Trouble in Noonworld, Take 2

Guest post by Bob Wentworth

I have been trying to understand and deconstruct the climate-modeling work of Philip Mulholland and Stephen Wilde (M&W). M&W seem to believe that the model they have developed explains planetary temperatures as a consequence of atmospheric mass movement, without any need to reference the radiative effects of greenhouse gases.

There have been a few ups and downs in my engagement with M&W’s work.

My essay Atmospheric Energy Recycling was stimulated by M&W’s work, yet it addressed their work only tangentially. I reviewed one of their papers in Deconstructing Wilde and Mulholland’s Analysis of Earth’s Energy Budget. I didn’t realize at the time that that particular paper did not reveal the full essence of M&W’s approach to modeling. So, while my critique reflected what it was like to try to make sense of one of M&W’s papers in isolation, it didn’t address their full model.

Then, I wrote about M&W’s “Noonworld” paper, which lays the foundation for M&W’s approach to climate modeling. Unfortunately, my first analysis of that paper was seriously flawed.

If you’re willing to give me another chance, I’d like to try again.

* * *

I want to talk about the paper Modelling the Climate of Noonworld: A New Look at Venus by Philip Mulholland and Stephen Wilde. This paper lays out a framework for planetary climate modeling which M&W call the Dynamic-Atmosphere Energy-Transport (DAET) model. In subsequent papers, M&W have applied their DAET model to Titan and Earth and to further study of Venus.

I’ve carefully examined the DAET model. My conclusions are:

  • Unfortunately, the planetary temperature predicted by the DAET model relies on an inappropriate method of calculating temperature. This invalidates all the key claims of the work.
  • Additionally, the DAET model isn’t rooted in physics, which means that there is little assurance that the predictions of the model will have much relationship to reality.

If you’d like details, read on.

* * *

“Noonworld” is the name M&W give to a hypothetical tidally-locked planet, with an

atmosphere (nominally pure nitrogen) which is fully transparent to both shortwave and

longwave radiation.

The Lit hemisphere always faces the Sun, and the Dark hemisphere is in perpetual darkness.  The temperature difference between the Lit side and the Dark side induces convection in the atmosphere. The result is a circulatory system (somewhat like a Hadley Cell on Earth) that transports heat from the warmer Lit side to the cooler Dark side.

The planetary energy flow model is illustrated above in M&W’s Figure 4 (reflecting model values tuned to correspond to the insolation and observed temperature of Venus).

The model involves five energy flows:

  1.  “Solar Insolation” (Insolation absorbed on the Lit side)
  2. R₊  “Lit Surface Radiant Loss to Space” (Thermal radiation from the surface of the Lit hemisphere)
  3. R₋  “Dark Surface Radiant Loss to Space” (Thermal radiation from the surface of the Dark hemisphere)
  4. Aₓ  “Top of Atmosphere Thermal Export” (High-altitude warm air flow from the Lit side to the Dark side)
  5. Aᵣ  “Surface Cold Air Thermal Return” (Low-altitude cool air flow from the Dark side to the Lit side)

I’ve used my own symbols to denote these flows. (M&W don’t provide any symbols, and I’d like to be able to express the mathematical relationships between the various quantities.)

M&W’s DAET model of Noonworld and Venus is rooted in a single premise:

  • When an energy flow enters a hemisphere, it is partitioned between the surface radiant heat loss and the atmospheric energy flow leaving the hemisphere, according to a fixed ratio.

Although M&W describe the energy partitioning in terms of a ratio, I’ll be describing it in an equivalent way, using an energy partition fraction, 𝛾. A fraction 𝛾 of the energy flow entering a hemisphere is assumed to enter (or continue in) the atmospheric circulation, and a fraction (1- 𝛾) is assumed to be radiated by the surface as thermal radiation.

Mathematical Details

Those put off by math (or eager to get to the critique) are encouraged to skip to the next section, “Incompatible Temperatures.”

W&M don’t unpack the technical meaning of the “energy flows” in their model. However, it seems clear what must be meant:

  • “Solar insolation” and “radiant loss” energy flows technically refer to “irradiance” and “radiant exitance” (“radiant emittance”) values for received and emitted radiation. (The jargon for measuring radiation is very technical and specific, with many similar-sounding terms that have distinct meanings.) These values are averaged over a hemisphere of the planet.
  • For consistency, the atmospheric “thermal export” and “thermal return” energy flows must refer to the total energy flow between hemispheres divided by the area of a hemisphere.

All energy flows are measured in units of watts per square meter. During many of their calculations, M&W further normalize energy flow values by expressing them relative to the average absorbed isolation, S.

The average insolation absorbed by the Lit hemisphere is given by S = (1-A₀)⋅S₀/2, where S₀ is solar irradiance at the relevant distance from the Sun and A₀ is the planetary albedo.

The energy-partitioning rule implies that energy flows are related as follows:

R₊ = (1 – 𝛾)⋅(S + Aᵣ)

Aₓ = 𝛾⋅(S + Aᵣ)

R₋ = (1 – 𝛾)⋅Aₓ

Aᵣ = 𝛾⋅Aₓ

Although M&W mainly present numerical solutions to the model, these equations can be solved analytically, yielding:

R₊ = S/(1+ 𝛾)

Aₓ = S⋅𝛾/(1−𝛾²)

R₋ = S⋅𝛾/(1+ 𝛾)

Aᵣ = S⋅𝛾²/(1−𝛾²)

If follows that the total thermal radiance is R₊+R₋ = S. It’s reassuring that the total thermal radiance is always S. That means the model conserves energy and guarantees radiant balance between the energy absorbed and emitted by the planet.

The net atmospheric heat flow from the Lit side to the Dark side is Aₓ−Aᵣ = S⋅𝛾/(1+𝛾). This means that, depending on 𝛾, up to half the absorbed solar flux may be transported from the Lit side to the Dark side. It’s reassuring that the heat transport is confined to that range.

* * *

M&W calculate what they call the Total Global Energy Budget (TGEB). This can be conceptualized two ways:

  1. The sum of all the thermal radiance and atmospheric energy flows, TGEB = R₊+R₋+Aₓ+Aᵣ
  2. The sum of the energy flows arriving at the Lit side, S+Aᵣ, and the energy flow arriving at the Dark side, Aₓ.

These two formulations are mathematically equivalent. Both yield TGEB = S/(1−𝛾).

As far as I can tell, W&M’s Total Global Energy Budget, TGEB, has no physical significance.

Perhaps M&W eventually came to realize this. When they reported the predictions of their model for 𝛾 = ½ (in Table 7), they reported TGEB and included a corresponding “temperature.”  However, when they reported the predictions for a larger value of 𝛾 (in Table 9), they reported TGEB, but didn’t report any corresponding “temperature.” Perhaps they noticed that the result was difficult to rationalize as being a meaningful temperature?

* * *

G&W distinguish what they call “Diabatic” and “Adiabatic” versions of their model. The “Diabatic Model” sets the energy partition fraction to 𝛾 = ½, yielding TGEB = 2⋅S. The Adiabatic Model” involves an energy partition fraction 𝛾 > ½, yielding TGEB > 2⋅S.

I’m not certain, but I imagine the names relate to a belief that the model’s behavior for 𝛾 > ½ is a consequence of adiabatic processes in the atmosphere.

G&W apply the “Diabatic” model to Noonworld, and conclude that the “Adiabatic” model is needed to explain the temperature of Venus

To fit their model to Venus, G&W use a partition ratio 𝛾 = 0.991138. This yields atmospheric “thermal Export” and “thermal Return” energy flows respectively 56.17 and 55.67 times larger than the Lit hemisphere absorbed insolation, S = 299.15 W/m². Based on the Stephan-Boltzmann law, G&W calculate corresponding “temperatures” and assert that the global mean air temperature is 737 K (464℃), which matches their assumed mean temperature for Venus.

* * *

That’s M&W’s DAET model. Does it provide a useful model of how planetary temperatures are established?

I don’t think so. There is a “temperature” problem that invalidates the model, and there are other issues which further render the model suspect, as I’ll explain.

Incompatible Temperatures

Above is M&W’s Table 9, which shows energy flows and associated temperatures as predicted by their DAET model of Venus.

The “radiant loss to space” values for the Lit and Dark sides are calculated as corresponding to temperatures of -46.1℃ and -46.6℃, respectively. These temperatures are calculated using the Stephan-Boltzmann law, j* = σT⁴ where j* is the radiant exitance (radiant emittance).

This means that, in the model, the surface of Venus has a temperature of about -46.

M&W also apply the Stephan-Boltzmann law to the atmospheric “thermal export” and “thermal return” values, concluding that the “average global air temperature” is 464℃.

Compare those two temperatures. The model says the surface of Venus is at -46℃, but that the atmosphere is 500℃ hotter than the surface!

Do I need to say that that is not thermodynamically possible?

Non-Physical “Temperature” Calculations

As one can see from Table 9, M&W are happy to calculate a corresponding “temperature” for nearly every energy flow “Power Intensity Flux” in their model.

The problem with this is: the Stephan-Boltzmann law is only relevant to thermal radiation!

It is meaningless to apply it to any other sort of energy flow.

Suppose an audio speaker is playing music with an average acoustic energy flux of 1 watt/m². Applying the Stefan-Boltzmann (S-B) law to that acoustic energy flux yields a temperature of 65 K (-208℃).

