I’m taking a blog holiday this weekend. Right now I’m watching the History channel 2 (H2) while some off the rails eco-scientist explains to us why we are all going to die because of “what might happen if a gigantic methane-gas explosion occurred in the Pacific.” Methane Explosion (2007) watch the video:
I had to laugh at the YouTube description (bold mine):
History Channel “Mega Disasters” series. This explores the controversial paper published by Northwestern University’s Gregory Ryskin. His thesis: the oceans periodically produce massive eruptions of explosive methane gas… enough to cause global catastrophe on a regular basis!
Discuss the methane explosions or whatever you like, within site policy. If you want to submit a guest post, flag a moderator.
WUWT will return to its regularly scheduled programming Sunday evening.
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From Doug Cotton on February 13, 2012 at 9:46 pm:
Huh? As previously presented here on WUWT in the Ira Glickstein, Phd series of guest posts, Visualizing the “Greenhouse Effect” (one, two, three, four, five), the important longwave infrared wavelength bands when contemplating “back radiation” (see #2) are around 7 microns (micrometer, 10^(-6) meters, 5 to 8𝜇 range, centered at about 7𝜇), 10𝜇 (8-13𝜇), and 15𝜇 (13-30𝜇). 10𝜇 is the “atmospheric window” through which the longwave is given a free pass on the way to outer space. 7𝜇 is a water vapor absorption band, and 15𝜇 is where both water vapor and CO₂ absorb. After absorption, energy can be emitted in the 7, 10, and 15𝜇 bands, irregardless of whether it was 7 or 15𝜇 that was absorbed (with the different possible modes of molecular vibration, multiple lower-energy longer-wavelength photons can be absorbed before the emitting of a higher-energy shorter-wavelength photon, a higher-energy photon can be absorbed with multiple lower-energy photons released, although it’s most likely there’ll be an absorption/re-emitting of a single photon at the same wavelength). Only 15 to 30% of the upgoing (upwelling) thermal radiation is transmitted through the atmosphere, nearly all through the 10𝜇 window.
I found a definition of emissivity here, pretty well matches the one in the Wikipedia entry:
I then dug up a paper, E. Péquignot et al, 2008, Infrared Continental Surface Emissivity Spectra Retrieved from AIRS Hyperspectral Sensor, pdf here. It looks at emissivity spectra from 3.7 to 14𝜇 in 11 selected regions in Africa. The emissivity spectra graphs are presented in Fig. 7 spanning pages 1630-31 (pdf pages 12 and 13). Emissivity values for the different wavelengths are generally around 0.95 with a noted dip roughly around 8.3 to 9.3𝜇, the deepest one being zone 03 which goes down to around 0.65 emissivity.
Since the paper is looking at thermal radiation going from the surface through the atmosphere as detected by satellites, your much-lower value of 20% emissivity through the atmosphere doesn’t make sense. As mentioned in the Wikipedia emissivity entry, you might be specifying the total emissivity (the “gray body assumption”) rather than that of the thermal radiation wavelengths under discussion. “20%” as a number is inline with Ira Glickstein’s presented value of 15 to 30% of the upwelling thermal energy being transmitted through the atmosphere, except as you’ve used the value it appears you believe 1 minus 20% (=80%) of the thermal energy is lost to space, in opposition to what he said.
Thus I must conclude that when you state “…the emissivity of the atmosphere is only about 20%…”, you really don’t know what the heck you’re talking about.
I was quoting a Professor of physics and of course was using the term as the fraction that is emitted v. that emitted by a perfect blackbody. If you wish to continue to believe that those white clouds and thin atmosphere act like a perfect blackbody producing 100% “cavity” emission, then keep to your beliefs and just consider the two points in today’s post which are independent of the amount of radiation from the atmosphere. But you would do well to remember that a perfect blackbody has to be perfectly insulated from its surrounds so that there can be no transfer of thermal energy by diffusion, conduction, convection, phase change etc. That means your carbon dioxide molecules can’t share their thermal energy with any other air molecules, because then they won’t have that energy to re-emit.
