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.
Joel Shore says:
May 18, 2011 at 9:27 am
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.
So you are a “Quantum Electrodynamics Denier”? I had that feeling. You do not think amplitudes and probabilities can never be zero? Really? See Joel, you have to prove it to all here before you thrash more minds.
mkelly (and all of the other commentators that have made it this far):
CAUTION: Only for the brave.
I highly suggest all take the time to go back to school, just for a moment, and listen to these four lectures by Dr. Richard (Dick) Feynman. Really? Really! Your view of the transfer of radiant energy will never be the same again.
It requires no complex math, no huge intellect, just a curious mind.
http://vega.org.uk/video/programme/45
http://vega.org.uk/video/programme/46
http://vega.org.uk/video/programme/47
http://vega.org.uk/video/programme/48
The last one and a half you can really skip if you don’t want to see how this applies to electrons, Feynman diagrams and how it applied to the zoo of particles that physics once was experiencing, but the first two are mandatory.
Try it. You’ll like it! Then come back and prove it to me that there are always photons passing back and forth within a smooth hollow steel sphere at any temperature. (hint: by cancelling amplitudes everywhere it seems it is zero photons inside).
Seems to me Joel Shore is the one passing pseudo-science here!
Steve says:
May 18, 2011 at 8:56 am
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?
Steve a negative sign would indicate you put it in the formula wrong. You must have missed the part about watt equal joule per second and joule is unit of energy so no energy gain by either object.
I have not said that cooler objects don’t radiate at warmer objects but here is no demonstrable proof that the hotter object absorbs radiation of a lower frequency.
If anyone takes me up on listening to Dr. Feynman’s lectures, pay particular attention when he teaches you that “there is only one kind or type of photon”. Did he really mean that? Really! Frequencies of radiation has to do with the “source matter”, not the photon. They just come from the “source matter” at different frequencies. There are not different frequency types of photons, all photons are identical! Now, listen to his lectures, two or three times if necessary (me? four times).
Quantum is a strange twist indeed, but all experiments seem to show in every case and to huge precision that it is what is REALLY happening everywhere in our universe, our Earth, out atmosphere, and even down to single atoms. Well, not in Joel case after reading his response to mkelly and Phil. in his response to Dave Springer just above. Not in their minds. Watch them dance all around reality to fit their “worldview” and it always ends up firmly planted in AGW pseudo-science.
mkelly says: “Steve a negative sign would indicate you put it in the formula wrong. You must have missed the part about watt equal joule per second and joule is unit of energy so no energy gain by either object.”
Begin with T2=T1. Then switch on a heating element within object 2. Soon T2>T1, and q/A is negative. What does that mean? It means the equation represents a net flow, with positive or negative indicating the direction of flow. Zero indicates no net flow.
“I have not said that cooler objects don’t radiate at warmer objects but here is no demonstrable proof that the hotter object absorbs radiation of a lower frequency.”
So what happens to the radiation? Does the hotter object reflect it or transmit it? Absorption, transmission and reflection are your three options. Think it through. Propose your hypothesis and how to falsify it in terms of the competing hypothesis.
Robert Clemenzi says on May 17, 2011 at 4:43 pm:
“ —————. 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.”
Thank you Robert for a nice reply, I shall have to take a bit longer to think about it in full, but for now let me just confirm that as far your interpretation of the data is concerned they are not too far removed from my own.
I know the data say that 99 % of all the atmospheric Water Vapor (WV) is to be found in the Troposphere, so yes it is very likely to be some in the “tropospause” and even maybe as high up as in the Stratosphere. After all, they say it takes WV for the formation of “Cirrus Cloud” or Cirro Stratus. – But, even so we must not forget that because temperatures fall to well below 0 deg. C. in the upper Throposphere, much of it will have some kind of “Up Down” circuit high up in the Troposphere, i.e. rise – freeze –sink (drop) – melt, warm –rise – freeze – etc. etc. – And yes, – it should be equally opaque all the way.
