In Part One we looked at the usefulness of “tau” = optical thickness of the atmosphere.

Miskolczi has done a calculation (under cloudless skies) of the total optical thickness of the atmosphere. The reason he is apparently the first to have done this in a paper is explained in Part One.

The 2010 paper referenced the 2007 paper, *Greenhouse Effect in Semi-Transparent Planetary Atmospheres*, Quarterly Journal of the Hungarian Meteorological Service.

The 2010 paper suggested an elementary flaw, but referenced the 2007 paper. The 2007 paper backed up the approach with the same apparently flawed claim.

*The flaw that I will explain doesn’t affect the calculation of optical thickness, τ. But it does appear to affect the theoretical basis for why optical thickness should be a constant.*

First, the graphic explaining the terms is here:

*Figure 1*

The 2010 paper said:

One of the first and most interesting discoveries was the relationship between the absorbed surface radiation and the downward atmospheric emittance. According to Ref. 4, for each radiosonde ascent the

E_{D}= A_{A}= S_{U}– S_{T}= S_{U}(1− exp(−τ_{A})) = S_{U}(1− T_{A}) = S_{U}.A (5)

relationships are closely satisfied. The concept of radiative exchange was the discovery of Prevost [17]. It will be convenient here to define the term radiative exchange equilibrium between two specified regions of space (or bodies) as meaning that for the two regions (or bodies) A and B, the rate of flow of radiation emitted by A and absorbed by B is equal to the rate of flow the other way, regardless of other forms of transport that may be occurring.

Ref. 4 is the 2007 paper, which said:

According to the Kirchhoff law, two systems in thermal equilibrium exchange energy by absorption and emission in equal amounts, therefore, the thermal energy of either system can not be changed. In case the atmosphere is in thermal equilibrium with the surface, we may write that..

What is “thermal equilibrium“?

It is when two bodies are in a closed system and have reached equilibrium. This means they are at the same temperature and no radiation can enter or leave the system. In this condition, energy emitted from body A and absorbed by body B = energy emitted from body B and absorbed by body A.

Kirchhoff showed this radiative exchange must be equal under the restrictive condition of thermal equilibrium. And he didn’t show it for any other condition. (Note 2).

However, the earth’s surface and the atmosphere are **not** in thermal equilibrium. And, therefore, energy exchanged between the surface and the atmosphere via radiation is not proven to be equal.

Dr. Roy Spencer has a good explanation of the fallacy and the real situation on his blog. One alleged Miskolczi supporter took him to task for misinterpreting something – here:

With respect, Dr Spencer, it is not reasonable, indeed it verges on the mischievous, to write an allegation that Miskolczi means that radiative exchange is independent of temperature. Miskolczi means no such thing. To make such an allegation is to ignore the fact that Miskolczi uses the proper laws of physics in his calculations. Of course radiative exchange depends on temperature, and of course Miskolczi is fully aware of that.

and here:

..Planck uses the term for a system in thermodynamic equilibrium, and the present system is far from thermodynamic equilibrium, but the definition of the term still carries over..

I couldn’t tell whether the claimed “misinterpretation” by Spencer was of the real law or the Miskolczi interpretation. And this article will demonstrate that the *proper laws of physics* have been ignored.

And I have no idea whether the Miskolczi supporter represented the real Miskolczi. However, a person of the same name is noted by Miskolczi for his valuable comments in producing the 2010 paper.

Generally when people claim to overturn decades of research in a field you expect them to take a bit of time to explain why everyone else got it wrong, but apparently Dr. Spencer was deliberately misinterpreting something.. and that “something” is very clear only to Miskolczi supporters.

After all, the premise in the referenced 2007 paper was:

According to the Kirchhoff law, two systems in thermal equilibrium exchange energy by absorption and emission in equal amounts, therefore, the thermal energy of either system can not be changed.

In case the atmosphere is in thermal equilibrium with the surface, we may write that..

Emphasis added.

So if the atmosphere is **not** in thermal equilibrium with the surface, we can’t write the above statement.

And as a result the whole paper falls down. Perhaps there are other gems which stand independently of this flaw and I look forward to a future paper from the author when he explains some new insights which don’t rely on thermodynamic equilibrium being applied to a world without thermodynamic equilibrium.

### Thermodynamic Equilibrium and the Second Law of Thermodynamics

If you put two bodies, A & B, at two different temperatures, T_{A} and T_{B}, into a closed system then over time they will reach the same temperature.

Let’s suppose that T_{A} > T_{B}. Therefore, A will radiate more energy towards B than the reverse. These bodies will reach equilibrium when T_{A} = T_{B} (note 1).

At this time, **and not before**, we can say that ” ..two systems in thermal equilibrium exchange energy by absorption and emission in equal amounts”. (Note 2).

Obviously, **before** equilibrium is reached more energy is flowing from A to B than the reverse.

### Non-Equilibrium

Let’s consider a case like the sun and the earth. The earth absorbs around 240 W/m² from the sun. The sun absorbs a lot less from the earth.

Let’s just say it is a lot less than 1 W/m². Someone with a calculator and a few minutes spare can do the sums and write the result in the comments.

No one (including of course the author of the paper) would suggest that the sun and earth exchange equal amounts of radiation.

However, they are in the condition of “radiative exchange”.

### The Earth’s Surface and the Atmosphere

The earth’s surface and the bottom of the atmosphere are at similar temperatures. Why is this?

It is temperature difference that drives heat flow. The larger the temperature difference the greater the heat flow (all other things remaining equal). So any closed system tends towards thermal equilibrium. If the earth and the atmosphere were left in a closed system, eventually both would be at the same temperature.

However, in the real world where the climate system is open to radiation, the sun is the source of energy that prevents thermal equilibrium being reached.

The bottom millimeter of the atmosphere will usually be at the same temperature as the earth’s surface directly below. If the bottom millimeter is stationary then it will be warmed by conduction until it reaches almost the surface temperature. But 10 meters up the temperature will probably reduce just a little. At 1 km above the surface the temperature will be between 4 K and 10 K cooler than the surface.

*Note*: Turbulent heat exchange near the surface is very complex. This doesn’t mean that there is confusion about the average temperature profile vs height through the atmosphere. On average, temperature reduces with height in a reasonably predictable manner.

### Energy Exchanges between the Earth’s Surface and the Atmosphere

According to Miskolczi:

A_{A} = E_{D} [4]

Referring to the diagram, A_{A} is energy absorbed by the atmosphere from the surface, and E_{D} is energy radiated from the atmosphere to the surface.

Why should this equality hold?

The energy from the surface to the atmosphere = A_{A}+ K (note 3), where K is convection.

The energy absorbed in total by the atmosphere = A_{A} + K + F, where F is absorbed solar radiation in the atmosphere.

The energy emitted by the atmosphere = E_{D} + E_{U} , where E_{U} is the energy radiated from the top of the atmosphere.

Therefore, using the First Law of Thermodynamics for the atmosphere:

A_{A} + K + F = E_{D} + E_{U} + energy retained

i.e., energy absorbed = energy lost – energy retained

No other equality relating to the atmospheric fluxes can be deduced from the fundamental laws of thermodynamics.

**In general**, because the atmosphere and the earth’s surface are very close in temperature, A_{A} will be very close to E_{D}.

It is important to understand that absorptivity for longwave radiation will be equal to emissivity for longwave radiation (see Planck, Stefan-Boltzmann, Kirchhoff and LTE), therefore, if the surface and the atmosphere are at the same temperature then the exchange of radiation will be equal.

Where does the atmosphere radiate from, on average? Well, not from the bottom meter. It depends on the emissivity of the atmosphere. This varies with the amount of water vapor in the atmosphere.

The atmospheric temperature reduces with height- by an average of around 6.5 K/km – and unless the atmospheric radiation was from the bottom few meters, the radiation from the atmosphere to the surface must be lower than the radiation absorbed from the surface by the atmosphere.

If radiation was emitted from an average of 100 m above the surface then the effective temperature of atmospheric radiation would be 0.7 K below the surface temperature. If radiation was emitted from an average of 200 m above the surface then the effective temperature of atmospheric radiation would be 1.3 K below the surface temperature.

### Mathematical Proof

For people still thinking about this subject, a simple mathematical proof.

Temperature of the atmosphere, from the average height of emission, T_{a}

Temperature of the surface, T_{s}

Emissivity of the atmosphere = ε_{a}

Absorptivity of the atmosphere for surface radiation = α_{a}

If T_{a} is similar to T_{s} then ε_{a} ≈ α_{a} (note 4).

(In the paper, the emissivity (and therefore absorptivity) of the earth’s surface is assumed = 1).

Surface radiation absorbed by the atmosphere, A_{A} = α_{a}σT_{s}^{4} .

Atmospheric radiation absorbed by the surface, E_{D} = ε_{a}σT_{a}^{4} .

**Therefore, unless T _{a} = T_{s}, A_{A} ≠ E_{D}** .

If Roy Spencer’s experience is anything to go by, I may now be accused of deliberately misunderstanding something.