Does anyone think such a calculation is meaningful?

I can hear the protests now: “But M&W are talking about convective air circulation carrying thermal energy, which relates to temperature, so surely it makes sense to apply S-B there, doesn’t it?”

No, it doesn’t.

What are the atmospheric “thermal export” and “thermal return” energy flows?

They don’t represent “heat flux” because heat can only flow from the hot Lit side to the cold Dark side, not in the other direction.

The energy fluxes are likely intended to represent the movement of the air’s total energy density, consisting of the sum of internal energy, U, and potential energy, PE.

So, the energy flux would be given by 𝚽 = (U + PE)⋅v where v is the velocity of air movement.

How does the energy flux 𝚽 relate to the temperature of the air, T?

Glossing over nuances to keep things simple, one could say U = C⋅T where C is heat capacity. Plugging this into the formula for energy flux and solving for T yields:

T = (𝚽/v – PE)/C

This formula has an interesting implication. The energy flows in the atmosphere increase as the partition fraction, 𝛾, is increased. Yet, this increase in energy flow need not reflect any increase in air temperature. A higher 𝛾 value might simply correspond to a higher value of atmospheric heat capacity, C, or circulation velocity, v.

Increasing either heat capacity or velocity would lead to a larger atmospheric energy flow and greater efficiency in transferring heat from the Lit hemisphere to the Dark hemisphere, without involving any increase in air temperature.

So, air temperature does have a relationship to the energy flows that M&W associate with circulation of the atmosphere. However, energy flux does not uniquely determine air temperature. And, calculating the temperature of the air has nothing to do with the Stephan-Boltzmann law.

* * *

Calculating a temperature from an energy flow via the Stephan-Boltzmann law is valid only if the energy flow refers to thermal radiation, or to a combination of energy flows that are logically known to be equal to the amount of thermal radiation. In addition, even when dealing with thermal radiation, one cannot simply add fluxes (rather than averaging them) and compute a temperature.

M&W seem to have no idea when it is or is not appropriate to apply the Stephan-Boltzmann law. As a result, most of the temperatures they calculate in Table 9 are nonsense.

In particular, all the air temperatures M&W compute are nonsense, the meaningless output of an inapplicable formula. Unfortunately, the “average global air temperature” is the central output of the DAET model, the result upon which M&W base all their conclusions.

The only physically meaningful temperatures that M&W compute in Table 9 are the ones saying that the surface is at -46.

Given that M&W are trying to explain a 464℃ near-surface temperature for Venus, this does not constitute a good fit between the model and reality.

* * *

My earlier review of a paper by Wilde and Mulholland revealed similar issues of temperature being calculated from energy flows in an inappropriate way. This appears to be an ongoing core flaw in M&W’s work.

* * *

This issue of inappropriate “temperature” calculations invalidates the predictions of the DAET model upon which M&W base their conclusions.

Given that, one might feel it is pointless to examine the model further. So, I’ll understand if you choose to stop reading at this point.

However, for completeness, I’ll comment on a few lesser issues.

Peculiar Energy Flows

Let’s look at the energy flows that M&W’s DAET model predicts.

The diagram above depicts the energy flows between the Sun, the Lit and Dark hemisphere surfaces, the “thermal export” and “thermal return” air currents, and space.

The numerical values are based on an energy partition ratio 𝛾 = 0.9. Energy is conserved, though numerical rounding might suggest small discrepancies.

In this system, the Lit hemisphere acts as the heat source, absorbing solar insolation. The Dark hemisphere acts as the heat sink, radiating energy to space.

Thermodynamically, the temperatures of the air in this system must be between the temperatures of the heat source and the heat sink.

In particular, it must be true that T₊ > Tₓ > Tᵣ > T₋ where T₊ is the temperature of the Lit hemisphere surface, Tₓ is the temperature of the “thermal export” air current (when at the altitude of the surface), Tᵣ is the temperature of the “thermal return” air current, and T₋ is the temperature of the Dark hemisphere surface.

I notice two peculiar things about the predicted energy flows:

  1. The energy flow from the Lit hemisphere surface to the air, 0.90, is much larger than the energy flow from the air to the Dark hemisphere surface, 0.47.

    This is surprising. Thermodynamically, heat is flowing from the Lit surface to the air to the Dark surface, where it is then radiated. The heat flux is the same throughout this process.

    So, one might expect that the amount of energy the surface transfers to the air on the Lit side would match the amount of energy the air transfers to the surface on the Dark side.

  2. Energy flows at a non-zero rate, 0.43, from the cool “thermal return” air flows to the hot surface of the Lit hemisphere.

    We know that heat doesn’t flow from cool to hot. So, why does the model predict that energy does flow from cool to hot?

These two peculiarities cancel out each other mathematically. Energy is conserved and there is a net heat transfer rate, 0.47, from the Lit side to the Dark side.

So, at a level of overall effect, one can’t say that the net result is non-physical.

Yet, to me, it seems decidedly peculiar that the model seems to conceptually rely on energy flows which don’t match what one would expect to see based on the underlying heat transfer mechanisms.

Peculiar Energy Partition Asymmetry

W&M’s DAET model energy partitioning rule has a peculiar asymmetry to it.

Suppose the energy partition fraction is 𝛾 = 0.9. According to the model:

  • Of the insolation absorbed by the surface on the Lit side, 90% of that energy flux will be transferred from the surface to the air. This suggests strong thermal coupling between the surface and the air.
  • Of the energy that warm air brings to the Dark side, only 10% of that energy will be transferred from the air to the surface. This suggests weak thermal coupling between the surface and the air.

This sort of coupling, in which there is a bias toward the energy always flowing to the same destination, regardless of where it comes from, is not typical of physical systems.

Consider a partially reflective mirror separating two rooms:

  • If the reflectivity is high, then light will mostly stay in the room where it originated.
  • If the transmissivity is high, then light will move between the rooms equally easily in both directions.

The rule M&W have adopted is analogous to a magic mirror that somehow traps light on one side of it, regardless of which side the light originated on. Mirrors don’t work that way.

Heat transport, too, is usually symmetric (assuming you reverse the temperature difference). So, by default, I’d expect thermal coupling on Noonworld to be symmetric, just as coupling is symmetric with mirrors.

In the DAET model, it is difficult to relate what is happening to anything real, like the relative temperatures of the surface and the air.

Given that the model’s connection to physics is mostly non-existent, it’s hard to prove that the physics of the model is wrong.

Even so, it seems unlikely that the asymmetric behavior posited by the DAET energy partitioning rule could correspond to any real physics.

No Enforcement of the Second Law of Thermodynamics

The DAET energy partitioning rule guarantees that the First Law of Thermodynamics (energy conservation) will be honored.

That’s good, as far as it goes.

However, there is nothing about the DAET model which clearly ensures that the Second Law of Thermodynamics will be honored.

The Second Law of Thermodynamics requires that heat only flows from hot to cold, not cold to hot (unless something like a heat pump is involved).

For their model representing Venus, M&W calculated that the atmosphere of the planet is at 464℃. They also assumed that over 99% of insolation energy absorbed by the surface is transferred to the atmosphere. Yet, their model said the surface is at -46℃. So, the predicted heat transfer seems to involve heat flowing from cold to hot.

Of course, M&W’s calculation of that 464℃ temperature was not valid. So, maybe the model doesn’t really predict net heat transfer from cold to hot.

But the point is, it’s not clear that the DAET energy-partitioning rule prevents such Second-Law-violating outcomes from happening. Maybe it does. Maybe it doesn’t.

Without the Second Law as a constraint, it’s all too easy for a model to predict outcomes that are incompatible with reality.

Game-World Physics

Expecting a model like DAET to predict real-world physics is a little like expecting a computer game to accurately predict how reality functions.

M&W’s DAET model is based on an arbitrary energy-partitioning rule with no particular connection to the physics that M&W are trying to model.

When a model relies on arbitrary assumptions, the model can yield arbitrary outcomes. There is no reason to expect the results to have much to do with reality.

Significance of the Model

M&W have asserted that their DAET model explains how a planet like Venus can be much warmer than the “vacuum planet” temperature (computed by balancing absorbed insolation with thermal radiation emitted, for a planet of uniform temperature with the relevant albedo and surface emissivity).

Because M&W used an inappropriate formula for calculating atmospheric temperature, M&W’s assertion is false.

Properly interpreted, the DAET model does not predict or explain any atmospheric enhancement of temperature beyond the “vacuum planet” value.

The model does show that, given sufficiently powerful convective heat transfer, the temperatures of the lit and dark sides of a tidally locked planet could be nearly equalized. This is a valid conclusion, but it is not a surprising one.

Conclusion

I admire the passion, creativity, and effort that Philip Mulholland and Stephen Wilde have given to thinking freshly about how planetary climates function.

Regrettably, their DAET model doesn’t have much to tell us about real planets.

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John Tillman
April 27, 2021 2:06 pm

Venus is hot because it rotates so slowly, and, unlike also leisurely Mercury, has a thick atmosphere, so that its night side doesn’t cool off. Some heat might also move through the lithosphere from lit to dark hemisphere.

