Guest Post by Willis Eschenbach
OK, a quick pop quiz. The average temperature of the planet is about 14°C (57°F). If the earth had no atmosphere, and if it were a blackbody at the same distance from the sun, how much cooler would it be than at present?
a) 33°C (59°F) cooler
b) 20°C (36°F) cooler
c) 8° C (15°F) cooler
The answer may come as a surprise. If the earth were a blackbody at its present distance from the sun, it would be only 8°C cooler than it is now. That is to say, the net gain from our entire complete system, including clouds, surface albedo, aerosols, evaporation losses, and all the rest, is only 8°C above blackbody no-atmosphere conditions.
Why is the temperature rise so small? Here’s a diagram of what is happening.
Figure 1. Global energy budget, adapted and expanded from Kiehl/Trenberth . Values are in Watts per square metre (W/m2). Note the top of atmosphere (TOA) emission of 147 W/m2. Tropopause is the altitude where temperature stops decreasing with altitude.
As you can see, the temperature doesn’t rise much because there are a variety of losses in the complete system. Some of the incoming solar radiation is absorbed by the atmosphere. Some is radiated into space through the “atmospheric window”. Some is lost through latent heat (evaporation/transpiration), and some is lost as sensible heat (conduction/convection). Finally, some of this loss is due to the surface albedo.
The surface reflects about 29 W/m2 back into space. This means that the surface albedo is about 0.15 (15% of the solar radiation hitting the ground is reflected by the surface back to space). So let’s take that into account. If the earth had no atmosphere and had an average albedo like the present earth of 0.15, it would be about 20°C cooler than it is at present.
This means that the warming due to the complete atmospheric system (greenhouse gases, clouds, aerosols, latent and sensible heat losses, and all the rest) is about 20°C over no-atmosphere earth albedo conditions.
Why is this important? Because it allows us to determine the overall net climate sensitivity of the entire system. Climate sensitivity is defined by the UN IPCC as “the climate system response to sustained radiative forcing.” It is measured as the change in temperature from a given change in TOA atmospheric forcing.
As is shown in the diagram above, the TOA radiation is about 150W/m2. This 150 W/m2 TOA radiation is responsible for the 20°C warming. So the net climate sensitivity is 20°C/150W-m2, or a temperature rise 0.13°C per W/m2. If we assume the UN IPCC canonical value of 3.7 W/m2 for a doubling of CO2, this would mean that a doubling of CO2 would lead to a temperature rise of about half a degree.
The UN IPCC Fourth Assessment Report gives a much higher value for climate sensitivity. They say it is from 2°C to 4.5°C for a CO2 doubling, or from four to nine times higher than what we see in the real climate system. Why is their number so much higher? Inter alia, the reasons are:
1. The climate models assume that there is a large positive feedback as the earth warms. This feedback has never been demonstrated, only assumed.
2. The climate models underestimate the increase in evaporation with temperature.
3. The climate models do not include the effect of thunderstorms, which act to cool the earth in a host of ways .
4. The climate models overestimate the effect of CO2. This is because they are tuned to a historical temperature record which contains a large UHI (urban heat island) component. Since the historical temperature rise is overestimated, the effect of CO2 is overestimated as well.
5. The sensitivity of the climate models depend on the assumed value of the aerosol forcing. This is not measured, but assumed. As in point 4 above, the assumed size depends on the historical record, which is contaminated by UHI. See Kiehl for a full discussion.
6. Wind increases with differential temperature. Increasing wind increases evaporation, ocean albedo, conductive/convective loss, ocean surface area, total evaporative area, and airborne dust and aerosols, all of which cool the system. But thunderstorm winds are not included in any of the models, and many models ignore one or more of the effects of wind.
Note that the climate sensitivity figure of half a degree per W/m2 is an average. It is not the equilibrium sensitivity. The equilibrium sensitivity has to be lower, since losses increase faster than TOA radiation. This is because both parasitic losses and albedo are temperature dependent, and rise faster than the increase in temperature:
a) Evaporation increases roughly exponentially with temperature, and linearly with wind speed.
b) Tropical cumulus clouds increase rapidly with increasing temperature, cutting down the incoming radiation.
c) Tropical thunderstorms also increase rapidly with increasing temperature, cooling the earth.
d) Sensible heat losses increase with the surface temperature.
e) Radiation losses increases proportional to the fourth power of temperature. This means that each additional degree of warming requires more and more input energy to achieve. To warm the earth from 13°C to 14°C requires 20% more energy than to warm it from minus 6°C (the current temperature less 20°C) to minus 5°C.