However when it comes to the “conundrum” as to why the Troposphere cools, with height, the Stratosphere warms, the Mesosphere cools and the Thermosphere warms I think we shall have to take into consideration that there are two types of “radiation” i.e. “Ionizing” and “Non Ionizing” Ionizing is “short Wave as in solar incoming radiation that has enough energy to dilsodge electrons, non ionizing (long wave) has not got enough energy for that kind of behaviour and is the kind of radiation discussed amongst the (us) global warming pundits.
Some part of the atmosphere must be warming from the top down, but instead of me telling you what I conclude from that I urge you to use your own logic and decide for yourself.
And of course it is only the very few N, O and O2 atoms & molecules in the “upper atmosphere that warm, – not the space between them, so you will still freeze to death up there.
Just remember, (I haven’t yet formulated it but, – “energy absorption must lead to heat production. – Conversely, – heat absorption must lead to energy conservation”) — Oh, maybe somebody already thought of that one when he said “X (1 J of Energy) has got to enter before X (1 J of Energy) can exit”
wayne says: “Try it. You’ll like it! Then come back and prove it to me that there are always photons passing back and forth within a smooth hollow steel sphere at any temperature. (hint: by cancelling amplitudes everywhere it seems it is zero photons inside).”
The key phrase is “seems”, as in the ability for the quantum components of the system to measure a photon within the sphere can drop to zero. Now go back to the other key phrase “by cancelling amplitudes”. Amplitudes cannot be cancelled if, in fact, there are no photons traveling 180 degrees out of phase with each other. So the cancellation occurs because of the fact that photons are passing back and forth.
This is cavity quantum electrodynamics, and not something with any ghost of a chance of being observed in macroscopic objects. We are talking about extremely small (micrometer scale) spheres.
Oh yes Robert, O2 becomes O3 when ionized by UV radiation. – O3 as you probably know is OZONE.
Wayne
regarding your post wayne says: May 18, 2011 at 1:04 pm
I have previously mentioned something similar, ie., not disputing that all objects above absolute zero radiate but that does not automaticlly mean that radiative energy emitted by a colder object is actually absorbed by the warmer object. I will review the links to which you refer.
I am not for one moment saying that I have the answers. I think that either Joel or Tim set out an analogy with tennis balls. I do not particularly like giving analogies of matters not well understood. As far as the tennis analogy goes, it may be that on one side of the net there is a weaker player (ie., the side of the net with ‘colder’ photons) and that player cannot consistently hit the ball over the net such that balls are generally flowing from the side with the stronger player (the ‘hotter’ photons).
Whilst a photon may not have a temperature, it would appear that it has a temperature signature. I say this since it appears common ground that photons radiating from an object at temperature T cannot warm another oject to a temperature exceeding T. If they had no temerature signature the work that could be extracted from them would simply be proportional to the watts per sq m and this would mean that back radiation from the sky could be harnessed to heat objects with similar results to that achieved by using solar radiation. A black car is noticeably warmer to touch than a white car parked nearby during the day in sunlight, but at night there is no obvious difference. The back radiation from the sky that is bombardding the black car does not (noticeably) heat it to a temperature higher than a white car parked nearby.
If photons radiating from the atmosphere at say 240K towards the ground cannot heat the ground to above 240K, whereas photons radiating from the ground at say 288K towards the sky could potential heat an object to 288K, I question whether it is fully correct that one source can simply be deducted from the other since effectively the energy (the ability to do work) is different.
Hence I question Tim’s alternative take on the Trenberth energy budget wherein he deducts the (240K) 333 W/m^2backradiating from the sky from the (288K) 399 W/m^2radiating from the ground saying effectively that the ground is radiating 66 W/m^2.
wayne:
“Did he really mean that? Really! Frequencies of radiation has to do with the “source matter”, not the photon. They just come from the “source matter” at different frequencies. ”
Whaaaat? Have not listened yet, but this seems very confusing, since Planck gives the energy of a photon as E=hv. So, how can all photons be equal?
http://en.wikipedia.org/wiki/Planck_constant
jae says:
May 18, 2011 at 6:35 pm
Whaaaat?
—–
I agree. Caused me halt, reset, and rethink (with a headache). On the surface it seems to imply that the frequency of light is merely the frequency that identical photons are emitted by the source. Seems wrong but that does bring back words of many years back of photon packets or trains, forgot that long ago. Or maybe he just meant photons have no attatched properties that man has found that can be altered. Then I start thinking, well, how do photon particales get doppler shifteded? Is that related to his satement?
