Guest Post by Ira Glickstein
Solar “light” radiation in = Earth “heat” radiation to Space out! That’s old news to those of us who understand all energy is fungible (may be converted to different forms of energy) and energy/mass is conserved (cannot be created nor destroyed).
My Visualizing series [Physical Analogy, Atmospheric Windows, Emission Spectra, and Molecules/Photons] has garnered almost 2000 comments, mostly positive. I’ve learned a lot from WUWT readers who know more than I do. However, some commenters seem to have been taken in by scientific-sounding objections to the basic science behind the Atmospheric “Greenhouse Effect”. Their objections seemed to add more heat than light to the discussion. This posting is designed to get back to basics and perhaps transform our heated arguments into more enlightened understanding :^)
As I’ve mentioned before, during my long career as a system engineer I’ve worked with many talented mathematical analysts who always provided precise results, mostly correct, but some precisely wrong, usually due to mistaken assumptions. I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on their results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. In those days my analysis was really done using a slide rule and scrap paper, but I now use spreadsheets.
The graphic above is based on an excellent spreadsheet from http://serc.carleton.edu/files/introgeo/models/mathematical/examples/XLPlanck.xls. It uses Planck’s Law to calculate the black body radiation spectrum from the Sun, as observed at the top of the Earth’s Atmosphere. It also may be used to calculate the radiation spectrum from the Earth System (Atmosphere and Surface, see below for explanation) at any assumed temperature. (I will refer to this spreadsheet as “Carleton” in this posting.)
I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).
Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?
The yellow curve (above left) shows that Solar radiation is in a tall, narrow “shortwave” range, from about 0.1μm (microns, or millionths of a meter) to about 4μm, which we call ultra-violet, visual, and near-infrared. The vertical axis is Intensity of the radiation, measured in Watts/m^2/μm, and the horizontal axis is Wavelength, measured in μm. If you divide the area under the yellow curve into vertical strips, and add up the total area, you get 240 Watts/m^2.
Since we humans sense the visual portion of this radiation as “light”, that is the name we give it, and that has led to the false assumption that it contains no “heat” (or “thermal”) energy.
The violet curve (above right) shows that, assuming a mean temperature of 255 K, Earth System radiation to Space is in a squat, wide “longwave” range, from about 5μm to beyond 40μm, which we call mid- and far-infrared. If you divide the area under the violet curve into vertical strips, and add up the total area, you get the same 240 Watts/m^2 as is under the yellow curve.
DETAILED EXPLANATION
The graph on the left shows the actual observed Solar radiation spectrum (in red) as measured at the top of the Atmosphere. It is superimposed on a black body model (in blue) showing very good correlation. Thus, while the Sun is not exactly a black body, it is OK to assume it is for this type of “sanity check” exercise.
If you calculate the area under the curve you get about 1366 Watts/m^2. That means that a square meter of perfect black body material, held perpendicular to the Sun, would absorb 1366 Watts.
However, the Earth is not a perfect black body, neither is it a flat surface perpendicular to the Sun! So, to plot the yellow curve at the top of this posting, I had to adjust that value accordingly. There are two adjustments:
- The Earth may be approximated as a sphere, with the Sun shining on only half of it at any given time. The adjustment factor for this correction is 0.25.
- The albedo (reflectiveness) of the Earth system, primarily clouds and light-colored areas on the Surface such as ice, causes some of the Solar radiation to be reflected back out to Space without contributing any energy to the Earth System. The adjustment factor for this correction is 0.7.
After applying these adjustments, the net Solar energy absorbed by the Earth System is 240 Watts/m^2.
The graph on the right shows the black body model for an Earth System at a mean temperature of 255 K, a temperature that results in the same 240 Watts/m^2 being emitted out to Space.
Of course, the Earth System is not a perfect black body, as shown by the graph in the upper panel of the illustration below, which plots actual observations from 20 km looking down. (Adapted from Grant Petty, A First Course in Atmospheric Radiation, Figure 8.2, http://www.sundogpublishing.com/AtmosRad/Excerpts/index.html.)
The actual measured radiation is the dark squiggly curve. Note that it jigs and jags up and down between the topmost dashed curve, which is the black body spectrum for a temperature of 270 K and a lower dashed curve which is the black body spectrum for 230 K. This data was taken over the Arctic, most likely during the daytime. The Petty book also has a graph looking down from over the Tropical Pacific which ranges from 300 K down to 210 K. Observations will vary by tens of degrees from day to night, summer to winter, and Tropical to Polar.
