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## The Mystery of Tau – Miskolczi – Part Four – Emissivity

In Part Two we looked at the claimed relationship ED=AA in Miskolczi’s 2007 paper.

• Ed = downward atmospheric radiation absorbed by the surface
• Aa = surface radiation absorbed in the atmosphere

I showed that they could not be exactly equal. Ferenc Miskolczi himself has just joined the discussion and confirmed:

I think I was the first who showed the AA≈ED relationship with reasonable quantitative accuracy.

That is, there is not a theoretical basis for equating AA=ED as an identity.

There is a world of difference between demonstrating a thermodynamic identity and an approximate experimental relationship. In the latter situation, it is customary to make some assessment of how close the values are and the determining factors in the relationship.

But in reviewing the 2007 paper again I noticed something very interesting:

From Miskolczi (2007)

Figure 1

Now the point I made in Part Two was that AA ≠ ED because the atmosphere is a little bit cooler than the surface – at the average height of emission of the atmosphere. So we would expect ED to be a a little less than AA.

Please review the full explanation in Part Two to understand this point.

Now take a look at the graph above. The straight line drawn on is the relationship ED=AA.

The black circles are for an assumption that the surface emissivity, εG = 1. (This is reasonably close to the actual emissivity of the surface, which varies with surface type. The oceans, for example, have an emissivity around 0.96).

In these calculated results you can see that Downwards Emittance, ED is a little less than AA. In fact, it looks to be about 5% less on average. (And note that is  ED = Absorbed Downwards Emittance)

Of course in practice, εG < 1. What happens then?

Well, in the graph above, with εG = 0.96 the points appear to lie very close to the line of ED=AA.

I think there is a calculation error in Miskolczi’s paper – and if this is true it is quite fundamental. Let me explain..

Here is the graphic for explaining Miskolczi’s terms:

From Miskolczi (2007)

Figure 2

When the surface is a blackbody (εG =1), SU = SG  – that is, the upwards radiation from the surface = the emitted radiation from the ground.

The terms and equations in his 2007 are derived with reference to the surface emitting as a blackbody.

When εG < 1, some care is needed in rewriting the equations. It looks like this care has not been taken and the open circles in his Fig 2 (my figure 1) closely matching the ED=AA line are an artifact of incorrectly rewriting the equations when ε< 1.

That’s how it looks anyway.

Here is my graphic for the terms needed for this problem:

Figure 3

As much as possible I have reused Miskolczi’s terms. Because the surface is not a blackbody, the downward radiation emitted by the atmosphere is not completely absorbed. So I created the term EDA for the emission of radiation by the atmosphere. Then some of this, Er, is reflected and added to SG to create the total upward surface radiation, SU.

Note as well that the relationship emissivity = absorptivity is only true for the same wavelengths. See note 4 in Part Two.

### Some Maths

Now for some necessary maths – it is very simple. All we are doing is balancing energy to calculate the two terms we need. (Updated note – some of the equations are approximations – the real equation for emission of radiation is a complex term needing all of the data, code and a powerful computer – but the approximate result should indicate that there is an issue in the paper that needs addressing – see comment).

And the objective is to get a formula for the ratio ED/AA – if ED=AA, this ratio = 1. And remember that in Figure 1, the relationship ED/AA=1 is shown as the straight line.

First, instead of having the term for atmospheric temperature, let’s replace it with:

TA = TS – ΔT      [1]

where ΔT represents the idea of a small change in temperature.

Second, the emitted atmospheric downward radiation comes from the Stefan-Boltzmann law:

EDA = εAσ(TS – ΔT)4      [2]

Third, downward atmospheric radiation absorbed by the surface:

ED = εGEDA      [3]

Fourth, the upward surface radiation is the emitted radiation plus the reflected atmospheric radiation. Emitted radiation is from the Stefan-Boltzmann law:

SU = εGσTS4 + (1-εG) EDA    [4]

Fifth, the absorbed surface radiation is the upward surface radiation multiplied by the absorptivity of the atmosphere (= emissivity at similar temperatures):

AA = εASU      [5]

So if we put [2] -> [3], we get:

ED = εGεAσ(TS – ΔT)4    [6]

And if we put [4] -> [5], we get:

AA = εGεAσTS4 + EDεA(1-εG)/εG   [7]

We are almost there. Remember that we wanted to find the ratio ED/AA. Unfortunately, the AA term includes ED and we can’t eliminate it (unless I missed something).

So let’s create the ratio and see what happens. This is equation 6 divided by equation 7 and we can eliminate εA that appears in each term:

ED/AA = [ εGσ(TS – ΔT)4 ] / [ εGσTS4 + ED(1-εG)/εG ]    [8]

And just to make it possibly a little clearer, we will divide top and bottom by εG and color code each part:

ED/AA = [ σ(TS – ΔT)4 ] / [ σTS4 + ED(1-εG)/εG2 ]      [8a]

And so the ratio = blackbody radiation at the atmospheric temperature divided by

( blackbody surface radiation plus a factor of downward atmospheric radiation that increases as εreduces )

We didn’t make a blackbody assumption, it is just that most of the emissivity terms canceled out.

### What Does the Maths Mean?

Take a look at the green term – if εG = 1 this term is zero (1-1=0) and the equation simplifies down to:

ED/AA = (TS – ΔT)4  /  TS4

Which is very simple. If ΔT = 0 then ED/AA = 1.

Let’s plot ED vs AA for a few different values of ΔT and for TS = 288K:

Figure 4

Compare this with figure 1 (Miskolczi’s fig 2).

Note: I could have just cited the ratios of ED/AA, which – in this graph – are constant for each value of ΔT.

And we can easily see that as ΔT →0, ED/AA →1. This is “obvious” from the maths for people more comfortable with equations.

That’s the simplest stuff out of the way. Now we want to see what happens when εG < 1. This is the interesting part, and when you see the graph, please note that the axes are not the same as figure 4. In figure 4, the graph is of ED vs AA, but now we will plot the ratio of ED/AA as other factors change.

Take a look back at equation 8a. To calculate the ratio we need a value of Ed, which we don’t have. So I use some typical values from Miskolczi – and it’s clear that the value of Ed chosen doesn’t affect the conclusion.

Figure 5

You can see that when  εG = 1 the ratio is almost at 0.99. This is the slope of the top line (ΔT=1) in figure 4.

But as surface emissivity reduces, ED/AA reduces

This is clear from equation 8a – as εG  reduces below 1, the second term in the denominator of equation 8a increases from zero. As this increases, the ratio must reduce.

In Miskolczi’s graph, as εG changed from 1.0 → 0.96 the calculated ratio increased. I believe this is impossible.

Here is another version with a different value of ΔT:

Figure 6

### Conclusion

Perhaps I made a mistake in the maths. It’s pretty simple – and out there in the open, so surely someone can quickly spot the mistake.

Of course I wouldn’t have published the article if I thought it had a mistake..

On conceptual grounds we can see that as the emissivity of the surface reduces, it absorbs less energy from the atmosphere and reflects more radiation back to the atmosphere.

This must reduce the value of ED and increase the value of AA. This reduces the ratio ED/AA.

In Miskolczi’s 2007 paper he shows that as emissivity is reduced from a blackbody to a more realistic value for the surface, the ratio goes in the other direction.

If my equations are correct then the equations of energy balance (for his paper) cannot have been correctly written for the case εG <1.

This one should be simple to clear up.

Update May 31st – Ken Gregory, a Miskolczi supporter appears to agree – and calculates ED/AA=0.94 for a real world surface emissivity.

Other articles in the series

The Mystery of Tau – Miskolczi – introduction to some of the issues around the calculation of optical thickness of the atmosphere, by Miskolczi, from his 2010 paper in E&E

Part Two – Kirchhoff – why Kirchhoff’s law is wrongly invoked, as the author himself later acknowledged, from his 2007 paper

Part Three – Kinetic Energy – why kinetic energy cannot be equated with flux (radiation in W/m²), and how equation 7 is invented out of thin air (with interesting author comment)

Part Five – Equation Soufflé – explaining why the “theory” in the 2007 paper is a complete dog’s breakfast

Part Six – Minor GHG’s – a less important aspect, but demonstrating the change in optical thickness due to the neglected gases N2O, CH4, CFC11 and CFC12.

### 293 Responses

1. As usual, very interesting; before looking at the calculations your assumption:

“the atmosphere is a little bit cooler than the surface – at the average height of emission of the atmosphere.”

Is this correct over a day or year on an averaged basis? Isn’t an LTE condition at the immediate surface/atmosphere boundary necessary for convection because if the immediate air above the surface was cooler the air would not rise?

• Cooler than what? Cooler than the surface below it would not matter; the surface does not convect. Cooler than the air above it would matter, but SoD says cooler than the surface.

2. cohenite:

Have a read of Part Two for more background.

..Is this correct over a day or year on an averaged basis?

This is correct on an averaged basis.

The atmosphere doesn’t radiate from the surface. Well, it does, but it is quite transparent in the first few meters – and the first few tens of meters. So the effective height of radiation is well above the surface and, therefore, from a colder temperature.

This is why the εG = 1 calculation in the paper shows a 5% in ED compared with AA. Because the effective temperature of atmospheric emission is colder than the surface.

3. Shouldn’t the reflected radiation be absorbed entirely by the atmosphere (you can’t multply that component with Atmosphere emittance)?

So the A(a) should be
A(a) =ε(A)*ε(G)σTS4 + [1-ε(G)]*E(DA)

not

A(a)=S(U)*ε(A)

4. Mait:

Shouldn’t the reflected radiation be absorbed entirely by the atmosphere (you can’t multply that component with Atmosphere emittance)?

Based on what principle?

In fact, absorptivity of the atmosphere is a wavelength dependent parameter.

The radiation emitted from the surface is centered on almost exactly the same wavelength as the reflected downwards atmospheric radiation.

Which means the absorptivity of the atmosphere for surface emitted radiation is almost exactly the same as for reflected atmospheric radiation.

And there is absolutely no thermodynamic reason why reflected atmospheric radiation should be entirely absorbed by the atmosphere.

If you have a reason, go ahead and explain it.

5. You are right that was a bit weird assumption.
It was based on the assumption that the reflected radiation is at the same wavelengths it is emitted and I assumed an entirely opaque atmosphere for some weird reason (that is for any wavelength the atmosphere is either opaque or entirely transparent). Which is ofcourse not a correct assumption to make.

I’d still imagine the ε(A) to be different for reflected and emitted radiation though (it should be higher for reflected in theory).

6. cohenite:

Have you read Part Two yet?

AA = the upwards radiation absorbed by the atmosphere.
ED = the emitted atmospheric radiation absorbed by the surface.

So AA > ED just by virtue of the fact that the atmospheric temperature is less than the surface temperature.

So even if reflected radiation, ER = 0, AA > ED.

7. Mait:

Maybe you make a valid point. I will give it some thought.

8. Note that Miskolczi includes data from Mars (simulated) in your Fig 1 – but the temperature profile on Mars is very different from Earth, in fact much closer to isothermal which makes his Ed Aa relationship actually valid.

9. on May 1, 2011 at 5:15 pm | Reply DeWitt Payne

Your Eqn. (2) is only correct if the atmosphere is gray, i.e. emissivity constant at all wavelengths, or the sky is covered with clouds. For clear sky, the actual temperature dependence will be less than T^4. More properly, the exponent will be less than 4. And I agree with Mait that the absorptivity for reflected radiation will be greater than for surface emitted radiation.

Still, the idea that Ed ≅ Aa logically follows from the fact that ~90% of DLR is emitted from within the first 500m of the atmosphere where the difference between the surface temperature and the emission temperature is small. Ed can even be greater than Aa if there’s a big temperature inversion like over the polar ice caps in winter. The MODTRAN temperature profile for Sub-Arctic Winter has the temperature at 1 km 1.9 C higher than the surface. Ed is higher than Aa for that case.

10. DeWitt Payne:

It’s an approximation and I should have been clearer (will add a note).

An approximation because I don’t have the code and data to do the full calculation.

But how is it possible for the ratio to move closer to 1 as emissivity is assumed reduced from 1 to 0.96 ?

And more on the approximation – of course, we can’t even say that the atmosphere emits from one temperature.

The total received surface radiation, EDA, is an integral through the atmosphere, eq 16 in Part Six – The Equations (without the surface term):

Iλ(0) = ∫ Bλ(T)e dτ    [16]

And then EDA = ∫Iλ(0)dλ

11. This discussion went on three or so years ago. Unfortunately. one at the Climate Audit Bulletin Board went down the memory hole. Much of the others at Landshape can be found (mostly at the bottom, be patient, this went on for a long time) at Landshape in this list. Some parts from CA and elsewhere are preserved at Rabett Run.

Much is deja vu all over again, but Eli’s favorite point is what is the physical mechanism that decreases water vapor when CO2 increases.

• on May 2, 2011 at 12:02 am | Reply DeWitt Payne

I think there’s an excuse for trying to summarize again. As you pointed out, one discussion got lost. The other has a fairly low signal to noise ratio as a lot of stuff has to be dug out of the comment thread.

I completely agree about the lack of mechanism. I believe Pielke, Jr. refers to this as magical thinking. Of course Wegener didn’t have a mechanism for continental drift either, or at least the one he had was fairly easy to dismiss. But I don’t think this is in that class as it isn’t something that was blindingly obvious to anyone with a globe of the Earth.

• Perhaps time to demand stridently of Steve McIntyre that he stop hiding the scintillating comments at the CA Bulletin Board?:)

12. So inspired by DeWitt Payne’s comment and also Mait’s earlier point..

I have rewritten the equations in spectral terms.
And now I wonder why I wrote AA – eqn 7 – in terms of ED rather than atmospheric temperature.

On Mait’s point, if we consider the atmospheric radiation emitted to the ground, reflected by the ground, and then absorbed by the atmosphere:

= ε2.(1- ε).B(λ,TA)

where B(λ,TA) = Planck function at wavelength λ, and atmospheric temperature, TA

I’m now creating a program to calculate the results for “semi-realistic” non-gray emissivity of the atmosphere..

13. “Eli’s favorite point is what is the physical mechanism that decreases water vapor when CO2 increases.”

No need to thank me eli.

• No thanks necessary.

Under conditions of equal energy, the presence of liquid water will keep the temperature lower than without because some of the energy will be used to convert liquid water to water vapor. Yeah, we know. See ‘heat of vaporization’.

That in no way means that hotter air will not cause more evaporation than cooler air.

Also, the last I heard, globally, the specific humidity of the atmosphere was not down in recent decades.

http://www.ncdc.noaa.gov/bams-state-of-the-climate/2009-time-series/humidity

• Cohenite,
So, you are not disputing that water vapor content is not down, which is the direct counter to Miskolczi’s assertion that it will decrease under his model. OK

• Really Chris, you are still prone to jumping to conclusions; I went back and looked over some of the SH stuff; Dessler is leading the charge on this; in respect of his 2009 paper David Stockwell had an interesting take:

http://landshape.org/enm/dessler-zhang-and-yang-fail-significance-tests/#more-1772

In respect of Dessler’s 2010 paper Paltridge has this to say:

http://joannenova.com.au/2010/11/dessler-2010-how-to-call-vast-amounts-of-data-spurious/comment-page-1/#comment-125086

And it is not just Soloman, hardly a sceptic, saying high atmospheric WV is falling:

Click to access Pierce_et_al_AIRS_vs_models_2006GL027060.pdf

• The distribution of water vapor is what is important. See page 263 of:

The Thermodynamic Relationship between Surface Temperature and Water Vapor Concentration in the Troposphere

• Thanks William; the Lacis and Schmidt paper extoling the dominant climate impact of CO2 due to its non-condensing property is here:

http://www.nasa.gov/topics/earth/features/co2-temperature.html

Your take on the latent heat distribution of water through condensation dovetails with the Makarieva paper which takes a different approach to the Lacis paper:

http://www.atmos-chem-phys-discuss.net/10/24015/2010/acpd-10-24015-2010.html

Your eqn 3, which incorporates work done by the phase change of water due to condensation, Ldq, is arguably the most important part of atmospheric energy expenditure:

dU = CvdT + Ldq + gdh – PdV

Apparently the models either ignore this or do it badly.

• Cohenite,

I’m constantly awed by the small paradigm world inhabited by some of these “scientists”. The Lacis paper wants us to believe the non-condensing greenhouse gases control the earth’s temperature and disaster will occur because of a 3.7 W/m2 forcing over the next 100 years or so. But never mind that pesky evaporating/condensing water thing which generates energy fluxes from Ldq and PdV that well exceeds 1000 W/m2 each and every day. Rounding error anyone?

The Makarieva paper is interesting and especially so if you are a math freak. However I began to have some doubts when I saw that equation (1) was incomplete/incorrect and approximately 41 other equations were derived from it. I also checked out some of her conclusions against my own radiosonde data and could not support them. But I do like the non-traditional approach and I think there is some novel thinking there that could break some paradigms.

The models do not handle convection and water vapor correctly. The models predict decreasing convection intensity with increasing surface temperature when the exact opposite is true. That’s why they keep looking for that non-existent “hot spot”. They do not handle equivalent potential temperature correctly, probably because they are forcing a constant humidity. They also underestimate vaporization flux by a factor of two or more. Contrary to the models, increasing surface temperature increases convection intensity which in turn increases condensation efficiency. That’s why the upper troposphere dried out during the 1976 – 1998 warming period. Eli, are you there?

It’s amazing that generations of physicists have provided us with a scientific platform that resembles a finely honed, ivory carved, chess set; and today’s Climate Scientists only know how to play checkers.

14. on May 2, 2011 at 2:57 am | Reply DeWitt Payne

I’m having trouble reproducing Figure 3 in M2007 using equation (20) with πBg = OLR/f. I don’t get a curve with a peak. I get a monotonically decreasing function with an asymptote at OLR/2. I could be doing something wrong, but I don’t think so.

15. I rewrote the equations in spectral form – that is, each parameter as a function of wavenumber,v.

This allows emissivity, εAv, to be a function of wavenumber, and I took a fairly simplistic approach to defining the parameter.

The emission of radiation at each wavenumber is the Planck function x εAv.

For the time being εG is a “gray value”, i.e. constant but <1. Emission of radiation from the surface is still done with the same approach – the Planck function x εAv.

Here is an example:

The value of ED/AA is on the graph – 0.93.

This particular graph is just to illustrate the process.

I will now run some calculations of how this ratio changes with εG and ΔT.

16. Here is the relationship, for the specific gray atmosphere (shown above) and a surface temperature of 288K:

As εG reduces, ED/AA reduces.

And as ΔT increases, ED/AA reduces.

• Correction, the 1st line should read “non-gray atmosphere”

17. The equations are:

ED = ∫ εGεAv.B(v,Ta).dv

AA = ∫ { εGεAv.B(v,Ts) + εAv2(1-εG).B(v,Ta)}dv

where B(v,T) = Planck function at wavenumber, v and temperature, T
Ts = surface temperature
Ta = atmospheric temperature
εG = surface emissivity
εAv = atmospheric emissivity, which is a function of v

And, of course, these are still approximations, as the real equations need to vertically integrate the contributions of radiation from each height as noted in the earlier comment.

18. on May 2, 2011 at 4:06 am | Reply DeWitt Payne

I’ve also done a brute force calculation by spreadsheet of a gray radiative only non-reflective atmosphere. Not to put too fine a point on it, but there is a large discontinuity between the surface and the atmosphere. I calculated at 10 hPa steps from 1000 down to zero, forcing radiative balance at each level. Equal pressure steps will have the same number of molecules. For a surface temperature of 288.2 and a total atmosphere optical depth of 1.87, the temperature difference between the surface and the 990 hPa layer is 15.1C. I haven’t calculated the temperature profile, but it sure looks like it’s going to be greater than 10 K/km in the lower part of the atmosphere. If I force a lapse rate, I’m not sure I’ll get convergence at all.

• on May 2, 2011 at 4:11 am | Reply DeWitt Payne

Forcing a lapse rate is equivalent to forcing convective heat transfer at a particular level so radiative balance is pretty much out the window.

19. Ferenc Miskolczi:

I saw from another comment that some of your code is in Matlab.
If your fig 2 (2007 paper) results are built from Matlab code then I would be happy to look at it and see if I can identify the apparent discrepancy in the εG <1 results:
scienceofdoom – you know what goes here – gmail.com

20. EDA = εAσ(TS – ΔT)4 [2]

I have to look this one up. Since, q/A would be the difference in the fourth power of the temperatures should not the formula be

EDa= eA(SB)(Ts^4-dT^4)

Then the emisivities have to be accounted for if they are different than 1 and each other.

21. “Now, the transmittance for the atmosphere as a function of wavelength. For 4μm and longer wavelengths, absorptance = 1 – transmittance. For shorter wavelengths, especially below 0.5 μm, scattering becomes significant, which means that absorptance = 1 – transmittance – reflectance.”

I was operating under the understanding that LW was not reflected per your above. Are you now saying that it is reflected. If so your statement above is not correct.

22. mkelly:

Take a look at the diagram.
EDA is the emission of radiation by the atmosphere. The temperature of the atmosphere, TA = TS – ΔT.
Therefore, EDAAσTA4 = εAσ(TS– ΔT)4.

And if you reread the statement you cited it says “atmosphere”.

We are talking about reflection from the ground. And in the case of the ground, transmittance = 0. Therefore, absorptance = 1 – reflectance. (updated this last equation)

• on May 3, 2011 at 5:35 pm | Reply DeWitt Payne

Did you mean reflectance = 1 – absorptance?

r + a + t = 1

t = 0

&there4; r + a = 1 or r = 1 – a

• Thanks, fixed it.

• The mnemonic letter codes appear to work in blogger whenever I have used them. Maybe there is a switch in the software somewhere?

• on May 3, 2011 at 5:37 pm | Reply DeWitt Payne

test

those probably won’t print either (alternate codes for ‘therefore’).

