The Science of Doom

This article will be a placeholder article to filter out a select group of people. The many people who arrive and confidently explain that atmospheric physics is fatally flawed (without the benefit of having read a textbook). They don’t think they are confused, in their minds they are helpfully explaining why the standard theory is wrong. There have been a lot of such people.

Almost none of them ever provides an equation. If on rare occasions they do provide a random equation, they never explain what is wrong with the 65-year old equation of radiative transfer (explained by Nobel prize winner Subrahmanyan Chandrasekhar, see note 1) which is derived from fundamental physics. Or an explanation for why observation matches the standard theory. For example (and I have lots of others), here is a graph produced nearly 50 years ago (referenced almost 30 years ago) of the observed spectrum at the top of atmosphere vs the calculated spectrum from the standard theory.

Why is it so accurate?

If it was me, and I thought the theory was wrong, I would read a textbook and try and explain why the textbook was wrong. But I’m old school and generally expect physics textbooks to be correct, short of some major revolution. Conventionally, when you “prove” textbook theory wrong you are expected to explain why everyone got it wrong before.

There is a simple reason why our many confident visitors never do that. They don’t know anything about the basic theory. Entertaining as that is, and I’ll be the first to admit that it has been highly entertaining, it’s time to prune comments from overconfident and confused visitors.

I am not trying to push away people with questions. If you have questions please ask. This article is just intended to limit the tsunami of comments from visitors with their overconfident non-textbook understanding of physics – that have often dominated comment threads. 

So here are my two questions for the many visitors with huge confidence in their physics knowledge. Dodging isn’t an option. You can say “not correct” and explain your alternative formulation with evidence, but you can’t dodge.

Answer these two questions:

1. Is the equation of radiative transfer correct or not?

I_λ(0) = I_λ(τ_m)e^-τ_m + ∫ B_λ(T)e^-τ dτ [16]

The intensity at the top of atmosphere equals.. The surface radiation attenuated by the transmittance of the atmosphere, plus.. The sum of all the contributions of atmospheric radiation – each contribution attenuated by the transmittance from that location to the top of atmosphere

Of course (and I’m sure I don’t even need to spell it out) we need to integrate across all wavelengths, λ, to get the flux value.

For the derivation see Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations. If you don’t agree it is correct then explain why.

[Note that other articles explain the basics. For example – The “Greenhouse” Effect Explained in Simple Terms, which has many links to other in depth articles].

If you don’t understand the equation you don’t understand the core of radiative atmospheric physics.

—-

2. Is this graphic with explanation from an undergraduate heat transfer textbook (Fundamentals of Heat and Mass Transfer, 6th edition, Incropera and DeWitt 2007) correct or not?

You can see that radiation is emitted from a hot surface and absorbed by a cool surface. And that radiation is emitted from a cool surface and absorbed by a hot surface. More examples of this principle, including equations, in Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics – scanned pages from six undergraduate heat transfer textbooks (seven textbooks if we include the one added in comments after entertaining commenter Bryan suggested the first six were “cherry-picked” and offered his preferred textbook which had exactly the same equations).

—-

What I will be doing for the subset of new visitors with their amazing and confident insights is to send them to this article and ask for answers. In the past I have never been able to get a single member of this group to commit. The reason why is obvious.

But – if you don’t answer, your comments may never be published.

Once again, this is not designed to stop regular visitors asking questions. Most people interested in climate don’t understand equations, calculus, radiative physics or thermodynamics – and that is totally fine.

Call it censorship if it makes you sleep better at night.

Notes

Note 1 – I believe the theory is older than Chandrasekhar but I don’t have older references. It derives from basic emission (Planck), absorption (Beer Lambert) and the first law of thermodynamics. Chandrasekhar published this in his 1952 book Radiative Transfer (the link is the 1960 reprint). This isn’t the “argument from authority”, I’m just pointing out that the theory has been long established. Punters are welcome to try and prove it wrong, just no one ever does.

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Long before the printing of money, golden eggs were the only currency.

In a deep cave, goose Day-Lewis, the last of the gold-laying geese, was still at work.

Day-Lewis lived in the country known affectionately as Utopia. Every day, Day-Lewis laid 10 perfect golden eggs, and was loved and revered for her service. Luckily, everyone had read Aesop’s fables, and no one tried to kill Day-Lewis to get all those extra eggs out. Still Utopia did pay a few armed guards to keep watch for the illiterates, just in case.

Utopia wasn’t into storing wealth because it wanted to run some important social programs to improve the education and health of the country. Thankfully they didn’t run a deficit and issue bonds so we don’t need to get into any political arguments about libertarianism.

This article is about golden eggs.

Utopia employed the service of bunny Fred to take the golden eggs to the nearby country of Capitalism in return for services of education and health. Every day, bunny Fred took 10 eggs out of the country. Every day, goose Day-Lewis produced 10 eggs. It was a perfect balance. The law of conservation of golden eggs was intact.

