Guest Post by Willis Eschenbach. [See Update at end]
The CERES dataset continues to yield new insights. My joy is to graph different relationships and then see what I can learn and find out from the graph.
Inter a lot of alia, the CERES dataset allows us to calculate the amount of the very poorly named “greenhouse radiation”. This is the downwelling longwave (thermal) radiation from the atmosphere. It’s commonly called “DLR”, which stands for downwelling longwave radiation.
(NOTE: This is not a thread for disputing the existence of downwelling longwave radiation from the atmosphere. DLR exists, it’s measured both manually and automatically all over the planet every day, and it doesn’t violate the 2nd Law. Get over it.
If you wish to dispute that, fine, I encourage you to do so … but this thread is not the place to do it. There are dozens of threads on the subject, the web is a large place, pick one and go. Please, don’t test me on this. I don’t want to have to snip comments claiming no DLR or that the “greenhouse effect” violates the 2nd Law, but I sure will … END NOTE.]
As first proposed by Ramanathan, the downwelling longwave greenhouse radiation (DLR) can be measured from satellite and ground measurements. It’s the amount of upwelling longwave surface radiation absorbed by the atmosphere. It’s calculated as the upwelling longwave radiation from the surface minus the upwelling longwave radiation measured at the “top of atmosphere” or TOA, meaning in this case at the satellite.
The CERES dataset has values for all-sky and clear-sky radiation. The difference between those is the effect of the clouds.
Now, the amount of downwelling longwave radiation changes over time. And in theory, part of this change is from an increase in CO2. We can calculate the theoretical change in DLR resulting from the change in CO2.
Putting all of that together, we get the following plot of the changes in downwelling “greenhouse” radiation since the turn of the 21st century.
Figure 1. Changes in downwelling “greenhouse” radiation by source.
Now, there are several things of note in this graph.
First, the total change in downwelling radiation from the atmosphere (“greenhouse radiation”, or DLR) is more than three times the change due to CO2. Presumably, this must be from a combination of changes in water vapor, latent/sensible heat loss, atmospheric solar absorption, and increased surface temperature.
Next, CO2 is the minor player in the greenhouse DLR game. Non-CO2 clear-sky DLR increase over the 2000-2022 period was ~ 3.2 W/m2. Cloud DLR decrease was ~ 1 W/m2. But the CO2 change was only 0.7 W/m2, the smallest of the three.
Next, I theorized a couple of decades ago that thunderstorms, clouds, and other emergent phenomena act to oppose temperature changes and thereby stabilize the temperature. Since then, I’ve provided a variety of evidence to back up that claim. Here, in Fig. 1 you can see that while greenhouse DLR from CO2 and from water vapor are increasing, to the contrary, the greenhouse DLR from the clouds is decreasing. This is evidence in favor of my theory.
In fact, the change in the clouds over the period has more than offset the change due to CO2 … who knew?
Finally, we can compare the change in greenhouse DLR to the change in temperature. This gives us the “TCR”, the transient climate response to a change in DLR.
In this case, the TCR is 0.2°C per W/m2, which would be equivalent to 0.7°C per doubling of CO2. This is markedly smaller than the usual value for the TCR, which is on the order of 1.5W/m2 per 2xCO2 or so. Looking at Figure 1 we can see why that is so. The total change in observed greenhouse DLR is ~ three times the theoretical change from CO2. Since the change is larger, the sensitivity to the change perforce must be smaller.
And since TCR values are typically about 55% of the equilibrium climate sensitivity (ECS), that would make this estimate of the ECS about 1.4°C per 2xCO2.
Finally, there’s been little change in total greenhouse DLR since ~ 2016 … another unsolved mystery of the sea.
[UPDATE:
I note that despite my polite request, some impolite folks want to hijack this thread to proclaim that the “greenhouse effect” isn’t real and downwelling radiation doesn’t exist.
I said I’d snip such comments, but I realized that if I do, the authors will likely style themselves as martyrs rather than just common thread hijackers. So I’m gonna pass.
In hopes of expanding their concepts of the greenhouse effect, however, let me recommend the following posts of mine:
The Steel Greenhouse 2009-11-17
There is a lot of misinformation floating around the web about the greenhouse effect works. It is variously described as a “blanket” that keeps the Earth warm, or a “mirror” that reflects part of the heat back to Earth, or “a pane of glass” that somehow keeps energy from escaping. It is none of these things.
People Living in Glass Planets 2010-11-27
Dr. Judith Curry notes in a posting at her excellent blog Climate Etc. that there are folks out there that claim the poorly named planetary “greenhouse effect” doesn’t exist. And she is right, some folks do think that. I took a shot at explaining that the “greenhouse effect” is a…
Let me also repeat my request. Please, guys. There are lots of places to debate these questions. But this is not one of them.
Sadly,
w.
Best to all,
w.
To Avoid Misunderstandings: When you comment please quote the exact words that you are discussing.
Correct, convection stabilises the temperature so DLR is not the cause of the greenhouse effect.
Then what is ?
Downward adiabatic convection converting atmospheric potential energy to surface kinetic energy.
Back in 2010 I pointed out that in the Trenberth diagram one can simply replace heating from DLR with heating from downward convection to obtain the observed greenhouse house effect.
Perfectly consistent with Willis’s observations.
Effectively proved by the published work of me and Philip Mulholland.
To somewhat oversimplify, DLR is just the negative part of Q=k [Tsurface^4 – Tsky^4]
You seem to be saying that Tsky has no effect on DLR. Are you saying Planck and Boltzmann are wrong ?
Correct, DLR from the atmosphere is negative. That is what the surface measurements tell us too.
(I think Willis’s position is actually that Tsurface has no effect on DLR power, which is equally bizarre)
Gases do not follow Boltzmann’s Law.
Seems like the boiling oceans have something to do with it.
“Boiling oceans” = No, fairyland stuff.
However the additional heat release into the atmosphere by way of water vapor as a result if increased energy into the oceans, that I can accept.
Math is hard,,,I should say: “Maths are hard, but climates are even harder“. According to a Willis Theory, which I can accept, the release of ocean water vapor can ultimately result in cooling. Water vapor is much less dense than dry air. Its release from oceans can cause updrafts which create thunder storms. The thunderstorms shoot large amounts of heat up into the upper atmosphere where it is ultimately radiated off into space. This results in a heating-cooling loop that can keep tropical temperatures stable.
Yes, and the evaporation itself cools the Sea surface, and the clouds formed as the moist air rises reflect much incoming solar back to outer space.
Such claim are enough to give one the ‘vapors.’
The most curious thing is a diagram showing earth surface absorbing nearly twice as much power from downwelling LWR as it does from sunshine. This ignores the large differences between high energy SW and low energy LW. If the diagram were true, they’d be covering the earth with panels to capture LWR rather than solar panels. And downtime problem at night solved!
Isn’t the alarmist argument that the large change is as a result of the the CO2 (ie feedbacks) and therefore sensitivity is larger?
In the alarmist world, CO2 is the only driver of the change.
off topic, but when I saw the word “alarmist” in your text, I for some reason, saw it as a different word-the word “terrorist”
In a way, the climate alarmist are terrorists as they seek to terrorize all of us. And they’re pushing a revolution on us yet making it sound like they’re doing us a favor- which is what all terrorists say. And, we’ll be broke in a broken civilization.
hmmmm…. terrorists….
Well, there is a subset that is specifically and objectively terrorists in the traditional sense of physically causing damage and alarm.
Off topic. Alarmism is a brand of terrorism. It induces panic.
Terrorism is done to induce fear to cause people to react. Overreact. The terrorism of 9/11, no matter who caused it, was to induce fear and it worked. We reacted and now search everyone at airports — needless expense.
The terrorism of “be afraid of CO2” has worked. The world has reacted — needless expense.
Terrorism of “there’s a severe pandemic underway” has worked. It has caused people to take experimental treatments which have not worked — needless exposure to danger of untested medications.
Politicians these days do not stand on a platform but instead induce fear of what could happen if the other person is elected. It has worked.
In the early days downwelling IR from CO2 was estimated using radiative equilibrium. This is wrong for so many reasons. If one wants to claim they know downwelling IR from CO2:
I want to see the percentage of the CO2 quantum excited states in a N2, O2, CO2 mixture. This will take into account the thermalization/de-thermalization(non-radiative) coefficients of the gas mixture over temperature and pressure over the entire troposphere. Not interested in extrapolated furnace data as estimate of CO2 emmisivity. Will also need IR path length.
I find the modtran model of downwelling CO2 IR unconvincing.
You want to know more but you don’t indicate you did any study at all then say MODTRAN isn’t convincing without explanation.
Your comment is incomplete and unconvincing, you need to do better than this.
The very first time I found a reference(2005 time frame) to the 3.8 watts from doubling CO2 the paper referenced a planetary atmospheric text section that was based on radiative equilibrium. The very next section in the text discussed why using this gave wrong answer. This is same answer modtran gives today.
I have been on a deep dive in Physical Chemistry and still do not feel I have a complete picture of the CO2 excited states over the troposphere. With out this you can not know the CO2 IR emmision rate. Nothing that I have seen indicates this is taken into account in modtran.
Thanks for the reply but you missed my intention that you need to make your case which you still haven’t done that 3.7 W/m2 is wrong or irrelevant.
The IPCC are the ones who is pushing the doubling CO2 formula scenario for years which is hilarious since it doesn’t support them at all as the postulate warm forcing increase is too small to matter.
With the still never found “Hot spot” and the mythical Positive Feedback Loop which never existed outside of those silly models in just last BILLION years or so is still not found thus a broken record in trying to make CO2 some kind of a super molecule continues.
Really this entire CO2 is some kind of a powerful warm forcing bullshit propaganda deserves radicle since it is STUPID as hell as their two biggest predictive tests has never showed up after 30 some years.
DT,
I’d like to see figures on the amount of energy that can be accepted by a CO2 molecule from photon activation. The must be a certain CO2 concentration below which it cannot do the required lifting. It is complicated by the time taken for events to happen before CO2 can discharge it’s energy and be ready for more photons. I did Physical Chem years ago before molecular spectroscopy was developed very far. Geoff S
Maybe not important, but eyeballing the graph shows the total change (red line) should be about 1.25 W/m2/decade, not 1.0
Would this affect your TCR/ECS estimates?
I used linear trend, not finish minus start.
Best
w.
Why is “the change from clouds” in Fig 1 twice as big as the difference between clear-sky change and total change?
Most observant. Because it’s clear-sky without CO2.
w.
In what world doesn’t radiation net out? The graph is deliberately misleading in order to highlight radiation. Sad.
Nelson, I am an honest and honorable man. You’ve never met me. Despite that, you falsely accuse me of “deliberately misleading” people. That is a base, scurrilous, and most importantly, an untrue accusation.
I may indeed be wrong, but I tell the exact truth as best I understand it.
Regards,
w.
Willis, I am complaining about the energy budget graph, not your work. The energy budget graph is just wrong. How to interpret the CERES data is a different matter. You always present accurate data.
Huh? Nelson… I believe that graph is also a previous work of Willis’s and has more accurate numbers than most….
What I’m never happy with in the lead pictorial is the lack of temporal resolution.
My understanding is that black-body radiation emittance and spectral radiance always increases with rising temperature, at every wavelength. Therefore, on the most simple analysis, the earth could not emit more to space in the CO2-sensitive region than it actually receives from the sun.
That first order approach is wrong, presumably because the dark places on the planet must continue radiating at night and in winter to make up for what is not achievable during a sunny day.
Much of that outgoing IR radiation is of course derived from the daylight sunshine on a nice day or season, strongly affected by clouds.
Further, many people notice the difference in cloud cover between day and night, summer and winter.
Do they actually measure the diurnal cloud variation in the tropics?
And, if so, how accurately and precisely?