This means that as the temperature rises, each additional W/m2 added to the system will result in a smaller and smaller temperature increase. As a result, the equilibrium value of the climate sensitivity (as defined by the IPCC) is certain to be smaller, and likely to be much smaller, than the half a degree per CO2 doubling as calculated above.
anna v (13:59:37) : edit
I tried that before. All that it downloads is a page saying
“This script should be referenced with a METHOD of POST.
If you don’t understand this, see this Forms overview.”
An attempt to go to the referenced web page gives me a 404 error.
I tried with both Safari and Firefox and I get nothing … what’s your secret?
w.
In this thread there has been reference to IPCC definitions like ‘radiative forcing’ and ‘climate sensitivity’ and some values attached to those definitions, therefore, those definitions and values have been incorporated in various papers.
But there is a problem with that: Namely, that a presupposition exists that these are accurate and useful definitions and values.
Given the track record of the IPCC and it’s agenda & outcome driven processes, one is left to question that accuracy of such standards.
Sure, don’t use or abide by what the IPCC sets up and your paper won’t get into the IPCC report, but if they are flawed requirements then being constrained by these requirements garantees the papers, while jumping through the IPCC’s hoops, won’t reflect actual physical reality:
And, isn’t reporting on what is actually happening more important than being properly constrained by arbitrary IPCC standards?
Or is it simply about playing the game as the rule master dictates…damn the reality?
Leif Svalgaard (14:24:31)
Two comments:
1. Contrary to your claim, temperatures are well correlated, particularly over short distances. It is very unusual to find two sites a mile apart where one is cooling and one is warming.
2. While the theory that “the data is wildly differing over short spatial scales but the average is valid” is an interesting one and is sometimes true, we cannot assume that it is true. I encourage people to take a look at the five adjacent Czechoslovakian boreholes here, and think about whether averaging those and a hundred more like them will give a valuable answer …
Willis Eschenbach (12:26:45) :
44.2 terawatts = 4.4 * 10^13 watts
Earth’s surface = 5.1 * 10^14 square metres
Heat flux = ~ 0.1 W/m2
================
Yes, assuming the 44 terawats are right. I have no idea how that is measured.
However that may be, the above conversion has lead me to another surprising perspective.
If total human energy consumption is about 15 terawatts, then this would translate to ~ 0.3 W/m2.
So let’s think about the meaning of this. If we dedicated ALL the energy we consume to the exclusive purpose of heating the earth’s surface *permanently* from underneath using the most efficient means at our disposal — the warming effect would be barely noticeable.
On the other hand, if we just take a gas that happens to be released as a byproduct of our energy use — the same gas that is exhaled by the biosphere and and the oceans in quantities that are orders of magnitude larger than our contribution — and let it go in the air, the result of the concentration increase in this gas putatively attributed to our modest exhalations, is supposed to be a surface warming many, many, many times larger than what we could ever achieve by using all that energy permanently and directly for heating the surface as hard as we could.
When I look at it in this perspective, I find that extremely bizarre. Which is why I would like to see some piece of empirical evidence, no matter how primitive, of this marvel at work.
Francisco (16:25:37)
Of course I meant 0.03 (not 0.3) W/m2.
Sorry.
The science of climate change
The clouds of unknowing
There are lots of uncertainties in climate science. But that does not mean it is fundamentally wrong
http://www.economist.com/displayStory.cfm?story_id=15719298
brent comments:
The Economist should reflect on the Wegman equation:
“Method Wrong + Answer Correct = Bad Science.”
Of course “answer correct” in this context means “politically correct”
brent’s amendment to the Wegman Equation follows”
Method Wrong + Answer (Politically) Correct =”Post Normal Science”
Post Normal Science = Sceance
cheers
brent
“Another point, Using Trenbeth ballpark figures, if we get 240 watts from the sun, and the greenhouse effect makes that an effective 390 watts and a 33 C increase, isn’t the increase in temperature per delta watt increase in the ballpark of
33 C/(390-240) = about 0.22C per watt increase?”- AMC
“No, no, no. I’ll keep going over this until everyone gets it. According to the UN IPCC, climate sensitivity is the change in surface temperature divided by the change in TOA forcing … yes, I know, it’s a really dumb measurement, but there it is. I suspect it was specifically chosen by AGW supporters because downwelling TOA radiation can’t be measured directly … but perhaps that’s just my paranoia speaking.”- Willis Eschenbach
I REALIZE climate sensitivity would be delta surface temperature divided by delta watts at surface, but I believe the 33c/150 watts gives an upper limit ballpark figure.