Thankfully, history does not record any comment on the value of the services received for these eggs, or on the benefit to society of those services, so we can focus on the eggs story.

Due to external circumstances outside of Utopia’s control, on January 1st, the year of Our Goose 150, a new international boundary was created between Utopia and Capitalism. History does not record the complex machinations behind the reasons for this new border control.

However, as always with government organizations, things never go according to plan. On the first day, January 1st, there were paperwork issues.

Bunny Fred showed up with 10 golden eggs, and, what he thought was the correct paperwork. Nothing got through. Luckily, unlike some government organizations with wafer-thin protections for citizens’ rights, they didn’t practice asset forfeiture for “possible criminal activity we might dream up and until you can prove you earned this honestly we are going to take it and never give it back”. Instead they told Fred to come back tomorrow.

On January 2nd, Bunny Fred had another run at the problem and brought another 10 eggs. The export paperwork for the supply of education and health only allowed for 10 golden eggs to be exported to Capitalism so border control sent on the 10 eggs from Jan 1st and insisted Bunny Fred take 10 eggs take back to Utopia.

On January 3rd, Bunny Fred, desperate to remedy the deficit of services in Utopia took 20 eggs – 10 from Day-Lewis and 10 he had brought back from border control the day before.

Insistent on following their new ad hoc processes, border control could only send on 10 eggs to Capitalism. As they had no approved paperwork for “storing” extra eggs, they insisted that Fred take back the excess eggs.

Every day, the same result:

• Day-Lewis produced 10 eggs, Bunny Fred took 20 eggs to border control
• Border control sent 10 eggs to Capitalism, Bunny Fred brought 10 eggs back

One day some people who thought they understood the law of conservation of golden eggs took a look at the current situation and declared:

Heretics! This is impossible. Day-Lewis, last of the gold-laying geese, only produces 10 eggs per day. How can Bunny Fred be taking 20 eggs to border control?

You can’t create golden eggs! The law of conservation of golden eggs has been violated.

You can’t get more than 100% efficiency. This is impossible.

And in other completely unrelated stories:

A Challenge for Bryan & A Challenge for Bryan – The Solution

Do Trenberth and Kiehl understand the First Law of Thermodynamics? & Part Two & Part Three – The Creation of Energy?

and recent comments in CO2- An Insignificant Trace Gas? – Part One

Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics

The Three Body Problem

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In A Challenge for Bryan I put up a simple heat transfer problem and asked for the equations. Bryan elected not to provide these equations. So I provide the answer, but also attempt some enlightenment for people who don’t think the answer can be correct.

As DeWitt Payne noted, a post with a similar problem posted on Wattsupwiththat managed to gather some (unintentionally) hilarious comments.

Here’s the problem again:

Case 1

Spherical body, A, of radius ra, with an emissivity, εa =1. The sphere is in the vacuum of space.

It is internally heated by a mystery power source (let’s say nuclear, but it doesn’t matter), with power input = P.

The sphere radiates into deep space, let’s say the temperature of deep space = 0K to make the maths simpler.

1. What is the equation for the equilibrium surface temperature of the sphere, Ta?

Case 2

The condition of case A, but now body A is surrounded by a slightly larger spherical shell, B, which of course is itself now surrounded by deep space at 0K.

B has a radius rb, with an emissivity, εb =1. This shell is highly conductive and very thin.

2a. What is the equation for the new equilibrium surface temperature, Ta’?

2b. What is the equation for the equilibrium temperature, Tb, of shell B?

 

Notes:

The reason for the “slightly larger shell” is to avoid “complex” view factor issues. Of course, I’m happy to relax the requirement for “slightly larger” and let Bryan provide the more general answer.

The reason for the “highly conductive” and “thin” outer shell, B, is to avoid any temperature difference between the inside and the outside surfaces of the shell. That is, we can assume the outside surface is at the same temperature as the inside surface – both at temperature, Tb.

This kind of problem is a staple of introductory heat transfer. This is a “find the equilibrium” problem.

How do we solve these kinds of problems? It’s pretty easy once you understand the tools.

The first tool is the first law of thermodynamics. Steady state means temperatures have stabilized and so energy in = energy out. We draw a “boundary” around each body and apply the “boundary condition” of the first law.

The second tool is the set of equations that govern the movement of energy. These are the equations for conduction, convection and radiation. In this case we just have radiation to consider.

For people who see the solution, shake their heads and say, this can’t be, stay on to the end and I will try and shed some light on possible conceptual problems. Of course, if it’s wrong, you should easily be able to provide the correct equations – or even if you can’t write equations you should be able to explain the flaw in the formulation of the equation.

In the original article I put some numbers down – “For anyone who wants to visualize some numbers: ra=1m, P=1000W, rb=1.01m“. I will use these to calculate an answer from the equations. I realize many readers aren’t comfortable with equations and so the answers will help illuminate the meaning of the equations.