The earth is not a black body radiator nor is it of uniform temperature.
One example is that insolation hits the ground as visible and UV light, converts to heat and the surface re-radiates Iin part) in the IR range. There is no reason to expect the IR in to equal IR out. Outgoing should always be larger.
Exactly! No one said that each individual wavelength has to balance.
A simple question that I can’t find the answer to:
How do they measure the surface radiation?
What do you mean by “surface radiation”? Radiation received? Radiation emitted? Just curious.
“It’s calculated as the upwelling longwave radiation from the surface minus the upwelling longwave radiation measured at the “top of atmosphere” or TOA, meaning in this case at the satellite.”
How do they measure the radiation from the surface?
Maybe you mean instruments ? Look up pyranometer (used for SW sunlight), pyrgeometers and net pyrgeometers (used for IR facing up or down). Manuals are on the i’net and are very informative.
Well, CERES doesn’t. But pyrgeometers do. They measure a negative amount of downwelling longwave IR power at night (averaging around -50 watts/m^2, strongly dependent on humidity). Of course, they don’t want to report this number, because it doesn’t support their alarmist cause, so they adjust it by about +400 W/m^2, and then report that. Don’t believe a word of it… pay attention to the unadjusted numbers only.
“More than offset the change. . .”. There is feedback, but negative feedback, not the runaway positive feedback postulated by Hansen.
Thanks, Willis
Brooks
“The CERES dataset continues to yield new insights. My joy is to graph different relationships and then see what I can learn and find out from the graph.”
Wow, sounds scientific but I thought the topic is settled so all you gotta do is ask the authorities! They have models that explain the climate and its great threat to us- and they’re trying hard to save the planet with clean and green and cheap energy. Trust them and panic whenever the weather ain’t perfect.
The only real need in the scheme is to bring the actual production of “clean and green” into balance with demand. It is demand that must be sharply adjusted and we already know what doing that requires.
What is CERES really telling us?
Surface LW up = 399
Surface LW down = 345
OLR = 240
Compute how much of surface upward flux is never coming back:
Transmitted Flux = Surface LW up – Surface LW down = 399 – 345 = 54
54 W/m2 transmitted up from the surface, never to come back.
Optical Depth = -ln(transmitted/Surface LW up) = -ln(54/399) = 1.999977
Optical Depth = 1.999977
It beggars belief, when computing direct from CERES output, that the LW opacity should be such a round, whole, spectacularly perfect depth = 1.999977. You can’t find it better in textbook, and you couldn’t dream of finding something so perfect in nature. That’s the stuff of mathematical deduction, rarely found for real outside.
What is the meaning of that? I suppose maybe CERES is toooo perfect, or by bizarre coincidence the CERES system just happened to be launched and collecting data specifically when optical thickness was passing an unimaginable round and perfect value. Alternatively, the data is telling something of an optical constant in semi-transparent condensing atmosphere that should have been obvious long ago.
Or, on a different tack, surface emissivity of 54/399=0.153.
No, surface LW emissivity is about .95, even for snow!. 54 is the net heat transfer between surface and cooler sky. Just because 2 parallel plates are both 289 degrees (399 Watts emitted), so no net heat transfer in the engineering thermodynamics sense between them, doesn’t mean their emissivities are zero.
By definition, emissivity is
The viewing position is above and normal to the surface. The black body power of the surface is 399W/m^2. The radiated power of the surface is 54W/m^2. So emissivity, by definition, is 0.153.
The idea that there is radiation going back and forth is nuts. There is no radiant heat energy transfer between two parallel plates at the same temperature. It is in the same boat as back and forth convection because there is no convection if there is no temperature difference.
You cannot tell anything about the emissivities of two surfaces at the same temperature. One could have an emissivity of 0.6 and the other 0.8 when viewed separately. By your definition you are suggesting the one at higher emissivity will warm the one at lower emissivity – doesn’t happen. They will equilibrate to the same temperature irrespective of the different emissivities when viewed separately.
“The idea that there is radiation going back and forth is nuts.”
I disagree. All matter above absolute zero radiates. That there is no “net” energy transfer doesn’t mean that there is no energy transfer occurring.
No it doesn’t. EMR only travels in one direction at any point in space and time. It is a requirement to satisfy Maxwell’s equations. Proof here:
https://opg.optica.org/oe/fulltext.cfm?uri=oe-18-19-19770&id=205485
So lets take two parallel plates; one rusted steel plate and the other polished gold. The emissivity of the steel plate is 0.6 while the emissivity of the gold plate is 0.02.
Both plates have their back side heavily insulated as well as the sealing chamber around the edges of the plates. The space between the plates is evacuated. Both plates have a start temperature of 200C. What is the initial and final radiation from each plate? What is the final temperature of each plate given they have identical thermal mass and enough time for the plates to settle to their final temperature?
“No it doesn’t.”
Heh! You are completely wrong.
Are you saying that no photons leave either plate just because they are at the same temperature??
I think you have radiation and heat transfer mixed up. But that’s ok, you’re still way ahead of most climate scientists.
Rick – This MIT heat transfer textbook: A Heat Transfer Textbook, 5th edition (mit.edu) addresses this exact problem on book page 563 (file page 575) and carefully derives through the principle of “radiative exchange” that there is no NET heat transfer between to such plates of identical temperature.
I find it mystifying that someone who claims the technical background that you do could be so confused about very basic concepts. For example, you completely misinterpret the definition of emissivity, and it goes downhill from there.
This shows how screwed texts books can be.
There is no radiative transfer because they are at the same temperature. That does not require any derivation. It is the basis of the second law.
Seriously, Rick? A body stops emitting thermal radiation when there is another body of equal temperature facing it? What is the mechanism for the body detecting this state of affairs?
Are you really claiming that MIT is teaching its engineering students fundamentally wrong physics? This explanation is the same as when I studied there almost 50 years ago. If they were fundamentally wrong on this, there would have been great pushback in the mean time.
Yes Rick is not recognizing (in my 2 plates of equal temp scenario) that if the one surface emits photons…it would cool down, so it must be receiving as much energy in the form of photons from the other plate.
I do not need to worry about your photons zooming about because there is no EMR exchange between two surfaces at equal temperature.. I know there is no heat transfer because they are at the same temperature.
You clearly didd not understand Mishchenko’s paper on the-directional nature of EMR.
You are completely wrong. Of course there is EMR exchange between surfaces at the same temp. There is no NET HEAT exchange.
Unidirectional nature of EMR…really ?….so two flashlights can’t shine on each other. Ha, ha, ha, ha, ha, ROFL
Exactly. They both responded to the presence of each other. The interference of light waves is well known but there is spectral dependence that produce interference patterns.
Two coherent beams, split from a single laser, can completely cancel each other so there is no energy transmission (no light).
https://www.youtube.com/watch?v=RRi4dv9KgCg
Your understanding of EMR is bedded in classic radiative transfer theory rather than the fundamental physics of EMR described by Maxwell.
In JCM’s case the surface LW up is 399. (not BB power as you state) Thus the surface was at a temp of 293.37 K. (assuming emissivity is .95)
If the surface was a BB to emit 399 it’s temp would be 289.63 K
Is it appropriate to be talking about emissivity when we’re discussing the atmosphere? It’s a complicated heat engine of mixed substances, and emissivity is only meant to deal with how efficiently a material can radiate. Doesn’t include convection and conduction, spreading the heat to the poles or the dark side.
It’s an apples and wrenches comparison.
It was RickWill who erroneously said the emissivity of the surface was .153
I have always found it biased to explode out the radiative terms into the ups and downs, and to illustrate the non-radiative terms as a net. Latent and Sensible heat have ups and downs too, I guarantee it. Hiding such things is inconsistent.
Anyways, sticking to the exploded radiative paradigm and adhering to the rules of the headpost:
Having established the supremely round and stable optical depth = 2 boundary condition for thermal IR, it can all be reduced to Surface LW up = 2x Atmospheric LW up.
Clouds all in. in fact, clouds are essential for making it so.
Surface Up / Atmosphere Up = 2
I have never seen physicists do this, breaking out power into “gross” and “net” parts. Only climate scientists do it. In regular physics, power is calculated as a single number, in a single direction, represented as the Poynting vector, and it is always developed from a hotter object to a colder one. Never the other way around.
it’s a consequence of two-stream approximations which is unique for describing observations in semi-transparent media, such as atmosphere. The atmosphere should be thought to ‘cool’ or emit radiatively using both upward and downward “beams” under this framework. It’s useful to explain why one can ‘observe’ light from a semi-transparent media whether looking up or down, even when the radiation source is only from one side.
The main factor for radiative transfer through the media is the attenuation coefficient. The integral of attenuation is the optical depth.
Similar logics can also apply to latent flux, whereupon condensation as frost at the surface is a downward latent flux from the air. Or in the case of a temperature inversion, which is common over polar ice sheets, the warm air aloft forces sensible heat downward to the cooler surface
Read any introductory text on heat transfer. Radiative heat transfer is always explained as the NET result of GROSS flows. See for example this MIT textbook: A Heat Transfer Textbook, 5th edition (mit.edu) which explains it in the concept of “radiative exchange.
The Poynting vector is an abstract concept that is the result of these fluxes.
Thanks, Ed. Here’s the relevant diagram from the MIT textbook at your link.
Best to you and yours,
w.
Thanks for the diagram, Willis. Note that “Q1 to Q2” and “Q2 to Q1” are described as “radiating energy to one another“, not power. This is the crux of your entire misunderstanding.
Steve, it’s an energy FLUX. A flow of energy.
Are you ever going to get off this silly nitpicking and talk about the actual scientific issues?
w.
Willis, the difference between “energy” and “power” is not “nitpicking”. Nor is it “pig wrestling” or “leading a horse to water” or “pond scum” or “Pass!” or whatever other excuse you are planning to make next. Are you ever going to learn some basic physics? You know, actual science?
Radiant energy (what they show in the diagram that you posted, and they clearly labeled it as such) is measured in Joules. We already established this, remember? You yourself wrote “radiant energy (SI unit: Joule, J)”. But that was a couple of weeks ago, so you probably forgot. And I thought we were making such good progress, too.
“Flux” (if you want to use that term as a synonym for “power”, for some reason) is a whole other animal. You can tell because it’s labeled in a whole other unit. We haven’t gotten there yet, and it’s not in this diagram at all. (It’s a bad word to use for photons anyway; they don’t “flux” like water or molten metal do. But never mind that for now.)
The Poynting vector is based on field theory developed by Maxwell, which defines the physical basis for electro-magnetic radiation.
There is but one electromagnet field and energy transport in the field is unidirectional at any point in space and time. The proof is here:
https://pdfs.semanticscholar.org/c03b/2b493f57e13d3c3e2b58d17c9656d2dee978.pdf
You need to get a grip on fields and energy transport in them.
Gravity is a field not a lot different to the E-M field. Gravity waves and EMR travel at the speed of light in a vacuum so magnetic permeability and electric permitivitty are both associated with the speed of gravity waves. You cannot determine the gravitation pull between two object without knowing their individual mass and the distance between them. Likewise you cannot determine the EMR power flux between two objects without knowing their temperature and distance between them.
A radiating source will radiate to space at the characteristic impedance of 360.7ohms. But if that radiation encounters matter, the field will adjust accordingly and the source will detect its presence in the time it takes for light to travel to the matter and back. The sun’s output is reduced slightly by the presence of planets because their existence alters the E-M field. If Earth instantly disappeared, the sun would experience it 8 minutes later in both gravitational and E-M fields.
And all textbooks never date.
Any text that purports to discuss electromagnetic radiation without reference to Maxwell field theory is lightweight indeed. It will provide very little physical understanding of energy transfer.