In the real atmosphere there is plenty of thermal energy transfer by diffusion and convection for a start. So it is totally impossible that a layer of the atmosphere acts like a perfect blackbody with emissivity 1.0.
Furthermore, if there is a warmer layer below it, then correct application of SBL (including multiplying by the emissivity) means that net radiation would be from the warmer layer to the cooler one above.
In short, there is no way that the atmosphere can radiate downwards anything like the amount shown in energy diagrams. No actual experiments have ever found any warming effect from backradiation. You cannot point me to one, yet the IPCC should have tested such before going off into Greenhouse Land – the title of the book I’m now publishing.
So, if you treat the atmosphere’s emissivity as being 1.0 (highly improbable) then absolutely no energy is left up there to do any warming of other air because you are calling it a perfect blackbody. In reality it is no more like a true blackbody than is the surface, which also sheds energy by all the above processes, as well as what’s left by radiation.
Your problem is that you work with formulas for which you don’t fully understand the prerequisites and conditions necessary for such formulas to apply. You appear to be indoctrinated by climatologists who do the same when they assume they can apply Stefan-Boltzmann Law to the Earth’s surface.
If you had had 50 years experience in studying and teaching physics as I have, then maybe you’d have a better feel for things.
For the moment, though, you have a lot to learn, so I suggest you start by thinking about the two issues in my earlier post today. I’m only trying to help you and others understand reality.
PS Kadaka: Your statement Since the paper is looking at thermal radiation going from the surface through the atmosphere as detected by satellites, your much-lower value of 20% emissivity through the atmosphere doesn’t make sense. shows me that you don’t think clearly about a physical situation. Sure, when you look at any planet from space it radiates very close to a perfect blackbody with emissivity 1.0. But that radiation includes some from its surface, plus all the reflection from its atmosphere and its surface. If reflection is, say, 30% then emissivity is reduced 0.7. But that includes all the elements in the surface radiating away, plus water vapour and at least 50 gases in the atmosphere. It is highly unlikely that the atmosphere would radiate with emissivity as high as 0.7, but even if it did at least half would be space-bound.
Doug Cotton says (February 14, 2012 at 1:59 pm): “How about addressing these issues ..
(1) The direction of net radiative energy flow can be the opposite of the direction of heat transfer. if you have a warmer object (say 310 K) with low emissivity (say 0.2) and a cooler object (say 300 K) with much higher emissivity (say 0.9) then net radiative energy flow is from the cooler to the warmer object. Yet the Second Law says heat transfer is from hot to cold.”
If emissivity is less than 1, then so is aborbance. Why not make it easy and calculate for black bodies, emissivity = 1?
Aaaargh! The word “aborbance” above should be “absorbance”.
Although “aborbance” does have a certain ear appeal…
Gary Hladik says:
February 14, 2012 at 6:51 pm
Aaaargh! The word “aborbance” above should be “absorbance” (sic)
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Actually it should have been “absorptivity.” But we already know that you don’t know much physics. Your post also indicates you don’t understand my point, as others will realise.
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@JDN says:
February 11, 2012 at 12:09 pm
“I keep reading that India is going to try thorium-based nuclear reactors. However, I don’t think GE or Westinghouse even have a prototype under design. This makes me think it’s not workable for some reason, even though India & China have allegedly approved construction. I did a little searching and couldn’t find much. Does anyone have something that looks like a technical analysis of a prototype thorium reactor?”
General Atomic in San Diego built the Fort St Vrain HTGR in Colorado. This high temperature gas cooled reactor used fuel containing both uranium and thorium. The reactor has been shut down and the fuel has yet to be reprocessed. Someday… thorium will power the world.