I go through the equations in tedious detail, again for people who would like to follow the maths but don’t find maths easy.

Case 1

Energy in, Ein = Energy out, Eout  :  in Watts (Joules per second).

Ein = P

Eout = emission of thermal radiation per unit area x area

The first part is given by the Stefan-Boltzmann equation (σT_a⁴, where σ = 5.67×10^-8), and the second part by the equation for the surface area of a sphere (4πr_a²)

Eout = 4πr_a² x σT_a⁴ …..[eqn 1]

Therefore, P = 4πr_a²σT_a⁴ ….[eqn 2]

We have to rearrange the equation to see how Ta changes with the other factors:

T_a = [P / (4πr_a²σ)]^1/4 ….[eqn 3]

If you aren’t comfortable with maths this might seem a little daunting. Let’s put the numbers in:

T_a = 194K (-80ºC)

Now we haven’t said anything about how long it takes to reach this temperature. We don’t have enough information for that. That’s the nice thing about steady state calculations, they are easier than dynamic calculations. We will look at that at the end.

Probably everyone is happy with this equation. Energy is conserved. No surprises and nothing controversial.

Now we will apply the exact same approach to the second case.

Case 2

First we consider “body A”. Given that it is enclosed by another “body” – the shell B – we have to consider any energy being transferred by radiation from B to A. If it turns out to be zero, of course it won’t affect the temperature of body A.

Ein(a) = P + E_b-a ….[eqn 4], where E_b-a is a value we don’t yet know. It is the radiation from B absorbed by A.

Eout(a) = 4πr_a² x σT_a⁴ ….[eqn 5]- this is the same as in case 1. Emission of radiation from a body only depends on its temperature (and emissivity and area but these aren’t changing between the two cases)

– we will look at shell B and come back to the last term in eqn 4.

Now the shell outer surface:

Radiates out to space

We set space at absolute zero so no radiation is received by the outer surface

Shell inner surface:

Radiates in to A (in fact almost all of the radiation emitted from the inner surface is absorbed by A and for now we will treat it as all) – this was the term E_b-a

Absorbs all of the radiation emitted by A, this is Eout(a)

And we made the shell thin and highly conductive so there is no temperature difference between the two surfaces. Let’s collect the heat transfer terms for shell B under steady state:

Ein(b) = Eout(a) + 0  …..[eqn 6] – energy in is all from the sphere A, and nothing from outside

             =  4πr_a² x σT_a⁴ ….[eqn 6a] – we just took the value from eqn 5

Eout(b) = 4πr_b² x σT_b⁴ + 4πr_b² x σT_b⁴ …..[eqn 7] – energy out is the emitted radiation from the inner surface + emitted radiation from the outer surface

                = 2 x 4πr_b² x σT_b⁴ ….[eqn 7a]

 And we know that for shell B, Ein = Eout so we equate 6a and 7a:

4πr_a² x σT_a⁴ = 2 x 4πr_b² x σT_b⁴ ….[eqn 8]

and now we can cancel a lot of the common terms:

r_a² x T_a⁴ = 2 x r_b² x T_b⁴ ….[eqn 8a]

and re-arrange to get Ta in terms of Tb:

T_a⁴ = 2r_b²/r_a² x T_b⁴ ….[eqn 8b]

T_a = [2r_b²/r_a²]^1/4 x T_b ….[eqn 8b]

or we can write it the other way round:

T_b = [r_a²/2r_b²]^1/4 x T_a ….[eqn 8c]

Using the numbers given, T_a = 1.2 T_b. So the sphere is 20% warmer than the shell (actually 2 to the power 1/4).

We need to use Ein=Eout for the sphere A to be able to get the full solution. We wrote down: Ein(a) = P + E_b-a ….[eqn 4]. Now we know “E_b-a” – this is one of the terms in eqn 7.

So:

Ein(a) = P + 4πr_b² x σT_b⁴ ….[eqn 9]

and Ein(a) = Eout(a), so:

P + 4πr_b² x σT_b⁴ = 4πr_a² x σT_a⁴  ….[eqn 9]

we can substitute the equation for T_b:

P + 4πr_a² /2 x σT_a⁴ = 4πr_a² x σT_a⁴  ….[eqn 9a]

the 2nd term on the left and the right hand side can be combined:

P = 2πr_a² x σT_a⁴  ….[eqn 9a]

And so, voila:

T’_a = [P / (2πr_a²σ)]^1/4 ….[eqn 10] – I added a dash to Ta so we can compare it with the original value before the shell arrived.

T’_a = 2^1/4 T_a   ….[eqn 11] – that is, the temperature of the sphere A is about 20% warmer in case 2 compared with case 1.

Using the numbers, T’_a = 230 K (-43ºC). And T_b = 193 K (-81ºC)

Explaining the Results

In case 2, the inner sphere, A, has its temperature increase by 36K even though the same energy production takes place inside. Obviously, this can’t be right because we have created energy??.. let’s come back to that shortly.