Here is a true giant of atmospheric EMR transmission:
https://www.youtube.com/watch?v=P_36ML1TuwA
Take the time to watch the lecture and then spend a few days on Maxwell’s equations to get a grip on EMR.
You do know that Maxwell’s equations are approximate. QED replaces Maxwell because QED is more correct.
Rick – The Poynting vector you are so enamored of is simply an abstract analytical method of describing the NET RESULT of multi-directional fluxes. It says nothing about the causes of these fluxes. It adds no information or insight into heat transfer calculations, which is why it is not used in these analyses.
This highlights your lack of understanding of EMR. Poynting vector is more more than an abstract. It provides an important understanding about the nature of EMR. There are no multi-directional fluxes in THE electro-magnetic field at any point in time and space – there is only one field. The field responds to each source at the speed of light. Coherent laser beams produced from a single source will completely cancel each other. There is no energy flow from the source:
https://www.youtube.com/watch?v=RRi4dv9KgCg
Classic RTE approaches were found to be inadequate when trying ti understand radiation transmission through the atmosphere hence the need to resort to the fundamental physics rather than approximations based on classic RTE.
Rick – What you call “RTE” approaches are now based on modern quantum physics and statistical mechanics, which were unknown in Maxwell’s time. We understand very well the mechanisms for thermal emission of individual photons, as well as their absorption. The fact that it is possible to compute a value associated with the resulting EMR field does not negate this!
Ed,
At last. Somebody who can tell us all what happens to the photons emitted by a block of ice totally submerged in water!
Assume the ice is emitting 300 W/m2, not terribly cold. The water is liquid, and you can pick the temperature.
Well, you did say “We understand very well the mechanisms for thermal emission of individual photons, as well as their absorption”
You weren’t just making stuff up, trying to appear intelligent, were you?
In what way and how is 1.999977 “an unimaginable round and perfect value” rather than just a result of displaying far too many non-significant digits?
I agree completely. Between measurement uncertainty and resolution, the number of significant digits is wayyyy to large.
OFTLOG it’s a 2. I was illustrating how remarkably two-ish it is. It’s the most two-ish a thing can possibly be.
In the Fig, the yellow curve is marked “excluding CO2”. Should that be excluding CO2 and clouds?
Thanks, Nick. The clue is in the “Clear-Sky” part of the name.
w.
Clear sky does NOT mean there is no water vapour.
B Nice, Nick asked about clouds, not water vapor.
Regards,
w.
Water vapour is still present often in significant amounts, so any calculation should not just dismiss it.
Very good point – how are the CERES satellites data parsing out what’s absorbed by water vapour vs by CO2 if the bands overlap?
Water vapor is not “dismissed”. I assumed people would know that It’s the main player in the amount of clear-sky “greenhouse” radiation. Here’s a graphic from a previous post of mine entitled “Wandering Water“.
w
https://wattsupwiththat.com/wp-admin/upload.php?item=7680349
Pretending “clear sky” means only CO2 and no H2O is complete nonsense.
But it is what the AGW cult does… don’t be one of them. !
You have a good point but you don’t have to be argumentative about it. Willis is one of the most level-headed people posting on any issue so he deserves better from us.
Not to mention dust – which absorbs 100% of incident LWR. Do the dustiest parts of our atmosphere show a “super greenhouse” effect from the dust absorbing a larger percentage of the LWR? No, in fact it shows the complete opposite.
Mmmm … Robert, you’ve left out the other half of the equation. That’s where the dust is also absorbing a larger percentage of the incoming sunshine.
And when the dust absorbs that solar energy, it gets radiated in all directions, with about half going down and half going back up.
The net result is a cooling of the surface.
Why cooling?
First, because instead of the full amount of solar energy striking the surface, it only gets ~ half that amount after the solar energy is absorbed and re-radiated by the aerosol. So the aerosol is cutting the incoming solar energy in half, a very large loss.
Second, depending on the aerosol, it may significantly increase the amount of energy reflected back to space. This happens with e.g. sulfur dioxide from volcanoes.
Note that CO2 does neither of those two things.
Finally, the loss of solar energy via reflection plus absorption is only partially offset by the much smaller increase in absorbed LW.
Result? Cooling.
Best regards,
w.
Yet the Sahara surface day time highs are 110+ in summer and 90+ in the winter. Venus also shows this cooling effect from only 10% of solar radiation reaching the surface. Okay
Good heavens, Be Nice. I don’t “pretend” things. I tell the truth as best I know it. And you might benefit by reading twice before commenting. It seems you might have missed this part. From the head post (emphasis mine):
So I’m sorry, but your assertion is neither true nor pleasant. I listed water vapor first in the list above for a reason.
w.
Me and another fellow on another site were discussing this. His view is that the total change in DLR noted in the D&V paper was due to the GHG effect. I disagreed. The decrease in clouds results in an increase in DSR that increases absorbed surface radiation and increases surface temp. Increased surface temp equals an increase in SLR, which by default will increase DLR and OLR simply because more energy is upwelling through the system.
The main thing I pointed out about this data is the huge jump between 2012-2016 that just so happened to coincide with a few events totally unrelated to radiative physics. First, you had the collapse of the Antarctic Ice Shelf, resulting in a decrease in SWO even in clear skies (presumably due to decreased ice at the S.Pole as well as any effect on decreasing albedo due to cloud changes). But you also had a big El Nino thrown on top of that, greatly increasing SST.
As I see it, there are two disconnected processes going on. Charging followed by Discharging. They are never in equilibrium but follow their own paths within defined limits. Temperature doesn’t necessarily have to increase when the system is charging, and there doesn’t necessarily have to be an increase in GHG effects etc during a discharge. Surface Temperature marches to the beat of its own drummer to a certain extent, (given we are talking tenths of a degree change), and is determined by lots of factors including heat capacity of the respective surface material. The time tables for temp change are different and so will be the changes in DLR and OLR. People always like to think of the Climate System as being some sort of “equilibrium”, but frankly, I don’t think equilibrium is ever achieved in climate. There’s just too many sinks, leaks, and uncertainties in it all.
I very much agree with this view of there being “ just to many sinks leaks and uncertainties” and chaotic non linear connected effects and so forth. But I’ve often wondered in this day and age , it just seems like we should be able to measure things better. I do understand even sat measurements have their issues etc . But still …?
We are limited to instrument measurement precision. At the moment we just can’t measure what we want with the precision that we want. Temperatures, at the very best, to 0.3-5°C, sea level, at best, to 32mm. Once you start looking at that it makes an absolute mockery of the 0.01° or smaller changes to temperatures, or the 0.2mm changes to SLR that are both, obviously, just a fabrication of statistics and nothing to do with the real world.
As always. It’s the dynamics that matter. It’s not about being at equilibrium.. it’s about the forces at work that drive systems toward an ever changing equilibrium.
of course DLR exists … it just can’t heat up hotter air … you just need to figure out what it DOES do …
The strength of DLR is zero.
Better to use sign convention: Upwelling average [hotter to colder] +ve; downwelling average [colder to hotter] -ve. If the difference is positive, we’re cooling, if negative, we’re warming.
I don’t understand why conservation of “photons” is built-in to so many analysis.
For a photon, E=hf, so a few high energy photons from a hot surface like the sun….absorbed by say the surface of the Earth….will result in many low energy IR photons being emitted to cold outer space by the lower temperature surface….so conservation of photons isn’t really a discussion topic.
I am sure Londoners will be slightly mystified as to why the Docklands Light Railway is having an effect on climate…
Thank you for pointing out one of the problems with using an acronym in the headline of the post, It doesnot add up.
Regards,
Bob
Ah. Mornington Crescent – I win.
The whole article is a mystery. I barely ever read such total nonsense!
1) You should learn what the GHE is. Here is a correct(!) definition by the IPCC..
2) There is NO “back radiation” or DLR included, for good reasons. The IPCC dropped this errouneous idea of “back radation” with AR5. It is completely unrelated to the GHE.
3) The “back radiation” driven GHE theory fails on EVERY account. The perspective of it violating the 2nd law of thermodynamics is just one of which, and more for the simple minded.
4) “greenhouse radiation” is indeed poorly named. No one uses this termin. That might be because it does not exist.
5) Ouch, ouch, ouch. All wrong here..
No, a satellite can not measure downwelling radiation. I mean maybe yes, if you abstain from shooting it into space, keep it on the ground and let it look up to the sky, but that is a different story.
The reason why the atmosphere emits less radiation into space, is because it is colder than the surface. This delta in emissions due to the atmosphere is the GHE. It is unrelated a) to the amount of upwelling surface radiation absorbed by the atmosphere and b) to DLR.
6) Accordingly it is totally pointless trying to relate changes in DLR to CO2 forcing. Again, they have nothing in common. DLR increased because the temperature increased, that is all.
7) The mystery you are talking about is only reflecting the fact, that you have no clue what the GHE is and what CO2 forcing and so on mean. Sorry, but you will have to learn the basics.
The only way you can measure CO2 radiation is with supercooled sensors.
ie, by creating the conditions (negative temperature gradient) for it to exist. (eg Feldman et al)
No handheld sensor I know of can pick up CO2 radiation. It is outside their frequency range.
All matter emits infrared, if above absolute zero. If you want to call the IR emitted by the atmosphere “DLR”, “backradiation” or any other jargon term, fine.
Nobody at all has ever managed to describe the GHE in any way which agrees with reality.
For example, where may the GHE be observed, documented and measured? What is it supposed to result in – heating, cooling, or perhaps both simultaneously?
Complete nonsense. The Earth has cooled since the surface was molten. The surface cools every night, losing all the heat of the day, plus a little of the Earth’s internal heat content.
No GHE needed. Just ordinary physical laws in action.
Cooling does not seem to happen “every night”. The easiest to observe example of this is that the days start of warmer, even hotter, as late spring and early summer progress, If earth lost “all the heat of the day” each dawn would start out equal. Maybe over the annual cycle all he heat of all the days is lost, but certainly not day by day.
AndyHce,
Well, if the temperature increases in the absence of sunlight (possibly a campfire in the vicinity), I suppose you would be right.
You may not be aware, but even when a low level inversion exists, making the atmosphere hotter than the surface, the surface still cools.
Baron Fourier agrees with me “Thus the earth gives out to celestial space all the heat which it receives from the sun, and adds a part of what is peculiar to itself.”
You might also be aware of seasons, where in summer, sunrise occurs earlier, leading to increased daytme temperatures, in general.
In any case, leave a bowl of water in the sun. See how much of the sun’s heat has been retained when you check its temperature just before dawn. None. All else being equal, of course.
You may find the attached chart of interest.
I have been revisiting the method of determining the lowest level of free convection for the atmosphere starting from saturated conditions.
The attached chart shows the long wave emissivity to space; the absorption of long wave radiation and the transmitting power. For the 290K surface and saturated atmosphere, the radiating power is 237W/m^2 and the effective altitude of emission is around 4700m.
The LFC forms at the altitude where outgoing emissivity and outgoing absorption are equal. For the conditions used in the chart, that occurs at a value of 0.56. Condensate that descends to this altitude will absorb as much long wave heat from the surface and air below as it emits to the air above and space without any requirement for convective heat transfer. Below this altitude, the air is heated from surface absorption. Above this level the short wave emission cools the surface. This is the altitude where adiabatic equilibrium no longer applies and the reason the atmosphere can partition to high altitude dehumidified zone and lower altitude saturated zone.
The optical depth of the water vapour used for this analysis equates to 38% of broad spectrum long wave transmission at 1km for the near surface conditions. This result is similar to what Lowtran produces for tropical atmosphere near surface level.
Willis is a smart dude but it seems like he has never studied this topic. He appears to still maintain the media view of the greenhouse effect. I think the concept of the enhanced greenhouse effect was first mentioned in the TAR. Of course, it is also junk science but requires far more knowledge to understand why. Here’s what is really going on in the atmosphere.