Well, words can be confused – even though they seem plain enough in the extract shown. But the paper also asserts the mathematical identity:

A_{A} = E_{D} [4]

I have demonstrated that:

A_{A} ≠ E_{D} [4]

I don’t think there is much to be misunderstood.

*Two bodies at different temperatures will NOT exchange exactly equal amounts of radiation. It is impossible unless the current laws of thermodynamics are wrong.*

*As a more technical side note*.. because ε_{a} ≈ α_{a} and not necessarily an exact equality, it is possible for the proposed equation to be asserted in the following way:

A_{A} = E_{D} if, and only if, the following identity is always true, α_{a}(T_{s})σT_{s}^{4} = ε_{a}(T_{a})σT_{a}^{4} .

Therefore:

T_{s}/T_{a} = (ε_{a}(T_{a})/α_{a}(T_{s}))^{1/4} [Equation B]

– must always be true for equation 4 of Miskolczi (2007) to be correct. *Or must be true over whatever time period and surface area his identity is claimed to be true.*

Another quote from the 2007 paper:

The popular explanation of the greenhouse effect as the result of the LW atmospheric absorption of the surface radiation and the surface heating by the atmospheric downward radiation is incorrect, since the involved flux terms (A

_{A}and E_{D}) arealwaysequal.

Emphasis added.

Note in Equation B that I have made explicit the dependence of emissivity on the temperature of the atmosphere at that time, and the dependence of absorptivity on the temperature of the surface.

Emissivity vs wavelength is a material property and doesn’t change with temperature. But because the emission wavelengths change with temperature the calculation of ε_{a}(T_{a}) is the measured value of ε_{a} at each wavelength weighted by the Planck function at T_{a}.

It is left as an exercise for the interested student to prove that this identity, Equation B, cannot always be correct.

### The “Almost” Identity

In Fig. 2 we present large scale simulation results of AA and ED for two measured diverse planetary atmospheric profile sets. Details of the simulation exercise above were reported in Miskolczi and Mlynczak (2004). This figure is a proof that the Kirchhoff law is in effect in real atmospheres. The direct consequences of the Kirchhoff law are the next two equations:

EU = F + K + P (5)

SU − (F0 + P0 ) = ED − EU (6)The physical interpretations of these two equations may fundamentally change the general concept of greenhouse theories.

*Figure 2*

This is not a proof of Kirchhoff’s law, which is already proven and is not a law that radiative exchanges are equal when temperatures are **not** equal.

Instead, this is a demonstration that the atmosphere and earth’s surface are very close in temperature.

Here is a simple calculation of the ratio of A_{A}:E_{D} for different downward emitting heights (note 5), and lapse rates (temperature profile of the atmosphere):

*Figure 3*

Essentially this graph is calculated from the formula in the maths section and a calculation of the atmospheric temperature, T_{a}, from the height of average downward radiation and the lapse rate.

### Oh No He’s Not Claiming This is Based on Kirchoff..

Reading the claims by the supporters of Miskolczi at Roy Spencer’s blog, you read that:

- Miskolczi is not claiming that A
_{A}= E_{D}by asserting (incorrectly) Kirchhoff’s law - Miskolczi is claiming that A
_{A}= E_{D}by experimental fact

So the supporters claim.

Read the paper, that’s my recommendation. The 2010 paper references the 2007 paper for equation 4. The 2007 paper says (see larger citation above):

..This figure is a proof that the Kirchhoff law is in effect in real atmospheres..

In fact, this is the important point:

Anyone who **didn’t** believe that it was a necessary consequence of Kirchhoff would be writing the equations in the maths section above (which come from well-proven radiation theory) and realizing that it is impossible for A_{A} = E_{D}.

And they wouldn’t be claiming that it demonstrated Kirchhoff’s law. (After all, Kirchhoff’s law is well-proven and foundational thermodynamics).

However, *it is* certain that on average E_{D} < A_{A} but very close to A_{A}.

### Hence the Atmospheric Window Cooling to Space Thing

From time to time, Miskolczi fans have appeared on this blog and written interesting comments. Why the continued fascination with the exact amount of radiation transmitted from the surface through the atmospheric window?

I have no idea whether this point is of interest to anyone else..

One of the comments highlighted the particular claim and intrigued me.

Yes, indeed, that’s right: Simpson discovered the atmospheric window in 1928. It was not till the work of Miskolczi in 2004 and 2007 that it was discovered that practically all the radiative cooling of the land-sea surface is by radiation direct to space.

Apart from the (unintentional?) humor inherent in the Messianic-style claim, the reason why this claim is a foundational point for Miskolczi-ism is now clear to me.

If exactly all of the radiation absorbed by the atmosphere is re-radiated to the surface and absorbed by the surface (A_{A} = E_{D}) then these points follow for certain:

- radiation emitted by the atmosphere to space = convective heat from the surface into the atmosphere + solar radiation absorbed by the atmosphere
- total radiative cooling to space = radiation transmitted through the atmospheric window + convective heat plus solar radiation absorbed by the atmosphere

A curiosity only.

### Changing the Fundamental View of the World

Miskolczi claims:

The physical interpretations of these two equations may fundamentally change the general concept of greenhouse theories.

He is being too modest.

If it turns out that A_{A} = E_{D} then it will overturn general radiative theory as well.

Or demonstrate that the atmosphere is much more opaque than has currently been calculated (for all of the downward atmospheric radiation to take place from within a few tens of meters of the surface).

This in turn will require the overturning of some parts of general radiative theory, or at least, a few decades of spectroscopic experiments, which consequently will surely require the overturning of..

### Conclusion

How is it possible to claim that A_{A} = E_{D} and not work through the basic consequences (e.g., the equations in the maths section above) to deal with the inevitable questions on thermodynamics basics?

Why claim that it has fundamentally changed the the general concept of the inappropriately-named “greenhouse” theory when it – if true – has overturned generally accepted radiation theory?

- Perhaps α(λ) ≠ ε(λ) and Kirchhoff’s law is wrong? This is a possible consequence. (
*In words, the equation says that absorptivity at wavelength λ is not equal to emissivity at wavelength λ, see note 4*). - Or perhaps the well-proven Stefan-Boltzmann law is wrong? This is another possible consequence.

Interested observers might wonder about the size of the error bars in Figure 2. (And for newcomers, the values in Figure 2 are not measured values of radiation, they are calculated absorption and emission).

As already suggested, perhaps there are useful gems somewhere in the 40 pages of the 2007 paper, but when someone is so clear about a foundational point for their paper that is so at odds with foundational thermodynamic theory and the author doesn’t think to deal with that.. well, it doesn’t generate hope.

*Update 31st May – the author comments in the ensuing discussion that Aa=Ed is an “experimental” conclusion. In Part Four I show that the “approximate equality” must be an error for real (non-black) surfaces, and Ken Gregory, armed with the Miskolczi spreadsheet, later confirms this.*

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 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 Four – a minor digression into another error that seems to have crept into the Aa=Ed relationship*

*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.*

*Further Reading:*

*New Theory Proves AGW Wrong! – a guide to the steady stream of new “disproofs” of the “greenhouse” effect or of AGW. And why you can usually only be a fan of – at most – one of these theories.*

### References

*Greenhouse Effect in Semi-Transparent Planetary Atmospheres*, Miskolczi , *Quarterly Journal of the Hungarian Meteorological Service* (2007)

*The Stable Stationary Value of the Earth’s Global Average Atmospheric Planck-Weighted Greenhouse-Gas Optical Thickness*, Miskolczi, *Energy & Environment*(2010)

*The Theory of Heat Radiation*, Max Planck, *P. Blakiston’s Son & Co* (1914) : a translation of *Waermestrahlung* (1913) by Max Planck.

### Notes

**Note 1** – Of course, in reality equilibrium is never actually reached. As the two temperatures approach each other, the difference in energy exchanged is continually reduced. However, at some point the two temperatures will be indistinguishable. Perhaps when the temperature difference is less than 0.1°C, or when it is less than 0.0000001°C..

Therefore, it is conventional to talk about “reaching equilibrium” and no one in thermodynamics is confused about the reality of the above point.

**Note 2** – Max Planck introduces thermodynamic equilibrium:

**Note 3** – Geothermal energy is included in the diagram (P^{0}). Given that it is less than 0.1 W/m² – below the noise level of most instruments measuring other fluxes in the climate – there is little point in cluttering up the equations here with this parameter.

**Note 4** – Emissivity and absorptivity are wavelength dependent parameters. For example, snow is highly reflective for solar radiation but highly absorbing (and therefore emitting) for terrestrial radiation.

**At the same wavelength, emissivity = absorptivity**. This is the result of Kirchhoff’s law.

If the temperature of the source radiation for which we need to know the absorptivity is different from the temperature of the emitting body then we cannot assume that emissivity = absorptivity.

However, when the temperature of source body for the radiation being absorbed is within a few Kelvin of the emitting body then to a quite accurate assumption, absorptivity = emissivity.

For example, the radiation from a source of 288K is centered on 10.06 μm, while for 287 K it is centered on 10.10 μm. Around this temperature, the central wavelength decreases by about 0.035 μm for each 1 K change in temperature.