However, it is clear that my result, based on matching 240 Watts/m^2, is within a reasonable range of the true mean temperature of the Earth System as viewed from Space.
NOTE ABOUT THE ABOVE ILLUSTRATION
WUWT readers will notice some apparent inconsistencies in the graphs above. The top and bottom panels, from Petty, peak at 15μm to 20μm, while the purple, blue, and black curves in the middle panel, and the Earth System curves from the Carleton spreadsheet I used (see above) peak in the 9μm to 11μm range. Also, the Petty black body curves peak at a “Radiance” around 100 mW/m^2/sr cm^-1 while the black body curves from Carleton peak at an “Intensity” of around 14 W/m^2/μm. Furthermore, if you look closely at the Petty curves, the labels on the black body curves are mirror image! What is going on?
Well, I know some of the reasons, but not all. (I hope commenters who are more fluent in this than I am will confirm my explanations and provide more information about the differences between “Radiance” and “Intensity”.) I have Googled and Wikied the Internet and am still somewhat confused. Here is what I know:
- The horizontal axis in Petty’s plots are what he calls “Wavenumber”, increasing from left to right, which is the number of waves that fit into a cm (centimeter, one hundredth of a meter).
- This is proportional to the frequency of the radiation, and the frequency is the inverse of the wavelength. Thus, his plots are the mirror image of plots based on wavelength increasing from left to right.
- The spreadsheet I used, and my previous experience with visual, and near-, mid-, and far-IR as used in military systems, always uses wavelength increasing from left to right.
- So, when I constructed the above illustration, I reversed Petty’s curves, which explains why the labels on the black body curves are mirror image.
- Fortunately, Petty also included a wavelength legend, which I faithfully reproduced, in non-mirror image, at the top of each plot.
But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet. I tried to reproduce Petty’s blackbody curves by multiplying the Carleton values by the wavelength (μm) and that did not move the peak to the right enough. So, I multiplied by the wavelength again (μm^2) and, voila, the peaks agreed! (I hope some WUWT reader will explain why the Petty graphs have this perverse effect. advTHANKSance!)
ANSWERING THE OBJECTIONS TO BASIC ATMOSPHERIC “GREENHOUSE EFFECT” SCIENCE
First of all, let me be clear where I am coming from. I’m a Lukewarmer-Skeptic who accepts that H2O, CO2 and other so-called “greenhouse gases” in the Atmosphere do cause the mean temperature of the Earth Surface and Atmosphere to be higher than they would be if everything was the same (Solar radiation, Earth System Albedo, …) but the Atmosphere was pure nitrogen. The main scientific question for me, is how much does the increase in human-caused CO2 and human-caused albedo reduction increase the mean temperature above what it would be with natural cycles and processes? My answer is “not much”, because perhaps 0.1ºC to 0.2ºC of the supposed 0.8ºC increase since 1880 is due to human activities. The rest is due to natural cycles and processes over which we humans have no control. The main public policy question for me, is how much should we (society) do about it? Again, my answer is “not much”, because the effect is small and a limited increase in temperatures and CO2 may turn out to have a net benefit.
So, my motivation for this Visualizing series is not to add to the Alarmist “the sky is falling” panic, but rather to help my fellow Skeptics avoid the natural temptation to fall into an “equal and opposite” falsehood, which some of those on my side, who I call “Disbelievers”, do when they fail to acknowledge the basic facts of the role of H2O and CO2 and other gases in helping to keep temperatures in a livable range.
Objection #1: Visual and near-visual radiation is merely “light” which lacks the “quality” or “oomph” to impart warmth to objects upon which it happens to fall.
Answer #1: A NASA webpage targeted at children is sometimes cited because they say the near-IR beam from a TV remote control is not warm to the touch. Of course, that is not because it is near-visual radiation, but rather because it is very low power. All energy is fungible, and can be changed from one form to another. Thus, the 240 Watts/m^2 of visible and near-visible Solar energy that reaches and is absorbed by the Earth System, has the effect of warming the Earth System exactly as much as an equal number of Watts/m^2 of “thermal” mid- and far-IR radiation.
Objection #2: The Atmosphere, which is cooler than the Earth Surface, cannot warm the Earth Surface.
Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.
Objection #3: The Atmospheric “Greenhouse Effect” is fictional. A glass greenhouse works mainly by preventing or reducing convection and the Atmosphere does not work that way at all.