• Therefore, EDA =εAσTA4 = εAσ(TS- ΔT)4.

I reviewed my heat transfer book and there is no allowance for this.

Ts is the surface and if you want to check q for the difference in the T of surface with the T of atmosphere you must use the fourth power of each temperature.

q= e(SB)(Ts^4-Ta^4) at best, but again emissivity must be reconciled.

I will continue to review my heat transfer book to see what more I can gleen.

So solids in the atmosphere can reflect LW?

• Yeah, but it’s called scattering because the shape determines the angle and you have to average over all particles and incident angles,. In general it goes to zilch as the wavelength exceeds the size of the particle

23. Van Dorland and Forster wrote a short rebuttal to Miskolczi (2010):

http://www.klimaatportaal.nl/pro1/general/show_document_general.asp?documentid=1439&GUID=%7B9B3C6C34%2D861C%2D45F4%2DB664%2D76A75CFDEFCB%7D

Conclusion:

“The alternative greenhouse theory of Miskolczi (2007,2010) results in a constant infrared
optical depth with time, meaning that there can be no increasing greenhouse effect with time.
Miskolczi suggests that observations show this ratio to be fixed. However, both observations
and calculations with physically sound radiative transfer models show that Miskolczi’s theory
does not stand up to scrutiny. Moreover, there is ample observational evidence that the most
important greenhouse gases, water vapour and carbon dioxide have increased in the last
four decades, meaning that the total infrared optical depth is indeed increasing. Finally, direct
satellite observations of the outgoing infrared spectrum show that the greenhouse effect has
been enhanced over this period. Even the calculations of Miskolczi show a change of optical
depth with time. Therefore, neither observations nor radiative transfer theory can support
Miskolczi;s conclusions.”

• van Dorland also says: “As a next step using his quasi radiative equilibrium model, Miskolczi calculates the relationship between outgoing longwave radiation (OLR) and the infrared flux originating from the Earths surface (Su). The relationship is a function of infrared optical depth (τA) only”

This appears to misunderstand Miskolczi. On page 17 of Miskolczi 2010, figure 10:

Click to access E&E_21_4_2010_08-miskolczi.pdf

Miskolczi measures changes in “the true greenhouse-gas optical thickness”. This is made up of two parts which are depicted in Figure 10. The first is τA which is defined as “the total IR flux optical depth” [page 5 Miskolczi 2007]. This is a measure of the total amount of infra-red or LW radiation which is absorbed between the surface and the TOA. The second is A which is the flux absorbance [page 3 Miskolczi 2010] and is a measure of what wavelengths of LW are being absorbed and transmitted in the atmosphere.

• As Van Dorland and Forster point out, the fact that observations over the last four decades indicate that both CO2 and H2O vapor concentrations are increasing shows that no matter what the mathematical mumbo jumbo, the optical depth of the atmosphere is increasing.

Bluntly put, Miskolczi cherry picked his data for water vapor trends from a set which is know to be hinky
————————–

Miskolczi (2010) claims that the amount of water vapour is declining in time supporting his theory. He used NOAA NCEP/NCAR reanalysis (2008) results to suggest this (see Miskolczi’s Figure 11). NCEP/NCAR reanalysis is known to have poor long-term trends. What Miskolczi analysis actually shows is that water vapour fluctuations are dominantly responsible for the changes in optical depth, which is a reasonable finding. More importantly, his Figure 11 is a good illustration of the fact that the optical depth is not constant, and is therefore inconsistent with his own theory.

A more robust analysis of water vapour changes by Mears et al. (2010) shows that total column water vapour is increasing over the oceans in the period 1988-2009 at a rate of 0.27 +/- 0.08 mm/decade. This corresponds to about 1.2%/decade (IPCC, 2007). Although observations of trends in relative humidity are uncertain, they suggest that it has remained about the same overall, from the surface throughout the troposphere, and hence increased temperatures will have resulted in increased water vapour. Over the 20th century, based on changes in sea surface temperatures, it is estimated that atmospheric water vapour increased by about 5% in the atmosphere over the oceans (IPCC, 2007).

• Eli, what is the reference for Mears et al (2010)?

• It’s in the special BAMS State of the Climate 2009 monograph as referenced by Dorland and Forster

http://lwf.ncdc.noaa.gov/bams-state-of-the-climate/2009.php

• on May 5, 2011 at 10:42 pm DeWitt Payne

cohenite,

τ, A and T are all related. T is the transmittance and equals St/Su. A is the absorbance and equals 1-T or Aa/Su. τ is the optical depth and is equal to -ln(T).

• Following DeWitt Payne’s comment, here is a graph of the relationships. These are all “by definition” of the properties (and for a non-scattering atmosphere, which is why Absorptance + Transmittance =1).

• Mears collaborates with Wentz and they produced a 2007 paper on evaporation and atmospheric water levels:

Click to access wentz_science_2007.pdf

I haven’t seen their 2010 paper but there is a 2010 p/p floating about which essentially says the same as their 2007 effort, which is the GCMs don’t properly equate the level of precipitation with the calculated level of evporation by the same GCMs based on assumptions about AGW. Soden and others weren’t happy about that:

Click to access comment_on_wentz_et_al_2007.pdf

At the end of the day this is, at one important level, a dispute about the relative reliability of the data, Paltridge vs Dessler, Miskolczi vs everyone. Having said that I wish Misklos or Ferenc would respond to SoD’s point, even if it is just in respect of the base assumption:

TA = TS – ΔT [1]

24. mkelly:

Therefore, EDA =εAσTA4 = εAσ(TS- ΔT)4.

I reviewed my heat transfer book and there is no allowance for this.

Ts is the surface and if you want to check q for the difference in the T of surface with the T of atmosphere you must use the fourth power of each temperature.

Do you agree that this equation is correct:

EDA = εAσTA4 [1] ?

If this equation is true and if TA = TS – ΔT [2] then it is proven that:

EDA = εAσ(TS– ΔT)4 [3]

This is just maths. If equation 1 is true & equation 2 is true, equation 3 by necessity MUST be true.

If c= a+b then you can replace “c” in any equation with “a+b”.

I’m not sure where your confusion lies?

• And all I am doing with eq 2 here is simply writing atmospheric temperature, TA as surface temperature, TS less a small amount, ΔT.

I am simply writing atmospheric temperature in a different way. It is still atmospheric temperature.

Perhaps you are thinking of exchange of energy between 2 bodies?

This equation 1, rewritten as equation 3, is simply emission of radiation from one body.

25. We appear to be getting away from van Dorland and Forster’s point, which is that you don’t need any math, Miskolczi’s paper assumes things to be true that are false and that the conclusions based on these false assumptions therefore fail.

26. Eli Rabett:

We appear to be getting away from van Dorland and Forster’s point, which is that you don’t need any math, Miskolczi’s paper assumes things to be true that are false and that the conclusions based on these false assumptions therefore fail.

You are correct. If, and only if, water vapor concentrations have been increasing over the past few decades, and Miskolczi has simply cherry-picked the dataset which shows the opposite from reality – then of course his paper is proven false.

I haven’t yet commented on this aspect – although I plan to write an article on the fascinating subject of decadal water vapor trends.

These articles have approached Miskolczi’s claims from a different point of view. The alleged theoretical foundations of “constant tau” are not theoretical foundations at all. Instead they are “results worked backwards giving the appearance of theory“. Or something like that. But not theory at all.

And I have cancelled my subscription to the Quarterly Journal of the Hungarian Meteorological Service.

• Not only. Consider Dorland and Forster’s initial point about Miscolczi’s best fit to the TIGR2 ascent data being lousy. This underpins the false Ansatz that optical depth is CONSTRAINED to be constant. That an average exists in no way constrains the value of the average, and if the constraint is based on the fit then the differences between the data points and the fit should be randomly distributed. They ain’t, the fit ain’t very good. End of story.

Another issue is why TIGR2 when there is TIGR3? It turns out to be an interesting question for pickers of cherries.

• on May 8, 2011 at 10:43 am | Reply Ferenc Miskolczi

SoD,

You wrote:

“And I have cancelled my subscription to the Quarterly Journal of the Hungarian Meteorological Service.”

This was a smart action. Congratulation.

I did the same after the articles from V. Toth and later from H. de Bruin….

27. Miklos Zagoni on May 6, 2011 at 7:52 am:

..Firstly, you did not mention that there is a successful theoretical prediction by him. Miskolczi’s theoretically derived greenhouse gas optical thickness is 1.868, while the measured data on very different data archives is practically the same, 1.869..

By the time we get to equation 10 of the 2007 theoretical paper we have 3 foundational equations that are not “from theory”, one of which is still awaiting comment from Miskolczi.

Equation 4: AA = ED, has been agreed by Miskolczi to be not an equation from theory but instead "an experimentally derived approximate quantity".

And there is still an unanswered point in this Part Four, not yet answered. Currently the adjusted claim by Miskolczi is “AA≈0.97ED” but I have demonstrated that with realistic values of surface emissivity, this approximation must be even less.

Equation 7confirmed as being not a theoretical equation.

Equating kinetic energy with flux – awaiting Miskolczi’s response, but already demonstrated to be not correct. Kinetic energy is proportional to temperature. Flux is proportional to the 4th power of temperature, or close to that for non-gray emitting bodies.

When a paper uses its experimental results to derive its theory it cannot claim that the result is a “theoretical predication”.

The Miskolczi club needs to regroup – and rewrite – before trotting out this kind of indefensible “stuff”. I can’t think of a more accurate description that will keep within the blog guidelines.

• Best summed up by Nick Stokes

I think Miskolczi’s paper could have been written in two sentences:

“The greenhouse gas theory that has been used for the last century is TOTALLY WRONG! The proof is left as an exercise for the reader.”

As Nick concluded, people just lose patience because nothing is clear and what is clear is wrong and eventually everyone just gets tired of cutting another arm off and walks away. This is characteristic of the notebooks full of math based on physically false assumptions school of science with a soupcon of dividing by zero often thrown in. There is a lot of it out there, some having recently been featured over at Curry’s place. If someone tells you they were constrained by only have forty pages, it is a strong indicator that you are in this territory. It is real Black Knight stuff.

• on May 6, 2011 at 1:39 pm | Reply DeWitt Payne

The whole logic of an optimum value of τ as delineated in Appendix B of M2007 is based on flawed math. The concept that there is an intermediate value of τ that maximizes cooling to space is wrong. It’s based on holding both surface emission and emission to space constant (OLR) while varying τ. But emission to space is a function of τ if surface emission is held constant or surface emission is a function of τ if emission to space is held constant. By holding both constant, you get a solution for τ where OLR equals incoming solar radiation (Fo). But that’s where you started. Of course that assumes that B0 corresponds to some real physical variable, which is highly questionable. One would think that it’s Eu, the net emission from the TOA. But that can’t be right because it’s easy to show that
B0 → OLR as τ → 0.
But we know that
Eu → 0 as τ → 0

Hand waving about the equations not being meant to apply to τ=0 is disingenuous. Either the theory is general and applies to all values of τ or it’s wrong.

• Fair points; I still hope Miklos and Ferenc take the bit between their teeth and respond. For your part SoD, what do you think is relevant from Miskolczi and I include the 2004 paper in that frame?

28. cohenite:

Fair points; I still hope Miklos and Ferenc take the bit between their teeth and respond.

For your part SoD, what do you think is relevant from Miskolczi and I include the 2004 paper in that frame?

On theory:

The papers don’t have much to recommend them. See the comments already made on what is theory and what is experiment – and why these shouldn’t be mixed up.

My opinion – worth little – is that the confusion demonstrated in these papers indicates little chance of a rewrite producing something worthwhile.

But what is my opinion worth?

The proof of the pudding is in the evidence. (The evidence here being better explained theory).

On the experimental results:

There are many papers showing positive decadal water vapor trends. At some stage I will write an article on this but so far I haven’t read enough of the papers on this subject.

The papers by Miskolczi don’t currently allow a reproduction of the results – no identification of the concentrations of each GHG used over the time period.

But by far the biggest experimental issue – why not report the results of tau including cloudy skies? This is something I can’t fathom. As already stated – it’s like reporting rainfall under cloudless skies and claiming it as a useful metric of trends in global averaged rainfall.

29. Because you can’t use a layer by layer code to calculate it, which is where Miscolczi has some expertise. Remember where this all started.

• on May 9, 2011 at 4:36 pm | Reply DeWitt Payne

You can, but it’s more difficult. The Archer MODTRAN site has the capability of adding a cloud layer. However, it’s only low clouds and the results all look very similar. But it’s certainly not isotropic on a per unit mass basis like a stellar atmosphere, which is Miskolczi’s starting point for theory.

30. A) Data.

Su and OLR are measurable, so G = Su – OLR is known.
The LW absorption and transmission, Aa and St are not measurable, hence the partitions Su = Aa+St and OLR=Eu+St are only computable.

There are several estimations for St in the global energy balance distributions: 40, 45, 52, 60, 66 ….

If anyone is interested in the greenhouse effect, she or he should decide somehow which one is valid. Miskolczi produced his St, Aa and Eu quantities and hoped they are good approximations of the realistic ones. You can challenge them by your own numbers, and we’ll be on the best way to make good science.

B) Equations.

On HIS numbers Miskolczi has found some relationships.

(1) Su – OLR = Ed – Eu (= G)

(2) Su – OLR + Ed – Eu = OLR

(=> g = 1/3)
(observation : g = 0.33)

(3) Ed + Eu = 2OLR

Equivalent form:

Su + OLR – (Ed + Eu) = G

You can challenge them by your equations on your numbers.

C) Interpretations.

As you like.

Let’s try these:

(1) Atmosphere: net energy in = net energy out = G

(The greenhouse factor in a water-abundant radiative-convective atmosphere is not constrained by the amount of the IR-active material (GHG’s), but is determined by the energy minimum principle (“most effective cooling”)). (Consequence : Eu = K+F)

(2) The sum of atmospheric net energy flows cannot be higher than their source (=Fo (=OLR)) (energy conservation principle)

(3) The gross atmospheric energy emission (Ed+Eu) cannot be higher than 2Fo (=2OLR). (Consequence: Su=2Eu).

Overall physical conditions for the effective dynamic control:
– sufficient water vapor reservoir in the free ocean surfaces; and
– turbulent atmospheric thermal flows.

The best theoretical numbers for a cloud-free atmosphere (with Su=396) are: OLR=264, St=66, Ed=330. (=> Eu=198, Aa=330). But a completely cloudless atmosphere on the Earth is dynamically unstable, so this is only an “in principle” game.

You can, as you do here, challenge these interpretations freely and endlessly. If you think they are wrong, suggest other ones. But I still think it would be more useful to do this AFTER your numbers or equations were produced.

There are other issues (D: Methods; E1: Theory of radiative transfer, E2: Theory of energetic principles etc).

E1: You can use f = OLR/Su=2/(1+tau) or 2/(2+tau) or what you can derive. Just show how your tau comes from your equations, and how fits to measurements. Put in the clouds, calculate individual profiles, create a global average, and check your results on different data archives.

E2: You seem still to think that the absorption depends on the amount of H2O, not on the amount AND the structure together.

[[SoD: Quine, Carnap or Popper would disagree with your kind of experiment/theory demarcation. What about the theory-ladenness of observations? – F: Philosophy of Science (of Doom)]]

BUT FIRST:
GOTO A

• ————————
The greenhouse factor in a water-abundant radiative-convective atmosphere is not constrained by the amount of the IR-active material (GHG’s), but is determined by the energy minimum principle (“most effective cooling”)).
———————

Even if one accepts this, the atmosphere is not uniformly water abundant, and at some locations and altitudes it is bone dry. This is meaningless hand waving

• Eli says: “Even if one accepts this, the atmosphere is not uniformly water abundant, and at some locations and altitudes it is bone dry.”

That’s what he said Eli:

“You seem still to think that the absorption depends on the amount of H2O, not on the amount AND the structure together.”

31. Miklos Zagoni on May 9, 2011 at 5:12 pm:

Are you or Miskolczi going to address the main point in this article (Part Four) that there appears to be a mistake in one of the basic calculations?

I took one sample black point (εG=1) from the graph, highlighted in red. Then I recalculated ED for εG=0.96 – this is the first transition in blue, then I made an approx estimate of AA as a result of this extra upward energy – this is the final transition in blue.

As you can see, there must be a basic error in Miskolczi’s calculation – the open circles representing εG=1 should be even further away from the ED=AA line.

Can either of you explain how reducing surface emissivity can increase the value of absorbed energy in the surface and reduce the value of absorbed energy in the atmosphere? The opposite must be true.

This moves the ED/AA ratio to around 0.94, and nothing like ≈1.

It also implies the calculations of tau have quite a large error – if the same program is used to do the tau calculation.

• Note graph above has an error in step 2. Correct maths is here.

32. SoD:

Why do you think absorbed energy in the atmosphere is reduced when the surface emissivity is reduced? Are you considering that the total emission is a constant regardless of the εG value? Otherwise the power output of the surface should reduce as the εG reduces and the amount absorbed should reduce accordingly.
There are scenarios where this would be true if we consider εG and εA across the spectrum by wavelengths, but I can’t see that we can say that it must be higher if εG reduces.

To be honest I don’t quite understand how the ratio of E(D)/A(A) can be derived from the surface emissivity at all if it is approached like this (it can be from 0 to infinity depending on the properties of the atmosphere and the surface – but I’m guessing Miskolczi didn’t include the reflected radiation into the value absorbed by A(A) in this context?

E(D)=A(A) still looks quite weird either way (I don’t see this ratio having much importance at all to be honest).

33. Mait:

Why do you think absorbed energy in the atmosphere is reduced when the surface emissivity is reduced?

No, I think that the absorbed surface radiation (ED) is reduced (because some atmospheric radiation is reflected).

But reading your comment I realize my diagram above has an error compared with what I calculated earlier – see equations on May 2, 2011 at 3:57 am.

As you point out the emitted surface radiation also reduces (which is also what I wrote in the equations earlier and simulated in the graph).

The 2nd blue point in the diagram won’t be to the right.

34. Mait:

E(D)=A(A) still looks quite weird either way (I don’t see this ratio having much importance at all to be honest).

I agree. Its importance is claimed by Miskolczi. In mainstream atmospheric physics this ratio has no more significance than any other ratio.

And as an experimentally calculated ratio (possibly incorrectly calculated) can’t be used as a foundation equation that proves the experimental results.

35. on May 10, 2011 at 2:27 am | Reply DeWitt Payne

Miklos Zagoni,

Question: Is B0 supposed to be equivalent to Eu? If not, what physical quantity does it represent? Miskolczi refers to it in M2007 as the skin temperature. For an optically thick atmosphere, that would mean B0 = Eu. But if that’s the case, then how can B0 be limited to a maximum of OLR/2 at high tau?

Deriving quantities from the Earth’s atmosphere in its current state tells nothing about what might happen if greenhouse gases continue to increase. For an optically thick atmosphere B0 must equal OLR, not OLR/2. Miskolczi’s solution to the Swarzschild-Milne equation is therefore incorrect.

• on May 10, 2011 at 11:36 am | Reply Ferenc Miskolczi

DeWitt,

“Question: Is B0 supposed to be equivalent to Eu?”

Your understanding is amazing. Bo is the integration constant for the source funcion profile in equation 13 (M2007). And OLR=St+Eu by definition. As you can see they Bo and Eu are not equivalent.

—-

I do not see yet your computation of tau for the NOAA R1 global average profile with your MDTRAN….Remember, we have the same tau for the usst76 profile…

• on May 10, 2011 at 2:03 pm DeWitt Payne

Ferenc,

Equation (20) has πB0 as a function of τ and Bg. But when you take the derivative of B0 with respect to τ, you treat Bg as a constant. You then use that result to derive an equation for Bg as a function of τ. You can’t have it both ways. Replace Bg with OLR/f in Equation (20) and Figure 3 looks a lot different. But of course, Bg ≠ OLR/f because that relationship was derived incorrectly.

• on May 10, 2011 at 11:53 am | Reply Ferenc Miskolczi

DeWitt,

“Miskolczi’s solution to the Swarzschild-Milne equation is therefore incorrect.”

You do not know what you are talking about. What the B(tau) is doing when tau is large (infinite) is given by Eq. 21 or B8 (in M2007). For
large tau Eq. 21 will be equivalent with semi-infitite solution of Eddington.

36. on May 10, 2011 at 3:30 pm | Reply DeWitt Payne

Ferenc,

I+(0) is the upward radiance at the top of the atmosphere, correct? That makes the two terms in equation (B1) look very much like St + Eu. πBg*exp(-3τ/2) certainly looks like St. That would seem to make the conclusion inevitable that πBo(1-exp(-τ_A)) = Eu. Except, of course, it isn’t. It isn’t because the definite integral in B3 involving Bo is evaluated incorrectly by treating Bo as a constant.

37. Folks, for the casual reader inserting the physical meaning of various Bos Eus etc occasionally would be a large help.

38. On Part One, Ferenc Miskolczi left kindly this comment, which I reproduce because many people might have stopped monitoring that thread:

Sorry SoD,

No mor time for this. I have many contacts with other physicists – who are not my supporters – but with deeper understanding and clearer thinking and better attitude.

I thanked him for stopping by and explaining a few things.

• on May 10, 2011 at 9:29 pm | Reply DeWitt Payne

I thanked him for stopping by and explaining a few things.

The operative word in the above sentence is few. The behavior is typical for M. Drive by disparagement of us lesser mortals understanding followed by vanishing when tough questions are asked.

Eli,

Folks, for the casual reader inserting the physical meaning of various Bos Eus etc occasionally would be a large help.

