## Ned Nikolov: Implications of DIVINER results for the S-B standard equation

Posted: May 1, 2012 by tallbloke in Astrophysics, atmosphere, data, Energy, Solar physics

DIVINER’s results show that the Moon’s actual mean surface temperature is much lower than the estimate derived from the standard Stefan-Boltzmann (S-B) equation:

$Te = [\frac{S*(1-A)}{(4*e*\sigma)}]^{0.25}$

This means we now need to accept the following distinctions:

1)   The term emission temperature (Te) is strictly reserved for estimates obtained by the above (S-B) equation. It refers to a temperature calculated using the assumption of a uniform (spatially homogeneous) distribution of the absorbed solar flux (given by the term S*(1-A)/4) over the surface of a planet. Such a uniform  distribution of solar flux over a sphere implies an infinite thermal conductivity of the regolith, which is a physical impossibility.  This gives rise to Holder’s inequality.  Hence, Te is not related (does not refer) to the absence or presence of an atmosphere. Thus, the 197K lunar average surface temperature value is not an emission temperature, but the actual average physical surface temperature (Ts) of the Moon. The emission temperature of the Moon is Te = 270.2K as calculated by the above S-B equation assuming S = 1361.7 W m-2, A = 0.13, e = 0.98.  In other words, the emission temperature is not a physical temperature, but a mathematical construct having the units of temperature. As such, Te is not theoretically compatible with any measurable real temperatures.

2)   The ~255K emission temperature of Earth is produced by the same equation above using A = 0.3 and e = 1.0. The average surface temperature on Earth is 287.6K, which is an actual physical temperature obtained via direct observations.  Because of that, this temperature is conceptually incompatible with any emission temperature. Hence, calculating Earth’s atmospheric greenhouse effect by subtracting the 255K emission temperature from the 287.6K physical temperature is like deducting apples from oranges to get a lemon!

3)   A physical temperature may only be compared to another physical temperature. The Earth without an atmosphere would have no oceans and its surface would be no different from that of the Moon (or Mercury), i.e. covered with fine porous regolith of a similar albedo as that of the Moon (since there would be no clouds). Therefore, the size of Earth’s atmospheric greenhouse effect ought to be evaluated with respect to Moon’s actual mean surface temperature since both planets orbit at the same distance from the Sun. Thus, GE = 287.6 – 197.3 = 90.3K.  Perhaps, a better measure of GE is the ratio of these temperatures, which gives the relative thermal enhancement due to the presence of atmosphere (ATE), i.e. ATE = 287.6/197.3 = 1.46.

Newer visitors may wish to refer to several previous threads. This is the main one, where you’ll find pingbacks from others with links.

1. Chris M says:

Welcome back Ned! You have a knack of making science – even equations – enjoyable to a non-scientist. I’m looking forward to the discussion.

2. Stephen Wilde says:

Neat.

I think this is just step 1 and I’m looking forward to the follow through.

3. Harriet Harridan says:

Great to have you here Ned! I’m surprised that the Diviner measurements are not getting more exposure – it is such an important result. And this is a great way of presenting your theory Ned:
Break it down into smaller chunks, and let each one sink in before moving onto the next one. Leading them by the hand to the eventual conclusion….

Moderator: Fwiw, I think substituting the s-b equation with this:
$latexTe = [\frac{S*(1-A)}{(4*e*\sigma)}]^{0.25}$
will bring it out in LaTex under wordpress.

4. Harriet Harridan says:

$Te = [\frac{S*(1-A)}{(4*e*\sigma)}]^{0.25}$

5. tallbloke says:

I’m collecting bloopers, before they vanish. Join in if you like

http://www.lunarpedia.org/index.php?title=Lunar_Temperature
The temperature drop is limited by conduction of heat from layers several meters below the surface, which maintain a roughly steady average temperature that can also be determined from the Stefan-Boltzmann law. In this case ‘I’ represents the incoming solar energy averaged over a full day-night cycle

Iave = 1366cos(θ) / πW / m2

so at the equator T is about 296 K, or a comfortable 23 degrees C if you bury yourself sufficiently.

6. tallbloke says:

Thanks Harriet!

Unfortunately, Ned is v.busy so may not find time to join in. But there’s plenty of fun to be had spotting misused SB equations at academic websites anyway.

There will be a talkshop prize for anyone who can find a NASA or IPCC blooper.

7. dp says:

Not a NASA blooper but: Using methods of mathematical reduction and substitution recently demonstrated at a well-known blog, we can further simplify the right hand side of Harriet’s equation giving us:

Te = Te

This is true for all values of Te.

8. tallbloke says:

dp: you are a

9. Harriet Harridan says:

Diviner Videos on Youtube:

Do I get a prize for this one: http://youtu.be/rLXHEC3cS20 ?

10. tallbloke says:

“Do I get a prize for this one”

Lol, most excellent, definitely a front runner.

11. davidmhoffer says:

dp;
Te = Te
This is true for all values of Te.
>>>>>>>>>>>>>

Wah. Beat me to it.

12. Back of the Envelope says:

I have been following the N&Z posts since January on both blogs. I realize that they published their Part I prior to the Diviner data being fully analyzed and released. As I understand it, they are working to make sure that their physics is consistent with actual observations, adding terms for heat capacity, etc.

But could someone who is in the know update us on what the master plan is? (I know N&Z have day jobs and are very busy.)

Is Part II still in the works? Are the modifications to equations bearing fruit? Perhaps this status information is posted somewhere that I haven’t seen. This thread, while still excellent, didn’t really seem to provide any new information. Thanks.

13. tallbloke says:

Ned and Karl have successfully added a term for heat retention in the lunar regolith which integrates with their original equations to give an accurate result consistent with the DIVINER measurements. The details are embargoed for now, as they are still working on the ‘Reply to Comments Part II’ which will be published here in due course.

This thread simply restates in simple terms the inadequacy of the form of the S-B equation which has been in common use for many years by AGW proponents, lukewarmers and many teaching institutions in the climate science arena.

14. tallbloke says:

The moon, which has no atmosphere, has an average surface temperature of -18°C

http://www.factmonster.com/dk/science/encyclopedia/moon.html
Table 12. ESSENTIAL DATA
Diameter at equator 3,476 km (2,160 miles)
Average distance from Earth 384,400 km (238,900 miles)
Orbital period 27.32 days
Rotation period 27.32 days
Time to go through phases 29.3 days
Mass (Earth=1) 0.01
Gravity (Earth=1) 0.17
Average surface temperature -20°C (-4°F)

And from the UK’s illustrious Open University:

http://openlearn.open.ac.uk/mod/oucontent/view.php?id=398696&section=3.14
the Moon’s average surface temperature is not much lower than the Earth’s,

15. tchannon says:

Handling the thermal conduction I think necessarily moves into proper math by needing to include time, conduction will be a complex term and is a bidirectional process.

Shifting this to eg. earth will be fun. I already know some of the implications and surprises but whether this will agree with N&Z remains to be seen.

16. The bottom line on this is that:

(1) The best observed value we now have for the mean surface temperature of the Moon is 197K from the NASA Diviner project (and NOT from N&Z’s original theoretical formula).

(2) The best observed value for the Earth’s mean surface temperature is 288K.

I doesn’t take a genius to subtract the two to show that the Earth without any atmosphere would be around 90k colder than it actually is (assuming, as is reasonable, that a rocky, airless, cloudless Earth would have approximately the same albedo and emissivity as the Moon).

That 90K warming effect is a truly astounding discovery, since the received opinion among most climate scientists for the last 30 years was that the difference was less than 30K – a factor of 3 too small.

Why does this difference matter? Because it throws out of kilter all the carefully contrived calculations that purport to show that GHGs and CO2 in particular are the cause of the observed warming.

But of course, if N&Zs Unified Theory of Climate is correct, the ~90K warming is wholly explained as the consequence of an entirely different physical mechanism – the density of the surface atmosphere.

N&Z did more than just postulate this alternative physical warming mechanism. As true scientists, they went on to demonstrate from observed evidence (data gleaned from 8 planetary bodies) – that surface pressure and distance from the Sun are the only two variables needed to determine a planet’s mean surface temperature.

This astonishing discovery leaves no room for any other significant causative agent needed to explain the observed warming, such as the so-called greenhouse gases.

So I guess it’s game, set and match to N&Z.

17. I will comment, I will, I will…. tomorrow…

Just got distracted ‘cos the math function at my wiki is now working. Plus soooooo many amazing recent posts. Plus I’ve been working a lot on Graeff. Plus……

Simply no time to visit the lovely lukewarm u-no-hoo after all that!

18. ozzieostrich says:

I can’t help myself, but serendipity and all that, Tallbloke.

The Earth is still a large molten blob, with a thin layer of slag forming he surface.

Bearing in mind that the temperature gradient of the solid(ish) Earth seems would appear to go from hottest (the centre, I guess) to the surface ( in general terms), a question arises.

If we don’t know how hot the Earth was at creation, we don’t know if the Sun’s output has changed from then to now, or by how much, we don’t know the rate at which the Earth lost heat in any detail, how can we say what temperature the crust should be right now? How much insulation has the atmosphere provided over the life of the Earth?

Quite apart from anything else, if the Earth had no atmosphere, would the surface be hotter than that of the Moon, and if so, by how much?

Of course, I am assuming the Moon lost heat at greater rate han the Earth, due to its greater surface volume ratio. On the other hand, I have no way of knowing if the Moon was originally part of the Earth. That would probably complicate the issue, I guess.

Sorry, no excuses – I’m retired.

Live well and prosper.

Mike Flynn.

19. wayne says:

Ned and Karl have done a very welcome job of keeping this analysis of the relationship between radiative solar input to the surface temperatures of various planetary level bodies. We can’t thank them enough.

I started to raise this about a week ago but waited till a post related to S-B.

Many here, rightfully so, keep asking how these new relationships relate to the Earth, specifically, and especially, how water plays into this since it covers and is held within all of the Earth’s oceans, lakes, rivers, soils and life forms. I’ll take a moment while we are speaking of the Stefan-Boltzmann relationship to show everyone something rather curious I have come across.

There is many ways to view the S-B relation by the rearranging the terms to get different views. Ned and Karl did this in their original Unified Climate Theory poster when the grouped and moved constant terms outside the fourth root radical to give a constant term of 25.3966 in their equation eight. Willis Eschenbach threw up over this manipulation, which was perfectly mathematically correct, and I will give him a bit more to throw-up over.