The atmosphere absorbs surface IR very close to the surface. Almost all of absorption occurs within 10-20 meters. Conduction also heats this part of the atmosphere. Almost all the energy is transferred to other molecules (as kinetic energy). A better analogy for our atmosphere is that of a baseboard heating system. Energy enters the atmosphere at the bottom and moves up through the atmosphere via conduction, radiation and convection.
As this net flow of energy moves upward, the atmosphere gets thinner and thinner. There are less molecules to harbor kinetic energy. So where is the energy going? Most people don’t realize that energy continues to be lost to space from all altitudes. Every now and then an upward directed photon never gets reabsorbed even when radiated only a few meters from the surface. It is lost to space.
The radiation events move energy in all directions so naturally some of it is directed downward. Unless that event occurs very low in the atmosphere, the DLR energy is almost always reabsorbed and stored back into kinetic energy pool. The concept of a DLR flux is based on seeing DLR at all altitudes, but that is a misconception. That energy is almost always reabsorbed. There is no DLR flux.
Radiation moves energy through the atmosphere in very short jumps. It only moves a few meters on average. It starts as part of the kinetic energy pool, the energy moves a short distance after a photon is emitted and is reabsorbed back into the pool by another molecule. The average distance upward is greater than the average distance downward due to the changing density. This is why the net flux is upward.
The bottom line is there is no greenhouse effect in the conventional sense. If you double CO2 you increase the number of radiation events, but the path length before reabsorption shortens. You have more energy moving a shorter distance. The net is that total energy flow is the same. It is based on the changing density, not the amount of CO2.
The only reason increasing CO2 produces a radiative warming effect is by absorbing additional energy at the wings of 15 micrometer band. It has nothing to do with any kind of greenhouse effect. In fact, the actual function of the DLR that does reach the surface is to induce evaporative cooling. That’s another story.
These things?
The photon is a convenient way to look at the energy flow. I also doubt it exists in reality. I sometimes refer to a photon cloud rather than individual photons and reduce the actions to an average photon to understand changes due to doubling CO2.
Yes. Photons are not particles. Radiation is via electromagnetic waves as Planck shows. Planck called the discreet energy that be absorbed from an EM wave, “quanta”. The discrete energy available in a single quanta/photon is defined as E = h•f. So, as frequency increases so does the energy in a single quanta.
Can an EM wave “act” as a particle? Of course, everyone should be familiar with the slit experiment. That is called duality. But, as the radiation travels through space, it does so as an EM wave.
EM waves do diminish by 1/r². Total power is the same, but it is spread over a larger and larger area. The total power available to a single molecule is reduced by factors of distance and temperature of the molecule.
Too many people assume that “downward” radiation always reaches the surface. It does not. From altitude, downward radiation is quickly absorbed. That is one reason most radiation diagrams are wrong.
Averages simply can’t do justice to the processes taking place in the atmosphere. They are dynamic and act under exponential and trigonometric influences.
Back when I was in school learning quantum mechanics as a double E student, the Planck-Einstein equation was E = n*h*nu. The Greek letter “nu” is frequency in physics. (I’d use LaTex, but I can’t get it to work on this site anymore.) Notice the “n” term that is usually left off. The n value is 1, 2, 3, 4, . . . .
You can collapse the quantum wave function by measuring it before it is incident with the matter it is absorbed by, and then it no longer acts like a wave but still tranfers energy in a discreet packet. How does it do this if there is no particle after wave function collapse?
You don’t really collapse the whole wave function by measuring it. Energy is extracted from the wave and the remainder keeps on. The EM wave is a plane wave and doesn’t entirely disappear from existence when a molecule or test probe extracts energy from it.
As to the absorption mechanism of a molecule, I would have to research it.
Energy is extracted from the wave?
Sure. How do you think an antenna and receiver works? Energy is extracted to move electrons.
“Willis is a smart dude”
Well, he sounds smart, like a badly trained Large Language Model, until you point out any of his erroneous physics assumptions, at which point he rapidly reverts to his true nature – which is that of an ignorant petulant fisherman, calling people names and sticking his fingers in his ears shouting “Pond scum! Pass! Pass! Pass!” Or, in this case, simply threatening to censor anyone who points out that he is wrong. You can’t teach him anything, and many of us have tried, for many years. That is not “smart”.
He admits that he is “not a theory guy”, which is true. But he nevertheless spouts nonsense about “data” all day long without understanding any of it. Is that “smart”?
Steve, if you wish to show that I’m wrong, you need to do a couple of things:
1) QUOTE what I said that you think was incorrect, and
2) SHOW (not claim but demonstrate) exactly how and why I’m wrong.
You do neither of these. Instead, just as in your comment above, you lash out with meaningless but curiously fact-free attacks on my character, my education, my employment history, in fact, anything but the science at hand.
Which is why I often just say “Pass” to your endless slimy attacks—there’s nothing there to respond to, just your pent-up bitterness.
Best regards, and I hope your unbridled and unfounded hatred of me subsides. It’s not doing either your health or your reputation any good.
w.
PS—If you wish to stop with the mudslinging and actually discuss the science, here’s how to show that I’m wrong. Hele on.
https://rosebyanyothernameblog.wordpress.com/2019/03/10/agreeing-to-disagree/
Willis, I have quoted many times what you said that was incorrect. In this case, it was “DLR exists, it’s measured both manually and automatically all over the planet every day, and it doesn’t violate the 2nd Law”. Yet somehow, quoting your errors back to you never increases your understanding of physics, does it?
The problem is that you say “DLR” and you mean “energy”, but then you write “Watts”. These are not the same concept. Not even close.
stevekj November 22, 2023 4:53 am
Sorry, that’s not clear. I made three statements there.
One was that DLR exists.
The second one was that it is measured both manually and automatically every day.
The third is that it doesn’t violate the Second Law.
Which one of these statements are you claiming is incorrect, and why?
I’ll wait.
DLR is an energy FLUX, a constant flow of energy, and as such is measured in either watts or watts per square meter. Here’s NOAA on the subject:
Don’t like it? Go argue with NOAA.
w.
Okay, Willis, we’ll take the 3 statements one at a time.
“DLR exists” is ambiguous. You haven’t specified whether you are referring to radiant energy or radiant power. DLR energy exists, DLR power does not. So this statement is simultaneously both true and false.
“it is measured both manually and automatically every day” That is true, but the measurement (which is negative) is not the number they report – so the implication that “what you see is what they measured” is false. If you don’t want to stand on that particular implication, which is me putting words into your mouth after all, then feel free to disclaim it, and I will agree with you.
“it doesn’t violate the Second Law” is true when referring to DLR energy, but not when referring to DLR power. So, again, ambiguous, and therefore simultaneously true and false.
All this ambiguity is, naturally, not great for carrying on a scientific debate. Therefore, in future, please be careful to distinguish between radiant energy and radiant power, so everyone is on the same page. The word “radiation” by itself is insufficiently precise.
NOAA isn’t being any more scientific than you are when they refuse to distinguish between radiant energy and radiant power. I can’t fix that for them. If you assume they are referring to power, which they didn’t specify, but we can probably safely assume it because they measure it in Watts, then sure, they can measure positive outgoing longwave radiant power at the top of the atmosphere, from the warmer atmosphere to the (maybe) colder instrument they are using situated in space. Nothing wrong with that as far as the 2nd Law goes. (They won’t tell us their instrument temperatures, though, so the whole operation is pretty dicey from a scientific standpoint, but it’s the government, after all, and we know they have an anti-scientific control and power-grabbing agenda, as you told us yourself, so you can’t expect much)
As long as time (t) and area (A) are constant the 1LOT still applies. Here is the derivation.
ΔE = Ein – Eout
ΔE / t / A = (Ein – Eout) / t / A
ΔE/tA = (Ein/tA) – (Eout/tA)
and since F = E/tA…
ΔF = Fin – Fout
Since the area of the 3 bodies core, surface, and shell are very nearly the same we can approximate and model all 3 as having the same area (A). And, of course, all fluxes are assumed to be in reference to the same time (t).
What this means is that fluxes in Willis’ diagram are conserved just like energy is conversed. We can now present the energy budget in terms of fluxes (W/m2). The reader is assumed to know that they can convert fluxes (W/m2) into energy (j) by simply multiplying them time (t) and area (A). It’s just that we leave the figures in flux (W/m2) form because in this particular context 1) the figures are more familiar and 2) they can then be used as inputs into other models (like the SB law assuming a negligible rectification effect) that accept fluxes.
“Conservation of power”? Who taught you your physics? Remember, power is developed from entropy differentials. Entropy is not conserved, therefore power isn’t either. At the heat death of the universe (or any isolated portion of it), the total energy in the universe/portion will be the same as it is now, but the power will be 0.
The power in Willis’s diagram is made up. It is not calculated from entropy differentials. It is fictional.
Yep…when the time (t) and area (A) is the same and constant for the system being analyzed.
College.
I didn’t say it was.
Sorry. That does not follow from the 1LOT when time (t) and area (A) are the same and constant for the system being analyzed.
Do you want to see the proof again?
In an equilibrium scenario you can define power (energy flow) as being conserved across a system boundary, yes. For the shell in Willis’s scheme, it will (at equilibrium) be absorbing a certain quantity of power from the sphere (I picked 100 Watts as a reference value in my analysis lower down in the thread) and therefore also dissipating 100 Watts to maintain equilibrium. But Willis is trying to say that it must be dissipating those 100 Watts in two different directions (inwards and outwards) at the same time. That doesn’t sound very conserved to me, it sounds multiplied – and of course power cannot be developed from a colder object to a warmer one in the first place. Instead, all of the power absorbed by the shell from the sphere is dissipated (conserved) in the other direction – outwards.
Have no idea which “Willis” you are talking about, but it’s not the guy who posted here.
The very same. If you think he knows his physics, then you know even less about the subject than he does. Remember that he told us himself he is “not a theory guy”. Are you?
Steve, as usual, you are twisting my words beyond recognition.
By “not a theory guy”, I meant that I will take actual observations over theories every time. I meant it in the sense of Richard Feynman, who famously said:
“It doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are. If it doesn’t agree with experiment, it’s wrong.”
Seems like Feynman isn’t a theory guy either … but heck, Steve, if you prefer going for beautiful theories, be my guest.
And you STILL haven’t pointed out one thing wrong with my physics, despite many attempts.
w.
Preferring observations over theories is great, and that’s what Feynman said too, as you pointed out. I think Einstein said something similar also. But there is a bit of a problem when you take other people’s word for their “observations” which are nothing of the sort. You don’t know enough of the underlying theory to spot the deception. That’s a serious flaw in your reasoning. You are not making your own observations; you are relying on other people’s. And they are lying to you.
No Richard, in the 8-14 micron atmospheric window, about 85% of photons in that wavelength range make it to outer space in a single jump (as you term it). That 8-14 microns covers the Earthly temperatures from about -60 C to +90C and that also represents about 1/3 of the energy emitted by pretty well any object with a surface temperature in that temp range. Yeah, sounds confusing until you figure out SB and Weins law.
https://www.tec-science.com/thermodynamics/temperature/plancks-law-of-blackbody-radiation/
I should add those 85% of 8-14 micron photons make it to outer space in one jump unless they hit a cloud, and clouds cover about 2/3 of the sky…..So one can make a very good case for clouds controlling the Earth’s temperature by how much leaving IR they absorb, and how much incoming sunlight they let strike the IR emitting surface.
Wow – so Modtran is basically useless in terms of understanding Earth’s energy balance.!
UChicago Modtran contains a number of cloud layer configurations that show you the changes in IR with differing cloud cover. So useful !
Modtran does not come close to producing the measured 180or so W/m^2 OLR over a tropical ocean at 303K for any of its cloud settings. So it is useless..