And that, of course is the problem. Some like Eu which is the upward emission from the atmosphere at the top of the atmosphere are fairly obvious from Miskolczi’s Figure 1 which is reproduced in the post above. Bo and H, not so much. B(τ) is apparently the Planck function for a gray atmosphere. H is the net radiative flux (up – down) at a given value of τ. He talks about radiative balance applying, which would seem to require that H = 0 everywhere, except that it doesn’t because he apparently assumes all short wave radiation is absorbed at the surface. Those are from his ‘solution’ to the Schwarzschild-Milne equation and their physical significance is what I was trying to pin down. I think Ramanathan introduced the H function. I don’t have a copy and I think my math skills are too rusty to follow it anyway. It seems to me that he plays fast and loose with his math nomenclature. It’s not at all clear if B(0) is the same as Bo, for example.

SoD,

Suggestion for a post topic is Appendix B in M2007. I can’t see how he gets from equation (B1) to equation (B3). It looks like H(τ) becomes H*τ. My weak brain can’t follow the apparently brilliant logic of this step. Variables become constants whenever it suits him as can be seen in Figure 3. in M2007.

• DeWitt,

I had lost interest due to the foundation equations for the theory not being theory, but you’ve rekindled my interest.

I’m not promising I can sustain it.

The paper’s author isn’t so interested in answering obvious questions – probably not challenging enough for him – so it’s not always clear what is assumed, what is invented, what is experimental work re-introduced as “theory” to support the experimental work, and what are approximations introduced as identities.

I will see what I can reproduce..

• SoD

It was a remarkable debate, nonetheless. Not only you went down to the details of his equations, but Miskolczi alctually showed up to answer “a few” questions.

Even the non-answers are telling.

It was a very interesting episode of your great blog. Please do it again if you feel like it.

• Thanks Alexandre.

39. H(τ) becomes H*τ”
It’s an error – H(τ) should have been H*τ – he’s citing (13). So it’s a straight substitution, and I think it is then correct.

• on May 11, 2011 at 12:58 pm | Reply DeWitt Payne

Thanks. That still leaves the problem that H is actually H(τ) so treating it as a constant would only be valid if radiative equilibrium holds everywhere. But of course, it doesn’t. The surface doesn’t emit enough radiation to maintain radiative equilibrium. There has to be convective energy transfer. If I’m not mistaken, that means dH(τ)/dτ ≠ 0.

• This seems to be a significant flaw.

There are other questionable claims.

Miskolczi claims that the solution to the “semi-gray” radiative model requires a semi-infinite solution and infinite optical thickness. Neither is the case. Perhaps some people originally developed the framework this way but it’s quite easy to develop the equations with neither of these conditions.

He cites two recent papers as failed attempts to “resolve the problems”. One of these is Weaver & Ramanathan (1995) which is a very nice paper and recommended reading: Deductions from a simple climate model: factors governing surface temperature and atmospheric thermal structure – freely available from the link.

Here’s an extract from the introduction which explains the point of the paper:

Simple models of complex systems have great heuristic value, in that their results illustrate fundamental principles without being obscured by details..

They go on to develop a slightly more sophisticated “semi-gray” atmospheric model which is quite interesting.

That is – the problem being addressed by Miskolczi is not actually a problem – only an issue with a simple model.

Still, I am only on page 14 (at equation 20) so perhaps this is all addressed before the end..

Must press on.

• The para that includes Eq 12 says the derivative of H is zero ( for each frequency). So I think it is meant to be constant.

FM and I had arguments about this in the now obliterated thread at CA (as I recall). (12) is on these assumptions a linear first-order differential equation, with straight line solutions. The boundary condition is set by flux balance at the top (Schwarzschild eq 15). Consequently, it’s likely not to match at the bottom. That’s where convection is important, and acts to bridge the temperature gap predicted by a pure radiative solution.

FM imposes a bizarre criterion which ensures that his solution (20) doesn’t satisfy boundary conditions at either top or bottom. Failure at the bottom is cushioned by convection, but at the top there’s nothing.

I could never see the justification for the term semi-infinite. It (15) is a one-sided solution, with the BC set at the top, but there’s no assumption that it is semi-infinite. It’s just a solution that continues to apply until conditions change (eg ground).

40. on May 11, 2011 at 1:44 pm | Reply DeWitt Payne

Miskolczi’s radiative transfer solution cannot be correct because it leaves out convective energy transfer, K in his terminology. The reason that there is a temperature discontinuity when an arbitrary lower boundary is placed in the semi-infinite gray atmosphere is that the effective temperature looking down is always going to be higher than the actual temperature at that level. The reason is that looking down, you’re seeing radiation emitted from levels with higher temperature. To maintain the same radiation levels, the boundary has to have its temperature equal to Teff at that level, not Ta.

You can’t just wish this away. Energy has to be transferred from the surface to the atmosphere by convection to raise the temperature at the bottom of the atmosphere to the same level as the surface. So why isn’t K included in the source function at the surface? Convective heat transfer is fundamental to determining the atmospheric temperature profile. The atmospheric temperature profile in turn determines radiative heat transfer. For Miskolczi, the atmospheric temperature profile appears to be a given.

For example, radiative-convective models usually assume that the surface upward convective flux is due to the temperature discontinuity at the surface.

It’s temperature gradient, not discontinuity. And that’s rather fundamental to heat transfer.

41. DeWitt, you had some questions and comments about equation 20, which I can’t find.
Can you explain where equation 20 comes from?

And is τA different from τ? (Both with ~ on top)

It seems yes (eqn 21), but where is τA defined?

• 2nd part answered, τA is the optical thickness of the whole atmosphere, whereas τ is the optical thickness as a function of altitude.

I.e., τ varies from 0 → τA as z varies from TOA → 0

• on May 12, 2011 at 4:57 am | Reply DeWitt Payne

Equation 20 and 21 are derived in Appendix B. Equation B7 is (20) and B8 is (21). Replace A by (1-exp(-τ)), T by exp(-τ) and f by 2/(1 + τ + exp(-τ) in Eq. (20) and (21) to get (B7) and B(8).

(B1)-(B3) are probably correct. It’s (B4) and (B5) where I have a problem. H and Bo are treated as constants when evaluating the integrals, but both are functions of τ. That means (B6)-(B8) are questionable. Then he differentiates (B8) and treats Bg as a constant. That’s particularly ironic as (B11) expresses the relationship of Bg to τ.

• Thanks. I had just caught up and am most of the way through the appendix.

B0 and H are constants (should be constants).

There appear to be serious problems in this appendix, and I will try and get my thoughts straight before explaining what I think they are.

42. Nick Stokes:

The para that includes Eq 12 says the derivative of H is zero ( for each frequency). So I think it is meant to be constant.

p11 of his 2007 paper says:

For monochromatic radiative equilibrium dHvv)/dτv=0..

But there is no such thing as monochromatic radiative equilibrium so it’s really irrelevant.

Then he “applies the grey approximation” and doesn’t state what he believes about dH/dτ

Of course, it must be zero under radiative equilibrium, but it’s not stated. His solution to the semi-gray approximation (equation 15) can only be derived for radiative equilibrium: dH/dτ = 0.

I am working my way through the intriguing appendix B at the moment. I see more confusion.

• Nick, Probably I am travelling over old ground for you?

• on May 13, 2011 at 1:10 am DeWitt Payne

And now for the other shoe:

Of course πBg probably isn’t actually equal to OLR/f. It’s a first approximation at best. That result was obtained by doing exactly the same wrong thing as Miskolczi did to generate Figure 3: Treating something as a constant that isn’t when integrating or taking a derivative.

• SoD,
Yes, there was a very long thread at CA BB – over 100 pages. Now lost.

It’s true that the claim of equilibrium at each wavelength is dubious. But the claim of constant H is really just conservation of energy, or radiative equilibrium, as you say. There’s no other heat transfer mechanism postulated.

It is also indeed the Schwarzschild approx.

DeWitt,
As said, H is constant for reasons of cons en, and B_0 is just H/(2pi). So the integrating of B3 is OK. (21) is just a line in tau like (15), but with a different offset. Since (15) was the unique solution of (12) with flux balance at TOA, that means (21), a different solution, fails this test.

I think the argument for saying deriv of B_G is zero is that B_G is just the S-B function of ground T, which is assumed constant while “optimising” optical depth.

• on May 13, 2011 at 1:13 am DeWitt Payne

Wrong reply button above.

And now for the other shoe:

Of course πBg probably isn’t actually equal to OLR/f. It’s a first approximation at best. That result was obtained by doing exactly the same wrong thing as Miskolczi did to generate Figure 3: Treating something as a constant that isn’t when integrating or taking a derivative.

• on May 12, 2011 at 8:32 am | Reply DeWitt Payne

Nick Stokes,

Groan. There’s a sign error in (B6). No wonder I was having so much trouble. It should be minus H/4π[…] not plus. (B4) has the correct sign but it didn’t translate into (B6). Equation (20) is correct. Now with πBo = H/2, the second two terms in B6 behave like Eu.

That also makes πBo = H*(1-exp(-2τ))/(2*(1-exp(-τ)))

Bo → H as τ → 0
Bo → H/2 as τ → ∞
so dBo/dτ → 0 as τ → ∞

There is no optimum τ.

• You are correct about the sign error in B6. This sign error gets corrected (without comment) in B7.

• DeWitt, I presume you’re using B7 and τ means τ _A? If so, then I think it should be:

Bo → H/(2π) as τ → ∞
Bo → (H/π – Bg)/τ as τ → 0
and if Bg>H/π, Bo gets large negative as τ → 0.
In that case, since Bo actually exceeds H/(2π)for very large τ, there must be a maximum. Otherwise not.

• on May 12, 2011 at 3:54 pm DeWitt Payne

Nick Stokes,

I left out the π in my equations. τ should be τ_A.

I don’t consider that maintaining πBg constant as τ_A is varied is valid. Equation B11:

πBg = OLR/f = H/f = H/2(1 + τ_A + exp(-τ_A))

The fact that πBo becomes large and negative if πBg is held constant proves my point. A negative ‘skin temperature’ is unphysical.

I can now calculate all the factors in Miskolczi’s energy balance diagram if I assume that Ed = Aa, which isn’t far wrong. Plugging the numbers into equation (7) (Su -Fo + Ed – Eu = OLR) is interesting. It’s only true at τ_A = 1.8414. The ratio of the left hand side to OLR varies from zero to 4 as τ_A varies from 0 to 5. It’s linear with τ_A at τ_A > 3. Equation (8) (Su = 3OLR/2) is also only valid at τ_A = 1.8414 as well. Equation (9) (Su + St/2 + Ed/10 = 3OLR/2) is approximately true for an optically thin atmosphere but is already equal to 1.75OLR/2 at τ_A = 1.8414 and increases from there.

• on May 12, 2011 at 4:35 pm DeWitt Payne

Here’s a graph of πBo using πBg = OLR/f (pink) in (B7) and with πBg = OLR (black). OLR =264.169 and Bg = 389 W/m2 are from the MODTRAN (corrected for the missing tails of the spectrum) 1976 US standard atmosphere with default conditions (375 ppmv CO2, Tsurf = 288.2 K). The blue line is OLR-St calculated using St = (OLR/f)*exp(-τ_A). The yellow points were calculated using MODTRAN by varying the absorber concentration (CO2 and water vapor scale) and adjusting the surface temperature to keep OLR constant (264 W/m2). OLR/f underestimates Bg. Maybe that’s the 1.5 vs 1.66 thing.

• on May 12, 2011 at 4:38 pm DeWitt Payne

Or maybe it’s MODTRAN.

• on May 12, 2011 at 8:18 pm DeWitt Payne

If that graph didn’t convince you that by fixing πBg and varying τ_A in the πBo formula you are just solving for the original τ_A, but don’t get it because the formula isn’t exactly comparable to data from a line-by-line model. Here’s a graph created using Bg calculated from Miskolczi’s formula (OLR/f where f = 2/(1 +τ_A + exp(-τ_A)) at constant OLR (264.169 W/m2). Note that each curve reaches it’s maximum and crosses the πBo curve at exactly the original value of τ_A.

There is no preferred value of τ_A. It’s just what it is based on the profiles of the concentrations of the ghg’s and the temperature.

• on May 12, 2011 at 8:24 pm DeWitt Payne

For reference, the surface temperatures for the four values of τ_A assuming unit surface emissivity are:

τ_A T(K)
0.5 264.67
1.0 272.53
1.5 282.22
2.0 292.34

• on May 13, 2011 at 1:15 am DeWitt Payne

If I keep trying I’ll get this in the right place eventually.

And now for the other shoe:

Of course πBg probably isn’t actually equal to OLR/f. It’s a first approximation at best. That result was obtained by doing exactly the same wrong thing as Miskolczi did to generate Figure 3: Treating something as a constant that isn’t when integrating or taking a derivative.

• Why can’t there be a monochromatic radiative equilibrium?

43. Nick, did you also comment on the diffusivity approximation?

Generally the best fit is 1.66 not 1.5.

• No, SoD, I didn’t comment on that, and I’m not sure what you mean. Is it the ratio between tau bar and tau squiggle stated following Eq 15?

44. Yes it is the ratio between tau bar and tau squiggle, and there isn’t an analytical solution. If you integrate over the hemisphere and then average the “best” solution is τ/0.6 (as explained in Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations.

There are bigger issues but I wondered about the impact of choosing the non-optimum value.

45. on May 13, 2011 at 3:00 am | Reply DeWitt Payne

If πBo is in fact a constant and equal to H/2, then:

πBg = H/2(2 + τ_A)

That looks to be too big, but what if that’s the purely radiative flux? Then Su = πBg – K.

Going back to the US standard atmosphere with τ_A = 1.437, OLR = 264.169, Ed = 280.97 and Su = 389, πBg = 3.437*264.169/2 = 453.97 and K = 453.97-389 = 64.97 W/m²

Fo-F is then πBg – Ed = 453.97 – 280.97 = 173 and F = 91.169

That almost seems reasonable, although F looks a little high for clear sky with relatively low humidity. Let’s see if it holds up for higher τ_A in the tropical atmosphere.

MODTRAN default Tropical atmosphere:

τ_A =1.9985, OLR = 293.18, Ed = 380.03, Su = 448
πBg = 586.14, K = 138.14, Fo-F = 206.11, F = 87.07

F seems a little low for the tropics, but perhaps the sun being more directly overhead reduces atmospheric absorption even though the humidity is higher. Anybody out there with a line-by-line program to calculate incoming solar atmospheric absorption?

Obviously, if πBo is a constant it can’t have an optimum meaning that there’s still no preferred value for τ_A and Figure 3 is even more wrong.

46. cohenite:

Why can’t there be a monochromatic radiative equilibrium?

Radiative equilibrium means that each level in the atmosphere has achieved an energy balance with the rest of the atmosphere.

What does this mean?

It means that one layer absorbs X W/m2 and also emits X W/m2.

Imagine for a second that it emitted 0.9X W/m2.

So it is emitting less than it is absorbing. And therefore heats up. Until its temperature increases to the point where it is emitting X W/m2.

Each part of the atmosphere (in the absence of convection) will do this, until finally equilibrium is reached.

It is “easy” to see how this must happen. It is inevitable – without convection.

How if we consider the radiation at 14.1um why should this specific wavelength be in balance? What is the mechanism?

There isn’t a mechanism to achieve “monochromatic radiative equilibrium”. It cannot exist (except under some very special non-real world conditions). Because there is no mechanism to ensure that 14.1um radiation balances at each height, along with 16.3um radiation and 18.6um radiation…

Total energy absorbed by a layer = total energy emitted. Because if this doesn’t happen then the layer heats up or cools down. But there is no mechanism for individual components of radiation to balance. No law of thermodynamics comes into play to enforce this.

Does this help?

• on May 14, 2011 at 12:14 am | Reply ClimateWatcher

>>>Radiative equilibrium means that each level in the atmosphere has achieved an energy balance with the rest of the atmosphere.<<<

Don't we mean in this context that Radiative Equilibrium means that outgoing energy balances with incoming energy ( for the whole earth system ) ?

47. SOD,

I’m confused. If you can have radiative equilibrium around a given layer, and the composition of the atmosphere around that layer is uniform, why would you not also have “monochromatic radiative equilibrium” by definition?

Bill Gilbert

• on May 14, 2011 at 1:40 am | Reply DeWitt Payne

If the region is small enough, then it will be isotropic. But for purposes of calculation you have to make the layer thickness big enough that there is some change in properties between layers, density and temperature being the big ones. Water vapor isn’t well mixed so the layers will have a different composition even on a constant mass basis and then you don’t have monochromatic radiative equilibrium.

• DeWitt,

But if the layer thickness is big enough for a “change in properties”, you would no longer have radiative equilibrium either – correct? I’m still confused.

• You can have radiative equilibrium when there is no convection. The stratosphere is close to radiative equilibrium.

Radiative equilibrium means that the net energy via radiation into a layer = 0. It doesn’t imply that upwards and downwards flux don’t change because the reduction in upwards flux is balanced by the increase in downward flux.

In the next article on Miskolczi I will explain this in more detail.

• on May 14, 2011 at 5:08 am DeWitt Payne

The upward and downward flux out of the layer should be equal but the downward flux from above will be less than the upward flux from below. The total of in-out will be zero or the layer will fairly rapidly warm or cool until it is zero. Actually, that gets tricky in the stratosphere as it’s a balance between the increasing temperature and the decreasing pressure with altitude.

48. Isn’t monochromatic radiative equilibrium [MRE] assumed for calibration of methodology to ascertain the spectral and radiation properties of planetary bodies?

I don’t think Miskolczi states MRE occurs for Earth, in fact on page 23 of 2007 he says:

“There is, however, a slight non-linear dependence on the surface temperature introduced by the weighting of the monochromatic flux transmittances with the spectral SU.”

Isn’t it a fact that there can still be radiative equilibrium without MRE at any particular wavelength?

• on May 14, 2011 at 1:54 pm | Reply DeWitt Payne

You calibrate an instrument for spectral power at a given wavelength or frequency. When you measure emission, you measure emission. An assumption of radiative equilibrium, whether monochromatic or full spectrum isn’t necessary. The AERI instrument for measuring atmospheric emission is calibrated by measuring the spectrum from two black body sources at different temperature. The spectral power at each frequency is known from the Planck function. If the detector is linear, then two points at each frequency gives you the calibration function.

• cohenite:

I don’t think Miskolczi states MRE occurs for Earth..

You are correct.

49. I still don’t see how a gaseous atmosphere can be in radiative equilibrium but not in monochromic radiative equilibrium. But that is a minor point and I will leave it here.

The major point is whether or not the troposphere is in general radiative equilibrium, at least in the lower to middle altitudes. Miskolczi’s empirical data indicate that this is so regardless of the GHG composition (at least over the time period of his radiosonde data analysis). This would be true with or without convection.

Tropospheric radiative equilibrium would then indicate that radiative heat transfer through the troposphere is insignificant and the thermodynamic state of the troposphere is determined by non-radiative processes. Thus OLR is not a function of Aa, Ed, Su, etc. but is a function of what happens with K. And Miskolczi and this forum have paid little attention to K. Noor van Andel recognized this and was making very good progress prior to his passing away. (A group of international scientists are trying to pull together van Andel’s work and get it into publishable form – but this could take awhile).

But if SOD is about to post an article on radiative equilibrium, I will wait for that before going into any more detail.

Bill Gilbert

50. I did not know that Noor van Andel had passed away; that is sad news; I found his articles very helpful; he had the common touch and could translate the concepts for us plebs.

William: my comment about MRE and radiatitive equilibrium was an attempt to understand at an individual wavelength the ‘compensatory’ role water plays in terms of M’s idea of it ‘adjusting’ for variations in the radiative amount from other GHGs.

• Cohenite

Can you have a ’10 cent’ equilibrium? So you look at you bank accounts and somehow your balance of 10 cent pieces remains the same? It may happen by chance occassionally, but it’s purely meaningless. It has no meaning to the overall bank balance.

51. Cohenite,

Noor van Andel pssed away on April 19. He brought the multi-varied tools of Engineering into the world of atmospheric physics and was very adept at using them in that arena.

As to water vapor, I don’t beleive Mislolczi developed a specific mechanism for the role of water vapor in acheiving a constant tau. He relied on the empirical data and assumed that the non-radiative processes (K) act in such a manner that would achieve the energy balances he observed. Atmospheric radiative properties were more of a result than a cause (I did not express that very well in my previous post).

52. williamcg on May 14, 2011 at 6:17 pm:

I still don’t see how a gaseous atmosphere can be in radiative equilibrium but not in monochromic radiative equilibrium. But that is a minor point and I will leave it here.

In the absence of convection, e.g. in the stratosphere under some circumstances, the atmosphere will be in radiative equilibrium.

The reason for this is the first law of thermodynamics – if more energy leaves than enters, then the layer will heat up and therefore emit more radiation.

Easy point to grasp.

No such constraint exists for radiation at 14.1 μm or 12.7 μm. Just to pick two random examples.

The only constraint is on total energy.

Energy might be absorbed from “layers” below at significantly higher temperatures which will relate to the temperature of those layers and the concentration of radiatively-active gases in those layers. But the emission of radiation from the “layer” in question will relate to its temperature and its concentration of radiatively-active gases.

Therefore, we will find that energy at (for example) 14.1 μm will not be conserved. As the temperature changes, the Planck function shifts wavelengths slightly, and as different gases increase or decrease in concentration the energy will be absorbed at one wavelength and re-emitted at slightly different wavelengths.

I wish I could think of a really good conceptual picture to paint.

Note: I write “layers” as a convenient mental device, the atmosphere is, of course, continuous.

• on May 18, 2011 at 3:48 pm | Reply DeWitt Payne

I’m beginning to think we’re using radiative equilibrium to mean different things. Strictly speaking, if two objects are in radiative equilibrium, then they are, by definition, at the same temperature. The atmosphere as a whole is, therefore, never in radiative equilibrium because it isn’t isothermal. In fact, it can never be isothermal.

If the objects have different emissivity spectra, then they could still be in radiative equilibrium, i.e. at the same temperature, without being in monochromatic radiative equilibrium. But for the atmosphere, the spectrum will only be significantly different for fairly large separations in altitude. Then radiative equilibrium won’t apply because the temperature changes with altitude. Well, maybe in the tropopause where the rate of temperature change can be quite small.

