log base-2 of the CO2 concentration C — see Fig 2 of post 3.
10
The timing of photon emission from a CO2 molecule that has previously absorbed a photon will be dependent on the amount of collisional activity (conduction) going on at the location of that CO2 molecule.
The amount of collisional activity is related to atmospheric density because the denser the concentration of molecules the more readily collisions (and thus conduction) will occur relative to the amount of photon emission. The same package of energy cannot be radiated and conducted at one and the same time so photon emission from a GHG declines as collisional activity with non GHGs increases.
Atmospheric density at any given height is a consequence of atmospheric mass and the strength of the gravitational field.
The vast majority of the energy held by an atmosphere held off a surface in hydrostatic equilibrium was initially acquired by conduction and convection from the heated surface below and does not come from radiative absorption by radiative gases. The decline in temperature with height is due primarily to the declining density gradient (set by gravity) and not by radiative loss to space from GHGs.
The S-B temperature is calculated on the basis of there being no absorption of energy by an atmosphere. It implies complete transparency (zero opacity). Note that all absorption contributes to the opacity of an atmosphere to radiative energy flowing through.
Absorption by conduction (and then convection) contributes to opacity just as does radiative absorption by GHGs.
The importance of David’s step in acknowledging multiple emissions heights for differing radiative components within the atmosphere is that it allows convective adjustments to move energy from one ‘pipe’ to another so that radiative imbalances caused by one component (such as the blocking of the 15u wavelength by CO2) can be negated by a change in radiative activity by another ‘pipe’ or by the surface.
If the vast bulk of the energy in an atmosphere was initially obtained from conduction and convection at the surface and if convective adjustments deal with radiative imbalances as per established science:
http://www.public.asu.edu/~hhuang38/mae578_lecture_06.pdf
then it was a profound error to assume significance for the radiative capabilities of GHGs.
1617
Dr Happer in his 2014 lecture answered my question about where CO2 energy is radiated instead of being handed off via collision. Experimental data shows barely any radiation at 11 KM and that radiating is in the stratosphere ~ 47 KM above the surface.
David Burton put up an audio video and slides of Dr Happer’s presentation at this. link
SLIDES: link
Slides 16, 22, 42, 43 and 44 are the critical slides.
80
While possibly not relevant to this part of the topic, two comments should be made about incoming solar radiation: 1. The reflectance of sunlight from clouds containing ice particles as was found by the CSIRO cloud physics group many decades ago, to be much greater than they had originally assumed.
And 2. When sunlight is incident on the near spherical earth, as the angle of incidence decreases towards its ‘edge,’ the amount of radiation reflected increases, so less reaches the surface, as in polar regions and at sunrise and sunset. This is from Lambert’s cosine law of diffuse reflectance.
So when determining the albedo of the earth, using a supposed average figure of 0.3 may be questionable, when reflectance varies considerably with place, time and angle of incidence and angle of measurement.
90
Dr. Happer’s lecture has slides of FTIR satellite spectrum of interest to this discussion.
In Slide 16, The thunderhead anvil FTIR satellite spectrum is top right graph. The BOTTOM CURVE just above the X axis. In other words water, as liquid and ice, blocks almost all outgoing IR. The other four spectra are cloud free spectra for the tropical west Pacific, Sahara Desert, Southern Iraq and Antarctica.
Audio and slides of the physics lecture
http://www.sealevel.info/Happer_UNC_2014-09-08/
SLIDES:
http://www.sealevel.info/Happer_UNC_2014-09-08/UNC-9-8-2014.pptx
A less-technical lecture for the lay person:
http://jlf.streamhammer.com/speakers/williamhapper090814.mp4
40
gai –
Tx for the Happer link.
The spectrum of Antarctica in Slide 16 shows that Antarctica with its atmosphere radiates more heat than it does without, and that it’s CO2 wot dun it. This is remarkable, and (it seems) not widely discussed. Clearly the CO2 “pipe” has interesting properties, and I hope Dr Evans has a better explanation than the standard narrative :).
20
O/T…
27 Sept: WUWT: Eric Worrall: Aussie Government Climate Skeptic Purge Continues
The Guardian reports that Australia’s new climate alarmist Prime Minister Malcolm Turnbull, has not renewed outspoken skeptic Maurice Newman’s appointment as head of the Prime Ministers Business Advisory Council…
http://wattsupwiththat.com/2015/09/27/aussie-government-climate-skeptic-purge-continues/
70
Thanks Pat,
But it is not news I wanted to hear,
Cheers,
Dave B
20
I don’t agree with part of this.
The magnitude of emission of energy received from a photon (radiatively) is NOT going to be affected by the height of the emmission layer. Just the range of frequencies.
Cold molecules have a low range of velocities, and so the doppler shift associated is lower, the absorbtion band is therefore lower, this means that this layer becomes more transparent to IR emitted by faster moving hotter CO2 molecules below. Also the emission surface expands by the cube of the radius. Remembering also that the density of CO2 has also increased (more emitting CO2). This all needs to be taken into account.
What WILL be affected is the probability that a collision with another molecule will result in excitation and therefore able to be emitted to space. But that has more to do with conduction and convection.
I think this needs to be computed properly from statistical distributions of emission. The “average” is not the same distribution of wavelengths, the new higher emission layer has a different spectrum, it’s emitted from a different surface area and from more molecules.
61
TSI heat transfer properties alter depending on the spectrums of distance and time, this explains my previous question on the wings of Icarus melting at such a high altitude.
01
David, would you please confine your radiation explanations strictly to electromagnetic wave theory, which as an electrical engineer, I can fully understand and absolutley verify. The dual theory of particles and photons is surely an area in which we need not stray. Electromagnetic radiation is always a vector field, which explains why radiant energy flux always flows one way from higher energy emitters to lower energy emitters, i.e. hotter to colder. The radiant flux is always a vector function of their potential difference, and ceases, when there is vector equalisation. Perhaps an explanation along these lines would clarify the CO2 question once and for all.
89
Welcome to the community of the still barely alive! 🙂
810
EForster
What you say here is exactly wrong and is NOT valid wave theory at all.
To see this is wrong just throw two stones into a pond. The ripple waves pass right on through each other and DO NOT CEASE. Standing wave peaks and nulls occur but neither the larger or smaller wave is inhibited at all. Both sets of wave rings continue to the edge of the pond.
Single sideband radio reception of multiple signals of different strength also works with a local strong AM carrier to act as beat source! I have done this simple experiment using CB radio. Put a rubber band around one of two AM transciever microphones to lock one on TX with the other nearby in RX mode on the EXACT same frequency provided the local TX does not fully saturate the RX input, distant faint SSB signals are beutifully decoded. As a kid that could not afford a SSB CB but only cheaper AM ones i used this trick to listen to those who could as a teenager. It only works because the outgoing carrier DOES NOT cause the incoming weak signal to cease.
21
If you start generating a continuous wave in a pond then a float in the pond will tend to start bobbing up and down. At first it bobs a little and then it acquires enough energy to bob continuously almost as much as the incoming wave amplitude. Getting it bobbing initially takes more energy from the wave than after it gets into its stride. That is how waves transfer energy, and radiating sources interact in a similar way.
00
Yes and this bobbing will result in the float distorting the wave shape and causing it’s own different resonant frequency to radiate less energy back toward the source. This will alter the behavior of the source slightly as the back radiation is absorbed.
00
“EForster What you say here is exactly wrong and is NOT valid wave theory at all.”
EForster is correct for electromagnetic fields and unidirectional flux at any frequency and position. Orthogonal superposition of equal flux results in the vector sum root 2 magnitude at a combined 45 degree angle to each orthogonal direction, still in a direction of lower radiance (normalized field strength). At a location of equal and opposite field strengths, flux ceases,yet modulation sidebands proceed in both directions. There is power in the sidebands but flipping in direction with each half modulation cycle. Such is similar to the current claim of bidirectional flux, yet all power is supplied only from the local demodulation scheme.
Even with wideband thermal EMR; amplitude modulation in both directions can be demonstrated while both ends remain at the same average temperature and results in no power transfer in either direction, but synchronous demodulation at both ends is easy. This again may sound like ‘back radiation’ but is not. The random thermal noise carrier that cannot be demodulated, not only has disappeared, it is not needed, so is not generated.
29
This is like saying that you get no power from a coal fired power station because the current flows both directions every half cycle. Such is similar to the current claim of bidirectional flux, yet all power is supplied only from the local bar radiator scheme.
00
Now place the two lamps 1 light year apart. The transmitting lamp can be switched off, dissasembled, melted down and recycled before the zero power travels to the other lamp. Does that zero figure still seem logical to you?
00
BTW: Would you check please:
Each CO2 molecule that actually absorbs a 15 micron unit of Planck action (h), results in an increase of 1/10^20 kelvins temperature. If absorbing and retaining, at the maximum of 2 x 10^13 cycles (action units) per second, the temperature would increase at the astounding rate of (2/10^7) kelvins per second.
This is one kelvins increase every two months. This is but clarification for folk who think a photon is like a cement mixer truck, off in any direction.
All the best! -will-
315
One could wander into the Electric Universe territory Will, which could offer alternate valid answers but incur a wrath even greater than CAGW denialism. 🙂
00
One wrath at a time 🙂
30
Was that one wrath or one wraith?
10
bobl unless you were quoting they got there first. ” Major Evan Lorne compared global warming with a Wraith attack.”
http://stargate.wikia.com/wiki/Global_warming
10
Darn
10
Beautiful Sliggy!
I had no idea. Major Evan Lorne is even a contemporary.
10
While I’m a bit leery of using analogies like ‘pipes’, due to the tendency to argue from analogy (the ‘Weak Analogy’ fallacy if the relationships between the analogy and the reality fail around the points of argument), and prefer to discuss the things themselves, it’s not a bad way of explaining the enhanced greenhouse effect.
However, I suspect you would gain clarity by explicitly mentioning the two major causal links, rather than the one implicit in your discussion. The way you have phrased the CO2 ‘squeeze’ seems to leave out energy imbalance and resulting energy accumulation that drives changes in the other pathways. It’s important enough in this discussion (esp. given the number of people here who don’t believe that CO2 can affect temperatures) to state that accumulation of energy explicitly.
If the CO2 ‘pipe’ is squeezed by, for example, adding CO2 and raising the altitude of emission to space across the CO2 frequencies, the first thing that happens is that less energy goes to space, an imbalance with incoming energy.
Only when excess energy accumulates in the climate system do output energies rise in _all_ pipes, including CO2, leading to equilibrium with the CO2 emissions somewhere between the original values and those immediately following perturbation, and all other ‘pipes’ at higher values, so that the sum of all outgoing energy again matches incoming.
The other pathways, the other ‘pipes’, don’t respond to CO2 changes alone, but rather to a change in the temperatures driving those pathways. A squeeze to the CO2 pathway only acts upon other pathways by changing the ‘pressure’ behind the other pathways, by first imbalancing energies and then leading to energy accumulation. Which shows as and acts through increases in temperature at the surface, in the oceans, and through the atmosphere.
05
Note that the basic model only describes the changes from one steady to the next (or assume that the Earth is kind of near a steady state most of the time, and apply it over a decade or more).
So if increased CO2 reduces the OLR in the CO2 pipe, the OLR in the other three pipes combined must rise — and yes, some feedbacks or “pressures” cause the relative attractiveness of the other pipes to increase. And if more OLR goes out the surface pipe, then the surface must have warmed.
“Pipe” is foremost a notation, for brevity, but it is also a valid analogy.
71
“So if increased CO2 reduces the OLR in the CO2 pipe, the OLR in the other three pipes combined must rise — and yes, some feedbacks or “pressures” cause the relative attractiveness of the other pipes to increase. And if more OLR goes out the surface pipe, then the surface must have warmed.”
This is where I get really confused.
The operative words being “the surface must have warmed.” When it seems to me that what is actually warming is the atmosphere (atoms/molecules) in direct contact with the excited CO2. And once the energy is transferred the warmer molecule would move up (convection) and not down (conduction)
If what Dr Robert Brown (Duke Univ) and Dr Happer are saying is correct.
All the outgoing long wave radiation (OLR ) is captured by CO2 within a couple of meters of the ground.
And both Dr Brown and Dr Happer say (Quoting Dr Brown again)
However Dr Brown then goes on to say:
Why would the energy have an equal probability of diffusing downward once it is transferred to the much more plentiful non-GHG that would have more of a tendency to rise since it is now warmer than the surrounding gases?
Primer on Convection:
http://study.com/academy/lesson/convection-in-science-definition-equation-examples.html
21
Once the CO2 has captured the radiation (become excited) it needs to dump that energy, which can happen either by collisions with nearby molecules (thus heating up the local patch of atmosphere) or by re-radiating to produce new radiation travelling in a random direction.
The theory is that we ignore collisions with nearby molecules, and presume it all gets re-radiated. Approximately half is then turned around and sent back towards the ground.
I think the point of that is not so much that radiation is “captured” but whatever percentage of radiation that is turned around and sent back towards the ground doesn’t change a whole lot by adding more layers of CO2 after the first few meters. Once the radiation has been scattered into a random direction, then making it double random or triple random, it still heads in a random direction. That’s the way I imagine it at any rate, there’s no doubt a more precise way to calculate it because the extra layers must have some effect.
30
0 – The radiation back to the ground hides reality. Before the upper atmosphere can warm enough to heat the surface, its warmed enough to compensate for the reduced OLR from the surface. Can it be any other way? Hot air rises. The concentration of CO2 and water increases as you go down. Somethings missing.
21
Gai: Outgoing longwave radiation (OLR) is, by definition, outgoing — it is the radiation that makes it to space.
This misunderstands the terminology. Yes, most all the 15 micron radiation from the surface is captured within a few meters of the surface by CO2, and virtually none of it gets to space — but if it is captured then it’s not OLR. So the statement I’m quoting is quite wrong — and in fact none of the OLR emitted by the surface (on the wavelengths at which CO2 absorbs and emits) is captured by the atmosphere (but there is virtually no such radiation anyway).
Notice that for our modeling purposes we are only focused on radiation that actually makes it to space — the OLR. Changes in OLR can (help) tell us how much the emission layers warmed, and thus the ECS.
50
You may need to add another pipe. Your line “But O2 and N2 are not greenhouse gas molecules — they can’t release the photon’s energy, so the heat sticks around for a while.” Please see
http://vixra.org/pdf/1504.0165v2.pdf and see the experiments and reasoning behind why this is not correct.
Reinterpreting and Augmenting John Tyndall’s 1859 Greenhouse Gas Experiment with Thermoelectric Theory and Raman Spectroscopy. Blair D. Macdonald
This paper reveals, by elementary physics, the (deceptive) role thermopiles play in this paradox. It was found: for a special group substances – all sharing (at least one) electric dipole moment – i.e. CO2, and the other greenhouse gases – thermopiles – via the thermoelectric (Seebeck) effect – generate electricity from the radiated IR. Devices using the thermopile as a detector (e.g. IR spectrographs) discriminate, and have misinterpreted IR absorption for anomalies of electricity production – between the sample gases and a control heat source. N2 and O2 were found to have (as all substances) predicted vibrational modes (derived by the Schrodinger quantum equation) at 1556cm–‐1 and 2330cm–‐1 respectively – well within the IR range of the EM spectrum and are clearly observed – as expected – with Raman Spectroscopy – IR spectroscopy’s complement instrument. The non–‐greenhouse gases N2 and O2 are relegated to greenhouse gases, and Earth’s atmospheric thermoelectric spectrum was produced (formally IR spectrum), and was augmented with the Raman observations. It was concluded the said greenhouses gases are not special, but typical; and all substances have thermal absorption properties, as measured by their respective heat capacities.
Think about it. If you were to take a sample of N2 & O2 heated to 80 C into an IR transparent ball. Now float it out in space. Since these gases supposedly will not radiate any IR, and there is no convection or conduction, it will stay at 80 degrees forever?
31
DD More: Quite possibly. The percentage of OLR from N2 and O2 is tiny, however. We pretty much ignore the smaller pipes, including the methane, ozone, and NO2 pipes.
The main four carry over 90% of the OLR.
20
But David, we are arguing about a small LED worth of heating, 1/2 a watt per square meter, about 0.05% of peak insolation at TOA and claiming that this imbalance is warming the earth. Excuse me for saying, but when I went to school that meant that you needed to know the incoming and outgoing energy to an accuracy an order of magnitude better than this, or to 99.995% this means that you can’t really call a source or sink negligible for 2 orders of magnitude ( that is 0.0005% ) or 6.8 mW and less if you use averages. That is, to get a reasonably good handle on ASR to OLR imbalance, you need to account for all sources or sinks with a contribution or loss of greater than about 7 mW otherwise you know nothing.
This is why Hansens 0.6W imbalance is 0.6W +/- 17 watts!
Incredibly the imbalance could be -16.4 W per square metre, and this is MAINSTREAM climate science.
30
“The main four carry over 90% of the OLR.”
How can you specify the sum of the four without specifying the value (somehow) for each? It will be interesting to find out from measurement that the surface radiates more to space than the atmospheric N2 O2, all 160 km of it. Are you reling on a model?
All the best! -will-
110
relingrelying00
Empirical, no model. Estimated from the emission spectrum measured by Nimbus, plus Trenberth’s figures. Later in the series.
20
KR says
NO! the first thing is the raising of altitude increased the radiating surface area. This would raise the outgoing energy by increasing the M^2 part of Watts per Metre squared. Also the distance to space is now shorter so that attenuation of the outgiong signal due to anything else has been reduced. Also what is the reflectivity of CO2? Does it block incoming radiation at frequencies it cannot absorb? Does it increase or deacrease inward refraction?
11
Going from an average tropopause height of about 12.2 km, the tropopause has an area only 0.35 % larger than the surface. So yes, there is some effect there, but it’s really small.
CO2 is all but transparent to the incoming solar spectrum.
20
Transparent things alter refraction. You did not address this but did at least admit with “all but” that the blocking is not zero.
My calculator says that 0.35% of 1365W/M^2 is 4.8Watts/M^2.
00
And has that ratio significantly changed? The tropopause has risen by about 200 meters over the last 20 years, but that’s entirely accounted for by changes in GHG effective emission altitudes that hold heat for convection (and hence the lapse rate and position of the tropopause) to higher altitudes.
Refraction doesn’t matter – the incoming light will still hit the ground, and changes in atmospheric refractive index with additional GHGs are, roughly, zero.
If you look at this image you will see the incoming solar and upwards thermal radiation spectra, including absorption spectra for various GHGs. The solar spectra contains almost no IR, and the GHG effects utterly dominate in emission to space.
10
Even if it did not manage to hit the ground in the past?
Light that hits the atmosphere at a near parrallel angle may not veer off course enough to hit the ground. Co2 could have reduced the amount that does get bend in to hit. If the effect is a reduction in refraction. However as you say a change from 300 to 400PPM is too small to change much of anything.
01
From the climatescam.se: http://www.hindawi.com/journals/ijas/2013/503727/ from Hermann Harde.
It is heavy in math.
It was taken from this post:
http://www.klimatupplysningen.se/2015/09/25/varmestralning-atmosfaren-ett-aterkommande-tema/
The general believe that CO2 should be the main culpritt is contradicted at a clear night when the temperature can drop to freezing point even at summer.
It is a bit funny to hear meteorologists warn for frost during night, and in the next moment explain how CO2 stops the heat from leaving the Earth. 🙂
92
Yes Svend, a long time ago I posted about this several times on this site, there is this thing called the atmospheric window, and enough energy pours out of it to turn a 25 degree surface into -3 degrees in less than 12 hours and freeze a fair volume of water costing 334J per gram to boot!
One would think that if the magic gas and it’s supposed feedbacks prevented radiative cooling then we would expect to see dramatically reduced frost. If anything frost has increased!
30
The emission from any layer of the atmosphere depends not just on the temperature of the emitting layer but also on the temperature of the object or layer that is absorbing that radiation. For the correct formulae see M. Hertzberg “The Nightime Radiative transport between the Earth’s Surface, Its Atmosphere, and Free Space.”, Energy and Environment, Vol 23, No. 2, (2012) PP 819-831.
Omitting rare inversion conditions, the temperature of most of the atmosphere decreases monotonically with
altitude,and radiation from the atmosphere must always be outward toward free space.
814
Dr. Hertzberg, thank you for standing up for science.
I have seen your contributions at WUWT and Junk Science. I and most here agree that it is tragic what has happened to the science of meteorology and climatology. The politicians and MSM have turned it into a three ring circus and thereby cause science as a whole great harm.
813
Martin, In this post and for the model building we are focused only on OLR — radiation to space. We can (and do) ignore radiation within the atmosphere (that is, absorbed in the atmosphere). So, for the radiation we are interested in, there is no “object or layer that is absorbing that radiation”.
80
“So, for the radiation we are interested in, there is no “object or layer that is absorbing that radiation”.”
Have you correctly accounted for the 90% cloud cover in that 8-14 micron window? From where do you claim the exit flux to space originates? Into what kind of solid angle does some cloud with a km^2 cross sectional area generate its exit flux to space?
All the best! -will-
111
Will: Will cover such issues later in the series, but the figures I will estimate come from empirical estimates, not models.
30
Thank you, Dr. Evans.
I would love to find someone else that has seriously tried to measure surface radiative exitance. And the techniques they attempted. My measurements exist but have my error bars greater than the measurement! 🙂
210
This is wrong!.
If the co2 tube is moving less energy out to space and one of the others have to send out more, it still will be the same amount of energy leaving the system. The total system will NOT be hotter since the total amount of energy comming in and leaving would be the same.
22
Mr Petterson,
You are right and KR is wrong only if there is a parameter other than surface temperature which enables switching of energy flow between pipes.
That is why changes in the lase rate slope are significant. They can influence convection from within the atmosphere off the surface to the tropopause without necessarily changing the temperature at the surface.
I contend that GHGs do affect the lapse rate slopes so that convection does switch energy flows between pipes without the average global surface temperature needing to rise though it does appear that surfaces below rising convective columns become warmer whilst surfaces beneath falling convective columns become cooler to an equal extent. The sign of the change in lapse rate slopes is reversed both low down as compared to high up and in descending columns as compared to ascending columns.
514
Stephen,
The HAVE to, let’s take a look a water, the main GHG. The dry lapse rate is about 6.4 deg per km while the moist lapse rate is around 5. So increasing water vapour, which is a supposed feedback of CO2 has the effect of reducing the lapse rate and cooling the surface.
I see no reason why increasing CO2 does not work exactly like water, that is the increase in the radiative gas CO2 should result in a reduction of the dry adiabatic lapse rate.
20
Both CO2 and water vapour reduce the lapse rate slope but only at heights lower than the emissions level. Above that height they increase the lapse rate slope again by radiating to space so we only need to consider any net difference between the two stages.
Then one must appreciate that the effect of reducing the lapse rate slope is reversed in descending air as compared to in ascending air.
When air is rising the reduced slope slows the rate of ascent (decompression of air) but when air is falling the reduced slope slows the rate of descent (compression of air) so that the thermal consequences of the reduction in compression offsets the thermal consequence of the reduction in decompression.
Net thermal effect is zero.
313
Can’t see that, any “effect” of a change can never completely cancel the change and remain causal.
If I modify the lapse rate by changing the composition of the atmosphere, then yes, the smaller slope will reduce energy flows by convection and conduction. But that’s a feedback, the reduced energy flows will tend to increase the lapse rate such that a new equilibrium is established. The outcome can’t be thermally neutral though, otherwise there nothing to drive the change in equilirium.
30
bobl,
A change in lapse rate slope, by definition, is not thermlly neutral since the temperature has changed along that slope.
It is just that there need be no overall net temperature change at the surface. It is just a little warmer at the surface beneath rising columns and a little cooler below falling columns at the surface with surface winds flowing between to equalise the surface imbalance.
213
I concede that could happen but in the general case I don’t see any mechanism to enforce it. This is not my speciality so I will think more about it.
20
Excuse the spelling, I’m using the tablet with the virtually unusable virtual keyboard
00
bobl September 29, 2015 at 10:02 pm
“Excuse the spelling, I’m using the tablet with the virtually unusable virtual keyboard”
Thank you, I must remember that; if ever I come upon some drokn desire to respond to the Stephan Wilde nonsense!
09
Mr Pettersen,
Although the total system is indeed at the same temperature for the purposes of radiating to space, some emission layers are hotter and some are colder.
Suppose increased CO2 means that the CO2 pipe carries less OLR, and the ASR remains constant, so the OLR is also constant (that is, no change except the Co2 increased, ignoring surface albedo feedbacks as minor). So the other three pipes must carry extra OLR, so those three emission layers must warm while the CO2 EL must cool.
Because the OLR remains constant, the temperature characterizing the Earth in the Stefan Boltzmann law also remains constant — and this is, in a sense, the temperature of the total system. However, the amount of energy on Earth may well have increased — it depends on which emission layers warmed and what extra energy that required. If, say, only the surface warmed (the water vapor emission layer (WVEL) and cloud tops stayed at the same temperature, emitting the same OLR), then the total amount of heat energy on Earth would have to increase, because it takes a relatively large amount of energy to warm the surface (especially in the longer term, think oceans).
By the way, notice that one way the WVEL and cloud tops can change temperature is for the lapse rate to change whilst the surface stays at the same temperature.
10
“By the way, notice that one way the WVEL and cloud tops can change temperature is for the lapse rate to change whilst the surface stays at the same temperature.”
Indeed! This true mostly-independent variable is not under control of temperature CO2, computer models or any identifiable property,it simply is at every location. The idea of any atmospheric model being valid while this is unresolved is futile!
All the best! -will-
29
a CO2 blanket? you are telling me 4ppm constitutes a blanket? REALLY?
28
400 ppm. Yes.
71
“400 ppm. Yes.”
You have nothing but Climate Clown Models (CCM) to show any effect of change in atmospheric CO2 between 100-800 ppmv. Where is some evidence, any evidence, that this claimed effect is applicable to this Earth’s atmosphere? The effective radiating temperature in your 15, or even 13-18 micron band remains firmly at the temperature of the tropopause throughout this range.
All the best! -will-
27
The effect of CO2 changes is a different matter. Stay tuned 🙂
30
OK!