Notice something very important – Tb in case 2 is almost identical to Ta in case 1. The difference is actually only due to the slight difference in surface area. Why?

The system has an energy production, P, in both cases.

• In case 1, the sphere A is the boundary transferring energy to space and so its equilibrium temperature must be determined by P
• In case 2, the shell B is the boundary transferring energy to space and so its equilibrium temperature must be determined by P

Now let’s confirm the mystery unphysical totally fake invented energy.

Let’s compare the flux emitted from A in case 1 and case 2. I’ll call it R.

• R(case 1) = 80 W/m²
• R(case 2) = 159 W/m²

This is obviously rubbish. The same energy source inside the sphere and we doubled the sphere’s energy production!!! Get this idiot to take down this post, he has no idea what he is writing..

Yet if we check the energy balance we find that 80 W/m² is being “created” by our power source, and the “extra mystery” energy of 79 W/m² is coming from our outer shell. In any given second no energy is created.

The Mystery Invented Energy – Revealed

When we snapped the outer shell over the sphere we made it harder for heat to get out of the system. Energy in = energy out, in steady state. When we are not in steady state: energy in – energy out = energy retained. Energy retained is internal energy which is manifested as temperature.

We made it hard for heat to get out, which accumulated energy, which increased temperature.. until finally the inner sphere A was hot enough for all of the internally generated energy, P, to get out of the system.

Let’s add some information about the system: the heat capacity of the sphere = 1000 J/K; the heat capacity of the shell = 100 J/K. It doesn’t much matter what they are, it’s just to calculate the transients. We snap the shell – originally at 0K – around the sphere at time t=100 seconds and see what happens.

The top graph shows temperature, the bottom graph shows change in energy of the two objects and how much energy is leaving the system:

At 100 seconds we see that instead of our steady state 1000W leaving the system, instead 0W leaves the system. This is the important part of the mystery energy puzzle.

We put a 0K shell around the sphere. This absorbs all the energy from the sphere. At time t=100s the shell is still at 0K so it emits 0W/m². It heats up pretty quickly, but remember that emission of radiation is not linear with temperature so you don’t see a linear relationship between the temperature of shell B and the energy leaving to space. For example at 100K, the outward emission is 6 W/m², at 150K it is 29 W/m² and at its final temperature of 193K, it is 79 W/m² (=1000 W in total).

As the shell heats up it emits more and more radiation inwards, heating up the sphere A.

The mystery energy has been revealed. The addition of a radiation barrier stopped energy leaving, which stored heat. The way equilibrium is finally restored is due to the temperature increase of the sphere.

Of course, for some strange reason an army of people thinks this is totally false. Well, produce your equations.. (this never happens)

All we have done here is used conservation of energy and the Stefan Boltzmann law of emission of thermal radiation.

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Bryan needs no introduction on this blog, but if we were to introduce him it would be as the fearless champion of Gerlich and Tscheuschner.

Bryan has been trying to teach me some basics on heat transfer from the Ladybird Book of Thermodynamics. In hilarious fashion we both already agree on that particular point.

So now here is a problem for Bryan to solve.

Of course, in Game of Thrones fashion, Bryan can nominate his own champion to solve the problem.

Case A

Spherical body, A, of radius ra, with an emissivity, εa =1. The sphere is in the vacuum of space.

It is internally heated by a mystery power source (let’s say nuclear, but it doesn’t matter), with power input = P.

The sphere radiates into deep space, let’s say the temperature of deep space = 0K to make the maths simpler.

1. What is the equation for the equilibrium surface temperature of the sphere, Ta?

Case B

The condition of case A, but now body A is surrounded by a slightly larger spherical shell, B, which of course is itself now surrounded by deep space at 0K.

B has a radius rb, with an emissivity, εb =1. This shell is highly conductive and very thin.

2a. What is the equation for the new equilibrium surface temperature, Ta’?

2b. What is the equation for the equilibrium temperature, Tb, of shell B?

 

Notes:

The reason for the “slightly larger shell” is to avoid “complex” view factor issues. Of course, I’m happy to relax the requirement for “slightly larger” and let Bryan provide the more general answer.

The reason for the “highly conductive” and “thin” outer shell, B, is to avoid any temperature difference between the inside and the outside surfaces of the shell. That is, we can assume the outside surface is at the same temperature as the inside surface – both at temperature, Tb.

For anyone who wants to visualize some numbers: ra=1m, P=1000W, rb=1.01m

This problem takes a couple of minutes to solve on a piece of paper. I suspect we will wait a decade for Bryan’s answer. But I love to be proved wrong!

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In previous articles in this series we looked at a number of issues first in Miskolczi’s 2010 paper and then in the 2007 paper.