With NO forcing, the earth’s temperature would be 33 C lower, radiating at 240 watts. Those additonal 150 watts have been pumped into earth’s atmosphere, gradually warming the earth at a lower and lower rate. The total warming was 33C, the total additional sensible wattage was 150 watts, so the average sensitivity over the full range of 240 to 390 watts was
33C/150 watts = 0.22C. It stands to reason that ACTUAL climate sensitivity should be less.
The Stefan-Boltzmann equation for a blackbody goes
T(degrees Kelvin) = S(constant)*(watts/square meter)^0.25. Our first step is to find that S constant.
Doing a google search, I find 1000 K implies a blackbody flux of 56790 watts/square meter.
1000K = S* (56790 watts/square meter)^ 0.25. Click on your calculator and use the scientific view. Plug in 56790
X^Y
0.25
=
and you get 15.43718 Divide 1000K by 15.43718 and you get S = 64.77867
We now know
T(kelvin) = 64.77867 ( watts/square meter)^0.25.
Let’s plug in some numbers.
At 100 watts per square meter flux,
T = 64.77867 * 100^0.25 = 204.848 K, at 400 watts per square meter,
T = 64.77867 * 400^0.25 = 289.699 K
T = 64.77867 W^0.25. From elementary calculus, the derivative of that
equation will give you the sensitivity.
dT/dW = 64.77867* 1/4 * (W^(-3/4)) = 64.77867*1/4 * (1/(W^0.75)).
At 100 watts/sq meter, you get a sensitivity of
64.77867*1/4*(100^-0.75) = 64.77867*0.25* 0.0316228= 0.512121 K/watt per meter squared
At 400 watts/sq meter, you get a sensitivity of
64.77867*1/4 *0.01118034 = 0.181062 degrees K per watt per meter squared.
My above example was equivalent to
taking (289.699-204.848)/(400-100) and getting an “AVERAGE” sensitivity of 0.283 K/watt- note that this is higher than the “actual” sensitivity of 0.181 once we’ve reached 400 watts. The earth is not a blackbody, but the reasoning should be the same, that “AVERAGE” sensitivity of 0.22 C must be larger than the current sensitivity with a surface 390 watts.
Willis Eschenbach (00:03:04):
You claim to have accounted for atmoshperic pressure. I’d like to see where! All the cartoonish schematics (not just yours) show only power fluxes–not enthalpy calculations–presented in an disconnected way, contradicting Kirchhoff’s current law. Furthermore, the figures for the flux of latent heat seem to be land-based, rather than accounting for the oceans, where evaporation exceeds upwelling IR radiation and conduction combined. Sorry, but that’s just not a physically cogent exposition of the “budget” of the Earth’s climate sytem.
daniel says:
I am not sure what you mean about “more successful” in regards to Arthur Smith’s paper. I think that his paper was very nice, but because at the time that he wrote it G&T had not yet gotten their paper published in a peer-reviewed journal, Arthur did not even try to submit his. (It is very hard to get a paper published in a peer reviewed journal that just rebuts nonsense floating around on the internet. Arthur pretty much admits in the introduction that his paper does not really break significant new ground but just clarifies well-understood science in light of G&T’s confusions.)
Also, Arthur focused on one aspect of G&T’s claims (the supposed “fictitiousness” of the calculation that the greenhouse effect warms the earth’s surface by about 33 C), whereas our comment on G&T addresses a broader array of aspects of their paper, including the central claim that the atmospheric greenhouse effect somehow violates the 2nd Law of Thermodynamics. Our paper is not yet available on the Arxiv although it may be up there before too long.
Sorry, I meant to agree with W.E. and state, “I realize forcing shold be delta T at surface divided by delta watts at TOP of atmosphere.
Willis Eschenbach (16:16:56) :
1. Contrary to your claim, temperatures are well correlated, particularly over short distances. It is very unusual to find two sites a mile apart where one is cooling and one is warming.
That’s where careful cherry picking comes in.
Boreholes: I saw this guy’s presentation: http://adsabs.harvard.edu/abs/2009AGUFMPP53A..10C
And couldn’t put a finger on where it was wrong, so he convinced me.