00
GPM: 400 ppm of CO2 doesn’t seem like much of a blanket, but the atmosphere is so different from everyday materials that our intuition isn’t very good. If we compressed the atmosphere until it became a liquid with the same density as water, that 14.5 lbs per sq inch (10,000 kg/m2) would make a layer 10 m high. (Try it. The calculation is easy.) 400 ppm is 0.04% of that 10 m or 4 mm or 1/6 of an inch. Similar to the thickness of a blanket or a pane of glass (that won’t let infrared escape from your car on a hot summer day..
10
David Evans,
After toughing through these posts, so far mostly in total ignorance, I now have a great curiosity about one aspect of the ‘Greenhouse Effect’, the CO2 based Climate Models and the statistical errors:
What concentration of CO2, I wonder, would actually produce a cooling trend in the IPCC climate models?
100
Cooling? Please explain?
00
Peter C,
The opposite of warming I would expect.
60
James,
In the IPCC models, pretty much only a falling concentration of atmospheric CO2 would produce a sustained cooling. (Short of the Sun spluttering out or something most unexpected. Volcanoes or atmospheric nuclear explosions might cool us for a while in the IPCC models. Or maybe some huge release of certain aerosols. All of these are unexpected events, however.)
Later in the series we predict a sustained cooling, despite increasing CO2. Stick around to see why 🙂
110
“so far mostly in total ignorance”
Seems we are on the same page.
KK
40
Yep, high five KK.
50
What about circulation (upwelling) of heated air? Is that different pipe or is it like a increase in the flow of the existing pipes?
00
DonG, no, a “pipe” is only for OLR or radiation to space. Convection is quite different.
10
Good post. Solar radiation is the main driver of the earth’s climatic systems. The earth is sensitive or shall we say, in tune with the Sun’s variable irradiative forcing. As the Absorbed Solar Radiation (ASR) increases so does the four main pipes as well as the minor ones increases in their emissions. It’s a bit like an exhaust system of your car engine except with the purpose of removing unwanted GHG’s and heat. This probably explains why the earth is a good convector of heat transfers and a poor insulator when it comes to night time temperatures.
In figure 3, your diagram shows Solar radiation (visible, UV, etc) for day time temperatures. Is the same true for night time temperatures when solar rays are replaced with gamma cosmic rays?
20
Bartender, the time scales for the basic models is between steady states, so we are talking at least a decade. Everything here applies on time scales of years or decades, and for global averages, so it not all applicable to day/night and regions of the Earth.
10
In my view, the main pipe is too large for C02 in fact one could argue 99% too large in volume. And besides, it would take a tremendous amount of warm air (convection energy) to shift already a minor gas upward that would make it even less detectable in space. My point to you is this; C02 is a minor gas (ozone, methane, nitrous oxide, etc.) Given the size and magnitude of the other three main pipes, C02 should therefore be set aside a minor pipe and be treated as such in order to make your model workable
00
The CO2 pipe carries about 20% of the OLR, that is, about 20% of the outgoing heat radiated by Earth is emitted from CO2 molecules.
00
Unless you are going to include this in a separate section, there’s one pipe you forgot which is the heat pipe, a highly non-linear Carnot machine that cranks heat away from the surface and up to the top of the troposphere. That’s not the “fixed altitude” top of the troposphere, that’s whatever altitude is necessary to dump the heat… warm moist air rises, and will continue to rise until the water condenses, whatever it takes.
There’s no “hot spot” because the concept of a hot spot would require warm moist air to sit without rising, cooling, and dispersing. This is mechanically impossible with Earth’s atmospheric structure. There is nothing to contain a hot spot.
Willis Eschenbach demonstrated the sharp non-linearity around about 300K as seen in the Argo ocean data.
http://wattsupwiththat.com/2012/02/12/argo-and-the-ocean-temperature-maximum/
I remember George White also demonstrated the same 300K turning point was seen in the monthly satellite data, I can probably find the graph with enough digging.
50
Tel, we are using the term “pipe” strictly as defined, for the outgoing longwave radiation (OLR) over a group of electromagnetic wavelengths and from a particular emissions layer (I beefed up the definition in the text to explicitly mention OLR).
That is, a “pipe” as we will use the term, is for OLR — radiation to space, only.
10
Dr. Evans,
Your reply to Tel is a bit confusing. As you can see from my reply to him below, my understanding is that the heat pipe he’s refering to is the water vapor pipe. After all, OLR represents the heat loss of the atmosphere to space. Did I miss something?
Abe
00
Abe, see Comment 21.3.1. where I replied to your other comment.
00
Well said Tel, The heat pipe is the convection cell which respond to any increase in surface temperature from any increase in energy flow rate. The 30 c limit so well addressed by Eschenbach is to be fully expected from the many observations of heat flow physics which he describes. From an old closed loop system engineer’s viewpoint it can be summarized simply by the positive feedback loop behavior of more heat, more water vaporization, capture more heat etc until the system hits its energy deliver/power limit. The atmospheric system has extracted all of the power it can from the power limit of heat transfer from the surface(mostly ocean). This is the well understood ‘hard’ limit phenomena when reaching the limit of the power supply in any system having positive feedback (here the thermodynamics at the surface). Note that this is not positive feedback in planet temperature. In fact any increase in temperature at the surface layer of the atmosphere including most importantly an increase in water vapor content will as you observe enter the ‘heat pipe’, carrying any increased in heat energy flow to the altitude at which the water vapor radiates this heat to space. There is no other significant radiator so the climate system must necessarily utilize water vapor to get rid of all of the heat to space including any increase in water vapor from the random complex thermal processes being discussed. Thus we have by definition water vapor is a negative feedback (since it is *the* cooling mechanism) in planet temperature regulation. As is so well pointed out by the IPCC, if there is no positive feedback (from water vapor in their conjecture) there is no significant man made driven global warming. This simple physics is the ‘kiss of death’ to the whole CAGW conjecture.
21
Tel,
At the point of condensation, water vapor releases large amounts of latent heat. That latent heat is released in the form of EM radiation. That radiation, then, is the OLR that is already accounted for in the water vapor pipe.
I too agree that this must be the major contributor to the sum of OLR for three reasons.
First: Water vapor rises well into the troposphere before condensing and so the EM radiation released from the latent heat has much less atmosphere to traverse on it’s way to space. This means that this EMR has less oportunities to be absorbed – reflected – scattered by the molecules of any of the other atmospheric gases.
Second: The amount of energy that is stored within each molecule of water vapor is greater than the amount of energy that is absorbed and re-radiated by each molecule of any of the other atmospheric gases.*
Third: There’s so much more water vapor than both carbon dioxide and methane combined that even if all three molecules were to move the same amount of energy per molecule from the surface up into the atmosphere, water vapor swamps the other two just by sheer numbers.
Abe
* This may be incorrect. I only have the memory of reading this in at least two sources but didn’t bookmark the links to those pages.
01
Just-A-Guy: The main mechanism by which Earth cools is heat from the surface is carried by convection and latent heat of water to the upper atmosphere, where it is emitted to space (as OLR) from the water vapor emission layer, from cloud tops, and from the CO2 emission layer.
The pipes as defined are for OLR, and there is one for each source of OLR — CO2, water vapor, surface, cloud tops, methane, ozone, NOx, etc. There is no “heat pipe”.
When the water vapor condenses, releasing its latent heat, it warms the atmosphere. The atmosphere then cools to space via the pipes just mentioned.
01
Just-A-Guy October 27, 2015 at 9:10 am · Reply
“Dr. Evans,
Your reply to Tel is a bit confusing. As you can see from my reply to him below, my understanding is that the heat pipe he’s refering to is the water vapor pipe. After all, OLR represents the heat loss of the atmosphere to space. Did I miss something? Abe”
You who seem to know so much; both you and Tel missed the topic of the series!! David writes of radiation, EMR which can never be heat! It is the dispatch of entropy from the atmosphere; the secondary reason for an atmosphere.
This series is not about some meteorological fantasy of convection that you peddle with no evidence! That is but rearrangement of accumulated power, within the earth atmosphere open system. It is about EMR exitance to space and where that EMR flux originates. The surface is not required!!
Just-A-Guy October 27, 2015 at 9:02 am
“At the point of condensation, water vapor releases large amounts of latent heat. That latent heat is released in the form of EM radiation. That radiation, then, is the OLR that is already accounted for in the water vapor pipe.”
What you fake meteorologists and CCC; that seem to know so much, fail at is understanding is that most of that latent heat in no way comes from surface evaporation. It is the airborne atmospheric water condensate that remains airborne. Nine times as much condensate, than that of the surface water cycle, is evaporating by day and condensing by night continuously! This is mostly lateral convection not radiation, and not the subject of this series.
“I too agree that this must be the major contributor to the sum of OLR for three reasons.”
You, Tel, and Kristain, have not a clue as to what you spout! Why must you try to disrupt what others are doing with your abject nonsense?
03
Is it not possible to absorb 4 at a high energy level and radiate 5 at a lower level if the total energy is the same? Thus allowing for multiple absorbtion and re-radiation wavelengths to work simultaneously give or take random noise.
01
Yes, absolutely. Absorption spectra for any substance are matched by emission spectra, and at radiative equilibrium are of equal energy – but that’s a statistical statement, what the energies add up to, and it’s far more likely that an absorbing molecule will either exchange energy with its surroundings, changing its own, or radiate from a different vibrational transition.
The match between absolution and emission spectra is statistical, based on the likelyhood of various energy transitions both up and (matching) down.
20
Siliggy, Are you asking whether maybe 4 photons are absorbed at once, or 5 emitted at once? As far as I know (and I am no expert in this area), absorptions and emissions only ever involve a single photon. In any case, non-single-photon events are at most very rare.
20
Dr Evans, think back to Engineering chemistry 1, it is possible for this to happen, if a photon raises an electron to a given excitation shell ( a high enough energy photon can actually liberate the electron and break the bond) the electron can either drop back a shell ( vibration mode) at a time, giving off a photon for each transition or it could transition directly multiple electron shell states (vibration modes) giving off a higher energy photon. So yes, it’s possible for many low energy photons to cause a high energy photon or the reverse.
Once the CO2 molecule is out of the base state (excited) however, to enter the next excitation state a photon of a different energy is required representing a different spectral line.
10
bobl,
IR radiation does not involve electron shell energies. 15 u IR is far below that energy level and only excites mechanical oscillation in the 0=C=O bending mode. Since this vibrational energy is immediately dispersed by millions of collisions to the O2, N2 molecules thus heating the surrounding gas infinitesimally, there is virtually zero probability of absorbing another IR ‘photon’ in addition to the one just absorbed even if such a mode did exist.
00
David KR’s answer makes good sense. My problem with it all is that to bring it all into familiar territory as a radio electronics guy it needs to be first converted in my head from particle theory to wave theory. So the head gets bent as the internal conversion dances around and avioding the interferometer experiments to make sense of it all. If you are considering a molecule as a unit. Try a very long chain polymer and imagine photons passing to and from individual atoms within a wavelengh time span of each other but NOT exactly inphase. I think that inphase instantaneous exactly equal re-radiation is NOT absorbtion at all but is reflection. Absorbtion and re-radiation MUST have a time delay and be free as KR says to “radiate from a different vibrational transition”. Thus the frequency or wavelength is NOT exactly the same other than remaining statisically within bandwidth limits. Likewise the energy is not exactly the same but must remain within the statisical equilibrium as it varies within noise bandwidth curves unless there is warming, cooling or another path like conduction.
00
And when you have something like 2.2*10^21 CO2 molecules per mole of gas, pretty much every transition is statistically represented, and you get the characteristic full CO2 emission spectra.
00
“Is it not possible to absorb 4 at a high energy level and radiate 5 at a lower level if the total energy is the same? Thus allowing for multiple absorbtion and re-radiation wavelengths to work simultaneously give or take random noise.”
An atmospheric gas molecule has a ‘noise power’ exactly equal to kT/t. This the molecular definition of ‘temperature’. Each CO2 molecule’s cycle of radiant 15 micron absorption/emission ‘must’ increase/decrease the molecular temperature by 1/10^20 kelvins. Maximum continuous absorption from a direction with no dissipation would take approximately 2 months to increase one kelvins. Each molecule ‘may’ receive such radiant power from almost 4 PI steradians,asymptotically increasing temperature to that of its environment. At that point the whole concept of radiant flux disappears.
All the best! -will-
16
I was looking to see if someone commented on this. Siliggy, you are right. A CO2 molecule can absorb and emit at a number of specific frequencies, each with its own absorption line. Each of those lines represents a specific change in quantum states and the associated energy difference. It is definitely possible for a CO2 molecule, for example, to absorb energy from a photon at one energy state (or multiple photons or mechanical energy from hitting another molecule, or a combination), and to emit a different amount of energy at a different frequency, leaving it at a different total energy state than it started.
In your example though, the 5 and 4 would have to happen at different times, because each one is a specific state change, and it would have to ‘get excited’ back into that state to emit another photon. Can’t happen at the same time.
As someone else stated, the energy transfer doesn’t have to all happen through photons, except for the last transfer into open space. It doesn’t matter how the energy gets to that last CO2 molecule, just that it has enough to make that quantum transition.
David,
Great series of posts. Can’t wait to read more…
30
David’s arguments do not convince me that greenhouse gases (i.e., gases that both absorb and radiate electromagnetic energy in sub-bands of the IR band) in the Earth’s atmosphere will necessarily produce a “greenhouse effect” (i.e., ensure an increase in the Earth’s surface temperature). I’m not arguing that greenhouse gases in the Earth’s atmosphere won’t increase the Earth’s surface temperature. I’m only arguing that David’s line of reasoning does not make the case.
[Note: In what follows I explicitly assume the rate the Earth absorbs energy is unaffected by the presence/absence of greenhouse gases.]
As I understand it, David’s basic argument is (a) CO2 prevents some of the photons emitted from the surface of the Earth from escaping to space; (b) atmospheric CO2 has an emission layer that is colder than the Earth’s surface, and thus radiates energy at a reduced rate relative to the rate the Earth’s surface radiates energy; (c) to maintain energy-rate-equilibrium in the presence of atmospheric CO2 (i.e., to maintain equal incoming and outgoing energy rates), the CO2-emission-layer-induced decrease in outgoing energy rate requires an increase in the outgoing energy rates from the other emission layers, and in particular, from the Earth surface emission layer; and (d) an increase in the rate energy is emitted from the Earth’s surface requires an increase in the surface temperature.
This argument is flawed because it only considers radiation as a means of energy transfer. Radiation may be the overwhelmingly dominant (if not only) way energy leaves the Earth/Earth-atmosphere system, but radiation is not the sole means of energy transfer within the Earth/Earth-atmosphere system. If internal forms of energy transfer (conduction, convection, evaporation) other than radiation exist, it can be shown that there are cases where an object with a constant-rate source of input energy is surrounded by a “photon absorbing blanket” whose energy-rate-equilibrium temperature is less than the surface temperature of the object in isolation, the surface temperature of the object decreases. Thus, the conditions (a) photons emitted from the object’s surface are absorbed by a “photon absorbing blanket,” (b) the “photon absorbing blanket has an emission layer that is at a colder temperature than the surface of the object, and (c) energy-rate-equilibrium is maintained can simultaneously exist with lower object surface temperatures. If the conditions can exist with lower object surface temperatures, how can anyone conclude that the presence of the three above conditions for a “CO2 blanket” guarantees an increase in Earth surface temperature?
Consider the case where (a) the object is an active sphere with an internal, constant-rate source of energy, and (b) the surrounding “photon absorbing blanket” is a thin, concentric, spherical shell with thermally conducting “internal radial arms” that connect the shell to the active sphere. The presence of the blanket has no effect on the internal energy rate of the sphere. Such a “shell blanket” not only prevents IR sub-band photons emitted from the sphere’s surface from escaping to space, the “shell blanket” prevents all photons emitted from the sphere’s surface from escaping to space. Thus, in the sense that “trapping heat” means the prevention of photons that originate from the sphere’s surface from escaping to space, the “shell blanket” traps all heat. If the outer surface of the shell has the same emissivity as the surface of the sphere, because the surface of a “concentric surrounding spherical shell” is larger than the surface of the sphere, the energy-rate-equilibrium temperature of the shell will be less than the temperature of the isolated sphere. With the proper (and realistic) thermal conduction properties, the “internal arms” can conduct heat from the sphere surface to the shell surface at a sufficient rate to produce a lower sphere surface temperature in the presence of the “blanket” than the sphere surface temperature in the absence of the “blanket”.
As I see it, the “shell blanket” meets David’s requirements for a “greenhouse effect:” (a) The shell blanket “traps” outgoing photons emitted from the sphere’s surface—i.e., the shell blanket prevents photons originating from the surface of the sphere from escaping to space; (b) the “shell blanket’s emission layer” is at a lower temperature than the surface of the sphere, and (c) energy-rate-equilibrium is established. If I understand David’s line of reasoning, these conditions are sufficient to produce a “greenhouse effect” in that in the presence of the “shell blanket” the surface temperature of the sphere should be higher than the surface temperature of the sphere in the absence of the “shell blanket.” That there are cases where the opposite is true, implies David’s three conditions are insufficient to claim an increased object surface temperature.
30
Reed, good comment. You understand my argument almost but not quite, and I agree with you. Let me explain using the briefer “pipes” terminology.
Suppose we increase CO2 and nothing else changes, ASR remains constant (ignore minor surface albedo feedbacks).
Therefore the OLR in the CO2 pipe decreases, yet the total OLR remains the same. Therefore the OLR in the other three pipes combined must increase. That’s all I “proved” above.
I then said that this could involve the surface pipe increasing its OLR (thus the surface warms). But it need not — the constraint is only that the three non-CO2 pipes combined carry more OLR. Yes, one can imagine situations where the surface pipe also carried less OLR, and you are correct that this is a theoretical possibility, and the post above does not prevent that possibility — in the logic of the post, the surface could either warm or cool.
As it happens, for reasons I’ll introduce later, I think the surface does warm but not nearly as much as the IPCC thinks. Let’s go a little bit further now, casting the problem into the language of pipes.
The establishment position is that the decrease of OLR in the CO2 pipe causes a sympathetic decrease of OLR in the water vapor pipe — because of the water vapor feedback (surface warms, more evaporation, more water vapor). Thus the surface and cloud top pipes combined must increase their OLR by the combined decreases in OLR through the CO2 and water vapor pipes. The cloud top pipe is thought not to change much, so the establishment loads nearly all the accommodation onto the surface pipe — the surface pipe must carry nearly all the decrease in the CO2 pipe and the water vapor pipe (which is the water vapor amplification)! It’s going to burn down here on the surface! The establishment has fairly good reasons for their belief, yet the end result is quite wrong IMHO — the upcoming posts resolve that situation.
40
I believe you are over-decomposing linked responses.
If the CO2 ‘pipe’, or rather the CO2 frequencies of the emission spectra to space, decrease, then there is an energy imbalance that will lead to warming. The water vapor portion of emission will, by the classic feedbacks (the Clausius–Clapeyron relation and observations indicating steady or near-steady relative humidity) also decrease.
As the climate warms, however, _all_ emission wavelengths will increase, including CO2 and water vapor. It’s just that the relative proportion of those portions of the spectra will be a smaller fraction of the total than before. Your description gives the impression of having one ‘pipe’ restricted and another separately expanding, and that isn’t accurate.
The ‘surface pipe’ won’t increase separately from the rest of the spectra – in fact, the IR ‘window’ will slightly narrow as GHG absorption bands widen. Rather, under a GHG induced imbalance the entire Earth emission spectra will increase across the board (with accompanying feedbacks both positive and negative) until the energy radiated once again equals energy incoming, just with a slightly different (more jagged, less efficient) shape than before.
00
KR, I agree (except for the “over-decomposing” part).
If CO2 doubles, the CO2 pipe is not restricted by D-sub-R,2X (the amount OLR drops due to a doubling of CO2, all else constant, see post 2), but by something a bit less, for the reasons you describe. (One might regard it as an initial drop of D-sub-R,2X followed by an increase as the surface warms and the lapse rate changes, but because the basic model only goes between steady states that is more a helpful fiction.) Later in the series we quantify it, taking surface warming and lapse rate changes into account.
20
David,
Thank you for your response.
I’ll now get to the crux of the problem as I see it. I didn’t mention the crux in my first comment because that comment was too long anyway.
You wrote: “Suppose we increase CO2 and nothing else changes, ASR remains constant (ignore minor surface albedo feedbacks). Therefore the OLR in the CO2 pipe decreases, yet the total OLR remains the same. Therefore the OLR in the other three pipes combined must increase. That’s all I ‘proved’ above.”
As I see it, crucial to your argument is the belief that increasing CO2 while changing nothing else implies the OLR in the CO2 pipe must decrease. I have two problems with this belief. First, changing the amount of CO2 will almost surely affect the atmospheric temperature profile and will likely affect the density profiles of the various atmospheric gases. Thus, I think it is an oversimplification and unrealistic to “increase CO2 and keep everything else constant. To put this point in perspective, return to the “shell blanket” analogy. Suppose we poke a few small holes in the shell. By doing so, we decrease the shell area that radiates to space; and if the shell’s temperature doesn’t change, we reduce the shell’s OLR. This drop in outgoing OLR is “recovered” when we consider radiation from the sphere surface that escapes to space through the holes in the shell. That is, the sphere temperature will adjust itself so that the total outgoing OLR (shell and sphere) is equal to the ASR. If we now add (or remove) additional thermally conducting rods between the shell and the sphere, the temperatures of both the shell and the sphere will change. It’s not at all clear to me what effect adding thermally conducting rods (the analogy to adding CO2 to the Earth’s atmosphere) will have on the sphere’s temperature. My intuition tells me that adding thermally conducting rods will result in a decreased sphere temperature because adding rods provides additional paths for heat conduction whose heat transfer rates are proportional to the difference temperature not the difference of the fourth powers of the temperature.
Second, and even more important, it may be true for CO2, but it’s far from clear for all greenhouse gases that increasing the amount of the greenhouse gas (increasing the number of greenhouse gas molecules) decreases the OLR for the corresponding “greenhouse gas pipe.” In my opinion almost all treatments of outgoing OLR assume the radiation from the CO2 emission layer can be modelled as emission from a surface—something like Planck’s blackbody surface radiation law. To me, such a treatment is nonsense. First, a surface is a well-defined two dimensional object. Curvature of a surface may extend the surface into three dimensions, but all surfaces possess a “normal”—i.e., a surface area is characterized by both an area and a direction. In Planck’s blackbody surface radiation law this effect is manifested by a COSINE(theta) factor where theta is the angle between the direction of propagation and the normal to the surface. I know of no such “directionality equivalent” for a differential volume. For radiation from a differential volume of a gas, what, if anything, is the equivalent of this directionality factor?
Emission of radiation from the CO2 emission layer (a finite, non-zero thickness) is emission from a volume, not emission from a surface. In my opinion, any computation of the rate radiation is emitted from a differential volume of a gas (or gases) must start with a formulation similar to Planck’s surface blackbody radiation law. Such a formulation should include any directionality terms (if they exist) and definitely must include things like temperature, gas density(ies), gas emission/absorption bands (if they exist), etc. A little over a year ago, I emailed a few people asking them if they knew of such a formulation. All responses were negative. This doesn’t mean that such a formulation doesn’t exist, only that if it does exist it is not well known.
Once you have such a formulation, then and only then can you attack the problem of the rate energy is radiated to space from a gas (or gases) surrounding an object. The problem has two components: (a) the rate energy is radiated from a differential volume of the gas, and (b) as the gas propagates from its origin, the fraction of the radiated gas that reaches space. The latter problem can likely be addressed by Beer’s law; but in my opinion, using Planck’s surface blackbody radiation law for the former is a major “stretch.” I’ll continue to be skeptical of all models that treat the rate radiation escapes to space from a finite thickness of a gas as radiation coming from a planar surface. At least at a quick look it makes sense that the more gas molecules (i.e., the greater the density) the greater the rate of energy radiated from a differential volume. It’s true that greater densities also imply greater absorption, so I don’t know what effect adding more gas molecules has on the sign of outgoing radiation rate. However, no one to date has convinced me that adding CO2 to the atmosphere when coupled with energy transfer from the Earth’s surface to the CO2 emission “layer” (finite thickness) via means other than radiation (conduction, convection, evaporation) results in a decreased CO2 OLR output.
11
Reed,
Yes it is unrealistic to allow CO2 to increase and keep everything else constant, but it is the only starting point we have in this case. Then we try to figure the effect of changing the lapse rate. It’s probably roughly correct, and until something better comes along we’ll have to go with it.
Yes, GHG emission layers are a volume of gas and not a surface as required for exact application of SB, so applying SB notions won’t be entirely accurate. No, I’ve never heard of any formulation taking directional radiation for such emissions into account. However, the observed emission spectra (e.g. the Nimbus spectra) seem to follow the blackbody spectral lines for the wavelength intervals that come from a single emissions layer, so the SB approximation for an emissions layer probably isn’t too bad. Again, we haven’t got anything better to work with.
In this series we are not after the last word in realism, so much as demonstrating that the conventional basic model is decidedly unrealistic and that we can do a lot better.
10
David,
The spectral nature (i.e., the distribution of energy as a function of frequency) of emissions from a volume of gas isn’t my primary concern. My primary concerns are the geometrical aspects and the total rate of radiation–specifically, (a) the factor for a differential planar surface area, dS, (b) the COSINE(theta) term, and (c) the radiation equation scaling constant. I will admit that lacking a differential volume radiation formula similar to Planck’s blackbody surface radiation law, a first approach would be to treat the radiation from a finite thickness emission layer as coming from a surface. It’s just that I’m uncomfortable placing any confidence in conclusions resulting from such a treatment–at least until someone can show theoretically or experimentally that such a treatment is a good approximation.
Best of luck with your efforts.
11
Reed Coray September 30, 2015 at 4:32 am
Now you are expressing all that is not only way to simplistic, but also backwards in these models. The detailed geometry for directional EMR exitance from Earth’s atmosphere has so far never been conceived let alone verified!
All so far from the surface or atmosphere is blatant fantasy. Why do Climate Clown Models (CCM) continue?
Even the fourth grader starts to understand the difference in Santa Clause and parents wishing me to be both happy and gleeful on Christmas day!