Answer #3: I always try to put “scare quotes” around the word “greenhouse” unless referring to the glass variety because the term is misleading. Yes, a glass greenhouse works by restricting convection, and the fact that glass passes shortwave radiation and not longwave makes only a minor contribution. Thus, I agree it is unfortunate that the established term for the Atmospheric warming effect is a bit of a misnomer. However, we are stuck with it. But, enough of semantics. Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.
Dave Springer says:
What? You must be confusing me with someone else who tries to correct your misstatements.
RJ says:
It’s Aussie-speak for a barbeque. See http://www.australianbeers.com/culture/barbies.htm
Joel Shore says on May 17, 2011 at 11:59 am:
“O H Dahlsveen ———— at least preface your nonsense with some statement that this is how you imagine it works. Don’t pretend you know, because you rarely do.”
So you see me as you see yourself then Joel? Are you really telling me that the so-called “Hotspot” has got nothing to do with “Greenhouse Gas Warming”? If so, please expand on your claim:
”It has to do with the physics of the moist adiabatic lapse rate.”
Dave Springer, I’ll try to respond to the few things left in your recent posts that are not unintelligible gobbley-gook (or are at least less-unintelligible gobbley-gook) .
So, now you have made a full transition to the pseudoscience side and are endorsing the notion that there is no way the earth’s surface can emit more longwave radiation than the shortwave radiation that it absorbs from the sun? Try doing the calculation for Venus and see what you come up with. I really think that jae must have taken over your keyboard or something.
And, silly me to interpret the quote from a paper that says “net longwave emission” to mean “NET longwave emission”. After all, it makes perfect sense that something that is nearly a blackbody at a temperature of about 15 C will emit only 50 W/m^2 of emission (gross)…at least once you repeal a few laws of physics that were never much use to us anyway! It’s not like we could use the Stefan-Boltzmann Equation to calculate what it actually emits (gross)!
And, what does that prove exactly? That albedo affects climate? I’ll alert the media.
You really have gone off the deep end and I have no clue what points you are even trying to make anymore. That oceans affect climate? That water vapor is important? That albedo matters? I really have no clue.
RJ says: “But what does he mean by Barbie. What is it? So what does the slayer think the reader is reading if it is not incoming radiation because I do not know what a Barbie is.”
Don’t ask me why they picked it as their example “hot object”, but by Barbie (capitalized) they are referring to a Barbie doll. If it had been “barbie” it would make more sense to me, being slang for a barbecue (which could quite understandably be a good example of a hot object).
In either case you can easily substitute “rock” for “Barbie/barbie”. They are saying that if you point the IR thermometer at a rock, the reading represents the device’s calculated temperature of the surface. This is correct. What is incorrect is their claim that the “thermometer is not measuring the radiant energy you are receiving, it is measuring the resonance of the [rock’s surface].” The thermometer is, in fact, measuring the radiant energy received (in the device’s IR capabilities). It then uses this measurement, along with other measurements (e.g. distance), to calculate a temperature for the object it is pointing at.
Anyway Joel if you stand by with what you say on May 17, 2011 at 11:59 am i.e. :
“———. As a result, there is a gradient. And, the top layer emits more both in the upward and downward directions than the bottom layer, as it must since it is colder.”
Then I must assume the above statement is part of your own great knowledge and you can certainly say to yourself: “it is not embarrassing to admit my ignorance but it is embarrassing when I parade it around as knowledge.”
RJ when you said “Your $60 REMOTE thermometer is not measuring the radiant energy you are receiving, it is measuring the resonance of the Barbie.”, I tend to agree with you even though I have never owned an IR thermometer. From what I have gathered from other commenters, it reads the maximum possible radiant power not the actual flux. Just because it reads 390 Wm-2 when aimed at the 59 ºF surface does not mean that 390 Wm-2 is actually flowing, just that it has the capability if that surface at that temperature were in the void of space with nothing else to attenuate it, THEN, and only then, would the actual flux also be 390 Wm-2.
In other words, the temperature of the Barbie (barbeque for those ‘up above’) you read with an IR thermometer is the same if you are 6 feet away or 12 feet away, when you know it should be reading 1/4th at 12 feet away than when 6 feet away.
Is that not what you are saying? That is some of the fallacy that is going on in minds, people are thinking 390 Wm-2 upward when the reality is close to 65 Wm-2 upward. The 390 is the maximum power, the 65 is the actual flux upward.