Take the normal gray body form (including emissivity) of P = ε/(1-A) σ T^4, epsilon(ε) being the emissivity, A being the unit-less albedo, sigma(σ) the Stefan-Boltzmann constant of 5.670373 × 10^-8, T the temperature and P the radiative power. This can solve for T to give the well known form used above Te = (P (1-A) / (ε σ))^(1/4). If not all, but most reading this know these well.

But I’m going to do some further rearranging. First lets equate the power P to the temperature T to extract a curious factor, called here the SB.bp or Stefan-Boltzmann balance point, where the temperature curve crosses the radiative power curve, or, vice-versa, they are the same.

P = T
T = ε σ / (1-A) T^4
T/T^4 = ε σ / (1-A)
T^-3 = ε σ / (1-A)
T at balance = (ε σ / (1-A))^-(1/3)

Using albedo and emissivity of one as commonly used (usually wrongly) in the SB common form you just get T = σ^-(1/3) = 260.29318… or roughly 260.3. Now what this temperature represents is the exact temperature where the power by SB is equal. Using a well know value the at 255 K the raw SB radiative output is σ × 255^4 or 239.7 W/m^2, or roughly the IR output at the top of the atmosphere.

Using this new form:
P = 260.3 * (255 / 260.3)^4 = 239.7 W/m^2

oppositely:
T = 260.3 * (239.7 / 260.3)^(1/4) = 255 K

If you ever use this symmetric form of S-B, you raise to ¼ when going from power to temperature and raise to 4 when going from temperature to power, that simple, identical equation but for the power term.

So instead of P = σ T^4 you can use the bit longer form of P = 260.3*(T/260.3)^4 form to get always the same power. I have been using this form for months simply because it easier on my calculator to key in, but I never looked deep enough in this curious form of SB until I needed to also include emissivity and albedo.

Before you laugh and say this is all but trivial, look what happens when you recalculate this SB.bp for the Earth as a whole as viewed by far, for instance, the Diviner probe. The input to the Earth is averaged at 1361.7/4 and we know the mean surface temperature by observations to be 287.6K as Ned said above so this curious SBbp factor pops out as (287.6^4 / (1361.7/4))^(1/3) = 271.88, K or W/m^2, in this instance it is either, they are equal and symmetrical.

Well I don’t know about anyone else but that shocked me a bit, the 271.88K… that is so very close to the triple-point of mean surface seawater. Using that new relation you get

T = 271.88 * ((1361.7/4) / 271.88)^(1/4) = 287.6 K
or
P = 271.88 * (287.6 / 271.88)^4 = 340.4 W/m^2 times 4 or 1361.7 W/m2 TSI

… all hinged on the SB crossing point of the SB relation at, get this, apx. mean triple-point of seawater…. now I call that rather curious! Above this balance point radiation rapidly (4th power) increases and water is liquid. Below this point radiation rapidly decreases when water is solid. I just don’t know if this stumbled upon relation has any deeper reason this is the value it is, but this curious form of S-B sure highlights it.

I guess, as in all real physics, all equivalent forms of all relationships must be equal, hold, and make sense and certain equivalent forms lead you to deeper, sometimes hidden, understanding of exactly what you are speaking about when using the equations. This is one such equivalent form of the Stefan-Boltzmann equation.

Sorry for this being just a tad off the prime thrust of this post.

20. tchannon says:

David,
Pressure has a role but so do several other factors.

21. tchannon says:

wayne,
Aside question: the lunar model I showed a little while ago, where is albedo or emissivity in there?
I haven’t put it in.
The surface conductance is a good match for published figure and the heat capacity has nothing to do with this. Result is an excellent match with diviner.

If it matters it must be somehow encoded in something else.

22. wayne says:

tchannon, I go back to that post and investigate that… I’m assuming the post was when you were doing the equivalent electric circuit matching the temperature curve. Just off-the-cuff, my first answer would be that anything, except radiating to blank empty space from a true blackbody (really, never), has emissivity involved whether ignored or not, so it is probably is inferred or embedded from other terms and values.

A few days ago I finally found the correct way to handle emissivity in a SB relation using T differentials:
$\displaystyle \large {\varepsilon}_{effective}=\frac{\varepsilon_1\varepsilon_2}{\varepsilon_1+\varepsilon_2-\varepsilon_1\varepsilon_2}$
When one of the two is truly one, as against void space, it collapses to just the emissivity of the other. I’m advancing though slowly!

23. ferd berple says:

dp says:
May 1, 2012 at 3:39 pm
Te = Te
This is true for all values of Te.

What about Te = n/0 ?

24. wayne says:

tchannon, an aside answer
http://tallbloke.wordpress.com/2012/04/02/a-model-of-lunar-temperature/

My guess any emissivity might be in the ARB1 voltage/current ratio for I don’t see any hard figures on this listed (or am just missing it). Since it is a current sink and the circuit is constantly in flux I assume the voltage across this is also in constant flux. Don’t know if you can do this, but, if you can get an instantaneous relation between the current (radiative flux power) to the voltage (temperature) and feed them into the equation I gave above (T^4 / P)^(1/3) and if the value is not 260.3 then there is either emissivity or albedo or both hidden within. Just a guess but I don’t have the tools (SPICE) or the electronics expertise, and you definitely do, to check this out.

25. tchannon says:

If I take out the power 1.3 term, which is unimportant it is pure SB as a rotating planet, not steady state.

The flux is constant but modulated. The sink is literally SB applied to itself. No hidden magic.

The mean temperature of the body is not the SB temperature. SB is applied to a thermal resistance which has a millstone of inertia connected. SB is a pure theoretic surface, an interface with a physical body.

The mean can be widely adjusted via the thermal conductivity. Highly conductive gives earth temperature.

If the body was perfectly conductive it would heat up fastest. Leave it infinite time it would attain a terminal temperature of whatever SB can manage when leaking away. Something like this is I guess what is generally written about.

Perhaps all this suggests there has to be a context if a real world is involved. Dunno if that makes sense.

26. tallbloke says:

tchannon says:
May 2, 2012 at 1:13 am
David,
Pressure has a role but so do several other factors.

What are they and what are the approx magnitudes?

27. tallbloke says:

Wayne: Very interesting. So what conclusions might we draw? Is it possible the ‘oceanic greenhouse effect’ gets us up to around freezing (your 271.88K) and N&Z’s ‘Atmospheric Thermal Enhancement’ (ATE) plus a few other effects gets us the rest of the way to 287.6K?

If that were the case, then out of the ~90K difference between the Moon’s average surface temperature and Earth’s, we could start trying to assign various quantities to various possible effects.

For example:

Oceanic heat retention and heat spreading
Land surface heat retention
Faster axial rotation
ATE due to higher near surface air density
ATE due to Loschmidt gravito-thermal gradient (Jelbring hypothesis)
Back radiation (possibly negative rather than positive considering evaporation)

I appreciate what David Socrates is saying about N&Z’s success in obtaining the surface temperatures of eight celestial bodies from just two quantities, distance from Sun and surface pressure, but given that Earth has oceans whereas Venus hasn’t, it think there is plenty more of interest to investigate within their overarching discovery.

28. dp says:

n/0 is undefined. Therefore, Te = n/0 is undefined. Do I have to do all the work around here?

29. wayne says:

TB, I don’t know what that is. It all started just as a mathematical rearrangement so keying SB and reverse SB calculations was faster on my trusty TI-36 calculator. (btw: I have more graphical calculators but that’s my quik & easy mainstay). Remember me telling you for that curious 260.3 calculation? At that time I hadn’t even taken the time to figure what the equation was, it came from a fit on an excel spreadsheet. Later I found what it was, sigma^-1/3. I hit my head… of course, this is just the point where temperature and flux are equal so T to flux is two squares and flux to T is two square roots, easy trick and I’ve used it for over a month without further thought.

Then on day I needed emissivity too, ok, how first can I adjust that 260.3 to slide up the scale and how do I calculate that balance point. Bit of mental crunch and it became apparent it was simply (epsilon*sigma)^-1/3. I’ve used that for a about a week, just for ease. After some more thinking it became apparent that if you take (T^4/flux)^(1/3) for any temperature-flux pair you can easily get the emissivity right back out. It that calculation is 260.3 … well, that calculation is just using SB without albedo or emissivity.

Take the maximum temperature on the moon listed in the Diviner paper. They measured 399K max and insolation is 1362 with no atmosphere so cuberoot(399^4/1362) is 264.99 so emissivity is indeed involved. What is the effective emissivity? (260.3/264.99)^3 or 0.947. It all flows so easily on my dippy calculator. That is what it “means” to me right now, easy keying. Scientifically I’m still searching to see if there is even ANY meaning but falling on the triple-point threw me.

See, I was not making a long ‘statement’, I want some help to see if that has any meaning, I’m asking the same question. There are many here, you too that could take a look at this. I just stumbled upon it.

And yes, Ned and Karl have done some fine work and I can find nothing yet that counters their theory. In fact, I have sent to Ned and Karl a pair of parameters that seem to both mathematically make sense and are just a tad from the two exponents in equation seven but have yet to hear back from them.

30. tallbloke says:

Stephen: Other way round I think. As pressure drops equivalent evaporation takes place at a lower temperature.

Wayne: “I have sent to Ned and Karl a pair of parameters that seem to both mathematically make sense and are just a tad from the two exponents in equation seven but have yet to hear back from them.”

Send it to me and I’ll see if I can get Ned and Karl to take a look and comment.

31. tallbloke says:

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32. Stephen Wilde says:

“Stephen: Other way round I think. As pressure drops equivalent evaporation takes place at a lower temperature.”

I think that relates to my post on the other thread.

Pressure drops so less than one unit of energy is then required to break the molecular bonds and take up the 5 units in the phase change so the ratio gets bigger than 5 to 1.

If pressure rises then more than 1 unit of energy is required to take up the 5 units in the phase change so the ratio gets smaller than 5 to 1

The energy required when the bonds between the water molecules break remains the same at 5 but the amount of energy needed to cause the break is pressure dependent.