After 30 years of Climate Clown Models (CCM) failure, such can no longer be excused as stupidity or incompetence! Such must now be considered intentional [snip] on the part of governmental entities in charge of funding incompetence.
All the best! -will-
07
“David’s line of reasoning, these conditions are sufficient to produce a “greenhouse effect” in that in the presence of the “shell blanket” the surface temperature of the sphere should be higher than the surface temperature of the sphere in the absence of the “shell blanket.” That there are cases where the opposite is true, implies David’s three conditions are insufficient to claim an increased object surface temperature.”
Reed,
I cannot find a case where the powered sphere can have a ‘lower’ temperature than the shell. It can only go to the ‘same’ temperature at zero thermal impedance.
With your added shell and zero conductivity, you have the Willis Echenbach steel greenhouse, with all that nonsense.
In that case the sphere has a power, area, and a radiative impedance of 377 ohms Stephan’s constant. add the shell and add another 377 ohms between the shell and space. There must exist a radiative potential difference for the sphere to radiate anything. In this atmosphere convection and latent heat reduce the lapse rate between surface and tropopause decreasing surface temperature. The Climate clowns can understand the need for a potential difference for thermal conductivity. But refuse to acknowledge that need for EMR. Now Dr. Evans seems to agree with them that exit EMR originates at the surface rather than in the atmosphere. The same 15 micron CO2 exitance originates from the tropopause, whether the CO2 fraction is 100 ppmv or 800ppmv.
All the best! -will-
09
Will, I never said “the powered sphere can have a ‘lower’ temperature than the shell“. What I said was the temperature of the [powered] sphere in the presence of the shell may be lower than the temperature of the [powered] sphere in the absence of the shell
00
Reed Coray September 30, 2015 at 4:38 am ·
“Will, I never said “the powered sphere can have a ‘lower’ temperature than the shell“. What I said was the temperature of the [powered] sphere in the presence of the shell may be lower than the temperature of the [powered] sphere in the absence of the shell”
OK Can you please explain such behavior unless your sphere is a ‘sink’ for some shell power rather than a ‘powered’ source?
Reed,
Especially in the radiative, the public must acquire the concept of difference in potential for actual (not virtual) power transfer, always and only, in the direction of lower potential. This induced SCAM of opposing radiative flux at any frequency has continued way longer than even the profit motive. This is now inculcated nonsense from the Clowns! 🙁
17
Will,
A complete theoretical development of lower powered-sphere temperatures in the presence of a shell with thermally conducting rods connecting the shell to the powered sphere can be found at:
https://www.dropbox.com/s/ejn20s9glptl2qv/solid-sphere-shell_temperatures_with_conduction_heat_current_02PDF.pdf?dl=0
Truth in advertising: I am the author of that paper. My being the author means that my referencing the paper adds no confidence to the paper’s results. However, in the end my feelings about the paper are irrelevant–the paper stands on its own.
An example given in that paper is the following (emissivity of one assume for all surfaces):
Powered-sphere properties:
radius: 0.1 meters, internal power: 60 Watts.
Shell properties:
radius: 0.2 meters
Connecting rod (cylindrical and oriented radially wrt the sphere center) properties:
Number of rods: 6, cylinder radius: 0.002582 meters, thermal conductivity: 401 Watts per meter per degree Kelvin–the thermal conductivity of copper.
When I do the calculations, for the above conditions the temperature of the powered sphere in the presence of the shell/connecting rods is lower than the temperature of the powered sphere in the absence of the shell/connecting rods. Of course, you are free to perform your own calculations. But when completed, the following must be true: (a) the rate radiation leaves the system (powered sphere, shell, connecting rods) must be 60 Watts, and (b) the sum of (i) the radiative heat transfer rate from the powered sphere to the shell and (ii) the conductive heat transfer rate from the powered sphere to the shell must also equal 60 Watts.
In isolation, the powered-sphere’s temperature is 302.923 Kelvin. If in the presence of the shell you can find a combination of powered-sphere temperature and shell temperature that (a) meets the conditions that both (i) the rate of heat flow from the system and (ii) the rate of heat flow (radiation plus conduction) from the powered sphere to the shell are 60 Watts, and (b) the temperature of the powered sphere is greater than 302.923 Kelvin, I’d be interested in your numbers.
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“If in the presence of the shell you can find a combination of powered-sphere temperature and shell temperature that (a) meets the conditions that both (i) the rate of heat flow from the system and (ii) the rate of heat flow (radiation plus conduction) from the powered sphere to the shell are 60 Watts, and (b) the temperature of the powered sphere is greater than 302.923 Kelvin, I’d be interested in your numbers.”
Reed,
You have both your sphere 303 K
and your shell 214 K somehow with emissivity of 1.0 and radiating to an opposing radiance of zero. Where oh where can you physically find such conditions? You claim a real demonstration, then return with a silly theoretical ming game. I will not play!! BTW your coupling via only rod conductivity would result in a sphere temperature of 324 K. With radiation only (no rods) the temperature would be 321 K. If you just moved the 60 Watts to the shell the shell and all contents would be at that same shell temperature of 214 K -60 C. under all internal transfers. This is trivial nonsense and never can be demonstrated!!
For a real repeatable demo! of how radiative transfer works (has been done physically) three thin aluminum shells splittable for nesting diameters of 14cm, 20cm, 28cm! in and out emissivity 0.9. power source 5 watt electrical resister mounted inside the 14cm shell 28cm shell sink held at 0 Celsius under a vacuum bell jar, thermistors as needed!
1.)With 14cm centered within 28cm calculate temperature of powered 14 cm shell, verify that 28cm stayed at 0C verify that increase in LCO2 coolant increased for the added 5 watt flux.
2.) insert 20cm instrumented shell centered correctly between 14cm and
28cm. Rerun #1. but also calculate temperature for 20cm shell.
3.) post calculations here!
4.) Actually do the measurements and report those measurements here.
5.) explain your best guess for differences here.
Reed Coray October 1, 2015 at 6:21 am
“Will, expanding the above comment. Using the powered-sphere/shell/connecting-rod system parameters….”
Yup! the combination of conduction put that temperature between the 303 K and the 214K that I listed above.
If you now move the 5 Watt source to the 20cm shell its calculated and measured temperature must both drop slightly. The now un-powered inner 14cm must go to that same temperature
Why? Is not the 20cm radiating 5 watts internally as well as externally, as Climate Clowns and Willis Echenbach claim? By splitting the 5 Watts between the 14cm and 20cm you can get the whole range of allowable 14cm temperatures, while the 20 cm stays almost constant almost like Earth’s tropopause and surface. This can be fudged to get to lapse rate situation if you futz with surface (14cm) emissivity to between 0.9 and 0.6 that all but Climate Clowns already know about.
Any demo/model that cannot physically demonstrate your errors is useless, as are conceptual or numerical.
All the best! -will-
04
Will,
Simple question. Will the temperature of the powered sphere (60 Watts, 0.1 meter radius, emissivity 1) in the presence of an inert, concentric shell (0.2 meter radius, emissivity 1) and connecting thermally conducting rods (10 rods, each having length 0.1 meters, radius 0.002582 meters, and thermal conductivity 401 Watts per meter per Kelvin) be less than, greater than, or equal to the temperature of the powered sphere in isolation?
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Reed Coray October 1, 2015 at 11:56 pm
“Will,
Simple question. Will the temperature of the powered sphere (60 Watts, 0.1 meter radius, emissivity 1) in the presence of an inert, concentric shell (0.2 meter radius, emissivity 1) and connecting thermally conducting rods (10 rods, each having length 0.1 meters, radius 0.002582 meters, and thermal conductivity 401 Watts per meter per Kelvin) be less than, greater than, or equal to the temperature of the powered sphere in isolation?”
Reed,
I think with 10 rods rather than the 6 rods, the temperature would be less than the 303K even with no EMR flux from sphere to shell.
Lets see (324-214) x (10/6) + 214 = 280 K. Yup, less than 303K. With internal EMR even lower yet. Left as an exercise for the student! 🙂 Do a iterative spread-sheet or write your one prog in any language, show results.
Now do the three shell one and demonstrate that the concept of back radiation is never needed when a closed system must have a finite lowest temperature sink, please!
All the best! -will-
04
Will, this comment is primarily in response to your 2 October 2:42 am comment, but also addresses something you wrote in your 1 Oct 11:01 am comment. First, I’ll deal with your 1 Oct. 11:01 comment. You wrote:
“Where oh where can you physically find such conditions? You claim a real demonstration, then return with a silly theoretical ming game. I will not play!!”
What’s unphysical about nested 10cm and 20cm concentric spheres held apart by one or more 10cm long copper rods of cross section radius 0.2582cm?
Your statement that I claimed “a real demonstration” and “returned to a silly theoretical ming game” is nonsense. I reread every comment I made on this thread. The only time I used the word “real” or a variant thereof was in the statement:
“With the proper (and realistic) thermal conduction properties, the “internal arms” can conduct heat from the sphere surface to the shell surface at a sufficient rate to produce a lower sphere surface temperature in the presence of the “blanket” than the sphere surface temperature in the absence of the “blanket”.
How anyone can interpret this statement to imply a “real demonstration” (i.e., an experiment) is beyond my comprehension. The thermal conductivity of copper is 401 watts per meter per Kelvin, so my use of this value in a theoretical analysis seems “realistic” to me.
Now to your 2 Oct comment. Two points. First, I will estimate the temperatures of your three thin aluminum foil spheres. When I have completed those estimations, I’ll report them via a later comment.
Second, the whole discussion started because in response to my statement:
““Will, I never said “the powered sphere can have a ‘lower’ temperature than the shell“. What I said was the temperature of the [powered] sphere in the presence of the shell may be lower than the temperature of the [powered] sphere in the absence of the shell”
You wrote:
“OK Can you please explain such behavior unless your sphere is a ‘sink’ for some shell power rather than a ‘powered’ source?”
I was trying to explain how the temperature of a “non-sink” powered sphere in the presence of a “photon blanket” could be lower than the temperature of the powered sphere in the absence of that blanket. Apparently you now agree. I say this because you wrote:
““I think with 10 rods rather than the 6 rods, the temperature would be less than the 303K [303K is the approximate temperature of the powered sphere in isolation] even with no EMR flux from sphere to shell.”
What am I missing?
00
OK Will, here are my results.
First a restatement of the conditions:
(1) Three concentric, thin, aluminum foil, spherical shells of radii 0.14cm, 0.20cm and 0.28cm.
(2) Outermost shell (28cm shell) everywhere in contact with a heat sink maintained at 0 celsius.
(3) Internal to innermost shell a 5 Watt source of thermal energy.
(4) For the case of the 14cm and 28cm shells only (no 20cm shell), the space between the 14cm shell and the 28cm shell is a vacuum.
(5) For the case of all three shells, the spaces between (a) the 14cm shell and the 20cm shell, and (b) the 20cm shell and the 28cm shell are vacuums.
(6) Absorptivity of all shell surfaces 0.9
(7) If there are any “spacers” between the shells keeping them in position, (a) the thermal conductivity of those spacers is assumed to be nil, and (b) the volume occupied by the spacers is assumed to have negligible effect on the radiation between the shells.
Results:
To avoid any misunderstanding, all results presented below are theoretical—i.e., no experiment was run. I don’t have the facilities to conduct such an experiment.
The results come in two parts. First, assuming energy incident on a shell that is not absorbed by the surface of the shell undergoes Lambertian reflection. Second, assuming energy incident on a shell that is not absorbed by the surface of the shell undergoes specular reflection. Specular reflection will likely be the case if the shell foil is smooth. Lambertian reflection will likely be the case if the shell foil is crinkled.
Lambertian Reflection Results:
For the 14cm / 28cm shells only, the temperature of the 14cm shell is 4.87 celsius.
For the 14cm / 20cm / 28cm shells, (a) the temperature of the 14cm shell is 7.33 celsius, and (b) the temperature of the 20cm shell is 2.48 celsius.
Specular Reflection Results:
For the 14cm / 28cm shells only, the temperature of the 14cm shell is 5.22 celsius.
For the 14cm / 20cm / 28cm shells, (a) the temperature of the 14cm shell is 7.67 celsius, and (b) the temperature of the 20cm shell is 2.59 celsius.
Average of Lambertian and Specular Reflection Temperatures.
For the 14cm / 28cm shells only, the temperature of the 14cm shell is 5.04 celsius.
For the 14cm / 20cm / 28cm shells, (a) the temperature of the 14cm shell is 7.50 celsius, and (b) the temperature of the 20cm shell is 2.54 celsius.
The format of this blog does not allow me to “post my calculations here.” If you’re still interested in seeing my calculations, maybe Joanne can act as a go between or give me your email address.
00
Oops.
The shell radii are 14cm, 20cm, and 28cm, not 0.14cm, 0.20cm and 0.28cm. I work in meter/kilogram/second units and forgot to remove the decimal point when converting to centimeters.
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Reed Coray October 2, 2015 at 9:52 am
Will, this comment is primarily in response to your 2 October 2:42 am comment, but also addresses something you wrote in your 1 Oct 11:01 am comment. First, I’ll deal with your 1 Oct. 11:01 comment. You wrote:
“Where oh where can you physically find such conditions? You claim a real demonstration, then return with a silly theoretical ming game. I will not play!!”
Where are you to get the zero radiance environment that contains your sphere/shell? This is but a mind game, you can never verify that you calculations were correct. Believe me with combination heat transfer they never will be the same as your first actual measurement.
*******************************
(“““I think with 10 rods rather than the 6 rods, the temperature would be less than the 303K [303K is the approximate temperature of the powered sphere in isolation] even with no EMR flux from sphere to shell.”)
“What am I missing?”
Missed what I wrote next:
Lets see (324-214) x (10/6) + 214 = 280 K. Yup, less than 303K. With internal EMR even lower yet.
The 60 Watt 303K went down to 214K because you quadrupled the surface area. Try that with a 100Km atmosphere around a 6400km radius sphere. If you just move to 60 watts to the shell everything is at 214K. I guess I miss read what you did write. I thought you meant that just adding something “interfering” with the sphere exitance could result in a decrease in sphere temperature. Your conductive rods added a different path for part of the 60 Watts.
All the best! -will-
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Reed CorayOctober 3, 2015 at 10:37 am
“OK Will, here are my results.
First a restatement of the conditions:
(1) Three concentric, thin, aluminum foil, spherical shells of radii 0.14cm, 0.20cm and 0.28cm.”
When I did this my shells had thickness of 0.75 mm but the were
14cm, 20cm, 28cm diameter! hard to find a bell jar with 56cm inside diameter. doesn’t really matter
“(2) Outermost shell (28cm shell) everywhere in contact with a heat sink maintained at 0 Celsius.
(3) Internal to innermost shell a 5 Watt source of thermal energy.
(4) For the case of the 14cm and 28cm shells only (no 20cm shell), the space between the 14cm shell and the 28cm shell is a vacuum.
(5) For the case of all three shells, the spaces between (a) the 14cm shell and the 20cm shell, and (b) the 20cm shell and the 28cm shell are vacuums.
(6) Absorptivity of all shell surfaces 0.9”
Okay fine!
04
Reed CorayOctober 3, 2015 at 10:37 am
“OK Will, here are my results.
First a restatement of the conditions:
(1) Three concentric, thin, aluminum foil, spherical shells of radii 0.14cm, 0.20cm and 0.28cm.”
When I did this my shells had thickness of 0.75 mm but the were
14cm, 20cm, 28cm diameter! hard to find a bell jar with 56cm inside diameter. doesn’t really matter
“(2) Outermost shell (28cm shell) everywhere in contact with a heat sink maintained at 0 Celsius.
(3) Internal to innermost shell a 5 Watt source of thermal energy.
(4) For the case of the 14cm and 28cm shells only (no 20cm shell), the space between the 14cm shell and the 28cm shell is a vacuum.
(5) For the case of all three shells, the spaces between (a) the 14cm shell and the 20cm shell, and (b) the 20cm shell and the 28cm shell are vacuums.
(6) Absorptivity of all shell surfaces 0.9”
Okay fine!
Lambertian Reflection Results:
For the 14cm / 28cm shells only, the temperature of the 14cm shell is 4.87 celsius.
For the 14cm / 20cm / 28cm shells, (a) the temperature of the 14cm shell is 7.33 celsius, and (b) the temperature of the 20cm shell is 2.48 celsius.
Specular Reflection Results:
For the 14cm / 28cm shells only, the temperature of the 14cm shell is 5.22 celsius.
For the 14cm / 20cm / 28cm shells, (a) the temperature of the 14cm shell is 7.67 celsius, and (b) the temperature of the 20cm shell is 2.59 celsius.
05
Reed Coray October 3, 2015 at 10:37 am
“OK Will, here are my results.”
This computer keeps posting before I click on anything!!
“Specular Reflection Results:”
This I do not get whatever is being reflected inside the sphere? Here is what I get!
For the 14cm/28cm radius shells only, the temperature of the 14cm shell is 5.1 Celsius.
For the 14cm/20cm/28cm radius shells, (a) the temperature of the 14cm shell is 9.95 Celsius, and (b) the temperature of the 20cm shell is 2.48 Celsius.
For the 14cm/28cm diameter shells only, the temperature of the 14cm shell is 18.5 Celsius.
For the 14cm/20cm/28cm diameter shells, (a) the temperature of the 14cm shell is 34.8 Celsius, and (b) the temperature of the 20cm shell is 9.8 Celsius.
Kick back!! this I do not know where this is from!
——————————
(“If you now move the 5 Watt source to the 20cm shell its calculated and measured temperature must both drop slightly. The now un-powered inner 14cm must go to that same temperature”)
“Why? Is not the 20cm radiating 5 watts internally as well as externally, as Climate Clowns and Willis Echenbach claim?”
????? No there is only one 5 Watt source of power. How can something unpowered inside a fixed environment go to a higher temperature than the environment. When the inner temperatures reach equilibrium all internal power goes to zero.
When the 5 Watts in in the 14cm all five watts go straight through the 20cm without restriction. However the 14cm increases in temperature by diameter 5C or radius 16.3C that the difference in radiance is to facilitate the 5 Watt transfer from sphere to shell and the necessary rethermalization of the shell.
All the best! -will-
05
Will,
The shell radii do make a difference to the shell temperatures. For shell radii of 7cm, 10cm and 14cm (diameters 14cm, 20cm and 28cm) my theoretical results are:
Lambertian Reflection Results:
For the 7cm / 14cm (radii) shells only, the temperature of the 7cm (radius) shell is 18.12 celsius.
For the 7cm / 10cm / 14cm (radii) shells, (a) the temperature of the 7cm (radius) shell is 26.44 celsius, and (b) the temperature of the 10cm (radius) shell is 9.54 celsius.
Specular Reflection Results:
For the 7cm / 14cm (radii) shells only, the temperature of the 7cm (radius) shell is 19.32 celsius.
For the 7cm / 10cm / 14cm (radii) shells, (a) the temperature of the 7cm (radius) shell is 27.54 celsius, and (b) the temperature of the 10cm (radius) shell is 9.96 celsius.
Average of Lambertian and Specular Reflection Temperatures.
For the 7cm / 14cm (radii) shells only, the temperature of the 7cm (radius) shell is 18.72 celsius.
For the 7cm / 10cm / 14cm (radii) shells, (a) the temperature of the 7cm (radius) shell is 26.99 celsius, and (b) the temperature of the 10cm (radius) shell is 9.75 celsius.
Specular reflection and Lambertian reflection affect the shell temperatures differently. In what follows, concentric spherical shells are assumed. For specular reflection, all of the energy that leaves the inner surface of an outer shell in the direction of the inner surface of the outer shell is reflected in the direction of the inner surface of the outer shell—i.e., none of this energy ever encounters the inner shell. For Lambertian reflection, some of the energy that leaves the inner surface of the outer shell in the direction of inner surface of the outer shell will be reflected towards the inner shell. In addition, for specular reflection, all of the energy leaving the outer surface of the inner shell that is reflected from the inner surface of the outer shell will be reflected back towards the inner shell. For Lambertian reflection, some of the energy leaving the outer surface of the inner shell that is reflected from the inner surface of the outer shell will be reflected towards the inner surface of the outer shell. These different reflection properties affect the total energy absorbed by the inner and outer shells, and hence can affect the temperatures of the shells.
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Reed Coray October 5, 2015 at 3:39 am
“Specular reflection and Lambertian reflection affect the shell temperatures differently. In what follows, concentric spherical shells are assumed.”
Interesting conjecture but can never be demonstrated as correct even once!
The three shells, innermost powered, must have decreasing temperature outward. This is but a consequence of projective geometry for an electromagnetic field (EMFS). Decreasing actual EMFS as 1/distance^2. This demands that there be no ‘back radiation’ ever!
This is easy to demonstrate, with a single thermistor within a isothermal shell. No-mater where you move the thermistor within the volume, its resistance (temperature) remains the same, that of the isotherm! The way you are applying your concept, at the exact center of the isothermal shell (focus), the temperature “must” go to infinity as the surface area of the thermistor goes to zero.
Please demonstrate such, or offer other physical demonstration of thermal EMR flux in the direction of higher radiance!
“For specular reflection, all of the energy that leaves the inner surface of an outer shell in the direction of the inner surface of the outer shell is reflected in the direction of the inner surface of the outer shell.”
How can this be true? the cross-sectional area of the inner shell is 1/2 of the entire view from any point on the inside of the outer shell. i.e. 5% first incident on original flux indecent on outer from inner shell. Exactly the same as for diffuse (Lambertian) reflection, in this case an increase in outer shell emissivity to 0.95.
With EMR, reflection, and what is reflected is critical, and why the S-B equation is never a value but only a upper limit of what may be. Reflectivity however is not allowed, except by specific geometry description. One of many changes from in observed and measured emissivity to be painfully learned via physical experiment.
BTW the 5% that could be reflected back to the inner sphere is never emitted in the first place. Such has already increased the apparent radiance of the inner surface of the outer shell. Are we to accept emissivities as emissivities, or with an unannounced (1-emissivity), as reflection, transmission, or transfer to elsewhen? 😉 Another example of how clueless are the CAGW Clowns of EMR!
All the best! -will-
03
Will,
You made three statements that are incorrect.
First, you wrote: “Interesting conjecture but can never be demonstrated as correct even once!”
Why can’t it be demonstrated? For large reflector telescopes, people routinely polish the surface of the reflector to a high degree of smoothness. The shape of the reflecting surface of a telescope is parabolic, not circular; but there’s no reason why two equal-radius hemispherical reflectors and a smaller radius sphere couldn’t be polished to the same degree of smoothness. Mount the sphere center at the geometrical center of one of the hemisphere rims and place the other hemisphere on top of the first hemisphere. You now have two concentric spherical surfaces whose reflection properties are overwhelmingly specular. Admittedly, a highly polished surface will have a high reflectivity (low emissivity), but that only changes the numbers, not the experimental concept.
Second, you wrote: “ The three shells, innermost powered, must have decreasing temperature outward. This is but a consequence of projective geometry for an electromagnetic field (EMFS). Decreasing actual EMFS as 1/distance^2. This demands that there be no ‘back radiation’ ever! This is easy to demonstrate, with a single thermistor within a isothermal shell. No-mater where you move the thermistor within the volume, its resistance (temperature) remains the same, that of the isotherm! The way you are applying your concept, at the exact center of the isothermal shell (focus), the temperature “must” go to infinity as the surface area of the thermistor goes to zero.”
There is nothing about the way I’m “applying the concept” that implies the temperature of the thermistor must go to infinity at the center of the innermost shell. Neither the radiation originating from the inner surface of the innermost shell nor any reflected radiation (specular or Lambertian) from the inner surface of the innermost shell is “focused” on the center of the shell. The radiation emitted from the inner surface of the innermost shell is Lambertian—i.e., decreases in strength (power) as the COSINE of the angle from the surface normal. This does imply that the direction of greatest emission is towards the center of the sphere, but does not imply that all emitted radiation is in that direction—i.e., is focused on the center of the innermost shell. In fact, as the size of a thermistor positioned at the center of the sphere goes to zero, the energy incident on the thermistor from the inner surface of the innermost shell also goes to zero. So there is no implication that the temperature of such a thermistor must be infinite.
Third, in response to my claim: “For specular reflection, all of the energy that leaves the inner surface of an outer shell in the direction of the inner surface of the outer shell is reflected in the direction of the inner surface of the outer shell” you wrote:
“How can this be true? the cross-sectional area of the inner shell is 1/2 of the entire view from any point on the inside of the outer shell.”
Simple. Draw a straight line from any point on the inner surface of the outer shell in any direction that misses (i.e., does not pass through) the concentric innermost sphere. This line will eventually intersect the inner surface of the outer shell at some point. At the point of intersection, draw a normal to the inner surface of the outer shell. Specular reflection implies that the angle the reflected ray makes with the normal to the reflecting surface must be (a) in the plane of the incident ray/surface normal, and (b) be equal to the incident angle. Using this rule, draw the reflected ray. You will see that the reflected ray will completely miss the inner sphere. [As a side note, it can be mathematically shown that such a reflected ray will not intersect the inner sphere.] The reflected ray possesses the same properties as the original ray—namely (a) it originates from a point on the inner surface of the outer shell, and (b) its direction of propagation misses the inner sphere. Thus, when the reflected ray is itself specularly reflected from the inner surface of the outer shell, the second specular reflection will also miss the inner sphere. This process continues indefinitely. That is, energy that originates from a point on the inner surface of the outer shell in a direction that misses the inner sphere will be such that all outer shell reflections of that energy will miss the inner sphere. The fact that the cross-sectional area of the inner sphere is a large fraction of the field of view from a point on the inner surface of the outer shell has nothing to do with specularly reflected energy that originates from a point on the inner surface of the outer shell and first encounters (is reflected) the inner surface of the outer shell.
Now a piece of unsolicited advice. If I were you, I stop spouting how “Clueless the CAGW clowns of EMR are” and give what they say some serious thought. Not everything they say is nonsense. Since Planck himself stated that radiation from surface “A” to surface “B” occurs whether surface “A” is hotter, cooler, or the same temperature as surface “B”, I wouldn’t pooh-pooh the concept of “backradiation.” Specifically, if the only source of energy for a cooler surface “A” is the absorption of radiation from a hotter surface “B”, then any radiation emanating from surface “A” in the direction of surface “B” can with some justification be called “backradiation.” To deny that such radiation exists is to deny a statement from Planck. You may be right and Planck may be wrong; but as it stands now, most people would side with Planck.