An example of when it is a totally incorrect assumption is for solar radiation absorbed by the earth. The solar radiation is from a source of about 5800 K and centered on 0.5 μm, whereas the terrestrial radiation is from a source of around 288 K and centered on 10 μm. Therefore, to assume that the absorptivity of the earth’s surface for solar radiation is equal to the emissivity of the earth’s surface is a huge mistake.

This would be the same as saying that absorptivity at 0.5 μm = emissivity at 10 μm. And, therefore, totally wrong.

**Note 5**: What exactly is meant by average emitting height? Emitted radiation varies as the 4th power of temperature and as a function of emissivity, which itself is a very non-linear function of quantity of absorbers. Average emitting height is more of a conceptual approach to illustrate the problem.

on April 25, 2011 at 1:18 pm |Miklos ZagoniSoD:

Let me ask some questions.

I. Are you able to compute Aa and Ed on realistic vertical profiles with a high-level accuracy? If yes, what are your numbers?

II. The surface and the lowest layer of the atmosphere (being in permanent dynamic contact with each other) are at the same global average temperature or not?

III. Did you see the newest “Earth radiation budget revisited”-kind articles and estimates, from very reliable sources? What do you think about their numbers? Here they are:

source (a) Su=397, K=108, F=80, Eu=188, OLR=240,

St = 52, Ed=345

source (b) Su=393, K=101, F=97, Eu=198, OLR=240,

St = 45, Ed=348

How much are their Aa?

IV. You wrote:

“Apart from the (unintentional?) humor inherent in the Messianic-style claim, the reason why this claim is a foundational point for Miskolczi-ism is now clear to me.

If exactly all of the radiation absorbed by the atmosphere is re-radiated to the surface and absorbed by the surface (AA = ED) then these points follow for certain:

1. radiation emitted by the atmosphere to space = convective heat from the surface into the atmosphere + solar radiation absorbed by the atmosphere

2. total radiative cooling to space = radiation transmitted through the atmospheric window + convective heat plus solar radiation absorbed by the atmosphere

A curiosity only.”

Yes, your 1. and 2. “curiosity” are there in the numbers. Their meaning is that Ramanathan’s greenhouse factor, G = Su-OLR, also equals to G = Ed-Eu , meaning that the greenhouse effect in our radiative-convective atmoshere is the difference of downward radiative heating and the upward convective cooling — see

http://miskolczi.webs.com/SUMMARY.htm

V. How much is the normalized clearsky greenhouse effect? Is it around g = 0.334 somewhere? What do you get from Miskolczi’s first rule (just analyzed above), applying some further direct physical principles? — Let mie give you the answer here: you will get Su = 3OLR / 2.

What does this mean? This means that g = 1/3, theoretically, and independently of the trace-gas concentration, maintanined by your “curious” system.

VI. Would you explain please the graph given in slide 11 of

VII. Would you explain please the two plots given in slide 12 of the same presentation? (I cab help: it shows the validity of the afore-mentioned equality on temperature-inversion profiles. We have 40 others from this type, with the same result.)

Would you repeat please the same kind of computations, showing that these results are not valid?

on April 25, 2011 at 1:29 pm |Miklos ZagoniFurther:

VIII. Textbooks in the radiative transfer literature are talking about the identity of LTE and Kirchhoff Law. One of them says: LTE is the name of those circumstances in the atmosphere when K.L. is valid. Is our atmosphere in LTE, up to some 60 kms, or not?

IX. The P° term is included in the diagram because there are other possibe heat sources (industrial heat generation, ocean-atmosphere heat exchange) that can play a role; and, on other planets, geothermal energy might be higher.

X. Did you know? The Aa=Ed equality is valid on the known NASA-measured Martian profiles as well.

on April 25, 2011 at 3:40 pm |DeWitt PayneSoD,

You may have found the fatal flaw that everyone else has missed. Without the assumption of equilibrium, none of M’s relationships are identities, but only approximations. If the atmosphere were in thermal equilibrium with the surface, there would be no convective heat transfer and no greenhouse effect because the atmosphere would be isothermal.

Ed can be greater than Aa if there’s a temperature inversion as in the Arctic winter and Ta > Ts.

on April 25, 2011 at 5:36 pm |Miklos ZagoniDeWitt Payne:

The atmosphere is in thermal equilibrium with the surface by the contribution of the convective heat transfer. It is of course accounted as a part of its energetic ins and outs.

on April 25, 2011 at 4:01 pm |james kennedy“The comings and goings of CO2 are lost

in the rain and the wind.”

The back and forth between “Climate of Doom”

and Mr. Zagoni is highly educational for me. This is all

the more true for me because I am, as I just

said above, more aligned with his thinking than I am with

that of the “Climate of Doom” site.

I have enjoyed reading both your objections

and Mr. Zagoni’s very thoughtful

and detailed replies to them.

.

.

.

.

I see your recent objections and Mr. Zagoni’s

responses to them as further evidence that,

generally speaking, this “Climate of Doom”

site misses the thermodynamic

forest for its studying of radiative trees.

Which is to say, the set point for earth’s

surface temp. has very nearly nothing

at all to do with the details of radiative transfer,

such as Schwarzchild’s Equation.

Which is to say, “Climate of Doom” still does

not seem to give enough weight to the thermodynamic

realities such as g = 1/3, and why details related

to CO2 cannot change that fact.

I see the works of Miskolczi/Zagoni as

asserting the dominance of water and

wind related thermodynamics

over the climate details deriving from CO2s

presence in the atmosphere.

As a test of what I have just said above, and as

a guard against it being seen as too broad and

general, I will see what I am able to do

with your “….0.035…” issue using the humidity

data Miscolczi has accumulated.”

on April 25, 2011 at 5:52 pm |Miklos ZagoniSoD,

You did not mention: Roy Spencer had accepted that tau=1.87 . It is free to you to get this number on your most trusted global average atmospheric profile by your favorite computational method. I just would like to note that if you want to derive this number theoretically, the Aa=Ed equality is necessary.

So really many thanks for your efforts to understand, but interpretations and explanations could come only after you showed up your own numbers.

on April 25, 2011 at 7:27 pm |james kennedyThe case I was trying to make in that recent

post that began with

“The comings and goings of CO2 are lost

in the rain and the wind.”

is very well elaborated here:

http://globalconnexus.net/blog/?p=372

.

.

.

The trend I see is an increasing appreciation

and understanding of the work of Miscolczi/Zagoni

and, correspondingly, a weakening belief that

CO2 plays a major role in the climate.

on April 25, 2011 at 10:23 pm |scienceofdoomMiklos Zagoni on April 25, 2011 at 1:29 pm:

The atmosphere is in LTE up to around 60km. The atmosphere is not in TE anywhere.

You can read the description of

Thermodynamics Equilibrium(TE) in the extract from Max Planck in Note 2.You can various descriptions of

Local Thermodynamic Equilibrium(LTE) in Planck, Stefan-Boltzmann, Kirchhoff and LTE.They are quite different conditions.

Some of Kirchhoff’s insights are still valid in non-TE conditions.

For example, at any given wavelength, absorptivity = emissivity. A body does not need to be in TE for this to be true.

However, if you claim that bodies at different temperatures exchange equal amounts of radiation due to Kirchhoff you are claiming that you have overturned basic thermodynamics. Because for well over 100 years everyone has known that it is NOT true.

Or demonstrating that you are confused about the very basics of heat transfer.This is demonstrated by the very simple equations (above) where I demonstrate:

A

_{A}≠ E_{D}This proof is trivial because this is elementary heat transfer.

And from your earlier comments at April 25, 2011 at 1:18 pm:

Simply demonstrates that you haven’t understood any of the content of the article.

on April 25, 2011 at 10:33 pm |scienceofdoomMiklos Zagoni:

Yes, that article already says that.

But the average atmospheric emission absorbed by the surface is

notemitted from the bottom millimeter, or the bottom meter.If we consider 667cm

^{-1}then the emission of radiation is effectively from the bottom meter.If we take an example of a different wavenumber.. e.g., 645cm

^{-1}then the emission is from around 100m (very approximately). At 645cm^{-1}the atmosphere is not nearly as opaque. The transmittance of 100m of atmosphere near the surface at 645cm^{-1}is around 0.5 due to CO2. Depending on the amount of water vapor in the atmosphere, this transmittance will reduce.This means that the atmospheric radiation absorbed by the surface comes from a range of heights, with an average around 100m. Or a little less, as water vapor increases.

Surely the point is clear?

If you can demonstrate that the atmosphere has a transmittance of less than 0.1 (or 0.2? in the bottom meter across the complete band of emission then you will achieved something amazing!

In fact, if you realized that this was a necessity you would have demonstrated it already.