So by KT97 if you let the surface be 289K as stated in TFK09 instead of 288K you can take:
67 Wm-2 absorbed SW by the atmosphere plus
24 Wm-2 carried upward by thermals (dry conduction/convection) plus
78 Wm-2 carried upward by evaporation (convection) plus
66 Wm-2 actual LW radiation flux upward (radiation)
———
235 Wm-2 detected LW upwelling by satellites above the TOA
Also 24+78+66 equals the 168 absorbed by the surface and 168+67 is the total SW absorbed by the system. See, the only fallacy numbers are the 390, 350 and 324 maximum radiant powers injected into the IR side of the graphic. So, scratch out 390 and put 66, scratch out 350 and replace with 26, scratch out 324 and replace with zero. Now all figures on the diagram are all FLUXES, not a mixture of fluxes and maximum radiant powers as if against empty space.
Tying in the other figures you have the 67+24+78+26−30 = 195 in the atmosphere less the 30 that K&T isolated and 165+30+40 = 235 or total flux to space. It all jives now and all numbers are now fluxes.
(RJ, I just realized that after reading your comment, it is maximum radiant powers mixed with actual fluxes that is the problem thrashing so many people’s minds and I owe it to your simple comment above)
I like simplicity, physics is simplicity, AGW proponents insist on complexity when it is not even needed. The 390, 350 and 324 should have been just a footnote on their graphic.
Joel – I’ve been thinking about a couple of things re statistical mechanics. Firstly, a comment you’ve made a few times I’ve noticed, that the equations are used, tried and tested by scientists and engineers who build things and you give this as a proof that this shows the science behind the approach is sound.
I’ve given before the example of the oximeter, that ‘uses’ the Beers-Lambert law, but then has to adjust it because the law is idealised and not of the real world, like using ideal gas laws. Because there is actually no such thing as an ideal gas, it is purely imaginary, various adjustments have to be made when making any calculations of real gases in the real world because real gases don’t obey the idea gas law at all temps and pressures. Scientists who actually build things, like engineers, know these limitations and know how to adjust the equations to include the missing parameters, so your comments on this aspect don’t actually relate back as proof that such laws are valid to describe the real world.
What I’ve seen often enough in AGW reasoning is Carbon Dioxide’s properties described as an ideal gas – which is how I came into these arguments – especially by one physics PhD who claimed that CO2 mixes thoroughly in the atmosphere because it behaves like an ideal gas; that a pool of CO2 on the floor in a room will therefore diffuse into the atmosphere of the room according to the random nature of the ideal gas without any work being done to move it, fan or open window and so on, and once thoroughly mixed cannot be unmixed without work being done. He said he would fail any of his students who said this was not how CO2 behaved in our atmosphere, as I was saying.
Secondly, in thinking about the descriptions of ‘heat as the net which is from hotter to colder because some heat is going from colder to hotter, how exactly?
What is the actual mechanism, what physical event decides that the ‘net’ will always come out as heat flowing from hotter to colder?
O H Dahlsveen says:
Well, you’ve got me there, although I don’t so much think it is a case of ignorance in that case but rather of typing something at variance to what I was thinking. Obviously, I meant to say that the top layer emits less.
Yes…The amplification of warming in the tropical troposphere is not specific to the mechanism of the warming being due to greenhouse gases. People ranging from Gavin Schmidt ( http://www.realclimate.org/index.php/archives/2007/12/tropical-troposphere-trends/ ) to Richard Lindzen (who says “The dominant role of cumulus convection in the tropics requires that temperature approximately follow what is called a moist adiabatic profile. This requires that warming in the tropical upper troposphere be 2-3 times greater than at the surface. Indeed, all models do show this, but the data doesn’t and this means that something is wrong with the data.” http://wattsupwiththat.com/2011/01/17/richard-lindzen-a-case-against-precipitous-climate-action/ )
Basically, the picture is this: Imagine two saturated parcels of air that ascend through the atmosphere, one being 1 deg warmer when it starts out at the surface. As it ascends, it expands and cools…but as it cools, some of the water vapor must also condense out and the warmer one will contain more water vapor and have more water vapor condense out as it cools than the colder one. The water vapor condensing out releases latent heat, which means that the air parcel no longer cools as quickly as if it were not saturated. However, the warmer one has more latent heat released than the colder one; hence, as it rises, it cools less rapidly than the colder one cools. This means that as you go up in the troposphere, the temperature difference you had between these two at the surface gets amplified.