The author himself has shown up and commented on some of these points, although not all, and sadly decided that we are not too bright and a little bit too critical and better pastures await him elsewhere.

Encouraged by one of our commenters I pressed on into the paper: Greenhouse Effect in Semi-Transparent Planetary Atmospheres, Quarterly Journal of the Hungarian Meteorological Service (2007), and now everything is a lot clearer.

The 2007 paper by Ferenc Miskolczi is a soufflé of confusion piled on confusion. Sorry to be blunt. If I was writing a paper I would say “..some clarity is needed in important sections..” but many readers unfamiliar with the actual meaning of this phrase might think that some clarity was needed in important sections rather than the real truth that the paper is a shambles.

I’ll refer to this paper as M2007. And to the equations in M2007 with an M prefix – so, for example, equation 15 will be [M15].

Some background is needed so first we need to take a look at something called The Semi-Gray Model. Regular readers will find a lot of this to be familiar ground, but it is necessary as there are always many new readers.

The SGM – Semi-Grey Model or Schwarzschild Grey Model

I’ll introduce this by quoting from an excellent paper referenced by M2007. This is a 1995 paper by Weaver and the great Ramanathan (free link in References):

Simple models of complex systems have great heuristic value, in that their results illustrate fundamental principles without being obscured by details. In particular, there exists a long history of simple climate models. Of these, radiative and radiative-convective equilibrium models have received great attention..

One of the simplest radiative equilibrium models involves the assumption of a so-called grey atmosphere, where the absorption coefficient is assumed to be independent of wavelength. This was first discussed by Schwarzschild [1906] in the context of stellar interiors. The grey gas model was adapted to studies of the Earth by assuming the atmosphere to be transparent to solar radiation and grey for thermal radiation. We will refer to this latter class as semigrey models.

And in the abstract they say:

Radiative equilibrium solutions are the starting point in our attempt to understand how the atmospheric composition governs the surface and atmospheric temperatures, and the greenhouse effect. The Schwarzschild analytical grey gas model (SGM) was the workhorse of such attempts. However, the solution suffered from serious deficiencies when applied to Earth’s atmosphere and were abandoned about 3 decades ago in favor of more sophisticated computer models..

[Emphasis added]

And they go on to present a slightly improved SGM as a useful illustrative tool.

Some clarity on a bit of terminology for new readers – a blackbody is a perfect emitter and absorber of radiation. In practice there are no blackbodies but some bodies come very close. A blackbody has an emissivity = 1 and absorptivity = 1.

In our atmosphere, the gases which absorb and emit significant radiation have very wavelength dependent properties, e.g.:

Figure 1

So the emissivity and absorptivity vary hugely from one wavelength to the next (note 1). However, as an educational tool, we can calculate the results for a grey atmosphere – this means that the emissivity is assumed to be constant across all wavelengths.

The term semi-grey means that the atmosphere is considered transparent for shortwave = solar wavelengths (<4 μm) and constant but not zero for longwave = terrestrial wavelengths (>4 μm).

Constructing the SGM

This model is very simple – and is not used to really calculate anything of significance for our climate. See Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations for the real equations.

We assume that the atmosphere is in radiative equilibrium – that is, convection does not exist and so only radiation moves heat around.

Here is a graphic showing the main elements of the model:

Figure 2

Once each layer in the atmosphere is in equilibrium, there is no heating or cooling – this is the definition of equilibrium. This means energy into the layer = energy out of the layer. So we can use simple calculus to write some equations of radiative transfer.

We define the TOA (top of atmosphere) to be where optical thickness, τ=0, and it increases through the atmosphere to reach a maximum of τ=τ_A at the surface. This is conventional.

We also know two boundary conditions, because at TOA (top of atmosphere) the downward longwave flux, F↓(τ=0) = 0 and the upwards longwave flux, F↑(τ=0) = F0, where F0 = absorbed solar radiation ≈ 240 W/m². This is because energy leaving the planet must be balanced by energy being absorbed by the planet from the sun.

We also have to consider the fact that energy is not just going directly up and down but is going up and down at every angle. We can deal with this via the dffusivity approximation which sums up the contributions from every angle and tells us that if we use τ*= τ . 5/3  (where τ is defined in the vertical direction) we get the complete contribution from all of the different directions. (Note 2). For those following M2007 I have used τ* to be his τ with a ˜ on top, and τ to be his τ with a ¯ on top.

With these conditions we can get a solution for the SGM (see derivation in the comments):

B(τ) = F0/2π . (τ+1)   [1]   cf eqn [M15]

where B is the spectrally integrated Planck function, and remember F0 is a constant.

And also:

F↑(τ) = F0/2 . (τ+2)    [2]

F↓(τ) = F0/2 . τ    [3]

A quick graphic might explain this a little more (with an arbitrary total optical thickness, τ_A* = 3):

Figure 3

Notice that the upward longwave flux at TOA is 240 W/m² – this balances the absorbed solar radiation. And the downward longwave flux at TOA is zero, because there is no atmosphere above from which to radiate. This graph also demonstrates that the difference between F↑ and F↓ is a constant as we move through the atmosphere, meaning that the heating rate is zero. The increase in downward flux, F↓, is matched by the decrease in upward flux, F↑.