Where does this disagree with anything I stated? I was discussing energy which is absorbed by the atmosphere. That is why I didn’t get into the atmospheric window.
You stated “Radiation moves energy through the atmosphere in very short jumps. It only moves a few meters on average. It starts as part of the kinetic energy pool, the energy moves a short distance after a photon is emitted and is reabsorbed back into the pool by another molecule.”
….Which I pointed out was a misconception with regards to the atmospheric window and a goodly percentage of all IR emitted by surfaces at Earthly temperatures.
This post is about DLR. Energy flowing through the atmosphere. It’s not about the atmospheric window. There was no “misconception”. It’s a different topic. In addition, I mentioned “energy continues to be lost to space from all altitudes“.
What is your problem?
Your statement remains incorrect.
Not surprised you cannot admit your error.
Did Trenberth’s diagram survive to AR5?
There is back radiation, just no back radiation flux. Almost all of the back radiation that reaches the surface is emitted very low in atmosphere. Because of this the view has shifted from the previous greenhouse effect definition (still quoted by the media) to the enhanced greenhouse effect. This might help.
In Mishchenko’s words – profoundly incorrect.
There is but ONE E-M field. So the energy flow is only in one direction at any point in time and space.
Maybe the way to think about it is an analogy with water surface and gravity waves on that surface. You can have multiple wave energy sources creating their own unique wave pattern but there remains just one surface. The gravitational potential of the surface has a single value as does the velocity of surface at any location at any time. Hence the wave energy, as measured by the gravitation potential and surface velocity, at any point on the surface at any time is in one direction.
There is only one water surface and that surface is defined by a single energy state. There can be a whole series of waves combined to create that surface but it remains a single surface. Similarly there is only one E-M field and its energy is defined by its magnetic and electric field strength.
RickWill November 22, 2023 3:55 pm
The folks at MIT beg to disagree …
The whole textbook is available online, I’ve added the URL at the top left.
Notice that there are TWO energy flows going in opposite directions, Q1to2 and Q2to1.
w.
The book was first published in 1981 and is rooted in Classic RTE. There is no mention of Maxwell field theory so is lightweight physics. It’s physical basis is not universal so is not fundamental science. It is like curve fitting without really understanding what is happening.
In fact there is a disclaimer at the start of the book:
Classic RTEs are not fit for the purpose of looking at EMR transmission through the atmosphere.
You need to get a basic understanding of field theory. Otherwise you are just promoting the crap in climate fissix.
Try to find any of the information in that text book to explain what is observed in this video:
https://www.youtube.com/watch?v=RRi4dv9KgCg&list=PL4E7FAAD67B171EBC&index=13
Same basic physics – energy transmission in the E-M field.
OK, here’s a 2018 reference, since you seem to think that the basic equations of radiative heat transfer have changed in the last few decades …
Radiation Heat Transfer. Nonlinear Systems in Heat Transfer, 105–151. doi:10.1016/b978-0-12-812024-8.00003-5
Note the two flows of energy, Q1 and Q2. Just like in the 1981 textbook. You know, the 1981 textbook that has been regularly updated and is now in its FIFTH EDITION and still used in college classes … which generally doesn’t happen unless it’s still valid and accurate.
And while we’re at it, here’s another one, from Colorado State University …
Again, two separate flows, Q1 and Q2 … are we starting to see a pattern here?
w.
You keep coming back to “basic” RTEs. They are not rooted in the basic physics of E-M transmission. They are “basic”, read simplistic, for basic applications. Not applicable to E-M transmission through the atmosphere.
Use the information in any of those text to explain how the split laser beam produces no light at the target when they are in phase. The energy is coming from a single source, spilt, then destructively interfered such that no energy arrives at the target despite the target not changing.
I guess you probably missed the beginning of this chapter, Willis, where it says “Heat transfer from a body with a high temperature to a body with a lower temperature”? This is what we’ve been trying to teach you. It never goes the other way, contrary to what you wrote in the head post.
The textbook chapter also says “All physical substances in solid, liquid, or gaseous states can emit energy”, which is the other thing we’ve been trying to teach you. Energy is not the same as power. You can tell because they’re measured in different units, so even when you don’t know what the units actually mean, this should be a clue that you should not try to conflate them.
Hate to tell you but the SI unit for energy is JOULE. The SI unit for power is WATT, or Joules/second. Radiative flux is Watts/m^2, or Joules/ sec•m^2.
Power is energy per unit of time. So, when you say “radiates energy”, that is correct. Radiating Watts is correct. Radiating Watts/m^2 is correct.
Jim, as the textbook says, “radiates energy” is correct. Radiating Watts can only occur if you also have an entropy (thermal) gradient to develop power across. And then power is only developed in one direction, down the gradient. Do you have a thermal gradient?
You are trying to equivocate. Energy is a magnitude, i.e., Joules. It is an instantaneous value. Power is energy delivered over a period of time, i.e., Joules/second.
When I transmit a radio signal, it is measured in Watts. It is the amount of energy available in the EM wave. IR radiation is no different. The EM IR wave has a certain amount of energy passing a point every second.
A thermal gradient is not necessary for a given amount of power to be radiated. The temperature at an instant in time will cause a given amount of energy (Joules/sec) to be emitted regardless of any other factor. The fact that the object cools to another temperature via a gradient is irrelevant.
It’s all in the context. Photonic radiation is a construct used to understand energy flow. Within that context you have back radiation. If reality is, as you say, a massive EM field with waves, then it very well could carry only the net flow.
If we use the context properly we should get the same results either way.
Willis, if you’re reading ES and RW comments….
It is a hobby of mine to try to figure out how guys like E. Schaffer and RickWill get so far off track with their understanding of radiant energy…Engineering degrees with key lectures missed it seems….
Yeah, DMac, I’ve given up on trying to rescue fools from their own foolishness. I just nod my head and keep rolling.
w.
On this topic they aren’t off track. You should also check out this link.
I read that when it was first published and Sabine bases her presentation on the “emission altitude” concept. We are way past that over-simplification here at WUWT and know that IR as seen from outer space is a mosaic of wavelengths that comes from sea surface, cloud tops, water vapor and other GHG at various elevations, NOT some average level of the atmosphere.
Sabine has a couple of pretty kooky climate vids. In one she claims we’re all going to die from waste heat buildup in 400 years….
As far as I can tell WUWT still pretty much accepts the original greenhouse effect concept. While your description is correct, I don’t think WUWT has caught on. I also think WUWT accepts CO2 increases will have a warming effect. It doesn’t.
Richard, as far as I can tell, there’s no such thing a “WUWT” which can “accept” some concept.
Me, I’m quite clear that there is indeed a greenhouse effect. However, I don’t think that CO2 has more than a marginal effect on temperature.
w.
Part of the problem is “the greenhouse effect” has many different definitions. I think to understand differences in opinions a person needs to separate out forcing vs. feedback.
My understanding is that you believe CO2 has a warming forcing and negative (cooling) feedback to that warming..
My own view is that CO2 has both a warming forcing and a cooling forcing which cancel out. Hence, no feedback.
While the end result is similar, the processes involved are completely different. Only one can be correct.
Take some time to understand Mishchenko’s derivation of EMR energy transfer using first principals rather than constantly displaying your ignorance:
https://pdfs.semanticscholar.org/c03b/2b493f57e13d3c3e2b58d17c9656d2dee978.pdf
The key words are (my bold):
And I will add the the link to image above from AndyHce:
?fit=744%2C417&ssl=1
It starts with “There is no such thing as a photon”. Really no use reading past that. Its like saying there is no such thing as gravity.
Again you are displaying your lack of knowledge on energy quanta and a giant of the science. Lamb was awarded the Nobel Prize for physics in 1955 for:
https://www.nobelprize.org/prizes/physics/1955/lamb/facts/
So if you pointed overhead at a cloud with one of those handheld IR temperature sensors you would get the same reading as when the cloud passes by and you take another measurement of a clear sky?
No. The hand held device will send less energy to the cloud than clear sky so it will produce a higher reading. Typically around -2C for cloud and maybe -30C for clear sky.
The device is calibrated using the S-B equation and is battery powered so quite capable of send energy out and giving a reading lower than its temperature.
“send less energy”????
Please specify what frequency or wavelength this energy is, how it is generated in the device, and how it is directed at the target.
I ask because as far as I know, other than the normal thermal IR emitted by most everything, an IR thermometer absorbs energy but doesn’t “send” any energy anywhere.
w.
Loeb disagrees. He says the shortwave increase is the result of GHG (including CO2) forcing feedbacks. In other words, all of the planetary energy imbalance has its roots in GHG forcing.
[Loeb et al. 2021]
[Hansen et al. 2023] Loeb is author.
LOL… Anyone mixed in with Hansen… probably ought to find a new partner . !!
Terminally WRONG !!
I’m going to let you pick that fight Willis alone.
Yawn !
Which doesn’t even pass the sniff test. The earth has been warming since well before CO2 was on the increase. Hell even the models are warming with their prescribed starting values based on real world observations (for what they’re worth) from before CO2 is added.
Means absolutely nothing to this discussion but I’m very good friends with one of the coauthors of the first paper. Went to high school and college with him and was in his wedding. I don’t discuss climate change or global warming with him because I want to remain friends.
It would be great if you could convince him to do an article for WUWT – a discussion article where the various issues could be explained and discussed in a back-and-forth between him and Willis and any other posters. To get the things described properly and fitted together in a way that everyone agrees, and then looking at the remainder that causes the debates.
But Loeb doesn’t know that and he is just guessing because the predicted effect of an increase in GHGs in the atmosphere is a decrease in OLR, not an increase in SWR.
Therefore, Loeb’s disagreement is just a “nullius in verba,” and your appeal to his authority is just a well-known fallacy.
Was he serious? I thought he was being sarcastic mentioning that Loeb pegged SWR increase on CO2. ShortWave Radiation, right? Light and UV, no? Or did I get my short forms mixed up?
How in the world could CO2 lead to increased SWR? Is the climate making the CO2 into a laser?😜
Ultimate Climate Doom!
People are going wrong because CERES is showing the system strives to maintain optical depth, not temperature. It does this via cloud unmasking called the cloud radiative effect CRE lambda. Greenhouse factor remains constant but the distribution of temperature is free to change. More SW in, more OLR out. It does not cause an energy imbalance.
In addition to the constancy of optical depth via CRE, there is an additional 0.4 Wm-2 “cloud forcing” on temperature which remains unexplained.
NetCRE = 0
Cloud Forcing = 0.4 W/m2/decade
In sum, the SW heating is +0.7 Wm/m2/decade and LW cooling is -0.3Wm-2/decade-1
It is this residual 0.4 Wm-2 SW that remains largely unexplained, as it’s not coming from the cloud feedback effect.
JCM: “CERES is showing the system strives to maintain optical depth (…)”
WR: Interesting idea.
Optical depth for solar SW reaching the surface mainly results from clouds and water vapor.
Optical depth for surface LW radiation radiated to space mainly depends on water vapor, clouds, and CO2.
Upward convection of latent and sensible heat ‘shortcuts’ surface radiation inconsistencies and diminishes the dependency of surface temperatures on optical depth. In convection water vapor plays a main role and clouds result from convection.
Looking at changes in water vapor and clouds, weather patterns come to mind. Changing weather patterns change both the quantity of incoming solar SW reaching the surface and the share of surface LW radiation reaching space. Could natural variation (in ocean patterns and weather patterns) play a role?
So far, no physical force finished the 4 billion-year-old natural variation.
The observed greenhouse effect intensity is the balance of opacity and overturning.
Turning off tropospheric overturning makes the greenhouse effect appear stronger.
Using round numbers in units of W/m2:
Greenhouse as Observed = Surface LW up – OLR = 400 – 240 = 160
Net heat transport by overturning = Latent Flux = 80
Now turning off the overturning heat transport: the greenhouse effect intensity increases by 80 units.