Rotation periods in Earth hours (Venus rotates “backwards”):

Mercury 1407.6
Venus -5832.5
Earth 23.9

Last edited 3 months ago by John Tillman
Bob Wentworth
Reply to  John Tillman
April 27, 2021 2:28 pm

Slow rotation does not result in a higher average planetary temperature. Generally, slow rotation results in large temperature variations which contribute to the average temperature being lower than the temperature of a similarly located planet with uniform temperature.

However, because of its atmosphere (combined with other effects that buffer temperature), Venus has very low surface temperature variations between the day and night sides, possibly less than 0.05 K. This temperature uniformity raises the average temperature, but still leaves 500 K in warming to be explained.

Because of Venus’s high albedo, it absorbs less sunlight than Earth, despite being closer to the Sun. Yet, Venus’s average temperature is hotter than that of Mercury.

Last edited 3 months ago by Bob Wentworth
John Tillman
Reply to  Bob Wentworth
April 27, 2021 2:41 pm

Yes, the surface of Venus recieves less insolation than the surface of Earth, thanks largely to high albedo of its SO2 clouds. However, it’s atmosphere also absorbs a lot of solar radiation.

But the planet rotates so slowly, that any one spot receives much more sunlight than Earth despite the high albedo. The calculation is straightforward. Again, unlike Mercury, Venus doesn’t lose heat during its 121.5-Earth day (~24 hours) night.

Consider how much heat any equatorial sea level spot on Earth gains during just an average of 12 hours insolation (dawn to dusk). Then adjust for albedo and distance from the Sun, and multiply by ten. QED.

Last edited 3 months ago by John Tillman
John Tillman
Reply to  John Tillman
April 27, 2021 3:14 pm

Its. Maldita autofill.

gbaikie
Reply to  John Tillman
April 27, 2021 3:50 pm

–Yes, the surface of Venus recieves less insolation than the surface of Earth, thanks largely to high albedo of its SO2 clouds. However, it’s atmosphere also absorbs a lot of solar radiation.–
An atmosphere includes clouds, but I would say the clouds absorbs some solar radiation and the gases of the atmosphere absorbs less radiation than the clouds.
Overall Venus absorbs less solar radiation than the further away Earth, but at Venus distance from the sun, the solar radiation is more intense {twice many watts per square meter} and cloud droplets can heated to higher temperature {as compared to Earth distance sunlight}. And heated droplets transfer heat to the air via convection and evaporative processes. And acid droplet gain heat when absorb water vapor and evaporatively cool when evaporating water vapor [and become stronger acid}. Venus is very dry, the clouds acid is much stronger the battery acid {battery acid has a lot water in it}. wiki: “Venusian clouds are thick and are composed mainly (75–96%) of sulfuric acid droplets”
Battery acid [between 15% and 35%}.

John Tillman
Reply to  gbaikie
April 27, 2021 3:53 pm

Yes, it’s not just insolation incident on the surface directly and through scattering, but heating of the atmosphere on the way down that matters.

To bed B
Reply to  John Tillman
April 27, 2021 3:59 pm

It’s effective surface is about 70 km up because the atmosphere is dense and opaque. It has a temperature gradient similar to Earth’s, despite clouds of sulphuric acid rather than water, because of circulation. Its surface should be 450 K hotter than the effective surface.

Despite Venus being 2/3 of the distance from the Earth , It’s surface is only 450 K warmer.

John Tillman
Reply to  To bed B
April 27, 2021 4:08 pm

Top of atmosphere irradiance is twice Earth’s.

But about three fourths of solar irradiance at the top of Venus’ atmosphere is reflected away, which is why it’s so bright.

Even so, its slow rotation means that its surface and most of its air get very hot. There are however pockets of cold air shaded by the SO2 cloud banks.

Last edited 3 months ago by John Tillman
Robert of Texas
Reply to  John Tillman
April 27, 2021 6:29 pm

It isn’t just visible light you have to account for, it’s all the light. Various wavelengths *might* be absorbed much more readily than the visible light that makes Venus so bright.

It’s these sorts of details that make models about Earth’s climate relatively useless – they just simplify and average (or even make up) anything they do not readily understand.

Bob Wentworth
Reply to  Robert of Texas
April 29, 2021 4:58 pm

It isn’t just visible light you have to account for, it’s all the light. Various wavelengths *might* be absorbed much more readily than the visible light that makes Venus so bright.

The Bond albedo of a planet is an integration over all the wavelengths that the Sun emits. They don’t just guess about that.

To bed B
Reply to  John Tillman
April 29, 2021 12:55 am

Yes. The last line is a bit meaningless without that. More than twice the albedo of Earth negates the 2/3^2

Bob Wentworth
Reply to  John Tillman
April 27, 2021 5:25 pm

It’s not just the surface of Venus that receives less sunlight. When you consider surface+atmosphere, Venus absorbs less sunlight per square meter than does Earth.

Accounting for albedo, Mercury absorbs 2072 W/m², Venus (surface+atmosphere) absorbs 156 W/m², and Earth (surface+atmosphere) absorbs 236 W/m², averaged over the globe and over a rotation.

Reply to  Bob Wentworth
April 27, 2021 4:46 pm

“Generally, slow rotation results in large temperature variations which contribute to the average temperature being lower than the temperature of a similarly located planet with uniform temperature.”

These two modelling studies say otherwise:
Hunt, B.G. 1979 The Influence of the Earth’s Rotation Rate on the General Circulation of the Atmosphere

Del Genio, A.D. & R. J. Suozzo 1987 A Comparative Study of Rapidly and Slowly Rotating Dynamical Regimes in a Terrestrial General Circulation Model

Michael S. Kelly
Reply to  Bob Wentworth
April 27, 2021 7:40 pm

There are a lot of factors present in Venus’ energy balance not equivalent to those of Earth. Yes, it has a high albedo. But how high? Venus is ~0.7, while Earth averages ~0.3. So Venus reflects (on average) 2.333 times the incoming solar radiation than does Earth. However, its obit is at 0.7 AU, which means the “top of the atmosphere” insolation at Venus is a little over 2 times that of Earth. The atmosphere of Venus is almost entirely CO2, at very high pressure (93 bar, or 1,350 psi). That Venus’ atmosphere would saturate all of the IR absorption bands should be evident to the casual observer. Plus, Venus has no magnetic field, and is thus subject to the full impact of both the Solar wind (at twice the flux as that at Earth’s orbit), and cosmic radiation. The CO2 and SO2 in Venus’ upper atmosphere would have much different chemistry in response to intense Solar radiation (especially UV, which is not a factor in Earth atmosphere).

It’s not a simple comparison.

Bob Wentworth
Reply to  Michael S. Kelly
April 27, 2021 7:57 pm

As I mentioned in another comment, the net effect is that Venus (surface+atmosphere) absorbs 156 W/m², and Earth (surface+atmosphere) absorbs 236 W/m², averaged over the globe and over a rotation.

That Venus’ atmosphere would saturate all of the IR absorption bands should be evident to the casual observer. 

As I pointed out in my Energy Recycling essay, having absorption saturated doesn’t mean that more of the same gas can’t further increase temperature.

One way of looking at it is that the “surface” of Venus for emission purposes is the altitude at which the CO₂ thins out enough to be transparent to radiation. That’s the altitude where radiative thermal balancing occurs. The higher that altitude is, the warmer the surface below will be.

DMacKenzie
Reply to  Bob Wentworth
April 27, 2021 10:40 pm

Tall columns of gas molecules in planetary atmospheres exhibit adiabatic lapse rate characteristics. They sort themselves into hot fast moving molecules at the bottom, and colder slower moving ones at the top. Climbing uphill against gravity in their journey up the column causes the upper molecules to slow down, which means their temperature is lower. Even Ph.D heads explode before they understand its just conservation of kinetic plus potential energy. The formula works out the same as if parcels of gas were ascending or descending as we often see for the explanation for Earth’s adiabatic lapse rate.
Venus has a very thick atmosphere of dense phase CO2, and a huge atmosphere as well. Venus’s 3.5% nitrogen atmosphere contains more nitrogen than Earth’s 80% nitrogen atmosphere.
The ideal gas adiabatic lapse rate is not accurate enough and is out a couple of hundred degrees from top to bottom of the Venusian “troposphere”. The following paper worked it out using real gas laws. See their figs 9 to 11. Lapse rate varies with altitude but is about 11 C per Km. Take the radiative temp altitude, and multiply by lapse rate times the 60 Km to surface, and voila, the answer is the Venus surface temp, well, approximately…
https://arxiv.org/pdf/1806.06835.pdf

Bob Wentworth
Reply to  DMacKenzie
April 27, 2021 10:53 pm

Tall columns of gas molecules in planetary atmospheres exhibit adiabatic lapse rate characteristics.

Do you know of places where this happens and there is no convection? I would think convection is pretty pervasive in the lower atmosphere of planets, so I’m not sure how one could identify a tall gas column that isn’t influenced by convection.

There are compelling thermodynamic arguments for why the argument “Climbing uphill against gravity in their journey up the column causes the upper molecules to slow down, which means their temperature is lower” is not necessarily right, under all conditions.

But, I think it’s largely moot, given how pervasive convection is.

And, I don’t think it ultimately tells us how planetary temperatures are determined. A lapse rate sets the slope, but doesn’t set the absolute level of the end points.

And yes, depending on what you mean by it, the “radiative temp” altitude may be what sets that overall level. What do you mean by that term?