It’s a very simple model.

By contrast, here are the heating/cooling rates from a comprehensive (= “standard”) radiative-convective model, plotted against height instead of optical thickness.

Heating from solar radiation, because the atmosphere is not completely transparent to solar radiation:

Figure 4

Cooling rates due to various “greenhouse” gases:

Figure 5

And the heating and cooling rates won’t match up because convection moves heat from the surface up into the atmosphere.

Note that if we plotted the heating rate vs altitude for the SGM it would be a vertical line on 0.0°C/day.

Let’s take a look at the atmospheric temperature profile implied by the semi-grey model:

Figure 6

Now a lot of readers are probably wondering what the τ really means, or more specifically, what the graph looks like as a function of height in the atmosphere. In this model it very much depends on the concentration of the absorbing gas and its absorption coefficient. Remember it is a fictitious (or “idealized”) atmosphere. But if we assume that the gas is well-mixed (like CO2 for example, but totally unlike water vapor), and the fact that pressure increases with depth then we can produce a graph vs height:

Figure 7

Important note – the values chosen here are not intended to represent our climate system. 

Figure 6 & 7, along with figure 3, are just to help readers “see” what a semi-grey model looks like. If we increase the total optical depth of the atmosphere the atmospheric temperature at the surface increases.

Note as well that once the temperature reduction vs height is too large a value, the atmosphere will become unstable to convection. E.g. for a typical adiabatic lapse rate of 6.5 K/km, if the radiative equilibrium implies a lapse rate > 6.5 K/km then convection will move heat to reduce the lapse rate.

Curious Comments on the SGM

Some comments from M2007:

p 11:

Note, that in obtaining B0 , the fact of the semi-infinite integration domain over the optical depth in the formal solution is widely used. For finite or optically thin atmosphere Eq. (15) is not valid. In other words, this equation does not contain the necessary boundary condition parameters for the finite atmosphere problem.

The B0 he is referring to is the constant in [M15]. This constant is H/2π – where H = F0 (absorbed solar radiation) in my earlier notation. This constant B0 later takes on magical properties.

p 12:

Eq. (15) assumes that at the lower boundary the total flux optical depth is infinite. Therefore, in cases, where a significant amount of surface transmitted radiative flux is present in the OLR , Eqs. (16) and (17) are inherently incorrect. In stellar atmospheres, where, within a relatively short distance from the surface of a star the optical depth grows tremendously, this could be a reasonable assumption, and Eq. (15) has great practical value in astrophysical applications. The semi-infinite solution is useful, because there is no need to specify any explicit lower boundary temperature or radiative flux parameter (Eddington, 1916).

[Emphasis added]

The equations can easily be derived without any requirement for the total optical depth being infinite. There is no semi-infinite assumption in the derivation. Whether or not some early derivations included it, I cannot say. But you can find the SGM derivation in many introductions to atmospheric physics and no assumption of infinite optical thickness exists.

When considering the clear-sky greenhouse effect in the Earth’s atmosphere or in optically thin planetary atmospheres, Eq. (16) is physically meaningless, since we know that the OLR is dependent on the surface temperature, which conflicts with the semi-infinite assumption that τ_A =∞..

..There were several attempts to resolve the above deficiencies by developing simple semi-empirical spectral models, see for example Weaver and Ramanathan (1995), but the fundamental theoretical problem was never resolved..

This is the reason why scientists have problems with a mysterious surface temperature discontinuity and unphysical solutions, as in Lorenz and McKay (2003). To accommodate the finite flux optical depth of the atmosphere and the existence of the transmitted radiative flux from the surface, the proper equations must be derived.

The deficiencies noted include the result in the semi-gray model of a surface air temperature less than the ground temperature. If you read Weaver and Ramanathan (1995) you can see that this isn’t an attempt to solve some “fundamental problem“, but simply an attempt to make a simple model slightly more useful without getting too complex.

The mysterious surface temperature discontinuity exists because the model is not “full bottle”. The model does not include any convection. This discontinuity is not a mystery and is not crying out for a solution. The solution exists. It is called the radiative-convective model and has been around for over 40 years.

Miskolczi makes some further comments on this, which I encourage people to read in the actual paper.

We now move into Appendix B to develop the equations further. The results from the appendix are the equations M20 and M21 on page 14.

Making Equation Soufflé

The highlighted equation is the general solution to the Schwzarschild equation. It is developed in Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations – the equation reproduced here from that article with explanation:

I_λ(0) = I_λ(τ_m)e^-τ_m + ∫ B_λ(T)e^-τ dτ 

The intensity at the top of atmosphere equals..