Optical Greenhouse* = 160 + 80 = 240*
Surface LW up (480) – OLR (240) = 240*
There the optical depth is revealed much more plainly in the absence of overturning; Surface LW up / Atmospheric Up = 2.
A true geometric constraint, irrespective of dynamics.
Notice that in the absence of overturning, the Greenhouse Effect is about 3/2 stronger (240 units vs 160 units)
Ratio of Optical Greenhouse vs the Real One = 1.5 or 3/2
This is equal to the ratio of the dry adiabatic lapse rate and the environmental one
Dry Adiabat 9.8 K per km vs 6.5 K per km = 1.5 or 3/2
Wonderful
Without overturning (atmospheric heat transport) the lapse rate is equal to the dry adiabatic one and the Greenhouse is 3/2 stronger. Only latent flux can impact the lapse rate. Only latent flux describes net heat transport. Latent flux is king when it comes to global climate variations. Nothing else.
So we can see clearly that heat transport and opacity both have strong influence on the observable greenhouse effect intensity. The perfectitude of the relations in simple ratios is revealing of natural geometric constraints and the beauty of numbers.
I am serious.
Loeb does peg the ASR increase on the shortwave feedback.
[Donohoe et al. 2014] provides a quick summary of how it works.
Nope. The CMIP models predict an increase in both OLR and ASR. See [Donohue et al. 2014] for why this happens.
Hence the term “perpetual motion machine of the second kind”.
bdgwx November 20, 2023 5:21 pm
Sorry for my lack of clarity, bdgwx.
The largest player in the greenhouse DLR game is clouds. When a cloud comes over, the DLR can increase by as much as 100 W/m2.
Second largest player in the greenhouse DLR game is water vapor. It changes the DLR by tens of W/m2.
Third largest is CO2. 3.7W/m2 per slow doubling.
Regards,
w.
Willis, you have the theoretical value for CO2, but should there not be additional effect from the other greenhouse gases? The total from CH4, NOx, and others is smaller than CO2, but not negligible.
Interesting info, as always.
In theory they shouldn’t be negligible. However in the real world do they have an effect or are they swamped by water vapour?
I frequently criticize professional alarmists, and those carrying water vapor for them, for not showing error bars or uncertainty ranges. How about including them in your articles? I have encountered such things as a one-sigma value being +/- 100%, which doesn’t give me a lot of faith in the conclusions drawn from such measurements. I’m not familiar with the precision of CERES data and I’d like to know just what sort of precision we are dealing with in using CERES data.
Good question Clyde. Here’s what the CERES tech note says:
Source:
https://ceres.larc.nasa.gov/documents/cmip5-data/Tech-Note_CERES-EBAF-Surface_L3B_Ed2-8.pdf
Thank you, Ron. If I read that table right, the reflected visible light from the oceans (excluding polar regions) is about 12 +/-11 W/m^2. It doesn’t specify whether that is climatology uncertainty (68%) or physics uncertainty (95%).
It is things like that which was my motivation for encouraging Willis to include uncertainties in his presentations. An uncertainty of approximately 90% in a fundamental climate modeling parameter, the albedo of the oceans, does not comfort me that things are being done properly by those earning good salaries.
Clyde, yes, that’s the way I read that. A more detailed NASA paper states that the uncertainty refers to discrepancies between satellite and surface observations (in particular a set of ocean buoys detecting radiation fluxes)
https://ceres.larc.nasa.gov/documents/DQ_summaries/CERES_EBAF-Surface_Ed4.1_DQS.pdf
I find it a little strange that someone would down-vote Ron for ostensibly providing official, vetted information regarding the over-arching question of the utility of CERES data. It says a lot about the mentality of someone who apparently has no use for facts — as in “sweets for the sweet.”
Willis thank you for making WUWT worth reading this week with real mentally digestible material.
“ I don’t want to have to snip comments claiming no DLR or that the “greenhouse effect” violates the 2nd Law . . . .”
I don’t understand Mr. Eschenbach. The Second Law only applies to isolated systems. Clearly the Earth and the Earth’s atmosphere are not isolated systems. There is no requirement for the Earth’s atmosphere to obey the Second Law–although it may. You should be able to shut down any argument on that fact alone.
I find this an odd comment. My thermo book uses the atmosphere as and example of a large reservoir for use in development of the second law.
As the second law also deals with entropy is Jim also saying the atmosphere doesn’t have follow the idea of entropy?
I watched a video where Neil deGrasse Tyson was explaining the Second Law. At the beginning of the video he firmly and correctly states that the Second Law only applies to isolated systems. Then all the examples in the show demonstrating the Second Law were not of isolated systems. What can I say?
Entropy is defined as dS = squiggly dQ divided by T. The equal sign only applies to reversible systems. If you can find an actual reversible system, then let me know.
DeGrasse differs from my thermo book so I think he is wrong.
You didn’t answer my question. You said the second law only applies to isolated systems and the atmosphere wasn’t isolated so according to your claim the atmosphere doesn’t follow the idea of entropy.
i don’t think you can have it both ways.
“i don’t think you can have it both ways.”
I have no idea what you are talking about. The 1st Law only applies to closed and isolated systems; the 2nd Law only applies to isolated systems; and the 3rd Law only applies to closed systems.
Technically that is true, Jim. However, nothing in our universe satisfies any of those conditions. Yet we still manage to use the 3 Laws to describe how the universe works. It as all about “close enough approximations”, like everything else in physics (there are no frictionless surfaces or ideal springs either, but we can still figure things out close enough for practical purposes).
“Technically that is true, Jim.”
Thanks.
“However, nothing in our universe satisfies any of those conditions.”
So, considering that the Universe is an isolated system doesn’t qualify the Universe as an isolated system? The Earth and the Sun aren’t open systems? TYS’s hermetically sealed water glass isn’t a closed system? I think you need to rethink that statement.
“Yet we still manage to use the 3 Laws to describe how the universe works.”
It’s four laws, but isn’t it amazing how we can do that with our stupid physics?
“close enough approximations”
Sounds like government work.
Four laws, yes, my bad, the 0th tends to escape my memory. Never mind, carry on!
I said “nothing in our universe satisfies any of those conditions”, and the universe isn’t really “in” the universe. It “is” the universe. But yes, the universe (as far as we know today) satisfies the condition of being an isolated system subject to the 2nd law.
Technically, we can’t use the 2nd law to refer to anything smaller than the universe. But the same reasoning that gives rise to the whole-universe law (statistical probability) also applies to more-or-less isolated subsets of it. Right?
Your original point was that the Earth’s atmosphere need not obey the 2nd law. Neither does Earth itself. That is true. But what do you think is the exact probability of a colder object the size of Earth’s atmosphere making an object the size of Earth (which is warmer to begin with) even warmer still? It’s not 0, for sure, but how close to 0 is it?
Do you think that Willis is trying to stake his case on “The laws of probability are against me, not just astronomically, but universally – but come on, it could happen” ???? He could say that, but who would fall for it if he did? Besides you, I mean?
“Sounds like government work” Yes, the whole climate scare is indeed based on government work. And most of it is false.
“Four laws, yes, my bad, the 0th tends to escape my memory. Never mind, carry on!”
Thanks. The 0th law allows us to measure thermodynamic systems. Unfortunately, climate science misuses the concept.
“Technically, we can’t use the 2nd law to refer to anything smaller than the universe. . . . Right?”
Right.
“Your original point was that the Earth’s atmosphere need not obey the 2nd law.”
Thanks for remembering my original point. I get annoyed when people say that radiation from a colder object can’t be absorbed by a warmer object. If that was true then how could we see things colder than our eye retina? I doubt that the light from a firefly–even on a warm summer evening–is coming from an object warmer than my retina.
There’s a difference from saying that radiation from a colder object can be absorbed by a warmer object and that a colder object can warm a warmer object. The “net” flow of radiation will be from warmer to colder.
The warmer object will lose entropy as it cools and the cooler object will gain entropy as it warms. And the total entropy will increase in an isolated system per the 2nd Law.
“And most of it is false.”
True. I agree.
Quantum radiation absorption is not the same as classically measured power, which I think might be the point you are making. Fireflies produce short-wavelength light through some quantum trickery that avoids having to heat their body up to 6000 K, fortunately for them. But if you try to measure thermal power from them via a sensor that is at human body temperature (around 38 C), you will get a zero or negative answer (the firefly isn’t warmer than you are, at least I don’t think so). The Earth-surface pyrgeometers say the same thing (negative power) when pointed at the colder atmosphere. That is what I keep trying to teach Willis, but he refuses to listen.
“The warmer object will lose entropy as it cools” Don’t you mean gain entropy? The cooler state is the more statistically likely one (given an environment to lose heat to), hence higher entropy. The entire universe gains entropy as it cools to a uniform temperature. That’s the 2nd Law.
““The warmer object will lose entropy as it cools” Don’t you mean gain entropy?”
Seriously? Heat added to a system is positive heat transfer.. Heat removed from a system is negative heat transfer. That’s the usual standard. Entropy is delta S = delta Q/T (for a reversible system). Negative delta Q = negative entropy as T is always positive–absolute temperature. You’re not trying to tell me that a crystalline solid has more entropy than a gas or liquid? That would violate Boltzmann’s formula.
Let’s say that there’s a net transfer of heat, delta Qnet, from object B to object A. Object B is warmer than object A, so Tb > Ta. Therefore delta Qnet/Ta > delta Qnet/Tb. The gain of entropy by the cooler object is greater than the loss of entropy by the warmer object. And that would satisfy the 2nd law for an isolated system.
“The entire universe gains entropy as it cools to a uniform temperature. That’s the 2nd Law.”
No. The entire universe gains entropy as it warms–the so-called heat-death. Realize that the universe is also expanding which cools it.
Right, my bad, I got my signs backward. In any case, that’s not going to solve Willis’s problem. He’s trying to claim that a hotter object is gaining energy directly from a colder one. The 2nd Law doesn’t address this directly in the case of a sub-universe-sized system, true, but the same statistical mechanics still apply, just with less certainty. (Not much less in this case – Earth and its atmosphere are pretty macroscopic.)
As you said, two objects of different temperatures placed together will tend to equalize their temperatures, not drift farther apart, due to entropy (via the same statistical mechanics calculations that give rise to the universe-spanning 2nd Law). But that’s not what Willis said in the head post. You wanted him to tell people to stop using the 2nd Law to refute his false claim, but if he instead follows your advice and states more accurately that “although the 2nd Law doesn’t directly refute this statement, it is nevertheless vastly more likely to be wrong than right, due to the same statistical mechanics calculations that the 2nd Law is based on”, do you think that (a) he would understand any of that or (b) anyone else would now give in and let him get away with continuing to be almost certainly wrong with a probability on the scale of the lifetime of the known universe and beyond?
I’m going to reply here, because the troll has ruined the other thread.
Notice in my example, I said “net” heat transfer. That doesn’t mean that the cooler body isn’t transferring heat energy to the warmer body—just not as much.
I know I’m a stupid double-E, but I learned about electromagnetic waves differently than you did apparently. An electromagnetic wave has two vectors: E and H. The E vector is the electric field strength, and the H vector is the magnetic field strength. The units of E are volts/meter and the units of H are amperes/meter. The dot product of E and H is zero. That means the two vectors are perpendicular or ninety degrees apart as they travel through space.
A volt in MKS/SI units is a joule/coulomb. An ampere in MKS/SI units is a coulomb/second.
Where exactly is the “joules” only in the units?
An important operation is the cross product of these two vectors. Specifically, it’s E x H. Multiplying volts/meter by amperes/meter gives watts/meter^2. Wow!—it’s not just joules.