Last edited 3 months ago by Bob Wentworth
Another Joe
Reply to  Bob Wentworth
April 27, 2021 11:18 pm

He means, that you can treat the atmosphere as a insulation layer, where the energy flux from the top of atmosphere satisfies the need of energy conservation, by radiating away the amount of solar energy received.
Below that, the atmosphere acts as normal insulation, with a resistivity to the flow of heat.
This resistivity and the heat flow through the atmosphere can be calculated by the radiation temperature at TOA, the temperature on the ground and the radiation height. This can normally also be expressed as the lapse rate.

Bob Wentworth
Reply to  Another Joe
April 28, 2021 12:02 am

I think I agree with this narrative, in general terms, if we can agree that:

  1. It’s the net lapse rate between the surface and the radiation height that matters.
  2. The atmosphere as a whole doesn’t follow a single, linear lapse rate.
  3. Treating the atmosphere as an “insulation layer” in this sense doesn’t mean that conduction is actually the dominant heat transfer mechanism.
  4. The “radiation height” is something that can’t be defined entirely rigorously, except via a bit of a circular argument. In particular, you could define it as the height where the temperature (and emissivity??) would lead to the observed amount of TOA radiation. The actual TOA radiation is made up of a mix of radiation from the surface and radiation at a variety of altitudes, depending on wavelength. So, it will inevitably be only a crude approximation to treat radiation as if it were coming from a single height in the atmosphere. (I’m not sure how one could rigorously figure atmospheric emissivity into all this.)
Another Joe
Reply to  Bob Wentworth
April 28, 2021 8:35 pm

Bob,

1) Correct
2) Does not have to, different insulation materials also display different temperature gradients following their material specific heat transfer properties
3) If you look at the gradient only, you do not know which is the dominant heat transfer mechanism. You only know that there is one and what it means to the temperatures in and out. For a solid insulation material you know its only Heat Conduction. In the atmosphere you know there is more. No need to justify it in particular of you know the sum of it, as displayed in the gradient. Keep it simple
4) This is correct for Earth atmosphere. If you are looking for a model where it is entirely valid, then use one where the atmosphere is a grey body emitter. Venus is a proxy for this if you ignore the little radiation that makes it to the ground.

What you need to take from this, is that the solar energy input has only to be offset by the radiation at emission height. Underneath there is no need or requirement for anything to have the same energybalance. This means the temperature of the surface is independent from solar radiation and therefore the temperature at the surface is rather set by the Lapse Rate and emission height.

DMacKenzie
Reply to  Another Joe
April 28, 2021 6:11 am

One must be careful with the term ‘lapse rate’. There is the ‘measured’ one by weather balloons on Earth or Magellan Venus probes, and the wet and dry adiabatic ones calculated from basic thermodynamic principles (BTW including gravity from the equation). Atmospheric thermodynamicists ponder the discrepancies, and there aren’t many of them, although there are many self declared climate scientists who study tree rings and icebergs, and other meter maid kind of stuff.

Another Joe
Reply to  DMacKenzie
April 29, 2021 3:27 am

To keep it simple you should refer to the lapse rate entirely by how it is calculated.
Why is the wet lapse rate different from the dry adiabatic one?
With gravity being the same, the difference can be found in the C of the heat capacity.
Wet means more heat can be stored due to the specifics of water and water vapour.
As the heat capacity changes so does the gradient of the lapse rate.
For planet Earth this complicates things a bit, for other atmospheres this is maybe less important.
At the core the take away is, that the lapse rate sets the gradient from the top to the bottom.

Solar insolation defines the top. The lapse rate the surface temperature.

Bob Wentworth
Reply to  Another Joe
April 29, 2021 5:05 pm

Why is the wet lapse rate different from the dry adiabatic one?

With gravity being the same, the difference can be found in the C of the heat capacity.

That doesn’t match my understanding.

The heat capacity of water vapor in humid air would likely somewhat affect the adiabatic lapse rate. But generally, the moist adiabatic lapse rate refers to what happens in the regime where the air is 100% saturated with water vapor, so that any cooling leads to condensation and the release of latent heat.

Wikipedia says “Before saturation, the rising air follows the dry adiabatic lapse rate. After saturation, the rising air follows the moist adiabatic lapse rate.”

Another Joe
Reply to  Bob Wentworth
April 29, 2021 7:00 pm

If you take the gradient from the top, the reverse effect is seen by the surface. The moist adiabatic lapse rate makes the surface cooler since you only get 6 odd deg Celsius temperature increase per kilometer, while the dry lapse rate gives you roughly 10 deg C.
Think of it that way, wet air transports heat faster than dry air. Dry air is the best insulation material.

DMacKenzie
Reply to  Bob Wentworth
April 28, 2021 5:46 am

“Do you know of of places where this happens and there is no convection?”

I think that puts you in the realm of stable versus unstable air parcels, a common topic of meteorology textbooks, and delved into when teaching pilots how to read skew-T diagrams and determine where cloud layers are likely to form.

”….the “radiative temp” altitude may be what sets that overall level. What do you mean by that term?…

I just mean the altitude at which T = [S/4α × (1-A)]^1/4. Yes it is a little circular as far as lapse rate extrapolated to the surface, because you have to measure the atmospheric temp vs. height and therefore know the answer before you start calculating….for cloudy planets like Venus this is likely to be the average cloud top temperature or an optical depth or 2 lower, for partly cloudy planets like Earth, a proration of cloud top temp and surface temp (to oversimplify greatly by ignoring radiative gases, and SW cloud reflection…oops, thats getting to be inaccurate)

Last edited 3 months ago by DMacKenzie
Bob Wentworth
Reply to  DMacKenzie
April 29, 2021 5:12 pm

I just mean the altitude at which T = [S/4α × (1-A)]^1/4. Yes it is a little circular as far as lapse rate extrapolated to the surface, because you have to measure the atmospheric temp vs. height and therefore know the answer before you start calculating

Given that circularity, I think the idea of “radiative temperature altitude” creates more an illusion of understanding than actual understanding of the atmospheric physics.

In an atmosphere with a near black-body absorption-emission characteristic, the “radiative temperature altitude” would nicely track the altitude at which the atmosphere became thin enough to be transparent to its own thermal emissions.

But, in a real atmosphere, in which the optical depth varies widely for different wavelengths, I’m not convinced that the “radiative temperature altitude” as you’ve defined us tells us much about what is going on to determine atmospheric structure and planetary temperature.

You still need to figure out all those details. Once you have, then the “radiative temperature altitude” is a kind of interesting but not really significant emergent property.

At least, that’s they way it seems based on my current thinking.

Sebastian Magee
Reply to  Bob Wentworth
April 28, 2021 4:34 am

The reason why the surface of venus is so homogeneous is because CO2 has a critical pressure of 73.8bar, which means that the lower levels of Venus atmosphere are more similar to a huge ocean than a typical atmosphere. The homogeneity of the temperature at the surface of Venus is similar to the homogeneity at the bottom of the oceans on Earth.

Robert W Turner
Reply to  Bob Wentworth
April 28, 2021 6:25 am

The surface of Venus receives very little solar insolation yet it is hottest at the surface. If the GHG back radiation hypothesis were correct then the hottest area of the planet would be at some midpoint in the atmosphere.

Bob Wentworth
Reply to  Robert W Turner
April 28, 2021 6:45 am

If the GHG back radiation hypothesis were correct then the hottest area of the planet would be at some midpoint in the atmosphere.

That’s not thermodynamically possible. Maximum heating may occur “at some midpoint in the atmosphere”, but that doesn’t imply maximum temperature.

The only way the surface could be cooler than “some midpoint in the atmosphere” would be if there were a heat sink in the surface or low atmosphere. And, I don’t see how there could be a heat sink there.

There is no way for the surface to cool itself, except via heat flowing upward through the atmosphere. For that to happen, it needs to be warmer than the air above it, if it is receiving any insolation at all that needs to be rejected to maintain energy balance.

Last edited 3 months ago by Bob Wentworth
Robert W Turner
Reply to  Bob Wentworth
April 28, 2021 8:17 pm

Lol yes thus you are simply showing how the GHG back radiation hypothesis is pseudoscience. Where are you proposing the additional heat go and how does it get to the surface?

Bob Wentworth
Reply to  Robert W Turner
April 28, 2021 8:45 pm

I don’t think anything I’m saying is dependent on GHG back-radiation. It’s simple thermodynamics that applies regardless of what heat transfer mechanism is involved. Could be conduction, convection, or radiation, and the answer would be the same.

Where are you proposing the additional heat go and how does it get to the surface?

Do you mean how does heat get to the surface?

Well the first random sources I find report of Venus: “a surprising amount of sunlight does reach the surface” and “The actual illumination of the surface is about 14,000 lux, comparable to that on Earth in the daytime with overcast clouds.”

So, it sounds like enough sunlight reaches the surface to provide surface heating.

I guess I’m not following what you think is problematic.

Last edited 3 months ago by Bob Wentworth
Another Joe
Reply to  Robert W Turner
April 28, 2021 9:52 pm

I propose it was already there!

Another Joe
Reply to  Bob Wentworth
April 28, 2021 9:52 pm

Bob,

you are correct that the surface is not a heat sink. It is a heat source. Even if it just provides 0.1 W of heat per square meter, it still means it has a temperature and it needs to be higher than the atmosphere.