33. wayne says:

“Your E-mail has been selected to receive the sum of $850,000.00USD from Chevron Texaco Oil Company Mega Millions jackpot …..” Just send us your account number and we’ll be sure to take care of you! Careful… that sure sounds like some enviromentalists are behind it! Haven’t we heard that same story before? Guessing the tax-the-world plan isn’t going to come through for them. 34. adolfogiurfa says: One thing is a “black body”, another that “black body” covered with 71% of seawater. We can forget about atmosphere, that´s peanuts, what really counts is water heat capacity: air cannot hold heat, its volumetric heat capacity, per cubic cemtimeter is 0.00192 joules, while water is 4.186, which means: 3227 times. Water is a very peculiar material: Prof.Giorgo Piccardi´s “Structure of Water … Subtle Properties”: http://www.rexresearch.com/piccardi/piccardi2.pdf 35. ferd berple says: dp says: May 2, 2012 at 8:30 am Therefore, Te = n/0 is undefined. undefined undefined Te Te for n/0 Therefore, Te = Te only for the set of defined numbers 36. steveta_uk says: Can someone explain please exactly what is meant by “actual average physical surface temperature of the Moon”? Is this an arithmetic average, gemoetric average, 4th power average, or what? The temp range reported by Diviner is approx 40K to 400K, but an average of just these two numbers could be 220 ((40+400)/2) or 336 ((40^4+400^4)/2)^0.25. Both are valid answers, but only the 2nd one can be compared to the SB values. 37. ferd berple says: aaargh, html dp says: May 2, 2012 at 8:30 am Therefore, Te = n/0 is undefined. undefined != undefined Te != Te for n/0 Therefore, Te = Te only for the set of defined numbers 38. tchannon says: steveta_uk, Lot of background, take weeks to wade through it all. 1) The surface of a sphere should be treated as a sphere and now read carefully, therefore S-B must be applied to the sphere as a spherical integration. It is not a flat disc. The surface area of a flat disc is smaller than the surface area of a sphere, so it is obvious the flux per unit area is lower and therefore temperature is lower. Account also needs to be taken of the thermal properties of what is a spinning sphere. 2) The DIVINER project has made the best attempt so far to measure the surface temperature of the entire moon. 3) The result of computation as (1) is in agreement with (2) and is disagreement with all the prior official moon temperature figures, usually circa 250K Footnote, I modelled the system using a completely different method and the result is also in agreement with (3). Others have computed the equivalent to the spherical integration by other methods such as monte carlo variants, agrees with (3). The comments about go capture the old published figures before they are silently changed without admitting to the mistake might now make sense. I add that there is still disagreement, plus what exactly is meant. Doesn’t seem to be disagreement the old figures are wrong. By implication this makes a mess of the calculations of earth temperature, which are based on the same incorrect math. The above is my understanding and should be treated as an opinion. 39. steveta_uk says: tchannon, my question was much simpler than your answer. I just want to know what is meant by “average” in this context. I’ve asked this many times, and not got an answer. As I pointed out in my question, there are many ways to calculate an average. Which method is used here? 40. tchannon says: Grin. I realised afterwards I hadn’t answered the question… do I edit or not. Bad call. I can’t answer for others. The form that gives the body temperature given the surface temperature. which seems with what I have been doing to be the common average but where I am handling area that means the sine weighted average. (I am well aware of the problem over what kind of average) Could be that the question is not as clear as it might appear. What exactly is being averaged and what is the answer supposed to represent. Good idea if you pursued this with others, some clarity. Don’t think anyone is hiding anything so much as getting things clear can be tricky. 41. Steveta Diviner effectively took the temperature of lots of small equal-area patches and the average was calculated from that kind of data AFAICT. Check here. Nikolov and Zeller’s genius was to take on that method mathematically for the theoretical temperature calculations, and develop appropriate equations from first principles, using calculus for a spherical surface. So we have the integral of the S-B temperature of tiny patches over the daylight hemisphere of the moon, averaged with the nighthemisphere which follows a completely different pattern. 42. Ned Nikolov says: To steveta_uk: The therm ‘average physical surface temperature of the Moon’simply means an area-weighted arithmetic average of surface temperatures. One cannot estimate a correct mean global temperature by simply averaging the observed minimum and maximum values, because, on the surface of a sphere, higher values tend to occur more frequently at lower latitudes that occupy larger surface area, while extreme cold temperatures tend to occur near the poles, which take a much less area. So, the correct procedure to calculating the physical mean temperature of a spherical body is this: 1) Compute the average annual temperature for several latitude (say at 5 or 10 degree increment) from the equator to the poles. 2) Fit a high-order polynomial through the mean latitudinal temperature values using a non-linear regression. A 6th-order polynomial usually provides a very precise fit. The purpose of this regression is to produce a continuous function T(L) that accurately and smoothly describes the variation of the average temperature with latitude (L). 3) The global mean temperature of the planet (Tg) is then calculated from the definite integral of the polynomial, i.e. Tg = INT (0 to pi/2) T(L)*cos(L) dL, where L is the latitude in radian. 43. tchannon, On May 2 at 1:35am in answer to my comment of May 2 at 12:23am pointing out that N&Z had shown that mean surface temperature was ONLY determined by surface pressure and distance from the Sun (leaving no role, in particular, for CO2 or any other so-called greenhouse gases), you replied: David, Pressure has a role but so do several other factors. …to which Tallbloke quite properly replied: What are they and what are their approx magnitudes? Your comment did not advance the argument one jot and I now note that you haven’t responded to Tallbloke’s dry comment. Likewise, on May 2, 2012 at 3:12 pm steveta_uk asked: Can someone explain please exactly what is meant by “actual average physical surface temperature of the Moon”? …to which you provided an entirely incomprehensible reply at 4.09pm to which steveta_uk quite properly responded: tchannon, my question was much simpler than your answer. I just want to know what is meant by “average” in this context. I’ve asked this many times, and not got an answer. As I pointed out in my question, there are many ways to calculate an average. Which method is used here? …to which you replied by first agreeing that you hadn’t answered his question and then by including the statement: Could be that the question is not as clear as it might appear. What exactly is being averaged and what is the answer supposed to represent? Since you are clearly out of your depth in both the above instances please let me relieve you of your misery and answer steveta_uk’s question directly and, I hope, succinctly: The actual average physical surface temperature of the Moon is correctly determined in one way, and one way only: by measuring the temperature in Kelvin at every infinitessimal point on its surface, adding all those temperature values together, and then dividing by the number of points measured. Note that the above is a physical description of what needs to be done, as one might achieve by sticking thermometers into the Moon’s surface at regularly spaced intervals. N&Z provided a theoretical mathematical integration using calculus but their underlying rationale was anchored in this physical reality. All other alternative theoretical formulations that you, or others, might dream up are not anchored in physical reality, and are therefore wrong. 44. Ned Nikolov says: I agree with David Socrates above. His description of the mean physical temperature of a planet is correct. See also my post (just preceding David’s reply) about a practical method for calculating a mean global temperature that was used to derive the 197K value for the Moon … The so-called emission temperature computed by the ‘standard’ SB equation (which uses a globally averaged absorbed solar flux) is a fiction as as far as physical temperatures are concerned. Comparing emission temperatures with any palatable (measurable) temperatures is like comparing apples with oranges. The fact that both temperatures have the same units does not make them physically compatible, and that’s because of Holder’s inequality! 45. wayne says: Ned, this might be asking too much, but it would behoove many here if you could help explain mathematically why Dr. Sweger was incorrect in the way he was using the integration by bringing the cosine integration outside the radical. I attempted to but my calculus lacks. See: http://tallbloke.wordpress.com/2012/03/30/daniel-m-sweger-response-to-nikolov-and-zellers-unified-theory-of-climate and my attempt at: http://tallbloke.wordpress.com/2012/03/30/daniel-m-sweger-response-to-nikolov-and-zellers-unified-theory-of-climate/#comment-21828 David Socrates was agreeing with me here on the numeric side: http://tallbloke.wordpress.com/2012/03/30/daniel-m-sweger-response-to-nikolov-and-zellers-unified-theory-of-climate/#comment-21633 but I fall on the explaining his error in the calculus, my lead term ends up 0.40528… (2/π)^2, not exactly 0.4 so I am also wrong to about ½%. Could you clarify so Dr. Sweger’s disagreement is put to bed? MODS– I think I just mistakenly posted this also on Dr. Sweger’s article, might remove it. Woops! [Had a look at your history can't see it anywhere. --Tim] 46. tchannon says: David Socrates, I’m struggling with language more than understanding, we are talking about the same thing. There are a variety of ways of calculating something like that. Ned commenting is welcome because only he knows what he means. The ignored question? As co-mod I need to make a calls. If I explained there would be an Oh. Takes thread off in undesirable directions. I will give one on the moon, more conductive surface, higher mean body temperature. 47. Ned Nikolov says: wayne (May 3, 2012 at 12:37 am): I went to the paper by Dr. Sweger at: but all equations are gone. The text shows the word ‘image’ in place of every single equation. So, I could not follow his derivations … [Reply] Thanks Ned, I’ll check that out. The pdf document is here: http://tallbloke.files.wordpress.com/2012/04/response-to-unified-theory-of-climate-update-1.pdf [ fixed --Tim] 48. RKS says: Has anybody calculated the contribution to surface temperature of the heat from the Earth’s core? I read somewhere it was low but I have seen no figures for it. 49. tallbloke says: RKS: around 0.1W/m^2. I think that is the land surface figure. The Ocean bed is thinner, and is continuously water cooled to ~2C so I would expect there to be a higher transfer of heat down there. Plus the continuous volcanism at the mid ocean rifts of course. 50. dp says: Ferd – data do not often contain useful undefined numbers. I appreciate your interest in mathematical preciseness but all accept the limits that divide by zero brings to numeric ranges. I could counter that n * infinity is also an awkward factor. We all know this. The equation was introduced as a tongue in cheek humor offering which you’ve elevated to pedantic status. Happy, Mr. Pedant? 51. RKS says: tallbloke says: May 3, 2012 at 7:26 am>>>> Many thanks and much appreciated. 52. steveta_uk says: DS said: The actual average physical surface temperature of the Moon is correctly determined in one way, and one way only: by measuring the temperature in Kelvin at every infinitessimal point on its surface, adding all those temperature values together, and then dividing by the number of points measured. And Ned seems to concur, so thanks for providing the answer. Now, back to my original point. I believe this method is not the correct way to determine an average temperature, as it will provide a much lower average than the apparent SB temperature (which is indeed the results everyone has found). As TC has requested a more detailed explanation, here goes (and I hope html pre tags works, else this format won’t make much sense). Here is my input data:  0 3 6 9 12 15 18 21 24 +----+----+----+----+----+----+----+----+NP | 60 | 50 | 40 | 90| 110| 100| 80 | 70 | +----+----+----+----+----+----+----+----+ | 90 | 85 | 70 | 200| 250| 225| 115| 95 | +----+----+----+----+----+----+----+----+ | 100| 90 | 88 | 320| 390| 350| 150| 120| +----+----+----+----+----+----+----+----+EQ | 100| 90 | 88 | 320| 390| 350| 150| 120| +----+----+----+----+----+----+----+----+ | 90 | 85 | 70 | 200| 250| 225| 115| 95 | +----+----+----+----+----+----+----+----+ | 60 | 50 | 40 | 90| 110| 100| 80 | 70 | +----+----+----+----+----+----+----+----+SP  This shows a lunar “day” split into 24 “hours” in 3 “hour” bands, so 0 is dark side, 12 is overhead sun, and back to dark side at 24. The temps are in 6 bands north(NP) to equator(EQ) to south(SP). The numbers are taken by eyeballing a Diviner map – the result will work demonstrate what I want even if the exact numbers are wrong, so don’t fuss about the values. I assign weights to the bands going 1, 2, 3, 3, 2, 1 from north to south, so give an approximation for the area weighting corrections required for a sphere. I then calculate two “average” temperatures. The first is a simple arithmetic mean of the weighted values, and the result is 160.08K. The second is a 4th power geometric mean, which I think is the correct way to average temperatures, and the result is 242.51K Given the total emissions from the moon, using the SB law, you apparently get a much higher “average” temperature than measured by Diviner. And I’m fairly sure this is nothing more than a difference in the use of the word “average”. Code below (bash + bc). #!/bin/bash NUMS0="60 50 40 90 110 100 80 70" NUMS30="90 85 70 200 250 225 115 95" NUMS60="100 90 88 320 390 350 150 120" SUM=0 CNT=0 for N in$NUMS0 $NUMS30$NUMS30 $NUMS60$NUMS60 $NUMS60$NUMS60 $NUMS60$NUMS60 $NUMS30$NUMS30 $NUMS0 do SUM=$((SUM+N))
CNT=$((CNT+1)) done A_AVG=echo -e "scale=2\n$SUM/$CNT" | bc SUM=0 CNT=0 for N in$NUMS0 $NUMS30$NUMS30 $NUMS60$NUMS60 $NUMS60$NUMS60 $NUMS60$NUMS60 $NUMS30$NUMS30 $NUMS0 do VAL=$((N * N))
VAL=$((VAL * VAL)) SUM=$((SUM+VAL))
CNT=$((CNT+1)) done G_AVG=echo -e "scale=2\nsqrt(sqrt($SUM/$CNT))" | bc echo A Ave =$A_AVG
echo G Ave = $G_AVG  53. wayne says: Hi steveta_uk. In your code I see no weighting being applied. Seems 4,11,15,15,11,4 would have been a bit closer but I don’t seem to find it at all. If it’s there, point to where in the code it is performed. It is making both of your cases to low. Second, why are you raising to the 4th power bit in the second example, like RMS but RMS squared? That just hugely weighting high temperature points over lower temperature points. Why not just squared to get the RMS mean? But even RMS mean of temperatures makes absolutely no senese to me at all though I guess there may exist some special views of the temperature field where that may make some sense. Explain please. I can’t see what you are getting to at all and would like to. 54. steveta_uk says: It’s a 4th power average because SB law uses 4th power. The weighting in the code is done by this line: for N in$NUMS0 $NUMS30$NUMS30 $NUMS60$NUMS60 $NUMS60$NUMS60 $NUMS60$NUMS60 $NUMS30$NUMS30 \$NUMS0