Local thermal equilibrium implies local radiative equilibrium.

• I’m beginning to think we’re using radiative equilibrium to mean different things..

You make a good point.

Here is how I am using it: “radiative equilibrium” = equilibrium temperatures have been reached where radiation is the only heat transfer mechanism operating.

Equilibrium temperatures don’t imply isothermal – instead that temperatures have ceased changing.

• on May 24, 2011 at 7:26 pm DeWitt Payne

Equilibrium temperatures don’t imply isothermal – instead that temperatures have ceased changing.

Nope. If the temperatures of the two objects aren’t the same, then you have the potential for a perpetual motion machine of the second kind (I think), i.e. the extraction of energy from a reservoir at a single temperature.

Consider two objects with arbitrary emissivity spectra inside a perfectly reflecting container. If the objects are at the same temperature, they must be at radiative equilibrium. Otherwise, one will warm and the other cool, a clear violation of the Second Law.

53. DeWitt Payne:
You said that two bodies are at the same temperature by definition when they are in radiative equilibrium. This might be a bit off topic, but could you please elaborate from what the definition of radiative equilibrium would be for that to hold true (I couldn’t quite think of a definition for radiative equlibrium from the top of my head that would require the same temperature and also make much sense).

• on May 24, 2011 at 7:30 pm | Reply DeWitt Payne

Radiative equilibrium means that each object (two objects inside a perfectly reflecting (insulating) container, e.g.) is emitting and absorbing the same amount of energy so the temperature of each object cannot change as there is no net heat gained or lost.

54. There is some continuing discussion about the subject of this article over on Part Two, starting with Ken Gregory’s comment here.

55. On my comment:

Equilibrium temperatures don’t imply isothermal – instead that temperatures have ceased changing.

DeWitt Payne responded:

Nope. If the temperatures of the two objects aren’t the same, then you have the potential for a perpetual motion machine of the second kind (I think), i.e. the extraction of energy from a reservoir at a single temperature.

Consider two objects with arbitrary emissivity spectra inside a perfectly reflecting container. If the objects are at the same temperature, they must be at radiative equilibrium. Otherwise, one will warm and the other cool, a clear violation of the Second Law.

If there were just two objects..

The reason why heat transfer textbooks are full of worked examples of “what is the equilibrium temperature and flux at boundary A” is because there is a usually a source of energy in these examples.

Like the earth’s surface, continually supplied by the sun. So in equilibrium (radiative equilibrium with no convection) the surface will be a higher temperature than 10km up in the atmosphere. The sun is what prevents it from being classified as a perpetual motion machine.

But as I know that you know this – because you have explained this concept to others – I’m not sure whether you have got yourself in a tangle, or I am just missing some point you are trying to explain..

• on May 24, 2011 at 11:36 pm | Reply DeWitt Payne

I’m not sure whether you have got yourself in a tangle, or I am just missing some point you are trying to explain..

Because I think you’re confusing steady state with equilibrium. They aren’t the same. The Earth and all its parts is in near steady-state. It’s cooler than the sun and warmer than deep space. Energy absorbed is close to energy emitted. However, it is not in equilibrium and never will be unless the sun goes out and sufficient time passes for all heat to be lost from the core because the entropy of the system continually increases.

Local thermal equilibrium exists in the lower atmosphere because radiative transfer of energy is negligibly small compared to collisional transfer. True radiative equilibrium, as opposed to steady-state, requires a Planck spectrum for the photon gas. Put bodies in a perfect reflector and unless the emissivity is identically zero at some wavelength, which isn’t going to happen for a solid body, the photon gas will have a Planck distribution at equilibrium. If the system isn’t closed, there can be no equilibrium because entropy is increasing.

56. SoD.

I can’t help thinking that this discussion is about a dry planet with some water vapour in the atmosphere. The surface of that planet would be a lot hotter than Earth’s surface and, perhaps, not include the ‘surface disconnect’ that’s perceived?

As an engineer, I can only see Earth’s atmosphere as a ‘cooling tower’ for the surface. Some convection in the atmosphere (tropo) is unavoidable (if only by virtue of the fact that water vapour [if it could exist as a singular gas] is a lighter than air gas), but surface temperatures are reduced by its evolution there and that energy is then carried aloft to disperse elsewhere with little alteration to temperature (latency).

IMHO it’s impossible to formulate a ‘radiative concept’ for Earth’s lower atmosphere (~tropo) without the inclusion of a ‘latency’ variable that tracks the variability of the atmospheric hydrocycle. Let’s face it, water vapour acts like a rechargeable battery in the atmosphere. How can this be shown as a ‘radiative quality’?

Best regards, Ray Dart.

• on May 25, 2011 at 2:12 am | Reply DeWitt Payne

And that’s a problem with Miskolczi’s analysis. K (convective heat transfer from the surface to the atmosphere) exists in the flow chart, but is almost completely ignored in the theoretical and empirical analysis.

• That’s why I referred to the ‘Clausius Clapyron’ (CC) relationship elsewhere on this site. M’s observations strongly suggest that the hysteresis point for ‘tau’ is attached to the CC relationship for H2O and ‘phase change’ can’t be identified in an ‘SGM’ that’s been discussed here without separating H2O as an ‘array’ of its energetic phase states within the model.

Best regards, Ray Dart.

• on May 25, 2011 at 10:40 am | Reply Ferenc Miskolczi

suricat

” IMHO it’s impossible to formulate a ‘radiative concept’ for Earth’s lower atmosphere (~tropo) without the inclusion of a ‘latency’ variable that tracks the variability of the atmospheric hydrocycle. Let’s face it, water vapour acts like a rechargeable battery in the atmosphere. How can this be shown as a ‘radiative quality’? ”

Can we discuss this question in detais? See the plot in my EGU 2011 presentation, slide 20:

Click to access EGU2011-13622_presentation.pdf

My e-mail address: fmiskolczi@cox.net

Thanks, Ferenc

• Hi Ferenc.

“IMHO it’s impossible to formulate a ‘radiative concept’ for Earth’s lower atmosphere (~tropo) without the inclusion of a ‘latency’ variable that tracks the variability of the atmospheric hydrocycle. Let’s face it, water vapour acts like a rechargeable battery in the atmosphere. How can this be shown as a ‘radiative quality’?”

That phrase was aimed at this site. It’s considering that your observations can be modelled by a ‘simple grey model’ (SGM) that can’t include the degrees of freedom that are necessary for the inclusion of ‘latency’ (hidden heat), to say the least. You have my apology if you thought that this phrase was, somehow, directed towards yourself. This was unintentional.

However, the connection between observation and theory continues to be unresolved. How can we fill that ‘gap’? I don’t know, but as my mum’s main carer (she’s 99 this month), my time is compromised. I’d be more disposed to interactions here than by e-mail.

Best regards, Ray Dart.

• Ferenc, I’ve reconsidered my decision. I’m not a scientist, but an engineer. Looking upon this situation as an engineer that may want to discuss a ‘patent proposal’ I’d expect full confidentiality. Thus, I’ll mail you.

I just hope I can be of help.

Best regards, Ray Dart.

• on May 26, 2011 at 10:39 pm DeWitt Payne

suricat,

That phrase was aimed at this site. It’s considering that your observations can be modelled by a ‘simple grey model’ (SGM) that can’t include the degrees of freedom that are necessary for the inclusion of ‘latency’ (hidden heat), to say the least. You have my apology if you thought that this phrase was, somehow, directed towards yourself. This was unintentional.

You clearly haven’t been paying attention. The entirety of Miskolczi’s theory is based on a simple gray model of the atmosphere. Everything else is empirical observation that gets elevated to thermodynamic identities. I’m specifically referring to page 13 in the EGU presentation. Note that it has been proved beyond a doubt that Miskolczi’s derivation, particularly of Bg = OLR/f is simply wrong.

• “The entirety of Miskolczi’s theory is based on a simple gray model of the atmosphere.”

Yes, but ‘which’ “simple grey model”? Perhaps this link will help:

In the ‘interest of confidentiality’ I need to ‘stay quiet’ here right now.

Best regards, Ray Dart.

• on May 29, 2011 at 2:48 pm DeWitt Payne

suricat,

Yes, but ‘which’ “simple grey model”? Perhaps this link will help:

As I said, you haven’t been paying attention. This whole series of posts has been about precisely that simple model in the 2007 paper as well as the follow-up paper in 2010. The 2007 paper is mentioned in part one and linked in part two. Part four specifically analyzes Equation (7) in the 2007 paper and part five analyzes the derivation of Equations (20) and (21) in Appendix B and shows that it is wrong.

I suggest you start at the beginning of the series and read it through again before assuming that SoD has invented an irrelevant simple model out of whole cloth.

57. on May 26, 2011 at 11:49 pm | Reply Ferenc Miskolczi

suricat

“I just hope I can be of help.”

We are on an uncharted area of the research in the planetary greenhouse effect and the clear rational thinking is the greatest help.

Ferenc

• Ferenc,

You are exactly correct about this area being uncharted. And I agree wholeheartedly with Ray’s instinct that a missing peice is latent heat/convection.

Understanding “K” is the key to fitting all the radiative relationships that you have measured into a theoretical whole. Good luck.

Bill Gilbert

58. I think people are not seeing the forest through the trees here. At each and every layer through the atmosphere where a surface emitted photon is absorbed by a GHG molecule (or a cloud), the re-emission is isotropic. Ultimately this means half of what is absorbed is emitted up out to space and half is emitted down to the surface. If the energy is transferred to other gases via collisions, the heated gases of the atmosphere will radiate according to the laws of black body radiation (i.e. equal in all directions). The fact that there are multiple exchanges many times over between radiative and kinetic energy components through the atmosphere will not change any of this.

Miskolczi’s measured data showing Aa = Ed just confirms what the physics dictates to be the case.

A lot of confusion seems to lie in not realizing that all the energy entering and leaving at the TOA is radiative, and as a result of this the effect of the non radiative fluxes from the surface (from latent heat of water and thermals) on the radiative budget has to be zero, because COE dictates that the atmosphere cannot create any energy of its own.

59. RW:

If you have a read of Part Two you will see why Aa≠Ed.

And the argument I put forward in this article has just been confirmed by a supporter of Miskolczi, Ken Gregory, who said:

I have updated this spreadsheet to include the reflected Ed here..

..Columns T to AF show the fluxes and flux ratios with emissivity = 0.967. This is the value given on page 8 of M2010. Su and St includes the reflected portion of Eda.

..The graph at cell AK10 shows Ed/Aa vs Su at emissivity = 0.967. The average value is 0.9365..

Your explanation is a jumble of different concepts. Isotropic means equal in all directions. This has no bearing on whether energy emitted from a layer must balance absorption from a completely different layer at a completely different temperature.

Energy absorbed is proportional to the absorptivity at the relevant wavelengths. Energy emitted is proportional to the emissivity at the relevant wavelengths AND the temperature of the body emitting.

If the two bodies are at different temperatures they will not exchange equal amounts of radiation.

To take an extreme case, if an atmospheric layer at 200’C absorbed radiation from a surface at 1000’C will it emit an equal amount of energy? No, it will emit according to its temperature (200’C) and its emissivity at this temperature. This will be less.

If you believe that there is some rule that means that two bodies must exchange equal amounts of radiation then you should spell it out in detail. I suggest you draw a diagram first, then write down the equations, and when you have a complete argument feel free to post it here.

• scienceofdoom says:

“Your explanation is a jumble of different concepts. Isotropic means equal in all directions. This has no bearing on whether energy emitted from a layer must balance absorption from a completely different layer at a completely different temperature.

Energy absorbed is proportional to the absorptivity at the relevant wavelengths. Energy emitted it proportional to the emissivity at the relevant wavelengths AND the temperature of the body emitting.

If the two bodies are at different temperatures they will not exchange equal amounts of radiation.”

But they will at the boundaries of the surface and the TOA where only radiation is emitted.

“To take an extreme case, if an atmospheric layer at 200′C absorbed radiation from a surface at 1000′C will it emit an equal amount of energy? No, it will emit according to its temperature (200′C) and its emissivity at this temperature. This will be less.

If you believe that there is some rule that means that two bodies must exchange equal amounts of radiation then you should spell it out in detail. I suggest you draw a diagram first, then write down the equations, and when you have a complete argument feel free to post it here.”

It’s not a special rule – it’s just Conservation of Energy as it’s applied to the laws of black body radiation.

I’ll break it down into a series of separate yes or no questions to see where our discrepancy lies:

1. Do you agree that all the energy entering and leaving at the Top of the Atmosphere is radiative (i.e. all photons)?

2. Do you agree that all the energy emitted at the surface is radiative also?

3. Do you agree that the net effect of the non-radiative fluxes from the surface (latent heat of water and thermals) on the energy budget have to be zero because the atmosphere cannot create any energy of its own?

If your answer is yes for all of these, I’ll move on the next series of questions. If it’s no, we’ll discuss if you’d like.

60. Of course energy is conserved within any body – in this case “a body” = “a layer of the atmosphere” or “the surface”.

This is the first law of thermodynamics, and everyone agrees with it.

Take a look at Fig 3 in Part Five. The vertical axis is just a proxy for height through the atmosphere.

This graph is for an atmosphere in radiative equilibrium (although perhaps “steady state” is a better choice) – with no convection.

The graph is calculated by using energy balance at each layer in the atmosphere.

So how can different flux exist in different layers?

The sun is maintaining the temperature differential between the top of atmosphere and the bottom of the atmosphere.

1. Do you agree that all the energy entering and leaving at the Top of the Atmosphere is radiative (i.e. all photons)?

Yes

2. Do you agree that all the energy emitted at the surface is radiative also?

No, convective flux is very significant (frequently written as sensible heat + latent heat).

3. Do you agree that the net effect of the non-radiative fluxes from the surface (latent heat of water and thermals) on the energy budget have to be zero because the atmosphere cannot create any energy of its own?

No. It’s true that the atmosphere cannot create any energy of its own.

The energy comes from the sun and because the atmosphere is colder than the surface, heat is transferred by convection (and by conduction at the boundary) from the surface into the atmosphere.

The net total energy flux at the surface is zero.

Downward atmospheric radiation absorbed + solar radiation absorbed = Upward terrestrial radiation emitted + convective heat

(Strictly speaking it doesn’t have to be equal if the temperature is not static, but over the long term it will be exactly equal if the long term temperature trend is zero, and even if the long term temperature trend is not zero, the imbalance will be very small).

61. scienceofdoom says:

“2. Do you agree that all the energy emitted at the surface is radiative also?

No, convective flux is very significant (frequently written as sensible heat + latent heat).”

Ah, I think this is where the discrepancy lies. Let me re-phrase the question:

Do you agree that the surface is emitting about 390 W/m^2 and that all of this is radiative?

If yes, do you agree that the surface is emitting about 390 W/m^2 solely due to its temperature and nothing else (assuming an emissivity of 1 or very close to 1)?

(*I’m not ignoring your other questions – I just want to take it one step at a time at this point).

62. scienceofdoom says:

“(Strictly speaking it doesn’t have to be equal if the temperature is not static, but over the long term it will be exactly equal if the long term temperature trend is zero, and even if the long term temperature trend is not zero, the imbalance will be very small)”

Yes, I fully understand this and agree.

63. scienceofdoom,

I’ll add one more question for you if the answer to the above two questions is yes:

Do you agree that if the surface is emitting 390 W/m^2 that it cannot be receiving more energy than this?

64. RW:

Ah, I think this is where the discrepancy lies. Let me re-phrase the question:

Do you agree that the surface is emitting about 390 W/m^2 and that all of this is radiative?

Yes, I agree that the surface is emitting (globally annually averaged) about 390 W/m2.

When people use the term “emitting” they are talking about emission of thermal radiation. So by definition all of the 390 W/m2 is radiative.

Also, heat transfer of about 100 W/m2 takes place by convection from the surface.

If yes, do you agree that the surface is emitting about 390 W/m^2 solely due to its temperature and nothing else (assuming an emissivity of 1 or very close to 1)?

Yes.

Do you agree that if the surface is emitting 390 W/m^2 that it cannot be receiving more energy than this?

The surface energy flux of radiation plus convection is more like 490 W/m2, so it is absorbing (globally annual averaged) 490 W/m2.

65. scienceofdoom,

“The surface energy flux of radiation plus convection is more like 490 W/m2, so it is absorbing (globally annual averaged) 490 W/m2.”

Ah, this is where the problem lies.

Do you agree that if the surface is emitting 390 W/m^2 that it cannot be receiving more energy than this in total?

I’m going to presume your answer is yes, even though you didn’t answer it directly. I do not disagree that there is an additional 100 W/m^2 of kinetic flux from latent heat and convection, but COE dictates that it must be a closed loop – meaning if 100 W/m@^2 is leaving the surface, 100 W/m^2 has to be returning at the same time, because all the energy leaving at the TOA is radiative and atmosphere cannot create any energy of its own.

The point is that the surface cannot be receiving more than 390 W/m^2. What leaves the surface in the kinetic form of latent heat and thermals is returned to the surface via precipitation, wind, weather, etc. If some of that kinetic energy is radiated into the atmosphere and eventually find its way radiated out to space, the amount of energy returning to the surface in kinetic form will be less than the amount that is leaving the surface. This net energy loss will cool the surface, reducing surface emitted radiation by an equal and opposite amount less, making no difference to the radiative budget.

Thus, latent heat and convection have no effect on the half up/half down effect of the surface emitted radiation that is absorbed by the atmosphere. They are just moving and redistributing energy around.

66. RW:

“The surface energy flux of radiation plus convection is more like 490 W/m2, so it is absorbing (globally annual averaged) 490 W/m2.”

Ah, this is where the problem lies.

Do you agree that if the surface is emitting 390 W/m^2 that it cannot be receiving more energy than this in total?

Yes it can be receiving more energy that this in total.

The first law of thermodynamics says:

Ein = Eout + Eretained

(where the terms have their obvious meaning).

Emission of thermal radiation, (let’s call it Rout is dependent on the temperature:

Rout = εσT4

Convective flux, (let’s call it Cout) is dependent on many factors but for the surface these are mainly:

-the wind speed
-the temperature of the wind
-the surface friction
-the availability of moisture at the surface
-the moisture content of air

There is no law of conservation of radiative flux

Ein = Rout + Cout + Eretained.

For total energy balance of the planet, only radiation can transfer energy between the climate system and space. So for the total planetary balance – radiation in = radiation out.

At the surface, conduction, convection and radiation can all play their respective roles, and there is no law that says each individually must balance.

And as an aside, convective flux can potentially be from the atmosphere to the surface, but as the surface is, on average, hotter than the atmosphere above it, convective flux is on average from the surface to the atmosphere.

67. on May 31, 2011 at 2:19 pm | Reply DeWitt Payne

RW,

Have you read Trenberth, Fasullo and Kiehl, 2009 published in BAMS and widely available on the web?

Click to access EarthsGlobalEnergyBudget.pdf

In it there is a diagram of the global average energy flows. The surface absorbs 161 W/m² from incoming solar radiation and 333 W/m² from atmospheric radiation for a total of 494 W/m². It then emits 396 W/m² IR radiation and transfers 97 W/m² by convection, 80 by latent and 17 by sensible for a total of 493 W/m². That leaves 0.9 W/m² causing increased heat content of the surface and the ocean. I think the absorption amount is too high, especially since ocean heat content has been flat for the last seven years. But for the purposes of your contention, it’s a minor quibble.

Now let’s look at the atmosphere. It absorbs 78 W/m² from incoming solar radiation, 356 W/m² from radiation emitted by the surface and 97 W/m² of convective energy for a total of 531 W/m². As I’ve shown above the atmosphere emits 333 W/m² downward and from the energy balance diagram emits 199 W/m² upward for a total of 532 W/m². The difference is caused by rounding errors.

Your contention that energy transfer by convection is not important is wrong. Without convection, the surface would be much hotter than it is now, ~305 K rather than 289 K (Teff assuming ε = 1).

68. science of doom says

“Yes it can be receiving more energy that this in total.”

Explain how. This is impossible, unless of course the surface is in the process of warming or cooling, which we have – for the purposes of discussion, assumed a steady-state condition.

You say:

“At the surface, conduction, convection and radiation can all play their respective roles, and there is no law that says each individually must balance. ”

Except when the only way energy enters and leaves the system is through radiation.

Let me ask you this question. The surface is emitting about 390 W/m^2, right? Without violating COE, explain where is this energy coming from?

69. Dewitt Payne says:

“Your contention that energy transfer by convection is not important is wrong. ”

I never said that is wasn’t important. It’s very important – it just doesn’t effect half up/half down at the boundaries of the surface and the TOA.

Convection is just moving kinetic energy around – from the surface to the atmosphere, from the atmosphere to the surface, or from the atmosphere to other parts of the atmosphere.

The point is convection is not moving energy from the surface to space. It and latent heat are closed loop circulation currents between the surface and the atmosphere. As Joules are being removed from the surface, Joules are also being returned at the same time. The net flux has to be zero if all the energy leaving at the TOA is radiative. Now, this does not mean the surface cannot exchange radiative for kinetic with the atmosphere and vice versa, but any exchange will be equally offset at the opposite end, making no difference to the energy budget, which is all radiative.

70. I wrote:

“Let me ask you this question. The surface is emitting about 390 W/m^2, right? Without violating COE, explain where is this energy coming from?”

And by this I mean show me the energy in = energy out calculations without counting any energy twice.

The surface is emitting 390 W/m^2. There is about 240 W/m^2 entering post albedo from the Sun and 240 W/m^2 leaving at the TOA.

• on June 1, 2011 at 1:09 am | Reply DeWitt Payne

RW,

You’ve apparently completely missed the basics of the greenhouse effect. What’s important is net energy transfer. The surface absorbs 161 W/m² coming in from solar radiation. Net emission is 396 up -333 down or 63 W/m² and 97 up from convection for a total of 160 W/m² + ~1 W/m² absorbed. I have counted nothing twice.