You just provided the answer to the question what else is hotter than the atmosphere, so it can be as hot on the bottom! Its the surface!

Izaak Walton
April 27, 2021 2:46 pm

It is also worth pointing out that noonworld is composed of a very strange substance that does not conduct heat yet somehow is able to transfer that heat to the surrounding atmosphere. On any realistic planet at least 50% of the absorbed energy would be transported through the surface of the lit side to the dark side via conduction resulting in significantly different temperatures when at equilibrium compared to the analysis presented above.

Reply to  Izaak Walton
April 27, 2021 3:17 pm

Our model relies on the surface being capable of conduction since conduction from surface to air on the lit side leads to convection and the cold radiating surface on the unlit side draws KE from the falling air by conduction.
Obfuscation of this nature is most unhelpful.

Izaak Walton
Reply to  Stephen Wilde
April 27, 2021 6:16 pm

Stephen,
But your planet only transfers heat from the lit surface to the air and not to the ground beneath the surface. Similarly your surface only emits radiation upwards and not downwards. If the ground beneath the surface is at a lower temperature then the hotter surface will transmit heat downwards making all of your calculations wrong. And if there is no temperature gradient through the interior of the planet then the unlit side is as hot as the lit side which again makes your calculations wrong.

In essence your choice of a surface that does not transmit energy through conduction except to the air and which does not radiate in all directions does not correspond to any physically plausible material and as such isn’t helpful in explaining anything.

Derg
Reply to  Izaak Walton
April 27, 2021 7:12 pm

Neither are climate models, but that doesn’t stop your clown show.

Izaak Walton
Reply to  Derg
April 27, 2021 9:00 pm

But this noon world model takes bits of physics and uses them in an inconsistent way. Take the Stefan-Boltzmann law, it only applies to a black body that is in thermal equilibrium. But if the planet itself is not at a constant temperature then you can’t apply it. For the simple reason that if the interior of the planet is colder than the surface then the blackbody radiation will be directed inwards. So for their calculations to be correct then the planet’s interior must be at the same temperature as the surface which means that the unlit side would be at the same temperature as the lit side.

Similarly for the convection mechanism to work the air of the planet must be able to warm up by conduction with the surface but once heated cannot warm the surrounding air by conduction but has to store all of that thermal energy as potential energy until it gets to the unlit side and then it can warm the surface through conduction but not the surrounding air.

So their entire analysis is internally inconsistent and so you can’t believe any of their results.

Granum Salis
Reply to  Izaak Walton
April 27, 2021 9:54 pm

Is it really your view that the Earth’s surface is furiously radiating into its nether regions, rather than towards the sun or the space station?

One of the favorite back radiation memes is that photons don’t know where they’re going, just where they came from.

How about the thing where collisions are half a gazillion times more frequent than emissions?

Most people think that the interior of the planet is quite a bit warmer than its surface. I will admit I used to dig a hole for my root cellar, but that was before the new normal.

Izaak Walton
Reply to  Granum Salis
April 27, 2021 10:06 pm

Granum,
There centre of the earth is hotter than the exterior and therefore there is no net flux of radiation from the surface to the centre. But an atom at the surface does not know which way is up and which way is down so 50% of the time it will emit radiation directed towards the centre of the earth.

Jim Gorman
Reply to  Granum Salis
April 28, 2021 5:31 am

Funny you would mentiion the SB law and the need for a black body. Do you think the current radiation theories for the earth use a similar assumption. Maybe they are wrong also about “back radation”. Explain how it’s wrong in one case but not another!

Derg
Reply to  Izaak Walton
April 28, 2021 1:49 am

My point is climate alarmists have created model after model after model and our understanding of temperature is no better in those nearly 30 years. But scam artists somehow morphed this into climate change, climate extinction and climate emergency. Meanwhile you say nothing and pile on with this idiocracy.

Robert W Turner
Reply to  Izaak Walton
April 28, 2021 6:30 am

It’s ironic that first you claim their model includes no conduction between surface and atmosphere and then go on about downward radiation within the solid planet itself.

April 27, 2021 2:53 pm

Looks like a complete inability to comprehend basic principles of meteorology to me plus multiple misrepresentations of the substance of our model.
We are clearly applying the Gas Laws so I fail to see how it is not rooted in physics.
I am not aware of any attempt to use the S-B equation in relation to convective processes, quite the opposite since our point is that S- B cannot apply there.
Perhaps Philip could make a limited response to a few of the technical points but I fear we may be wasting our time.

Reply to  Stephen Wilde
April 27, 2021 2:59 pm

And I fail to see how one could possibly assert that the reappearance of KE from P E in descending air could be represented as a flow of cold to hot in breach of the Second Law.

Stevek
Reply to  Stephen Wilde
April 27, 2021 4:04 pm

As far as my limited understanding I think you are correct and that is not a breach of 2nd law. One thing I don’t understand is isn’t the whole KE / PE gravity induced temperature gradient just convection ?

If there was no sun and no atmosphere for earth and then a giant space ship came along and pumped in a whole atmosphere of warm air to earth that gravity would cause the lower atmosphere to be warmer. Not sure how quick it would take but gradient would happen. Of course with no sun the air would start cooling but there would be temperature gradient.

Stevek
Reply to  Stevek
April 27, 2021 4:05 pm

Meant “then gravity would cause”

Reply to  Stevek
April 27, 2021 5:11 pm

Yes, just convection No convection, no greenhouse effect.
A static atmosphere would behave like a solid and become isothermal via conduction alone and the temperature from surface to top of atmosphere would then match the prediction of the S-B equation.
It is the weak bonds between gas molecules that allow it all to happen as per the Gas Laws.

Granum Salis
Reply to  Stephen Wilde
April 27, 2021 8:54 pm

AFAIK there are no interactive forces between gas molecules; ergo no bonds between molecules.
The bending and stretching of intra-molecular bonds permit radiative transfer.

JamesD
Reply to  Stephen Wilde
April 28, 2021 11:19 am

For a perfect reflector planet surface and a transparent atmosphere, the lapse rate would be zero and the atmosphere and surface would have the same temperature.

If the atmosphere contains GHG’s, then the top of atmosphere would be cooler due to optical depth and you would get a lapse rate.

Bob Wentworth
Reply to  Stevek
April 27, 2021 6:59 pm

If there was no sun and no atmosphere for earth and then a giant space ship came along and pumped in a whole atmosphere of warm air to earth that gravity would cause the lower atmosphere to be warmer. Not sure how quick it would take but gradient would happen. Of course with no sun the air would start cooling but there would be temperature gradient.

There would be a temperature gradient due to adiabatic compressive heating, and the temperature profile would follow an adiabatic lapse rate.

However, this initial gravitational heating is entirely a transient effect. It has nothing to do with the steady-state configuration of the system.

As you note, without a Sun, the whole thing would cool, eventually to about 3 K.

If a Sun is present, what happens depends on the presence or absence of greenhouse gases.

In the absence of any greenhouse gases, the only mechanism the planet would have to cool itself would be via radiation from the surface. The radiation absorbed and emitted by the surface would come into balance at some temperature. The atmosphere would come into thermal equilibrium with the surface, with no net heat flowing between the surface and the atmosphere (given that there is no way for the atmosphere to get rid of heat). There would be convection because of temperature differences between the tropics and the poles, and this would create an adiabatic lapse rate temperature profile within at least some portion of the atmosphere.

But, I repeat, there would be no net heat flow into or out of that atmosphere, in the absence of greenhouse gases.

With greenhouse gases present, the atmosphere has an ability to cool itself, and there will be a net heat flow from the surface into the atmosphere. This scenario also leads to an adiabatic lapse rate in the troposphere, with different temperature profiles in the upper atmosphere.

Pablo
Reply to  Bob Wentworth
April 27, 2021 11:28 pm

“As you note, without a Sun, the whole thing would cool, eventually to about 3 K.”

“Millions of years after that, our planet would reach a stable -400°, the temperature at which the heat radiating from the planet’s core would equal the heat that the Earth radiates into space, explains David Stevenson, a professor of planetary science at the California Institute of Technology.”

Minus 400ºF equals 33K. So Earth has a 33ºC head start on anything the sun can provide.

https://www.businessinsider.com/how-long-life-could-survive-if-the-sun-went-out-2013-7?r=US&IR=T

Bob Wentworth
Reply to  Pablo
April 27, 2021 11:43 pm

our planet would reach a stable -400°, the temperature at which the heat radiating from the planet’s core would equal the heat that the Earth radiates into space… Minus 400ºF equals 33K. So Earth has a 33ºC head start on anything the sun can provide.

Fascinating. Thanks for that tid bit.

The part about a “33ºC head start” is rather misleading, though.

Temperatures are not additive, energies are.

So, to figure out how much that much energy would warm Earth above 255 K (-18℃) one can calculate it as T = [255^4 + 33^4]^(1/4) = 255.018 K.

In other words, that “head start” only amounts to an extra 0.018℃ when combined with the heat of the Sun.

Pablo
Reply to  Bob Wentworth
April 28, 2021 12:06 am

Drat..thought I was onto something there!

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 4:43 pm

Nobody asserted “that the reappearance of KE from P E in descending air could be represented as a flow of cold to hot in breach of the Second Law.”

What lead you believe I was asserting that?