Myrrh says:
For all forms of heat transfer, it basically has to do with the statistics of large numbers of events. For conduction, which occurs via molecular (or electronic) collisions, you will have collisions between molecules…and the molecules will have a range of velocities but ON AVERAGE, those coming from the hotter side of the object will have slightly higher speeds than those coming from the cooler side and so, in a large number of collisions, the molecules from the cooler side will gain speed on average and those from the hotter side will lose speed on average in the collisions.
For radiation, it is basically that the hotter object emits more radiation than the cooler object. That, coupled with Kirchhoff’s Law ( http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation ), guarantees that the heat flow will be from the hotter object to the colder object.
O H Dahlsveen says:
May 17, 2011 at 6:13 am
Do not make the mistaken assumption that the Atmosphere can be likened to a solid piece of, say steel which can easily be warmed more on one side than the other.
I agree, it is more like a pile of blankets – cold in the middle and warm on both the top and bottom. The question is – Why is it coldest in the middle? The interesting thing is that water vapor stays fairly high until it reaches the cold middle. My interpretation of the data is that water vapor is IR opaque until the tropopause and that CO2 is opaque until a bit higher. This model completely explains why the middle is colder than both the air above and below the tropopause.
Otherwise, I agree with your arguments. The problem is that they do not explain why the stratosphere gets warmer with height.
Dave Springer, THANKS for confirming that downwelling radiation from the Atmosphere to the Surface is real and may be conveniently measured.
On the other hand, I am surprised you referred to water vapor, CO2 and other so-called “greenhouse” gases as polluting the Atmosphere. I suppose you were being sarcastic about, for example, the EPA ruling that CO2 was a toxin, but I know you agree that CO2 and water vapor are natural components of the Atmospheric gases and, within reasonable limits, are beneficial.
Dave springer:
“If you can’t accept all these things as proof that back radiation is real then you’re a cause and if this was my blog I’d throw your ass out so to reduce the dimwitted clutter that pollutes the more informed converstations.”
Nice comment. You have all the markings of a “civil, tolerant progressive,” which is just as mythological as CAGW!
BTW, you will never find a place where I state that backradiation is not real, so your comment is not only insulting, but also dishonest.
Dave Springer says:
What you have referenced is not an actual measurement, Dave. It is a schematic plot. And you can’t take numbers off of a schematic plot. The mixed layer can extend for anywhere from 1 metre to a hundred metres. And yes, it’s called the mixed layer because a good chunk of it mixes every day. The citation you referenced says:
Note that the mixed layer is defined as the layer having no vertical temperature gradient, and it is defined as 1 to 100 metres deep. Now Dave, that mixed layer is being warmed at the top both day and night. In the day there’s a temperature gradient. In the night there’s no temperature gradient.
If there is no daily overturning in that mixed layer, and the mixed layer can only lose energy from the top … they why is there no temperature gradient in the night but there is a gradient in the day? The answer, of course is that the overturning actually is diurnal.
Yes, there is also a slow seasonal component to the mixed layer. It mixes deeper during one season and less during another, equalizing the entire mixed layer over the period of a year. But what that means is that some times of the year, the daily mixing is deeper on average.
So despite your claim that the diurnal mixed layer is only a metre thick, any oceanographer is going to laugh in your face about that one. For me, as an inveterate night diver, I knew your statement about the nocturnal overturning being only one metre deep was terribly wrong because I’ve regularly dived through the descending columns of cool water during the nocturnal overturning … down ten metres or so …
w.
Dave Springer says:
May 17, 2011 at 12:38 pm
@Phil
Your numbers are so far from reality it’s ridiculous.
The troical atlantic ocean receives an average of 200w/m2 (see Tropical Atlantic Mixed Layer Heat Budget) yet you’d have me believe it gets 400w/m2 at night. Must be some really bright moon on your planet. Full moon every night that shines down with almost half the intensity of the unobstructed high noon tropical summer sun. ?That’s some trick. What’s your moon made of – radioactive cheese?
Right.
Your comment has been duly noted.
In order for that to be true you’d have to have read it and understood it, you apparently did neither!