At the TOA, the atmosphere absorbs 78 W/m² from incoming sunlight. So total sunlight absorbed is 161 surface + 78 atmosphere = 239 W/m² total absorbed. Of the 160 W/m² net emission by the surface, 40 w/m² is emitted directly to space, the rest is absorbed by the atmosphere for a total of 78 + 120 = 198 W/m², which, when added to the 40 W/m² transmitted to space gives 238 W/m² out and 1 W/m² absorbed by the system.

There’s a whole series of articles here explaining in far more detail how this works. I suggest you start here as it’s a complete waste of bandwidth to do this all over again:

• Dewitt Payne,

I assure you I understand the GHE. I’m also very familiar with Trenberth diagram and I have read the paper more than once.

Trenberth’s diagram counts the energy designated as “absorbed by the atmosphere” twice, which is a violation of COE.

He has the surface receiving 161 W/m^2 directly from the Sun. He absorbs the remainder of the post albedo of 78 W/m^2 by the atmosphere and brings it to the surface as part of the 333 W/m^2 designated as ‘back radiation’. Only the 78 W/m^2 portion of this is not ‘back radiation’ but ‘forward radiation’ that last originated from the Sun yet to reach the surface (key distinction).

This is highly misleading, and I don’t know where he’s getting 78 W/m^2 absorbed by the atmosphere as that figure is not referenced in the paper and is way to high. The main point however is 239 W/m^2 from the Sun gets to the surface.

Another problem is he has 97 W/m^2 leaving the surface in the form of latent heat and thermals. He then lumps this in as part of the 333 W/m^2 return path of ‘back radiation, which is also misleading because it makes it look like of the 396 W/m^2 emitted at the surface, 333 W/m^2 is coming back from the atmosphere when in reality COE dictates that only 157 W/m^2 is coming back (157 + 169 + 70 = 396 and 239 + 157 = 396).

Of the 396 emitted at the surface, he has 70 W/m^2 passing straight through to space as if the atmosphere wasn’t even there (40 through the clear sky and 30 W/m^2 through the cloudy sky) and 169 W/m^2 emitted to space by the atmosphere. 70 + 169 = 239 leaving.

The 333 designated as ‘back radiation is composed of 78 W/m^2 from the Sun, 97 W/m^2 from latent heat and thermals and 157 from downward emitted LW that last originated from surface emitted (78 + 97 + 157 + 332). Trenberth purposefully has an extra watt in there.

In short, the diagram is very misleading and obfuscatory. Also, the 70 W/m^2 transmittance or ‘window’ is not referenced in the paper either, and appears to be just a rough estimate or guess. It also seems low, as other sources put this number more like 80-90 W/m^2. However, even using this number gives very close to half up half down (specifically, 169 W/m^2 up and 157 W/m^2 down or 52% up and 48% down). A window of 82 W/m^2 gives exactly half up/half down.

His surface emitted of 396 W/m^2 is also high. The satellite data is more like 287K or about 385 W/m^2 emitted at the surface.

Furthermore, 82 W/m^2 divided by 396 = 0.207 and (1-0.207)/2 = 0.397; 0.397 x 396 = 157 W/m^2, exactly the downward emitted amount. Is this just a coincidence? If you think so, show why.

• Also, 1-0.397 = 0.603 and 0.603 x 396 = 239, exactly the post albedo amount from the Sun. Is this another coincidence?

71. Take a look at this. Another researcher, using what appears to be much more accurate methodology than Trenberth or Miskolczi, calculates a transmittance of 93 W/m^2 and gets exactly half up and half down:

72. RW:

You have made a few comments since my comment of May 31, 2011 at 5:34 am.

Let’s go back to your initial response to it.

You had said:”Do you agree that if the surface is emitting 390 W/m^2 that it cannot be receiving more energy than this in total?

I had responded, of course it can.

Explain how. This is impossible, unless of course the surface is in the process of warming or cooling, which we have – for the purposes of discussion, assumed a steady-state condition.

You seem confused about something. But I’m not sure what it is.

The surface is absorbing 490 W/m2 and 490 W/m2 is leaving the surface by convection + radiation.

That means it is not cooling or warming.

So in answer to “Explain how..“.

Because energy in = energy out. That’s how.

If the surface was only absorbing 390 W/m2 then the convected flux + radiative flux = 390 W/m2.

I can’t fathom which part you don’t understand or disagree with.

• OK, let me explain.

I want to go back to one of my original questions:

“Do you agree that the surface is emitting about 390 W/m^2 solely due to its temperature and nothing else (assuming an emissivity of 1 or very close to 1)?”

You answered yes to this question, which I asked for a very specific reason.

Now, think about this very carefully. If the surface is emitting 390 W/m^2, then it must have an incoming energy flux of 390 W/m^2 in the steady-state, right? There is 240 W/m^2 coming in from the Sun and 240 W/m^2 leaving at the TOA, right? Where is the additional incoming flux of 150 W/m^2 at the surface coming from to get the total of 390 W/m^2 (240 + 150 = 390)? This has to be coming from somwhere, right?

Your convection/latent heat flux of 100 W/m^2 is a net zero flux at the surface, because 100 W/m^2 is leaving the surface and 100 W/m^2 is returned to surface. In other words, 100 W/m^2 is entering the atmosphere from the surface and 100 W/m^2 entering the surface from the atmosphere.

However, at the TOA, only 240 W/m^2 is entering and leaving, so yes energy in = energy out, but where is the additional incoming 150 W/m^2 flux at the surface coming from?

73. on June 1, 2011 at 5:00 am | Reply DeWitt Payne

RW,

I assure you I understand the GHE. I’m also very familiar with Trenberth diagram and I have read the paper more than once.

Clearly you don’t understand the diagram or the GHE as evidenced by the following quote.

Trenberth’s diagram counts the energy designated as “absorbed by the atmosphere” twice, which is a violation of COE.

Nope. There is no double counting or violation of the First Law.

He absorbs the remainder of the post albedo of 78 W/m^2 by the atmosphere and brings it to the surface as part of the 333 W/m^2 designated as ‘back radiation’. Only the 78 W/m^2 portion of this is not ‘back radiation’ but ‘forward radiation’ that last originated from the Sun yet to reach the surface (key distinction).

Nope. All the energy for the 333 W/m² downward radiation comes from the 356 W/m² (396-40) surface upward radiation absorbed by the atmosphere.

The surface absorbs 494 W/m² and emits 493 W/m² for a net absorption of ~1 W/m². At the top of the atmosphere 239 W/m² is absorbed 238 W/m² is emitted. Energy is conserved and energy flows balance. This is really basic stuff and I don’t feel the need to spend any more time trying to make you understand this rather simple concept.

74. RW:

I think that one issue about the climate system is pushing your judgement “out of whack”. But I will return to that in a later comment.

The surface emits according to the equation:

Rout = εσT4

Let’s suppose – as another quick diversion to establish an important principle – that we have heated up a surface to 15’C and now all sources of energy are removed (it’s floating in the vastness of space, for example).

Ein = 0.
Rout = 390 W/m2 (assuming ε = 1).

So no energy is required for the surface to emit at 390 W/m2. (Of course in this case it will cool down).

Equally, we can have a situation when the surface, with a high heat capacity, is absorbing well above 390 W/m2. But Rout = 390 W/m2, until the surface temperature increases, which might take seconds, years or centuries.

Emission of thermal radiation in any given second is only dependent on the surface temperature (and emissivity).

Let’s confirm that you agree with these points.

End of quick diversion.

I have to ask it because your statement is still wrong:

If the surface is emitting 390 W/m^2, then it must have an incoming energy flux of 390 W/m^2 in the steady-state, right?

As I have already stated and you haven’t directly contradicted – Energy in – Energy out = Energy retained.

And, therefore, in steady state, Energy in = Energy out.

This is not the same equation you are implicitly stating.

You are stating, for steady state, Radiation in = Radiation out.

This is NOT the first law of thermodynamics. But you seem to believe it.

Please comment on this specific point so I can be clear what exactly you are thinking.

• scienceofdoom says:

“The surface emits according to the equation:

Rout = εσT4

Let’s suppose – as another quick diversion to establish an important principle – that we have heated up a surface to 15′C and now all sources of energy are removed (it’s floating in the vastness of space, for example).

Ein = 0.
Rout = 390 W/m2 (assuming ε = 1).

So no energy is required for the surface to emit at 390 W/m2. (Of course in this case it will cool down).

Equally, we can have a situation when the surface, with a high heat capacity, is absorbing well above 390 W/m2. But Rout = 390 W/m2, until the surface temperature increases, which might take seconds, years or centuries.

Emission of thermal radiation in any given second is only dependent on the surface temperature (and emissivity).”

Yes I agree, except to the extent that as the surface cools down or warms up it will emit less or more than 390 W/m^2.

You also say:

“As I have already stated and you haven’t directly contradicted – Energy in – Energy out = Energy retained.

And, therefore, in steady state, Energy in = Energy out.

This is not the same equation you are implicitly stating.

You are stating, for steady state, Radiation in = Radiation out.

This is NOT the first law of thermodynamics. But you seem to believe it.”

Now you seem to be playing a semantics game. Yes, ‘steady-state’ requires energy in = energy out (i.e. radiation in = radiation out), but in the context here it also implies a constant surface temperature or surface emitted radiative flux, which in this case is 390 W/m^2.

So my defined ‘steady-state’ is 240 W/m^2 entering at the TOA, 240 W/m^2 leaving at the TOA, and 390 W/m^2 emitted at the surface (all of which are radiative fluxes).

Does this clarify everything?

75. scienceofdoom says:

“You had said:”Do you agree that if the surface is emitting 390 W/m^2 that it cannot be receiving more energy than this in total?”

I had responded, of course it can.”

Sorry, I should have said “…cannot be receiving a NET total of more energy than this.”

76. You also say:

“As I have already stated and you haven’t directly contradicted – Energy in – Energy out = Energy retained.

And, therefore, in steady state, Energy in = Energy out.

This is not the same equation you are implicitly stating.

You are stating, for steady state, Radiation in = Radiation out.

This is NOT the first law of thermodynamics. But you seem to believe it.”

Now you seem to be playing a semantics game. Yes, ‘steady-state’ requires energy in = energy out (i.e. radiation in = radiation out), but in the context here it also implies a constant surface temperature or surface emitted radiative flux, which in this case is 390 W/m^2.

What is semantic about trying to make sure we agree about definitions? If you believe it is semantic then perhaps you haven’t understood what it is I am asking – or why I am asking it.

You still haven’t actually answered my question, which is not at all “semantic games”.

Does the First Law of Thermodynamics require Radiation in = Radiation out for steady state? (“steady state” = no change in temperature).

If you believe it does, you need to find a reference like a textbook, because if you every definition and explanation of the first law I have found only refers to conservation of energy, not conservation of radiation.

If you don’t believe it does as a general rule, then:

First, please confirm that point so I am sure of this important common ground, and

Second you need to explain why it is that energy in = energy out means that radiation in = radiation out.
By what general rule or specific circumstance is convection = 0?

So my defined ‘steady-state’ is 240 W/m^2 entering at the TOA, 240 W/m^2 leaving at the TOA, and 390 W/m^2 emitted at the surface (all of which are radiative fluxes).

Does this clarify everything?

No it doesn’t clarify everything, we happily agree about TOA, but not about the surface.

I keep saying that “radiation in” is not required to equal “radiation out” when convection also takes place.

You haven’t explained why you believe it is required by the first law of thermodynamics, so I don’t know whether you don’t understand this first law, or are misapplying it for this application.

Nothing is clear until you explain what is wrong with the equation I wrote earlier:

Ein = Rout + Cout + Eretained.

In steady state, Eretained=0.

⇒ Ein = Rout + Cout

⇒ if Cout ≠ 0, Ein ≠ Rout

This is very simple and not semantics.

77. on June 1, 2011 at 4:52 pm | Reply DeWitt Payne

RW,

Is part of your problem that you have the idea that if the atmosphere absorbs 78 W/m² then it must emit 34 up and 34 down? If so, then you’re wrong. The atmosphere is not a thin isothermal shell. It’s optically thick and the temperature varies. In the troposphere, which is the most important region for radiative energy transfer, the temperature decreases with altitude. That means for a given total radiative flux, the top of the troposphere will emit less than the bottom.

But for accounting purposes in an energy balance diagram, it really doesn’t matter. Let’s say that emission from atmospheric solar absorption is split equally. So now we have 195 (161 + 34 ) W/m² of solar radiation absorbed by the surface. But we have to subtract the 34 W/m² from the rest of the downwelling IR and so the surface absorbs an additional 299 W/m² of IR from the atmosphere for a total of 195 + 299 = 494 W/m² and we’re back to where we started. The same thing can be done at the top of the atmosphere.

78. on June 1, 2011 at 6:37 pm | Reply DeWitt Payne

RW,

For completeness, let’s look at just the atmosphere:

Incoming solar absorption: 78 W/m²
Upward surface radiation absorption: 356 (396-40) W/m²
Convective energy absorption: 97 W/m²
Total absorption: 531 W/m²

Upward atmospheric emission: 198 (238-40) W/m²
Downward atmospheric emission: 333 W/m²
Total atmospheric emission: 531 W/m²

Ein = Eout

Teff TOA looking down (atmosphere only) =(198/σ)^0.25 = 243 K
Teff surface looking up = (333/σ)^0.25 = 277 K

Where’s the problem?

• The problem is the atmosphere cannot be absorbing or emitting 531 W/m^2 as the surface is only emitting 396 W/m^2 and only 239 W/m^2 is entering and leaving at the TOA.

Where is the extra energy coming from? With regard to the energy flows in and out, all the energy in or flowing through the atmosphere either last originated from the Sun or last originated from the surface in the past.

The atmosphere emits 169 W/m^2 to space and 70 W/m^2 from the surface emitted of 396 W/m^2 passes straight through the atmosphere to space (169 + 70 = 239 W/m^2 leaving at the TOA).

The surface emits 396 W/m^2 and 70 W/m^2 passes straight through the atmosphere to space. The difference of 326 W/m^2 (396 – 70 = 326) is the amount of surface emitted radiation that is absorbed by the atmosphere. Of the 326 W/m^2 absorbed by the atmosphere, 169 W/m^2 is emitted to space and the remainder of 157 W/m^2 is emitted down to the surface (70 + 169 + 157 = 396).

The way Trenberth counts energy twice is by absorbing 78 W/m^2 of the 239 W/m^2 from the Sun by the atmosphere and then brings it to the surface lumped in as part of the 333 W/m^2 of ‘back radiation’. Only it’s not ‘back radiation’, but ‘forward radiation’ that last originated from the Sun yet to reach the surface. 78 + 97 + 157 = 332 W/m^2 all lumped in the return path the surface designated as ‘back radiation’. Doing this violates COE, because if 78 W/m^2 is absorbed by the atmosphere, it has to be going somewhere – it can’t be staying in the atmosphere, nor can it be radiated out to space, because the energy sources of those flows are already accounted for.

79. scienceofdoom says:

“What is semantic about trying to make sure we agree about definitions? If you believe it is semantic then perhaps you haven’t understood what it is I am asking – or why I am asking it.

You still haven’t actually answered my question, which is not at all “semantic games”.”

OK, I apologize about the ‘semantics’ remark. I clearly see that you are just looking for clarification.

80. scienceofdoom says:

“Does the First Law of Thermodynamics require Radiation in = Radiation out for steady state? (“steady state” = no change in temperature).

If you believe it does, you need to find a reference like a textbook, because if you every definition and explanation of the first law I have found only refers to conservation of energy, not conservation of radiation.

If you don’t believe it does as a general rule, then:

First, please confirm that point so I am sure of this important common ground”

I’m not sure I understand the question or your point. The first law says energy must be conserved.

My point is energy can only enter and leave the system by radiation. Energy cannot be convected or conducted out to space.

81. scienceofdoom says:

“Second you need to explain why it is that energy in = energy out means that radiation in = radiation out.

By energy in = energy out, I mean the energy entering and leaving the earth’s thermal mass at the TOA. The reason why in this particular case it means radiation in = radiation out is because the only way energy can enter and leave the thermal mass of the Earth is by radiation.

You also write:

“By what general rule or specific circumstance is convection = 0?”

By Conservation of Energy. We have agreed that energy cannot be convected out to space at TOA, right? All the energy entering and leaving is radiative.

If energy is convected from the surface into the atmosphere and some of that energy ends up radiated out to space, the amount of energy returning to the surface by convection will be less, right? Because energy has to be conserved, this will reduce the energy at the surface, cooling the surface and reducing surface emitted by an equal and opposite amount less than the more that was radiated out at the TOA. The same is true in reverse.

Latent heat of water is no different. Whatever amount leaves the surface and does not return (i.e. ends up radiated to space), will have a cooling effect, which will reduce surface emitted by and equal and opposite amount less.

In the case of latent heat, virtually all of this is returned in the form of precipitation, weather, etc. If there is an imbalanced, it will be equally offset at the opposite end (either at the surface or the TOA).

Does this explain it better?

82. The point of Aa=Ed or half up/half down, is that at the boundaries of the surface and TOA, the physics requires the split to be 50/50. Meaning whatever amount of outgoing surface radiation is absorbed by the atmosphere, half will go to the surface and half will go to space. This is because when a GHG molecule or cloud absorbs and re-emits a photon, it’s flung equally in all directions. Or if the photons absorbed energy is transferred via collisions by other gases in the atmosphere, those heated gases will radiate according the laws of black body radiation – also equally in all directions. Ultimately, this means half of what the atmosphere absorbs goes up out to space and half goes down to the surface.

Now, in between boundaries of the surface and TOA , the exchange of energy does not have to be a 50/50 split, because other transfers of energy by convection and conduction are moving energy around non-radiatively.

83. Also, if you haven’t already, take a look at the results derived in this paper:

If you think these results are just coincidence, explain why.

84. Some of the confusion may lie in the fact that the kinetic energy in the form of latent heat of water moved from the surface into the atmosphere acts primarily to redistribute energy from the tropics to the higher latitudes. What goes up must come down. The water condenses in the atmosphere to form clouds and is eventually returned to the surface in the form of precipitation. The key thing to understand is that as joules are being removed from the surface, joules are also being returned to the surface at the same time.

85. RW on June 1, 2011 at 9:42 pm:

As best as I can tell, you have invented a new law of thermodynamics. I summarize it in 3 points.

1. For every body space is the ultimate boundary of the encompassing system.
2. Radiation can only leave and enter the ultimate boundary via radiation.
3. Therefore, convective heat flux from any body within the system must be zero.

Is this your hypothesis?

I agree with steps 1 & 2.

You have invented step 3 and it appears you think it is obvious. But it’s not true and this is why heat transfer textbooks are full of conduction and convection calculations.

The surface of the earth has a net convective loss to the atmosphere. This is returned via radiation.

We can measure the radiation imbalance. We can demonstrate the convective heat transfer from the surface.

Yes, the surface would be hotter if there was no convection. That is because convection is a net heat transfer from the surface to the atmosphere. If convection = 0, the surface would be hotter.

The fact that, at the ultimate boundary of the climate system, heat is only transferred via radiation has no bearing on what heat transfer mechanisms operate at the surface.

You need to draw a diagram with the surface, some layers of the atmosphere and a TOA boundary, and then write out some equations for each layer.

It will make it easier to explain where your flaw lies. It will be easier to demonstrate your conceptual problem to you. Try writing it down with arrows, values and equations. And more than one atmospheric layer.

• scienceofdoom,

I mentioned several times in this thread that COE dictates that atmosphere cannot create any energy of its own. You dismissed this as something obvious and that you already knew, but it is apparent to me that you don’t understand the constraint COE puts on the system and the energy flows at the boundaries of the surface and the TOA.

I have not invented any new laws of thermodynamics. I have simply applied the first law to the system we’re discussing.

I asked this question before and you didn’t answer. If 240 W/m^2 are entering and leaving at the TOA and the surface is emitting 390 W/m^2, where is the 150 W/m^2 coming from (390 – 239 = 150)???

• In other words, where is this 150 W/m^2 flux entering the surface coming from if only 240 W/m^2 are entering the surface from the Sun?

• Technically, the atmosphere has virtually an infinite number of ‘layers’. I make no contention that between any two particular layers in the atmosphere, the the total energy exchange or flow will be half up and half down. Of course, there are large amounts of non-radiative fluxes between layers exchanged by convection.

What I’m saying is at the boundaries of the surface and the TOA, the split must be 50/50 if energy is conserved and the laws of black body radiation are adhered to. Is this what you are disagreeing with? Are you sure you understand it.

Might I ask you to examine the paper I linked and explain – if it’s not true, why the numbers work out as the do?

86. scienceofdoom,

Let me try to approach this from a different angle. The effective emissivity of the Earth is referenced as being about 0.60, correct? What is the physical meaning of this quantitatively?

87. RW

I mentioned several times in this thread that COE dictates that atmosphere cannot create any energy of its own. You dismissed this as something obvious and that you already knew, but it is apparent to me that you don’t understand the constraint COE puts on the system and the energy flows at the boundaries of the surface and the TOA.

I have not invented any new laws of thermodynamics. I have simply applied the first law to the system we’re discussing.

To apply these laws you need to prove why convective flux = 0 at the surface.

Hand waving arguments are not demonstrations.

I have written down an equation and you just keep claiming it can’t be true because of something happening at a totally different boundary.

Saying “you don’t understand the constraint COE puts on the system” is not proof.

You haven’t proved or even attempted to prove to convective flux = 0. You keep asserting it.

I asked this question before and you didn’t answer. If 240 W/m^2 are entering and leaving at the TOA and the surface is emitting 390 W/m^2, where is the 150 W/m^2 coming from (390 – 239 = 150)???

It’s simple.

There is no “conservation of power” law in thermodynamics. There is a conservation of energy law.