10
“Will,You made three statements that are incorrect.”
“First, you wrote: “Interesting conjecture but can never be demonstrated as correct even once!””
“Why can’t it be demonstrated?
Try to demonstrate your fantasy even once!
You insanely insist that thermal EMR exitance (flux) from a surface or cross-sectional is independent of any opposing radiance. Where is the evidence of such.
You seem to be only a paid troll in opposition to what is or may be.
02
Will,
Tone it down, or we’re done. Phrases like “insanely insist” and “You seem to be only a paid troll in opposition to what is or may be” don’t help. However, in response to your question: “Where is the evidence of such?” On pages 6 and 7 of a translation of Max Planck’s 1912 paper
(see https://archive.org/stream/theoryofheatradi00planrich#page/6/mode/2up)
the translator wrote: “A body A at 100 degree C emits towards a body B at 0 degrees C exactly the same amount of radiation as toward an equally large and similarly situated body B’ at 1000 degrees C. The fact that the body A is cooled by B and heated by B’ is due entirely to the fact that B is a weaker, B’ a stronger emitter than A. ”
Plank’s statement is not experimental evidence, but in a court of law most people would agree it qualifies as an “expert opinion.” Since to the best of my knowledge, experiments to date confirm rather than nullify Planck’s statement above, I’d say I have “pretty good evidence.”
P.S. To date no one has given me so much as a single penny in support of or in disagreement with the AGW and CAGW claims or with the skeptics rebuttals of those claims. As to my being “in opposition to what is or may be“, well I’d describe that statement coming from you as the “pot calling the kettle black.”
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Reed Coray October 5, 2015 at 1:45 pm
“Now a piece of unsolicited advice. If I were you, I stop spouting how “Clueless the CAGW clowns of EMR are” and give what they say some serious thought. Not everything they say is nonsense.”
Take your unsolicited (illicit) crap and shove it, troll!!
“Since Planck himself stated that radiation from surface “A” to surface “B” occurs whether surface “A” is hotter, cooler, or the same temperature as surface “B”, I wouldn’t pooh-pooh the concept of “backradiation.”
As far as I have found Dr Max Planck only wrote or spoke of thermal radiative ‘specific intensity’, now called ‘spectral radiance’ Can you give evidence where ‘he’ wrote of radiative flux, the actual power transfer?
“Specifically, if the only source of energy for a cooler surface “A” is the absorption of radiation from a hotter surface “B”, then any radiation emanating from surface “A” in the direction of surface “B” can with some justification be called “backradiation.””
All you have to do is demonstrate the actual thermal emission from cold ‘A’ in the direction of higher temperature ‘B’! You cannot, nor can anyone else! All fantasy!
“To deny that such radiation exists is to deny a statement from Planck. You may be right and Planck may be wrong; but as it stands now, most people would side with Planck.”
Again just show the writing with sufficient prior to indicate a meaning of ‘power transfer in a direction of higher temperature’!
Planck, Boltzmann, and Maxwell all very well understood the difference between field strength and flux. Only you and the other warmist trolls. insist that ‘radiation’ fits all with deliberate intent to deceive!
“Simple. Draw a straight line from any point on the inner surface of the outer shell in any direction that misses (i.e., does not pass through) the concentric innermost sphere.’
Go ahead and draw such with surface specular reflection in the geometry given! Only in the case of diffuse reflection can such ‘miss’, if your fantasy reflection does not mess the inner than it was not ever radiated., If it does miss than all such would be 90% absorbed at the next inner surface of the outer shell. Besides just who allows the troll to assign arbitrary reflectance to my problem?
“The radiation emitted from the inner surface of the innermost shell is Lambertian—i.e., decreases in strength (power) as the COSINE of the angle from the surface normal.”
The troll, with no concept of any geometry assigns ‘Lambertian’ to the concave surface of a sphere! There is absolutely no radiation emitted from the inner surface of the innermost shell. The whole thing is isothermal. No internal flux whatsoever!
I learn much from Joanne’s blog. You however, have nothing to offer, please go somewhere and buy a clue! 🙁
02
Reed Coray October 6, 2015 at 7:00 am
“However, in response to your question: “Where is the evidence of such?” On pages 6 and 7 of a translation of Max Planck’s 1912 paper”
(‘the translator wrote: “A body A at 100 degree C emits towards a body B at 0 degrees C exactly the same amount of radiation as toward an equally large and similarly situated body B’ at 1000 degrees C. The fact that the body A is cooled by B and heated by B’ is due entirely to the fact that B is a weaker, B’ a stronger emitter than A. ”
What you and others claim Max Planck wrote, is the translators clarification. Immediately before your intentionally abbreviated quote are the words (Prevost’s principle).!!! Clearly Planck was referring to the current, at that time, false empirical law of Prevost! In that same Max Planck’s 1912 paper, Planck completely trashes that concept.
That trashing was the purpose of the paper!!
The ‘specific intensity’ (spectral radiance) of B is less than that of A while that of B’ is greater than that of A. Is the actual reading of what Max did write beyond your comprehension?
“Since to the best of my knowledge, experiments to date confirm rather than nullify Planck’s statement above, I’d say I have “pretty good evidence.””
Can you please name such experimental evidence? Many claim radiance is actual flux, but no one has ever witnessed or measured such thermal flux in a direction of higher temperature. It remains complete fantasy!
12
Will,
If I decide to “go somewhere and buy a clue! 🙁, I’ll be sure to avoid the Will Janoschka store of “cram-it-down-your-throat” Physics. Even at bargain basement prices, you only get junk from a know-it-all.
Any continuation of a discussion between you and me is pointless. I’m signing off.
10
Will, expanding the above comment. Using the powered-sphere/shell/connecting-rod system parameters in the above comment, the outer surface of the shell is the only surface from which energy leaves the system (sphere/shell/connecting rods) to space. Thus, for a powered-sphere’s internal energy rate of 60 Watts, independent of the number/properties of the connecting rods, in energy-rate-equilibrium the outer surface of the shell must radiate energy to space at a rate of 60 Watts. For the assumed emissivity of one and the assumed radius of the shell of 0.2 meters, the temperature of the shell will be 214.199 Kelvin. For an assumed emissivity of one and an assumed radius of 0.1 meters, the temperature of the powered-sphere in isolation will be 302.923 Kelvin.
Assume that in the presence of the shell with connecting rods the temperature of the powered-sphere must be greater than the temperature of the powered-sphere in isolation. Then at a minimum, in the presence of the shell with connecting rods the temperature of the powered sphere must be greater than 302.923 Kelvin. For the purposes of computation, assume the temperature of the powered sphere is exactly 302.923 Kelvin.
Give or take a small variation arising from the curvature of the ends of a rod that connects the powered sphere and the shell, a radially oriented connecting rod will have a length of 0.1 meters. Let the connecting rod be a cylinder with radius 0.002582 meters. The cross-sectional area of the connecting rod will thus be 0.000020944 square meters–a small fraction of the surface area, 0.12566 square meters, of the powered sphere. If “K” is the thermal conductivity of the connecting rod, then the heat current, H, through a rod joining two objects whose temperature difference is deltaT is
H = K * A * deltaT / L
where “A” is the area of the rod, and “L” is the length of the rod. Thus for a thermal conductivity of K = 401 Watts per meter per degree Kelvin (the thermal conductivity of copper), the conduction heat current through a single rod connecting the powered sphere at temperature 302.923 Kelvin to the shell at temperature 214.199 Kelvin is 7.452 Watts. Thus if the temperatures of the powered sphere and shell don’t change, for 10 such connecting rods the conduction heat current from the powered sphere to the shell is 74.52 Watts. Note that because the powered sphere’s temperature is greater than the shell’s temperature, any radiative heat flow between the two must be in the direction of the powered sphere to the shell. Thus independent of any radiative heat transfer from the powered sphere to the shell (which must be positive), the powered sphere is losing energy via thermal conduction at a rate greater than the rate energy is being input to the powered shell. This situation can’t exist in energy-rate-equilibrium. Since (a) the length/area of each connecting rod is fixed, and (b) the temperature of the shell is fixed at 214.119 Kelvin by the system energy-rate-equilibrium condition, the only way to establish energy rate equilibrium for the solid sphere is to lower its temperature. But if we lower the surface temperature of the powered sphere to establish energy-rate-equilibrium, then the temperature of the powered sphere in the presence of the shell/connecting rods (10) will be less than the temperature of the isolated sphere.
10
The molecular weight of CO2 is 44.01 and the molecular weight of Nitrogen (N2) is 28.01 so why does CO2 migrate to the upper atmosphere since it is much heavier and so I would expect it to stay close to the ground?
11
I believe the CO2 does tend to hug the bottom somewhat, not extending up as far as the N2. But even so the CO2 extends well into the stratosphere in concentrations not much lower than at the surface.
00
“But even so the CO2 extends well into the stratosphere in concentrations not much lower than at the surface.”
The ppmv is approximately the same but there is very little radiative flux that originates in the stratosphere at any wavelength. The lateral optical depth thus temperature, is hundreds of kilometers. 🙂
04
Dave: The atmosphere is turblulently mixed and homogeneous up to the turbopause – 100 km. See turbopause in Wikipedia. Above there is some separation by molecular weight.
00
“A blanket of Co2 surrounds the Earth, trapping heat as follows” you say David.
If this atmosphere (or CO2) trapped some heat,..because the planet Earth is ripping along through space..wouldn’t we expect to see (measure) a heat trail left behind the planet?
10
Space is empty (almost). Nothing to leave the heat trail in.
10
Mack worse we do have a tail of another sort that would erase that one. Our planet pumps a huge amount of mostly DC energy into it (Let Will calculate the power of a DC photon). It even causes changes in the weather on the moon.
“During the crossing, the Moon comes in contact with a gigantic “plasma sheet” of hot charged particles trapped in the tail.”
“Two million degree electrons in the plasma sheet race around like crazy and many of them hit the Moon’s surface.”
“Researchers now believe the glow is sunlight scattered from electrically-charged moondust floating just above the lunar surface.”
http://science.nasa.gov/science-news/science-at-nasa/2008/17apr_magnetotail/
00
from a reply of mine above
The radiation back to the ground hides reality. Before the upper atmosphere can warm enough to heat the surface, its warmed enough to compensate for the reduced OLR from the surface. Can it be any other way? Hot air rises. The concentration of CO2 and water increases as you go down. Somethings missing.
I think that the amount of OLR from water condensing in the stratosphere atmosphere is swept under the carpet. Page 8 shows how water vapour follows closely the temperature.
There is roughly as much water vapour up there that if it were to all condense over a year, that’s about 100W/m2 of latent heat for the surface area of the Earth. If just 2% condenses a day (night) that’s 365W/m2 of water droplets emitting OLR.
It seems to be the perfect pump that is temperature controlled and affected by meridonal air currents.
20
There are a few issues in there, which we will be dealing with in detail later, starting with the next post.
20
Introducing this nonsense on OLR pipes gives credence to the nonsense of “greenhouse gases”. The only influence of significance IR-responsive gases can have on climate is to alter cloud formation and affect the absorption/reflectance of the atmosphere.
As long as some solar radiation gets to the surface it is the specific heat of the atmosphere that controls the lapse rate and makes the surface warmer than it would be if there was no atmosphere. It has nothing to do with the IR-responsiveness of the atmospheric gasses.
As long as some solar radiation gets to the surface, the solid earth surface provides the heating plate and convection dominates the heat transfer into the atmosphere. All heavenly bodies with a substantial atmosphere, meaning pressure grater than 100mb, exhibit lapse rate that can be determined by the isentropic expansion with increasing elevation that is a function of the specific heat of the gasses and the gravitational constant – no need to consider diabetic processes:
http://www.nature.com/ngeo/journal/v7/n1/images/ngeo2020-f1.jpg
Diabatic processes have negligible influence in the troposphere. By definition it is the well mixed zone.
If Earth had a fully transparent atmosphere, without IR-responsive gasses, that atmosphere would still exhibit a lapse rate. In fact the surface temperature would be little different to what it is now. With the equatorial zone dominated by oceans the radiative thermal balance occurs at an overall global SST of 289K using measured values of ocean albedo and ice caps.
42
Rick,
IMHO that is largely correct. However, the concept of pipes is helpful in explaining how the thermal balancing act is really effected despite changing the amount of GHGs in an atmosphere.
Changing lapse rate slopes is at the heart of it because that allows convection to change and in turn redistribute energy flows between pipes.
212
Unlike a human body which creates its own warmth and which is, of course, kept warm by a blanket which stops it from losing the heat the body creates, the earth’s surface has essentially no self created heat. Thus, IMHO, the blanket analogy fails to pass the sniff test!!
The earth’s surface has only the heat it receives from the sun, and since no cold body can direct heat (rather than solely energy) to a body at a higher temperature, once the sun’s energy stops warming the surface of the earth, the surface can only cool. It can never be warmed by the colder atmosphere above it.
To those of you who disagree with the above, I pose a simple solution.
Describe to me an experiment, including a description of the apparatus and the replicable empirical results of the experiment, which demonstrates that a cold body can increase the temperature of a nearby hotter body.
Good luck with that! I won’t accept theoretical arguments about steel greenhouses or incandescent light bulbs, for example, only properly constructed replicable experiments.
14
This is exactly right, now the lapse rate absent energy transfers in the atmosphere would be 9.8 deg per Km from about 10 Km and that would put earths average surface temp at about 49 degrees as opposed to the current 15, all the pipes then reduce that lapse rate to about 6.4 (dry) and 5 wet, and combining these we get an average temperature of 15, all because hot flows to cold, the total energy (KE + PE) increases with height because heat wants to flow from hot to cold. Thus the temp at the surface is roughly 49 deg – 10 × (9.8 – lapse rate). Hope you follow this.
Climate science attempts to say that CO2 reduces the ability of heat to flow from hot to cold, this would INCREASE the lapse rate slightly and increase the difference in temp between the
Tropopause and the earth, remember the higher the lapse rate, the hotter it is. The lapse rate however can never be higher than the zero energy motion -9.8 deg per km
Do you see here that we start out at 49 degrees, and then cool down to 15, if we reduce the cooling the temperature can rise up to 49 degrees at most.
Irritatingly this is one of the good reasons why climageddon is impossible. The IPCCs position is that the average temperature of the world with a CO2 atmosphere at 1 ATM pressure (13 doublings) would be from 16 to 65 degrees warmer (between 31 and 80 C)
That would require a lapse rate of between 8 and 12.9 if the stratosphere remained at the same temperature, but the same theory claims the stratosphere will cool, which raises the required lapse rate again. Remember then that 9.8 is the maximum lapse rate possible greater lapse rates require energy to flow from cold to hot ! Or the surface to SOURCE energy.
Even on Venus, the lapse rate is not above the theoretical no loss maximum.
CO2 warming cannot be unbounded it MUST saturate. There is other math that helps confirm this idea.
02
By my understanding the Earth’s atmosphere is not directing heat back down to the surface thereby raising its temperature. Instead it directs electromagnetic radiation back down to the surface and those two things are different. Electromagnetic radiation is just an undifferentiated blob of radiation that can be directed anywhere, it only becomes heat when it is absorbed by a body and converted into kinetic energy. Once the surface absorbs radiation from whatever source (even if that radiation comes from a colder body) its internal temperature should increase in accordance with the 1st law of thermodynamics. Not so? If people want to argue against the greenhouse effect then they’ll have to present an alternative source of energy for raising the mean temperature of the Earth by 33C above its effective temperature of 255k. Can anyone here do that?
The self-created heat from the Earth comes in the form of the energy provided by the Sun. Greenhouse gases are transparent to shortwave radiation and so the Sun’s radiation does not heat them directly. The gases are heated by the longwave radiation reemitted by the Earth’s surface. Thus the Earth is not required to heat itself. It is heated by the Sun and this means that, in theory, the Earth’s maximum possible temperature can be raised to the level of the primary source-radiation, which I gather is about 1368W/sq.m. The blanket analogy that David talks about is a useful aid in understanding the mechanism that makes the greenhouse effect possible. The blanket acts as a barrier to prevent warm air surrounding the person’s body from being lost and by its recycling the body’s radiation back to it the energy of that radiation warms the body. Where does the energy of that returning radiation go if not into warming the body?
There is no need to describe a thought experiment in my view. The principle at work here is the same as that used in a metal-foil survival-blanket when applied to an accident victim or someone who tightly cocoons themselves in a sleeping-bag at night to keep warm.
21
I didn’t ask for a thought experiment, rather a real honest to goodness experiment, with which anyone with the requisite experience and equipment could produce real empirical evidence that a cold body can warm a hotter one!
01
“Once the surface absorbs radiation from whatever source (even if that radiation comes from a colder body) its internal temperature should increase in accordance with the 1st law of thermodynamics.”
The requirement is for some demonstration that such radiative flux is ever generated in the first place. Have you such a demonstration, not fantasy, that the rest of us can demolish with little effort? None so far has ever been offered. 🙂
210
David Wood September 29, 2015 at 2:01 pm
“Describe to me an experiment, including a description of the apparatus and the replicable empirical results of the experiment, which demonstrates that a cold body can increase the temperature of a nearby hotter body.
Good luck with that! I won’t accept theoretical arguments about steel greenhouses or incandescent light bulbs, for example, only properly constructed replicable experiments.”
David,
I have done such demonstration easy to do, but means nothing. Any interference with spontaneous thermal radiative exitance from a powered source to space or any lower temperature must result in a higher temperature of that powered source.
See my response @ 23.2.
If there is some passive opaque matter in the path of powered thermal EMR to a lower temperature sink that opaque must re-thermalize to a temperature between that of the powered source and that of the sink (lower temperature than the source, but no back radiation). The opaque must have a temperature sufficient to radiate all power to sink. The ‘powered’ however must increase in temperature to provide sufficient difference in radiative potential to maintain the exact power transfer to the opaque to be passed unhindered through the opaque to the sink. This is precisely what Gus Kirchhoff, extensively wrote about in 1859!
All the best! -will-
29
One of the most interesting things about this conversation is how political it is. Yes, it’s about science, but since it involves human beings (or some approximation thereof) it is inherently political. There are way too many commenters who appear to be grinding their own axe or defending their own position (often by rhetorically attacking an imaginary agonist), entirely preoccupied with their pet content instead of watching the overall context of the individual posts accumulate. The whole is greater than the sum of the parts, especially when the parts are so saturated in indeterminacy and complexity. David’s straw-dog “establishment model” is for some a dread paper tiger to be wrangled to the ground and put out of their own misery.
Since those of us who have some mature perspective know that whatever comes of all this will be just one more small step on our drunkards walk toward a more perfect experience of all that’s always happening on the planet, we needn’t make it a clash of egos or an opportunity for status-seeking posturing. And some understand the pitfalls of prioritizing symbol-bound mentation over experiential contemplation. One of the key strategies for dealing with coupled chaotic systems is to pay more honor to intuition and “feel” than to reason and logic.
Please, fellas, slow down, lighten up, pay attention, and discover what you get. Part of what you’ll get is some paradigm-shift dissonance and discomfort, and that is in the long run a splendid reward indeed. Along with that, you get to live in a larger domain of the world and become more wise about its political nature.
40
Well put Alexander !!
00
I disagree, what you are seeing here is a contest of ideas, this is the melting pot of sciemce. All the comments fitting their particular ideas to the basic model are finding the chinks in the armour. David will still get to say what he wants and others will chip in, and either succeed or be corrected.
This is the problem on the other side of the debate, too much agreement. As a manager I have always said that I welcome disagreement among my staff because from disagreement we will get real solutions, from “yes men” you get only groupthink.
Too many managers sacrifice debate for “peace” and misassociate disagreement with disloyalty. I say to my people you are free to disagree, but once the decision is made, you may continue to disagree with me privately, however you must respect the decision.
I once had a manager that insisted that you actually agreed with him, despite the fact that I said that while I disagree wigh it I’ll respect his decision and implement it to the best of my ability. That wasn’t good enough, I HAD to actually agree with the decision. Of course, since integrity matters I never did – didn’t help my career much…
Oh well, it takes all sorts to make a world
30
“Since those of us who have some mature perspective know that whatever comes of all this will be just one more small step on our drunkards walk toward a more perfect experience of all that’s always happening on the planet, we needn’t make it a clash of egos or an opportunity for status-seeking posturing. And some understand the pitfalls of prioritizing symbol-bound mentation over experiential contemplation.”
Good words, but only words, You seem to be writing only of what you wish, with no attention to what must be!
“One of the key strategies for dealing with coupled chaotic systems is to pay more honor to intuition and “feel” than to reason and logic.”
You seem to insist that Earth’s weather is somehow chaotic! If you would bother to examine, Earth’s weather at any location is mostly cyclic, many multi-cyclic, with well defined dispersion to prevent resonant destruction.
“Please, fellas, slow down, lighten up, pay attention, and discover what you get. Part of what you’ll get is some paradigm-shift dissonance and discomfort, and that is in the long run a splendid reward indeed. Along with that, you get to live in a larger domain of the world and become more wise about its political nature.”
WoW! The great prophet for profit, demanding what others must do! Wise indeed!
09
Correct me if I am wrong but by this explanation of the greenhouse effect, changes in concentration of CO2 should have no effect on temperature at all because the atmosphere is opaque to ~15μm radiation into the stratosphere.
The logic is as follows:
The greenhouse effect depends on the fact that the effective emission level of the relevant frequency is at a high altitude. At high altitudes the temperature is lower than at the surface. Therefore the total rate of energy emitted is less, because lower temperatures radiate with lower power.
The concentration of a gas changes the altitude from which the radiation starts to escape to space because that is the altitude at which the concentration of that gas falls below a certain level. Therefore the rate of energy radiation depends on the temperature at this new altitude.
Thus if the concentration increases then the altitude of the emission layer increases, and if the temperature at this level is lower then the power of emission at that frequency reduces.
However the lower stratosphere is roughly isothermal. In the International Standard Atmosphere it remains at -56.5 C from 11km to 20km. From there it actually increases in temperature with altitude. While the real atmosphere does not behave quite like this it is a reasonable average.
The Tropopause is higher at low latitudes, more than 16 km, so it might be that CO2 might increase temperatures in that region. I am not sure how high this emission level is for carbon dioxide.
So what is the error in my reasoning?
00
You just need to go a little bit further and consider the wings of the CO2 blockage.
Yes, at around 15 microns the CO2 EL is well into the stratosphere (40 km? don’t quote me), where it gets slightly warmer as one ascends. In fact, increasing the CO2 concentration warms the CO2 EL at 15 microns, and causes more emissions at this wavelength (see the spike right at 15 microns in a Nimbus emissions spectrum; later in the series).
However the atmosphere is not nearly as opaque around 13 and 17 microns, out in the wings of the CO2 blockage or well. At these wavelengths the CO2 EL is in the troposphere, so increasing the CO2 concentration causes the Co2 EL to ascend. Nearly all the difference in the emission spectrum as CO2 increases occurs well out in the wings — and, taking the entire spectrum into account, less OLR is emitted by CO2 molecules as the CO2 concentration increases. (For each CO2 doubling, a decrease of about 3.7 W/m2; see post 2.)
30
That doesn’t make sense, you say there are more emitting molecules yet the emission is less – nup, can’t happen.
A molecule that is excited, is not the same as one that is hot, one is energised through a raised bond energy while the other has greater translational velocity. The emission band at any temperature is dependent on the temperature as velocity can be be plus or minus the doppler shift dependant on the velocity of the molecules hotter=faster. This means that all the atmosphere will emit IR at the base wavelength ( from stationary molecules). But only the warmest layers will emit at the extremes.
When you look at this from a quantum point of view, its clear what must happen. Radiation temperature is NOT the same as thermal temperature.
12
Just to correct something in my last, where I use the word “concentration” twice in one sentence with different meanings, which is terribly imprecise.
“The concentration of a gas changes the altitude from which the radiation starts to escape to space because that is the altitude at which the concentration of that gas falls below a certain level”
should have read
“The proportion of a gas changes the altitude from which the radiation starts to escape to space because that is the altitude at which the absolute concentration of that gas falls below a certain level”.
10
I’m not sure if anyone else has covered this but OLR should be a function of both temp and density (concentration) If you increase the concentration of CO2 at the emission layer then more photons will be emitted to space at the same frequency. The number of Watts is equal to the number of emitted photons (concentration dependent) times the energy of each photon (temperature dependant.) think of Watts=Voltage x Amps.
00
“I’m not sure if anyone else has covered this but OLR should be a function of both temp and density (concentration)”
It is covered, but not properly, Exitance at every wavelength is determined by optical depth (63%) emissivity and the apparent temperature of that emissivity. Outside the emission bands of H2O and CO2 the optical depth can be greater than the whole atmosphere. In the CO2 15 micron band optical depth is the least, 5 to 40 meters right at the temperature of the tropopause. In the WV window 8-13 microns cloud tops have an optical depth of five meters right at the temperature of those cloud tops. from above, throughout the rest of the WV band one optical depth forms at about 6km altitude with the temperature of 5km level.
The only real variables are the cloud top level and the WV continuum value. No one on or about this Earth has any idea of what may determine these two variables. 🙁
All the best! -will-
011
Dr. Evans, I wish to thank you for talking this down to a level where I can basically understand it. No, I am not attempting the math since I hated it in college in the first place, seeing no real correlation between higher math and the world – it really does help in learning a tool if you can actually see a use for it, and at the time – about 1968 – not a lot of uses for it was being promoted by the math department in the school I went to, if any. But again, I want to thank you for bringing the writing down to a level that I can “wrap my mind around.” I gave up on the comments because most of them are shots at the same material because, I guess, they feel it isn’t at a higher level of discussion. I look forward to each new post, realizing you are now trying to bring understanding to what, as you put it, we instinctively knew was wrong, but couldn’t exactly express how it was.
10
Dear Dr. Evans, what seems to be missing is the emission pipe of dust and dirt. I suggest that dirt and dust and water droplets and ice crystals are like the laser etchings in “Forever Bubbling Champagne Flutes” http://toms.homeip.net/global_warming/w-champagne-flute-bubbles-c.jpg They are points of very high LR. What is the ratio of the laser etched surface to the smooth glass surface and what is the ratio of effect on CO2 coming out of solution. In the air What is the ratio of a solid’s IR emission to the trace green house gases and other neutrally charged molecules.