The fact that the paper doesn’t comment on this necessary topic simply means the author of the paper has misunderstood the very basics.

on April 25, 2011 at 10:39 pm |scienceofdoomAnd just as a semi-technical note on the above comment -strictly speaking the transmittance of the atmosphere doesn’t need to be very low across the

entirelongwave band for the atmosphere to emit from the bottom meter.The atmosphere could be totally transparent (transmittance = 1) in some bands. This wouldn’t affect the average height of emission, because in these bands there would be zero emission (and zero absorption from the surface).

on April 25, 2011 at 10:47 pm |scienceofdoomjames kennedy on April 25, 2011 at 4:01 pm:

If you are going to comment, try and read the article you are allegedly commenting on.

The article discusses thermodynamics basics and shows a trivial proof that A

_{A}≠ E_{D}.If you just want to cheer for a side then you have already done that, so no need to post a new comment.

If you want to actually discuss something scientific please go ahead and deal with an

actual pointfrom the article.on April 26, 2011 at 12:29 am |DeWitt PayneMiklos Zagoni,

It is possible to calculate Ed and Aa using MODTRAN. I’ve done it. Ed is never exactly equal to Aa for clear sky conditions. Aa is always higher than Ed except for the sub-Arctic winter conditions where a temperature inversion exists. For cloud covered sky, Aa is very close to Ed, but that’s because for a cloud covered sky, you have what amounts to two parallel planes at effectively the same temperature.

atmosphere Aa W/m2 Ed W/m2

1976 US 292 281

Tropical 388 380

Midlat Summer 344 337

Midlat Winter 223 221

SubArctic Summer 299 292

SubArctic Winter 167 172

on April 26, 2011 at 12:42 am |DeWitt PayneMiklos Zagoni,

At the risk of piling on, thermal equilibrium (TE) has a very specific meaning. It means that all objects are at the same temperature and there is no net heat transfer and no gain or loss of energy from the system. The Earth is not in thermal equilibrium with the atmosphere. It’s not in thermal equilibrium with the sun and it’s not in equilibrium with deep space. Even the energy content isn’t constant over long periods.

Local thermal equilibrium in the atmosphere also has a specific meaning. It means that energy transfer is dominated by collision and the distribution of kinetic energy of the molecules follows the Maxwell-Boltzmann distribution. Kirchhoff’s Law applies to systems in LTE or TE. LTE does not mean isothermal on a macro scale. TE does.

on April 26, 2011 at 1:02 am |scienceofdoomMiklos Zagoni on April 25, 2011 at 5:52 pm:

I doubt that you understand the subject you claim to be supporting.

A

_{A}= E_{D}is not necessary to calculate the optical thickness of the atmosphere. Nor is A_{A}≠ E_{D}necessary.The optical thickness is calculated via the weighted integral of the path lengths x absorption coefficients.

There are no fluxes involved in this calculation.

And no assumption about fluxes.

You can see this on p244 (2nd page) of Miskolczi’s 2010 paper, eqs 1 & 2.

on April 26, 2011 at 7:02 am |Miklos ZagoniSod:

This kind of arguments that “I doubt that you understand the subject you claim to be supporting” are not too useful. I know pretty well how tau is calculated, where Aa=Ed is necessary and where not.

It would be a wrong direction if I followed your way, saying your quoted sentence on you.

* * *

DeWitt Payne:

I think we agreed once that USST76 is not a good representation of the global average, not even a realistic one.

The question is how the real atmosphere works. This needs accurate approach, not only approximations.

I would like to call your attention to two most recent global energy budget estimates from very highly respected soruces:

(a) Su=397, K=108, F=80, Eu=188, OLR=240,

St = 52, Ed=345 .

(b) Su=393, K=101, F=97, Eu=198, OLR=240,

St = 45, Ed=348 .

The Aa=Ed equation is exact in each. (They do not refer to M.)

on April 26, 2011 at 5:20 am |The Mystery of Tau – Miskolczi – Part Three – Kinetic Energy « The Science of Doom[…] Comments « The Mystery of Tau – Miskolczi – Part Two – Kirchhoff […]

on April 26, 2011 at 7:11 am |scienceofdoomMiklos Zagoni on April 26, 2011 at 7:02 am:

Claiming it doesn’t prove it. The paper says differently.

If you

believeit, that’s fine with me. I’m sure you do.But if you want to

proveit, then either you need to show where the paper demonstrates it. Or you need to quote from the author of the paper where he says it includes some unstated points.And in that case, please provide an updated equation for how tau is

really calculated.on April 26, 2011 at 7:36 am |Miklos ZagoniTo calculate the empirical tau on the measured profiles, eqs 1 and 2 in M2010 are needed. The result is about 1.87, both on TIGR and NOAA.

Is is possible to derive theoretically an equilibrium tau, simply from the four new relationships (here Aa=Ed is one of them). The solution of the resulting equation is tau=1.867561…..

on April 26, 2011 at 7:20 am |Miklos ZagoniSod:

With all respect: you are fighting against interpretations, not against the equations.

M. partitioned the components of Ramanathan’s G as Su=Aa+St and OLR=Eu+St. He computed them spectrally and integrally by reproducable methods on verifiable databases. I think such a reproduction is needed.

He realized some empirical relationships amongst them.

Are these relationships valid or not? If not, what are the correct relations? If yes, what are the consequences?

[He did not assume Aa=Ed as a condition of the tau-calculation. On the contrary, after calculating “tau” (that is, Aa, St, Eu, and Ed) he realized this relation, as you can see from the 2010 paper. (I still wonder why do you think the opposite.)]

He used to say that the theoretically sound approach to the greenhouse problem is to get Aa, St, Eu, and Ed as a function of the thermal and chemical distribution and composition of the atmosphere — that is, to find the proper Aa(T, p, GHG …) etc. In short, to find g(tau). He gave one.

What kind of physical principles are there in its background? Prévost? Kirchhoff? Planck? Boltzmann? Helmholtz? Hamilton? Radiative exchange equilibrium? Local thermodynamic equilibrium? Entropy maximum? Energy minimum? Most effective cooling? Principle of least action? Radiation pressure? Virial rule? The minimum of gravitational potential energy? None of them? All of them? (Max Planck wrote that Kirchhoff was not a good teacher. He seems to be correct also in this…)

Schrodinger wrote his equation in 1926. What they mean? DeBroglie’s material waves? Heisenberg’s matrices? Bohr’s complementarity? David Bohm’s causality? Max Born’s probability? All of them are paradoxical, according to EPR? Wrong, as said by Wigner’s Cat? Is it only an approximation of Dirac’s relativistic case? The debate is still open, after eight decades, but the equation is there — and works.

on April 26, 2011 at 8:04 am |scienceofdoomMiklos Zagoni:

I don’t think you have understood my comment of April 26, 2011 at 1:02 am at all.

In any case my original comment in the article was not a central point, it was an unimportant explanation.

So I look forward to you pointing out the flaws on my central point.

So far you have not addressed my central point, except to repeat that Miskolczi did a calculation. Which we already know.

I have said that A

_{A}is close to E_{D}but not equal. It cannot be equal and I have demonstrated that from basic thermodynamic equations.If Miskolczi has proven that two bodies at different temperatures exchange equal amounts of radiation

then this is a thermodynamic revolution.The fact that neither of you have realized this suggests a major problem..

The fact that this was claimed as supporting Kirchhoff’s law suggests a major problem, seeing as his law is actually under a very restrictive condition that is not seen in the atmosphere..

on April 26, 2011 at 8:17 am |Miklos ZagoniTwo recent “official” energy budget estimates:

— Su=397, K=108, F=80, Eu=188, OLR=240, St = 52, Ed=345 .

— Su=393, K=101, F=97, Eu=198, OLR=240,

St = 45, Ed=348 .

The Aa=Ed equation is exact in each. (They do not refer to M.)

Two bodies in permanent physical connection NOT in temperature eqauilibrium IS the thermodynamic revolution.

on April 26, 2011 at 8:38 am |scienceofdoomMiklos Zagoni:

Can you address the points I made?

Height of average emission?

Temperature drop with altitude?

Is the atmosphere at 100m high in permanent physical connection with the surface?

Is the atmosphere at 200m high in permanent physical connection with the surface?

Are they at the same temperature?

Do you understand the relevance of these points?

Do you understand why I said:

Emphasis added.

So my suggestion if you want to engage in scientific debate – either

a) show why the equation is wrong, OR

b) show

whyTa = Ts, with reference to the height of emission and the lapse rateOr carry on ignoring the points..

on April 26, 2011 at 9:16 am |Miklos ZagoniThe altitude (temperature) dependent contribution density functions to Aa and Ed are given in M2010. Look at eq 7 for example. To all layer, the total absorption above that layer equals to Ed coming down to that layer, and the contribution to absoprtion and emission for each layers are also given. Look at the GAT (global average TIGR) lapse rate profile in Figs. 3, 4, and 6, and the four different specific profiles in Fig 5. The computational global average differenece is about 3 %.

You are not an average blogger, you are the site-owner. Your style about “understanding, ignorance and scientific debate” of your readers should keep a standard. That’s unique enough in the blogosphere where the blogkeeper insults his readers. I got the strong feeling you did not understand a word from the article you are just commenting.

on April 26, 2011 at 10:13 am |scienceofdoomMiklos Zagoni:

You claim to represent a scientific paper that is a revolution.