The surface radiation attenuated by the transmittance of the atmosphere, plus..

The sum of all the contributions of atmospheric radiation – each contribution attenuated by the transmittance from that location to the top of atmosphere

For those wanting to understand the maths a little bit, the 3/2 factor that appears everywhere in Miskolczi’s equation B1 is the diffusivity approximation already mentioned (and see note 2) where we need to sum the radiances over all directions to get flux.

Now this equation is the complete equation of radiative transfer. If we combine it with a simple convective model it is very effective at calculating the flux and spectral intensity through the atmosphere – see Theory and Experiment – Atmospheric Radiation.

So equation B1 in M2007 cannot be solved analytically. This means we have to solve it numerically. This is “simple” in concept but computationally expensive because in the HITRAN database there are 2.7 million individual absorption lines, each one with a different absorption coefficient and a different line width.

However, it can be solved and that is what everyone does. Once you have the database of absorption coefficients and the temperature profile of the atmosphere you can calculate the solution. And band models exist to speed up the process.

And now the rabbit..

The author now takes the equation for the “source function” (B) from the simple model and inserts it into the “complete” solution.

The “source function” in the complete solution can already be calculated – that’s the whole point of the equation B1. But now instead, the source function from the simple model is inserted in its place. This equation assumes that the atmosphere has no convection, has no variation in emissivity with wavelength, has no solar absorption in the atmosphere, and where the heating rate at each level in the atmosphere = zero.

The origin of equation B3 is the equation you see above it:

B(τ) = 3H(τ)/4π + B0   [M13]

Actually, if you check equation M13 on p.11 it is:

B(τ) = 3H.τ/4π + B0   [M13]

This appears to be one of the sources of confusion for Miskolczi, for later comment.

Equation M13 is derived for zero heating rates throughout the atmosphere, and therefore constant H. With this simple assumption – and only for this simple assumption – the equation M13 is a valid solution to “the source function”, ie the atmospheric temperature and radiance.

If you have the complete solution you get one result. If you have the simple model you get a different result. If you take the result from one and stick it in the other how can you expect to get an accurate outcome?

If you want to see how various atmospheric factors are affected by changing τ, then just change τ in the general equation and see what happens. You have to do this via numerical analysis but it can easily be done..

As we continue on working through the appendix, B6 has a sign error in the 2nd term on the right hand side, which is fixed by B7.

This B0 is the constant in the semi-gray solution. The constant appears because we had a differential equation that we integrated. And the value of the constant was obtained via the boundary condition: upward flux from the climate system must balance solar radiation.

So we know what B0 is.. and we know it is a constant..

Yet now the author differentiates the constant with respect to τ. If you differentiate a constant it is always zero. Yet the explanation is something that sounds like it might be thermodynamics, but isn’t:

If someone wants to explain what thermodynamic principle create the first statement – I would be delighted. Without any explanation it is a jumble of words that doesn’t represent any thermodynamic principle.

Anyway B0 is a constant and is equal to approximately 240 W/m². Therefore, if we differentiate it, yes the value dB0/dτ=0.

Unfortunately, the result in B10 is wrong.

If we differentiate a variable we can’t assume it is a constant. The variable in question is B_G. This is the “source function” for the ground, which gives us the radiance and surface temperature. Clearly the surface temperature is a function of many factors especially including optical thickness. Of course, if somewhere else we have proven that B_G is a constant then dB_G/dτ=0.

It has to be proven.

[And thanks to DeWitt Payne for originally highlighting this issue with B_G, as well as explaining my calculus mistakes in an email].

A quick digression on basic calculus for the many readers who don’t like maths – just so you can see what I am getting at.. (you are the ones who should read it)

Digression

We will consider just the last term in equation [B9]. This term = B_G/(e^τ-1). I have dropped the π from the term to make it simpler to read what is happening.

Generally, if you differentiate two terms multiplied together, this is what happens:

d(fg)/dx = g.df/dx + f.dg/dx   [4]

This assumes that f and g are both functions of x. If, for example, f is not a function of x, then df/dx=0 (this just means that f does not change as x changes). And so the result reduces to d(fg)/dx = f.dg/dx.

So, using [4] :

d/dτ [B_G/(e^τ-1)] = [1/(e^τ-1)] . dB_G/dτ + B_G . d [1/(e^τ-1)]/dτ  [5]

We can look up:

d [1/(e^τ-1)]/dτ = -e^τ/(e^τ-1)²  [6]

So substituting [6] into [5], last term in [B9]:

= [1/(e^τ-1)] . dB_G/dτ – e^τ.B_G /(e^τ-1)²   [7]

You can see the 2nd half of this expression as the first term in [B10], multiplied by π of course.

But the term for how the surface radiance changes with optical thickness of the atmosphere has vanished.

end of digression

Soufflé Continued

So the equation should read:

Where the red text is my correction (see eqn 7 in the digression).