As a result of the surface of the planet heating up (long ago) due an imbalance in energy in – energy out, it has retained more heat.

Now that various parts of the planet are approximately in equilibrium, energy flows per second (power) balance.

At the surface, energy in per second = energy out per second.

Energy in = 170 W/m2 from the sun + 320 W/m2 from the atmosphere = 490 W/m2

Energy out = 390 W/m2 radiated + 100 W/m2 convected.

The original energy that has increased the surface temperature has come from the sun.

Now the surface is in balance, the atmosphere is in balance and the top of atmosphere (ie the whole climate system) is in balance.

And no – energy has not been created.

If you want an over-simplified soundbite (aimed at educating rather than technical accuracy) – energy just can’t get out.

Perhaps you might join the throng of people convinced it’s wrong but unable to demonstrate it.

And before commenting, ask yourself why so many people convinced this simple example is wrong all have different explanations as to why (take a look at the comments). And why no one is able to write out the simple equations and solve them to get a different answer.

Except for the guy who wrote his article with equations where radiation magically transferred through the PVC sphere. Then retracted his article.

• Sorry, I replied to the wrong post, so I’m putting my reply here (feel free to delete the one in the incorrect location):

scienceofdoom says:

“To apply these laws you need to prove why convective flux = 0 at the surface.

Hand waving arguments are not demonstrations.”

I don’t think I’m hand waving. Technically, it’s probably not zero relative to the surface to the atmosphere and to the atmosphere back to the surface. Is this what you mean? The point is relative to the radiative fluxes of 390 W/m^2 from the surface and 240 W/m^2 at the TOA, the net energy flux must be zero at the surface if energy is to be conserved.

Why do you not see that if more energy is convected from the surface to the atmosphere than is convected back to the surface, the surface will lose energy and subsequently radiate less?

We are talking about the system in the steady-state, are we not? That means the surface is not cooling or warming and the energy entering and leaving at the TOA is equal (or that the system is in equilibrium).

• So where is the problem?

88. Let me try to approach this from a different angle. The effective emissivity of the Earth is referenced as being about 0.60, correct? What is the physical meaning of this quantitatively?

Not correct.

Emissivity is a wavelength dependent parameter. Assuming that you are talking about the wavelengths at which the earth emits (longwave) it is more like 0.9 – 0.96 depending on which surface we are considering.

The oceans are around 0.96.

The physical meaning of it is explained in Planck, Stefan-Boltzmann, Kirchhoff and LTE.

In essence it is the proportion of radiation emitted compared with a blackbody.

• I’m referring to the effective emissivity of Earth, not the emissivity of the surface of the Earth. Wikipedia references the effective emissivity of the Earth as 0.612:

http://en.wikipedia.org/wiki/Climate_model#Zero-dimensional_models

Do you know what this means?

• In other words, do you know what the physical meaning of this is quantitatively?

• on June 2, 2011 at 2:56 am DeWitt Payne

RW,

A zero dimensional model has no physical meaning. Therefore the ‘effective emissivity’ of 0.612 also has no physical meaning. It’s a mathematical construct to get a surface temperature of 288 K from the Stefan-Boltzmann equation. The emissivity of clouds is 1 in the IR not 0.5. Assigning a value of 0.5 is necessary to obtain an average of 0.612 for the planet. It also is a mathematical construct that has no physical meaning.

As SoD has pointed out, there is no law of conservation of power. No energy needs to be created to have the atmosphere absorb and emit 531 W/m². Energy in = energy out so energy content doesn’t change. There is no violation of conservation of energy because the total energy doesn’t change.

89. Dewitt Payne says:

“A zero dimensional model has no physical meaning. Therefore the ‘effective emissivity’ of 0.612 also has no physical meaning.”

No, it most certainly has a very specific physical meaning, as there is a very specific physical reason why the effective emissivity is what it is.

90. scienceofdoom says:

“Now that various parts of the planet are approximately in equilibrium, energy flows per second (power) balance.

At the surface, energy in per second = energy out per second.

Energy in = 170 W/m2 from the sun + 320 W/m2 from the atmosphere = 490 W/m2

Energy out = 390 W/m2 radiated + 100 W/m2 convected.”

With all due respect, I do not think you understand the Law of Conservation of Energy as it applies to the boundary between the surface and the TOA.

I’m running out of angles to try to explain it. At this point, I think I’ll invite Miskolczi to this thread and see if he can shed any light on these things.

91. “Now that various parts of the planet are approximately in equilibrium, energy flows per second (power) balance.

At the surface, energy in per second = energy out per second.

Energy in = 170 W/m2 from the sun + 320 W/m2 from the atmosphere = 490 W/m2

Energy out = 390 W/m2 radiated + 100 W/m2 convected.”

Another shot: Where is the 150 W/m^2 coming from (240 – 170 = 70; 320 – 70 = 250; 250 – 100 = 150)???

• on June 2, 2011 at 4:31 am | Reply DeWitt Payne

You do understand that the downward flux of 333 W/m² is not invented out of whole cloth to make up the energy balance. It can be and is measured by a variety of instruments as well as calculated by radiative transfer programs including Miskolczi’s HARTCODE. The same goes for upward radiation and convection. Convection is harder to measure, but the latent part can be confirmed by comparison to total annual rainfall and the heat of vaporization of water.

Your supposed missing 150 W/m² comes from the 326 W/m² absorbed upward emission plus 78 W/m\$#178; solar absorbed and 97 W/m&#178 convection. 501 W/m² in and 168 + 333 = 501 W/m² out.

92. on June 2, 2011 at 4:14 am | Reply DeWitt Payne

RW,

The atmosphere emits 169 W/m^2 to space and 70 W/m^2 from the surface emitted of 396 W/m^2 passes straight through the atmosphere to space (169 + 70 = 239 W/m^2 leaving at the TOA).

Good so far.

The surface emits 396 W/m^2 and 70 W/m^2 passes straight through the atmosphere to space. The difference of 326 W/m^2 (396 – 70 = 326) is the amount of surface emitted radiation that is absorbed by the atmosphere. Of the 326 W/m^2 absorbed by the atmosphere, 169 W/m^2 is emitted to space and the remainder of 157 W/m^2 is emitted down to the surface (70 + 169 + 157 = 396).

You’re starting to get into trouble here by assuming you can assign energy flows and not doing it proportionally. But the accounting is still not completely wrong.

The way Trenberth counts energy twice is by absorbing 78 W/m^2 of the 239 W/m^2 from the Sun by the atmosphere and then brings it to the surface lumped in as part of the 333 W/m^2 of ‘back radiation’. Only it’s not ‘back radiation’, but ‘forward radiation’ that last originated from the Sun yet to reach the surface. 78 + 97 + 157 = 332 W/m^2 all lumped in the return path the surface designated as ‘back radiation’. Doing this violates COE, because if 78 W/m^2 is absorbed by the atmosphere, it has to be going somewhere – it can’t be staying in the atmosphere, nor can it be radiated out to space, because the energy sources of those flows are already accounted for.

No, they’re not. You have 396 W/m² radiating from the surface plus 97 W/m² convecting for a total of 493 W/m², but have as input only 157 W/m² downward radiation from upward absorbed and 161 W/m² from solar radiation absorbed for a total of 318 W/m². Energy balance requires an additional 175 W/m². Instead of TF&K double counting, you’re not counting enough. Not coincidentally the 78 W/m² of solar radiation absorbed by the atmosphere plus the 97 W/m² convected from the surface provides exactly this amount.

Trying to tag individual energy flows is not a good idea. It’s better to just look at the totals. But if you do want to tag, you should keep the proper ratios. Atmospheric radiation up is 169 W/m² and down is 333 W/m². So any energy absorbed by the atmosphere should be emitted in the same proportion: 0.337 up and 0.663 down.

So doing it that way, of the 326 W/m² absorbed from the surface, 109.7 W/m² goes up and 216.3 W/m² goes down.
That leaves 169-109.7 =59.3 W/m² of upward radiation unaccounted for and 333 – 216.3 = 116.7 W/m² of downward radiation. Now let’s add the 78 W/m² of solar radiation absorbed by the atmosphere. That’s 26.3 W/m² up and 51.7 W/m² down, leaving 33 W/m² up and 65 W/m² down. Finally, add the 97 W/m² of convection and get 32.7 W/m² up and 64.3 W/m² down or a near perfect balance. No double counting of anything and no violation of conservation of energy.

93. RW:

I don’t think I’m hand waving.

You’re not writing down an equation for a body or surface, explaining what physical law requires this equation and then providing a solution.

You are handwaving.

Conservation of energy refers to a body. You can pick any “body” that you like.

If we refer to the surface as our body of choice (perhaps with the inclusion of everything inside), conservation of energy must be written for this surface.

I have written the equation and you keep claiming that I am not understanding conservation of energy.

Write down your equation then it will clearer where your problem lies.

Technically, it’s probably not zero relative to the surface to the atmosphere and to the atmosphere back to the surface. Is this what you mean? The point is relative to the radiative fluxes of 390 W/m^2 from the surface and 240 W/m^2 at the TOA, the net energy flux must be zero at the surface if energy is to be conserved.

You have written a meaningless jumble of words.

Net energy for conservation of energy relates to “a body” = a defined system.

⇒ Energy in = energy out – energy retained.

Apply it to the surface (I’ve already done this).

You are making some statement about the surface of the earth “relative to the 240W/m^2 at the TOA” – which doesn’t really mean anything.

If we include the whole climate system then we have a body in radiative steady state. Energy in = energy out and radiation in = radiation out.

If we include a subset of the whole climate system we have a body in steady state, where energy in = energy out and radiation in ≠ radiation out unless convection and conduction are zero for that subset.

Why do you not see that if more energy is convected from the surface to the atmosphere than is convected back to the surface, the surface will lose energy and subsequently radiate less?

Because it’s not true as I have already explained.

I said: “The surface of the earth has a net convective loss to the atmosphere. This is returned via radiation.

This is called conservation of energy.

If energy is convected from the surface it can be supplied back by other means.

This is basic heat transfer theory. Take a look at Heat Transfer Basics.

Any heat transfer textbook will help you understand this, they have lots of worked examples and all of the worked examples rely on conservation of energy, not conservation of radiation, and not “conservation of radiation of one body because of a completely different system boundary requires conservation of radiation there”.

The end.

If you find a textbook with something different let me know. Otherwise not much more for me to say.

• scienceofdoom,

It seems we are at or very close to an impasse. I’ve tried to explain it from a multitude of angles, but I’m running out of angles to try.

Let’s take this of yours:

[quote]”I said: “The surface of the earth has a net convective loss to the atmosphere. This is returned via radiation.”

This is called conservation of energy.”[/quote]

OK, I do not dispute that the surface of the Earth has net convective loss to the atmosphere. However, how is it that all of this loss is returned to the surface via radiation?

Are you saying the convected energy into the atmosphere does not radiate according to the laws of black body radiation (i.e. equal in all directions)?

And even it were all returned via radiation, as you claim, this would make the net energy flux zero, right?

Let me rephrase the initial question you quoted this way:

“Why do you not see that if more energy is convected from the surface to the atmosphere than is convected from the atmosphere back to the surface, and some of that convected energy from the surface to the atmosphere is radiated out to space, the surface will lose energy, cool down, and subsequently emit less radiation?

• BTW, how do you do those tabed in quotes in your replies? Can you fix my post to display your quote like that?

94. RW:

No, it most certainly has a very specific physical meaning, as there is a very specific physical reason why the effective emissivity is what it is.

“Effective emissivity” as you referencing it is not a general term in use.

Emissivity is real term which measures real material properties. It varies with wavelength and, in some cases, direction.

If someone has created a term “effective emissivity” they are welcome to do so but it’s usefulness will be decided by how many people take it on in the scientific community, which will be determined by the physical insight it gives.

“Effective emissivity” is a ratio unrelated to the real property called emissivity. Because it is in Wikipedia doesn’t mean it is correct, or useful.

• The Earth’s “effective emissivity” of about 0.61 or 0.62 is not a controversial or disputed thing. It is an empirically derived figure that has a specific physical meaning, just as the Earth’s “effective temperature” of 255K has a very specific physical meaning.

I just did a quick search in Wiki to get one source for the figure. If you Google it, you’ll find multiple sources in the literature giving roughly the same number.

That you don’t seem to know what it is or specifically means may be the reason why you don’t seem to understand the constraints COE puts on the boundary between the surface and the TOA.

• on June 2, 2011 at 6:28 am DeWitt Payne

Actually, effective emissivity means many things. In the scientific literature it’s usually the emissivity averaged over a range of wavelengths. In one paper on clouds, it was 1-transmissivity because thin clouds can be partially transparent to IR. But that’s just Kirchhoff’s Law, (1-t) = absorptivity = emissivity. Calculating an emissivity to give a T of 288 K when the measured Teff is 254 K is simply a mathematical exercise and the resulting emissivity does not, in fact, have a physical meaning because it’s not the real emissivity. If it were, then the Earth would have to have a reflectivity of 0.388 in the IR since
a + r + t ≡ 1 a = ε and the Earth isn’t transparent to IR so t = 0.

• I’m good with average emissivity. Just not so happy with “invented emissivity” which is totally different from emissivity.

Of course, that definitely means I don’t understand conservation of energy and its constraints.

RW has definitely nailed it there.

Of course an equation that demonstrates that convection from the surface has to be zero (on average) would be nice but probably not really necessary now the whole invented emissivity thing has come to light.

95. RW on June 2, 2011 at 4:58 am:

The atmosphere is not a blackbody.

Any given portion of the atmosphere does radiate equally in all directions. It doesn’t need to be a blackbody to do that.

Yet the atmosphere at different levels radiates different flux dependent on the concentration of “greenhouse” gases and the temperature. Outgoing longwave radiation at TOA is very different from longwave radiation at the surface.

You can see some examples in Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Ten.

When you write down some equations I will have a look at them.

96. on June 2, 2011 at 5:34 am | Reply DeWitt Payne

RW,

With all due respect, I do not think you understand the Law of Conservation of Energy as it applies to the boundary between the surface and the TOA.

There is no boundary between the surface and the TOA. There’s a boundary between the surface and the atmosphere and another boundary between the TOA and space.

I’m running out of angles to try to explain it. At this point, I think I’ll invite Miskolczi to this thread and see if he can shed any light on these things.

Don’t look to him for help. He disputes the specific values of TF&K, but not the overall thrust. I suggest you look at M’s Figure 1 reproduced above and compare TF&K’s numbers with the numbers in Table 2 of M2010.

Here, I’ll help:

variable M2010(GAT) TF&K09

ta 286.04K 288.2K
Su 379.64 396 W/m²
Eu 192.7 169
Ed 310.5 333
St 58.54 70 (40 surface + 30 cloud)
OLR 251.25 238.5
K + F 182.1(=Eu-(Aa-Ed) 175
Aa 321.1 326
If we assign the value of K equal to TF&K of 97 W/m² then
F 85.1 78
Aa + F 503.2 501
Assuming radiative steady state, OLR = Fo so
Fo-F 166.15 160.5
Ed+(Fo-F) 476.65 493.5

Note that Miskolzci’s numbers don’t follow his own rules. Aa = Su -St = 321.1 ≠ Ed = 310.5, which required a correction to K + F which M claims is equal to Eu but isn’t in this case.

Unfortunately, the formatting is going to make the data hard to read. Still, it’s pretty obvious that the numbers are similar and in some cases quite close.

• on June 2, 2011 at 6:08 am | Reply DeWitt Payne

Thanks for fixing the blockquote thing.

My second sentence should read: There’s a boundary between the surface and the atmosphere and another boundary between the TOA and space. [inserted text italicized]

• I’ve asked the great people at WordPress when they will be bringing in comment preview for hosted blogs. I’m lucky, I can fix my own typos when I see them..

97. DeWitt Payne helpfully pointed out to RW:

You do understand that the downward flux of 333 W/m² is not invented out of whole cloth to make up the energy balance. It can be and is measured by a variety of instruments..

Lots of measurements of downward longwave radiation from the atmosphere can be seen in The Amazing Case of “Back Radiation” -Part One.

It is usually around this point (for those who take a look at the data presented) that people confused about the first law of thermodynamics, or the second law, suddenly decide that pyrgeometers are not to be trusted.

And not just by a percentage point or two. The data is just totally wrong. “It has to be. After all, I understand thermodynamics and I believe different results would be obtained. The measurements are wrong.

98. on June 2, 2011 at 7:05 am | Reply DeWitt Payne

As far as preview, I went back and looked at CA Assistant (Greasemonkey script for Firefox) and discovered there were instructions on adding new sites. I put yours in and it works, sort of. There are duplicate buttons and the preview button is partially hidden. If I knew anything about scripting, I could probably fix that, but I don’t. Still something is better than nothing. Too bad I didn’t think of this earlier.

Let’s see which buttons work and which don’t:

bold

italic

Blockquote

X2

X2

less than <

strikethrough

underline

I’m pretty sure you don’t have TeX and I can’t insert an image so I won’t try those.

Underline doesn’t work in preview, but I’ll try it anyway.

99. on June 2, 2011 at 7:11 am | Reply DeWitt Payne

So no superscript, subscript or underline. I didn’t expect super- or subscript to work. The good thing is that after I posted, the duplicate buttons went away. There’s also a nice reply to feature that isn’t working here either.

Could you possible change the color of links or underline them or something. Maybe it’s my monitor, but the green color of a link isn’t very different from the black of the text.

100. You can insert an image.

‘less than’ img src=”link” ‘greater than’ : works.
‘less than’ =

I can’t change the green underline link – although I agree with you – without changing the whole theme.

Or maybe someone knows how to adjust the theme in a hosted WordPress blog..

101. scienceofdoom says:

“Of course an equation that demonstrates that convection from the surface has to be zero (on average) would be nice but probably not really necessary now the whole invented emissivity thing has come to light.”

There are some very basic equations in that paper I referenced. Have you looked at them? Tell me why you think they’re wrong and we’ll go from there.

BTW, the ‘effective emissivity’ of the Earth just means the actual emissivity or the emissivity of the entirety of the Earth’s thermal mass, including the atmosphere.

I’ll give you hint. 240/390 = 0.615

That you don’t know what the physical meaning of this is – is why you’re apparently still confused.

• If it makes you feel better to make these kind of childish comments you are on the wrong blog.

I have pointed out that effective emissivity is something quite different from real emissivity which is a measured parameter. Effective emissivity as you are using the term is not a real material property, it is a ratio which is not related to the actual emissivity of any part of the system.

102. scienceofdoom says:

“The atmosphere is not a blackbody.

Any given portion of the atmosphere does radiate equally in all directions. It doesn’t need to be a blackbody to do that.”

Yes, you’re correct the atmosphere is not a black body – but a grey body. Sorry, the point is its radiates isotropically.

103. scienceofdoom says:

“Yet the atmosphere at different levels radiates different flux dependent on the concentration of “greenhouse” gases and the temperature. Outgoing longwave radiation at TOA is very different from longwave radiation at the surface.”

The atmosphere simply radiates isotropically according to its temperature and emissivity. As I said before, there are other transports of energy through the atmosphere and through various layers of the atmosphere by convection, conduction and latent heat of water. The point is at every level in the atmosphere when energy is transferred or released via radiation, the split is half up half down. Ultimately this means the average of all the surface emitted radiation absorbed by the atmosphere will be half to space and half to the surface.

104. scienceofdoom says:

“I have pointed out that effective emissivity is something quite different from real emissivity which is a measured parameter. Effective emissivity as you are using the term is not a real material property, it is a ratio which is not related to the actual emissivity of any part of the system.”

Oh, but it most certainly is. The physical meaning of the effective emissivity of the Earth is net transmittance to space, or for each 1 W/m^2 emitted at the surface, 0.6 W/m^2 escapes to space and 0.4 W/m^2 are returned to the surface.

Put another way, it is the reciprocal of the surface response to solar forcing – meaning it takes 1.6 W/m^2 emitted at the surface to allow 1 W/m^2 to leave at the TOA, offsetting each 1 W/m^2 entering the surface from the Sun.

It’s an energy flux loop between the boundary of the surface and TOA, where each pass through the loop about 60% of the surface emitted radiation escapes to space and about 40% is returned or recycled back to the surface.

• Since the only energy source in the system is the Sun, all the energy at the surface ultimately has to have come from the Sun or the TOA incoming flux of about 240 W/m^2, according to the first law (excluding of course the infinitesimal amount from geothermal).

The reason why or how the surface is emitting more than the incoming 240 W/m^2 (390 W/m^2) is because the atmosphere is acting as a ‘filter’ between the surface and space, where each pass through the ‘filter’ only 60% (240 W/m^2) of the 390 W/m^2 emitted is allowed to leave the system. The remaining 150 W/m^2 are emitted back to the surface due to the GHGs and clouds in the atmosphere, which absorb and re-emit a portion of the 390 W/m^2 emitted at the surface, only half of which is emitted to back to the surface.

In other words, the rate the incoming energy from the Sun, which is mostly transparent through the atmosphere, is faster than the rate the absorbed energy re-emitted at the surface is allowed to leave at the TOA. The result is the surface accumulates energy until it radiates enough energy to push through atmosphere out the TOA the same amount of energy entering at the TOA from the Sun.

• on June 2, 2011 at 11:31 pm DeWitt Payne

The result is the surface accumulates energy until it radiates enough energy to push through atmosphere out the TOA the same amount of energy entering at the TOA from the Sun.

More or less. So why do you have a problem with energy balance diagrams which show exactly that?

105. scienceofdoom says:

“DeWitt Payne helpfully pointed out to RW:

You do understand that the downward flux of 333 W/m² is not invented out of whole cloth to make up the energy balance. It can be and is measured by a variety of instruments.”

Yes, it can be measured locally, but there is no real accurate way to get a global average for this figure, because you’d need measuring instruments looking up all over the globe 24/7. Unlike like satellites in space that can look down on the whole of the Earth’s surface and get accurate global data for average energy fluxes.