I think of any matter at a temp higher than zero K as supersaturated with heat, molecular energy that wants to emit IR just as CO2 in beer or champagne is a supersaturated solution and the CO2 will escape at points of discontinuity in the medium.
The IR emission spectrum of said solids would be in general generic black body radiation of which most emitted up would be gone in an instant. The part of the spectrum randomly emitted down would hit the earth except of that which is or can be recaptured by whatever green house gas it in it path. Recaptured within a few feet and then conducting to more dust and dirt particles …..
It is postulated that volcanic dust adding to said dust cools the Earth by blocking incoming solar radiation, but is it really that? The denser the dust the more speed of emit recapture emit increases where 50%?? going up is mostly gone at a faster rate.
It could be your water pipe deals with water as droplets and water is ice or maybe that just handled in clouds. But to me it seems much of the atmopheric IR comes from the dust and dirt as most CO2 gas escapes from the laser created defects in bubbling Champagne Flutes.
To the vacuum of space all molecules at temp greater than 3K are supersaturated with mechanical heat energy looking for a lower state.
01
David Evans
In #16.2 you state that one or more of the other pipes to space has to becom hotter.
That only thrue if you have a limited number of pipes.
In the atmosphere every singel molecule is a different path towards space. IR is not the only way to transport heat in the atmosphere.
Having infinite ways to space heat will always escape.
Its like a higway with 100 lanes and 50 cars.
No mather how many speed bumps you put up in one lane, none of them will be affected by the speed bumps.
The pipes in the figure are not seperat ways, the energy can jump between them, just like the cars can change lanes.
11
Mr Pettersen,
The “pipes” are the routes to space for OLR, one pipe for each type of OLR emitter (characterized by a range of wavelengths). There are only a few pipes. While energy can switch from molecule to molecule in the atmosphere, an individual photon escapes to space from the one molecule that emitted it, so it belongs to one pipe (only).
Nearly all photon escape Earth on the wavelengths at which CO2 emits (the CO2 pipe), water vapor emits (the water vapor pipe), cloud tops emit thru the atmospheric window (the cloud topics pipe), or the surface/water continuum emits through the atmospheric window (the surface pipe). Less OLR through one pipe means more OLR through others.
00
“Nearly all photon escape Earth on the wavelengths at which CO2 emits (the CO2 pipe), water vapor emits (the water vapor pipe), cloud tops emit thru the atmospheric window (the cloud topics pipe), or the surface/water continuum emits through the atmospheric window (the surface pipe). Less OLR through one pipe means more OLR through others.”
David,
Your use of the word ‘photon’, as some sort of directionally random bullet, with energy, completely trashes the concept of wave/particle duality, the photoelectric effect, or anything in quantum electrodynamics. Please stop.
Your pipes are fine, except that the top CO2 pipe does not change at all with over a magnitude change in atmospheric CO2 ppmv. The bottom surface pipe is non existent to space as 90% is only to 200 K cloud bottoms. The middle two are quite interesting as no one knows how they work.
All the best! -will-
29
Whew. At least the pipes are ok! Yes, the middle two are the interesting ones.
00
In my comment number 35 I tried to relate my belief the dust and dirt are a major source of shifting CO2 and H20 and other GHG absorbed energy and then re-emits that energy in what you call the surface pipe or the surface blackbody spectrum, some up, some down, the spectrum of this radiation that is the GHG pipes or spectrum does not get very far, but is then re-emmitted in the surface spectrum. The surface will see no spectrum of the GHG gas emission as they go nowhere. And in space very little energy will be seen in the GHG absorbing spectrum. Only emission in the spectrum of the atmospheric window can escape or come back to transfer energy to the Earth’s surface.
I believe The emissivity of a solid varies by orders of magnitude based upon the surface smoothness. The emissivity of any solid is orders of magnitude higher than that of a gas. In he super smoothness of air, dust and dirt are the point imperfection. The nooks and crannies and pointies of dust and dirt are the molecular cradle for electrons giving up and energy states as photons.
So I guess your model will likely be better than the current bogus models being used, but I am not convinced you are modeling what is really going on correctly. This is my current impression of what I understand you are saying.
00
Tom, I suppose dust and dirt might be like surface in the sky, emitting as you say much like a generic black body. Mountains ditto. Yes, I’m ignoring it as small (an assumption).
Yes, the albedo effect however can be quite large, especially for dust up in the stratosphere. We touch on this later in the series.
Climate is full of ever more detail. In trying to find simple or tractable models that capture the essence of the situation so they are mainly right, it is hard to know when to stop expanding into more detail. I sometimes liken it to the hierarchical file system on a computer, with an endless depth of sub-folders within sub-folders, each also containing files — how far do you have to expand the tree to get most of the bytes?
00
I can see or understand that clouds AKA ice crystals would lower solar energy by reflection. But dust and dirt I believe can be seeds for crystals for clouds at possibly warmer temps and lower humidity. But in clear sky it absorbs and warm and re-emits but half goes back up. But all of the time is it is and orders of magnitude more effective black body emitter that is aways sending half up and away. That is I believe a substantial cooling effect as well as cloud seeding that may produce more clouds with less humidity.
My beliefs are based upon all my real world observations of what cause state transitions. And where there is a discontinuity in the medium there are orders of magnitude more transitions. The randon spinning of electron in any solid are far more likely to randomly vibrate together and produce a photon over even slightly charged gas molecules. And several solid atoms will damp out the energy of individual gas molecules. Dust and will quench the heat energy of all gas molecules.
00
When volcanoes kick up a lot of dust the effects persist for a few years. The climate models seem to account for that reasonably well, presumably by fitting the measured effect on temperature to estimates of the amount of dust.
00
I believe we both agree the models do things wrongly in many area. It does appear that for volcanic dust and dirt the models execute a rational proportional effect. But do the models really subtract or add energy in the correct places or in the correct way? Who knows. But estimating the amount of cooling effect correctly is not the same as correctly adding it to albedo or as more radiation black bodies. Decreasing warming or increasing cooling are DC offsets having the same effect. As I thought or visualized the process I would suggest in clear air where the dust is not part of ice that the surface is ragged and it would both absorb energy and not reflect it and it would also have a very high emissivity. Once a particle has seeded an ice crystal or a water droplet it is in your H20 pipe I guess.
I have seen no info on measurement methods that can quantify the effects by spectral photon counts. The models play with numbers and come up with a number that represents an average temperature.
I wonder what if my proffer is correct that in clear air dust and dirt are a or the major player in generating black body IR.
Would it suggest a different or modified set of equations to quantify the effect?
I guess I don’t see how your pipes idea does it correctly, It may give better answer and results. It may demonstrate an even lower sensitivity of CO2 driven AGW.
I feel your pipes idea and the AGW models as like the various dew point curve fitting equations that add terms to make the output fit the curve and do not really identify the player of effect. And I may be all wet.
00
tom watson October 2, 2015 at 12:26 am
“I believe we both agree the models do things wrongly in many area. It does appear that for volcanic dust and dirt the models execute a rational proportional effect. But do the models really subtract or add energy in the correct places or in the correct way?”
This is why an even more simple “physical” model is always needed to check for unknowable gotchas! Even the guys at CalTech blew smoke on a spinning sphere to observe forces not included for the original fluid dynamics of such sphere spinning only in self contained (gravity) compressible fluid. When your computation can fully represent everything you can try with the physical model you can have some confidence that the computation is not just spitting back your fantasy. This only lasts until the new guy tries something else with the stupid physical model! 🙂
04
Thanks to Jo and Dr. Evans for what is a superb series.
One can’t help thinking that there are discontinuities in the Climate Model’s assumptions concerning the relationship between CO2 and it’s influence on temperature.
perhaps the climate models need to be re-designed to take into account scale, rather than “gluing” the small known physical behaviors and relationships together in the hope that the overall effect would yield reliable results.
Perhaps we ask too much of current day Climate scientists and should allow them more time (50 – 100 years) to develop their models, especially considering that it can take weeks to calculate a single “run” climate scenario.
00
Why is there no pipe for those light wavelengths for which the atmosphere is transparent? That would include colors from the surface of Earth that can be seen in detail from the moon (“Earth rising”, f’rinstance). That represents quite a lot of radiated energy and it counts in the radiated budget.
00
What you see from the moon is _reflected_ visible light, not absorbed. And that’s already accounted for by visible light albedo.
10
Albedo is missing from the right-hand side in his energy budget diagram at the top of this article. Energy lost to albedo is included in the left-hand side (energy in). His “T” doesn’t balance without including albedo.
00
NM – I put on my glasses and found the incoming is pre-cast as absorbed, not all energy. My bad.
10
David,
So far both very easy to grasp and very enlightening. You’ve cleared up some confusion in my mind about how the greenhouse effect works.
I’ll have to go over it more than once to be sure I remember it all. But at first reading I already have some useful info.
And for the first time that I can remember, someone has included water and other GHGs in an explanation that makes it all make sense.
20
The example of surface temperatures actually recorded on the moon, a body averaging about the same distance from the sun as the Earth, makes it clear that our atmosphere keeps us both warmer and cooler than we would otherwise be, assuming of course that we could be here at all without an atmosphere. But the details of it have always seemed too simple to my analytical mind because it wasn’t explained completely.
Now a question: Does convection enter into the greenhouse effect with either a plus or minus sign? That still isn’t clear unless I’ve missed something.
10
“Now a question: Does convection enter into the greenhouse effect with either a plus or minus sign? That still isn’t clear unless I’ve missed something.”
Free atmospheric motion (advection) carrying FRESH from cold poles, or stinky heat radially outward (convection), is encouraged by the gravitation extension to include atmosphere. This atmosphere is a thermodynamic isopotential. It takes no work to move any atmosphere to any other place within this atmosphere. Work is still required to accelerate/decelerate the mass of atmosphere so moved. This means that the slightest force (differential pressure) applied over the vast area of this atmosphere results in much heat flux directed outward quite independent of any surface radiative effects. Such outward convective flux lowers the lapse rate increase in temperature from the tropopause to the surface. This in turn greatly reduces the surface temperature compared to radiative effects alone. Gravity establishes the maximum lapse rate temperature increase from tropopause to surface. Convection can only reduce such surface temperature increase. Your question was about some greenhouse effect!! There is no such thing! Gravity alone establishes the maximum surface temperature increase from the tropopause.
Convection can only reduce that temperature increase. It is the atmosphere itself, not the surface, that is involved in EMR exitance to space. The opposite is the SCAM!
All the best! -will-
110
Roy,
Convection only enters the greenhouse picture by providing the lapse rate (or temperature gradient) in the troposphere that makes the atmosphere cooler as one goes higher. This temperature differential is necessary for there to be a greenhouse effect. The temperature gradient is affected by the OLR and radiation, but is primarily set by convection in the troposphere.
In the stratosphere, the gradient switches around (it gets warmer with more height) and there is no convection — the stratosphere is governed by radiation alone.
20
OK David,
Which one is it? Your convection providing a lapse rate within some greenhouse picture? Or gravity, independent of EMR determining the maximum atmospheric lapse rate, always lowered by any convection? What type of demonstration can falsify one of these? Must we remain mired in opposing fantasy? Thank you for your consideration!!
19
Thanks David. That leads to another question, perhaps a “dumb one” but does convection in the troposphere remain more or less constant or does it change over a wide range and if so, exactly why?
I ask because it’s obvious from your answer that if convection varies a lot, so does the lapse rate and with it, the greenhouse affect. Or am I all wet here?
00
Not sure what you mean Roy. Convection has always (for millennia) been roughly what it is today, AFAIK.
00
David,
I think you answered my question even without understanding it, convection is nearly constant.
It occurred to me that if convection varied widely for some reason, seasonally for instance, then the lapse rate would also vary widely. Maybe that’s not even worth worrying about?
00
Lapse rate is determined by gravity not convection. Lapse rate is lowered by the sensible heat produced by condensation. The WV does not have to be surface evaporation. Most WV is produced via insolation of existing airborne water condensate.
07
The decline in density with height produced by the force of gravity creates the lapse rate slope. The surface temperature rises above S-B via conduction from the surface to ALL atmospheric molecules, not just the radiative ones. That lapse rate slope marks the increasing amount of collisional activity as one descends deeper into the increasing density of atmospheric mass. If there is no radiative capability in an atmosphere then you still have a lapse rate slope and convective overturning but all radiation to space goes out from the surface. If there is radiative capability within the body of the atmosphere then one gets lapse rate distortions and convective adjustments to deal with those distortions.
The more molecules there are (denser) the more surface energy can be taken by conduction so that less photons are released and the higher surface temperature must rise.
The decline in density with height also causes convection because warmer lighter groups of molecules at an unevenly heated surface can then rise above colder heavier groups of molecules situated alongside them.
Convection both upwards and downwards requires energy and to ascertain the total of that energy one has to add together the energy engaged in moving mass vertically in both columns. You cannot just set uplift against descent and announce that the energy involved in all that movement is zero. Movement requires energy whether upward or downward.
If convective overturning is to be maintained that energy needed to drive it must remain present at the surface over and above the energy required to match radiation in from space with radiation out to space.
Depleting that energy store at the surface by radiating it to space would cause the atmosphere to cool and fall to the ground.
All that is a consequence of conduction to atmospheric mass from the surface and not radiation from the surface to GHGs.
Gravity causes the decline in density with height, uneven surface heating is inevitable and all else follows from that. GHGs not needed.
512
Outstanding piece Stephen.
A full outline of mass, heat and momentum transfer.
KK
36
Actually, no.
True, uneven heating would probably cause convection even if the atmosphere were totally non-radiative, so a lapse rate would result. (It’s also theoretically true that gravity would cause a temperature gradient even without the uneven-heating-caused convection, but that gradient would be so small that even theoretically it would probably take something on the order of a lifetime to detect.)
However, the convection-caused lapse rate would not make the earth’s surface as warm as it is now if the atmosphere were not radiative. Over time, average radiative power out must equal average radiative power in, and radiation would be emitted only by the surface if the atmosphere were not radiative, but a black body whose temperature is the surface temperature we now enjoy emits more radiation than we receive from the sun.
Convection-caused lapse rate notwithstanding, therefore, the surface would cool from its present level if the atmosphere somehow became completely non-radiative.
73
Joe Born said:
“a black body whose temperature is the surface temperature we now enjoy emits more radiation than we receive from the sun.”
It doesn’t get out to space, though, does it ?
The reason it doesn’t get out to space is conductive absorption and subsequent convection diverting it from photon emission proportionately to density (a consequence only of mass and gravity).
and he said:
“that (temperature) gradient would be so small that even theoretically it would probably take something on the order of a lifetime to detect.”
Joe is assuming that the density gradient is as small as the decline in power of the gravitational field with height but that is not the case. The weight of atmospheric mass bearing down on the surface leads to a much steeper density gradient than the gradual fall off of gravity with height as we see in the real world. I referred to the actual observable density gradient and not the gradient in power of the gravitational field.
Our surface would be the same temperature with no radiative gases at all. The only thing that changes when an atmosphere acquires radiative capability is the location from which the radiation is able to exit to space and David’s recognition of the existence of multiple pipes which can be adjusted as necessary relative to one another by convective adjustments is spot on.
412
Again Stephen,
That does sound very coherent and understandable given my limited knowledge of the atmosphere as a real world element.
Only 10 km up we get down to minus 38 deg C and that is a temperature at which no one can survive long term
Not much farther out into interplanetary space we get down to just 1.6 C deg above ABSOLUTE ZERO.
Whether Earths stored energy (from yesterday’s solar Input) moves around by conduction, convection, photon movement or EMR, one
thing is for sure: IT WILL MOVE TOWARDS DEEP SPACE BECAUSE IT MUST FOLLOW THE TEMPERATURE GRADIENT. (or energy gradient)
KK
00
Forgot to add;
All we are arguing about is how long it takes to escape and what mechanisms may be at work affecting that delay.
00
Thanks, KK.
10
Tanks for that Joe
I am well educated scientifically but as you might suspect I have no real workable knowledge as to what happens in the finer details of some areas.
All I can work on the the expression, the coherence and apparent logic. These I found in Stephen’s piece.
As to the truth; well you have called him out. I don’t know.
I am however, concerned by your use of the term “black body” when in my understanding it is a totally theoretical concept which may only be achieved in highly controlled laboratory conditions.
The use of BB calculation outside such lab is fraught with control problems.
Whatever.
David is only going through all of this to DISPROVE the existing climate models.
My main view of all this is that the CAGW by MM CO2 Meme does not hold water quantitatively since Man’s contribution is swamped by water and Natural Origin CO2.
The warmers have seen the faults in their position and tried every indecent scientific obfuscation possible to cover up
AND THEY GOT AWAY WITH IT.
look at the avalanche of money pouring into the UN.
🙂
KK
01
If earth did have a non-radiative atmosphere, meaning no water vapour present, but still had oceans dominating the equatorial surface regions, the surface temperature would be very little different to what it is now. Clouds play the dominant role in regulating earth’s temperature. Clouds have a huge influence on the amount of incoming radiation and likewise the amount of outgoing radiation. Their net balance is not much though, nothing like the 30C cooler figure without them quoted in the greenhouse gas fairy tale.
The hero of climate modellers, Syukuro Manabe, concedes this weakness regarding cloud formation in climate models during an interview this year:
http://www.carbonbrief.org/blog/2015/07/the-carbon-brief-interview-syukuro-manabe/
” One of the most challenging tasks of climate science is to determine the sensitivity of climate. It is often defined as the response of the global mean surface temperature to the doubling of atmospheric concentration of carbon dioxide, given sufficient time. Unfortunately, there is a large spread among the sensitivity of climate models. The spread is attributable in no small part to the parameterisation of cloud process that has become increasingly detailed, introducing many parameters that are difficult to determine either theoretically or observationally.”
Essentially climate models are no more than elaborate extrapolations of trends from the period taken for calibrating the parameters. They are still fundamentally primitive in terms of the physical and biological processes that influence Earth’s climate. Climate models are based on weather models. Weather forecasts one week out are suspect so what are the odds of the models producing an accurate forecast 100 years out.
31
No. Elementary physics tells me that even in the absence of convection atmospheric density would decay exponentially with altitude. But statistical mechanics tells me that the gravity-caused mean-molecular-translational-kinetic-energy (temperature) decay with altitude would instead be linear and minuscule, on the order of picokelvins per parsec.
It’s hard to follow Mr. Wilde’s logic, but the core of the disagreement may lie in the following. I think temperature has less to do with how often molecular collisions occur and more with how violent they are. I may be misinterpreting him, but my take is that he thinks just the opposite.
12
Joe,
I think the core of the disagreement may be that you are referring to the process of simply lifting mass vertically against gravity. That is indeed linear and miniscule and contributes little to the decline in density with height.
However, density is a matter of how far apart the molecules are and as one goes up in the atmosphere the distance between molecules increases exponentially with altitude.
It is that increase in distance between molecules (expansion) that constitutes the density gradient and that gradient falls off far more sharply (exponentially) than the power of gravity during the process of simple uplift (linear).
It is the former that determines the temperature gradient rather than the latter.
39
Perhaps you could quote where I said anything about lifting; I don’t remember doing so.
I believe I said that: density decays exponentially with altitude. This has almost nothing to do with lifting. It’s the result of random particle motion in a gravitational field, which because of the low ratio of the height of the atmosphere to the radius of the earth can be taken as uniform for our purposes.
I assume that by “power of gravity” you mean gravitational force. I can’t fathom exactly what you’re trying to get at with “simple uplift.” What’s linear is the decay of equilibrium temperature with altitude (if gravitational force is taken as uniform).
That’s where the disagreement is. You say density determines the temperature gradient. I say it’s the (for our purposes, uniform) gravitational force that, together with total kinetic energy and degrees of molecular-motion freedom, determines both density and temperature, making the former decay exponentially and the latter decay linearly.
That is the point on which we will have to agree to disagree. I don’t expect to convince you; I just wanted to help any readers new to the issue avoid being misled.
11
Joe Born said:
“You say density determines the temperature gradient. I say it’s the (for our purposes, uniform) gravitational force that, together with total kinetic energy and degrees of molecular-motion freedom, determines both density and temperature, making the former decay exponentially and the latter decay linearly.”
If that is the point of disagreement then I don’t see it as a significant disagreement.
The density gradient and the temperature gradient that follows it are both a product of gravity working on the kinetic energy and degrees of molecular freedom of mass so we are saying the same thing in different ways. However, there is no doubt that if density at the surface were less then the temperature at the surface would be less so I still prefer my choice of density as determining temperature rather than gravity determining it directly. The thermal effect is gravity working via density and not gravity working independently of density.
As regards the rest we may have been at cross purposes.
On that basis KK’s comment would appear to have been correct.
29
The linear decay of equilibrium temperature that I’m talking about is too small to be detected; it’s orders of magnitude less than, for example, the current atmospheric lapse rate. Also, for the reasons I gave above, I believe that the surface would become markedly colder if the atmosphere somehow suddenly lost its ability to radiate.
If you agree with that, then maybe we are in agreement–but I don’t think you do.
So, again, we’ll just have to agree to disagree.
00
I erred above. It should be:
I say it’s the (for our purposes, uniform) gravitational force that, together with total potential plus kinetic energy and degrees of molecular-motion freedom, determines both density and temperature, making the former decay exponentially and the latter decay linearly.
00
I’m trying to understand.
Please could you explain why the expansion with height is not enough to produce the observed lapse rate.
Thanks.
00
No, I can’t begin to do justice to the explanation of why the expansion doesn’t end up in the lapse rate you suppose.
If you look for a paper by Coombes & Laue, though, you’ll see that they explain the lapse rate of zero better than I could, by assuming a gas in which no collisions occur and the molecule-speed distribution at every altitude is the Gaussian corresponding to the same temperature, but with different magnitude in accordance with the exponential density decay. They show that the kinetic energy that molecules gain by falling and lose by rising is just right to maintain that same-temperature Gaussian at every altitude.
Of course, I said the lapse rate is not quite the zero value they show, and that’s because there comes an altitude at which you can no longer have a Gaussian distribution if the number of molecules is finite. Some intimidating calculus in a paper by Roman, White, and Velasco, and a follow-up paper, Velasco et al., by the same authors, show the effect of the finite particle count and arrive at a linear molecular-kinetic-energy decay with altitude.
Sorry I can’t be more help.
00
Well, Joe, at least you are honest.
You seem to be relying on third party assertions without understanding them so perhaps your reaction to KK’s post was unwise?
It seems to me that anyone who is resorting to complex contortions to try and deal with my simple question (such as Coombes and Laue) is merely trying to justify the unlikely assertion that without GHGs an atmosphere must be isothermal (zero lapse rate).
I think you and maybe others have been misled if you subscribe to the concept of an isothermal atmosphere despite decreasing density with height.
In my view it just cannot happen.
17
Hi Joe
All of your posts seem to be finely focused on the micro effects of radiation and such.
That is a sure way to lose contact with what actually happens in situ.
Whatever mechanisms are at work certain things must happen so?
My earlier comment sums up my way of looking at it.
Always approach from a distance so that you fully understand each step.
I have admitted that some aspects of micro detail are above my skill level but I still can see the important overview.
KK
http://joannenova.com.au/2015/09/new-science-6-how-the-greenhouse-effect-works-and-four-pipes-to-space/#comment-1749358
00
Actually, no.
I do understand those “assertions” with the exception of one single calculus identity in the exposition in Roman, White, and Velasco, which I haven’t derived myself. I could write it up, but it’s too technical for a blogger to accept as a post, and my sell-by date is too imminent for me to make that futile effort.
Also, with all due respect, I don’t think you’d get it.
As to Coombes and Laue, I certainly did understand every bit of it; I just don’t have the time to explain it to you. I’m open to helping those who are can be helped, but, no offense intended, I’ve seen no evidence that you are among them.
I’m pretty sure I’m right about Roman et al., although, again, there’s that one identity that I haven’t taken the time to verify. If you can find any step in those papers that is incorrect—I have, but it’s not relevant to this issue—we may have something to discuss. Otherwise, I’m afraid that my response to KinkyKeith’s comment was entirely justified.
I’ve been patient here; I think you actually believe what you are saying. But this question requires more analytical ability than your comments have so far betrayed. So I’m going to sign off. Again, I’m fairly sure you’re sincere. But there comes a point when each of us has to recognize his limitations. I’m sure you think I haven’t recognized mine.
I reciprocate.
02
A perfect example of “Argument From Assholeity*”
Believe me, no intended offense would be sufficient.
*Yes you may quote that so long as credit is given.
20
“No. Elementary physics tells me that even in the absence of convection atmospheric density would decay exponentially with altitude. But statistical mechanics tells me that the gravity-caused mean-molecular-translational-kinetic-energy (temperature) decay with altitude would instead be linear and minuscule, on the order of picokelvins per parsec.”
Both atmospheric pressure and density decay exponentially with altitude. The ratio of the two P/rho is a constant called gamma by chemists and isentropic exponent by engineers for Earth’s atmosphere the value is 1.4 this ratio also sets the gravitationally induced lapse rate to -9.8 Celsius per kilometer. Your misuse of statistical mechanics leaves out the PV energy gradient of this atmosphere. The molecular noise power of gas is kT/t for each molecule.
Molecular power is not velocity but rate of change in momentum.
26
Correction to the above which should read:
I think the core of the disagreement may be that you are referring to the process of simply lifting mass vertically against gravity. That is indeed linear and miniscule and contributes little to the decline in temperature with height.
However, density is a matter of how far apart the molecules are and as one goes up in the atmosphere the distance between molecules increases exponentially with altitude.
It is that increase in distance between molecules (expansion) that constitutes the density (and temperature) gradient and that gradient falls off far more sharply (exponentially) than the power of gravity during the process of simple uplift (linear).
28
Hi Stephen,
There is no excuse for Joe Born’s rudeness but I am convinced that he is right when he maintains that in the absence of any GHGs at all, the atmosphere would be isothermal. The reason is simple and obvious: All of the radiation to space would be from the earth’s surface. The atmosphere would be warmed up to that surface temperature, just like air would be if it were contained inside a vertical cylinder (with perfectly insulated sides and top) that is placed upon a constant temperature surface.