Doug Cotton says (February 14, 2012 at 7:35 pm): “Actually it should have been “absorptivity.” But we already know that you don’t know much physics. Your post also indicates you don’t understand my point, as others will realise.”
http://en.wikipedia.org/wiki/Absorptivity
Doug, I don’t think God Himself understands your point.
Or should that be “pointity”?
Gary Hladik says:
February 14, 2012 at 10:11 pm
Doug, I don’t think God Himself understands your point.
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Well I’ll try to explain it better to you, or God, or whoever.
For the “hot” and “cold” bodies respectively, insert these values for temperature (310K and 300K) and emissivity (0.2 and 0.9) into the Stefan-Boltzmann equation. Then explain how we can have a situation in which the colder body radiates more energy than the hot one (as you should have calculated) but heat transfer has to be the other way from the hot body to the cold one in order to obey the Second Law of Thermodynamics
Whenever AGW proponents post their “explanation” as to why it’s OK for a colder atmosphere to warm the surface, or slow the rate of cooling, they assume that net radiative flux is always in the same direction as net heat transfer, and I have just shown you why it may not always be so. They say something like, “We know the flux from the atmosphere to the surface is positive but the surface is cooling and so net flux is outwards and heat transfer is outwards so the Law has not been broken.” You have only to ask them what happens every sunny morning when the surface is already warming. When you add more flux from the cold atmosphere does that then cause extra warming? If it did it would break the Second Law.
You see, they can’t get out of their heads the concept that “heat” always flows in the direction of (most of) their lovely little mass-less photons which can’t help but heat up anything they collide with. That is not what the laws of physics say will happen, and there is absolutely no empirical evidence showing it happen anywhere in the World.
And that’s why the atmospheric radiative greenhouse conjecture is mistaken.
Gary Hladik says:
February 14, 2012 at 10:11 pm
Doug Cotton says (February 14, 2012 at 7:35 pm): “Actually it should have been “absorptivity.”
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I stand by my statement because it is preferable to use “absorptivity” as the opposite of “emissivity.”
Absorbance (with a “b”) has a different meaning altogether, so the item in Wikipedia which you linked is in error. In fact it is contradictory to the Wikipedia entry they linked which reads…
Absorptance[5] (not absorbance) is defined as: The ratio of the radiant flux absorbed by a body to that incident upon it. Also called [absorption] factor. Compare absorptivity.
Note the all-important “p” in absorptance which gives it an altogether different meaning to absorbance. (See Wikipedia “absorbance’)
Doug Cotton says (February 15, 2012 at 1:26 am): “Absorbance (with a “b”) has a different meaning altogether, so the item in Wikipedia which you linked is in error.”
OK, I stand corrected.
Doug Cotton says (February 15, 2012 at 1:08 am): “Well I’ll try to explain it better to you, or God, or whoever.
For the ‘hot’ and ‘cold’ bodies respectively, insert these values for temperature (310K and 300K) and emissivity (0.2 and 0.9) into the Stefan-Boltzmann equation. Then explain how we can have a situation in which the colder body radiates more energy than the hot one (as you should have calculated) but heat transfer has to be the other way from the hot body to the cold one in order to obey the Second Law of Thermodynamics”
I used the calculator here:
http://www.endmemo.com/physics/radenergy.php
1 m^2 radiator at 300 degrees K, emissivity 0.9 radiates 413.343 W
1 m^2 radiator at 310 degrees K, emissivity 0.2 radiates 104.727 W
Assuming for simplicity that all of each body’s radiation is directed toward the other,
Cooler body of emissivity 0.9 absorbs 0.9 x 104.727 W = 94.254 W
Warmer body of emissivity 0.2 absorbs 0.2 X 413.343 W = 82.669 W
i.e. the cooler body does not transfer more energy to the warmer than it receives, and God and I still don’t understand your point.