Yet your responses to my points – which are very simple – are very confused.

I said:

And your response is that the paper agrees that the temperature changes with height. After before saying that the atmosphere and surface were at the same temperature.

Don’t you think that demonstrates my point?

If you can’t make sense out of the choice between the two points – or why I raised them, then you can claim I understand nothing if you like.

That’s one possibility.

I will let the readers decide.

Not a lot of point discussing this further with you.

on April 26, 2011 at 2:38 pm |james kennedyClimate of Doom said:

“If you want to actually discuss something scientific please go ahead and deal with an actual point from the article.”

James Kennedy replies:

I could of course be wrong, but I thought

I had made it more than clear that

I was NOT discussing a SPECIFIC point in the article.

I thought that I was discussing SEVERAL points in

the article where you seem to think that certain

details of magnitudes are relevant to M’s

line of reasoning.

Let me see if I can be clearer by referring to

a recent writing

“I doubt that you understand the subject you claim to be supporting.

AA = ED is not necessary to calculate the optical thickness of the atmosphere. Nor is AA ≠ ED necessary.

”

.

.

from the recent post by Z.

If I have begun to understand Z and M, and I can certainly

not say so with confidence, I interpret what I just

quoted as meaning that

.

whether AA = ED, or, not,

.

is an issue of detail that is not relevant to the

basic argument of M.

That is to say, if I do begin to understand

M and Z, you find fault with them by raising

an issue of magnitudes detail regarding

AA and Ed which is moot.

Other details of magnitudes which you raise

with regard to radiative transfers are, if

I do, in fact, begin to understand M and Z, largely moot.

on April 26, 2011 at 3:11 pm |DeWitt PayneMiklos Zagoni,

Cite please.

K&T97 has Ed = 324 W/m2 and Aa = 350 W/m2.

TFK09 has Ed = 333W/m2 and Aa = 356W/m2

You do know that τ for the energy balances you cited is nowhere near 1.87 don’t you?

Example 1 τ = -ln(52/397) = 2.03

Example 2 τ = -ln(45/393) = 2.17

on April 26, 2011 at 3:22 pm |DeWitt PayneAnd that Su + Ed – Eu ≠ 2Fo for either example as well.

on April 26, 2011 at 3:42 pm |DeWitt PayneMiklos Zagoni,

In slide number 9 of EGU2011, the minimum St is 22.2 W/m2. Does this mean that the subset of TIGR2 included only clear days? Or was the calculation done assuming that there were no clouds? St = 0 for a cloud covered sky. If clouds were ignored for the calculations then they are simply wrong. If only clear days were used then the calculations only apply to 40% of the Earth’s surface.

on April 26, 2011 at 11:18 pm |DeWitt PayneI think I can answer that. The TIGR-2 atmospheres are a set of initial guess profiles for inverting satellite IR readings (TOVS, ATOVS, etc.) to temperature and water vapor profiles. The problem is ill-posed so you have to have a pretty good guess as a starting point for the inversion. If you look into the details, you find that the satellite has to be able to see to nearly the surface, i.e. a pressure level of 950 mbar or so. That means no clouds. So τ = 1.87 only applies to the 40% of the Earth’s surface not covered by clouds. Somehow, I don’t think that means much.

on April 26, 2011 at 11:05 pm |DeWitt PayneMiklos Zagoni,

Then there’s the precipitable water problem. There’s no way you can have 2.61 cm precipitable water vapor at a surface temperature of 288.9 K. Most of that water is in the tropical atmosphere. Looking at the global average change in precipitable water doesn’t really tell you much. It’s kind of like looking at the global average temperature, you throw out the most important information.

on April 27, 2011 at 8:56 am |BryanMiklos Zagoni says of SoD

You are not an average blogger, you are the site-owner. Your style about “understanding, ignorance and scientific debate” of your readers should keep a standard. That’s unique enough in the blogosphere where the blogkeeper insults his readers. I got the strong feeling you did not understand a word from the article you are just commenting.

Well said Miklos, a number of former posters would, I’m sure, agree!

on April 27, 2011 at 3:23 pm |DeWitt PayneBryan,

We’re still waiting for your explication of where KT97 and TFK09 went wrong, not to mention the energy balances that Zagoni quoted but didn’t reference.

on April 30, 2011 at 12:34 amFerenc MiskolcziHi DeWitt,

Let us forget about Mark Twain – I want to help you. In the EGU2011 I had a presentation with the title:

F. Miskolczi: The stable stationary value of the Earth’s global average atmospheric infrared optical thickness.

Poster-presentation at the European Geosciences Union General Assembly, 7 April 2011, Vienna, presented by M. Zagoni.

Maybe you have seen this, but look at slide 19, and compare the TIGR 2 and the Kiehl-Trenberth tau curves. If you do not understand this, then my articles were not written to you (or SoD or Neal or Alexandra).

Anyway, it looks that my mystery tau has some role in the climate system, no matter the 3% deviation from the Kirchhoff law (if you know what it means). The Aa=Ed approximation is still far better than the ad-hoc feedback parameters of the GCMs.

The main point is that tau=1.87 and the AGW due to CO2 greenhouse effect is non existent. To fight with this you should show that the atmosphere absorbed more IR radiation 60 years ago, and come up with your own mystry tau – as Zagoni already suggested.

But be careful, once I told you that the MODTRAN or even other LBL codes are not good enough, you should write your own radiative transfer code which riogorously obey first principles (like monochromatic Beer-Lambert law)….best wishes for that.

on April 30, 2011 at 5:49 am |scienceofdoomFerenc Miskolczi:

Welcome to the discussion of your paper.

Hopefully, you can clear up the many questions asked in these three articles (and in the comments).

on April 30, 2011 at 11:28 am |Ferenc MiskolcziCorrection – in my comment on April 30 :

….atmosphere absorbed less IR radiation 60 years ago,….

on April 30, 2011 at 12:24 pm |Ferenc MiskolcziSoD you say:

“I have said that AA is close to ED but not equal. It cannot be equal and I have demonstrated that from basic thermodynamic equations.”

Maybe I am missing something, but forgive me, I shall not go through your elementary radiative transfer and thermodynamics.

I think I was the first who showed the Aa~=Ed relationship with reasonable quantitative accuracy. If not, you may reference to some published Aa and Ed. I showed when Aa must be larger than Ed and when Aa must be smaller than Ed. If interested I can show you some structures when they are equal. The reason why they are not exacty equal is simply the anisotropy in the thermal structure. Just study carefully slide 8 in http://presentations.copernicus.org/EGU2011-13622_presentation.pdf . (Here E_D^i is the downward emittance from a homogeneous atmosphere with t=ts temperature).

Your problem is, that Aa (and St and Eu) can not be measured, so you must compute St=Su*exp(-tau). The reason why most

of the energy budget cartoons fail is the incorrect assumption how to interpret (or calibrate) stellite broadband (windows) radiance measurements, and the lack of a coherent greenhouse theory which relates Su to OLR.

Regarding your complaint about the excessive referencing to my own articles I can only tell, that as soon as somebody else is willing to compute and write articles on the tau and the analytical relationships among the atmospheric radiative fluxes, I shall be happy to reference them. In fact I am wating for this for more than seven years…BTW, my equation for tau is in slide 6.

on April 30, 2011 at 8:21 pm |DeWitt PayneFerenc Miskolczi,

TIGR 2 (Thermodynamic Initial Guess Retrieval) is only representative of the cloud free atmosphere. It’s a set of temperature and humidity profiles to be used as first guesses when inverting satellite measurements. An initial guess is needed because the inversion calculation is ill-posed. It’s cloud free because the satellite data can’t be inverted to temperature and humidity at and below the cloud level. It’s also obvious that it’s cloud free because your minimum St > 0.

on April 30, 2011 at 8:26 pm |DeWitt PayneFerenc Miskolczi,

Good enough for what? And your proof of this is? How about some data comparing your HARTCODE program with MODTRAN, SpectralCalc or LBLRTM.

on May 3, 2011 at 10:43 am |Ferenc MiskolcziDeWitt,

You may compare two codes if they are computing the same quantity.

You can not extract accurate spectrally integrated directional transmittances from any other LBL code, therefore you can not compare anything.

And BTW, the true planck weighted tau was first defined and introduced in the Appendix of my 2007 paper, and so far it is not computed by any other LBL code but HARTCODE. (The old Planck mean absorption coefficient used by astrophysicist is also not appropriate.) And here the problem is that an LBL code is some 10000 code line and no one is going into the details of the code structure and operation of other people’s code. You may find some radiance comparisons here:

AO_V41_No6_2002: Simulation of uplooking and downlooking

high-resolution radiance spectra with two different radiative transfer models;

JQSRT_Vol90_2005:An inter-comparison of far-infrared line-by-line radiative transfer models.

Band averaged MODTRAN or MAKMUS type RT codes are are not suitable even for spectral radiance comparisons.

on April 30, 2011 at 8:52 pm |Neal J. KingHello Ferenc,

I am going to open up a parallel discussion.