Your mistake, which Joe pointed out, is to assume that in a very tall cylinder, where the air density at the top is markedly lower than the air density at the bottom, the larger average distance between the molecules at the top than at the bottom would result in the top layer being cooler than the bottom. That is NOT correct because the definition of temperature in a particular region of gas is proportional to the average speed of the molecules in that region. And the average speed of molecules would be the same at all levels, according to thermodynamic theory.
This might seem un-intuitive but your error is due to your knowledge that in the atmosphere the temperature falls off with height, from which you have drawn the wrong conclusion. In the case of the atmosphere, the real reason for the drop-off in temperature with height is NOTHING to do with the reduction in air density with height and EVERYTHING to do with the fact that heat is flowing from a heated end (the earth’s surface) to a cooled end (due to radiation to space from GHGs towards the top). In other words it is not like a cylinder with perfectly insulated side and top.
Cheers, David C°
05
How does the radiative theory deal with adiabatic cooling as air moves upward and expands ?
There is no radiative component to that cooling process at all.
01
I am not sure which case you are referring to, the actual atmosphere which is radiating energy to space or my hypothetical atmosphere without GHGs which is not radiating energy to space.
But as far as I can see in either case, when some air is moving upwards adiabatically, other air will be moving downwards adiabatically and the two processes will cancel out in the aggregate.
01
The two processes do not cancel out in the aggregate because energy is required to move air upwards AND energy is required to nove air downwards so actually the two processes must be summed to find the amount of energy required at the surface to maintain the complete overturning cycle.
That is applicable whether there are radiative gases or not.
Radiative capability does allow leakage from the adiabatic processes to space but the remaining adiabatic portion (by far the majority) remains within the system.
02
Steve wrote: “How does the radiative theory deal with adiabatic cooling as air moves upward and expands? There is no radiative component to that cooling process at all.”
Adiabatic, of course, means no transfer of radiation or other form of heat into or out of the moving air. In the real world, radiation transfer occurs and transfers heat from hot to cold. So we figure out the adiabatic cooling of rising air over a 15 minute period and the radiative transfer of energy into and out of the rising air for the same 15 minute period assuming no temperature change. Then we calculate the combined effect of both on the temperature 15 minutes later. Such calculations produce accurate weather forecasts for about a week (700 15-minute periods) and we know that chaotic behavior limits complete predictability for longer periods
30
Franktoo,
I’m not aware of any such calculation actually used in practice in weather forecasting. Do you have any evidence ?
Aeronautics and rocketry using the standard atmosphere ignores radiation completely.
Any net radiative exchange is far too small to be relevant simply because the cooling of rising air during adiabatic uplift causes radiation from the cooler air to reduce.
04
Steve: If weather forecast programs did not perform accurate radiative transfer calculations, how do they accurately predict the rise in temperature when the sun is shining and the fall at night, with both modulated by cloud cover?
20
That isn’t what we are talking about.
You mentioned calculations that separate the kinetic energy moving into convective available potential energy (known as CAPE) during adiabatic uplift from the kinetic energy able to be radiated and then you suggest that somehow the net balance of radiative flows can usefully be calculated separately from the adiabatic effect at any given height.
In reality, the Standard Atmosphere finds it adequate for aeronautics and rocketry (both very precise disciplines)to simply rely on the adiabatically driven dry and moist lapse rates with no need to consider net radiative balances from one height to the next.
04
David Cosserat October 3, 2015 at 9:20 pm
“Your mistake is to assume that in a very tall cylinder, where the air density at the top is markedly lower than the air density at the bottom, the larger average distance between the molecules at the top than at the bottom would result in the top layer being cooler than the bottom. That is NOT correct because the definition of temperature in a particular region of gas is proportional to the average speed of the molecules in that region. And the average speed of molecules would be the same at all levels, according to thermodynamic theory.”
But not according to the gas laws, Maxwell/Boltzmann, or Loschmidt. In every gas, or gas mixture, each molecule has a noise power equal to kT/t. That is the definition of molecular temperature. Part of that us due to the random velocity of some molecule (constant) until it intercepts some other mass. At that point that molecules changes its momentum from 0 to a maximum of 2mv (direction reversal). This is the actual kinetic of Kinetic Theory. Energy is proportional to the first dirivitve of momentum with respect to time, thus kT, after change a new (perhaps) direction and temperature! This is a miserable concept for a gas mixture of He and SF6. In Earth’s atmosphere much more simple as N2 and O2 predominate with similar molar mass. The real kicker for any atmosphere is the introduction of gravity, which expresses in an atmosphere, as a logarithmic change in density with altitude. This forces, at the same velocity, linear increase in the first derivative of momentum with respect to time at decreasing altitudes, and a corrosponding linear increase in temperature, with absolutly no work! Location change in any direction of atmosphere within the atmosphere is always isentropic. An isentropic lapse rate. Atmospheres are isopotential; but seldom isobaric or isothermal.
All the best! -will-
24
Sorry about asking difficult questions but is there an important layer missing?
The layer known as the Exosphere. This layer presents a problem for those who say backradiation cannot exist because it is hotter than the surface. It is part of the reason space vehicles need heat re-entry shields. Numbers around the net have it at 1000 (at the exobase) to 1800 degrees K but it oscillates. Before this layer can be dismissed it needs to have at least a known approximate effect.
The third most abundant element is hydrogen. Hydrogen if liberated by any process will travel up as fast as 3 to 5 kilometers per second. So heat can be carried all the way from the surface to the Exobase in just a few minutes. This velocity is below escape velocity but only just.
http://faculty.washington.edu/dcatling/Catling2009_SciAm.pdf
The breakdown of methane is one source of free hydrogen.
00
Siliggy, the exosphere radiates very little OLR, AFAIK, so I neglect it for the modeling here.
00
David: If you double CO2, a photon emitted upward from the current “emission” level is less likely to reach space, BUT twice as many CO2 molecules are emitting photons. Which factor dominate? I found it tough to come up with a convincing answer to this dilemma. Later I learned about the Schwarzschild equation (which I discussed above). It provides a simple explanation.
00
Frank: When CO2 doubles the old emission layer is of little interest, but the new emission layer emits slightly less than the old one because crucial parts of it are in the troposphere and are thus higher and cooler. (Yes, some other parts of the EL, at different wavelengths, are in the stratosphere and are slightly warmer, so there are countervailing tendencies to be weighed up. But, according to AR5, doubling CO2 reduces OLR from CO2 molecules by about 3.7 W/m2, so the new emission layer emits less.) Sure there are more CO2 molecules, but for OLR the only thing that matters is the temperature of the emissions layer — what happens underneath the emission layer is of little interest to OLR.
00
David: Thanks for the reply. In practice, the “emission layer” for any GHG is spread over many kilometers of altitude and is ill-defined. Optical depth at a particular wavelength defines an altitude where the average photon of a particular wavelength is emitted, but GHGs emit at many wavelengths and therefore many altitudes. Any discussion of emission layers involves a lot of hand waving.
You then cite a very tangible result: a doubling of CO2 will reduce OLR by 3.7 W/m2. That result is calculated by numerically integrating the Schwarzschild equation from the surface to space. (The process is a little more complicated: They use paths that start in the tropics, temperate zone and poles in summer and winter, starting from the surface and cloud tops and ending at the tropopause.) Instead of waving our hands about a doubling of photons emitted by 2XCO2 and a reduction in the distance they travel or their chances of reaching space, we can consult the equation that quantifies what really happens at any altitude. A 3.7 W/m2 change is a 1.5% reduction – the net result of a 100% increase in emission and a 101.5% decrease in the effectiveness of absorption. Hand waving arguments are unreliable when two factors nearly compensate for each other. The chances of getting the right answer are far better if one thinks in terms of thie Schwarzschild equation equation rather than more nebulous concepts. The need for numerical integration is sometimes a problem, but MODTRAN does the work online.
10
Frank, the mean path length to absorption around 15 microns is only a few tens of meters, even well up into the atmospere, so the emission layer might be relatively compact at some wavelengths? In any case, the notion of an average temperature of emission layer is a useful one, a step up in realism from a single “characteristic emission layer” used in some modeling.
The emission layer itself can be well-defined, as one optical depth as seen from space, or (slightly different) the average height of emission of OLR.
The 3.7 W/m2 is also viewable as a simple move up by the emission layer (at each wavelength), which is not the difference of two similar quantities. Sure the number of emitting molecules doubles, but we don’t care about those under the top layer for computing OLR, and the number of molecules near the top layer hardly changes.
00
David: Everything you wrote above is as accurate as words can describe phenomena that are best described using mathematical equations. An alternative, but not necessarily better, description can be derived from this version of the Schwarzschild equation:
dI/ds = n*o*{B(lambda,T) – I_0}
When radiation passes far enough through a homogeneous gas so that emission and absorption have come into equilibrium, the intensity of the radiation at any wavelength is given by Planck’s Law. (This phenomena produces the blackbody radiation emitted by denser materials and scattering at interfaces produces emissivity/absorptivity.) The denser or more strongly absorbing the gas, the quicker equilibrium (blackbody intensity) is reached. In all cases, the change is towards blackbody intensity/equilibrium, but equilibrium may not be reached when the temperature of the gas changes with distance. As radiation travels higher in the atmosphere and the local temperature changes, eventually the density of GHG (n) becomes low enough that the radiation no longer has an intensity reflecting equilibrium at the local temperature – ie blackbody intensity. Radiation in the strongest lines for CO2 equilibrates with local temperature at the tropopause and into the stratosphere. Less strongly absorbed lines and minor GHG’s maintain equilibrium and blackbody intensity to lower altitudes (higher temperatures). In the atmospheric window, there is negligible equilibration with the temperature of any part of the atmosphere and upward radiation retains the intensity of surface emission.
I find it hard to associate an “emission level” with this description of surface emission being changed as it passes upward through the atmosphere, but I do understand the concept you are trying to convey. For a given wavelength, I usually discuss the altitude at which the average photon escaping to space was emitted.
00
Frank: Perhaps we are converging on this point. The emission layer is about one optical depth looking from space, so it is the height where the equilibrium is starting to get pretty unbalanced as we ascend — the downward radiation has become substantially less than the upward radiation, because there are no longer enough emitters above.
The leap from that observation to an “emission layer” is just to note that the emission layer is the average (or maybe median is better) height of emission of OLR (at the given wavelength). Furthermore, the standard deviation of the emission heights is quite small for many wavelengths, especially where absroptivity is strong.
So, effectively, for many purposes, one can think of the emissions as coming from the emission layer and for simplicity neglect the distribution. One can also speak of a “temperature” of the emission slayer, as the weighted temperature of the heights of emissions, which hopefully would be reasonably close to the temperature of the height designated as “the emission layer”.
The “one-optical depth” is rough: the precise statistical handling of it depends on your purpose, and for the purpose in this blog series it does not matter. We are just going to be estimating how changes to various climate variables affect OLR from the various emission layers (in the various pipes).
00
David: Words are a poor substitute for the mathematical equations quantifying how radiation changes as it passes through the atmosphere. In part 8, you discussed brightness temperature (or a related concept). As long as one recognizes that this “temperature” is the result of averaging, this concept may be the best way to convey the information about where photons escaping to space are emitted.
If I go to Archer’s online MODTRAN and use no GHGs and the US Standard atmosphere with no clouds, OLR looking down from 70 km is 358.3 W/m2. If I add 400 ppm CO2, it drops to 323.7 (34.6 less). Looking down from 5 km, 341.0 (17.3 less). Looking down from 10 km, 326.6 (31.7 less). So about half of the reduction in OLR by CO2 develops below 5 km and 90% below 10 km. Above 20 km, CO2 in the warmer stratosphere adds about 1 W/m2 to OLR. This shows how the intensity of OLR in the CO2 channel evolves as it rises through the atmosphere. The picture this information conveys is quite different from the picture of an emission layer describing where the photons escaping to space are emitted. Neither picture is “wrong”.
Water vapor decreases with altitude, so its major effects develop rapidly lower in the atmosphere.
The “right answer” comes from quantitative integration of the Schwarzschild equation, not our imperfect attempts to describe those results with words – especially with the misinformation pervading comments on skeptical blogs. I suspect we may be in substantial agreement about the math*. The tricky part comes from expressing the results as concepts (forcing, feedback, pipes …) and reaching the right conclusions without doing the math.
*Of course, the equations need to be accurate and complete. If there is a mechanism that takes energy from CO2 directly to water vapor without going through warmer water vapor (your rerouting feedback), that mechanism needs to be in the equations.
The Schwarzschild eqn tells us how radiation changes, but can’t predict temperature change because it doesn’t predict how convection changes.
20
Frantoo,
David’s rerouting feedback does involve going through warmer water vapour via the distorted lapse rate.
00
Frank: Yes, there is a lot of averaging in this emissions layer concept. It is only one step up from the characteristic emissions layer (CEL) concept, which characterizes all OLR as coming from one layer at 255 K at a height of about 5 or 6 km — too crude to be useful.
In this series I am saying that the minimum complexity required to estimate the effect of increasing CO2 is a four pipe model — four main emission layers, which variously cool or warm as CO2 increases. I believe I have found a way to estimate ECS that is fairly light on math but is generally faithful to the realities of radiation.
00
David,
With respect, CO2 can absorb radiation of any frequency. If it absorbs energy at a faster rate than it emits it, its temperature will rise. Conversely, it will fall. Remove all EMR, and the CO2 will eventually reach absolute zero, emitting progressively less energy at ever decreasing wavelengths.
Nothing except a vacuum, (the absence of any matter at all), is totally transparent to EMR.
CO2, O2, iron, or pixie dust, at identical temperatures radiate wavelengths in line with temperature, rather than composition, at least at temperatures below excitation.
Please correct me if I am wrong, but my understanding is that objects at absolute zero emit no radiation at all. Raise the temperature to 288K, and in the dark, from a distance, you have no idea what is around you, if you are measuring temperature only.
So, you can heat CO2 with your body as you breathe it out. It has exactly the same temperature as the oxygen, nitrogen, H2O, and other gases you exhale. No CO2 created warming to be found. As with the atmosphere in general. No global or other warming due to any greenhouse effect.
After 4.5 billion years of heat trapping by GHG’s, the Earth has managed to cool quite a lot. Just look down at what you are standing on – no molten rock to be seen!
End of story. Have fun!
18
Mike, CO2 can absorb only certain amounts of energy, or equivalently, only photons at certain wavelengths. This is a quantum effect; such observations led to quantum mechanics, a major advance in physics about a century ago.
Emphatically no; a CO2 molecule cannot absorb radiation of any frequency.
40
David,
Somenpeople who might disagree with you, (apart from myself), are Professor John Tyndall, Richard Feynman, and others.
However, we would all be stunned into submission by your reference to an actual experiment backing up your unsupported assertions.
Of course, you have none, any more than support for Uri Geller’s assertions that spoons could be bent by the power of the mind.
Possibly, after reading John Tyndall’s “Heat as a form of motion”, (try the 1905 edition), or Feynman’s “QED – the strange theory of light and matter.”, you might care to reconsider your position. Or maybe not.
Passionate belief may not represent reality. Enjoy life. I think you’re wrong, but what do I know?
Regards.
08
The Earth’s atmosphere is the necessary ‘actual experiment’.
Observations fit David’s description very nicely. 🙂
10
I admit that I did not red all the comments but I have not found explanation for the phenomenon of IR absorption by GH gases. Here is quote from JoNova’s text:
“If the photon is absorbed, the photon ceases to exist and its energy is transferred to the molecule or atom, which is now in a higher energy state (i.e. it vibrates more).Conversely, a molecule or atom can emit a photon, creating it from scratch and firing it off in a random direction with energy equal to the difference between the stored energy of the atom or molecule in its initial state and its lower energy level after the emission. Notice that the energy and thus the wavelength of an emitted photon are exactly the same as that of an absorbed photon that changed the energy level of the atom or molecule previously — it’s a reversible process, as the energy stored in the atom or molecule moves up or down with the absorption or emission of a photon. Thus a molecule or atom absorbs and emits photons on exactly the same wavelengths.”
According to this description, a photon is absorbed by a GH molecule, it ceases to exist, because it makes a GH molecule to vibrate, and then a GH molecule will emit a photon with exactly the same wavelength (=same energy). This is a description of perpetual motion machine: mechanical movements have been created without losing energy – not possible. On the other hand we know by real measurements that H2O and CO2 can absorb practically totally the IR radiation of certain wavelengths emitted by the Earth’s surface. For me there are two options: 1) GH molecules absorb IR radiation creating only vibrations of molecules (higher temperature) without emitting photons or 2) a GH molecule emit a photon with longer wavelength (=lower energy) because a part of the energy is lost for mechanical vibrations.
20
Each gas molecule always has noise power of kT/t, energy depends on Cv. The power is split but not evenly between states. The IR power is in the resonant bending and elongation moments of the particular molecule. This gives rise to the specific absorption bands of all trimers and above. All of the rest is contained in the up to 6 degrees of freedom for molecular motion. This is called molecular Kinetic energy, made up of; linear and angular momentum mv, and the energy of the rate of change of momentum d(mv)/dt. With gas most of the change of momentum is from a change in direction upon collision, not necessary a change in velocity. This troposphere has a linear lapse rate because density increases logarithmically with decreasing altitude, and the increasing atmospheric collision rate, increasing d(mv)/dt.
For a clear POV of the EMR effect, that 15 micron Photon (conceptual only) upon absorption/emission by a CO2 molecule does (we think) increase/decrease that molecules temperature by 1/10^20 kelvins. Consider, so far the lowest temperature produced is a whopin 0.45/10^6 K. __ 0.45/10^6 K + 1/10^20 kelvins is what?
Your claim of: “This is a description of perpetual motion machine: mechanical movements have been created without losing energy – not possible.”, is not correct (as far as we know), the energy transfer is simply immeasurably subsumed into the huge sensible heat of that molecule.
Any atmospheric CO2 molecule is continually dispatching EMR power to space as best it can from its own temperature. This temperature is maintained not from surface EMR flux but instead by the gravitationally induced lapse rate, and collisions with its many N2, O2 neighbors. Adding CO2 changes nothing.
All the best! -will-
03
Antero wrote: “According to [JoNova’s above] description, a photon is absorbed by a GH molecule, it ceases to exist, because it makes a GH molecule to vibrate, and then a GH molecule will emit a photon with exactly the same wavelength (=same energy). This is a description of perpetual motion machine: mechanical movements have been created without losing energy – not possible.”
No, this is a statement of conservation of energy. If you think of a molecule as atoms linked by springs, the energy contained in the photon has been converted to the energy in a simple harmonic oscillator. (The energy stored in a simple harmonic oscillator is potential energy when the spring is stretched or compressed to the maximum extent and kinetic energy when the spring is relaxed.) However, in quantum mechanics, the oscillators are allow to possess only discrete amounts of energy.
In the troposphere and the stratosphere, GHG molecules are in local thermodynamic equilibrium – which means that collisions relax excited vibrational and rotational states of a GHG much faster than the average time it takes for the excited state to emit a photon. (Higher in the atmosphere this isn’t true.) “Re-emission” of absorbed photons is negligible. Consequently, essentially all photons are emitted by GHG molecules that have been excited by a collision. The Boltzmann distribution determines what fractions of molecules are found deltaE above ground state and the Planck function B(lambda,T) takes into this into account. The emission rate from GHGs is proportional to their density, absorption cross-section and B(lambda,T) and independent of absorption. Sadly, many skeptics don’t understand collisional excitation and relaxation of GHGs, and think the only thing that happens in the atmosphere is “re-emission” of absorbed photons.
After absorption of a photon, collisional relaxation of the resulting excited is sometimes described as “thermalization”. The energy of the absorbed photon becomes part of the mean kinetic energy of a large group of nearby molecules – i.e. their temperature.
You can see a simulation of the excitation and relaxation of vibrational and rotational states of gases caused collisions at the following website. Use the PRESETs pull down menu to select Figure 7 diatomic and press START. The simulation does not include photons.
http://physics.weber.edu/schroeder/md/InteractiveMD.html
50
You are still pushing the falsified Schuster Schwarzschild two stream approximation for Earth’s atmosphere. Please go back to Prevost’s principle of the caloric and radiant heat fluid motion in opposing directions.
EMR power both in and out works nothing like that! If it were, modern Doppler Radar Warning of, must be powered by oxen proceeding in a circle, encouraged by whips! Please go somewhere and buy a clue!
03
Will: Why would anyone go back to Prevost’s view of heat as a self-repelling fluid? (See Caloric theory.) Prevost’s world literally was powered by oxen! Prevost worked 50 years before Maxwell’s discoveries about electromagnetic radiation and Clausius’s formulation of the 2LoT and entropy. He worked almost a century before the atomic and molecular nature of matter became clear, before we knew about the Boltzmann distribution and the kinetic theory of gases, and before we understood temperature as being proportional to the mean kinetic energy of molecules. He worked more than a century before the full development of QM (especially the particulate nature of EMR) and – most importantly – before statistical mechanics explained how large numbers of molecules following the laws of quantum mechanics could result in macroscopic behavior that agreed with the empirical laws of thermodynamics.
Yes, I am still pushing the Schwarzschild equation: The one discussed in Chapter 8 (atmospheric emission) and Chapter 11 (scattering) of Grant Petty’s 2006 textbook “A First Course in Atmospheric Physics” and Section 4.5.3 of Wallace & Hobbs 2006 textbook “Atmospheric Science” and (for those without access to either of these texts) http://barrettbellamyclimate.com/page47.htm The textbooks are written for meteorologists, not climate scientist, and have nothing to do with global warming. Variants of the equation are associated with Milne and Schuster. The latter considers the radiation flux through an infinite plane parallel to the surface of the earth. Since emission by molecules in the plane occurs in all directions, the flux is broken up into three components: upward and downward components perpendicular to the plane and a component parallel to the plane (which doesn’t effect the earth’s energy balance). To my knowledge, this equation has never been invalidated, but it has been shown to give the SAME RESULTS as a one-way flux by the folks at Principia Scientific. The Schwarzschild equation makes the same predictions about heat transfer (net radiative flux) as the alternatives proposing one-way flux.
http://www.tech-know-group.com/papers/Planckabsorption.pdf
Your previous comment dismissed the Schwarzschild equation without citing any authority, because you claimed it wasn’t suitable for “luminous” atmospheres. After I pointed out that the earth’s atmosphere was luminous in the infrared, you claim the equation had been falsified, again failing to cite any authority. Next time, please be honest with other readers and admit that I am citing generally-accepted physics that has been challenged by a small community of skeptics who can’t get their work published in a traditional journal. That doesn’t mean their work must be wrong – it is simply useful information that enables others to understand the basis for our disagreement.
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Frank,
Agreed with all of that. But see my reconciliation with the “traditionalist” position at 48.1.1 and then at 49.1 and then at 49.3
Will prattles on regardless, of course. I seek not to convince him (impossible it seems) but to urge other readers not to venture into the muddy waters of “back radiation” by rising to the usual traditionalist bait. 🙂
00
Dear Will,
Thank you for your comments. Still I am not happy with the explanations what happens at the molecular level in the IR absorption by GH gas molecules. I have used the application Spectral Calculator in analysing the impacts of GH gases in the atmosphere. The results are logical and in line with the real measurements carried out at the surface and at the top of the atmosphere (TOA). For example the emitted radiation by the Earth’s surface is 395 W/m2 according to the surface temperature 15.9 C. The outgoing LW radiation at TOA is only 239…240 W/m2 corresponding exactly to the incoming SW radiation from the Sun. What is the reason for the difference 395 – 240 = 155 W/m2? Has it disappeared without leaving any impacts? No, there are impacts and that is the absorption of IR radiation in the atmosphere causing the greenhouse phenomenon. Another evidence is that in the spectrum measured at TOA, certain wavelengths cannot be found. For example the frequencies above 13 micrometers are almost totally missing and the reason is the absorption caused by H2O and CO2 exactly as calculated by the means of spectral analysis.
The only problem for me is that I find several descriptions at the molecular level, which cannot be correct. If a photon hits a GH gas molecular, creates “mechanical” vibrations and a GH gas molecule emits a photon with the same wavelength, it must be wrong. If this would be true, IR radiation would travel through the atmosphere without losing any wavelengths and without any absorption. The absorption is a fact, and therefore something is wrong in the description above. If a GH gas molecule emits a photon with a longer wavelength (with a lower energy), it would make sense. I have found somewhere also this kind of description.
There is something wrong with IPCC’s models, and we need not to abandon the results of spectral analysis. So I decided to analyse by myself, what is the real RF value of CO2 concentration. I used two spectral analysis tools and two methods; one being the exact definition of climate sensitivity (CS). The main result is that RF value of CO2 can be calculated by the equation RF = 3.12 * ln(CO2 conc./280). RF value of 560 ppm is 2.16 W/m2 and not 3.7 W/m2 as commonly used. Using this RF value, the CS = 0.58 K. Also the spectral analyses give CS values between 0.58 and 0.56 K. The different calculation methods give the same result, when the RF value of CO2 is 2.16 W/m2. There is a simple explanation between 3.7 and 2.16 and that is that the value of 3.7 W/m2 has been calculated in the constant relative humidity (RH) conditions, where water feedback almost doubles the real RF value of CO2.
So IPCC shows that the transient CS is 1.85 C using the CS parameter 0.5 (0.5 * 3.7 = 1.85 C). There is the same error used twice: 1) RF value of CO2 and 2)transformation of RF into temperature. Positive water feedback has been applied twice because of constant RH conditions, which are not true. The CS is 0.6 C in the constant absolute humidity conditions of the atmosphere. There are evidences that there could be a negative water feedback bringing 0.6 C to 0 degrees.
More information in the paper “The potency of Carbon Dioxide (CO2) as Greenhouse Gas”.
Climate change issues generally on my web pages “http://www.climatexam.com/”.
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Antero Ollila October 7, 2015 at 4:44 pm ·
” Dear Will, Thank you for your comments. Still I am not happy with the explanations what happens at the molecular level in the IR absorption by GH gas molecules.”
Antero,
My explanation of what happens at any molecular level is but my best guess from weak measurement. The claimed photon flux or EMR flux influence from the surface remains immeasurable, at any location in this atmosphere. This atmosphere dispatches via molecular EMR at every possible wavelength to space all that it can to reduce accumulated entropy ((sensible heat)/temperature). Fortunately this is limited by the 377 ohm impedance of space.