I’m totally open for help on exactly what Dr. Feynman could have meant by that statement. That is why I threw it out there. HELP.
Do watch at least the first two, may be three, videos. You should enjoy, he’s very entertaining. (BTW: there is some static in video #2 but they do correct it some twenty minutes in)
jae, whoa, excuse me for not running that through a spell checker! Don’t know about you but the words are appearing two or three behind where I am typing. So slow…
Response to O H Dahlsveen – May 18, 2011 at 2:23 pm
There are several, somewhat conflicting, definitions of the tropopause. However, the data indicates that the temperature decreases until water vapor drops to 5 ppm. (Some sources say zero.) For the stratosphere, the water vapor stays about 5 ppm until the stratopause. Above that, there is no detectable water vapor. It is my opinion that this distribution is somehow related to the temperature profile.
O2 becomes O3 when ionized by UV radiation
Well, yes and no. Apparently, UV radiation is more likely to convert O3 into O2 and an oxygen free radical than the other way around. From what I’ve read, UV converts H2O and O2 into OH radicals. Then UV combines the OH radicals with O2 to make ozone. When an additional UV photon destroys the O3 molecule, the excess energy becomes heat. The reason that there is a hole over Antarctica is because it is too cold for the OH radicals to form. I know this disagrees with what is in the popular press, but it appears to more correct than what most sources say.
At any rate, once the water vapor goes to zero, there is no longer a hydrogen source for the OH radicals and the temperature decreases above that point. The weird thing is that there is almost no ozone where the temperature is highest, which strongly argues against my theory.
jae, I’m getting way OT so I’ll stop on the quantum aspect here.
I did find his insisting that all photons being identical, #3@23min, but of course he is not invalidating E=hv, seems he is only speaking of when you look at existence (amplitudes) in spacetime. Later he shows that even electrons and photons are identical except for the mass, more, even all point-like particles are identical but mass (muons, etc.). That is what experiments show really happens. That in itself is getting too way too deep for here, any further and I’ll really get myself in trouble by wording something wrong.
My only point in starting this dive into the depths is sometimes all things are not as they first appear (or how others think they must appear). That’s all.
“Then I start thinking, well, how do photon particales get doppler shifteded?”
If photons were baseballs and spectrometers were baseball bats then the doppler effect is the difference between a bunt and a swing.
So I’ve been looking over the Trenberth diagram Joel pointed me to from the point of view of my new-found understanding of the bi-directionality of energy exchanges.
It has been well established here that it is the net energy flow which heats and that it always flows from hot to cold. Therefore it is only the net energy flows which need be considered when estimating the radiative heat transfers in the diagram.
Based on that there are a few things that still puzzle me about this diagram:
1- The total net radiative energy from the Sun to the Earth is approximately 239W/m^2 (16W/m^2 to the surface + 78W/m^2 to the atmosphere) and the total net radiative energy from the Earth to space is 239W/m^2 (169W/m^2 +30W/m^2 + 40W/m^2).
Therefore, as I understand it, based on the repeated assertions made here, there is no net heat and energy transfer between the Sun and the Earth. Which means that the Earth is neither cooling nor warming.
Is this correct?
2- The surface radiates 453W/m^2 (17 + 80 + 356) to the atmosphere and the atmosphere radiates 333W/m^2 towards the surface, for a net heat transfer of 120W/m^2 from the surface to the atmosphere. Additionaly the atmosphere receives 78W/m^2 radiative energy transfer from the sun, for a total energy gain of 198W/m^2. The atmosphere also radiates 199W/m^2 (169 + 30) to space, for a net loss of 1W/m^2 or approximately ) 0.505% of the energy input.
So this suggests that the atmosphere is cooling very slightly, as Tim pointed out, 1 m^2 of atmosphere may contain 30,000,000 J of energy. However, unless I am mistaken that 1W/m^2 = 1J/second/m^2/K and therefore over the course of 100 years the atmosphere would have lost 3,155,760,000 J or 3.15576x10e9J which should represent significant cooling.