NUMS0 is in there twice, NUMS30 4 times, NUMS60 6 times, hence to 1:2:3 weigts (4:11:15 is close).

“That just hugely weighting high temperature points over lower temperature points.”

Exactly – that’s the whole point. If you use a simple average, the result is MUCH LOWER than the SB calculated value for the known radiation flux.

55. tchannon says,May 3, 2012 at 3:03 am: David Socrates, I’m struggling with language more than understanding, we are talking about the same thing. There are a variety of ways of calculating something like that.

OK, my apologies if I seemed too harsh and I hope there are no hard feelings. I do appreciate that you are trying as hard as I am to get to the truth.

The reason for my frustration with all of this talk about alternative ways of calculating the mean surface temperature of an airless rocky planetary body at Earth-distance from the Sun is that it is only the method offered by N&Z (namely, measuring the temperature at all infinitesimally small points on the sphere’s surface, adding up all the temperature values, and dividing by the number of points) that matches exactly the empirical method used to determine the Earth’s actual measured mean surface temperature of 288K (namely, taking temperature readings from thousands of weather stations all over the Earth’s surface, adding up all the temperature values, and dividing by the number of points). [Note: Unfortunately the weather stations are not conveniently positioned at equal spacing, so some additional area normalisation also has to take place but in principle it is exactly the same averaging process.]

N&Z’s original formula for the mean surface temperature of an airless Moon (a good proxy for an airless Earth) delivered a value of 154K. But since then, the empirical data from the Diviner project has produced an observed mean surface temperature of 197K. The difference is due to the heating of the Moon’s surface, a factor that was not taken into account in N&Zs original formula (but this has now been corrected).

Subtracting 197K from the Earth’s actual measured mean temperature of 288K gives a temperature enhancement due to the addition of an atmosphere of 91K.

In contrast, using the Classic formula for the mean surface temperature of an airless Earth, the temperature enhancement due to the addition of an atmosphere is only 33K (if we use Earth’s current albedo) or around 15K (if we use the Moon’s albedo, arguably more representative of an airless Earth).

So the conclusion is that the temperature enhancement due to the atmosphere is between three and six times larger using the N&Z method than hitherto calculated using the Classic formula.

Why the huge difference beween the two methods of calculation? It is because the Classic method uses a different method of averaging from the method used for the measurement of the Earth’s real surface mean temperature. It should be very obvious that this is a complete nonsense: if you are going to subtract two mean surface temperature estimates to produce a meaningful temperature difference, both original numbers must have been produced using the same averaging method. That is simply and solely what N&Z did. In contrast it is what the users of the Classic formula failed to do over the past 30 years.

For the above reasons it is not simply a matter of opinion as to which averaging method one should use to calculate the surface temperature of an airless Earth, assuming of course that one’s objective is to determine the temperature enhancement resulting from the addition of its atmosphere. One method is correct and the other is wrong.

56. steveta_uk says, May 3, 2012 at 10:33 am: David Socrates said: “The actual average physical surface temperature of the Moon is correctly determined in one way, and one way only: by measuring the temperature in Kelvin at every infinitessimal point on its surface, adding all those temperature values together, and then dividing by the number of points measured.”

And Ned seems to concur, so thanks for providing the answer.

Now, back to my original point. I believe this method is not the correct way to determine an average temperature, as it will provide a much lower average than the apparent SB temperature (which is indeed the results everyone has found).

steveta_uk, you are entitled to your opinion but please read the detailed explaination I have provided above to tchannon. The method of determining the average surface temperature of an airless Earth MUST be the same method as is used to determine the average temperature of the real Earth (using weather stations). Otherwise when you subtract the two numbers to find out the temperature difference due to the addition of an atmosphere you get a meaningless number. It is like subtracting apples from oranges.

That is an inescapable logical conclusion.

As far as your calculations are concerned, I am afraid they are simply wrong because you have not performed the calculations correctly according to the rule I specified, namely to measure the temperature at every infinitessimal point, add all the values together, and divide by the number of points. Unless you do it that way you will go on and on getting the wrong answer how ever hard you try!

57. steveta_uk says:

DS, please answer this very simple question.

If the entire system consists of only two point samples, point A has a temperature of 100K and point B has a temperature of 500K, what is the physical meaning of your claim that “the average temperature is 300K”?

I know of no justication for the assumption that a simple arithmetic average of two temperatures has any physical meaning whatsoever.

If you can prove you are correct, then fair enough, but arrogant shite like “they are simply wrong because you have not performed the calculations correctly according to the rule I specified” doesn’t cut it with me.

58. tchannon says:

Not a problem David, things look different from this side where I get it wrong sometimes.

The problem of how computation and measurement is done keeps turning up. My background is very unusual so I come to all this with a different mind which includes not understanding the incomplete ghg arguments.

The reason I created a model of lunar temperature using a thermonics dual of electronics was because it must agree, same math done differently. It matches within likely error: Diviner is a best attempt, things like TSI are only approximate.

Essentially a plane sq metre is illuminated with the sun going around the moon, same effect via modulation. The surface conduction was twiddled to match the Diviner equatorial profile (digitised plot), giving a value within many published guesses and akin to Apollo.

I then took latitudinal bands with the illumination further modified to simulate the double curvature. (plane is tilted)

Finally the latitudinal band spot values are common averaged, and the latitude averages weighted averaged. This gives the ~200K figure. (hold off knee jerk)

I already know that this math is identical to computing earth global from published gridded, have code which does that.

So we have all giving the same figure
N&Z
Diviner
A dual model

Others have reported agreeing with N&Z where they have for example used random spot temperatures instead of direct integral.

It does not follow this is correct!

For example, is there double accounting of a geometry factor? (can knee jerk now)

In my view the temperature of a planet is the body temperature, which is also the entropy. Ignoring internal heat I expect for steady state this to be the same as the mean surface temperature. We have no measured figure, is unknowable. (for earth it is even harder)

Clarity is needed. Some of this means being extremely clear on meaning, no ambiguity in the minds of the reader, a very hard trick for a writer.

59. RKS says:
May 3, 2012 at 6:27 am | Reply w/ Link

Has anybody calculated the contribution to surface temperature of the heat from the Earth’s core?

One interesting thought following on from Graeff and also from Ned and Karl, is that Earth is hot inside because of gravitational pressure… getting hotter and hotter from the top of the troposphere down, with just a change of phase for Earth’s surface. Therefore… Earth’s interior heat will not generally rise, as unlike the atmosphere it is generally too constrained by the solidity of substance, to convect. And that would then be the real reason our feet are not scorching hot.

60. Harriet Harridan says:

Lucy:
“One interesting thought following on from Graeff and also from Ned and Karl, is that Earth is hot inside because of gravitational pressure…”

Lightbulb moment! Thanks Lucy.