This study looks solid too: http://home.badc.rl.ac.uk/mjuckes/mitrie_files/docs/mitrie_borehole.pdf
Unfortunately it shows global warming so many would reject it just on that, regardless of the merit of the method.
To a point; even energy in the 52 or 146 or 432 MHz ranges can become affected or susceptible to ‘action’ by the atmosphere *and* the ionosphere. I won’t go into other effects e.g. Faraday rotation: rotation of the plane of polarization of a linearly polarized wave propagating through a magnetized dielectric medium – that can be seen at GPS frequencies in the L-band (1.2 and 1.5 GHz) .
THEN there is an ‘opaqueness’ in the atmosphere that sets in at wavelengths shorter than 1 cm (above about 30 GHz) .
Click my name to see a chart.
.
.
Leif,
You’ve over constrained the system or at least implied that the emissivity of the two objects is identical even though one of them clearly isn’t a black body. Take out the nuclear reactor and start with identical temperature spheres but with different emissivities because one of the spheres is not a black body. We’ll return to the case where there’s internal heat generation, but for now simply rely on the internal energy of the two spheres giving them equal temperatures. BTW, I don’t really understand the electron acceleration comment. That sounds like bremsstrahlung to me which isn’t black body radiation at all. Anyway…
Here’s a simple way to look at the problem. Let’s turn the spheres into two parallel semi-infinite walls separated by a vacuum. The only way these walls can lose heat is to radiate it to each other. Let’s also say that the white wall has an emissivity and therefore absorptivity of 0.5 and the black wall is a perfect black body with emissivity of 1. And yes, let’s make the emissivity a wavelength independent quantity because I’m lazy. To avoid messing around with constants and T^4 let’s also say that the black wall is of such a temperature as to radiate 10 W/m^2 at its initial state. Since the white wall has the same temperature but half the emissivity, it emits 5W/m^2. Now let’s look at the energy balance for the two walls. The white wall is exposed to 10W/m^2 incoming radiation of which it absorbs half, 5W/m^2. The other half is reflected back toward the black wall It also emits 5W/m^2 of its own radiation for a net heat transfer of: 5-5=0W/m^2. The black wall emits 10W/m^2. It receives 5W/m^2 generated from the white wall and 5W/m^2 of its own radiation reflected back for a net heat transfer of: 5+5-10=0W/m^2. So, two bodies at the same temperature and different emissivities do not have to radiate the same amount and can still be in radiative equilibrium at the same temperature. No work can be done. Entropy is constant. Carnot is happy. Maxwell is happy. And his demon sleeps.
To return to your example, unless the emissivities are the same, then the problem you framed is not in steady state. If the same amount of energy is being produced in both spheres and they have different emissivities, which they will since they are different colors and hence absorb/emit differently, then the white sphere is only temporarily at 400C on its way to its final temperature. Put another way your white/green/blue sphere can’t be a black body, but it will still emit according to Planck’s law albeit with a sub-black body emissivity.
Francisco says:
Why is that all that surprising? As many people have noted, the amount of energy that we use in miniscule compared to that supplied by the sun. Hence, a much more efficient way of causing heating of the earth is to change the atmospheric chemistry so that more of that energy from the sun is retained, rather than simply heating it directly ourselves.
By the way, when you say, “the same gas that is exhaled by the biosphere and and the oceans in quantities that are orders of magnitude larger than our contribution”, you are neglecting the other side of the equation – that it is also inhaled by the biosphere and oceans in these large quantities. And, the exchange is only large with the mixed layer of the ocean. The exchange between the mixed layer and the deep ocean is quite a bit smaller…and that is the rate-limiting step. The mixed layer + biosphere + atmosphere are just passing the same carbon around and around. We are taking a quantity of carbon that has long been locked away from the atmosphere and rapidly liberating it into this system.
willis,
sorry but i’m out of town again and have limitations on access to things. I tried to go through the erbe etc. paper you referenced looking for the 0.123 clear ssky albedo. It occurred to me that perhaps there is more clear sky over land than ocean or something to that effect. Putting the number 0.123 in to the spread sheet indicates essentially a requirement of 0.32 land surface albedo and a 0.38 cloud albedo. 0.32 seems rather high, only snow and sand produce this and there’s not that much of those around. the 0.38 (or even 0.44 assuming surface is 0.085 or so) seems rather realistic. Hopefully, I can get some time to go through that erbe paper in detail.
Sorry…I didn’t mean to be impatient. I just wanted to make sure that you saw my post amidst everything else.