Is that correct?
I am not trying to deny anything here, just trying to understand.
wayne says:
May 18, 2011 at 1:04 pm
You seem to have missed the beginning lesson that theory must explain reality. If it fails then the theory is wrong. Reality trumps theory in all cases.
In the particular instance of so-called back radiation the reality is that if you point an IR spectrometer up at the night sky photons of far higher energy than the cosmic microwave background are hitting it. Theory must explain these. In fact theory does explain it. If your understanding of the theory denies reality in this case your understanding is flawed.
In each and every instance where someone denies the reality of back radiation they are wrong. To say this back radiation might be real but a warmer surface cannot absorb photons emitted by a colder surface is also wrong as the reality is that if you measure the rate of energy loss between two surfaces of different temperatures the rate decreases as the temperature difference decreases and increases as the temperature difference increases. Thus when the two surfaces reach the same temperature there is no loss or gain by either of them. But just because the two surfaces are the same temperature doesn’t mean they stop emitting energy. It only means they are absorbing as much energy as they are emitting. This is the reality of the situation.
Alleyne, you are basically correct, except that a difference of 1 W/m2 is less than what can actually be determined. As a result, it is not correct to assume that the atmosphere is actually cooling that much. Based on actual measurements, it is not currently possible to determine if the net balance is other than zero.
@Alleyne
“Therefore, as I understand it, based on the repeated assertions made here, there is no net heat and energy transfer between the Sun and the Earth. Which means that the Earth is neither cooling nor warming.”
That is correct for a system in equilibrium. The earth/sun system is never in perfect equilibrium but it will always seek to attain equilibrium and the farther out of equilibrium the harder it tries to reach equilibrium.
The AGW thesis is that additional CO2 moves the system further out of equilibrium and that in moving back towards equilibrium the surface temperature will rise.
That thesis is fine in theory up to the point where the claim is made that surface temperature must rise. The surface temperature is hypothesis not theory. There are other possible responses the system might make to move back towards equilibrium.
Two possible responses are:
1. An increased rate of evaporation and convection will move the additional energy at the surface to a higher layer in the atmosphere and because evaporation carries energy in what’s called “latent heat” there will be no measurable rise in temperature near the surface as thermometers measure what’s called “sensible heat”.
2. An increased amout of energy at the surface may cause an increase in albedo higher in the atmosphere through change in type and/or amount of cloud cover which simply reflects more energy away before it ever reaches the surface.
The reality of the situation is that no matter how much CO2 is in the atmosphere at least up to the point of 10x today’s level surface temperature correlates poorly with it. The reality is also that anywhere up to at least 10x today’s level of CO2 there is no runaway greenhouse.
The reasonable interpretation of the reality of the situation is that there is a negative feedback loop involved with increased energy at the surface. The negative feedback is almost certainly one or both of the equilibrium responses enumerated above.
Dave, I accept that analogy of a bat, but it was more questioned in relation to the statement by Feynman that all photons are identical. Now I regret even mentioning it. As I clarified later, that seems to have been said by him only as to how amplitudes were being calculated in relation to time. That’s the catch that I missed. ☺
By the way, Feynman in his dated lectures insists reality is non-deterministic. Some physicists every bit as smart believe otherwise. Indeterminism implies that information (not the layman’s concept of “information”) can be destroyed. This concept of information being non-destructable is a basic tenet of modern physics. It reaches so deeply that a rather notorious claim by Stephen Hawking was argued for years by the most highly respected theoretical living physists and recently Hawking admitted defeat.
Hawking proposed that if a set of encyclopias fell through the event horizon of a black hole that the information in them could never be recovered. He didn’t even propose that the information was destoyed but merely lost forever to the universe outside the black hole’s event horizon.
Othere physicists claimed it was not possible to remove information from the universe even in the most exotic situation imagineable – falling into a black hole – because conservation of information is as inviolable as conservation of energy and that both conservation laws are in fact different manifestations of the same phenomenon.