This is my best guess. I do not sell anything, or have spiffy Power Point presentation. I only guess and deliberately poke holes in ALL that of those that claim to know! I do not know, I observe, peering out from bushes and wondering.
All the best! -will-
12
Dear Frank,
I posted my earlier comments without noticing your comments. Yes, you can classify me as skeptic concerning the IPCC’s climate models. At the same time I do trust on the spectral analysis. As I said, I am not a molecular physician and that is why I try to find an explanation what happens at the molecular level. Your comments are almost satisfactory for me. Here is a quote from your comments: “Re-emission” of absorbed photons is negligible”. For me it means that because of the absorption, a GHG molecule does not emit always a photon as it was described on this web page originally. I also make a conclusion that a GHG molecule does not emit a photon, because the energy provided with an original photon has been transformed into vibrations and for possible collisions with other molecules. The additional energy provided by a photon has been transformed into heat. If you can also explain, why the absorbed wavelengths above 13 micrometers are missing at the TOA based on the processes at the molecular level, then it would make the whole picture clear.
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Antero Ollila
October 7, 2015 at 5:05 pm · Reply
Dear Frank,
“The additional energy provided by a photon has been transformed into heat. If you can also explain, why the absorbed wavelengths above 13 micrometers are missing at the TOA based on the processes at the molecular level, then it would make the whole picture clear.”
Good God this is hard! CO2 molecules have a huge elongational resonance at 15.2 microns, with certain temperatures and pressures. At the tropopause 200K at 15 microns wavelength/frequency,(Davids 13-18 microns) The CO2 are screaming radiative exitance to space. The observed specular radiance from space indicates that the CO2 are at a temperature of 200K not the 288K of the surface. This exact 15 micron radiance from the 200K tropopause does not change between 100 and 800 ppmv atmospheric CO2.
All the best! -will-
03
Antillo,
As Frank says above, a CO2 molecule that absorbs (‘annihilates’) a quantum of electromagnetic energy (a ‘photon’) thereby acquires a quantum of additional internal energy. In the bulk of the atmosphere this internal energy is almost always lost by collision with another molecule or molecules (the vast bulk of which are NOT GHGs) before it has time to lose any energy by emitting a photon.
In this way the whole of the atmosphere in a particular region tends towards the same temperature (i.e. average kinetic energy) because there is a ‘soup’ of molecules (GHGs and non-GHGs) all exchanging energy with each other by collision (or, comparatively rarely, by photon exchange).
However, increasingly towards the top of the atmosphere, where molecules are fewer and further apart, the reverse becomes progressively the norm. As the average interval between collisions becomes longer and longer, an increasing proportion of GHG molecules have time to emit photons before they encounter collisions. And an increasing proportion of those emitted photons (at least the half that are emitted upwards!) will not encounter another GHG molecule at all and so will be lost to space.
00
Anteros: FWIW, I am a “molecular physician” by training, though my specialty was not physical chemistry. Let’s go to MODTRAN model that calculates outgoing radiation from the earth at:
http://climatemodels.uchicago.edu/modtran/
Observations from space agree with this model. Make the concentrations of CO2, CH4, both ozones, and water vapor scale = 0. Locality: Choose 1976 US Standard Atmosphere, no rain or clouds. Looking down from 70 km (which is as high as the model goes = TOA). The TOA OLR is the blue curve between the curves for blackbody radiation at 280 degK and 300 degK. What you are seeing is the blackbody radiation emitted by the surface of the earth at 288 degK, because no GHG’s are in the way.
Now make the CO2 concentration 400 ppm. The big reduction in TOA OLR from 13-17 um is obviously caused by CO2 absorbing the radiation at these wavelengths emitted by the surface. Notice that on either side you are still seeing radiation of blackbody intensity traveling from the surface directly to space without interference.
Now try 40 ppm and 4000 ppm of CO2. In all three cases, the intensity of the radiation emitted by through the TOA in the center of the band is the same – on the blackbody curve for 220 K. Why doesn’t more CO2 reduce the output at the center of the CO2 band? In all three cases, CO2 absorbs essentially all of the radiation emitted by the surface near 15 um. The remain intensity comes from photons emitted by CO2 from high in the atmosphere where the average temperature is 220 K (the tropopause).
You must remember that GHGs both absorb and emit photons of the same wavelength. The intensity of radiation at around 15 um is changed by both absorption AND emission as it travels upward from the surface through the atmosphere. The atmosphere gets less dense with altitude, slowing down the rate of both absorption and emission.
The rate of emission of photons depends on the density of GHGs and the temperature. The rate of absorption of photons depends on the density of GHGs and how many photons are traveling through them. When the the density of GHG molecules is high enough, absorption and emission reach a temperature-dependent equilibrium and the radiation has blackbody intensity. The curves for 220, 240, 260, 280, and 300 give you an idea of what temperature (altitude) it was when absorption and emission were last in equilibrium.
Now try 4 ppm of CO2. Essentially ll of the photons emitted by the surface near 15 um are still being, but now the average photon escaping to space is emitted by CO2 molecules around 250 K, because there are fewer CO2 molecules above to absorb them.
Now try 40000 ppm. The average photon in the center of the CO2 band is being emitted from the stratosphere, which is warmer than the tropopause.
Notice that the width of the band around 15 um grew wider as CO2 increased. The center of the band is “saturated” and increasing CO2 doesn’t reduce the number of photons reaching space. But on the sides of the band, more CO2 does reduce the flux to space. The radiative forcing for 2XCO2 occurs on the sides of the main band.
30
Antero: You also seem slightly confused about the role of collisions.
Suppose I shine a 15 um laser pulse lasting 1 picosecond into a sample of air to excite the CO2 molecules it contains. Nanoseconds to microseconds later, I measure how many CO2 molecules are still excited (using a second laser pulse). I also measure how many CO2 molecules are normally in an excited state. (None of these times are meant to be accurate. They should be the right order of magnitude.)
At 1 atmosphere of pressure, the number of excited CO2 molecules returns to a background level in several nanoseconds. Let’s call this relaxation. If I reduce the total pressure (but leave the amount of CO2 unchanged), the time needed for the excited CO2 molecules to relax increases (inversely with pressure). Relaxation is also faster at higher temperature. The rate at which excited CO2 molecules relax clearly depends on how often they collide with other molecules. At atmospheric pressure, the average time between collisions is a little less than 1 nanosecond, but not all collisions result in relaxation.
When an excited state is relaxed by a collision with another molecule, the energy from the photon that produced the excited state has become kinetic energy. This process is called thermalization. The photon’s energy has become thermal energy (heat).
The opposite of collisional relaxation – collisional excitation – also occurs. This process creates the background amount of excited CO2 molecules in any sample. The fraction of excited molecules at any temperature is determined by the Boltzmann distribution.
If I continue to reduce the pressure, it takes longer and longer for the excited CO2 molecules to relax – until the relaxation time is about 1 microsecond. Then, further reduction of pressure (and collision rate) do not increase the rate of relaxation. Now relaxation is not caused by collision – is caused by emission of a photon. This process is sometimes called re-emission. Photons are not stored in the excited state and Re-emitted. They are destroyed by absorption and recreated when an excited state emits a photon. The average time between excitation and emission of a photon is roughly 1 microsecond – unless collisions with other molecules relax the excited state.
In the troposphere and lower stratosphere, a CO2 molecule excited by absorbing a photon is relaxed by collisions long before it is likely be relaxed by “re-emission” of a photon. The number of excited CO2 molecules (capable of emitting a photon is determined by the local temperature via the Boltzmann distribution. Any given CO2 molecule may undergo hundred of collisional excitations and relaxations before it emits a photon. We say that the lower atmosphere is in “local thermodynamic equilibrium” and the number of molecules in an excited state depends only on temperature.
30
Dear Will,
With respect, I know it very well that the CO2 absorption happens below 2 km height in the region 13-18 micrometers and in the stratosphere its portion from the total absorption will actually decrease because ozone absorbs so strongly there. I have carried out numerous spectral analyses and found this fact.
Anyway, nobody has explained so far, what happen at the molecular level, when the IR radiation is emitted by the Earth’s surface. If a GHG molecule is hit by a photon, it vibrates and then re-emits a same kind of photon into any possible direction, the end result would be that there is no absorption but the total IR radiation of 395 W/m2 from the surface would disappear into the space. That is not the case as we know. So who can explain this at the molecular level? So far there has not been a complete explanation. Frank is very near.
20
Hi Antero,
I have tried to explain this at 14.72 above (but, as I write, that is still stuck in moderation).
The absorption of a photon by a GHG molecule does NOT cause it to immediately emit a photon. You need to think about what happens in two stages:
(1) The GHG molecule stores the photon’s worth of energy internally; and then either…
(2a) It collides with another molecule and so loses some internal energy; or…
(2b) It doesn’t collide with another molecule before it is able to successfully emit a photon.
In the bulk of the troposphere where the mean free path between molecules is comparatively short, (2a) happens predominantly. Towards the top of the troposphere where the mean free path between molecules is increasingly long, (2b) happens more frequently.
You have to average what is happening over zillions of collisions to appreciate the macroscopic effect… which is that only towards the top of the troposphere does the probability of photons escaping to space become predominant, so that is where the energy balance with the incoming Sun’s absorbed radiation is increasingly achieved.
00
Antero,
…also, you said:
“…the end result would be that there is no absorption but the total IR radiation of 395 W/m2 from the surface would disappear into the space.”
That number you are using – actually 396W/m2 if taken from the Trenberth et. al (2009) energy balance diagram – is a radiative potential not a flow of radiative energy. It needs to be netted off against the opposing radiative potential from atmosphere to surface – which Trenberth gives as 333W/m2. This yields a real net energy flow of 63W/m2, which you will note is upwards, from surface to atmosphere, in full accordance with the Second Law of Thermodynamics.
I mention this NOT because it has any bearing at all on your query about the photon creation/annihilation mechanism but just to warn you about the upset this often causes with people who don’t believe in the two-way flow of photons between radiating bodies, but instead use the classical electromagnetic radiation approach of netting off radiative potentials. Some of them get very upset at talk of the 396W/m2 being a real flow even though, actually, contra-flowing photon streams are just a different mathematical way of looking at the same phenomenon. Anyway, I suggest we should indulge them for the sake of peace amongst all climate skeptics. 🙂
03
Antero wrote: “Anyway, nobody has explained so far, what happen at the molecular level, when the IR radiation is emitted by the Earth’s surface. If a GHG molecule is hit by a photon, it vibrates and then re-emits a same kind of photon into any possible direction, …”
More precisely: The photon is absorbed – it no longer exists. It was a massless particle; only its energy existed. That energy is part of the molecule that absorbed the photon. Classically, we would say that the energy is in simple harmonic oscillation: Two masses attached by a spring constantly vibrating and interconverting kinetic energy and potential energy. In QM, that oscillator must have a discrete amount of energy and when a molecule has more than the minimum vibrational energy, we say it is in vibrational excited state. There are also rotational excited states.
HOWEVER, excited vibrational and rotational states are created (excited) and destroyed (relaxed) by more than just absorption or emission of a photon. They are also created and destroyed – usually much, much FASTER – by collisions between two molecules. During relaxation, the energy difference between the ground and excited state is added to the kinetic energy the colliding molecules – it becomes heat. During excitation, some of the kinetic energy possessed by the colliding molecules “disappears” and “re-appears” as the energy difference between the ground and excited state. The colliding molecules are “colder” in the sense that average kinetic energy of the molecules near them is lower. Individual molecules don’t have a temperature.
Since collisional excitation and relaxation of GHGs is so much faster than absorption and emission of a photon in the troposphere and lower stratosphere, the fraction of molecules in an excited state depends only on collisions and could be calculated with the Boltzmann distribution. Planck’s Law is derived relying to the Boltzmann distribution to predict what fraction of molecules are in an excited state capable of emitting a photon. The Planck function, B(lambda,T), controls how often a GHGs emit photons.
Does the absorption of photons by GHGs increase the rate at which they emit photons? NOT in the troposphere! Excited states in the lower atmosphere are usually relaxed by collisions long before they can emit. The heat released by relaxation does raise the temperature – BUT the temperature can also be changed by latent heat (condensation or evaporation of water) and radiative cooling. In the lower atmosphere, you only need to know the local temperature – not the rate photons are being absorbed – to know fast photons will be emitted. Emission is proportional to the density of the GHG, its absorption cross-section, and B(lambda,T).
As for the 390 W/m2 of OLR and 333 W/m2 of DLR, those are ENERGY fluxes – radiant energy fluxes. The amount of HEAT transferred is the NET energy flux. Heat transfer must obey the 2LoT. Unfortunately, Trenberth’s diagram does not show heat transfer, it shows energy transfers. To prove that the Trenberth diagram is compatible with the 2LoT, calculate the net energy flux BY ALL ROUTES from: the surface to the atmosphere, the atmosphere to space, the sun to the earth, the sun to the atmosphere and the earth to space. In each case, the HEAT transfer – the only thing constrained by the 2LoT – is from hot to cold. Trenberth’s diagram confuses some skeptics because it doesn’t show HEAT transfer.
Those skeptics who don’t believe in two way transfer of radiant energy can replace opposing radiative fluxes with a single one-way radiative flux. That one-way flux is still not the HEAT flux. The flux of HEAT from the earth to the atmosphere includes latent heat and conduction as well as net radiation. (:))
20
Franktoo(Frank):
In light of that account would you agree that the dry adiabatic lapse rate slope represents the change in balance between radiation and collisional activity as one descends through the mass of an atmosphere?
It seems to follow from your narrative that collisional activity reduces photon emission so that would explain how a surface at 288k beneath an atmosphere with the mass of Earth’s atmosphere can only lose 255k to space once the absorption effects of atmospheric mass via conduction and convection have done their work.
You confirm that the flux of HEAT from the earth to the atmosphere includes latent heat and conduction as well as radiation BUT is it not the case that descending warming air RETURNS heat energy to the surface from potential energy higher up?
That return of kinetic/heat energy to the surface in descending columns is missing from the radiation budget and that is why the concept of back radiation needed to be invoked to replace the missing component.
It is true that descending columns do not transfer energy back to the surface by conduction but what they can do is inhibit convection from the surface beneath the descending column so that incoming radiation can raise that surface above the S-B expectation.
You said:
“Two masses attached by a spring constantly vibrating and interconverting kinetic energy and potential energy”
Well I suggest that you apply that to the macro scale whereby the surface and the atmosphere are the two masses which are constantly interconverting kinetic energy and potential energy via convection (both up AND down) in accordance with the ideal gas law.
Ascending air converts KE to PE whilst falling air converts PE to KE and the descending air inhibits convection so that the surface heats up. That is where the enhanced surface temperature is coming from, not from back radiation.
The thing is that to establish the amount of energy involved one must sum the total energy in BOTH columns and NOT set off one against the other.
Both rising and falling columns require an energy supply to sustain them. Ascending columns require a KE store at the surface and descending columns require a PE store at the top.
The net effect is a surface temperature enhancement above S-B.
04
Steve: You appear to be attributing things to me that I didn’t mean to imply:
You asked: “In light of that account would you agree that the dry adiabatic lapse rate slope represents the change in balance between radiation and collisional activity as one descends through the mass of an atmosphere?”
From my perspective, the concept of buoyancy-driven convection dominates my thinking. Convection should stop when the environmental lapse rate drops below the MALR, leaving convective regions on the AVERAGE with the MALR. (Radiosondes, however, show that the real world with weather and diurnal change is far more complicated in the short run.) The role of radiation in the lapse rate is complicated. High in the atmosphere, there is little DLR and the air loses more heat by emission than it gains by absorption. This produces an unstable lapse rate and convection makes up the difference.
You say I said (in part): “It seems to follow from your narrative that collisional activity reduces photon emission …”
Definitely not in the (relevant) lower atmosphere where LTE exists. The rate of emission is determined by B(lambda,T). I discuss “collisional excitation” and “collisional relaxation” (and the Boltzmann distribution which is links these processes), not “collisional activity”.
You comment: “You confirm that the flux of HEAT from the earth to the atmosphere includes latent heat and conduction as well as radiation BUT is it not the case that descending warming air RETURNS heat energy to the surface from potential energy higher up?”
A good question I’ve wondered about myself. How does the heat of hot dry air descending over a desert reach the surface? DLR (especially at night) works for me. Conduction of simple heat works for me too – analogous to the simple heat in Trenberth’s diagram, but going the oppose way. Of course, simple heat is a fudge factor to Trenberth.
The concept of back radiation doesn’t have to be “invoked”; it arises from analyzing the components of radiative fluxes perpendicular and parallel to a layer of atmosphere parallel to the surface. GHGs emit in all directions, so the radiation emitted by the layer must contribute to both DLR and OLR. (Needless to say, both are also absorbed by any layer too.)
You say: “Well I suggest that you apply that to the macro scale whereby the surface and the atmosphere are the two masses which are constantly interconverting kinetic energy and potential energy via convection (both up AND down) in accordance with the ideal gas law.”
I think there is little potential energy in the atmosphere because all packets air are floating, slowly sinking or slowly rising in an atmosphere with similar density at one altitude. For a packet of air to rise, another packet somewhere else must fall. The force is associated with the difference in density between the two packet, not the weight/mass of the packet itself. A scuba diver with a weight belt converts potential to kinetic energy falling through the air, but hovers in the ocean despite his apparent potential energy. There is a great deal about fluid mechanics that I don’t know, so I’ll stop here
30
Dear David,
The same thing may be interpreted in different ways and some times they are all true. The value of 395 or 396 is based on the black body radiation from the surface of 15.9 C. This radiation flux has been also measured from the surface as well as the reflected SW radiation upwards. So there is no uncertainty that this flux exists. Also the downward LW flux to the surface has been measured. If some skeptics do not approve measurement results, they should think it again.By the way, Trenberth has not the right value for the downward flux. It is about 345 for the all-sky conditions.My analysis show the value 344.7 and Stephens et al have shown the value 344-350. IPCC do not use Trenberth’s energy balance presentation anymore.
20
Antero,
You say there is no uncertainty that the upward and downward fluxes exist. That is fine if you believe in photons (as you and I do) because the difference between the two opposing fluxes is always from warmer to cooler surface, so the 2LT is not violated.
However, some people simply don’t believe in separate photonic fluxes and are clearly not going to be shifted from that opinion. The instrumentation that measures such fluxes contains an opposing surface at a definred temperature – so such people simply say that what is actually being measured is always the net flow, from which the ‘fictitious’ uni-directional flow is then calculated.
My suggestion is that it is best to leave them to their interpretation – otherwise we will get involved in endless arguments in which they conclude that ‘back radiation’ is the work of the devil and contradicts the 2LT, which would be very boring and unproductive. 🙂
21
“The value of 395 or 396 is based on the black body radiation from “the surface of 15.9 C. This radiation flux has been also measured from the surface as well as the reflected SW radiation upwards.”
Can you show any evidence that radiative flux was ever measured from the surface? All measurements published are for a calculated radiance (field strength) never flux.
“So there is no uncertainty that this flux exists.”
Show the evidence!!
“Also the downward LW flux to the surface has been measured.”
Again the field strength of the sky, never any flux. there is none downward at night!
“If some skeptics do not approve measurement results, they should think it again.”
Do you even know how to measure radiative flux? Can you please describe the equipment and procedures used for distinguishing radiative flux from conductive and convective fluxes? Can you quantify the accuracy of your measurements? How? Meters to measure surface and sky radiance are now $20 but do not measure flux.
23
Will (and Antero and David Cosserat): If you believe in one-way flux, any detector you place between a warmer object and a cooler object to measure the weaker radiative flux from cooler to hotter probably will interfere with you ability to measure two fluxes. Two-way and one-way fluxes make the same predictions concerning what we should observe.
The fundament question is why should we believe that photons are incapable of traveling from a colder object to a hotter object – as long as more photons are traveling the other way. Planck’s Law or the S-B eqn tells us how much radiation is emitted by the cooler object. Will asserts that this radiation can’t be absorbed by the warmer object BECAUSE OF THE 2LoT. This reflects his misunderstanding of the 2LoT – which is based on formulations of the 2LoT made before we understood molecules, temperature!, photons, heat, QM and statistical mechanics. Will and others are forced to cite Prevost or Clausius as authorities, because 20th-century physicists know about these later discoveries and make careful statements about the 2LoT and heat in serious textbooks.
It is absurd to discuss the emission and absorption of a single photon in terms of the 2LoT, because neither the emitting molecule nor the absorbing molecule have a temperature! Temperature is defined only for a large group of rapidly colliding molecules and is proportional to their mean kinetic energy. Furthermore, the existence of a Boltzmann distribution of kinetic energies DEMANDS energy transfer from slower-moving to faster-moving molecules – either by collision or radiation. If not, all molecules would soon be moving at the same speed. Either: 1) The 2LoT doesn’t apply to energy transfer between two molecules – OR 2) the Boltzmann distribution, Planck’s Law, and much other accepted physics is flawed. Will: Do you believe that the 2LoT doesn’t apply to energy transfer between two molecules?
The wavelength of any emitted photon contains information about the energy difference between the excited and ground states in the molecule that emitted it, but no information about local temperature where it was emitted. So when an emitted photon arrives at a warmer or cooler surface, it has no way of “knowing” whether to be absorbed or reflected.
Information about local temperature is ONLY available from the total flux of photons at a given wavelength emitted by a large GROUP of molecules! Ignoring stimulated emission, the rate photons are emitted by an excited state depends on the Einstein A21 coefficient for that transition. The number of molecules in an excited state depends on the local temperature and the Boltzmann distribution. The radiative flux from a warmer group towards a colder group will always be larger than in the opposite direction because the Boltzmann distribution predicts fewer excited molecules where it is colder. This ensures that the net flux (heat transfer) is from hot to cold and that the 2LoT will be obeyed.
So it all works out neatly: Temperature is only defined for large groups of molecules and heat flux is defined as NET energy transfer between those groups of molecules. The 2LoT is obeyed – as long as we remember that heat is NET energy transfer and that individual molecules are not hot or cold.
Statistical mechanics is the branch of physics that derives the macroscopic behavior of large groups of molecules from the laws of quantum mechanics. Mean molecular kinetic energy becomes macroscopic temperature. Molecular disorder becomes macroscopic entropy. The impulse delivered by molecular collisions becomes macroscopic pressure. Boltzmann’s constant (k) used with individual molecules becomes the universal gas constant (R). The 2LoT is a consequence of the behavior of individual molecules that is not governed by the 2LoT.
It is logical to ask whether all photons emitted from one surface actually reach a second surface when other photons are traveling the opposite direction. Perhaps the flux from cold to hot is cancelled by interference from the flux from hot to cold. Waves certainly interfere with each other, but light is composed of photons. The laws of quantum mechanics produce wave-like phenomena, but only when the radiation is COHERENT and the amplitudes for alternative pathways interfere. Radiation emitted by two different sources is not coherent. Shining two flashlights at each other or the same spot on a wall doesn’t produce interference. Signals simultaneously travel in both directions in a fiber optic cable without interference. Two-way flux is NOT converted to a one-way flux by interference.
In some situations (lasers, fluorescent lights, the thermosphere), the number of excited states is not determined by the local temperature and the Boltzmann distribution. In these situations, rate of photon absorption is faster than the rate at which molecular collisions can redistribute energy. In these situations, the concept of temperature becomes vague. Technically, thermodynamic temperature is defined as the mean kinetic energy of a group of molecules in local thermodynamic equilibrium – meaning they are exchanging energy by collision more rapidly than they gain or loss energy by any other process. The 2LoT applies to thermodynamic temperature.
50
Frank,
Very, very well said!
You have provided a perfect renderering of the position from the point of view of QM and statistical mechanics. I particularly liked: “The 2LoT is a consequence of the behavior of individual molecules that is not governed by the 2LoT. “.
My only point that I made to Antero is that even people such as Will who appear to reject QM and statistical mechanics and instead talk in terms of opposing radiative potentials (as per 18th century Provost!) get the right answer by subtracting one potential from the other potential to produce a net energy flow between the surfaces, always from hotter to colder.
However having successfully come to the same conclusion as everybody else, they then still manage to rant on about their radiative potential from atmosphere to earth AS IF it were a real and unopposed flow and AS IF it violated the 2LT. It seems to me they want to have their cake and eat it.
My strong advice to everyone is to indulge them and then to ignore them.
10
” Technically, thermodynamic temperature is defined as the mean kinetic energy of a group of molecules in local thermodynamic equilibrium – meaning they are exchanging energy by collision more rapidly than they gain or lose energy by any other process”
So, increasing density results in an increase in collisional activity at the expense of photon emission and thus a higher temperature?
That means that lower rate of photon emission leads to a gain in kinetic energy and a higher temperature.
Yet AGW theory proposes that a higher temperature implies a higher rate of photon emission.
Some contradiction there, surely?
25
To resolve that apparent contradiction for the Earth’s surface this is what must happen:
i) Density at the surface is such that the initial receipt of radiation at 255k is split between radiation out at 222k and collisional activity at 33k. At that point the temperature as viewed from space would appear for a while to be 222k.
ii) The surface temperature then rises to 288k, photon emission increases accordingly and viewed from space the temperature rises back up to 255k so that radiative equilibrium is regained.
iii) That 33k then provides the energy required to maintain hydrostatic balance via convective overturning.
The point that AGW theory overlooks is that a surface beneath an atmosphere can be at 288k yet only release radiation to space at 255k simply because collisional activity diverts potential photon emission to ongoing conduction and convection.
The amount of collisional activity is related to mass density and not radiative capability which is why temperature increases in line with mass density as one moves down the dry adiabatic lapse rate slope.
That lapse rate slope marks the changing balance between photon emission and collisiomal activity as one descends through the depth of an atmosphere.
15
Steve wrote: The point that AGW theory overlooks is that a surface beneath an atmosphere can be at 288k yet only release radiation to space at 255k simply because collisional activity diverts potential photon emission to ongoing conduction and convection.”
I disagree. Where collisional excitation and relaxation are much faster than absorption and emission (as they are up to about 50 km), we say LTE exists. Where LTE exists, the emission rate depends one the local temperature via B(lambda,T), not the local radiation field.