61. Harriet Harridan says:

Lucy, further to my last – then why are the deep oceans cold?

62. Ned Nikolov says:

To All,

There were questions raised by some of you about the status of the Unified Theory of Climate (UTC) and if there will be a Reply part 2 to address comments put forward by bloggers. I would like to answer this question directly to put to rest rumors or speculations as the ones spread by Joel Shore for example at other blogs.

We are currently working on submitting several papers to the peer-reviewed journals. We thought it is critical for the successful acceptance of this new paradigm that we publish our findings in the formal science literature before continuing with the blog discussions. To this end, we have devised a publication strategy based on a piecewise approach (i.e. publishing different aspects of the theory in separate papers). This is dictated by the sheer amount of new information and by the novelty (non-conventionality) of the concept.

We have done a series of additional analysis and theoretical derivations beyond what was originally presented in our first blog article and the follow-up Reply Part 1:

The result is that our UTC has become even more robust due to a much improved gray-body temperature formula and a new better (tighter) relationship between atmospheric pressure and Nte (the relative thermal enhancement factor) across planets of the solar system… Physical concepts and real data have converged into a beautiful relationship that reveals a qualitatively new insight about the nature of the so-called ‘Greenhouse Effect’.

The new relationships and theoretical implications will be presented on this blog in due course…

63. tchannon says:

Thank you Ned.

64. Harriet Harridan says:

Ned: “We are currently working on submitting several papers to the peer-reviewed journals.”

That is excellent news Ned, really excellent.

I suspect certain quarters may try to hamper you with that They don’t like it up ‘em you know.

65. Harriet Harridan says:
May 3, 2012 at 8:50 pm | Reply w/ Link

Lucy, further to my last – then why are the deep oceans cold?

Fascinating isn’t it? I forgot to include the oceans.

But this is my personal feeling FWIW. In atmospheres up to the tropopause (ALWAYS it is at 0.1bar, on all atmos planets), there is a dynamic balance between gravitational warming causing expansion/ lower density, and the convective factor causing air to rise and expand further to cool again. By night this can happen so stealthily we kid ourselves that no convective forces are at work. But Graeff’s experiments clearly demonstrate the power of convection to partly neutralize gravitational warming.

This is why the adiabatic lapse rate is lower than the gravitational lapse rate calculated and measured by Graeff.

Oceans: we have to remember various factors here. (1) Cold water sinks at about 4°C – miraculous anomaly (2) Really cold water at the poles sinks continually to drift equator-wise slowly (3) The temperature/density profile of water, plus its viscosity when compared with air, keeps the cold water sluggish at the bottom…

The facts are before us, like Harriet says. But though I’m clear that Graeff has shown beyond a doubt the existence of a gravity-induced temperature gradient in water in the lab, it is invisible in the oceans and I don’t feel I fully understand why.

Anyone with thoughts here?

66. wayne says:

Great news Ned, enough said. Forget that question above, it’s not that important and can wait. Hope to soon give you both a big congrats!

67. wayne says:

@steveta_uk:
I ran you numbers and also came up with ~162K using 4,11,15 weighted even though those are just your ‘eyeball’ figures. And yes, it is that cold on the average. I now see where your weighting was applied.

See, that is basically a proportion to the mean kinetic energy contained at any point on the moon, averaged, and can be understood by looking into either PV=NkT or KE=3/2kT relations with constant adjustments for units. T is proportion to that energy. Taking that back to the fourth power is the average power that any average point can radiate that energy away (sb). If you don’t then take two square roots in your example two, even that does have some meaning, to me anyway (applying all proper constants and unit adjustments of course). The climatology branch of science has had this wrong for so long has contaminated everyone’s view of the correct science.

I just don’t understand why you think this proper average as given by N&Z is meaningless (the 162K).

I also got:
n75 = mean({ 60 , 50 , 40 , 90, 110, 100, 80 , 70 }), s15 = n15
n45 = mean({ 90 , 85 , 70 , 200, 250, 225, 115, 95 }), s45=n45
n15 = mean({ 100, 90 , 88 , 320, 390, 350, 150, 120}), s75 = n75
avg1 = mean({n75,n45,n15,s15,s45,s75})
avg2 = sum({4*n75, 11*n45, 15*n15, 15*s15, 11*s45, 4*s75})/60

avg1 is the un-weighted average of 139K.
avg2 is the proper sine weighted average of 162K.

Ned, Karl and now Diviner have shown it is not quite that cold mainly due to the surface vertical diurnal conduction. Seems you need to break it into smaller sections which would raise that rough average as you approach a section size of zero.

68. dang Harriet you got me out of bed as the answer about oceans flashed in.

It’s to do with the energetic differences between the solid, liquid and gaseous states, and the comparative effects of gravity. In the solid state, I expect the pressure-induced heat to just be there: hot core of Earth. In the gaseous state, Graeff has shown it is there and measurable to fit theory when convection is suppressed; moreover we see it at work in the adiabatic lapse rate, although partially negated by convection. Because the gaseous state is so rarefied, gravity’s effect becomes far more significant and noticeable per unit mass (far further to fall) than in the other states. Liquids conversely are far denser and so take up less vertical space per unit mass and so have less gravity effect compared to convection, and as Graeff showed, if there is nothing to inhibit convection, the gravity effect does not show at all, being neutralized by convection.

Different factors win out in different circumstances.

And personally, I suspect that the greenhouse gas factor DOES win out… in the rarefied (pressure<0.1 bar) stratosphere-mesosphere temperature inversions, the centre kink of Earth's "W" temperature profile.

69. well, I should append to all the above “it seems…” “it appears” since this is largely my intuition at work, albeit an intuition based on wide info.

70. Harriet Harridan says:

Lucy: “dang Harriet you got me out of bed as the answer about oceans flashed in.”
:-O sorry Lucy!

Makes sense that this small force is more noticeable in a gas. But then again in should be less noticeable in a liquid and even less in a solid.

I’m far from an expert but is this how it works? Each molecule has energy:
Gas = molecules in high energetic state. In the atmosphere lots of these high energetic molecules are packed together, by gravity, at the surface of the earth. Hence surface is warmer and we have our lapse rate. High pressure = high temperature = more energetic molecules per m^3.

Liquid = medium energetic state: In the seas the situation is confused because the maximum density of water is at 4C. So packing more water molecules together, by gravity, at the ocean floor can’t make them more dense (molecules closer together) as they are already as dense as they can get, so we can’t increase the energetic molecules per m^3.

Solid = low energetic state. can’t readily be packed together molecules per m^3 already close to the maximum. Takes a whole pile of gravity, and thus exceptional circumstances only found in the centre of a planet or a star. This tiny amount accumulates over the millennia and heats and heats

71. Harriet Harridan says:

Scratch my last. Re-reading yours you explained it much better than I did. (even at 1.23 am!) Thanks Lucy.

72. paulinuk says:

It’s been stated that the only atmospheric gases that radiate thermal energy back to the surface are the greenhouse gases because nitrogen and oxygen don’t radiate (which is nonsense otherwise balls of nitrogen or hydrogen gas in space would never cool down). It’s true that GHG’s are the only ones able to absorb infrared to the greatest extent but molecules in the air get some or all of their energy from collisions with other air molecules or the ground.

I found an excellent description Thermal radiation at the The National Radio Astronomy Observatory website:

Thermal emission is perhaps the most basic form of emission for EM radiation. Any object or particle that has a temperature above absolute zero emits thermal radiation. The temperature of the object causes the atoms and molecules within the object to move around. For example, the molecules of a gas, as in a planet’s atmosphere, spin around and bump into one another. When the molecules bump into each other, they change direction. A change in direction is equivalent to acceleration. As stated above, when charged particles accelerate, they emit electromagnetic radiation. So each time a molecule changes direction, it emits radiation across the spectrum, just not equally. As a result, the amount of motion within an object is directly related to its temperature.”

So according to the desription, when molecules of gas bump into each other and they suddenly change direction or start spinning violently, this will cause the electrons and protons within the molecule to emit thermal (black body) radiation because they have been accelerated, irrespective of what type of molecule it is.

Another similar form of thermal radiation is that emitted by the sun: Due to nuclear fusion taking place, the Suns’ plasma has accelerating electrons and protons of mainly hydrogen and helium emiting electromagnetic thermal radiation called free-free emission, which is descibed as:

“Another form of thermal emission comes from gas which has been ionized. Atoms in the gas become ionized when their electrons become stripped or dislodged. This results in charged particles moving around in an ionized gas or “plasma”, which is a fourth state of matter, after solid, liquid, and gas. As this happens, the electrons are accelerated by the charged particles, and the gas cloud emits radiation continuously. This type of radiation is called “free-free” emission or “bremsstrahlung.”

For some reason climate scientists have insisted that only GHG’s can radiate thermal energy because some molecules i.e. the GHG’s H2O,Carbon dioxide, Methane etc which have dipole oscillations which allow them to absorb/ emit radiation from the Earth’s surface and the Sun in the infra red region. This gives the molecules dipole oscillations which gives the molecule internal motion, and not directly translational or rotational motion; they’ll get that when coming into contact with another molecule. As I stated above, air molecules don’t get all their energy through absorsorbtion of radiation, some get their energy from collisions at the surface or from other air molecules.

Therfore, unless someone can point to measurements that show otherwise, Nitrogen, Oxygen, Clouds in the atmosphere must be assumed to emit basic Black Body Thermal radiation back to the surface and since they outnumber CO2, H2O by several orders of magnitude this may explain the ~90K difference between grey body temperature and observed temperature of the Earth.

Sorry, but all this talk about gravity inducing temperature change is a bit “way out” there. I’ll stick to what I know about pumping up my bike tyres.

73. tallbloke says:

Great stuff Ned,
I know it’s a long process from submission of paper to final publication, so we’ll have to be patient. Thanks for confirming that you’ll continue to collaborate with the Talkshop as your base in the blogosphere, we look forward to publishing your science as soon as you give us the go-ahead.
meantime, keep pitching in on threads here as and when you have spare moments.

Cheers – TB

74. Paulinuk: …”all this talk about gravity inducing temperature change is a bit “way out” there…”

Can I tempt you to study Graeff’s paper? – because it is not about talk, it is about experimental measurements and this experimental work backs up Nikolov in a way I regard as crucial.