Hawking conceded when sufficiently convincing mathematics showed that the information in the enclopedia would reemerge as patterns in so-called Hawking radiation emitted by the blackhole. Hawking radiation is based on the well established fact of quantuum tunneling where a particle may disappear at one point in space and reappear at another point without enough energy to have moved across a barrier from point A to point B. Flash memory chips work by quantuum tunneling where an electron is raised to an energy level just short of being able to cross a barrier into a holding pen. The electron then tunnels through the barrier and its energy level is reduced such that the probability of it tunnelling back in the other direction is so small you count on it never happening. When power is removed from the flash memory device the electron stays trapped indefinitely.
So there you have it. At the very deepest level the universe still appears to be deterministic and Feyman’s dated claim that it isn’t was premature at best and probably wrong at worst.
Not to get too much farther off-topic but the law of conservation of information reaches deep into theology. If indeed an inviolable law then it implies all the information in the universe today, which includes all the information encoded into each and every human brain, was present in the universe since the instant of the big bang. This then raises the question “where did that information come from?”. Or if the information wasn’t present at that time then “where, when, and how was the information introduced into the universe?”. Science has no answer, or rather the best current answer offered by science is the so-called multiverse which postulates a reality of an infinite or near infinite number of universes each born with different information in them and so in an infinite series there must inevitably be a universe just like ours – in fact an infinite number of them just like ours. By implication there would also be an infinite number of them that are infinetely close to the same as ours but not quite.
The multiverse hypothesis of course doesn’t pass the giggle test and theologians answer it by saying God made the universe the way it is. I don’t have a position on the matter either way but the multiverse explanation seems just as contrived, if not moreso, than the God hypothesis.
Mr. Springer says: “To say this back radiation might be real but a warmer surface cannot absorb photons emitted by a colder surface is also wrong …”
Harkening back to my heat transfer classes it was explained that higher temperature object does not absorb lower temperture emissions. The object at the higher temperature is already at a state higher than what the lower temerature object is or so there is no “where to go” with the lower energy.
wayne says:
May 18, 2011 at 1:47 pm
“If anyone takes me up on listening to Dr. Feynman’s lectures, pay particular attention when he teaches you that “there is only one kind or type of photon”. Did he really mean that? Really!”
Sure he meant it. Going back to the baseball analogy where the rules sanction only one baseball so that all baseballs used in the games are as identical as possible. The only difference as it approaches the batter is thus reduced to the spin and speed imparted to it by the pitcher. Photons are pretty much like that. The rules of the universe make them all the same except for the amount of energy in them which is determined by the pitcher (the object which emitted it).
Now that I think about it Feynman got something else wrong in his lecture to the best of my knowlege. He insisted that electromagnetic radiation is entirely a particle phenomenon. I believe that’s incorrect and the modern interpretation is that EMR exhibits wave-particle duality (in fact ALL matter has wave-particle duality) and this was verified by the famous double slit experiment.
http://en.wikipedia.org/wiki/Double-slit_experiment
The truest thing Feynman said was the reason this is all so confusing to students is because the physics professors trying to explain it don’t fully understand reality either. That includes Richard Feynman. The theory of reality is woefully incomplete. We still don’t have, for instance, a quantum theory of gravity which is required to reconcile relativistic physics with quantum physics. Essentially we don’t know the properties of gravity at the quantum scale. We only know it at the bulk scale and have no quantum theory which predicts or explains gravity at the macroscopic scale.
There’s no explantion for all kinds of observations. My favorite is so-called “dark energy” which somehow, in some way, overcomes gravity across vast distances and is accelerating the expansion of the universe. The standard model offers no explanation but the observations reveal that dark energy must comprise some 70% of the “stuff” which makes up the universe. In other words 70% of what makes up you, me, and the rest of the universe is a complete mystery to modern physics. We think we know a lot but in reality we know about as much about the universe as you can know about the ocean just from observing the waves & froth at the surface.
I lost track of who said the lapse rate & hence temperature structure of atmospheric layers can’t exceed dry adiabatic lapse rate.
This is a nonsense claim. There are three kinds of lapse rates: dry, saturated, and environmental. The environmental lapse rate brings convective processes into the picture and explains the reality of temperature inversions where a warmer layer of air is riding on top of a cooler one. Inversions don’t just exceed the dry lapse rate, they change its polarity!