Doug Cotton’s 4 questions are (1) What happens to the extra radiation that the warmer body does not absorb because of its low absorptivity? (2) Does the warmer body convert to thermal energy any of the remaining radiation from the cooler body? If not all, what happens to that radiation? (3) Is the Second Law of Thermodynamics obeyed or does the warmer body get warmer still? (4) Does heat transfer from the cooler one to the warmer one, or vice versa?
As pointed out above for (1), the absorption is based on the spectrum of the radiation and the surface of the body that the radiation is impacting. Extra radiation is reflected which does not depend on the temperature of the body. For (2) some is absorbed and some reflected. (3) the Second Law is obeyed since the warmer body does not warm but cools more slowly. (4) Net heat transfer is from warmer to cooler, but slower transfer due to the cooler body being above 0K.
Doug Cotton says “You have only to ask them what happens every sunny morning when the surface is already warming. When you add more flux from the cold atmosphere does that then cause extra warming? If it did it would break the Second Law.”
The flux from the warm atmosphere (relative to space) lowers the net heat outflow from the surface.
Eric (skeptic) says:
February 15, 2012 at 3:42 am
The flux from the warm atmosphere (relative to space) lowers the net heat outflow from the surface.
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What “net heat outflow” from the surface? How can there be a net heat outflow when the surface is getting hotter at 11am? Go for a walk on the beach and see if the sand gets hotter and hotter on a warm sunny morning. It does in my part of the world.
Next time, when you don’t understand my points, don’t bother to write about something different and irrelevant. I think like a physicist – obviously you don’t.
The Second Law of Thermodynamics cannot work anywhere where there is matter in the universe unless the radiation from cold to hot has absolutely no effect and only the radiation from hot to cold leads to the conversion of the energy in that radiation to thermal energy in the colder body. There is no such conversion when radiation from a colder body meets a warmer one. “Heat” does not “travel” in the direction of net radiative flux, as I have demonstrated above, and because it does not do so, the veracity of the first statement is confirmed.
Gary Hladik says:
February 15, 2012 at 3:14 am
1 m^2 radiator at 300 degrees K, emissivity 0.9 radiates 413.343 W
1 m^2 radiator at 310 degrees K, emissivity 0.2 radiates 104.727 W
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And yet you can’t subtract these two values and see that there is net radiation from cold to hot?
My point was just that. It is possible to have a situation like this where net radiation is in the opposite direction to heat transfer. The energy diagrams always assume heat transfer is in the direction of net radiative flux, now don’t they? So they are wrong. Simple as that.
I don’t dispute that heat transfer is from hot to cold, but I do dispute your subsequent calculations, because there is no component of heat transfer due to the radiation from cold to hot. If you blocked that radiation with some filter somehow, the radiation from hot to cold would still be as great and the hot body would cool just as fast and no faster – as does the Earth’s surface when I block “backradiation” in my backyard experiment. You may not like my experiment, but you cannot point me to any contrary empirical evidence, and you could always try it yourself with sand in two identical wide necked vacuum flasks, one shielded from the sky at night and the other not,
You will note that in my earlier post today (which I now refer you to) I am careful to talk about conversion of radiated energy to thermal energy, rather than use the word “absorb.” It may well be debatable as to whether “scattering” amounts to absorbing and immediately re-radiating, but the effect is the same, and no extra thermal energy is left behind.
When considering heat transfer, as I have explained in the first post today, only the radiation from hot to cold is relevant and the following is the physical process (which Prof Claes Johnson also confirmed in a different way computationally):-
Look at the first plot here http://scienceworld.wolfram.com/physics/WiensDisplacementLaw.html
and note that the distribution for a lower temperature is always contained within that for a higher temperature. Thus radiation from a cooler object will always be able to resonate with a warmer object and it does so. It is scattered by this resonating process and no energy left behind. In contrast, when radiation from a warmer body hits a cooler one, there will always be frequencies above the upper limit of those which will resonate in the cooler one. Because of this, the energy in those higher frequencies is converted to thermal energy, as happens when solar insolation hits the surface. (There will be more in my book, Greenhouse Land but this is a brief outline.)