As we have communicated a few years ago, I got stuck in trying to understand your (2nd?) paper, and sent you my analysis and questions on that; as far as I had gotten. You were on travel and did not have time to reply for many weeks; and after that, I did not have time for a few months. So we did not proceed with the discussion.

The background and questions can be found at:

I am still interested in trying to clarify these points. Anyone interested is welcome to download the file as well, of course; so one can say that this earlier discussion can be considered to be incorporated into this website by reference.

Regards,

Neal J. King

on May 1, 2011 at 1:49 am |Neal J. KingERRATA

With the distance of 3 years, I notice a few typographical errors in my equations in the letter linked above:

– page 2, bottom line: Should be

[KE] = (3/2) ∫n(z)*k*T dz = (3/2) ∫P(z) dz

I think that’s all.

Unfortunately, for some reason I don’t seem to have a Word version of this anymore, so it’s not easy to alter without retyping the whole thing.

on May 1, 2011 at 11:41 amNeal J. KingOops, I forgot another one:

– page 1, 4th line from bottom: Should be

F = -mg z^

(where z^ is the unit vector in the z-direction)

on May 3, 2011 at 8:08 pm |Neal J. KingFerenc and I are having some off-line discussion on the Virial Theorem applications.

I will post back here if we get to any definite results.

I have improved the clarity of my mathematical argument; so if someone else is interested in reading it, let me know.

on May 1, 2011 at 12:11 am |scienceofdoomFerenc Miskolczi on April 30, 2011 at 12:24 pm:

Glad that you have clarified that this is not a thermodynamic equality. It’s clear that they cannot be exactly equal on any globally averaged basis.

Your paper implies otherwise, and your two supporters that I have read so far – here and on Roy Spencer’s blog – have asserted that Aa=Ed.

Now that this is “official”, don’t you see that your paper now needs (apart from this clarification) some reworking with the the actual relationship?

For example, “

we show by calculation that globally averaged, A..”_{A}> E_{d}> 0.98 A_{A}etc etcThen it’s clear what the error is, or the unknowns.

I’m looking forward to you clearing up some other questions on the theory in Part Three.

on May 2, 2011 at 1:34 am |Ferenc MiskolcziSoD,

What I am saying is one thing, what you are saying is another thing and what the figure is showing is the reality. If you do not understand what the figure says and you call textbooks for that, I can not help aboaut that. You may even write an article about Aa and Ed, but I am sure if you compute the quantities correctly, you will end up with the same results, and I shall be very happy that you prove the Aa~=Ed relationship.

To spend a lifetime arguing on the Aa=Ed approximate equality is useless, and the primary objective of my three papers were quite different. And by the way, do not think that there are many LBL codes out there with +/- 3% accuracy – especially not in oversimplified GCM radiative transfer modules (see Quiang Fu’s code for example). And further on, do not forget that the atmosphere is a stochastic environment, it is also not trivial how you create you global average structure.

Here the primary finding was not the 3 % deviation (which is demonstrated, known, and correctly evaluated), but the overall close agreement. The point is, that this result contradict the classic greenhose theory, that is, the atmosphere is not warming the surface by back radiation. Now you may explain me how the AGW works – in the hypothetical case, where tau is increasing with increasing co2.

Think of the IPCC microwave owen joke….

I have not stated (yet) where the ~10 w/m2 is going, but I never denied the existence of the difference. I am wondering why people were not talking about this 3% difference before, and why nobody estimated even the magnitude of this difference? If I am wrong, show me the relevant references.

If you can show that the average Aa-Ed~10 w/m2, and you know precisely where it is going (statements that it is dissipated in the system is useless, not good enough). Look at slides 8 and 12 in EGU2011-1262, calculate the related contribution density functions and you will know what is going on…In any way we are there again where Roy Spencer, Pierrehumbert, Gavin, Eli, Levenson, DeWitt and others gave up …you must compute the mystery tau, or St, or Aa, (and perhaps they dependence on the altitude). There is no way out – if you want to continue your critics.

on May 1, 2011 at 1:08 am |scienceofdoomFerenc Miskolczi:

In part one I asked:

I am hoping you will be able to supply this.

I would like to play around with energy balance numbers from OLR, reflected solar, surface emitted radiation and see whether global tau presents itself as a useful number in practice.

Also, I hope you can answer the question about the “greenhouse” gas concentrations assumed in the time-series of tau over 60 years.

on May 1, 2011 at 4:30 am |scienceofdoomIt looks like there is a problem with fig. 2 in the 2007 paper:

Perhaps it goes to the heart of the problems in the claimed relationship of A

_{A}= E_{D}.I will write up a short article about it.

on May 1, 2011 at 8:15 am |scienceofdoomPart Four now published.

on May 1, 2011 at 8:14 am |The Mystery of Tau – Miskolczi – Part Four – Emissivity « The Science of Doom[…] Part Two we looked at the claimed relationship ED=AA in Miskolczi’s 2007 […]

on May 1, 2011 at 9:09 am |scienceofdoomFerenc Miskolczi on April 30, 2011 at 12:24 pm:

There is a coherent theory of the inappropriately-named “greenhouse” – generally under the term

the radiative transfer equations. This accurately describes the relationship between OLR and surface temperature as a function of atmospheric temperature profile and concentration of radiatively-active gases.Can you demonstrate your claim “..

most of the energy budget cartoons fail..” with reference to:a) which ones

b) by how much

c) your proof of this

on May 2, 2011 at 2:09 am |Ferenc MiskolcziSoD,

I have the Su=OLR/f relationship. What do you have from the general radiative transfer theory?

on May 10, 2011 at 12:31 pm |Ferenc MiskolcziSoD,

-There is a coherent theory of the inappropriately-named “greenhouse” – generally under the term the radiative transfer equations.-

You will never be able to explain the greenhouse effect by applying the Beer-Lambert law, semi-infinite atmosphere, classic radiative transfer

equations, or thermodynamics alone. What is left for you is to believe

what others are telling, or in toy games with GCMs. Perhaps you may tell

that Miskolczi is wrong, because Miskolczi is wrong, because Miskolczi is wrong – like deWitt does. I do not have time for such discussions any more.

Until I see deWitt’s miracle tau I have no further comment here. You know that somehow he reproduced the USST 76 tau with pretty good accuracy using MODTRAN, therefore he can prove my miracle tau. Let us wait.

on May 1, 2011 at 7:37 pm |DeWitt PayneFerenc or Miklos,

Here’s another one:

In Table 2 in M2010, OLR is calculated to be 251.25 W/m2 GAT. That would imply a radiative imbalance of at least -10 W/m2 or an albedo of 0.27 compared to the usual 0.30 Given that OHC over 60 years is up, indicating a positive imbalance, how do you reconcile this?

on May 2, 2011 at 2:13 am |Ferenc MiskolcziDeWitt,

251 w/m2 is the clear sky OLR. What imbalance you are talking about?

This must be balanced by Fo….

on May 2, 2011 at 2:06 pmDeWitt PayneBut the ERBE clear sky global OLR is 266 W/m2. You can only get 250 if you hand wave in a 6% error. And yet somehow ERBE gets the correct total global average OLR. Your assertion of a 15 W positive bias in the overall ERBE clear sky flux is unfounded. A quick search on ERBE errors shows that while there is a significant positive bias in regions of high humidity, it isn’t 15 W and those regions are only a fraction of the Earth’s surface. ERBE also underestimates flux for regions of low humidity, so the positive systematic error is partly compensated in the global average. Therefore, your calculation of a clear sky flux of 250 W/m2 is wrong.

on May 3, 2011 at 2:30 am |Ferenc MiskolcziDeWitt,

Why do you think that my clear-sky TIGR global average OLR of 251 w/m2 is wrong?

Here is Trenberth’s OLR from KT97:

—-

TABLE 2. Longwave fluxes at the top of the atmosphere (TOA) and at the

surface (SRF). Fluxes are in W m-2. Here, Fd is the downward flux and Fu is the upward flux.

Level Clear Cloudy

W m-2 Fd Fu Net Fd Fu Net

TOA 0 265 265 0 235 235

SRF 278 390 112 324 390 66

—–

I think Trenberth is wrong, his computation is based on the Malkmus model and he used the USST76 atmosphere. (BTW, Kevin told me personally that he was wrong, and he admitted this to Noor van Andel as well ).

on May 2, 2011 at 2:00 am |scienceofdoomFerenc Miskolczi on May 2, 2011 at 1:34 am:

I think your supporters have confused the readers on this point. Your paper also confuses the readers on this point.

Now you have clarified this point we can have a more fruitful discussion.

I have identified another problem with this relationship which seems more fundamental.

You can see it explained in Part Four.

I am just running some other calculations with spectrally dependent emissivity to verify that E

_{D}/A_{A}reduces as ε_{G}reduces. Yet your graph shows the opposite.This appears to be impossible.