I did assert that having an “average global air temperature” of 464℃ when your model clearly indicates a surface temperature of -46℃ is a breach of the Second Law.

Reply to  Bob Wentworth
April 27, 2021 5:02 pm

That -46C is the emission temperature to space and NOT the surface temperature.
You are not following the narrative adequately enough to present any sort of opinion or judgment.
Kindly withdraw your adverse conclusions.

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 5:09 pm

That -46C is the emission temperature to space and NOT the surface temperature.

How could it possibly not be the surface temperature?

Your model assumes an atmosphere transparent to radiation. The only thing that can radiate is the surface. Therefore, if -46℃ radiation is being emitted, it is being emitted by the surface. Therefore, the surface is at -46℃.

Reply to  Bob Wentworth
April 27, 2021 5:29 pm

So, how does the surface obtain the additional KE needed to maintain continuing convective overturning?
If all the KE at the surface is radiated to space then it is gone, nothing left for convection.
How does PE get into the atmosphere at all ?

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 6:43 pm

Your model says that of the 299 W/m² absorbed by the surface, 150 W/m² are radiated into space on the Lit side, while 149 W/m² are convectively transported by the air to the Dark side, where they are radiated into space.

So, that 149 W/m² going into the air is what powers convection.

Reply to  Bob Wentworth
April 28, 2021 12:59 am

Yes, and it warms the surface on the dark side where it partially offsets the temperature fall due to radiation to space and a residue then circulates back to the lit side so the whole globe gets warmer.
Not all of it can be radiated away from the dark side because it is a continuous circulatory flow.

Bob Wentworth
Reply to  Stephen Wilde
April 28, 2021 1:36 am

Well, there is a baseline energy flow of 16,654 watts per square meter that goes around and around in the atmosphere. I only included the delta above that. The portion that I mentioned goes entirely into the Dark side and is radiated.

JamesD
Reply to  Stephen Wilde
April 28, 2021 11:15 am

PE is pressure in a gas. The upper atmosphere is lower pressure than the surface. Gibbs Free energy decreases as the air rises and expands. As I went over, your planet heat balances, but your process description doesn’t make any sense.

For convection to occur, you will need higher pressure and more atmospheric mass on the unlit side, which you probably do. The entropy and energy sink is the unlit surface, which will increase the density of the unlit atmosphere..

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 4:36 pm

I am not aware of any attempt to use the S-B equation in relation to convective processes, quite the opposite since our point is that S- B cannot apply there.

The term “S-B equation” does not refer to the “vacuum planet” calculation. It refers to j* = σT⁴, which is what you clearly used to calculate the temperature of the air.

In Table 9 of your paper, you say:

  1. For “Retention of energy by the lit air” you say the Power Intensity Flux of 16,953 W/m² corresponds to a temperature of 736.2 K (463.2℃)
  2. For “Return flow of colder air from the dark side” you say the Power Intensity Flux of 16,654 W/m² corresponds to a temperature of 739.5 K (466.5℃)
  3. You average the results of #1 and #2 to calculate an “Average global air temperature” of 737.0 K (464.0℃).

The equation j* = σT⁴ explains the temperatures that you report in steps #1 and #2.

Are you trying to tell me that you calculated #1 and #2 without using the equation j* = σT⁴?

If you didn’t use the equation j* = σT⁴, how did you calculate the temperatures in steps #1 and #2?

Reply to  Bob Wentworth
April 27, 2021 5:07 pm

There is no heat flowing from cold to hot involved in those numbers. You misread our paper so as to wrongly assume we set the surface at -46C when that is the emission to space temperature from top of atmosphere.

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 5:13 pm

In the comment you’re responding to, I didn’t say anything about “heat flowing from cold to hot” or about -46℃. Those issues are unimportant compared to this one question:

In calculating the temperatures 736.2 K and 739.5 K (in steps #1 and #2 of my comment), did you use the equation j* = σT⁴?

JamesD
Reply to  Stephen Wilde
April 28, 2021 11:24 am

If the gas is transparent, then the -46C has to coincide with the temperature of the emitter, which is the surface.

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 4:47 pm

We are clearly applying the Gas Laws so I fail to see how it is not rooted in physics.

The only rule in your model is the principle of energy being partitioned in a fixed ratio between the air and the surface.

How is this rule “rooted in physics”?

Could you point me to anything at all in your paper which constitutes “clearly applying the Gas Laws”? Where?

Reply to  Bob Wentworth
April 27, 2021 4:56 pm

Adiabatic expansion and contraction are governed by the Gas Laws.
A single unit of K E at the surface cannot be in two places at once or carry out two jobs at once. KE radiated to space is gone and cannot contribute to conduction as well.
Therefore the two processes must be partitioned in such a way that the outcome matches real world observations.
Otherwise there is a breach of conservation of energy.
That is what we have achieved.
You should apologise for the confusion you are causing and for the misleading representations of our work.

Derg
Reply to  Stephen Wilde
April 27, 2021 7:14 pm

Professional obfuscators?

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 8:10 pm

Adiabatic expansion and contraction are governed by the Gas Laws.

Yes. But “Adiabatic expansion and contraction” do not show up in the rules of your model.

The only rule relates to energy partitioning. Energy partitioning guarantees conservation of energy, but doesn’t inherently relate to any other physics.

If what was going on involved herds of cattle running from one hemisphere to the other, transporting hot coals on their backs, your equations could be the same.

There is no explicit evidence in your equations that they have anything to do with “gas laws.”

Yes, your model conserves energy.

But there is no evidence in you equations of any other specific physics.

Reply to  Bob Wentworth
April 28, 2021 12:55 am

It is implicit and taken as a given via the presence of the lapse rate and the circulation in the vertical plane.
It followed the Gas Laws correctly.

Rud Istvan
April 27, 2021 3:08 pm

This is not a ‘fight’ I would have chosen to join, even tho capable. But I admire your having done so repeatedly, addressing past analytic deficiencies. This third time seems pretty conclusive.

My reasoning is simple. We do not need alternative models (no matter good or bad) to explain why consensus climate models are ‘wrong’. (Jo Nova’s husband Dave Evans attempted a similar attack using electronics feedback delay metaphors, and has possibly also floundered.)

Rather, we can attack the prevailing ‘official’ climate models directly by showing their deficiencies and failures. That suffices. No tropical troposphere hotspot=>excessive WVF from improperly parameterized Tstorm humidity washout. Unavoidable parameterization dragging in the attribution problem=>natural variation. Model ECS ~2x observational energy budget estimation methods=>Feynman dictum. And so on.

Reply to  Rud Istvan
April 27, 2021 3:21 pm

It is Bob who has repeatedly had to correct deficiencies in his analyses as he admits.
We have been consistent throughout.

Rud Istvan
Reply to  Stephen Wilde
April 27, 2021 3:36 pm

Stephen, consistent does not mean right. Mann has been consistent, also. As said above, in my opinion this whole thingy is a side show distraction not worth the price of admission, and at a detail level useless for general public persuasion. The KISS principle (taught at HBS and definitely needed for my many Fortune 500 board presentations as a very senior consultant—Keep It Simple, Stupid) has many virtues concerning the bigger picture dialogue. KISS is hard to do. I try sometimes here, as in my comment to which you responded.

Nelson
Reply to  Rud Istvan
April 27, 2021 4:00 pm

It’s not if you want to get the physics right. It might be a sideshow if all you want to do is show the faults in the climate hysteria.

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 4:53 pm

I’ve demonstrated that I’m willing to say so, when I’m wrong. I’m not wrong, this time around.

Are you willing to take a look at the way that you’ve been calculating temperatures?

Reply to  Bob Wentworth
April 27, 2021 5:35 pm

No.

commieBob
Reply to  Rud Istvan
April 27, 2021 4:57 pm

Hey Rud or Bob …

If all the energy the Earth receives from the sun were perfectly distributed, what temperature would the planet have to be at to re-radiate all that energy?

I did the calculation here and got the answer: 279K (or 6C or 43F).

Could someone please tell me if I got the basic physics and math correct.

Bob Wentworth
Reply to  commieBob
April 27, 2021 5:40 pm

It looks like you’ve got the solar constant a bit higher than what I’ve typically seen, e.g., 1367 W/m² and are failing to take into account Earth’s albedo, 0.306. Taking those into account yields 255 K (-18℃).

commieBob
Reply to  Bob Wentworth
April 27, 2021 6:13 pm

Indeed, one of my stated assumptions was that the planet is a black body.

Indeed, albedo is a can of worms. My first temptation is to try the calculation treating the planet as a grey body.

Dealing with the part of the albedo due to the clouds, could we approximate that the energy the clouds reflect back to space is matched at some ratio by energy reflected back to the planet’s surface?

Bob Wentworth
Reply to  commieBob
April 27, 2021 6:29 pm

Indeed, one of my stated assumptions was that the planet is a black body.

Well, the temperature would be higher for a black body.

But, that has no relevance to a legitimate calculation for Earth. The albedo is an indication of how much energy is reflected out into space. That lost energy is irrelevant to any legitimate calculation of Earth’s temperature. It’s as if it never existed, so why would one include it?

Dealing with the part of the albedo due to the clouds, could we approximate that the energy the clouds reflect back to space is matched at some ratio by energy reflected back to the planet’s surface?