If this were not so the Second Law of Thermodynamics could not operate in all cases.
Doug, I think like an engineer. There is not much point in discussing a warm sunny morning since back radiation doesn’t really matter then. The better example is at night. Do you agree that clouds generally make it warmer at night? The clouds are almost always colder than the surface. The sign of that difference never matters although the cloud temperatures matter. If downward radiation is not what is keeping it warmer, what is?
When solar radiation (UV, visible and IR etc) travels through space we do not know what its end effect will be until it strikes something. We will observe its effect and say – there’s some light from the Sun – but it may be more light if it hits a white surface than a dark surface, as a camera exposure meter will confirm. It may generate thermal energy (more or less depending on what it strikes) or it may appear as light as it starts to penetrate the oceans, but end up as thermal energy in the deeper depths. Of course some will be reflected or scattered and strike another target sooner or later, and another etc.
My point is, “heat” is the transfer of thermal energy, but thermal energy is not a fixed amount of energy travelling along with radiation. The energy in the radiation has to go through a physical process of being converted to thermal energy. This happens only for those frequencies in the radiation which are above the natural frequencies that can be emitted by the target. (The hotter the source of spontaneous radiation, the higher will be the peak frequency.) So solar radiation can be converted to thermal energy in the Earth’s surface, but radiation emitted from a cooler atmosphere cannot be converted to thermal energy in a warmer surface. “Heat” only appears to be transferred (and only from hot to cold) because only radiation from hot to cold will be converted to extra thermal energy in the target.
It does not matter whether you are increasing the rate of warming in the morning or decreasing the rate of cooling later in the day, you still need extra thermal energy to do this. You cannot get this extra thermal energy from a cooler atmosphere, morning or evening. You cannot say the Second Law is not broken because of the direction of net radiation or net heat flow. All that matters is, what actually happens between any two points – one point on the surface and one point in a cooler atmosphere. What goes on between other “points” – a point on the Sun and another point on the surface is irrelevant. The Second Law must apply between any two points.
Doug, regarding your statement “This happens only for those frequencies in the radiation which are above the natural frequencies that can be emitted by the target. (The hotter the source of spontaneous radiation, the higher will be the peak frequency.)”
What happens when a frequency from the source exactly matches the natural frequency of the target? What happens when the source frequency is 0.000001% above the natural frequency of the target and what happens when it is 0.000001% below? Will there be a discontinuity?
Also please answer if clouds at night will keep the surface warmer and if so, how.
Doug Cotton says (February 15, 2012 at 2:26 pm): “And yet you can’t subtract these two values and see that there is net radiation from cold to hot?”
Um, you mean “from cold toward hot”. The warm body only absorbs 20% of the incident radiation, so the net flow is warm to cool. Your experimental conditions, not mine.
“You will note that in my earlier post today (which I now refer you to) I am careful to talk about conversion of radiated energy to thermal energy, rather than use the word “absorb.” It may well be debatable as to whether “scattering” amounts to absorbing and immediately re-radiating, but the effect is the same, and no extra thermal energy is left behind.”
It doesn’t matter if you used the word “absorb” or not. You specified emissivities in your thought experiment, and thus specified absorptivities. Your own experiment requires that the warm body absorb 20% of the incident radiation, after scattering, reflection, etc. Once the energy is absorbed, you’re telling us that it has no effect on the warm body. You have, in fact, destroyed energy. Not good, Doug.
“…and you could always try it yourself with sand in two identical wide necked vacuum flasks, one shielded from the sky at night and the other not,”
Except that I have nothing to gain by proving (again) what (nearly) everybody already knows. You, on the other hand, have everything to gain. Why not perform Dr. Spencer’s “Yes, Virginia” thought experiment for real? If it comes out as you predict, it’s pretty much a guaranteed Nobel Prize. I look forward to your upcoming publication in a peer-reviewed journal.