Perhaps, with a realistic values for ε

_{G}=0.96 the ratio E_{D}/A_{A}is between 0.90-0.95, rather than ≈ 1.00.on May 2, 2011 at 2:30 pm |DeWitt PayneFerenc Miskolczi,

I would like to calculate πBo from your Equation (20) in your 2007 paper. Unfortunately, the information in the paper is insufficient to allow me to do this and obtain anything that looks like your Figure 3. The form of the equation implies the subtraction of two large numbers which, at low τ are divided by a very small number. Clearly H isn’t equal to OLR everywhere or πBg isn’t OLR/f everywhere or both, so what are they and how are they calculated? How do you vary τ for the US 1976 standard atmosphere to produce a πBo curve as a function of τ?

on May 3, 2011 at 1:02 am |DeWitt PayneAre you taking the temperature and density profile of whatever atmosphere you’re testing, varying tau and calculating emission? Is tau a linear function of pressure or are you treating water vapor separately as its scale height is about 1/4 that of the non-condensable gases?

on May 3, 2011 at 3:02 amFerenc MiskolcziDeWitt,

Here first you calculate OLR and tau for your atmosphere with your RT code, then put the OLR(=H) and Su(=pi*Bg) into Eq. 20, and plot the pi*Bo for an arbitrary tau vector of your choice. You will see that pi*Bo will have the maximum at your computed tau. That is, the atmaosphere try to maximize Bo. If you need, I can send you the matlab script…

on May 3, 2011 at 6:04 pmDeWitt PaynePlease send me a copy of the Matlab script. Do you know if it’s general enough to run on Octave? If not, I’ll ask SoD to run it as he has Matlab.

Please send to payne dot dewitt usual symbol gmail.com.

I’m still not clear about the arbitrary tau vector, though. Don’t you have to change the composition of the atmosphere to change the tau vector? Perhaps it will be clearer when I see the code.

on May 3, 2011 at 7:17 pmDeWitt PayneAt the limit of τ = 0, isn’t Su = OLR? If you plug that into Equation 20, you get πBo → OLR as τ → 0.

on May 3, 2011 at 8:03 pmNeal J. KingDeWitt’s last question is related to one of mine, from a few years ago:

– Eqn(8) seems to derive a ratio, apparently based on some conservation-of-energy argument:

S_u/OLR = (3/2)

– I assume this is based on some non-zero interaction between the gas molecules and the IR photons.

– Now, if we weaken the interaction to zero (e.g., by reducing the concentration of the relevant molecular species), the factor MUST change, because S_u = F_o= OLR in that case. Therefore, the ratio becomes 1.

– EITHER: the logic of the derivation must change discontinuously as the concentration goes to 0;

OR: the ratio must gradually change from (3/2) to 1 as the concentration goes to 0. Which is it?

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

I believe this is related to DeWitt’s question, since I guess that:

tau = 0 0-concentration of IR-absorbing molecules

on May 3, 2011 at 8:14 pmNeal J. KingAnother typo, due to the way this blogging system handles special characters:

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

I believe this is related to DeWitt’s question, since I guess that:

“tau = 0”

is equivalent to

“0-concentration of IR-absorbing molecules”

on May 4, 2011 at 5:24 pm |Ferenc MiskolcziDeWitt, I know nothing of OCTAVE, but Jan Pompe used to easily convert my matlab scripts for use with OCTAVE… it should not be that difficult …

Here are my HARTCODE results (U means usst-76 atmosphere), the tau vector, and the plot command:

OLR_U=260.8;SU_U=391.2;TAU_U=1.432;B0_U=148;xtau=[1:0.001:3]’;

plot(xtau,(0.5*OLR_U.*(1+(xtau+1).*exp(-xtau))-SU_U.*exp(-xtau))./(1-exp(-xtau)),TAU_U,B0_U,’*’);

Do you need anything else? It is obvious that the usst 76 is not good for any global average flux computation….

on May 4, 2011 at 8:06 pm |DeWitt PayneUsing MODTRAN US76 375 ppmv CO2, I get OLR_U = 264.169; SU_U = 389 (ε = 0.98 for the surface + 0.02 * 280.97 (E_D)); TAU_U = 1.437; B0_U = 152.3 W/m². So we’re in the same ballpark. What I don’t see is leaving SU_U constant while varying τ in (20). OLR_U should indeed be constant, but SU_U must vary with τ. At τ = 0 SU_U = OLR_U by definition because E_D goes to zero. Either that or you have a massive radiative imbalance at the surface. When holding SU_U constant, B0_U goes negative at τ < 0.43, which also seems odd. On the other hand, if you do vary SU_U, say by assuming SU_U = OLR/f, then B0 vs τ is monotonic with no optimum value.

I’m also not clear on the physical meaning of B0_U. It’s not E_U because for MODTRAN, that’s 173 W/m².

on May 4, 2011 at 11:01 pmDeWitt PayneOn further consideration, B0 going negative at low τ is exactly what you would expect to see if Bg is allowed to remain constant rather than decreasing as it should at low τ. At high τ the Bg term becomes small compared to the OLR/2 term and B0 converges to OLR/2 at high τ.

on May 5, 2011 at 2:24 amFerenc MiskolcziDeWitt, if your MODTRAN tau is 1.43 just like mine, why do not you run MODTRAN with a real global average NOAA R1 profile and see

how much is your MODTRAN tau. This is what I am waiting for. If

you get a tau close to 1.87 then you proved that I am right. If you get

something different then we may look for the source of the difference.

But I am optimistic, the taus are pretty close….

on May 5, 2011 at 5:27 am |DeWitt PayneNeal J. King,

I’ve done some quick and dirty calculations on the ratio Su/OLR. For τ < 2, the ratio varies smoothly and close to linearly with τ. We can fix the intercept at a value of 1. We also know that at very high τ the ratio will be close to zero. The fact that the ratio is approximately 3/2 in the Earth’s atmosphere is coincidence.

on May 5, 2011 at 5:31 am |DeWitt PayneNeal,

Steve Short says he remembers discussing the problem of keeping Bg constant when differentiating equation (20) in M7 wrt τ, but I haven’t been able to find it. Do you by any chance remember when and where? I’ve tried searching at Niche Modeling with no luck.

on May 5, 2011 at 9:19 amNeal J. KingDeWitt:

I’m sorry, but I stopped before getting that far, waiting for clarification on what were, for me, logically prior matters. So I didn’t discuss this matter myself, at Niche Modeling or anywhere else.

Also, I do not recall the name, Steve Short: my apologies.

on May 15, 2011 at 7:15 am |The Mystery of Tau – Miskolczi – Part Five – Equation Soufflé « The Science of Doom[…] tells us that Ed=Aa due to Kirchhoff’s law. (See Part Two). His 2010 paper revised this claim as to due to Prevost.. However, the author himself recently […]

on May 23, 2011 at 2:33 am |Ken GregorySoD requested a spreadsheet with 61 annual values of the NOAA radiosonde observations.

Only the H2O and CO2 gases were changed. Other minor GHG were held constant.

The radionsonde inputs are here:

http://members.shaw.ca/sch25/Ken/Optical%20Depth%20Data.xls

There are graphs of specific humidity below the data.

The HARTCODE output is here:

http://members.shaw.ca/sch25/Ken/hartcode_61yearNOAA2.xls

I understand that the HARTCODE outputs of Su, St and OLR assume a surface emissivity = 1.

Calculated Aa, Ed/Aa, Su/Eu, Su/OLR, all with emissivity =1, are in columns M to P. The value of tau (optical depth) and Ed/Eu do not depend on emissivity.

M2010 states “From the TIGR2 data the global average hemispheric emissivity is e = 0.967.” I therefore calculated Su and St using this emissivity in columns T & U. Aa, OLR and the various ratios are recalculated in columns V to AA.

I was interested to see how these ratios varied with temperature or Su. Graphs of various parameters versus Su are shown starting at column AF. I am not a scientist, so let me know if I got anything wrong here.

There has been much discussion of if Aa = Ed exactly, or just approximately. Obviously, to the extent that the part of the atmosphere where Ed comes from is cooler than the surface, Ed must be slightly less that Aa. Fig 5 of M2010 says the Aa-Ed bias is about 3% of Su. The Aa is set equal to Ed for theoretical calculation purposes. I don’t see a problem with that. Even with emissivity set to 0.967, the average Aa/Ed ratio is 0.9953.

The second graph shows Su/Eu and Su/OLR are indeed very constant with Su (temperature). The average Su/Eu is 1.9545. Miskolczi used 2.0 in M2007. The average Su/OLR is 1.5016, very close to the 1.5 in M2007.

In M2010, Fig 8 shows the global average Ed/Eu is very close to 5/3, equation 10. I did not think the paper sufficiently showed that the ratio stays constant with changing temperatures or Su. This is necessary because equation 10 is used to derive equation 11, which is supposed to be true for all Su. But in fact, using the 61 years data, Ed/Eu does stay very constant.

The atmospheric transmittance appears to have a slight decline with Su at -0.0004918/(W/m2).