Some typical estimates are that for every 100 units of insolation, 46 are absorbed by the surface, 23 are absorbed by the atmosphere (water vapor, dust, ozone) and 31 are reflected into space.

Any reflections of visible light back to the surface are already taken into account in those numbers.

commieBob
Reply to  Bob Wentworth
April 27, 2021 7:18 pm

The problem is that something that reflects also does not emit. If you approximate the planet as a grey body, changing the absorbtivity does not change the surface temperature because the emissivity also changes. link

Bob Wentworth
Reply to  commieBob
April 27, 2021 7:28 pm

At a given wavelength:

  • absorptivity + transmissivity + reflectivity = 1
  • absorptivity = emissivity

But what you are missing is that the values of these quantities for shortwave and longwave radiation are different.

So, the reflectivity at shortwave wavelengths, relevant to sunlight, has no relationship whatsoever to the emissivity at longwave wavelengths, relevant to thermal radiation.

Jim Gorman
Reply to  commieBob
April 28, 2021 6:47 am

Once you get into the real world there are numerous things to consider. Bob’s equation is only for an ideal situation. Even S-B requires one to use emissivity and surface area to determine the power transmitted. Basically, P = AεσT⁴. Both surface area and emissivity vary by substance. CO2 -> ε = 0.04 and H2O -> ε = 0.95. There is about 100 times more H2O than CO2. That means surface area also has a considerable difference.

whiten
Reply to  Bob Wentworth
April 27, 2021 11:54 pm

Interesting. 🙂

April 27, 2021 3:16 pm

GH Effect on Venus makes no sense either. The surface is supposed to emit ~16900 W/m^2. A lot of that is in the visible range … something that CO2 blocks poorly. We should see Venus glowing in the visible all on its own … even when its shadow faces us.

The only valid theory is mine:
http://phzoe.com/2019/12/25/why-is-venus-so-hot/

The GHG funny math ignores the actual EM spectrum.

Nelson
Reply to  Zoe Phin
April 27, 2021 3:40 pm

Zoe, I have a hard time wrapping my head around the assumption that geothermic heat plays no role in determining surface temperatures.

gbaikie
Reply to  Zoe Phin
April 27, 2021 4:04 pm

http://parkersolarprobe.jhuapl.edu/News-Center/Show-Article.php?articleID=160
That aspect of the image took the team by surprise, said Angelos Vourlidas, the WISPR project scientist from the Johns Hopkins Applied Physics Laboratory (APL) in Laurel, Maryland, who coordinated a WISPR imaging campaign with Japan’s Venus-orbiting Akatsuki mission. “WISPR is tailored and tested for visible light observations. We expected to see clouds, but the camera peered right through to the surface.”

Reply to  gbaikie
April 27, 2021 5:10 pm

That is of course interesting, but Venus is said to be ~-42C as viewed from space. If more is coming out, this would not be true. This is confusing. Note the contradiction:

“WISPR effectively captured the thermal emission of the Venusian surface”

“At Venus, the camera detected a bright rim around the edge of the planet that may be nightglow — light emitted by oxygen atoms high in the atmosphere that recombine into molecules in the nightside.”

So high in the atmosphere or the surface? Which is it?

I don’t know what to make of this.

gbaikie
Reply to  Zoe Phin
April 27, 2021 6:38 pm

I posted it cause it was interesting. And it seems a spectrum of light is radiating at much higher temperature then that -42 C.
And I also think the clouds have a part of spectrum radiating around 70 C. But Venus emits a low amount of watts.
And if imagined it was blackbody surface {and it’s not} then could equal to a blackbody radiating in vacuum at around -42 C- of course this new amount detected light, should bump it up a bit. Say couple of degrees.
Of course, if we looked more spectrum, we might/could find other parts of electromagnetic spectrum not yet detected.

Reply to  gbaikie
April 27, 2021 7:00 pm

Yeah, but this is only a limb effect. We don’t see it head-on. Who knows?

To bed B
April 27, 2021 3:23 pm

I haven’t had time, or inclination, to read through this thoroughly. It appears to be a theoretical exercise pointing that a simple rock should be much cooler than a planet that redistributes the heat from the day side to the night side. The physics of how it does it is irrelevant.

Our real planet spins with an ocean that stores heat, conveys it to cooler latitudes and warms the atmosphere over land at nigh. The Moon merely stores some heat at the surface, only cooling to 90 K after two weeks of night at the equator while only to 25K where it’s permanently dark. So the Moon is not quite like a surface of individual blackbodies and Earth is not a blackbody that is a constant temperature over it’s whole surface regardless of whether it’s uniformly illuminated or not.

The planet that is like a surface of individual blackbodies should have about 40% of the mean temperature of a real blackbody, with everything else equal. Half the sphere is 0 K, with the other half ranging from 0 to what the temperature of a one sided disk would perpendicular to the incoming light would be, and because the loss is dependent on T^4, the bigger range of surface temperatures leads to a much lower mean T.

Neither the Earth nor Moon are like this, with a difference of 10% in mean (using the Moon’s equator). It’s just the question of why start from Earth’s mean temperature is 30°C more than it should be without an atmosphere when the simple model of what it should be is probably off by tens of degrees.

Bob Wentworth
Reply to  To bed B
April 27, 2021 7:09 pm

It appears to be a theoretical exercise pointing that a simple rock should be much cooler than a planet that redistributes the heat from the day side to the night side. 

Nope, not at all what the essay is about.

The main point of the essay is that one can’t take an energy flow for convective air movement and use that to calculate the temperature of the air using the Stephan-Boltzmann equation, j* = σT⁴, as W&M routinely do in their modeling.

Nick Schroeder
April 27, 2021 3:37 pm

Yes, by George, they’ve got it!!!!!

Global Heat Balance.jpg
Reply to  Nick Schroeder
April 27, 2021 4:57 pm
Reply to  Philip Mulholland
April 28, 2021 3:07 am

Since that is not our graphic could Nick be engaging in sarcasm ?

Nick Schroeder
April 27, 2021 3:38 pm

Another graphic.

Earth Heating PPt Video 021518.jpg
Nick Schroeder
April 27, 2021 3:38 pm

And one more.

Radiation & Emissivity Explained.jpg
Nick Schroeder
April 27, 2021 3:40 pm

 “…like Venus can be much warmer than the “vacuum planet” temperature (computed by balancing absorbed insolation with thermal radiation emitted, for a planet of uniform temperature with the relevant albedo and surface emissivity).”

Venus, we are told, has an atmosphere that is almost pure carbon dioxide and an extremely high surface temperature, 750 K, and this is allegedly due to the radiative greenhouse effect, RGHE. But the only apparent defense is, “Well, WHAT else could it BE?!” (besides/also molten core volcanism)

Well, what follows is the else it could be: (Q = U * A * ΔT) aka a contiguous participating media.

Venus is 70% of the Earth’s distance to the sun, its average solar constant/irradiance is about twice as intense as that of earth, 2,602 W/m^2 as opposed to 1,361 W/m^2.

But the albedo of Venus is 0.77 compared to 0.31 for the Earth – or – Venus 601.5 W/m^2 net ASR (absorbed solar radiation) compared to Earth 943.9 W/m^2 net ASR.

The Venusian atmosphere is 250 km thick as opposed to Earth’s at 100 km. Picture how hot you would get stacking 1.5 more blankets on your bed. RGHE’s got jack to do with it, it’s all Q = U * A * ΔT.

The thermal conductivity of carbon dioxide is about half that of air, 0.0146 W/m-K as opposed to 0.0240 W/m-K so it takes twice the ΔT/m to move the same kJ from surface to ToA.

Put the higher irradiance & albedo (lower Q = lower ΔT), thickness (greater thickness increases ΔT) and conductivity (lower conductivity raises ΔT) all together: 601.5/943.9 * 250/100 * 0.0240/0.0146 = 2.61.

So, Q = U * A * ΔT suggests that the Venusian ΔT would be 2.61 times greater than that of Earth. If the surface of the Earth is 15C/288K and ToA is effectively 0K then Earth ΔT = 288K. Venus ΔT would be 2.61 * 288 K = 748.8 K surface temperature.

All explained, no need for any S-B BB LWIR RGHE hocus pocus.

Simplest explanation for the observation. 

John Tillman
Reply to  Nick Schroeder
April 27, 2021 3:49 pm

Please see discussion above of Venus’ excruciatingly low rotation rate.

gbaikie
Reply to  John Tillman
April 27, 2021 4:16 pm

The rocks of Venus rotate very slowly but the sky rotates once every 4 to 5 days.
Venus has global wind, and terminator line is suppose to make a lot of noise. Some thing like Niagara falls.

John Tillman
Reply to  gbaikie
April 27, 2021 4:19 pm

Atmosphere does super-rotate, hence the little difference between temperature on the 121.5 Earth day-long lit side and the dark hemisphere.

Reply to  Nick Schroeder
April 27, 2021 4:00 pm

Thanks Nick.
I hope others will step up to the plate in relation to what I see as a bad faith attempt at total destruction of a valid explanation for the surface temperature of planets with atmospheres.
The only Venus surface temperature used by our model is the observed surface temperature of around 747K from which our model derived a perfectly accurate emission to space temperature of -46C
Bob’s comments are wholly erroneous and frankly embarrassing for him.
Our model has predictive utility for Earth, Venus and Titan and no doubt for others too.