I gather you believe that gravity creates the lapse rate in the atmosphere, while it is my understanding that large scale convection produces our lapse rate and the lapse rate plus the Schwarzschild eqn produce the GHE.
A lapse rate can be created by any process that moves heat through the atmosphere. Let’s consider candidate processes in isolation.
1) Radiative transfer creates a lapse rate: It would be isothermal (due to the 2LoT) if energy weren’t escaping to space, but is curved because of escape.
http://scienceofdoom.com/2010/08/08/vanishing-nets/
2) Molecular collisions (thermal conductivity) create an isothermal lapse rate.
3) Molecular diffusion plus interconversion of kinetic and potential energy apparently creates a lapse rate of g/Cp.
4) Convection produces a MALR.
The lapse rate that we observe is the result of the fastest process. Based on the lapse rate we observe in our atmosphere, convection dominates over radiation until the tropopause.
In thought experiments, we can imagine a non-radiating atmosphere and postulate that bulk convection does not occur. In that case, molecular collisions and molecular diffusion would compete to produce a lapse rate. If you calculate the tiny amount of kinetic energy that is converted to potential energy between collisions and the larger (1,000,000X) amount of kinetic energy in being exchanged by collisions between gas molecules, thermal conductivity should dominate and produce an isothermal lapse rate.
40
Franktoo
You may find the linked paper of interest with regard to isentropic lapse rate:
http://www.csc.kth.se/~cgjoh/climatethermoslayer.pdf
Section 7 establishes the conditions for isentropic or isothermal solutions for lapse rate. In the troposphere the lapse rate is well defined by the isentropic state. It does not matter if radiative gasses are present or not. Energy transport is dominated by the large scale convection rather than molecular scale heat transfer with conduction and radiation. The atmosphere in the troposphere is essentially in hydrostatic balance although there is never equilibrium because of the variation in heat input across the globe.
Above the troposphere, convective energy transport is no longer dominant and the thin atmosphere of the stratosphere is closer to isothermal.
30
Rick Will suggested: “You may find the linked paper of interest with regard to isentropic lapse rate”.
I was disappointed to find that the author was Claes Johnson and even more disappointed when I read it. Professor Johnson IS a very accomplished applied mathematician who has worked extensively with the Navier-Stokes and related equations. However, physics and other hard sciences approach knowledge from very different perspectives: Axioms/Logical Deduction and Hypothesis/Experiment. Euclidean and non-Euclidean geometry are both logically valid – pick whichever you prefer. That approach doesn’t work for physics.
For more on the difference between mathematicians and physicists, see Fenymann:
https://www.youtube.com/watch?v=obCjODeoLVw
In section 4 Claes says: “Climate alarmism as advocated by IPCC is based on the assumption that radiation alone sets an initial lapse rate of 10 C/km, which then in reality is moderated by thermodynamics to an observed 6.5 C/km. Doubled CO2 would then increase the initial lapse rate and with further positive thermodynamic feedback it is by IPCC predicted to reach an alarming climate sensitivity or global warming of 3 C. Climate alarmism skeptics like Richard Lindzen and Roy Spencer buy the argument of an intial rate of 10 C/km determined by radiation, but suggest that negative thermodynamic feedback effectively reduces climate sensitivity to a harmless 0.5 C.
No one – certainly not Linden and Spencer – believes in that radiation controls the lapse rate. (When it does, the lapse rate isn’t linear). So, needless to say, there is no citation for these baseless assertions. The standard derivation for the DALR and MALR can be found amid Claes’s equations (7) and (8) and that derivation is simple enough that I am embarrassed when I need to look it up instead of deriving it from first principles.
In Section 9 (basic data), we are told that 4 cm of water evaporates every day. This value is at least an order of magnitude too big. Mean global precipitation is about 1 m, year or 2.8 mm/day. If all the evaporation needed to produce 1 m of rainfall occurred from the ocean (it doesn’t), that would be 4 mm/day. However, Claes is discussing heat transfer from the surface to the atmosphere and condensation occurs over both ocean and land. So 2.8 mm/day is the appropriate figure. Claes calculates the power released by condensation as shown:
2.2*10^6 J/kg * 10^2 kg/m3 * 0.04 m/day * 1/(24*3600) day/s = 103 (J/s)/m2 = 102 W per m2
Anyone who has glanced at the KT diagram knows that latent heat is 80 W/m2. Somehow, Claes has cancelled his 10-fold error in the rate of evaporation with a 10-fold error in the density of water. (:)) He has also used the heat of vaporization of water at its boiling point (2,260 kJ/kg) rather than surface temperature (2444 kJ/kg at 25 degC). When water vapor falls as snow, the heat of fusion (335 kJ/kg) should be added to the heat of vaporization. Let’s round up to 2.5*10^6 J/kg. Then, the answer would be:
2.5*10^6 J/kg * 10^3 kg/m3 * 0.00273 m/day * 1/(24*3600) day/s = 103 (J/s)/m2 = 79 W/m2
Now let’s check Claes’s calculation for the amount of power it takes to raise a column of air 5000 m tall at a rate of 0.01 m/s. Claes tells us the density of air is 0.6 kg/m3 and uses 0.7 kg/m3; but the density actually varies with altitude. The pressure at 5 km is about 0.49 atm, so 51% of the atmosphere lies below with a weight/area ratio of 5266 kg/m2 and an average density of 1.05 kg/m3.
PE = mgh
d(PE)/dt = mg*(dh/dt) = 5266 kg/m2 * 9.81 m/s2 * 0.01 m/s
d(PE)/dt = 517 (kg-m2/s3)/m2 = 517 W/m2
Claes found 350 W/m2. However, that ascending column of air is being replaced by air that is falling elsewhere. For example, suppose the rising air averages 10 degC and the descending air 0 degC. The descending air will weigh 5266*(283/273) = 5459 kg/m2.
d(PE)/dt = mg*(dh/dt) = 5459 kg/m2 * 9.81 m/s2 * -0.01 m/s
= -536 (kg-m2/s3)/m2 = -536 W/m2
The net result is that convection releases (not consumes) 19 W/m2. This makes sense – buoyancy driven convection is a spontaneous process that doesn’t require input of energy.
Finally, Claes calculates the change in the internal energy of the rising air from the change in temperature using the lapse rate. He appears to be using this method:
dE/dt = dE/dT * dT/dz * dz/dt
dE/dt = (1000 (J/kg)/K * 5266 kg/m2) * -6.5 K/1000 m * 0.01 m/s
dE/dt = -342 (J/s)/m2 = -342 W/m2
Since Claes has used a different value for the average density of air in the column, he gets -230 W/m2. Checking showed that Claes used the correct value for heat capacity (Cp), but that reminded me that the upward motion isn’t under CONSTANT PRESSURE – or constant volume. The air rises via an ADIABATIC process where no energy is transferred. dE/dt is really zero! The energy released by falling temperature is used to perform PV work when the rising air expands.
So bad, it isn’t even wrong – except for the parts that are merely wrong. Sad.
Assuming my analysis is correct (blog commentators can always be wrong!), this is precisely what Feynman is talking about in the above video – mathematicians applying equations to physical situations they don’t fully understand. If this material had passed through some type of peer review, these putative mistakes should be caught. .
50
Steve asked: “So, increasing density results in an increase in collisional activity at the expense of photon emission and thus a higher temperature?”
No, where collisional relaxation and excitation are much faster than absorption and emission – i.e. where LTE exists – the rate of emission at a given wavelength is proportional to B(lambda,T) (and the density of the GHG and its absorption cross-section).
In the thermosphere, where collisions are infrequent, the fraction of GHGs in an excited state can depend on the rate at which photons are absorbed and the rate of collisional excitation. The concept of temperature becomes nebulous when a group of rapidly colliding molecules doesn’t exist
20
Franktoo (Frank) October 9, 2015 at 5:11 am ·
Sorry for the delayed response!
“Will (and Antero and David Cosserat): If you believe in one-way flux, any detector you place between a warmer object and a cooler object to measure the weaker radiative flux from cooler to hotter probably will interfere with you ability to measure two fluxes.”
As in all EMR, a fluxmeter between the absorbing/emitting surfaces any the power source/sink maintaining radiance (T^4) will clearly show the direction of all power/m^2! With a two body isolated system the higher temperature body supplies the same amount of power that is sinked by the lower temperature temperature maintenance. The S-B equation is extremely sensitive to both temperatures down to the microwatt/m^2 range. Remember the emissivity for both surfaces is both frequency and angle dependent.
“Two-way and one-way fluxes make the same predictions concerning what we should observe.”
Only if there is no dissipative translucent or opaque material or surfaces between the two isotherm surfaces. If there is such material it will always be at a temperature between that of the two isotherms. Never some summation. (proof #1).
If such material has a measurable thermal mass the time series of temperature change of that mass must indicate a flux exactly like displayed by two independent (and whole) S-B equations; each with only a one way flux to/from each isotherm surface with both isotherm maintainers each indicating the exact flux profile, and an ending with a total one way flux lower than that without such dissipative surface (proof #2)
“The fundament question is why should we believe that photons are incapable of traveling from a colder object to a hotter object – as long as more photons are traveling the other way. Planck’s Law or the S-B eqn tells us how much radiation is emitted by the cooler object.”
Plank’s equation only indicates maximum spectral radiance never any flux. The equation uses the spectral difference of two Planck formula and integrates over all wavelengths multiplies by PI steradians to achieve the maximum possible radiant flux from/to parallel infinite flat surfaces. at any two temperatures and any distance more than 5 wavelengths at the lowest frequency.
“Will asserts that this radiation can’t be absorbed by the warmer object BECAUSE OF THE 2LoT. This reflects his misunderstanding of the 2LoT – which is based on formulations of the 2LoT made before we understood molecules, temperature!, photons, heat, QM and statistical mechanics.”
I have never claimed such!!! My only claim is that thermally generated EMR is never emitted in a direction of higher radiance at any frequency. Such would be a direct contradiction to all of Dr. Maxwell’s field equations. EMR requires no mass and is completely independent of any thermodynamics. Your mass if accumulating that EMR power as heat can do what it must.
“Will and others are forced to cite Prevost or Clausius as authorities, because 20th-century physicists know about these later discoveries and make careful statements about the 2LoT and heat in serious textbooks.”
I generally cite nether! Provost has always been false, no caloric, a fantasy!! Rudy has never been falsified with his 2LTD of “Spontaneous don’t go uphill” 😉
“… Temperature is defined only for a large group of rapidly colliding molecules and is proportional to their mean kinetic energy.”
In the gas laws each molecule has the noise power of kT/t as the definition of temperature.
“Furthermore, the existence of a Boltzmann distribution of kinetic energies DEMANDS energy transfer from slower-moving to faster-moving molecules – either by collision or radiation.”
Never!! The rebound never adds energy to more energy. Else equilibrium can never establish.
“If not, all molecules would soon be moving at the same speed. Either: 1) The 2LoT doesn’t apply to energy transfer between two molecules – OR 2) the Boltzmann distribution, Planck’s Law, and much other accepted physics is flawed.”
What complete physical nonsense!!! The kinetic theory of gas is currently very misinterpreted. In the books all sensible heat is defined as mean squared velocity. This is nonsense!! Newtons kinetic is the derivative of momentum with respect to time. I.E. every change in molecular direction is mivh more important than original velocity. A gas then can only have an energy density proportional to mass, but never an mass energy o r conventional specific heat. Gas has two but leaves out atmosphere which relates to neither case.
“Will: Do you believe that the 2LoT doesn’t apply to energy transfer between two molecules?”
Sure, just as I explained above!
“The wavelength of any emitted photon contains information about the energy difference between the excited and ground states in the molecule that emitted it, but no information about local temperature where it was emitted.”
The wavelength spectrum in all absorbed/emitted wavelets from a gas contains some information about molecular resonances of that gas. Nothing of ground or excited states! The shape of the spectrum can (but hard to do) indicate temperature!
“So when an emitted photon arrives at a warmer or cooler surface, it has no way of “knowing” whether to be absorbed or reflected.”
Flux arriving from a higher radiance at a surface must be absorbed, reflected, or transmitted as per the material/surface properties The gauge boson ‘PHOTON” is the mediator in full Planck action units.
You go on and on about the nonsence of your religion!!
Your religion seems to think that some EMR spectrum can provide knowledge of thermometric temperature of an emitter. Go try that with your microwave oven, or an AEGIS radar! Why should I go on?
12
Antero,
See what I mean?
Will ignores my analysis where I say: “The instrumentation that measures such fluxes contains an opposing surface at a defined temperature – so such people simply say that what is actually being measured is always the net flow, from which the ‘fictitious’ uni-directional flow is then calculated., but he still continues to challenge… 🙂
21
Hi,
If we do not accept the measured values of radiation fluxes, then it is no use to continue conversation. If we do not accept the results and models of IPCC, we have to find out what is wrong in their calculations based on the common scientific methods. That is what I have tried to do.
One of the basic issues in the climate change science is the climate sensitivity. IPCC uses the assumption of constant relative humidity. If you just look at the trend from 1948, you can see by bear eye that this assumption is not valid. We have to find out reasons like this, if we want to show what is wrong in IPCC’s models.
20
Antillo,
You say: “If we do not accept the measured values of radiation fluxes, then it is no use to continue conversation.”
Almost all skeptics accept the photonic two-way flux theory – it is only a very few people like Will that don’t. So I suggest you should just continue the conversation with us mainstream sceptics…
31
David Cosserat October 9, 2015 at 9:19 am
Antero Ollila October 8, 2015 at 7:09 pm
Hi,
(“If we do not accept the measured values of radiation fluxes, then it is no use to continue conversation.”)
Please identify even one case of a measurement of any EMR flux in a direction of higher radiance at any frequency!
“If we do not accept the results and models of IPCC, we have to find out what is wrong in their calculations based on the common scientific methods. That is what I have tried to do.”
The calculations are based on fantasy like some thermal EMR flux in opposing directions through a dispersive media!
DC_the_troll
“Antillo,
You say: (“If we do not accept the measured values of radiation fluxes, then it is no use to continue conversation.”)
“Almost all skeptics accept the photonic two-way flux theory – it is only a very few people like Will that don’t.”
You make all claims only counting your own floters/sinkers!
There is no photonic there has never been a two way EMR flux. This is only a fantasy invented by DC and the rest of like CCC as a means to promote the CAGW scam!
“So I suggest you should just continue the conversation with us mainstream sceptics…”
That would be great even here an Joanne’s blog so we can all see the BS of the CCC!
04
Will – “Please identify even one case of a measurement of any EMR flux in a direction of higher radiance at any frequency!”
I filled an 8oz. common kitchen water glass with some refrigerator ice and tap water. I inserted a digital thermometer in the mix and its readout: 32F.
I pointed my room temperature (~70F) $30 Ryobi IR002 digital, laser directed IR brightness thermometer (emissivity set point=0.95) laser dot on the glass surface, through a dispersive media, and its readout: 32F. Same!
I repeated this test with boiling water in my painted surface tea kettle, both digital readouts were 212F. Same again! Easily, rather cheaply, replicated too. There’s two cases! Two way incoherent photon transfer between objects at different temperatures is easily proven.
This demonstrates a calibrated, accurate brightness temperature measurement of EMR flux with known emissivity in a direction of higher radiance at any frequency; and was so very easily demonstrated proving Will wrong. These tests help explain why Will is so often embarrassingly wrong and obtains so many red thumbs around here.
Ok, I expect that deer in headlights look from Will. I would expect nothing less.
40
“This demonstrates a calibrated, accurate brightness temperature measurement of EMR flux with known emissivity in a direction of higher radiance at any frequency; and was so very easily demonstrated proving Will wrong. These tests help explain why Will is so often embarrassingly wrong and obtains so many red thumbs around here.”
Have you even bothered to check how a higher temperature bolometer can determine the radiance of a lower temperature surface with the ‘only flux’ in the direction of the lower temperature? The actual S-B equation as done by Ludwig clearly gives the answer. Or you can ask anyone that has actually constructed the things, or you can read all about the details from the manual for recalibration. A wee calibrated thermistor on the reference side of the bolometer, or thermopile is needed as the higher temperature in the actual S-B equation.The measurement of outgoing flux in W/m^2 give all that is needed for a solution for the one way flux in the direction of opposing field strength.
“Ok, I expect that deer in headlights look from Will. I would expect nothing less.”
That is my avatar!! His name is Ralph at 9 months! The eyes are reflections at 0.9 microns from the illuminator on the trail camera. What is your excuse? Did you not even ask mommy for help?
All the best! -will-
16
Will – “..give all that is needed for a solution for the one way flux in the direction of opposing field strength.”
If so, the test using my $30 Ryobi IR002 digital, laser directed IR brightness thermometer opposing field strength flux at ~70F to measure the brightness temperature of the glass of ice water should have overpowered the IR flux from that ice water at 32F, yet it did not, the IR thermometer produced an accurate brightness temperature of the ice water from the two way incoherent photon flux compared to the digital thermometer reading . Thus proving that two way photon flux exists, the net of which results in an accurate temperature measurement.
Heck, eyes at 98F scan see the ice water too Will. Maybe Will has never noticed & applied physics to seeing things colder than ones eyes. Like that deer can see and avoid trees in the winter Will.
Original experiments ref.d in Planck’s writings verified the two streams of photons Will, with the net being shown on the instrumentation used at the time. You post here so often you might as well get it right Will. I expect no correction from you though, caught in headlights, Will is incorrigible in the face of proper test, that next red thumb will be mine.
30
Will, you are starting to be as rude as you became when you and I had precisely the same discussion under your other pseudonym over on the PSI site.
You are perfectly entitled not to believe in photons if that is your wish. Just as the majority of scientists who find the photon theory useful are perfectly entitled to continue to use it where appropriate.
What you are not entitled to do is to become progressively offensive and incoherent when you find you have nothing better to say on a subject.
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“You are perfectly entitled not to believe in photons if that is your wish. Just as the majority of scientists who find the photon theory useful are perfectly entitled to continue to use it where appropriate.”
A Photon is fine as a gauge boson mediating the one way flux transfer between absorption, reflection, and transmission in units of Planck action. What they certainly are not are particle bullets in all directions at the speed of light! Where is the power coming from? OH yah, from the lower temperature surround, for a perpetuum mobile of the second kind! Nice police work there Lou! 🙁
“What you are not entitled to do is to become progressively offensive and incoherent when you find you have nothing better to say on a subject.”
Please stop reading what I post! It is definitely not for your enlightenment! Hopeless one!
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I have a major problem with this portion of this article:
This is WRONG: “..if we thicken the blanket, the layer at the top that can emit is raised, and is also colder. A colder body can’t emit as much…” — Jo Nova
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CORRECTED:
The problem is the stratosphere WARMS as the height above the surface increases! In the stratosphere, the air temperature rises with altitude. This warming effect is the result of exothermic reactions triggered by UV radiation from the sun. This is exactly opposite of what Jo is saying. Therefore increasing the amount of CO2 causes the layer at the top that can emit to be raised. Since this is occurring within the stratosphere raising the CO2 emitting layer means MORE OLR with an increase in CO2 since a warmer body can emit more OLR.
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Dr Evans is more explicit saying:
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Dr Robert Brown agrees with Dr Evans first paragraph:
To paraphrase:
So the question becomes where does CO2 radiation instead of handing off the energy via collision?
Dr Happer in his UNC lecture (fall 2014) answered this question about where CO2 energy is radiated. Experimental data shows barely any radiation at 11 KM and that radiating is in the stratosphere ~ 47 KM above the surface.
This Figure is from Uherek, 2006 and shows CO2 radiating just above the tropopause in the stratosphere ~ 47 KM above the surface.
http://www2.sunysuffolk.edu/mandias/global_warming/images/stratospheric_cooling.jpg
The legend with the illustration:
So the stratosphere WARMS as the height increases.
http://www.weather-climate.org.uk/images/layers.jpg
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Yes Gai, the stratosphere warms as you go higher. This can even been seen in the OLR spectrum on say a Nimbus diagram (Fig 2 of post 14) — see the spike up at exactly 15 microns, indicating a higher emission temperature around 230 or even 240 K (hard to tell, Nimbus resolution limited).
So adding more CO2 raises the CO2 emission layer in the stratosphere so it is slightly warmer. But this scarcely changes the spectrum of OLR, because it is already at the temperature of the lower stratosphere at the CO2 wavelengths emitted from the stratosphere, and the stratosphere temperature gradient is quite slight near the tropopause.
However, as the CO2 concentration increases from around current levels, the main changes in the emission spectrum are in the wings of the 13 – 18 micron CO2 “well”, where the CO2 emission layer (viewed as wavelength dependent) are in the troposphere. See the last diagram of this page on the Barrett-Bellamy website. As the CO2 concentration increases from around current levels, this part of the CO2 emission layer, the crucial bit that outweighs the changes elsewhere in the emissions spectrum from CO2, ascends to a colder place and emits less — which is why increasing the CO2 concentration decreases OLR emissions form the CO2 molecules.
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“Yes Gai, the stratosphere warms as you go higher. This can even been seen in the OLR spectrum on say a Nimbus diagram (Fig 2 of post 14) — see the spike up at exactly 15 microns, indicating a higher emission temperature around 230 or even 240 K (hard to tell, Nimbus resolution limited”
The 35km of stratosphere above the ‘pause add little to exitance at any wavelength. Small fraction of an optical depth. No mass! The radiometer looks laterally through 300km of the stuff to get a radiance/temperature.
Very deceitful!! The strato, meso and thermosphere together add less than 8%.
All the best! -will-
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Sorry, Dr. Evans, but ‘These are two MODTRAN generated emission spectra’
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MODTRAN is an atmospheric radiative transfer model developed by Spectral Sciences Inc. and the US Air Force Research Laboratory.
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You are basing your statement on a MODEL, but that model is not what the real experimental data is actually saying!
“The [MODTRAN] spectra represent the emission towards space at an altitude of 15 km” That’s ABOVE the TROPOPAUSE!
Pressure measures the number of molecules per unit volume. Adding CO2 is not going to do much about the height (pressure) at which the captured energy is being released as OLR, since the time to radiate is about ten times slower than the time to the next collision in the troposphere. As the pressure goes down the number of molecules per unit volume goes down until fimally the number of molecules is low enough to allow energy transfer via radiation to occur instead of by collision. This I think is the critical point, since the modeled wings are based on PRESSURE BROADENING. As I mentioned before, Dr Happer found the wing shape in the climate models was all wrong.
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Slide 22: Lorentzian line shape nor Voigt line shapes are correct in the far wings! — the experimental data shows less broadening.
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When you think about it, Dr. Happer’s findings makes sense, since you are talking about radiating at 47 km above the surface. 80% of the atmosphere is below 16 km. The stratosphere is 1/100 of the pressure at sea level above 30 km in the area CO2 is radiating. I wonder if the models used standard sea level presure to come up with the line broading formulas.
http://www.atoptics.co.uk/highsky/images1/shim18.gif
from
http://www.atoptics.co.uk/highsky/htrop.htm
Dr. Happer’s experimental data shows barely any radiation at 11 KM (at 400 ppm CO2) and that CO2 is mainly radiating in the stratosphere ~ 47 KM above the surface, well above the tropopause. The Uherek, 2006 figure shows this as well.
So this comes down to the height of the tropopause which varies between 8 km near the poles (cold air) and 18 km over the monsoon areas during the monsoon (warm moist air).
Again it would seem the correlation is for warmer, not colder, if the height (11 km ) does not change–which of course it would since it is dependent on pressure (the number of molecules per unit volume), and cold air is going to be more dense in the troposphere.
http://www-das.uwyo.edu/~geerts/cwx/notes/chap01/trop_height1.gif
Sure makes you wonder just how well they are describing this critical part of the CO2 modeling where the oh so important CO2 ‘wings’ is playing ducks and drakes with the tropopause.
I do understand that you are trying to show that as the CO2 pipe is constricted by adding more CO2 the other three pipes will take up the slack. This comment on the lack of additional pressure broadening from Dr. Happer and the location at which those ‘wings’ are emitting (in the tropopause or stratosphere) is just another couple of nails in the coffin for this climate model.
It is amazing how sloppy the work on this all important basic model is.
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Gai, in this series, as explained in the Introduction, I’m going with establishment (AR5) values for all parameters.
AR5 gives the all-important figure of 3.7 W/m2 less OLR from CO2 per CO2 doubling from a 1998 paper, and the precision is low — 3.5 to 4.1 W/m2. Doesn’t sound like it’s very settled or being worked on. Wouldn’t exactly surprise me if it was lower than that.
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Thanks Dr. Evans. I do realize you are doing as Lord Monckton does, use their numbers to show where their logic is all mucked-up.
I am just adding a couple more nails to the coffin. Not that that seems to keep the monster contained. I think we need a steak (T-bone?) and some garlic to really kill it.
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Zombie science indeed.
The whole episode of MMGW is an excellent demonstration about how a refuted idea can be sustained by government funding.
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gai October 29, 2015 at 10:11 pm
“Sorry, Dr. Evans, but ‘These are two MODTRAN generated emission spectra’”
Extremely deceitful spectra
“MODTRAN is an atmospheric radiative transfer model developed by Spectral Sciences Inc. and the US Air Force Research Laboratory.”
This is a complete deliberate lie for monetary gain! The collected and verified HiTran database at a truly obscene price paid by all the armed forces of the US! This was all done to define atmospheric seeing at wavelengths from 0.1 to 200 microns. The seeing is limited by both absorption and scattering of any amplitude or spatial modulation of EMR from the far source. The total flux at all wavelengths from Germany’s Fulda Gap give no information as to what can be detected there. The spectral, spatial, and amplitude modulations have all the information. The HiTran database contains that physically verified, not theoretical, information in painful, deadly, detail. So sad that it is now now uses for scam and profit. Spectral Sciences Inc had absolutely nothing to do with it. Both ModTran and FasTran were developed at the Air Force Cambridge labs as LowTran (the band database), while fast had but little band edge detail, gave clumsy numbers, While Hitran, very detailed, would never finish before the computer crapped out! NEVER EVER USE THE WORD CRASHED WITH THE AIR FORCE More later maybe!
All the best! -will-
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Darn it Will,
You toss out bits and pieces and give us a glimpse but never a whole story!
How about a more coherent comment that those of us who do not have your base knowledge can follow?
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