I’ve also made no claim that the total downard LW flux from the atmosphere to the surface is only 150 W/m^2. I’m sure it’s more than this. What you apparently don’t understand is that the total downward LW at the surface last originated from 3 potential sources: Some of it last originated from the Sun, some of it last originated from surface emitted radiation, and some of it last originated from the kinetic energy (convection & latent heat) moved from the surface into the atmosphere.

The point is COE dictates that of the 390 W/m^2 LW emitted at the surface, only 150 W/m^2 of it can be coming back from the atmosphere. Any amounts more than 150 W/m^2 have to have last originated from the Sun or last originated from the kinetic energy moved from the surface into the atmosphere, which too radiates isotropically in the LW infrared. And yes, a small percentage of the post albedo from the Sun is absorbed by the atmosphere and emitted to the surface as LW.

• I meant to say:

“Unlike like satellites in space that can look down on the whole of the Earth’s surface and get accurate global data for average energy fluxes leaving at the TOA.”

• The key point here is that not all the downward emitted LW received by the surface is ‘back radiation’ as depicted in the Trenberth diagram. The amount from the Sun is ‘foward radiation’ yet to reach the surface. The amount from the kinetic energy in the atmosphere has to be part of the closed loop net zero flux from the surface to the atmosphere and back to the surface (in Trenberth’s diagram 97 W/m^2 leaving the surface and 97 W/m^2 returning to the surface).

106. on June 2, 2011 at 11:39 pm | Reply DeWitt Payne

But not all the energy is transmitted from the surface by radiation. Approximately 20% (97/493) of the gross flux and 60% (97/161) of the net flux is by convection from the surface. Without convection, the fluxes don’t balance unless the surface becomes a lot hotter than the atmosphere immediately above it. And the atmosphere would be cooler than it is now.

• Dewitt Payne says:

“But not all the energy is transmitted from the surface by radiation.”

I never said it was, but the 390 W/m^2 emitted is, which represents the net energy flux entering the surface in the steady-state. What you don’t seem to understand is that the effect of convection is cooling the surface – it’s just that the net energy loss at the surface from convection has reduced surface emitted radiation to the 390 W/m^2 it is. If there were no convection, the surface would certainly have accumulated more energy and subsequently be emitting more than 390 W/m^2.

• scienceofdoom says:

“Any given portion of the atmosphere radiates equally in all directions.

This does not mean that the top of atmosphere radiates up the same amount as the the bottom of the atmosphere radiates down. ”

I’ve never made any such claim. That you think I have shows me you haven’t been paying attention.

107. If this George White article is where you have learnt your stuff (or you have written it?) no wonder you are confused.

Any given portion of the atmosphere radiates equally in all directions.

This does not mean that the top of atmosphere radiates up the same amount as the the bottom of the atmosphere radiates down. This is not the consequence of an isotropically emitting atmosphere. It is what you get when you assume that the atmosphere is isothermal (all at the same temperature).

At the bottom of the atmosphere, for any parcel/layer the emitted upward and downward radiation is equal. At TOA the emitted upward and downward radiation is equal. But the atmosphere interacts with radiation in its journey from top to bottom and vice versa.

This is why downward longwave radiation at the top of atmosphere is zero. Yet downward longwave radiation at the bottom of the atmosphere is (globally averaged) around 330 W/m2.

The diagram shows the surface receiving 239 W/m2 from the sun, which it doesn’t. The climate system does but around 170 W/m2 is absorbed by the surface while the balance is absorbed in the atmosphere.

He writes the same confused reasons that you have for why convection = 0, in laymans terms “..if something that is not true was actually true then this other thing will be true

He says “A further constraint of Conservation Of Energy is that the global net non radiative flux between the atmosphere and the surface must be zero in the steady state if the radiative flux is also zero.

And yet if net radiative flux is not zero then non-radiative flux is also not zero. As I have already explained, numerous times.

Conservation of energy = “conservation of energy” ≠ “conservation of radiation”.

He says “if” and offers no proof.

Here are the correct equations, using his notation and adding the missing terms:

1. For TOA:
F0 = solar radiation absorbed by the climate system, therefore at TOA, radiation leaving must balance

F0 = 239 = 385.T + (1-F).A ….[eq 1]

so far so good.

2. Surface:

Energy out = energy in

⇒ 385 + K = 239(1-Fs) + F.A ….[eq 2]

where K is net convected flux from the surface
and Fs = fraction of solar radiation absorbed by the atmosphere.

3. Atmosphere:

Energy out = energy in

⇒ A = 385(1-T) + K + 239.Fs ….[eq 3]

To prove K = 0 requires proving it.
To prove Fs=0 requires proving it.

108. scienceofdoom says:

“Any given portion of the atmosphere radiates equally in all directions.

This does not mean that the top of atmosphere radiates up the same amount as the the bottom of the atmosphere radiates down. ”

I’ve never made any such claim. That you think I have shows me you haven’t been paying attention.

• “The diagram shows the surface receiving 239 W/m2 from the sun, which it doesn’t.”

Are you forgetting that the Sun is the only source of energy? All the energy in the system either at the surface or anywhere else originated from the Sun. What you don’t understand is that while 239 W/m^2 may not be directly received by the surface from the Sun, indirectly it must get to the surface because COE dictates that the atmosphere cannot create any energy of its own.

If 239 W/m^2 from the Sun does not get to the surface, then where is it going? Just because some of it is absorbed by the atmosphere, does not mean it never gets to the surface.

• What I said was the at the boundaries of the surface and the TOA, the split of surface emitted radiation absorbed by the atmosphere is half to the surface and half to space.

109. RW:

I’ve never made any such claim. That you think I have shows me you haven’t been paying attention.

You referred me to your article in lieu of writing equations yourself.

As the other confused statements you have made originate there it seems that this statement:

From the physics of black bodies, the atmosphere should behave as an isotropic radiator with half of it’s emitted power going up and half down, thus we can say that the transparent window of the atmosphere must be about 24% since physics dictates that F must be one half.

– is also something you endorse.

Physics does not dictate that F must be one half (ref the article for explanation of F). Unless the atmosphere is isothermal. Which it isn’t.

He makes the same incorrect statement that you have been making and draws an incorrect conclusion.

Like I said, write down your equations and then it will be easier to explain to you where you are wrong. You pointed me to this guy’s equations. I have demonstrated that they are wrong.

110. scienceofdoom says:

“And yet if net radiative flux is not zero then non-radiative flux is also not zero. ”

Of course. The key word he uses is ‘further constraint’.

Are you forgetting the original questions I asked that you answered ‘yes’ to?

You’ve said you agree that all of the 390 W/m^2 emitted from the surface is radiative, and you agree that the only way energy can enter in leave at the TOA is by radiation. You also agree that the 390 W/m^2 emitted at the surface is solely due to its temperature and nothing else (assuming an emissivity of 1 or very near 1).

Why do you think I asked these questions?

111. RW:

Are you forgetting that the Sun is the only source of energy? All the energy in the system either at the surface or anywhere else originated from the Sun. What you don’t understand is that while 239 W/m^2 may not be directly received by the surface from the Sun, indirectly it must get to the surface because COE dictates that the atmosphere cannot create any energy of its own.

If 239 W/m^2 from the Sun does not get to the surface, then where is it going? Just because some of it is absorbed by the atmosphere, does not mean it never gets to the surface.

You don’t understand how to write equations of energy balance.

This is the surface energy balance (my version):

385 + K = 239(1-Fs) + F.A ….[eq 2]

The last term in bold is the energy emitted from the atmosphere to the surface.

This is how the solar energy absorbed in the atmosphere has an effect on the surface.

We can’t pretend the atmosphere doesn’t absorb solar energy and get the correct equations.

I’m not sure I can help you any further in your quest.

I suggest you buy or borrow a heat transfer textbook (I recommend Incropera and DeWitt, Fundamentals of Heat and Mass Transfer) and work through the first few chapters.

By the time you have followed a few worked examples and tried to do a few exercises yourself you will understand how heat transfer calculations are done and how to apply the first law of thermodynamics.

I would explain it to you, but you are convinced you are right even though you haven’t understood the absolute basics so there’s little point.

Good luck in your journey.

112. scienceofdoom says:

“Like I said, write down your equations and then it will be easier to explain to you where you are wrong. You pointed me to this guy’s equations. I have demonstrated that they are wrong.”

No, you’ve demonstrated that you don’t understand the first law as it applies to the variables in those equations.

“To prove K = 0 requires proving it.
To prove Fs=0 requires proving it.”

The ‘proof’ is that energy can only enter and leave at the TOA by radiation. The surface cannot be receiving more than a net of 390 W/m^2 in the steady-state and all of the 390 W/m^2 emitted is radiation.

113. Also, if the author is wrong, explain why the numbers work out as they do. Or if it’s just a coincidence, explain why?

BTW, the author has a post-up and comments section on this paper over at Joanna Nova where he has answered many of your questions and more in elaborate detail in the comments section. His moniker is in the comments section is ‘co2isnotevil’:

http://joannenova.com.au/2011/01/half-of-the-energy-is-flung-out-to-space-along-with-the-model-projections/

114. scienceofdoom says:

“We can’t pretend the atmosphere doesn’t absorb solar energy and get the correct equations.”

We can if COE dictates that the atmosphere creates no energy of its own. The solar energy absorbed by the atmosphere has to be going either to the surface or back out to space. The amount emitted up back out to space in the form of LW is already accounted for in the albedo of 102 W/m^2. The actual measured albedo is more like 0.27 in the SW infrared. The 0.03 more to get full 0.3 is the LW from the solar energy absorbed by the atmosphere emitted back out to space.

Even Trenberth has the full amount designated as ‘absorbed by the atmosphere’ getting to the surface (78 + 97 + 157 = 332). But again there is no reference where the 78 W/m^2 figure is coming from. It’s way too high. It’s more like about 20 W/m^2 absorbed by the atmosphere.

115. scienceofdoom says:

“I’m not sure I can help you any further in your quest.

I suggest you buy or borrow a heat transfer textbook (I recommend Incropera and DeWitt, Fundamentals of Heat and Mass Transfer) and work through the first few chapters.

By the time you have followed a few worked examples and tried to do a few exercises yourself you will understand how heat transfer calculations are done and how to apply the first law of thermodynamics.

I would explain it to you, but you are convinced you are right even though you haven’t understood the absolute basics so there’s little point.”

You don’t seem to understand that heat and radiation are not the same thing and subsequently not governed by the same physical rules or constraints. Radiation has no temperature – it’s not thermal energy. Maybe this is why you’re confused – I don’t know.

116. Ah, I think I may know where the confusion is coming from. In the paper, the author says:

“While radiative flux to and from the surface can be traded off against non radiative flux, it makes no difference to the overall radiative balance.”

I think this is being misinterpreted as saying “no difference to the energy balance” or “no difference to the net energy flux at the surface” or even “no difference to the surface temperature”.

This is most definitely NOT what the author is saying or implying, though I can see how this could be interpreted this way.

117. Let’s try looking at this from the perspective of Trenberth’s numbers again – only from a different angle:

He has the atmosphere emitting 169 W/m^2 to space and 70 W/m^2 passing from the surface straight to space for 239 W/m^2 leaving.

Where is the 169 W/m^2 from the atmosphere coming from?

He has the surface emitting 396 W/m^2 and 70 W/m^2 of it passes straight through the atmosphere to space. The difference of 326 W/m^2 is the amount of absorbed by the atmosphere, right? He also has 333 W/m^2 designated as ‘back radiation’ returning to the surface, right?

Where is the 333 W/m^2 coming from? Is this depicting that the surface emitted radiation absorbed by the atmosphere never leaves at the TOA?

Also, where is the 78 W/m^2 designated at “absorbed by the atmosphere going? Back out to space? If this is true, where is the additional 91 W/m^2 leaving at the TOA coming from (169 – 78 = 91)???

• on June 3, 2011 at 9:33 pm | Reply DeWitt Payne

For the last time:

Try reading my comment above again. ‘ my comment above’ is a link to the comment.

Where is the 169 W/m^2 from the atmosphere coming from?

It’s coming from the 501 W/m² (396 + 97 + 78 -70) of absorbed radiative and convective flux from the surface plus solar absorption.

Where is the 333 W/m^2 coming from?

It’s coming from the remainder of the 501 W/m² not emitted to space (501-169=332, rounding error again).

Also, where is the 78 W/m^2 designated at “absorbed by the atmosphere going?

See above. Part of it is emitted to space and part to the surface. It’s not a 50:50 split because the top of the atmosphere is colder than the bottom.

118. on June 3, 2011 at 9:12 pm | Reply DeWitt Payne

RW,

What I said was the at the boundaries of the surface and the TOA, the split of surface emitted radiation absorbed by the atmosphere is half to the surface and half to space.

No, it’s half up and half down. The distinction is important. And the amount absorbed at any given level is only a small fraction of the total. You also ignore that the same thing applies to the downward flux. At the 500 mbar pressure level, for example, about half of the radiation emitted upward escapes to space while only about 10% reaches the surface directly. And the emission at the 500 mbar level is much less than at the 1000 mbar level because the temperature is lower.

You’re also forgetting the basic implication of Kirchhoff’s Law and local thermodynamic equilibrium. Emission is determined by temperature. Temperature is determined by heat content. Molecular kinetic energy doesn’t know or care where it comes from. A molecule that absorbs radiation transfers the energy by collision with other molecules rather than radiation at least 99.99% of the time. Radiative emission is due to collisional activation at least 99.99% of the time.

• on June 3, 2011 at 10:47 pm | Reply DeWitt Payne

At the ~500 mbar level for an atmosphere adjusted to 2.6 cm precipitable water and a surface temperature of 289.4 K, radiation up (100-1500 cm-1) is 299.1 W/m² and down is 99.4 W/m². The total energy absorbed for any altitude is emitted equally in both directions, but the end result of an atmosphere that cools with altitude is that total energy absorbed/kg (and hence also emitted) decreases with altitude and the ratio (absorbed Ed)/(absorbed Eup) goes to zero with altitude. Absorptivity/emissivity also goes down with altitude as pressure broadening of the lines decreases. Water vapor continuum absorption/emission goes down even faster as it is proportional to the square of the water vapor partial pressure.

119. on June 3, 2011 at 9:41 pm | Reply DeWitt Payne

In TF&K09 it’s not 239 W/m² in and 239 W/m² out. In round numbers, it’s 239 W/m² in and 238 W/m² out leaving 1 W/m² absorbed by the surface and retained. So emission from the atmosphere at the TOA is 168 W/m², not 169 W/m². So 168 W/m² up plus 333 W/m² down = 501 W/m².

• Yeah I know. He purposefully put an extra watt in there to make it look like the system is accumulating energy for future warming.

120. Dewitt Payne,

“Where is the 169 W/m^2 from the atmosphere coming from?

It’s coming from the 501 W/m² (396 + 97 + 78 -70) of absorbed radiative and convective flux from the surface plus solar absorption.”

You can’t have 169 W/m^2 emitted from the atmosphere to space and 333 W/m^2 emitted from the atmosphere to the surface if the atmosphere is only absorbing 326 W/m^2. There is only a net flux incoming flux at the surface of 396 W/m^2 in the steady-state and 239 W/m^2 leaving and entering at the TOA.

If you want to create an additional 105 W/m^2 or so (501 – 396) out of thin air to try justify Trenberth’s depiction of the energy flows go ahead.

121. DeWitt Payne says:

“You’re also forgetting the basic implication of Kirchhoff’s Law and local thermodynamic equilibrium. Emission is determined by temperature. Temperature is determined by heat content. Molecular kinetic energy doesn’t know or care where it comes from. A molecule that absorbs radiation transfers the energy by collision with other molecules rather than radiation at least 99.99% of the time. Radiative emission is due to collisional activation at least 99.99% of the time.”

I know most of the energy transfer is by collision rather than radiation, especially lower in the atmosphere. The point is everywhere in the atmosphere where energy is emitted via radiation it will be isotropic. When, as you say, absorbed energy is transferred via collision to other gas molecules in the atmosphere, the heated gases also emit radiation isotropically. Wherever there is radiative emission it is equal and in all directions. It’s just conservation of radiated energy that half of what’s absorbed ends up radiated to space and half ends up at the surface.

Now, I should add that the half absorbed does not necessarily have to be coming back to the surface in the form of downward emitted LW infrared. Some of it could be absorbed by the atmosphere and returned kinetically in the form of latent heat (precipitation), for example, but this just offsets energy that what would otherwise need to be radiated back to the surface.

122. on June 5, 2011 at 6:10 pm | Reply DeWitt Payne

RW,

I made the point about collisional activation/deactivation because you seem to think you can ignore convective energy transfer because it isn’t radiation. Kinetic energy is kinetic energy. It doesn’t matter whether it comes from absorption of radiation, collision with the warmer Earth’s surface or collision with a drop of water warmed by the heat of condensation.

You can’t have 169 W/m^2 emitted from the atmosphere to space and 333 W/m^2 emitted from the atmosphere to the surface if the atmosphere is only absorbing 326 W/m^2. There is only a net flux incoming flux at the surface of 396 W/m^2 in the steady-state and 239 W/m^2 leaving and entering at the TOA.

Why do you keep ignoring absorbed solar and convection? The atmosphere doesn’t just absorb 326 W/m² net emitted from the surface, it also absorbs 78 W/m² of incoming solar and 97 W/m² from convection from the surface for a total of 501 W/m² . The total upward energy flux at the surface is 493 W/m² , not 396 W/m² .

123. Dewitt Payne says:

“I made the point about collisional activation/deactivation because you seem to think you can ignore convective energy transfer because it isn’t radiation. Kinetic energy is kinetic energy. It doesn’t matter whether it comes from absorption of radiation, collision with the warmer Earth’s surface or collision with a drop of water warmed by the heat of condensation.”

I don’t claim otherwise. All I was saying is that whenever energy is emitted by radiation in the atmosphere it’s flung equally in all directions.

• on June 6, 2011 at 12:25 am | Reply DeWitt Payne

I don’t claim otherwise. All I was saying is that whenever energy is emitted by radiation in the atmosphere it’s flung equally in all directions.

That’s only true locally, not for the atmosphere as a whole.

You’ve exhausted my patience.

Bye.

• Dewitt Payne says:

“…not for the atmosphere as a whole.”

I’ve not claimed this either.

124. Dewitt Payne says”

“Why do you keep ignoring absorbed solar and convection? The atmosphere doesn’t just absorb 326 W/m² net emitted from the surface, it also absorbs 78 W/m² of incoming solar and 97 W/m² from convection from the surface for a total of 501 W/m² . The total upward energy flux at the surface is 493 W/m² , not 396 W/m² .”

I’m not ignoring the 78 W/m^2 designated as “absorbed by the atmosphere”. I’m simply saying it does not all have to be getting to the surface in the form of downward emitted LW infrared. Some of it could be returned kinetically in the form of latent heat via precipitation, for example, but this just offsets energy that what would otherwise be radiated to the surface.

Even if you want to assume there is net upward flux of 493 W/m^2 from the surface, you need to have a net downward flux of 493 W/m^2 coming into the surface in the steady-state. Without the 78 W/m^2 absorbed by the atmosphere getting the surface, you don’t have enough energy: 161 + 97 + 157 = 415 W/m^2, which is 78 W/m^2 short of the 493 W/m^2 you need coming into the surface.

125. Dewitt Payne says:

“…not for the atmosphere as a whole.”

Or least I’ve not claimed this in the context you’re implying.

• RW.

I sympathise RW and sorry I couldn’t add to the debate earlier.

No COE rule for ‘power’? There absolutely is, isn’t there! 🙂

‘Power = a measured quantity of energy’ and it’s nominated ‘SI Unit’ for this particular subject is the ‘W/m^2’. It could equally be the ‘kWhr/m^2’, ‘kW/m^2’ (no time unit means 1 second duration), or even ‘BTU/m^2’.

The more use we make of SI Units during our chosen careers it seems the more we tend to forget what the ‘unwritten’ terms mean. If a term for time duration is missing it doesn’t mean that the ‘flux’ is instantaneous, it means that the ‘flux’ transports the stated quantity of energy during ‘1 second’. However, I think that SoD was hinting at Plank’s rule for ‘radiation in “doesn’t equal” radiation out’.

Here’s another:

Top of Atmosphere ~ = 100 km (some say ~70 km).

http://earthobservatory.nasa.gov/IOTD/view.php?id=7373

What’s the temperature there? Trick question!

http://www.windows2universe.org/earth/Atmosphere/layers_activity_print.html

As can be seen, different [?]-spheres have different temperatures within them. The real boundary between Earth’s atmosphere and space is the thermosphere/exosphere boundary, but it’s more than 500°C there due to insolation activity that gets nowhere near to lower altitudes and radiates almost directly to space at ~500 km.

Is there a ‘universally’ accepted standard here? If not, one should be agreed across the board for the discipline of ‘Atmospheric Science’ and the forum for this should only be connected with academia (this would exclude the IPCC).

Hope this supports your POV, but I do need to limit my posts here just now.

Best regards, Ray Dart.

126. Dewitt Payne,

Table 2a in Trenberth’s paper under “this paper” has ‘Net Down’ of 0.9 W/m^2.

What do you think this means?

127. RW,

I think this whole discussion could possibly be cleared up by standing in front of a whiteboard version on the Trenberth diagram. To properly write the equations SoD is asking for, and to get all the energy balances correctly, you have to define your system boundaries correctly. For this 2D diagram, that means drawing a closed 2D polygon (square, circle or whatever) around an entity so that you capture ALL of the energy flows into and out of the entity. If you only capture a portion of the energy flows, they won’t balance. If you only capture a portion of the entity, you’ll also be missing some flows, and the energy won’t balance.