So I shine 1 watt on a 1 by 1 meter square surface. In one second, it’s 1 joule. In one minute, it’s 60 joules. In one hour, it’s 3.6 kilojoules. In one solar day, it’s 86.4 kilojoules. And in one non-leap year, it’s 31.563 megajoules.
Using your “only joules” criteria, let’s say it’s 60 joules. Now explain climate with that value–and good luck.
Did they say that the Stefan-Boltzmann law usually gives a watts/meter^2 value? The correct units for doing this climate nonsense IS watts/meter^2.
Well, Jim, you may have learned some EE, but you haven’t learned enough theoretical physics yet. You can’t just “shine 1 watt on a 1 by 1 meter square surface”. It doesn’t work like that. What temperatures are you dealing with? When you know the temperatures (energy differential), then you can calculate the power developed. Not before. This is what the S-B equation tells us.
And once you have your temperatures, and you calculate the power, the temperatures start to change over time as energy is transferred, and then the power developed will decline accordingly, until equilibrium is reached – which may take a long time of course depending on how your system is set up. (If your system includes the entire universe as a heat sink, then equilibrium won’t occur until the heat death of the entire universe.) Then at equilibrium you will develop 0 Watts.
Remember, you started this part of the thread by trying to defend Willis’s claim that a colder body can develop power (transfer Joules) to a warmer one. That doesn’t happen. And pointing out that the 2nd Law only applies to the entire universe (as written) isn’t going to help you, or him. The same calculations that go into it are also applied at smaller scales, with practically the same result, until you get down to individual atoms. But that’s not what we’re discussing here.
“Well, Jim, you may have learned some EE, but you haven’t learned enough theoretical physics yet.”
Yeah, I’m an engineer that doesn’t know physics. You’re on something, right?
“You can’t just “shine 1 watt on a 1 by 1 meter square surface”. It doesn’t work like that.”
Since when? You’re really off-base with that statement. I can shine a light/radiation source anywhere on any surface.
“When you know the temperatures (energy differential), then you can calculate the power developed. Not before. This is what the S-B equation tells us.”
No it doesn’t. The Stefan-Boltzmann Law is the integral of Planck’s Law. Wein’s displacement Law is the derivative of Planck’s Law. The temperature is only to define the specific curve of the radiating object, and the total radiation emitted. The radiation leaving the object contains no Kelvin term. There is no thermometer in the radiation. The Stefan-Boltzmann constant completely cancels out the Kelvin units.
“And once you have . . . Watts.”
Nonsense! Complete nonsense!
Remember, you started this part . . . that’s not what we’re discussing here.
I remember. I also know some classical thermodynamics. The 2nd Law only applies to isolated systems. The Earth, the Sun, and the Earth’s atmosphere are not isolated systems. If you want to get into statistical mechanics and statistical thermodynamics, then you may have an advantage. I’m not well versed in those subjects.
You are correct that the Earth and its atmosphere are not isolated systems. However, although the 2nd Law as written is descriptive of the entire universe, the probabilistic calculations on which it is based are equally applicable to any subset of it. The only difference is that as the subsets get smaller, the probability of following the expected entropy gradient gets smaller too. But those probabilities aren’t going to decrease to human lifetime scales until you’re talking about a few atoms. On the scale of Earth and its atmosphere, the probabilities involved are indistinguishable from the values that we would calculate for the entire universe. (i.e. either 0 or 1, depending on which direction you’re thinking in)
At night, with the Sun’s hot influence temporarily removed, the Earth’s surface, its atmosphere, and outer space are essentially 3 isolated (from the rest of the universe) thermally interacting bodies. Nothing else is going to add or remove a significant amount of energy to or from these 3.
So your earlier description of the entropy gradient when a warmer object is allowed to exchange energy with a colder one still apply to Earth and its atmosphere (and outer space). And as you said, warmer objects are required by entropy to transfer thermal energy to colder ones, because that is the direction that results in an overall entropy increase, in the absence of other significant influences. Since Earth’s surface is warmer than its atmosphere (which is warmer than outer space), the direction of energy transfer, work, and power is, as you said it must be, Earth -> atmosphere -> space. Not the other way around.
(Yes, because this subset of the universe is less than the entire universe, the probability of energy flowing in the direction we both think it should is not quite 1. It is 1 – ε. But ε is a tiny number indeed. 0 is not a better guess than 1 – ε, if you’re going to bet on it.)
So this obviously contradicts what Willis wrote in the head post. You are agreeing with me, not with him… so why are you blowing raspberries at me? Are you just trolling, or really unable to grasp the significance of what you yourself wrote?
(Your complete inability to grasp what the S-B law means is a separate topic, and isn’t necessary in order to follow what I wrote here, so we’ll leave that aside – you’ve already presented enough correct physics that you obviously agree with me and not with Willis)
Your comment differs from your last one as I seem to have learned more physics in the interim. I must be a fast learner.
Okay, a new presentation. There are three black bodies that I will label “A”, “B”, and “C”. A is colder than B and C is warmer than B. Since they are black bodies, they radiate at all frequencies. It’s possible for A and C to emit photons with identical frequency/energy. And if they arrive at B, then how does B know which is which? Also, if B refuses one of the photons, that would violate the property of a black body.
Now let’s say they are gray bodies (more realistic). With gray bodies you have an emissivity term in the S-B equation. Emissivity is frequency dependent. However, it’s still possible for gray body A and gray body C to emit photons with identical frequency/energy. If those photons arrive at gray body B (and assuming gray body B likes to absorb those specific photons), how does gray body B know which photon is which?
In other words, the 2nd Law does not prevent B from absorbing photons from a colder object.
“So your earlier description of the entropy gradient when a warmer object is allowed to exchange energy with a colder one still apply to Earth and its atmosphere (and outer space).”
I said “net” heat transfer. That doesn’t disallow heat transfers back and forth between the two objects.
“So this obviously contradicts what Willis wrote in the head post.”
No it doesn’t.
“You are agreeing with me, not with him… so why are you blowing raspberries at me?”
I like making noises with my tongue when I read nonsense.
“Are you just trolling, or really unable to grasp the significance of what you yourself wrote?”
Hmmm.
“Your complete inability to grasp what the S-B law means is a separate topic . . . .”
Yeah, it’s my double-E coming out. The units of S-B are watts/meter^2. The cross product units of the E and H vectors is watts/meter^2. I simply don’t understand your silly obsession with joules. Radiation isn’t just joules.
“But ε is a tiny number indeed. 0 is not a better guess than 1 – ε, if you’re going to bet on it.”
Did you know that according to statistical mechanics there’s a non-zero probability that all the air molecules in the room will gather in a corner and leave you gasping for breath?
I need to clarify a couple of things. When I said frequency/energy, I mean either frequency and/or energy not frequency divided by energy.
And the last statement should include the words: “But I wouldn’t bet on it happening anytime soon.”
Your latest scenario went off the rails when you got to this point:
“In other words, the 2nd Law does not prevent B from absorbing photons from a colder object.”
Now you are attempting to conflate classical thermodynamics with quantum physics. No one has managed to square that circle yet, and I doubt you’ll be able to either. I won’t say you can’t, of course – maybe you know more than you’re letting on, and you’re welcome to give it a shot. But it’s going to take a lot more heavy lifting than what you’ve done here so far. There’s a Nobel Prize in it for you, too, if you manage to accomplish it. Possibly more than one.
“Did you know that according to statistical mechanics there’s a non-zero probability that all the air molecules in the room will gather in a corner and leave you gasping for breath?”
I did know that, yes. And my entire point about your attempt to discredit the universal 2nd Law as a way to disprove Willis’s claim is that this probability exists, and is non-zero, but again, as you yourself said, “I wouldn’t bet on it happening anytime soon.” That would be my point exactly.
Your earlier description of the entropic motivation for a hotter object to lose heat to a colder one is indeed a direct contradiction of Willis’s claim. This is what we mean when we say that his scenario is a violation of the 2nd Law. That’s a bit of a shorthand, yes, and so the longer and more precise version would be more like “the odds are against it, and staggeringly so – don’t hold your breath”.
“Your latest scenario went off the rails . . . .”
Raspberries!
Keep on blowing those raspberries, Jim. So much fun, isn’t it? My daughters thought blowing raspberries was hilarious too – when they were about 3.
“Keep on blowing those raspberries, Jim. So much fun, isn’t it? My daughters thought blowing raspberries was hilarious too – when they were about 3.”
KIRK: What are you?
GUARDIAN: I am the Guardian of Forever.
KIRK: Are you machine or being?
GUARDIAN: I am both and neither. I am my own beginning, my own ending.
SPOCK: I see no reason for answers to be couched in riddles.
GUARDIAN: I answer as simply as your level of understanding makes possible.
SPOCK: A time portal, Captain. A gateway to other times and dimensions, if I’m correct.
GUARDIAN: As correct as possible for you. Your science knowledge is obviously primitive.
SPOCK: Really.
KIRK: Annoyed, Spock?
–Star Trek–TOS–The City on the Edge of Forever
Apparently blowing raspberries is all you understand.
“Now you are attempting to conflate classical thermodynamics with quantum physics. No one has managed to square that circle yet, and I doubt you’ll be able to either.”
This is from the back cover of my text on “An Introduction to Statistical Thermodynamics” by Terrell L. Hill:
“The book is divided into four major sections. Part I deals with the principles of quantum statistical mechanics and includes discussions of energy levels, states and eigenfunctions, degeneracy and other topics. Part Il examines systems composed of independent molecules or of other independent subsystems. Topics range from ideal monatomic gas and monatomic crystals to polyatomic gas and configuration of polymer molecules and rubber elasticity. An examination of systems of interacting molecules comprises the nine chapters in Part Ill, reviewing such subjects as lattice statistics, imperfect gases and dilute liquid solutions. Part IV covers quantum statistics and includes sections on Fermi-Dirac and Bose-Einstein statistics, photon gas and free-volume theories of quantum liquids.”
You incorrectly changed the subject so you wouldn’t have to answer my questions. Apparently, you don’t know about Statistical Mechanics and Statistical Thermodynamics, or you wouldn’t have made such an ignorant statement.
Someone several months (maybe years) ago tried to claim that the Sun would “reflect” any radiation from the Earth. Apparently he and others don’t realize that when a surface reflects radiation, it must first absorb it.
I also don’t think you have a clue about the vector cross product.
Which part of your textbook tells you that you can deduce the principles of classical entropy (the four laws of thermodynamics) from the absorption of a quantum of energy transmitted by a single photon?
There is no “entropy” in the universal quantum wavefunction. There is no arrow of time either, which is closely related to entropy. So these two domains are separate, and you cannot mix and match them to try to make a fake point.
You still haven’t explained how your accurate description of energy flow (heat) from a hotter object to a colder one, due to probabilistic maximization of entropy, i.e. what the 2nd Law is based on, fails to disprove Willis’s claim. He claimed exactly the opposite in the head post. (Measurements don’t support his claim either.)
“Which part of your textbook tells you that you can deduce the principles of classical entropy (the four laws of thermodynamics) from the absorption of a quantum of energy transmitted by a single photon?”
It was two photons actually. But we can increase the number of photons to millions from each source if that will make you happy. I’m sure it would be “theoretically” possible to find millions of photons from object C that all wouldn’t necessarily have the same frequency and/or energy and millions of photons from object A that would match up with the same frequency and/or energy—there would be a one-to-one correspondence in each group. Now please tell me how object B could tell the difference between photons from A and photons from C?
I do see your problem. If you answer that object B knows which-is-which, then you’ll have to explain how that magic occurs. If you answer correctly and say object B has no idea, then that destroys your argument. It’s best to try to discount my question and send me off on a “hopefully” wasteful hunting trip.