The flux density ratio equation 11 of M2010, OLR/Su = 0.4Ta + 0.6, is shown on the graph at column AN. The best fit slope of the 61 years of data is 0.479, quite close to the 0.4 of equation 11.

on May 23, 2011 at 12:45 pm |scienceofdoomKen Gregory:

When you say:

Other minor GHG were held constant?, do you mean:a) Were included at 1948 values and held constant?

b) Were included at 2010 values and held constant?

c) Were ignored and not included in calculations of tau?

Supporter Miklos Zagoni earlier stated:

Perhaps you can clear up the mystery.

on May 23, 2011 at 12:58 pm |scienceofdoomAnd also note my response to Miklos Zagoni on this point.

on May 23, 2011 at 2:06 pm |scienceofdoomKen Gregory:

This comment of yours agrees with Figure 2 in Miskolczi’s 2007 paper, which I questioned in Part Four.

It’s good to see the spreadsheet with the calculations.

If we take the first year, 1949, for the idealized surface emissivity, ε = 1.00:

Ed=319.86

Aa=332.64

where Aa = upwards surface radiation absorbed by the atmosphere, Ed = downwards atmospheric radiation absorbed by the surface

And so Ed/Aa = 0.9616

These are calculated values using the radiative transfer program HARTCODE.

Aa is calculated from Aa = Su-St (Absorbed flux = upward surface flux – transmitted with Su = 393.56 & St = 60.92

So now let’s consider what happens when we set emissivity, ε, to a more realistic surface value = 0.967 (this is the value in the spreadsheet – in the paper it is set to ε = 0.96).

I will use Ed for the value at ε = 1.00, Ed’ for the value in the spreadsheet at ε = 0.967, and Ed” for the valueit really should beat ε = 0.967, and likewise for other valuesIn the spreadsheet, Ed’/Aa’ = 0.9944 (for ε = 0.967).

This is because Su’ = Su x 0.967 (incorrect).

But in fact, upward surface flux, Su” = Su x 0.967

+ reflected atmospheric radiation.So Su’ is incorrect. Su’ (spreadsheet value) < Su'' (the real value). However, as Aa = Su – St, this has a smaller impact (as both Su and St suffer the same calculation flaw).

And the spreadsheet calculates Ed'/Aa' using the original Ed –

but the surface now reflects over 3% of the radiation.So Ed” = Ed * 0.967.

The final calculation isn’t super-simple, but should be simple for the person who did the original HARTCODE calculations. My equation is shown in an earlier comment on Part Four.

In simple terms Ed/Aa for ε = 0.967 is clearly not 0.9944 and should be less than the calculated value at ε = 1.00.

This should be evident to anyone who takes the time to look through it.

on May 23, 2011 at 9:09 pm |Ken GregorySoD says:

“When you say: Other minor GHG were held constant?, do you mean:

a) Were included at 1948 values and held constant?

b) Were included at 2010 values and held constant?

c) Were ignored and not included in calculations of tau?”

I don’t know what year the minor GHG correspond to. Here is the inputs for 2007:

1 1 H2O-01 0.3295921355D+04 0.2654498805D+01

2 2 CO2-02 0.3069882964D+03

3 3 O3-03 0.3069082957D+00

4 4 N2O-04 0.2408237656D+00

5 5 CO-05 0.8795448828D-01

6 6 CH4-06 0.1293575859D+01

7 7 O2-07 0.1668081961D+06

8 9 SO2-09 0.1094307925D-03

9 22 N2-22 0.6233129094D+06

10 39 f11-39 0.2190033276D-03

11 40 f12-40 0.4146156667D-03

12 41 ccl4-41 0.8587012795D-04

The second number for H20 is prcm.

The CO2 number is in atmos*cm at STP.

This come from a text file here:

http://members.shaw.ca/sch25/Ken/p60.log

The 1956 inputs are:

1 1 H2O-01 0.3221200981D+04 0.2594319837D+01

2 2 CO2-02 0.2520237660D+03

3 3 O3-03 0.3069027274D+00

4 4 N2O-04 0.2408222459D+00

5 5 CO-05 0.8795410852D-01

6 6 CH4-06 0.1293568245D+01

7 7 O2-07 0.1668072522D+06

8 9 SO2-09 0.1094311773D-03

9 22 N2-22 0.6233093880D+06

10 39 f11-39 0.2190020924D-03

11 40 f12-40 0.4146133214D-03

12 41 ccl4-41 0.8586964003D-04

http://members.shaw.ca/sch25/Ken/p60.log

Note that the H2O and CO2 values correspond exactly to the hartcode_61yearNOAA2.xls file.

on May 23, 2011 at 9:27 pm |Ken GregorySod says:

“But in fact, upward surface flux, Su” = Su x 0.967 + reflected atmospheric radiation. ”

The calculation by HARTCODE and my Excel spreadsheet does not consider reflected longwave radiation.

I don’t want to be changing definitions. M2007 and M2007 defines Su by Su = σ T^4. Su by definition is just the surface radiation due to its temperature.

Ed is defined as the downward atmospheric emittance flux.

I recognize that you think a portion of the Ed will be reflected, adding to St and Aa.

I guess that Miskolczi thinks this is a minor effect.

on May 23, 2011 at 10:09 pm |scienceofdoomKen Gregory:

Actually no.

In M2007: “

Sg is the LW upward radiation from the ground” and Su is not defined.From deduction we believe Su = total upward longwave radiation.

In M2010: “

Su = σ.Ta” and “^{4}is the total surface upward fluxIt is explicitly assumed that temperatures Ta and Tg are equal, the surface is black, and therefore, Su = Sg = σ.Ta”^{4}= σ.Tg^{4}My clarifications:

Sg is the emission of radiation by the ground.

Su is the total upward longwave flux from the ground = Sg + reflected longwave radiation

Now, when ε = 1.00:

Sg=Su = σT

^{4}When ε ≠ 1.00, Sg cannot be “defined” as σT

^{4}, it must be εσT^{4}Sg ≠ Su

And because the atmosphere also radiates, there must be a reflected portion:

Therefore, Su > Sg.

I thought this was science? If it is “a minor effect” it needs to be quantified so it can be demonstrated.

The paper says, Aa=Ed.

The author says on this blog: “

I think I was the first who showed the Aa~=Ed relationship with reasonable quantitative accuracy.And I am pointing out that because there are apparently

unaddressed errorsAa~=Ed might mean Ed = 0.94 Aa.Perhaps the author thinks it is a minor effect because he hasn’t realized his mistake.

Perhaps the author thinks it is a minor effect because he has redone his calculations for Ed = 0.94 Aa or even = 0.90 Aa and the results come out the same.

But if you are serious about science, please confirm with the author of the paper:

Eitherthat fig 2 is correct for ε = 0.96 – so therefore explain the mistakes I have made here and in Part Four. It should be very easy to clear up.Orif fig 2 IS wrong, and the author has made a mistake, please confirm, calculate the true ratio of Ed/Aa for real world emissivities (e.g., ε=0.96) and demonstrate that the results of the paper do not change with the corrected values propagating through all of the equations and calculations.on May 24, 2011 at 5:22 pm |FrankoSoD

Do your own version of the equations, in a format that I can cut and paste into an online solver, such as QuickMath

It would also help other casual readers to understand

on May 24, 2011 at 8:17 pm |scienceofdoomFranko, can you explain further?

a) What form do you need them in? I am basically typing them into wordpress (painful, adding the subscripts and superscripts in html code) from my notes.

b) What does QuickMath do?

on May 24, 2011 at 5:52 pm |Neal J. KingFranko,

A couple of years ago, I believe you had some detailed questions about the Virial Theorem, at Niche Modeling. I came across them later, but didn’t know how to contact you. I believe I have full answers to these questions, if you are still interested.

on May 28, 2011 at 8:54 am |The Mystery of Tau – Miskolczi – Part Six – Minor GHG’s « The Science of Doom[…] supporter Ken Gregory said: Only the H2O and CO2 gases were changed. Other minor GHG were held […]

on May 30, 2011 at 5:13 pm |RWI 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 through the atmosphere will not change any of this.

Miskolczi’s measured data of 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.

on May 30, 2011 at 11:29 pm |scienceofdoomRW:

See my response to your same comment in Part Four.

on August 27, 2011 at 7:14 pm |A. VanagsI am having a little bit of trouble following the basics. I figure 1 above (Miskolczi 2007) There seems to be only short wave radiation downwards (I asume solar) to the ground, but I cant find any upwards. Does this mean the the radiation upwards (ground reflection, albedo etc) are not taken into account? or it is that it is all already lumped into F0 and F0 is a net radiation down, already including earth reflection and albedo?. Does this means that the assumption is no absorption at all of shortwave by the atmosphere? (otherwise it would not be quite correct to do it as a net at the top of the atmosphere as it would absorb the reflection as well)

on August 29, 2011 at 3:24 pm |DeWitt PayneNo. Fo is the incoming radiation at the TOA corrected for albedo. F is the amount of incoming radiation absorbed by the atmosphere. Fo-F is the incoming radiation absorbed by the surface. Reflected radiation has already passed through the atmosphere so wavelengths readily absorbed have already been removed. Absorption of the reflected radiation should be small enough to be safely ignored.