75. steveta_uk says:

Paulinuk: when charged particles accelerate, they emit electromagnetic radiation. So each time a molecule changes direction, it emits radiation across the spectrum

O2 and N2 are not charged particles, they are charge-neutral molecules.

76. paulinuk says:

steveta_uk says:….

“O2 and N2 are not charged particles, they are charge-neutral molecules”

That confused me as well but I thought that If I get onto a bus and the bus accelerates then I accelerate too. So electrons and protons within a molecule must accelerate if the molecule accelerates.
Your refering to free-free emission which is free ions, electrons in motion. Basic thermal emission is any particle in motion.

77. tallbloke says:

Paul and Steve, I pursued this question of the capability of Nitrogen and Oxygen to absorb and emit radiation in the relevant wave bands with SWK (Might have been Leonard Weinstein – can’t remember). Compared to co2 the cabability is around 5 orders of magnitude less. On the other hand, there are 5 orders of magnitude more N2 and O2 molecules in the atmosphere than there are co2 molecules.

Both of the magnitudes of emission are small compared to what water vapour gets up to.

However, none of them are particularly good at transmitting energy back to Earth’s surface via long wave radiation, because that radiation is being re-absorbed very frequently as it fights buoyant convection all the way back to the surface. Even if it reaches the surface, it can’t penetrate 7/10ths of the surface (ocean) by much more than its own wavelength, where it will assist evaporation (cooling) more than heating. So Paul’s contention that “this may explain the ~90K difference between grey body temperature and observed temperature of the Earth. ” Isn’t going to fly, so far as I can tell.

78. paulinuk says:

TB “Paul and Steve, I pursued this question of the capability of Nitrogen and Oxygen to absorb and emit radiation in the relevant wave bands with SWK ”

As far as I can understand blackbody radiation has exactly the same emission spectrum curve shape no matter what the stuff is made of, but the curve is red shifted as the temperature of the object goes down.

Thanks for looking into it though.

79. tallbloke says:

Paul: “As far as I can understand blackbody radiation has exactly the same emission spectrum curve shape no matter what the stuff is made of,”

Yes, but it’s the size of the emission and absorption band ‘spikes’ along that curve which are at issue. (Along with how much they broaden at different temperatures and pressures).

80. paulinuk says:

…so if I were at the same temperature as the air or background I would be invisible to infrared detectors – that’s another way of saying the thermal emission spectrum is the same for all matter at the same temperature.

[Reply] No, because emissivities differ. Real materials are never perfect black-bodies.

81. paulinuk says:

Maybe I should have said pure Black Body radiation is the same for all matter. Note, the website I referenced states that Thermal radiation can be of 3 distinct types, Blackbody emission,Spectral Line Emission and Free-Free emission.

82. steveta_uk says, May 3, 2012 at 6:36 pm:

David Socrates, please answer this very simple question.

If the entire system consists of only two point samples, point A has a temperature of 100K and point B has a temperature of 500K, what is the physical meaning of your claim that “the average temperature is 300K”?

I know of no justication for the assumption that a simple arithmetic average of two temperatures has any physical meaning whatsoever.

If you can prove you are correct, then fair enough, but arrogant shite like “they are simply wrong because you have not performed the calculations correctly according to the rule I specified” doesn’t cut it with me.

1. On your first point, yes the arithmetic average temperature of only two temperature measurements is indeed the sum of those two temperatures divided by 2.

2. On your second point, whether using the arithmetic average has any physical meaning depends entirely on the circumstances. There is indeed a whole school of thought that the concept of an average temperature is a nonsense anyway, however you calculate it. I don’t happen to subscribe to that view because I live in the real world where we sometimes have to produce aggregate measures of collections of intensive variables in order to move forwards. For example you cannot go out and buy a FTSE share, but the FTSE stock exchange index is still widely useful. Similarly, you cannot go out and buy an ‘averagely priced house’ but knowing how that average house price figure has changed over the years may be useful.

3. On your third point (and setting aside the totally unwarranted rudeness in your comment), the “rule I specified” was not invented by me. And I don’t have to “prove I am correct” because I never actually claimed that there was a ‘correct’ method of measuring average temperature.

If you read more carefully my original reply to tchannon (and my earlier reply to you), you will see that all I was insisting is that the same averaging method must be used to determine the mean surface temperature of an airless, rocky Earth as the averaging method that has been used when measuring the mean surface temperature of the Earth as it is today – otherwise subtracting the two values most certainly gives you a meaningless result.

The Earth’s mean surface temperature of 288K has been determined empirically using the arithmetic average of Weather Station temperature readings from around the world. So it follows logically that one must use the arithmetic average when estimating the mean surface temperature of a rocky, airless Earth. And when you do this latter computation correctly using an iterative computational technique, taking care to calculate the arithmetic average, or indeed when you use N&Z’s (revised) formula which is of course also based on calculating the arithmetic average, the result comes out to around 197K. And this is exactly in line with the figure that the Diviner project has now found by performing, guess what, the arithmetical average on the real observational Moon data.

Do I now make myself clear?

83. tchannon says:

This thread has now departed from the original intent, particularly Ned in effect closing, therefore I will now say more.

SB acts on/as an interface between a real world mass and an unreal process. This is why thermal conduction and thermal mass are a critical part of the process.

If I raise the thermal conductivity the equatorial mean rises to around 289K but the computed global does not rise very much. All rather odd and more investigation might lead to insight.
Yes I did notice the 288, probably chance because there is so much else involved.

Trying to add other insight, the properties of a thermal flux is outside of normal human experience, is counter intuitive. There is an exact dual in a constant current. It is also energy in flight.

Given the above it has no power and there is infinite source impedance, no power loss there.

When a flux hits a thermal resistance it dissipates power, raising the entropy of the thermal mass, which is measured in units such as kelvin, is a temperature.
For a dual, thermal flux is electrical current, units the same size. Temperature is voltage, units same size. Thermal resistance is the reciprocal of thermal conductance, same for electrical except one is the reciprocal of the other, resistance one system is conductance the other system. All very simple.

One of the ways of looking at this problem which might shed some light is
What load gives maximum power loss in the load, which is maximum energy transfer?

For a system where humans are more familiar, say conduction from a source and equivalently a normal voltage source the standard answer is when the loss in the source and the load are equal, a point being that real sources of this kind have internal losses. Also key is that the load affects the sources, it does matter to the source what the load does.

In the case of thermal flux/constant current things are wildly different. There is a complete disconnect, the sun doesn’t notice whatever we do, partly caused by the speed of light, no time travel on offer.

In this case the maximum power in the load occurs with the maximum (as in short circuit) load possible. In the case of deep space it swallows the lot, which is consistent with all the flux being consumed, complete absorption of all the power.

Something like the moon or earth can’t go as far, neither is it at 0K.

So, what is the effect of the most complete absorption of the flux turning it into power?
This pumps up dayside as far as possible and since these are spinning bodies this gives the maximum power holdup during the night.

This gives highest SB night temperature and minimum SB day temperature (because of the conduction into the surface)

I’ll stop there. Might bring in the other factors to do with an atmosphere but I have no complete explanation there, is far too complex, far too many unknowns.
Point out any mistakes please. Garbled, probably.

84. paulinuk says:

So there are 3 types of thermal radiation:
TypeI: Black body emission;
TypeII: Spectral Line emission;
TypeIII: Free-Free emission;

and all matter above absolute zero emits TypeI radiation.

According to Wiens’ Displacement Law ( λmax=b/T where T is in degrees Kelvin and b=0.0028977685), the peak wavelength of TypeI black body emission for any object at 15C is 10.1 μm which places it firmly in the infra red band of the electromagnetic spectrum. So all air in the atmosphere emits basic TypeI blackbody radiation in the infra red band peaking at ~10μm.

Superimposed upon this radiation is TypeII spectral line emission for each individual molecule and a maybe small amount of TypeIII free-free emission from the ionosphere? So pointing a spectrometer up at the sky you’d be able to separate the TypeII emission lines for each element but unable to tell the difference between elements for TypeI emission .

85. tchannon,

Very interesting observations. As an electrical engineer I warm to your analogies. Keep them going.

The Sun’s radiation does appear exactly analogous to a constant current in an electrical circuit delivered from a zero impedance source. That is, its radiation flow rate does not change (even in principle, one would assume) when we intercept it with a body (such as the Earth) that absorbs part of the energy by thermal conduction into its mass.

As you say, the Earth is not an infinite impedance sink because it has non-zero thermal resistance, so its temperature rises, analogous to the voltage rise in an electrical circuit when current flows through it. In the latter case, if the source is a constant current, its voltage will rise exactly in proportion to the sink’s resistance.

My first thought was that when the Earth’s surface temperature rises, the Sun’s temperature (presumably!) does not rise. Whereas in the case of your constant current electrical analogy, the voltage of the current source must rise exactly in line with the additional ‘back voltage’ generated by the sink due to the current flowing through it. In other words there is a feedback from sink to source that gives the source the necessary information that it needs to raise its output voltage just sufficiently to maintain its constant current.

Then my second thought was that a small proportion of the energy absorbed by the Earth is indeed re-radiated back to the Sun, so perhaps (in principle although obviously not in practice) there is a feedback loop after all.

And then I realised we were into ‘back radiation’ territory, so I decided to stop!

86. tallbloke says:

Tim: conduction is one aspect of the heat spreading, but I suspect advection accounts for a lot more. So that might explain why the equatorial temp goes up to 289K but the global doesn’t increase much in your model. You’d need to start progamming in electrical analogies to the complexities introduced by rotation, vortex and eddy diffusion, centripetal force etc. Big job. For now, and in the spirit of keeping it simple, we just need a term we can assign to that complexity which produces the right amount of spreading. Surface energy? That has a known electrical effect on conductors. http://en.wikipedia.org/wiki/Surface_energy

David: On the contrary, keep going! Feedback from the planets to the Sun is a major theme on this blog. Energetic flow back down the ‘flux tubes’ identified by NASA that connect the magnetospheres of the planets to the Sun is important, perhaps more as a control loop than in terms of raw magnitude of energy flow.

However, in the context of improving Tim’s model, we should probably disregard it and treat the Sun’s output as being constant for now. We don’t want to make the job more difficult than it is already.

87. TB: Yes, I didn’t want to divert Tim’s important line of thought. Maybe it’s time for another (separate) slog at back radiation (deep breath…!)