One of the failures of “classical physics” was trying to explain the heat capacity of diatomic gases. The idea was to assign (1/2)*k*T to each degree of freedom. A monatomic gas has only three degrees of freedom and that means the heat capacity of monatomic gases should be (3/2)*k*T–and it was. For diatomic gases, the degrees of freedom were 7 and so the heat capacity of diatomic should be (7/2)*k*T—and it wasn’t exactly.
The (1/2)*k*T represents the energy assigned to how many particles? I’ll give you a hint: it’s greater than zero and less than two. There’s a relationship between R, the ideal gas constant, and k, Boltzmann’s constant. That relationship is R = N0*k, where N0 is usually referred to as Avogadro’s number. If you see k in an expression, then they are talking about individual particles. The number of particles is usually large, but they don’t have to be.
“There is no “entropy” in the universal quantum wavefunction. There is no arrow of time either, which is closely related to entropy. So these two domains are separate, and you cannot mix and match them to try to make a fake point.”
Yet, there it is in the text book. The fake point is all yours.
“You still haven’t explained how your accurate description of energy flow (heat) from a hotter object to a colder one, due to probabilistic maximization of entropy, i.e. what the 2nd Law is based on, fails to disprove Willis’s claim.”
I said “net” heat flow. You keep leaving that out.
“He claimed exactly the opposite in the head post. (Measurements don’t support his claim either.)”
What measurements? The ones that show “net” heat transfer or the ones that show instantaneous heat transfer? I disagree with Mr. Eschenbach’s claim that heat energy isn’t energy. The units of heat are joules—energy. In the past they used calories and BTU’s, but those are energy units too.
Your word salad is getting far beyond relevance. Maybe you should show some math derived from the universal quantum wave function that proves there is no entropy and no time in our unique universe.
Keep in mind that a photon and quantum of energy are the same thing. As far as I know the universal quantum wavefunction does not disprove that.
Jim, the 2nd law allows a local entropy decrease at the expense of a wider entropy increase. But that’s not what the head diagram is showing, is it? It’s showing the colder atmosphere applying work (Joule heating) and therefore power to the warmer surface, without any other energy input from anywhere else. That is a direct violation of the 2nd law, no two ways about it.
It’s not applying work – the downwelling crap is just shorthand for the insulative (and other) effects of the STUFF between the ground and outer space.
Really it should be modelled as depending on the solar radiation coming in and all the physical properties of the materials involved, instead of a mystery heat source in the sky.
But that would mean spending all that juicy climate change grant money on proper scientific research instead of jet-setting to climate conferences.
Knowing the specific heats and mass of CO2 vs H2O, the latent heat transfer on the diagram appears off to me. N2/O2 seem to be a heat reservoir until the sun falls to the point where they begin to supply energy to more radiative gases.
As I said before, averages just don’t cut it in a complicated system. My thermodynamics professor would have failed us.
That’s not exactly the case. A colder object DOES radiate toward a warmer object. And, it will radiate more and more until equilibrium is reached. But that really only applies to similar, ideal objects. What’s missing are gradients with time, mass, absorption/emission, specific heats, source varying as a sinusoid, conduction, etc.
Radiates energy, yes. Power, no. Please study the relationship between these two concepts very carefully and then get back to us.
I’m not sure what you call “power”, but Watts per square meter, i.e., Joules per second per square meter is a power term to me.
Ideal bodies radiate based only on their temperature. Other bodies do not change this. Granted there are lots of other variables in the real world that complicate the amount of radiation emitted, but radiation is still based on temperature.
Now if you are talking about the NET exchange of power, I’ll agree that a cold body can not generate sufficient power to add additional energy to a hot body.
Ideal bodies radiate energy based on their temperature. This is measured in Joules. Bodies do not radiate power. The conditions for developing power are more complex than that. And there is no such thing as “net” power. There is only power, or no power.
Are you saying that radiation from a colder black body can’t be absorbed by a warmer black body? How would the warmer black body know that the radiation is coming from a colder black body? Obviously theoretical objects, black bodies have three basic properties: they absorb all frequencies, they radiate all frequencies according to Planck’s Law, and they adjust their temperature depending on the total energy received and radiated each instant.
Gases are not black bodies…
I thought the Sun was an extremely close example of a black body.
Strictly speaking, the sun is mostly plasma.
All objects above absolute 0 radiate energy; but developing power is a very specific process that cannot occur from a colder body to a warmer one, due to the 2nd law.
…in an isolated system. A lot of people ignore the most important part of the 2LOT.
Sure. In a non-isolated system, you can calculate the probability that energy will flow against the entropy gradient. What is that probability if we are talking about Earth and Earth’s atmosphere? It’s not 0, but how close to 0 is it?
Energy flows against the entropy gradient even in an isolated system with 100% certainty. It is heat that does not flow against the entropy gradient in an isolated system. Remember, heat is the net flow of energy. Using contemporary 1LOT nomenclature ΔE = Ein – Eout the terms Ein and Eout are the energy transfer components while ΔE is the heat transfer.
Anyway the point is that in non-isolated systems heat can transfer from cold to hot. The most obvious obvious examples are evaporator/condenser/compressor systems like what is found in refrigerators and HVAC systems. This does not violate the 2LOT because the system is not isolated.
Energy does not “flow” against the entropy gradient, ever. Energy is the “potential to do work”. It is present as a field, when dealing with an EM radiation context. The field doesn’t “flow”. And work is only done “down” the entropy gradient. Statistical probability tells you how unlikely it is for the opposite to happen, which depends on the size of the system you are looking at. For human-scale systems, the likelihood is essentially 0.
At no point in an HVAC system is a cold part of the system making a hot part hotter through direct energy transfer. Energy always flows from hot to cold.
Patently False. And you can easily convince yourself of this with nothing more than an IR gun.
Correct. But don’t conflate energy with heat. Energy absolutely moves from cold to hot. It is heat that does not move from cold to hot spontaneously.
Patently False. Heat (the net transfer of energy) is moving from the evaporator (cold) to the condenser (hot). The cold evaporator definitely plays a role in making the condenser hot. This is not a violation of the 2LOT, however, because the system is not acting by its own means (isolated). Work is being performed on the system to make the heat go against the entropy gradient.
Patenly False. First, I think you are conflating energy and heat. Remember energy is the ability to do work. Heat is the net transfer of energy. Look at the 1LOT closely. ΔE = Ein – Eout. In this equation ΔE is heat while Ein and Eout are just energy. If Ein = Eout then ΔE = 0. In that scenario there is no heat transfer despite there being energy transfers.
Look at the radiant heat transfer equation Q = εσ(Th^4 – Tc^4)A . It is derived from the 1LOT and SB law. Start with the 1LOT ΔE = Ein – Eout. Multiple both sides by time (t) and area (A) to get ΔE/tA = Ein/tA – Eout/tA. Rewrite in flux form ΔF = Fin – Fout. Swap in the SB law to get ΔF = εσTa^4 – εσTb^4. Notice that ΔF is the net transfer of flux between bodies A and B so we can rewrite this as Q = εσ(Th^4 – Tc^4)tA where Q is the heat gain, body A is hot with temperature Th, body B is cold with temperature Tc, t is the time of heat movement, and A is the area for the view factor being analyzed.
Notice in the derivation of the radiant heat transfer equation that the heat transfer is from hot to cold despite there being an energy transfer from cold to hot. It is because the energy transfer from hot to cold Ehtoc is greater than the energy transfer from cold to hot Ectoh such that Ehtoc > Ectoh that you have a net transfer (or heat) moving from hot to cold. The fact that heat moves from hot to cold spontaneously does not mean that energy is not transferring across the entropy gradient.
“Energy absolutely moves from cold to hot.”
No it doesn’t.
“Heat (the net transfer of energy) is moving from the evaporator (cold) to the condenser (hot)”
Not directly. The compressor (using energy from elsewhere, such that total entropy will increase when that energy usage is accounted for) creates a series of intermediary entropy gradients. At each step, heat is flowing from a hotter object to a colder one, in accordance with entropy and statistical mechanics (the same calculations that the 2nd Law is based on).
Energy cannot be transferred in two directions along a single entropy gradient at a given time. That’s not how power works. You are inventing imaginary flows of energy while calculating the amount of power being developed. As a mathematical tool, that is fine, if it makes your calculation simpler, which it really doesn’t – but it is not physics.
“with nothing more than an IR gun.” How do you think an IR gun works?
“Heat (the net transfer of energy) is moving from the evaporator (cold) to the condenser (hot)”
You are trying to confuse Joule heating with isentropic compression, and hoping that no-one notices. These are different thermodynamic processes, but you knew that, right? What makes you think that when I said “heat flows only from hot to cold”, that I was talking about any process other than Joule heating? Willis certainly wasn’t, nor was I when I pointed out that he was wrong. Are you deliberately trying to deflect from the central point of the argument?
“Energy does not “flow” against the entropy gradient, ever.”
Raspberries!
Yummy! But what have raspberries got to do with anything?
Truly, you can’t be THAT ignorant!
It is 0. But that is moot because in the sun/atmosphere/surface/space system, which is effectively isolated, the heat flow is sun (hot) ==> surface (warm) ==> atmosphere (cool) ==> space (cold).
If we then put a thermal barrier between the lower and upper atmosphere the heat flow is still sun (hot) ==> surface (warmer) ==> loweratm (warmer) ==> barrier ==> upperatm (cooler) ==> space (cold). The fact that the surface and loweratm warmed while the upperatm cooled does not change the direction of heat flow. What it does is change rate of heat flow.
That description of energy flow based on temperature gradients is correct. But that’s not what Willis wrote, is it? He wrote that energy (transfer of Joules, i.e. Watts) flows from the atmosphere to the surface. It doesn’t.
I believe stevekj has a conceptual problem in expressing what goes on. I hope he is trying to say that a hot body can not absorb and RETAIN energy from a cold body, thereby making it “hotter”.
Absorption and emission are two separate phenomena. If absorption is less than emission, a body may cool at a slower rate, but it will not reverse the gradient.
Isn’t there a Kirchhoff law about emission and absorption?
What’s funny is that if the 2nd Law always applied, then nothing could cool down. Objects that lose heat are also reducing their entropy. Negative heat transfer is a negative entropy change.
Jim may also be having conceptual problems distinguishing between quantum and classical (macroscopic humanly observable) descriptions of a situation. They are not the same. Which one are you dealing with?
Jim knows the difference. Of course, I forgot about adiabatic systems. They can cool down without losing heat.
You should do some research about what you are talking about. It has nothing to do with classical vs quantum. In fact you will find little to nothing about the actual mechanism whereby an atom or molecule absorbs a quanta (photon) of energy. The results of absorption are well described by quantum theory but the actual physical mechanism is still only theoretical. Kinda like entanglement. We know it occurs and can describe the effects. Exactly how it works is only theoretical.
Radiation of energy occurs by electromagnetic waves. The power contained in a EM wave is accurately described using Maxwell’s equations. An EM wave can contain the power of one quanta (photon) to many millions of watts. A single atom or molecule can intercept the EM wave and absorb some of that energy depending on various factors such as frequency. One atom or molecule does not have to absorb the entire power contained in a given wave, it only has to extract the amount of energy needed to move electrons into a new energy level(s). But let there be no doubt power is transmitted based upon the temperature of an object. The object doesn’t need to be the either hot or cold, each emits radiation.
“But let there be no doubt power is transmitted based upon the temperature of an object.”
This is what I keep trying to teach Willis, and he’s not listening…
stevekj,
Willis is both incompetent and ignorant. I see a similarity to Einstein, though – both Einstein and Eschenbach start with the same letter.
That is one similarity, indeed, Mike. I think it ends there though!
I did manage to get him to write that objects emit radiant energy, not radiant power, and that radiant energy is measured in Joules, not Watts. I count this as significant progress. But I don’t think he really understands any of this either, even though he wrote it himself. I keep trying…