Okay…I will have to think about this more.
Ah…but therein lies the problem. All those enter into what you are saying the forcing is…but then when it comes time to calculate how this relates to CO2, you say that doubling CO2 produces a forcing of 3.7 W/m^2. That is not correct. In your first calculation (the 20 C divided by 150 W/m^2), you have called everything a forcing whereas in the second calculation you have called everything a feedback except for the change in CO2.
Let me give you a concrete example: Suppose that the increase in CO2 directly is responsible for 30 W/m^2 of forcing in the current climate due to CO2’s IR absorption and the water vapor is responsible for the other 120 W/m^2 due to its IR absorption (and, for simplicity, we just ignore everything else). Now, suppose that the water vapor feedback works in such a way that if we removed all the CO2, the feedback processes would be such that all of the water vapor also disappears (again, I am not claiming this is true…but just supposing so).
In that case, then to get the climate sensitivity due to changing CO2, I think you would agree that you should divide the 20 C change by 30 W/m^2, not the full 150 W/m^2. But, alas, that is not what you have done.
To put it another way, when the IPCC says that the central estimate for climate sensitivity is 3.0 C (with an assumption that doubling CO2 produces a TOA forcing of 3.7 W/m^2), they are not predicting that the TOA downwelling radiation will be increased by only 3.7 W/m^2 if you double CO2. What they are saying is that the direct effect of the IR absorption by CO2 will be to increase the TOA downwelling radiation by 3.7 W/m^2 but that the total increase, once you consider all the feedback effects will be an increase in the TOA downwelling radiation by almost 3 times this (where I am using the result that a 3.7 W/m^2 increase in TOA downwelling radiation produces a temperature rise of about 1 C, as calculated by the S-B Equation).
It would be a violation of radiative balance to claim that a 3.7 W/m^2 increase in downwelling radiation would produce a 3 C increase in temperature! We know from the S-B Equation that this is not the case.
The one thing that admittedly remains a mystery to me is why your calculation of the climate sensitivity in the way that you did it does not just come up with the value for climate sensitivity determined by the S-B Equation in the absence of any feedbacks, which is what I think it ought to get. So, I am still confused about whether the 150 W/m^2 is what you actually wanted to use, but I will have to think more about that.
Thanks, Willis (and anna v). I appreciate it.
Tsk Tsk (18:34:25) :
then the white sphere is only temporarily at 400C on its way to its final temperature.
The 400C is its final temperature, once the white and the black sphere have reached a final same temperature, they both have the same thermal spectrum. Every opaque body radiates a thermal spectrum determined by its temperature. The balls are in empty space far VERY from each other and from anything else. And do not exchange [measurable] radiation with each other or with anything else.
Tsk Tsk (18:34:25) :
I don’t really understand the electron acceleration comment
From a different view [more modern and more ‘correct’ view], electrons in atoms are ‘excited’ by thermal motions; when the atom relaxes [excited states have finite – short – life times], the electrons return to a lower state emitting photons in the process.
Jim writes, “Can you point to an article or source where this is actually measured (the quantitative measurement of atmospheric radiation of ‘heat’ energy ostensibly removed from the surface via convection)?”
Thank you for pointing that out, Jim, I have been trying myself to address that, I am sorry I don’t have a reference for you. That’s a good observation
Brian W, that’s a good comparison. Note that the ionosphere (Kenelly-Heaviside layer) acts as a deus ex machina to overcome the curvature of the Earth so far as short wave radio transmission is described by the laws of electromagnetic propagation; there’s no such associated “outside influence” as far as thermal IR transmission is concerned (1-30 micrometer radiation), and Gerlich and Tscheushner must be correct. (Look at the “wrong” description of how the “greenhouse” effect supposedly “works”.)
The Trenberth “analysis” is nothing more than some Enron-type bookkeeping to give an extra 1 w/m**2 or so (on the disc) that he wants to present as “overheating” the world from Greenhouse gases. It isn’t observed by satellites measuring incoming and outgoing radiation (which balance), but what the hell. It serves the purpose of overheating some temperaments and some egos, which the what the construct was really ever designed for
The moon couldn’t possibly be a better experimental setup for isolating the atmosphere as a variable in average surface temperature, we put temperature sensing instruments on it decades ago, have years of data from them, and yet almost everyone here is ignoring this hard data.