I chose a downwelling IR value of 300W/m^2, a fairly modest value for planet Earth as shown by measurement (300W/m^2+ has been measured during summer at the North pole). The value of 400W/m^2 referred to the blackbody radiation leaving the surface, what would be expected for a sea surface around 290K. So my numbers are quite reasonable for planet Earth.
Dave Springer says:
May 17, 2011 at 12:48 pm
@Phil
Hey thanks for digging up the diurnal turnover depth in that graph you linked. See how it only extends to a depth of 1 meter? Joel was trying to tell me the whole mixed layer turns over every day when I knew only the top 1% turns daily. The entire mixed layer turns seasonally.
I’m not at all sure what point you wanted to make with it. Did you accidently make the wrong link or just didn’t know what you were looking at or what?
Perhaps you should read it again since you evidently didn’t understand it the first time?
“what does happen is that the water surface radiates away at about 400W/m^2, in fact the top few microns will be cooler than the water a mm deeper because the transfer of heat from lower down (conduction, convection, diffusion etc.) is slower than the loss from the surface.”
Dave Springer says:
May 17, 2011 at 1:01 pm
@Phil
re; “Also the water vapor doesn’t convect up as you state, humid air will do so.”
Huh?
Of course it won’t rise if it condenses close to the surface. But then it isn’t water vapor. We call that “fog” on my planet. Again, what’s your point? As long as it remains a vapor it rises until adiabatic lapse cools it below the dew point.
Until it condenses the vast majority of the heat if lifted out of the ocean remains hidden away in latent form.
Did I lose you anywhere there? What part don’t you understand? I like you and am willing to spoon feed this to you if you’d just stop making faces and spitting it out.
You apparently fail to understand that water vapor doesn’t convect up on its own but does so along with the air that it’s part of, i.e. humid air.
Dave Springer says:
May 17, 2011 at 1:15 pm
@Phil
Maybe I figured out the soruce of your confusion.
The confusion is all yours!
Joel Shore says:
May 17, 2011 at 4:38 pm
Re my: “What is the actual mechanism, what physical event decides that the ‘net’ will always come out as heat flowing from hotter to colder?”
For all forms of heat transfer, it basically has to do with the statistics of large numbers of events. For conduction, which occurs via molecular (or electronic) collisions, you will have collisions between molecules..and the molecules will have a range of velocities but ON AVERAGE, those coming from the hotter side of the object will have slightly higher speeds than those coming from the cooler side and so, in a large number of collisions, the molecules from the cooler side will gain speed on average and those from the hotter side will lose speed on average in the collisions.
For radiation its basically that the hotter object emits more radiation than the cooler object.
Where in this process does it show that the radiation from the colder to the hotter is ever taking place?
As I’ve tried to explain what I mean here before, the ‘net’ in ‘statistics of large numbers’ has nothing to say about a colder’s heat flowing to the hotter.
And so can be fully true to the 2nd Law that heat ALWAYS flows from hotter to colder in that the ‘net’ can be simply explained as I described somewhere above – that the colder gaining energy becoming hotter which then in turn energises the colder, thereby losing energy and again being ready to be energised by the hotter. This is the real ‘net’ mechanism being described.
The ON AVERAGE then, is not about, or rather including, some idea of energy flowing from the colder to the hotter, but the average of the exchanges as the hotter energises the colder thereby itself losing energy in any amount where this is happening, the gains and losses in these exchanges.
In other words, the ‘net’ is irrelevant to the 2nd Law, as this refers only to the process of these exchanges depending on such things as amount of energy available and materials and their properties relative to each other.
The “statistics of large numbers” is no proof whatsoever that heat flows from the colder to the hotter, ever. That’s just been added in without any logical antecedents.
If the colder could add its energy to the hotter, then statistically in the ‘large number of events’ the colder could always be giving up its energy to the hotter, and losing what energy it has so itself getting colder; so putting something hotter on ice the hotter will get hotter not colder if there is enough colder to hotter events.
Unless, you can provide a mechanism which switches this off.
Myrrh says: “The “statistics of large numbers” is no proof whatsoever that heat flows from the colder to the hotter, ever. That’s just been added in without any logical antecedents.”
Heat doesn’t flow from the colder to the hotter, energy flows from the colder to the hotter. The heat flow from the hotter to the colder is the net energy flow.
“If the colder could add its energy to the hotter, then statistically in the ‘large number of events’ the colder could always be giving up its energy to the hotter, and losing what energy it has so itself getting colder…”
… and gaining what energy the hotter object transfers to it, for a net effect of getting warmer. The hotter object is transferring more energy to the cooler object than the cooler object is transferring back to it, for a net effect of getting cooler.