88. tallbloke says:

David: Perhaps so. Maybe we could deal with the issue of ‘coupled radiative-convective’ modeling and the fact that it makes no sense to treat the the atmosphere and surface as oppositely facing radiative surfaces with a vacuum between in order to misapply the standard S-B law…

89. ozzieostrich says:

All bodies above 0K emit EMR. It doesn’t matter to the Sun what happens to the EMR it emits – unless of course it is intercepted by a body at a lower temperature. This body’s temperature is then raised, which increase the amount of EMR that it radiates.

Obviously, (or maybe not so obviously), the Sun is receiving EMR that was not present before. Its temperature will rise, assuming the energy output internally generated does not change.

This is probably what,the lunatic fringe refer to as “back radiation”. There is no difference between “front radiation”, “back radiation”, or “twirly whirly side radiation”. There is radiation, there is lack of radiation – 0K. Ice at 270K emits EMR, at wavelengths in accordance with its temperature.

My cup of tea, at about 353K, emits EMR at its appropriate wavelengths. However, all the heat energy in 1,000,000 kgs of ice at 270K will not warm my tea one jot, even though it contains far more heat energy.

I notice that someone has made an analogy somehere along the way with “back EMF” or something similar. Although I normally hate analogies, (based on my assertion that if you have to use analogies, your explanation is not clear enough), the back EMF is valid, as it only arises as a consequence of EMF generated by the conversion of energy (usually of another form) moving a conductor across the flux lines of a magnetic field. Hence EMF – electromotive force.

Anyway, a motor generates a back EMF very close to the applied EMF under conditions of no load. Thus, the motor consumes little power. As the load increases, the back EMF reduces, the power consumed increases, until imposed EMF less back EMF times current (assuming DC – AC gets a little more complex) equals the power dissipated in the load.

This is why your body needs to generate less heat to maintain its 37C or so, at say, 30C, than 15C. Insulators work by slowing down heat loss. If the environment surrounding your body is radiating less heat than your body needs to maintain core temperature, even while oxidising fuel at maximum, you will die.

Conversely, if your environment is is sufficiently radiative (and I know someone will attempt to confuse the issue by mentioning convection, evaporative cooling etc, but bear in mind I stated “sufficiently”,), then you will also die. In his case, your body cannot radiate heat away fast enough to keep internal cell temperatures low enough to sustain continued life. You will die of “heat stroke”.

Calling radiation from a cooler body which has been warmed by a hotter body “back radiation” merely obscures the fact that all bodies radiate EMR. Yes, I said all, in a practical sense. I am told the universe background temperature is around 2.7K, therefore 0K is unsustainable. An object at that temperature cannot help but be warmed by its surroundings.

I agree with Tallbloke that experimental verification of a supposed observation trumps all sorts of “thought experiments”. I think that Einstein said something to the effect that only one person was necessary to prove that one of his ideas was wrong. He also said something to the effect that an explanation needed to be no more complex than necessary.

If there is a difference between a theoretical calculation of the Earth’s temperature, and the observed temperature, which would you believe? Why create a slew of increasingly complicated explanations for something which probably doesn’t exist in the first place? At the risk of becoming a serial poster, consider the following proposal -

The Earth originally had a surface temperature of some thousands K.

It cooled.

The surface temperature reduced to 500K.

It cooled. And so on.

It is now whatever it is – 288K if you like.

It is still cooling – fairly obvious. It’s a molten blob sitting in space, with the nearest external heat source a few hundred million kilometres away. The Earth is still radiating energy to space, and getting cooler, just as inevitably as my cup of tea. It will finally reach an equilibrium temperature when the core has cooled to thermal equilibrium with its environment in accordance with radiative physics as,we understand it.

Global warming? Really? Why not accept reality? Why waste a good worry?

Cheers.

Live well and prosper,

Mike Flynn.

90. Mike Flynn, I agree. You have said pretty much all that needs to be said about so-called ‘back radiation’. And your electrical analogies are excellent.

All we have to contend with now are the ‘slayer’ crowd who feel they need to impose on us an elaborate new theory in which photons originating from a cooler body simply do not get absorbed by a warmer body, so that the simple subtraction to obtain the net balance of radiation isn’t necessary.

91. tchannon says:

DS,
The analogy is an exact dual, no approximation.
Not very much is available about this but a list of duals is figure 1 here
http://www.stanford.edu/class/ee311/NOTES/InterconnectThermalModeling.pdf

92. ozzieostrich says:

David Socrates,

Thank you for your comments. It’s nice to have someone who understands what I am trying to say.
Unfortunately, I didn’t finish high school, so I am never sure if my thoughts are communicated clearly. You have made my day. I’ll shut up now.

tchannon,

Thank you also. I didn’t realise that the analogy was as good as it seems. Its always seemed logical to me. I downloaded the paper you linked to. It might prove useful the next time somebody implies that I’m a complete idiot, and don’t understand anything!

And a very good day to you both!

Live well and prosper.

Mike Flynn.

93. tchannon says:

I hope I don’t get too much flak for this conceptual drawing, which is incorrect in some respects. The idea I am trying to show is why the normal radiation ideas about interchange of energy are no problem, how it is.

94. tchannon says:

I came across a comment in a 2003 thread elsewhere which fits with a lot of my thinking.

“The author has an interesting thesis. I have written in the past a similar argument with respect to CO2 having an absorbing function for incoming radiation (sun light) but not a radiation blocking function at night because it absorbs and emits preferentially at 4.26um where the earths radiates back to space (cools)preferentially in the range of 8.0 – 12.0um due to the Black Body curve and laws according to Planck. Therefore, CO2 in this simple example would not have a definitive effect. Think of this, because there is strong 4.26um radiation from the sun, CO2 gets hot and the heat is radiated and conducted to all gasses and to the surface. At night the radiating temperature is now 300K degrees and the peak Black Body emission wavelength for that temperature is right at 10.0um. 10.0um light (radiation) goes right through a 4.26um absorber (CO2) but because water has such a broad absorption-emission band then water interferes with the radiation back to space (like on a cloudy night). Other strong green house gasses include methane (natural gas) and would behave similarly. It can’t be this simple.”
– Final Authority
http://www.freerepublic.com/focus/f-news/900129/posts

Particularly the top radiation on GHG is something where I always get a runaround, silence. This has critical effects because it blocks the incoming radiation channel so that solar input to the surface in that channel is muted, with the corrolory this ought to be subtracted from the usual sums but isn’t. There again nightside is whole different and ignored world.

The evidence of clear nightside sky having a gnat’s fart of incoming is very clear. Cloud or heavy humidity is a different matter.

Sigh, and I usually keep out of radiative stuff, say nothing.

95. RKS says:

I posted this in error on the N&Z part 1 thread so I’ll try again here.
[will remove that, not that it matters terribly, this is a currently active thread --mod]

I hope this is sufficiently on thread to deserve posting:-

We are told that surface radiation at the wavelengths absorbed [and re-emitted] by CO2 is absent when measured from space. The inference being that this energy is somehow ALL radiated back down to the surface as if from a mirror [IPCC claiming 40% back radiation], although nobody explains what happens to this radiated energy, at this wavelength, after that.

Are there any empirical data showing the percentage of surface radiation falling within this wavelength, how it was measured, where on the planet it was measured and at how many locations and is it uniform across the whole of the Earth’s surface.

And are they attempting to measure something which is simply not present at the surface at the levels they expected.

I ask this because there are a lot of contributors to Tallbloke who seem to have a firm grasp of the statistics.

96. paulinuk says:

The real greenhouse effect ?

1: H2O (1%), CO2 (.04%) get energy from:

a intermolecular collisions with the surface and other air molecules.
b: absorbtion of radiation from the ground in a FEW discrete bands of the infrared.
c; absorbtion of infrared from the sun, partially blocking the path to the surface.

They re-radiate ~50% of the energy back to Earth at the same FEW discrete bands in the infrared.

S-B radiation law means the Earth heats up a bit more by Delta X (in the past assumed to be 33C) to reach an equilibrium temperature.

2: Nitrogen (78%) and Oxygen(21%) get energy from

a: intermolecular collisions with the surface and other air molecules.

They radiate ~50% of the energy back to Earth across a BROAD spectrum in the infra red, peaking at ~10μm.

Planck’s law means the Earth heats up more by Delta Y to reach an equilibrium temperature.

The actual temperature of the Earth, T= Greybody Temperture + DeltaX + DeltaY

I’d need to ACTUALLY SEE measurements and calculations to know what DeltaY is, as it seems to have been conveniently ignored.

97. ozzieostrich says:

Paulinuk,

Please correct me if I’m wrong, but the Earth cools a bit (Delta something or other) as it radiates the energy which will be subsequently reradiated to raise the temperature to less than it was when it was reduced by Delta something or other.

The actual temperature of the Earth = whatever it has cooled to since its creation.

It hasn’t finished cooling yet. Obvious, or what?

Any and all corrections welcomed. I never stop learning, as pointless as it is!

Live well and prosper,

Mike Flynn.

98. tallbloke says:

Hi Paul: I think you’ve missed out quite a few things. I’ll add a few to your equation:

The actual temperature of the Earth, T= Greybody Temperture + DeltaX + DeltaY – energy of evaporation caused by back radiation – reduction of back radiation reaching surface due to convection + enhancement of near surface temperature by the effect of sunlight passing through denser near surface air + ‘greenhouse fluid’ effect of oceans which have to rise in temperature to lose energy as fast as they gain it.

99. RKS says:

paulinuk says:
May 7, 2012 at 12:56 pm
The real greenhouse effect ?

1: H2O (1%), CO2 (.04%) get energy from:

a intermolecular collisions with the surface and other air molecules.
b: absorbtion of radiation from the ground in a FEW discrete bands of the infrared.
c; absorbtion of infrared from the sun, partially blocking the path to the surface.

They re-radiate ~50% of the energy back to Earth at the same FEW discrete bands in the infrared.>>>>

How can 0.04% 0f the atmosphere radiate ~50% of the Earth’s radiated energy in those discrete bands? Or are we saying that CO2 can, at most, ‘back radiate’ ~50% of .04% of surface radiation within certain bandwidths – in other word not much more than background noise.

100. tallbloke says:

Mike, although the Earth might be cooling, it only cools through the crust at the rate of 0.1W/m^2. We can pretty much ignore this in climate energy balance calculations, because compared to the 1360W/m^2 arriving from the Sun at the top of the atmosphere, it is negligible.