In this diagram, there are three main systems that you can encircle: the earth, the atmosphere, and the earth plus atmosphere. If you do any one of those, and count up all the ins and outs, they will balance (i.e. fulfill COE). When you say you’ve found a violation of COE, and especially when talking about the boundary at earth’s surface and TOA, it tells me you’re drawing an incomplete boundary or a boundary around an incomplete system. It would be like having a lake with one river coming in, and two going out, and drawing your system boundary around the inlet and only one outlet, calculating mass flows for your system, and then complaining that the map maker must be wrong in his diagram because there’s a violation of conservation of mass.

• Did you read the whole discussion? All the issues you raise here were addressed.

The overwhelming majority of the Earth’s thermal mass is below the surface, so setting the boundary there is quite reasonable. The atmosphere is really just acting as a filter between the surface and space, where at the TOA energy only enters and leaves via radiation.

As I stated on another thread, technically absorbing 78 W/m^2 by the atmosphere and bringing it to the surface via ‘back radiation’ is not a violation of COE. It is however, highly misleading and why so many people are confused.

I’ve elaborated on this here in multiple posts:

https://scienceofdoom.com/2011/06/21/whats-the-palaver-kiehl-and-trenberth-1997/#comment-12206

• Are there more than one person writing under the name RW? Because I’ve ready pretty much the whole conversation, including the comment linked above, and I see you claiming violations of COE over and over again. And every time you do, you justify it by picking and choose energy flows so that balance is never achieved. It’s exactly analagous to the example I gave of the lake and rivers. You (or someone writing under your moniker) are ‘gerrymandering’, drawing odd boundary shapes to get the result you want. You do that whenever you pick a group of disparate numbers to add up and show a lack of COE. If you really think what you’re doing is defensible, then draw the relevant boundary on the diagram and upload it. A picture is worth a thousand words.

128. Science of Doom does not think the full 239 W/m^2 from the Sun gets to the surface even though the tables in Trenberth’s paper clearly says it does. And the numbers in the diagram only work out if it does too.

He writes: “The diagram shows the surface receiving 239 W/m2 from the sun, which it doesn’t. The climate system does but around 170 W/m2 is absorbed by the surface while the balance is absorbed in the atmosphere.”

This is crux of the disagreement between us in this thread. I’m operating under the assumption that the full 239 W/m^2 gets to the surface, albeit not all directly. He is arguing the full 239 W/m^2 does NOT all get the surface, either directly or indirectly.

• RW,

I’ve read enough of the conversation to know that both SoD and DeWitt have answered this question a number of times. I think I even saw you agree with the answer on at least one occasion. So that’s what makes me think there are two people using your name. I don’t see how you can give such self contradictory posts.

Look at it this way. The total entering at TOA is 341, but 102 is reflected and heads straight back into space. The 102 never gets absorbed at the surface. That leaves 239 to be absorbed in the earth system. In almost exactly the same way, 78 of the 239 goes into the atmosphere and is absorbed. A portion of that is re-emitted, and goes back into space. It never makes it to the surface. So the amount reaching the surface is less than 78, which makes the total less than 239. The diagram doesn’t attempt to say how much of the 78 makes it to the surface, and how much goes directly back to space. As you’ve said yourself, the portion of the 78 that makes it to the surface is lumped in and is part of the 333 called “back radiation.” And as stated in my earlier post, lumping these together makes sense since the photons actually originate in the atmosphere, just like the rest of the 333.

129. Science of Doom, if the 239 W/m^2 from the Sun does not get the surface, show me how the numbers work out. Where is the 78 W/m^2 going?

130. Are there more than one person writing under the name RW? Because I’ve read pretty much the whole conversation, including the comment linked above, and I see you claiming violations of COE over and over again. And every time you do, you justify it by picking and choosing energy flows so that balance is never achieved. It’s exactly analagous to the example I gave of the lake and rivers. You (or someone writing under your moniker) are ‘gerrymandering’, drawing odd boundary shapes to get the result you want. You do that whenever you pick a group of disparate numbers to add up and show a lack of COE. If you really think what you’re doing is defensible, then draw the relevant boundary on the diagram and upload it. A picture is worth a thousand words.

131. Jimwit says:

“Because I’ve ready pretty much the whole conversation, including the comment linked above, and I see you claiming violations of COE over and over again.”

What I mean is if it doesn’t get the surface, it’s a violation of COE and the numbers don’t work out.

132. Jimwit says:

“If you really think what you’re doing is defensible, then draw the relevant boundary on the diagram and upload it. A picture is worth a thousand words.”

If you haven’t already, look here and here:

133. Jimwit says:

“The 102 never gets absorbed at the surface. That leaves 239 to be absorbed in the earth system. In almost exactly the same way, 78 of the 239 goes into the atmosphere and is absorbed. A portion of that is re-emitted, and goes back into space. It never makes it to the surface. So the amount reaching the surface is less than 78, which makes the total less than 239.”

Again, this is the crux of the whole issue. Have you carefully read the tables in the Trenberth paper? All of the 239 W/m^2 gets to the surface because the “NET Down” to the surface is 0.9 W/m^2. You are correct that 102 never gets absorbed at the surface, but you are not correct that all of this is reflected SW. A portion of it is SW absorbed by the atmosphere and emitted up back out to space as LW infrared that never reaches the surface. What is happening is 78 W/m^2 of the 239 is being absorbed by the atmosphere (mostly by clouds), some of it is re-emitted down as LW infrared and some of it gets transferred into the kinetic energy in the form of latent heat of water and gets to the surface in kinetic form (probably mostly via precipitation). Again, this just offsets energy that would otherwise get radiated to the surface.

The point is all of the 239 W/m^2 gets to the surface. If it didn’t, then ‘NET Down’ in table 2a would be a negative number. -78 W/m^2 if none of the 78 gets the surface or some negative number in between -78 and zero if only part of it gets the surface.

If you doubt my interpretation of the data, contact Trenberth and ask him to clarify. Or you might try to show how the numbers work out with less than the full 78 W/m^2 ‘absorbed by the atmosphere’ getting the surface.

• RW,

I think you should keep the diagram in front of you. You say part of the 102 is absorbed. But the diagram shows that clearly as non-absorbed, completely reflected sw solar. A simple mistake, but adds to the potential confusion and lack of communication. It’s the 78 that gets partly absorbed by the atmosphere.

More troubling are the references to latent heat and kinetic energy. The latent heat comes when water absorbs heat at the surface, which leads to evaporation. The water vapor rises through the atmosphere, gives off heat, and condenses back to water. That’s the 80 W/m2 of evapotranspiration. This much colder water then falls back to the warmer earth in the form of rain. When cold rain hits the earth, it doesn’t bring latent heat back with it. It starts the cycle all over again by cooling the surface through absorbing heat, evaporating, etc. etc.

As for rain bringing energy back to earth in the form of kinetic energy, it’s very minimal. Raindrops typically travel about 2-9 m/sec. For convenience, imagine 1 gm of water falling at 5 m/sec. Using KE = 1/2 mv^2, you can calculate it carries .0125 Joules (Kgm^2/s^2) or about 0.003 calories. The latent heat of evaporation of water is 540 calories/g. So the Evapotranspiration carries about 180,000 times as much heat upward as the rain brings back in the form of kinetic energy. So kinetic energy can be ignored.

134. I might also add that the 78 W/m^2 figure is not a directly measured value. It’s derived from large number of assumptions and deductions, so it should be taken with a grain of salt.

135. RW,

Wow, the links explain a lot in terms of your thinking. Unfortunately, Mr. White makes some serious mistakes that will lead to confusion. Let’s deal with the major ones

1. His diagram shows a “measured” value of 239 W/m2 of solar (not long wave) radiation reaching the surface,. He doesn’t provide a reference for that measurement, and I don’t think he can. Solar radiation is almost entirely shorter than 4 micron. It’s easy to measure and distinguish from long wave (> 4 micron). Trenberth on the other hand shows 161+23 = 184 of short wave solar reaching the surface. This is a big difference. At least one of them has to be very, very wrong. Note that for White’s diagram to be correct, none of the 239 can be absorbed in the atmosphere and re-radiated downward or upward as longwave. And his equations assume this. If you begin by assuming that all of the non-reflected shortwave solar passes through to the surface, then of course you’re going to end up with answers at the end that say all the non-reflected solar ends up at the surface. It’s like saying “Let’s assume A is true. Then it is fairly simple to prove after a bunch of math that A is true.” But you need to ask yourself – does it make sense to assume that none of the solar is absorbed by the atmosphere? Or is it just possible that Mr. White has made a significant mistake from the outset?

Once Mr. White makes these two mistakes, then the rest of his numbers also have to be off by quite a bit.

• One clarification. Mr. White shows all 239 W/m2 being absorbed by the surface, as opposed to only 161 per Trenberth (for apples to apples, I shouldn’t have included Trenberth’s 23 reflected from the surface). So the delta between the two is even bigger.

136. Jimwit,

I don’t think you’ve been paying attention. What Mr. White is showing is that 239 W/m^2 from the Sun is getting the surface. He makes no distinction that all of this gets to the surface via SW radiation. Also, he’s not ignoring Thermals and Evapotranspiration.

Mr. White has a post up on this paper at Joanne Nova. Why don’t you pose these questions to him there:

http://joannenova.com.au/2011/01/half-of-the-energy-is-flung-out-to-space-along-with-the-model-projections/

• Yes, he is effectively ignoring non-radiatives. In the quote I provided earlier, he agrees that there can be tradeoffs between radiative and non-radiative, but then contradicts that by flippanty and wrongly saying that all the non-radiatives balance each other out, so don’t need to be considered, and you can just pay attention to making the radiative components balance. That’s gerrymandering – drawing system boundaries that ignore significant heat flows so you can get the result you want.

By his text and equations, he’s effectively presuming that the 97 W/m2 of non-radiative transfer from surface to atmosphere must be balanced by 97 W/m2 of non-radiative from atmosphere to surface somewhere else. But there’s no valid explanation of what that mechanism is. You’ve tried to say that it’s by virtue of latent heat and kinetic energy coming back to the surface in rain. But my earlier post shows how wrong that is. Cold rain doesn’t bring heat back to the surface, neither do cold winds from up in the clouds, and the kinetic energy transmitted by rain is miniscule compared to the energy carried up by evaporation. You can wave your hands and say there’s a balance of non-radiative transfer to and from the surface, but that doesn’t make it so.

And yes, I’ve paid careful attention to what Mr. White is saying. His diagram makes a division between short wave visible radiation on the left, and long wave on the right. There’s even a dotted vertical line between the two sections to show the division. He also makes a very clear point of showing how upward lw from the surface gets absorbed, then splits into downward and upward emissions of lw. He shows no such mechanism for any of the downward sw solar.

137. Jimwit,

Take a look at this:

• RW,

I’m not sure what you mean by posting this link. Do you mean to show that Mr. White’s treatment actually has none of the problems I’ve mentioned? If so, you need to go to the earlier link you provided at joannenova.com. What you’ll see there is the development and explanation of the original diagram I critiqued. Once Mr. White finishes his explanation of his faulty offering, he then goes on to show the one you’ve linked in this post, but he introduces it by saying “This shows how to obfuscate this view of the radiative balance to look like Trenberth’s.” He clearly shows his as different than Trenberth’s and thinks his is right and Trenberth’s is wrong. And as I’ve pointed out, White’s development and depiction has serious problems.

138. RW,

Can you elaborate on what you mean exactly by the energy absorbed “reaching the surface”?

Do you mean that if the atmosphere didn’t absorb the short wave – the downward longwave radiation figure would be smaller (because all the shortwave reaches the surface directly and isn’t converted to longwave)?

139. Jimwit, you say:

“By his text and equations, he’s effectively presuming that the 97 W/m2 of non-radiative transfer from surface to atmosphere must be balanced by 97 W/m2 of non-radiative from atmosphere to surface somewhere else. But there’s no valid explanation of what that mechanism is.”

If 97 W/m^2 of non-radiative transfer from the surface to the atmosphere is not balanced by 97 W/m^2 of non-radiative from the atmosphere to the surface, where is the energy going?

140. RW,

The 97 W/m2 from the surface warms the atmosphere, which then, by virtue of its temperature, radiates the heat away, some to space, some back to the surface in the bucket called “back radiation”.

Regardless of how warming occurs (conductive, radiative, convective, resistive, kinetic), a body will emit radiation based on its temperature. If you pour heat into a body, some of that will radiate away.

I could turn the question around, as I did already. If the heat doesn’t radiate away, then where does it go? How does the atmosphere get rid of it, as it must if maintaining steady state conditions?

• Yes. What then happens if a portion of the 97 W/m^2 is radiated into the atmosphere and ultimately radiated out to space?

The amount returned to the surface will be less than 97 W/m^2, right? What happens when there is net energy loss at the surface?

The surface loses energy and subsequently radiates less, right?

141. Jimwit says:

“I could turn the question around, as I did already. If the heat doesn’t radiate away, then where does it go?”

It either goes back to the surface or finds its way radiated out to space. I disagree that most of kinetic energy is radiated away – I think most of it stays with the water and falls back as precipitation, but regardless it doesn’t matter relative to what we’re discussion. As latent heat of water is removed from the surface, latent heat of water is returned to the surface at the same time. If the net kinetic flux at the surface is not zero, it just offsets energy that would otherwise need to be radiated at the opposite end (either at the surface or the TOA).

• Another thing that may not be clear is the surface flux of 385 W/m^2 and the TOA flux of 239 W/m^2 are both all radiative fluxes (i.e. all photons).

142. Even another thing that may not be clear is that the value of A = 292 W/m^2 is largely the result of latent heat of water in the form of water vapor and clouds, which absorb outgoing LW emitted from the surface and re-emit it.

Because energy only leaves the system by radiation, the 146 W/m^2 emitted to space is all radiative; however, the 146 W/m^2 returned to the surface is not – it’s just the equivalent of 146 W/m^2 radiated to the surface. No doubt some of it returns in kinetic form, but again this just offsets energy that would otherwise be radiated to the surface.

Fundamentally, what the author is trying to demonstrate is that only half of what the atmosphere absorbs relative to the surface emitted of 385 W/m^2 is ultimately returned to the surface to affect its temperature. The other half is radiated out to space and is essentially the same as LW infrared that passes straight through the atmosphere as if the atmosphere wasn’t even there.

So the author is not ignoring latent heat and thermals, as they are no doubt contributing greatly to the energy balance as a whole (i.e. the net flux of 385 W/m^2 at the surface), but relative to the radiative fluxes the net non-radiative fluxes are zero in the steady-state, because energy only enters and leaves the system at the TOA as radiation.

143. Jimwit says:

“You’ve tried to say that it’s by virtue of latent heat and kinetic energy coming back to the surface in rain. But my earlier post shows how wrong that is. Cold rain doesn’t bring heat back to the surface, neither do cold winds from up in the clouds, and the kinetic energy transmitted by rain is miniscule compared to the energy carried up by evaporation.”

How do you think energy is primarily transported from the tropics to the higher latitudes? If not via latent heat of water via precipitation systems, how? If the bulk of the latent heat didn’t stay with the water, how could this occur?

• RW,

I have no background or education to be able to answer your question. I can only offer a guess that makes sense to me – the heat you’re speaking of moves from surface to atmosphere, then travels by wind within the atmosphere from tropic to higher latitudes. At that point, the return of heat to the surface is still primarily radiative. Again, this is just a complete guess on my part. I don’t pretend to know anything about this aspect.

What I do feel comfortable with is saying that Trenberth’s energy balance diagram makes complete sense to me. My studies of heat transfer in chemical engineering forced me to be very conversant with energy balance calculations. This is a very simple system where all the mechanisms make sense, all the numbers balance, despite your attempts to make them unbalance. I don’t see any chance of us reaching a common understanding.

• Fair enough. You can’t say I didn’t try.

144. RW,

I don’t see how the DLR value could be lower when the atmosphere was transparent to shortwave. In order to reduce DLR you’d need to lower the temperature of the atmosphere, but I can’t see how this could happen if you absorb all the shortwave on ground level instead of partly in the atmosphere.

145. Jimwit and/or SoD,

Take a look at this paper, specifically on pages 19 and 20. G. White is not the only researcher deriving essentially the same energy flux relationships, where 239 W/m^2 gets to the surface and half of what’s absorbed from the surface emitted of 390 W/m^2 goes to space and half goes back to the surface.

I think we can at least agree that all of the energy at and below the surface originated from the Sun, correct (excluding an infinitesimal amount from geothermal)???

Also, if the surface is emitting 390 W/m^2 and 70 W/m^2 of this goes to space through the ‘window’, there is only 320 W/m^2 left to be absorbed by the atmosphere (390 – 70 = 320). If only 98 W/m^2 (165 – 67 = 98) of the 320 W/m^2 is emitted to space, that leaves only 222 W/m^2 available for downward emitted LW. If 102 W/m^2 from latent heat and thermals is subtracted from 222 W/m^2, there is only 120 W/m^2 left, which is short of the net flux of 390 W/m^2 required at the surface; 168 directly from the Sun + 122 from the atmosphere = 290 W/m^2 (390 W/m^2 required). There is not enough energy to account for all the fluxes.

On the other hand, if the 67 W/m^2 gets to the surface, it all balances out perfectly. 168 W/m^2 directly from the Sun arrives at the surface. 324 W/m^2 of ‘back radiation’ minus 102 from latent heat and thermals = 222 W/m^2 net arriving at the surface (168 + 222 = 390); 222 – 67 from the Sun = 155, and 155 + 168 + 67 = 390 W/m^2, the exact amount emitted at the surface. 390 – 70 – 165 = 155 W/m^2, the exact amount coming back from the atmosphere, and of course 165 + 70 = 235 W/m^2, the exact amount leaving. And finally, 168 + 67 = 235, the exact amount arriving from the Sun.

146. SoD says:

“[“He says “A further constraint of Conservation Of Energy is that the global net non radiative flux between the atmosphere and the surface must be zero in the steady state if the radiative flux is also zero.”]

And yet if net radiative flux is not zero then non-radiative flux is also not zero. As I have already explained, numerous times.

Conservation of energy = “conservation of energy” ≠ “conservation of radiation”.

He says “if” and offers no proof.

Here are the correct equations, using his notation and adding the missing terms:

1. For TOA:
F0 = solar radiation absorbed by the climate system, therefore at TOA, radiation leaving must balance

F0 = 239 = 385.T + (1-F).A ….[eq 1]

so far so good.

2. Surface:

Energy out = energy in

⇒ 385 + K = 239(1-Fs) + F.A ….[eq 2]

where K is net convected flux from the surface
and Fs = fraction of solar radiation absorbed by the atmosphere.

3. Atmosphere:

Energy out = energy in

⇒ A = 385(1-T) + K + 239.Fs ….[eq 3]

To prove K = 0 requires proving it.
To prove Fs=0 requires proving it.”

I’ve spent some time re-examing this issue. I realize that maybe it’s not clear that G White’s model is not a depiction of the actual behavior, but the net result equivalent of the actual behavior at the boundaries of the surface and the TOA. Do you know this? Perhaps you do.

At any rate, you seem to have agreed that the 385 W/m^2 emitted from the surface and 239 W/m^2 at the TOA are all radiative fluxes, yet you object to K and Fs equaling zero in the steady-state. How could they not be zero? Where is the energy from K or FS going? It’s either going back to the surface in some form as part of the net incoming flux of 385 W/m^2, or it’s being radiated out to space as part of the 239 W/m^2 flux leaving at the TOA.

Post albedo energy, Fs, “absorbed by the atmosphere” (mostly by clouds) that ends up being radiated out to space without ever reaching the surface is simply trading off energy emitted from the surface absorbed by clouds that would otherwise be leaving the planet, and subsequently falling back to the surface kinetically mostly in the form of precipitation. So indirectly, the full post albedo gets to the surface by requiring less of what’s absorbed by the atmosphere from the surface to be leaving the planet.

How else do you explain the graph below, which clearly shows that even when absorption is 100%, half is still escaping to space. If the full post albedo did not ultimately get to the surface one way or another, this would not be possible, would it?

147. SoD says:

“As best as I can tell, you have invented a new law of thermodynamics. I summarize it in 3 points.

1. For every body space is the ultimate boundary of the encompassing system.
2. Radiation can only leave and enter the ultimate boundary via radiation.
3. Therefore, convective heat flux from any body within the system must be zero.

Is this your hypothesis?

I agree with steps 1 & 2.

You have invented step 3 and it appears you think it is obvious. But it’s not true and this is why heat transfer textbooks are full of conduction and convection calculations.

The surface of the earth has a net convective loss to the atmosphere. This is returned via radiation.

We can measure the radiation imbalance. We can demonstrate the convective heat transfer from the surface.

Yes, the surface would be hotter if there was no convection. That is because convection is a net heat transfer from the surface to the atmosphere. If convection = 0, the surface would be hotter.

The fact that, at the ultimate boundary of the climate system, heat is only transferred via radiation has no bearing on what heat transfer mechanisms operate at the surface.”

I agree the surface has net convective loss to the atmosphere, which significantly cools the surface, but if the surface is cooler because of convection, how is it also returned via radiation as you claim above? Unless there is net energy loss, I don’t see how the surface can be cooler as a result of convection. Can you clarify what you mean by this?

The reason why the surface is cooler due to convection is kinetic energy from the surface is being traded off for radiative energy emitted at the surface, requiring the surface to emit less radiation to maintain equilibrium output power (239 W/m^2) at the TOA, as a lot of the convected energy from the surface to the atmosphere is ultimately being radiated out to space. In other words, without convection, the surface would need to emit more and subsequently be warmer as a result to maintain equilibrium. Because we are talking about the system in the steady-state, all of the kinetic energy trade offs from the surface to the atmosphere are automatically embodied in the surface emission of 385 W/m^2, which is just the net surface energy flux in and of itself.

So in regards to no. 3 above, NO that is not ‘my hypothesis’.

148. SoD,

Have you read Ray Pierrehumbert’s book?

• No.

• Are you aware the book is endorsed by RealClimate.org?