Perhaps the idea is that if we assume that surface temperature doesn’t change with optical thickness then we can prove that surface temperature doesn’t change with optical thickness.

This (flawed) equation is now used to prove B11:

Well, we can see that B11 isn’t true. In fact, even apart from the missing term in B10, the equation has been derived by combining two equations which were derived under different conditions.

As we head back into the body of the paper from the appendix, equations B7 and B8 are rewritten as equations [M20] and [M21].

Miskolczi adds:

We could not find any references to the above equations in the meteorological literature or in basic astrophysical monographs, however, the importance of this equation is obvious and its application in modeling the greenhouse effect in planetary atmospheres may have far reaching consequences.

Readers who have made it this far might realize why he is the first with this derivation.

Continuing on, more statements are made which reveal some of the author’s confusion with one part of his derivation. The SGM model is derived by integrating a simple differential equation, which produces a constant. The boundary conditions tell us the constant.

Equation [M13] is written:

B(τ) = 3H/4π + B0   [M13]

Then [M14] is written:

H(τ) = π (I+ – I-)    [M14]

So now H is a function of optical depth in the atmosphere?

In [M15]:

B(τ*) = H (1 + τ*)/2π    [M15]

Refer to my equation 1 and you will see they are the same. The only way this equation can be derived is with H as a constant, because the atmosphere is in radiative equilibrium. If H isn’t constant you have a different equation – M13 and 15 are no longer valid.

..The fact that the new B0 (skin temperature) changes with the surface temperature and total optical depth, can seriously alter the convective flux estimates of previous radiative-convective model computations. Mathematical details on obtaining equations 20 and 21 are summarized in appendix B.

Miskolczi has confused himself (and his readers).

Conclusion

There is an equation of radiative transfer and it is equation B1 in the appendix of M2007. This equation is successfully used to calculate flux and spectral intensity in the atmosphere.

There is a very simple equation of radiative transfer which is used to illustrate the subject at a basic level and it is called the semi-grey model (or the Schwarzschild grey model). With the last few decades of ever increasing computing power the simple models have less and less practical use, although they still have educational value.

Miskolczi has inserted the simple result into the general model, which means, at best, it can only be applied to a “grey” atmosphere in radiative equilibrium, and at worst he has just created an equation soufflé.

The constant in the simple model has become a variable. Without any proof, or re-derivation of the simple model.

One of the important variables in the simple model has become a constant and therefore vanished from an equation where it should still reside.

Many flawed thermodynamic concepts are presented in the paper, some of which we have already seen in earlier articles.

M2007 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 stated:

I think I was the first who showed the Aa~=Ed relationship with reasonable quantitative accuracy.

And doesn’t understand why I think it is important to differentiate between fundamental thermodynamic identities and approximate experimental results in the theory section of a paper. “My experiments back up my experiments..”

M2007 introduces equation [M7] with:

In Eq. (6) S_U − (F⁰ + P0 ) and E_D − E_U represent two flux terms of equal magnitude, propagating into opposite directions, while using the same F0 and P0 as energy sources. The first term heats the atmosphere and the second term maintains the surface energy balance. The principle of conservation of energy dictates that:
S_U − (F⁰) + E_D − E_U = F⁰ = OLR

Note the pseudo-thermodynamic explanation. The author himself recently said:

Eq. 7 simply states, that the sum of the Su-OLR and Ed-Eu terms – in ideal greenhause case – must be equal to Fo. I assume that the complex dynamics of the system may support this assumption, and will explain the Su=3OLR/2 (global average) observed relationship.

[Emphasis added]

And later entertainingly commented:

You are right, I should have told that, and in my new article I shall pay more attantion to the full explanations. However, some scientists figured it out without any problem.

Party people who got the joke right off the bat..

M07 also introduces the idea that kinetic energy can be equated with the flux from the atmosphere to space. See Part Three. Introductory books on statistical thermodynamics tell us that flux is proportional to the 4th power of temperature, while kinetic energy is linearly proportional to temperature. We have no comment from the author on this basic contradiction.

This pattern indicates an obvious problem.

In summary – this paper does not contain a theory. Just because someone writes lots of equations down in attempt to back up some experimental work, it is not theory.

If the author has some experimental work and no theory, that is what he should present – look what I have found, I have a few ideas but can someone help develop a theory to explain these results.

Obviously the author believes he does have a theory. But it’s just equation soufflé.

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

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)

Deductions from a simple climate model: factors governing surface temperature and atmospheric thermal structure, Weaver & Ramanathan, JGR (1995)

Notes

Note 1 – emissivity = absorptivity for the same wavelength or range of wavelengths

Note 2 – this diffusivity approximation is explained further in Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations. In M2007 he uses a different factor, τ* = τ . 3/2 – this differences are not large but they exist. The problems in M2007 are so great that finding the changes that result from using different values of τ* is not really interesting.

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