Eric (skeptic) says
February 15, 2012 at 4:36 pm
What happens when a frequency from the source exactly matches the natural frequency of the target? What happens when the source frequency is 0.000001% above the natural frequency of the target and what happens when it is 0.000001% below? Will there be a discontinuity?
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Good question. There is resonance or near resonance. The concept of near resonance is explained in more detail by Prof Claes Johnson – see link to his paper on the ‘Radiation’ page on my site http://climate-change-theory.com
Basically that’s what it’s all aboout. The radiation resonates with the target and is not absorbed in the sense that there is no conversion to thermal energy. Higher frequencies that cannot resonate cause a type of “chaos” which leads to conversion to thermal energy, rather like when x-rays cause ionization.
Gary Hladik says:
February 15, 2012 at 5:27 pm
it is you who assumed absorptivity = emissivity in this case, but there is no equilibrium while the objects are both still cooling.
In any event, the empirical measure of absorptivity in any particular situation already takes into account the fact that the portion of incoming radiation which is from a cooler source is scattered if not already reflected. You are assuming you can apply the ratio which applies for the total of all radiation (SW and LW) to just the LW from a cooler source. But that is a portion of the total which has already been deducted in determining the absorptivity.
If the source of radiation is cooler there is resonance and no energy from that radiation is converted to thermal energy. (See post above re resonance and near resonance.)
You may consider it absorbed and re-emitted with the same energy and frequency spectrum if you wish. I will consider it just scattered. But either way there is no effect on thermal energy levels in the target.
Your concept of absorbing implies conversion to thermal energy. This only happens if the source is warmer than the target. How many times do I have to repeat this? If you disagree with the mathematical proof of this (by a Professor of Applied Mathematics) then go and argue with Claes Johnson on his website blog.
The Earth’s surface absorbs incident solar radiation and converts it to thermal energy. That energy is not necessarily re-radiated immediately by the surface. In fact it may never be radiated because it may exit by evaporation, diffusion, conduction etc.
The surface converts SW to thermal energy (but not LW from a cooler atmosphere) and then some of that thermal energy subsequently exits as LW radiation if it hasn’t already done so by other processes, which more than half does.
The conjecture that “backradiation” (or even reflected radiation sent back to its source) can warm the original source (which is warmer than the source of the backradiation) is not to be found anywhere in conventional physics papers or textbooks.
It is something invented by climatologists. Yet, despite its lack of status as any kind of established theory, the IPCC launched it upon the world without so much as a single empirical experiment to confirm it happened. Yet they call it science, and say the science is settled.
Now even if people like Gary are not prepared to test it themselves (as I have and found it failed) then at least you would think there ought to be some experiment somewhere that the IPCC was hanging its hat on. But there isn’t.
End of story. End of Greenhouse fantasy.
PS Gary: This paper found a temperature dependence for absorptivity of 1% / K … http://icecube.berkeley.edu/~bprice/publications/T-dependence.pdf
Doug Cotton says (February 15, 2012 at 8:49 pm): ‘[snip more handwaving than the Rose Parade]
If you disagree with the mathematical proof of this (by a Professor of Applied Mathematics) then go and argue with Claes Johnson on his website blog.”
Math isn’t proof. Experiments are proof. Do the experiment, Doug, or help Johnson do it, and put an end to the arguments. Publish the results, and claim your Nobel Prizes. Until then you’re just flapping your gums.
I have done the experiment, Gary, and it failed to show any slowing of the cooling process due to backradiation. It will be published in an Appendix in my book, but Johnson hasall the credit and I am not taking anything away from him. He has solved a problem which baffled Planck and Einstein.
Now, you try it. Fill two wide necked identical vacuum flasks with sand and, throughout the night, shield one from nearly all the backradiation – just allowing air to escape by convection around the edge of the shield which should be at a slight slope to ensure this. My temperatures were measured to 0.1 deg.C and were identical just before dawn.