Joel regarding yout post
Joel Shore says:May 14, 2011 at 7:48 pm richard verney: Claim 3. Some caution is required. Theoretical black body calculations do not accord with actual measurements for Venus, Mars Titan or even the moon.,,,,,,”
Your response: You consider yourself a skeptic, I assume, and yet you accept any piece of garbage that you find on the internet? Hertzberg, Schreuder, and Siddons are well-known for being purveyors of pseudo-scientific nonsense. Why would you possibly believe their nonsense when you won’t accept real science that has been tested thoroughly? It seems like your skepticism is very selective. Here is a detailed debunking of their paper: http://scienceofdoom.com/2010/06/03/lunar-madness-and-physics-basics
//////////////////////
I have tried half a dozen times to open that link (including going directly to science of doom and searching for their article). Unfortunately, it merely causes my computer to freeze.
If you had read my comments you would know that I was citing the article that I referred to merely in support of the statement that that actual temperature measurements taken on the moon do not correspond with theoretical BB calculations. I was not citing the paper in support of their conclusions that there is no GHE.
I presume that their empirical data (ie., the BB calculations performed by NASA and the actual temperature measurements performed by NASA) are correct. If you consider that empirical data to be wrong, please refer me to the correct data.
If their empirical data is correct, then it follows that the moon does not behave in accordance with BB calculations. As you are aware, I consider that it is unlikely that the Earth would behave as if it were a BB. There are many reasons for this (the composition of the atmosphere with gases that have different absoptive and emissitivity, the vast stored heat capacity of the oceans etc). The upshot of this is that the Earth’s daytime temperatures are not as warm as BB calculations suggest that they should be nor are the night time temperatures as low as BB calculations suggest that they should be. The type of lag and dampening seen with respect to the moon would be exacerbated with the Earth.
I do not consider it likely that the Earth behaves as a black body, it is not a BB but a Greybody and we do not have a proper handle on this such that and assessments based upon the Earth being a BB are likely misconceived and erroneous.
Steve says:
May 18, 2011 at 7:14 am
“… energy flows from the colder to the hotter. The heat flow from the hotter to the colder is the net energy flow.”
Steve this is wrong. Let me do this again basic formula q/A = e (SB)(T1^4-T2^4) if T1=T2 then zero energy flows anywhere. Zero on right side of equation means zero on left side of equations. q/A is watts per meter squared a watt is juole per second and a joule is a unit of energy. Use any heat transfer formula you want but if T1=T2 no energy flows. There must be a path and a temperature gradient for q/A to have a value.
Dave Springer added a new comment to the post Visualizing the “Greenhouse Effect” – Light and Heat.
Dave Springer said on Visualizing the “Greenhouse Effect” – Light and Heat
May 16, 2011 at 8:06 pm
In response to Ira Glickstein, PhD on May 7, 2011 at 7:55 pm:
Guest Post by Ira Glickstein Solar “light” radiation in = Earth “heat” radiation to Space out! That’s old news to those of us who understand all energy is fungible (may be converted to different forms of energy) and energy/mass is conserved (cannot be created nor destroyed). My Visualizing series [Physical Analogy, Atmospheric Windows, Emission Spectra, […]
JAE
Nowhere did I say anything about any net warming caused by back radiation from greenhouse gases. All I claimed was that GHGs do indeed absorb radiation and redirect a portion of the absorbed energy back towards the source. This is the basis upon which most electronic CO2 sensors used in commercial ventilation systems work. This so-called back radiation has been physically measured many times by infrared spectrometers looking up from the surface and down from above. It is even verifiable by anyone who beg, borrow, or steal a $50 infrared thermometer
http://www.amazon.com/gp/product/B0017L9Q9C/ref=pd_lpo_k2_dp_sr_2?pf_rd_p=486539851&pf_rd_s=lpo-top-stripe-1&pf_rd_t=201&pf_rd_i=B000MX5Y9C&pf_rd_m=ATVPDKIKX0DER&pf_rd_r=0A2YZ0RNVN98DZFNB1EG
and verify for themselves that when pointed upward through clear sky it still reads a temperature far closer to the 200-something K of the ground you’re standing on than the 3K temperature of empty space.