Would someone please put forward an hypothesis which explains subsurface lunar regolith temperature. Any hypothesis must explain these observed temperatures to be at all credible. From what I can determine from experimental evidence the earth’s atmosphere accounts for a minimum of 30C warmer surface temperature.
The Moon’s reflectivity is known to be partly specular in the visible, since the Moon appears uniformly bright from center to limb as viewed from the Earth. Is it this reflectivity in the visible is the same reflectivity through the thermal IR? I don’t know.
I can’t answer your question about the Moon, Dave, but I can give you a calculation that shows the thermal exchange temperature of the Earth with the Sun is about 279.5K, and and we also know that the Earth may be viewed as a radiator at a temperature of about 288K, other than that, I don’t know what to say about the 30K warmer figure
Leif,
Again, you’ve over constrained the problem. The ball cannot be white and a black body. What makes it white? Clearly it does not absorb all incident light and therefore it cannot emit perfect black body radiation. You’re right that every body radiates a spectrum based on its temperature and based on its emissivity. You keep leaving that part out. By your definition every object is a perfect black body (emissivity of 1) and the only thing that matters is its temperature. That’s simply not correct. Fluke has some pretty graphs to demonstrate the impact of emissivity on measured temperatures using an IR thermometer: http://www.hartscientific.com/publications/pdfs/3187781_A_w.pdf
I placed the two spheres/walls in proximity to counter your claim that different amounts of radiation from the two objects would allow you to extract work and therefore that proved that the two objects must be emitting the same amount of radiation. What I showed was that there is no net heat transfer and no way to extract work from the transfer of heat from two bodies at the same temperature but different emissivities emitting different amounts of radiation.
I understand the emission of photons comes from electrons transitioning between different states. I’ve just never heard acceleration used to describe the process, but then again quantum was a long time ago and I spend most of my time on magnons and not phonons these days.
The 3.0C per doubling sensitivity comes from the following:
– doubled CO2/GHGs add 3.7 to 4.0 W/m2 of forcing at the tropopause emission layer where it is 255K (sometimes they call this the TOA forcing but the top of the atmosphere is much warmer than this level – it is really the tropopause at about 5 kms up but they often mix this up to keep people off balance);
– then there is feedbacks from increased water vapour and albedo which add another 6.5 to 7.4 W/m^2 after this initial forcing;
– the layer now increases in temperature by 3.0C (11.5 extra watts on top of the original 240 watts).
– the layer where the equilibrium emission temperature of 255K occurs is now 461 Metres higher.
– the adiabatic lapse rate of 6.5C/km stays intact and the surface warms by the same 3.0C.
So, the sensivitiy is calculated as 4.0 extra watts of GHG forcing eventually results in an increase of 11.5 watts (once all feedbacks occur) and temperatures at the surface rise 3.0C.
So, the theory calculates 0.75C/W/m2. It takes the Planck Response of 0.265C/W/m2 (the Stefan-Boltzmann Dtemp/Dforcing for 255K or 240 W/m2) at the tropopause and multiplies it by another 134% in feedbacks to arrive at the final sensitivity.
Its a theory based on some principles. But theories need to have factual empirical data behind them.
Re: Willis Eschenbach (Mar 18 16:02),
The two links connecting directly to the images I gave in my first post give this message you quote. I did not put a correction because the main page was linked also in the same post.
anyway:
I use Firefox with windows XP
I click on
http://isccp.giss.nasa.gov/products/browsesurf1.html
sometimes it might give a “not found”, but it is from traffic or bad width.
“select variable” example: mean surface skin
select a time period : Mean annual
select a format: EQ grid; Ascii text
and “download” .
I find the file in my download area labeled with sequential numbers.
No other secret except in persisting if I get a “not found”.
the differences. between skin and air are large in the cold and hot places. I do not know if they would integrate out. I have not learned to use EXCEL for plot making 🙁 . Maybe I should start.
Tsk Tsk (21:25:20) :
By your definition every object is a perfect black body (emissivity of 1) and the only thing that matters is its temperature.
Did not say that, I said that maintained at the same temperature the radiations would be the same.
I’ve just never heard acceleration used to describe the process, but then again quantum was a long time ago
Since we were talking blackbody and S-B laws in a 19th Century [pre-quantum] setting I stayed in that framework. You can even say that electron transitioning [to stay in a Bohr model] from one orbital to another is a change of velocity and hence an acceleration, but hose details are not really necessary and at a deeper level that is not how it works anyway [but we often use pictures based on incomplete models if they are useful]