101. ozzieostrich says:

Hi Tallbloke,

Your blog, so I understand if you pull my comments.

Actual temperature of Earth = impossible to calculate.

Or should I say, without a useful definition of the term “temperature of the Earth’, the term is by definition undefined, I would think.

If anybody could define what they mean by the “temperature of the Earth”, we could have a sensible and possibly useful discussion. When I asked someone on another blog to define the “Earth’s surface”, my interlocutor informed me that I was behind the times, as the new paradigm referred not to the surface, but rather to the heat content of the oceans.

I give up. The goalposts keep shifting. I suspect that the whole GHG/global warming thing is balderdash, but I can never pin the buggers down to anything concrete.

Live well and prosper,

Mike Flynn.

102. RKS says:

An addendum to my previous post:-

When we consider that for system Earth Energy flux IN = Energy flux OUT, what happens to the back radiated energy once it reaches the surface. It can’t just hang about there taking a rest.

i think by trying to be too creative with the accounting we’re in danger of losing sight of the big picture.

103. ozzieostrich says:

Hi Tallbloke,

Our comments have crossed. I saw your 1:37 pm after I posted my 1:46 pm.

My “problem”, if you wish, is that if, as you say, ” . . the Earth might be cooling . . .”, I need a little more certainty. If it’s a “maybe”, “possibly”, or “might be”, I have difficulty framing a point of view.

Now I don’t think I am asking too much. When I board an aircraft in Australia, I accept that the probability of arriving at say Heathrow, is not 1. However, I literally bet my life that the probability of a safe arrival on the ground is close enough to 1 as to be not worth wasting a good worry.

With climate probabilities, should I accept a lesser probability before all and sundry demand that I pay, (in one way or another – I live in Australia,) for their possibly bizarre theories.

As you say, the conventional wisdom is that the Earth is cooling, albeit by a small amount numerically. It doesn’t matter, surely. Even if it is cooling by an arbitrarily small amount, it sure as Hell ain’t getting warmer.

So there you go. No GHG warming. No gravitational warming. A certain amount of anthropogenic warming due to Man – living, working, oxidising this and that.

Eventually, isothermal universe – entropy wins. What a pisser, but I’ll be long gone.

Thank you so much for allowing my posts. I’ve been kicked off more blogs than you might think, for pointing out what I think is blindingly obvious. Freedom of speech? Yeah, right!

Live well and prosper

104. paulinuk says:

Ozzie…

For a body to be in Thermal Equilibrium the energy going in Ein has to equal the energy going out Eout (otherwise it’s not in equilibrium) . You can then use the Stephan- Boltzman equation to work out the temp using Eout=SBconstant*Surface Area*T^4 .
In the daytime there’s something going from the sun plus an additional bit from atmospheric radiation so the temp goes up by the fourth root of the energy in etc . Don’t think the internal temperature of the Earth affects things much.

105. ozzieostrich says:

RKS,

Crossed posts again. The warmists claim Energy flux IN x 1.2 (or thereabouts) = Energy flux IN.

I dont know whether it’s been “vanished”, but Chris Colose, (warmist acolyte in training,) claimed on his “educational” blog (for the mentally challenged, I would think,) that 10 in to the atmosphere = 12 at the surface. If you believe that, I have a nice tower, built by Mr Eiffel, that you might be interested in buying.

I agree with you. You can’t have more EMR at the surface than was intercepted from the Sun.

No more to be said, really. I think I’m on your side.

Live well and prosper,

Mike Flynn.

106. ozzieostrich says:

Hi paulinuk,

If the Earth is warming or cooling, it is obviously not in thermal equilibrium. I can’t think of any reason it should be so, anyway.

The “thermal radiation” to which you refer comes from the elevated temperature of the atmosphere resulting from radiation absorbed from the Earth’s surface, causing a concomitant and inevitable reduction in the temperature of the radiating body (the Earth).

Now, if you believe that an earth sized lump of molten rock, suspended in outer space in a vacuum, some hundreds of millions kilometers from a Sun sized source of radiation, which has not heretofore exhibited the ability to stop the aforementioned blob of molten rock from cooling, can be warmed by surrounding it with gas of any sort, then it must be so.

I defer to your superior knowledge.

Live well and prosper.

Mike Flynn.

107. tallbloke says:

Mike: “As you say, the conventional wisdom is that the Earth is cooling, albeit by a small amount numerically. It doesn’t matter, surely. Even if it is cooling by an arbitrarily small amount, it sure as Hell ain’t getting warmer”

The bulk temperature of the planet is irrelevant to the issue of what causes variation of the temperature where we live. Local and regional variation is very noisy data and unpredictable. So the climate debate tends to revolve around the causes of variation in globally averaged metrics.

Ocean heat content would be the best metric for working out what is going on, if the data custodians weren’t tweaking the data to fit their theory. This is because the ocean is the biggest heat capacity part of the system this side of the crust. Nonetheless, atmospheric variations are also of interest, because those have the most immediate effect on our crops, holidays, and fuel bills.

108. ozzieostrich says:

Hi Tallbloke,

If the bulk temperature (I assume you mean the average temperature so beloved of the warmists) is irrelevant, why all the fuss about it? Why all the money spent on research?

The local temperature varies for reasons which are not clearly understood – obviously.

What’s going on is simple. Physics in action. The warmists have given up on the lithosphere, and the atmosphere, and have adopted the “it’s all down to the aquasphere” paradigm, if you are correct.

Atmospheric (weather) conditions cannot be predicted by anybody better than you or I can do with a half hour training, historical records and a straight edge. The absurd notion that the future can be predicted from the past is just that, an absurd notion. We have assumptions, no more, no less.

Explanations of what happened in the past allow us to base our assumptions about the future, taking the past into account. As you have probably observed, basic physics combined with a small knowledge of chaos theory, allow me to explain everything, but, alas, to “predict” nothing.

Nature is paradoxical – the complex sometimes has a simple explanation, while the apparently simple proves intractable. I am lucky to have no schooling, as I have no preconceptions. I would surmise that in your field (whatever it is) you would have found examples of what I assume.

I will try to refrain from further annoying comment, but I seem to have inadvertently activated email advices when posts are made. You can inactivate me if you wish.

Live well and prosper,

Mike Flynn.

109. tallbloke says:

Mike: The email thing is a new wordpress ‘feature’ You need to remember to untick the ‘email notification of follow up comments’ tickbox when you comment on a new thread, damned annoying. Try right clicking on the email and see if your email client allows you to set a rule to send them to a spam folder.

By ‘bulk temperature of the planet’ I meant the whole planet, including the lithosphere. It is not relevant, for the reason I already outlined.

” basic physics combined with a small knowledge of chaos theory, allow me to explain everything, but, alas, to “predict” nothing.”

Well, maybe we can predict solar activity reasonably well. If we can better quantify the Sun-Earth climate link, we can then make more useful predictions concerning climate. On this blog, we have tried to develop methods to do those things. Now we wait and see if our predictions are matched by observations.

110. RKS says:

Just come over here for a bit of fresh air.

Some character at Bishop Hill is arguing that CO2 becomes opaque to radiation at certain wavelengths without anyone challenging him.

What is the wavelength for maximum absorption by CO2 and what is the emissivity at that wavelength please? I’ve tried looking it up on Wiki with no luck.

111. Larry Ledwick (hotrod ) says:

@RKS says:
May 7, 2012 at 7:03 pm

I think you should probably start by looking up optical thickness and Beer’s law.

What you are really dealing with is a process of extinction as the energy is absorbed along the path it travels through the atmosphere. With enough thickness even apparently transparent materials become opaque to radiation.

For example the tinted windows on a car only absorb perhaps 18% of the light that passes through them but even at that low absorption two passes through the window drastically reduces the light transmitted out to a viewer to levels low enough that they window is practically black in most light conditions. The outside light is reduced by 18% when it enters the car window, then a small percentage of that light is reflected off the objects inside the car and 18% of that reflected fraction is absorbed again as it passes out through the window to an outside viewer.

The same thing happens with common window glass. Typical window glass is slightly green in color but this is not noticed until you look at a thick layer of glass. At typical thicknesses this slight color cast and absorption is not apparent.

In the same way a gas like CO2 has an absorption spectrum and if the optical path through the atmosphere is long enough, the accumulated absorption of those frequencies over the length of the optical path “approaches” total absorption. Re-radiation, pressure broadening and all those other issues complicate it even further and the fact that much of the CO2 absorption spectrum overlaps the absorption spectrum of water vapor makes it something of a moot point.

I found this paper with a quick google search (have not read it in detail) but it discusses the effective optical thickness of CO2 in the atmosphere for total absorption in its frequency bands as about 10 meters at current concentrations.

Larry

112. RKS says:

Larry Ledwick (hotrod ) says:
May 9, 2012 at 7:59 pm
@RKS says:
May 7, 2012 at 7:03 pm

I think you should probably start by looking up optical thickness and Beer’s law.

Thanks for the links.

I’ll follow them through in a couple of days when I’m a bit more free to do so.

113. wayne says:

RKS: “In the same way a gas like CO2 has an absorption spectrum and if the optical path through the atmosphere is long enough, the accumulated absorption of those frequencies over the length of the optical path “approaches” total absorption.”

RKS, actually all of the frequency lines of CO2 are absorbed within about 10 meters of the surface and that energy just becomes part of the local Prevost photon gas. Look it up and always make it multiple sites. CO2’s influence, if any at all, still being discussed, occurs above the ½ mass (500 mb) level and into the stratosphere to aid in the shedding of Earth’s heat absorbed each day. This same total absorption close to the surface occurs for water vapor also to an even greater degree but water vapor influence rapidly decreases above cloud bases. The atmosphere is totally opaque to CO2 and H2O frequencies many times over. As Larry said, see optical thickness or depth, Beer law, Lambert law, and Prevost.

Now if you are speaking of the optical path over the entire atmosphere column upwards in the IR frequencies the transmittance has been measured via radiosonde data, hundreds of thousands of them, to be right at 0.1541 which translates to a tau (optical depth) of ~1.87 from -ln(0.1541). That figure has never changed in the last sixty years at all and that is why CO2’s influence may actually be zero or immeasurably close.

114. Ulric Lyons says:

“Therefore, the size of Earth’s atmospheric greenhouse effect ought to be evaluated with respect to Moon’s actual mean surface..”

How about comparing the absolute equatorial temperature at noon ?