If you can’t accept all these things as proof that back radiation is real then you’re a cause and if this was my blog I’d throw your ass out so to reduce the dimwitted clutter that pollutes the more informed converstations.
Just a quick note to thank Dave Springer for providing the empirical evidence of ‘back radiation” which cemented my understanding of the subject. I had missed your previous comments on the handheld infrared thermometer, but noticed it today while browsing previous posts.
FWIW most cheap handheld IRTs have the emissivity set by default to 0.95, so the temperature it read of the sky and earth would need to be corrected in order to account for their different emissivities. More expensive ones allow adjustment of the emissivity, or rough & ready calibration by measuring the temperature of masking tape (emissivity ~0.95) on the object to be measured, or by drilling a hole at least 6 times as deep as its diameter which approximates a blackbody (emissivity 0.98+). Problem with that approach is that the FOV of most cheap ones is so broad that a fairly large hole is required 🙂
There are also FLIR cameras which use the radiated LWIR energy to image things, a technology with applications in veterinary medicine, industry & many other fields.
All these technologies use the IR radiation emitted from the objects being scanned, at least according to all the various manufacturers I am familiar with.
I have known and used these devices for years, which got me thinking about the difficulty I had accepting the concept when applied to climate, and the fact I had not considered the implications of the science upon which the technology was built. I concluded that it had to do with semantics, at least in my case. All the textbooks I have read and all the books I have read state categorically that heat is the manifestation of the transfer of energy, and this is true. In the case of two objects, one colder and one hotter, there is an EXCHANGE of energy as well as a TRANSFER of energy. The energy exchanged amounts to the total energy radiated by the cold object and the energy transfered is always from hot to cold and is simply the difference between the amount the hot object radiates and the amount the cold object radiates – ie the net difference in their energies. This respects the 2nd law & requires no mechanism or volition, no little ‘Maxwell’s Demons” as Joel scarcastically suggested. In effect, the colder object transfers all it can to the hotter object, but it will always have less to give than the hotter object, therefore the transfer of energy (flow of heat) must always be from hot to cold – unless external work is applied or some other change in universal entropy compensates.
So with respect to the surface and the atmosphere, the atmosphere doesn’t heat the surface, or really make it warmer, all it does is slow radiative heat loss, instead of losing 356W/m^2, with a 333W/m^2 atmosphere all it really loses is 23W/m^2.
I also wanted to say that I agree with you about CO2 being a beneficial trace gas (plants wouldn’t grow and a lot of people would die of apnea), up to a fairly high level anyway. Witness the levels of CO2 used in commercial greenhouses… While the oceans must represent the largest single influence on global and regional climate, Land Use & Land Cover changes also appear to have significant regional forcing effect on land.
http://wattsupwiththat.com/2011/04/18/more-on-land-use-change-affecting-temperatures/
mkelly says: “Steve this is wrong. Let me do this again basic formula q/A = e (SB)(T1^4-T2^4) if T1=T2 then zero energy flows anywhere.”
Did you think that through? You are saying that, in the one case where T1=T2, the equation shows that there is no energy flow between the objects. But in all other cases where T1 is not equal to T2, you can have either a positive or negative energy flow between the objects. What would it mean to have an energy flow less than zero?
Are you positive this equation is representing individual energy flows? I think it represents net energy flow, hence it can be either zero or even negative.
Are you saying that if I surround a 500 degree ball bearing by a 600 degree sphere, the 500 degree ball bearing stops radiating energy away from itself, and the 600 degree sphere will begin radiating a less energy towards its center?
richard verney says:
Richard: The problem is in their analysis of the temperature data. So, it is not NASA’s data that is incorrect but their analysis that led them to conclude that the moon is not, to a very good approximation, obeying the expectations of a blackbody emitter. Basically, they did not understand how to deal with the case where temperature is not constant.
Emissivities of terrestrial materials in the wavelength range of interest are extremely close to 1 ( http://en.wikipedia.org/wiki/Climate_model ), so much so that the earth not being a perfect blackbody could only account for two or three degrees at the very most. (Because of the T^4 dependence of radiative power on temperature, an emissivity of 0.99 rather than 1 leads to ~0.7 deg K of change in the predicted temperature.)
mkelly says:
In this equation, q is the rate of heat transfer, which is the NET rate of energy transfer. So, it being zero does not mean that there are no energy transfers back and forth, but merely that they cancel out to give no heat transfer.
Not that it matters…The equation that you wrote down gives the greenhouse effect regardless.