Gary Ashe
Reply to  Stephen Wilde
April 27, 2021 4:21 pm

They love bob over at skeptical science for his so called debunking the deneirs skills, Watts should be ashamed to have the sophist GATEKEEPING tw@t any where near here.

Reply to  Gary Ashe
April 28, 2021 12:50 am

Is he a regular there, known for his so called debunking work ?
Until this latest effort I had been assuming good faith.
Was I wrong ?

Gary Ashe
Reply to  Stephen Wilde
April 28, 2021 3:23 am

He attacks all alternative papers to the RGHE Stephen, i could list them but i wont [go to his site], yeah they love him over at john crooks gaff.

You have to remember most here are firm believers in climate change via co2 and the enhanced radiative greenhouse effect, and attack anyone going against the consensus, all they pretend to disagree with is the severity.

They dont like radical’s and anti-science climate change deniers here, radical ideas like solar intensity controlling the globes climates are frowned upon here.

This site always has been a gatekeeper site for the consensus science of co2 being the control knob of everything to do with the atmosphere, they will fiercely defend it, and always have done.

Dont forget spencer watts monckton et al have been jetting around the world and dinning out on this shiit for 2 decades now, they never wanted to win the debate and still don’t.

They just want it to carry on, the lectures the after dinner speeches the symposiums and meeting all around the world, its become a way of life, give them notoriety, they dont want to win, perish the very thought of it.

Last edited 3 months ago by Gary Ashe
Tom Abbott
Reply to  Gary Ashe
April 28, 2021 7:31 am

“You have to remember most here are firm believers in climate change via co2 and the enhanced radiative greenhouse effect, and attack anyone going against the consensus, all they pretend to disagree with is the severity.”

I’m glad you said “most”. That leaves room for me.

I believe we don’t know enough to settle on one theory or another about how the climate works.

I think all theories should be aired and let the chips fall where they may.

CO2 may be overrated.

Gary Ashe
Reply to  Tom Abbott
April 28, 2021 8:43 am

Correct that is why one of the pretend skeptics has down voted you.
Cant have anyone believing the real intensity of a 1000 joules a square meter of short wave energy being absorbed by the earths surface in the tropics and the perturbation that actually causes to the air above with the rising heat runs the atmosphere, heaven forbid such radicalism.

No no its the consensus flat earth cold sun fisher-price RGHE physic’s for these boys.

Last edited 3 months ago by Gary Ashe
Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 5:01 pm

The only Venus surface temperature used by our model is the observed surface temperature of around 747K from which our model derived a perfectly accurate emission to space temperature of -46C

An “emission to space temperature of -46C” means that the surface of the planet is at -46℃.

You derived that 747K temperature from the “thermal export” and “thermal return” energy flows in the air. A valid way to derive air temperature from those flows would have been to use the equation T = (𝚽/v – PE)/C. However, instead, you used the Stephan-Boltzmann equation, j* = σT⁴, which is not a valid way of calculating air temperature.

Reply to  Bob Wentworth
April 27, 2021 5:23 pm

For a convecting atmosphere top of atmosphere radiation out is not the same as radiation at the surface because conduction and convection in the vertical column attenuate the radiation flow from surface to space.
The point is that our model demonstrates the truth of that for Venus, Earth and Titan where the figures are broadly known.
It has the same effect as is assumed from back radiation but it is actually achieved by non radiative energy transfers so the radiative models are demonstrably wrong once one sees that it can happen with a model based on a transparent atmosphere.

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 5:54 pm

“The model has a free-flowing atmosphere of pure Nitrogen gas that connects the two hemispheres. Consequently, because the model atmosphere is fully transparent to all wavelengths, it can only gain or lose heat from the solid surface at its base.” (p. 7)  

Could you clarify: In your model of Venus, are you assuming an atmosphere free of greenhouse gases?

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 5:55 pm

You’re ignoring my main point, which has nothing to do with radiation. My main point was:

You derived that 747K temperature from the “thermal export” and “thermal return” energy flows in the air. A valid way to derive air temperature from those flows would have been to use the equation T = (𝚽/v – PE)/C. However, instead, you used the Stephan-Boltzmann equation, j* = σT⁴, which is not a valid way of calculating air temperature.

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 5:59 pm

For a convecting atmosphere top of atmosphere radiation out is not the same as radiation at the surface because conduction and convection in the vertical column attenuate the radiation flow from surface to space… it can happen with a model based on a transparent atmosphere.

Do you understand what the phrase “transparent atmosphere” means? It means that what leaves the surface is exactly the same as what reaches space.

Gases transparent to radiation have no power to “attenuate the radiation flow from surface to space” regardless of whether or not they are convecting. They do not interact with radiation. So, they cannot attenuate it.

Reply to  Bob Wentworth
April 28, 2021 12:47 am

Transparent gases take KE from the surface via conduction and convection and convert it to PE. How can the same KE simultaneously be radiated to space without breaching conservation of energy ?

Bob Wentworth
Reply to  Stephen Wilde
April 28, 2021 1:03 am

It’s a flow, in which energy travels from one place ro another.

Your question is like asking “How can a flow of water be in a river and the same amount also be emptied into the sea?” Or “How can the same rate of flow of water both be entering a pipe and exiting the pipe?”

The energy starts at the Sun, goes to the Lit surface, goes into the air, the air rises gaining PE and losing KE, circulates to the other hemisphere, descends losing PE and restoring KE, goes into the Dark surface, and then radiates to space.

You’ll measure the same flow of energy in each place, without any contradiction.

Last edited 3 months ago by Bob Wentworth
Bob Wentworth
Reply to  Bob Wentworth
April 28, 2021 1:14 am

What I meant to say is that energy in the atmosphere or radiating away is the same energy at different points along its journey.

This omits the flow from Sun to Lit surface then radiated to space from the Lit side.

Also, If you’re thinking energy not heat, there is some energy that just circulates around and around in the atmosphere.

Last edited 3 months ago by Bob Wentworth
ferd berple
Reply to  Stephen Wilde
April 27, 2021 5:57 pm

I hope others will step up
=======
What bothers me is the groupthink behind GHG theory. It blinds people to examining other alternatives such as this.

There are numerous problems with GHG theory. One of these is back radiation. Back radiation takes energy from the surface that would otherwise radiate to space and returns it back to the surface. This flux is supposedly larger than what we receive from the sun, yet we have no machines that can derive work from back radiation. This is problematic.

Now consider convection. Convection coupled with conduction (C-C) also takes energy from the surface that would otherwise radiate to space and returns it back to the the surface. In the process work is done, with the sun as the power source. Windmills are but one example of a machine that can derive work from C-C. No matter how many IR reflectors you put out at night, you cannot heat your house with back radiation.

While this C-C process routinely moves energy from hot surfaces to cold surfaces, the alternating expansion and compression of air during convection is almost identical to mechanical heat-pumps. The Lapse Rate is derived from convection in a gravitational field, so in theory there should be no difficulty in C-C warming the surface by pumping energy from the upper troposphere back to the surface. More than a few times I’ve noted on WUWT that this heat pump could provide about 33C warming.

There are many other problems with GHG Theory. The hot-spot, CO2 lags warming, UHI effect, repeated cooling of the past via adjustments, experimenter expectation bias, LIA, MWP, models running hot, etc.

Assuming that GHG Theory is correct and then arguing that it isn’t a problem strikes me as highly problematic on both scientific and economic grounds. There is much too little scientific evidence for anyone to concluded that CO2 is warming the surface and by extension much too little evidence that GHG Theory is correct.

“It doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are. If it doesn’t agree with experiment, it’s wrong.” ― Richard P. Feynman

Izaak Walton
Reply to  ferd berple
April 27, 2021 6:34 pm

Ferd,
You can’t extract useful work from the back radiation because it has a high entropy /low temperature. In space you could take the emitted radiation and use it to make a heat engine since the temperature of space is about 3K and the temperature of the IR emitted by the earth is about 260k giving you a large temperature difference. On earth the temperature difference is much smaller so any work produced is correspondingly smaller as well.

Jim Gorman
Reply to  Izaak Walton
April 28, 2021 8:18 am

I am sorry. Back radiation is measured in watts, I.e. power. It is an EM wave carrying a given power at a specific frequency. Whatever absorbs it doesn’t know what temperature or surface area it was radiated from. It is simply energy. Your explanation of entropy & low temperature makes no sense when it comes to deriving work from an EM wave. If you want to argue that there is little power to be derived, feel free, but know that would upset your CAGW brethren.

Izaak Walton
Reply to  Jim Gorman
April 28, 2021 12:37 pm

Hi Jim,
Firstly back radiation does not have a specific frequency but rather a range centred at roughly 10 microns (mixing up frequency and wavelength but that is irrelevant). Secondly at a specific frequency the intensity of a EM wave is measured in Watts/m^2. You can then if you like integrate over the surface of the earth to find to total power (i.e. Watts).

If you want to get useful work out of the back radiation then you need to make a heat engine and the laws of thermodynamics tells you that the efficiency and the amount of useful work depends on the ratio of the initial and final temperatures. So you need a reservoir that is significantly colder than the effective temperature of the back radiation. In space that would be easy since the temperature of space is set by the cosmic microwave background at about 3K but on earth that is generally not possible.

Bob Wentworth
Reply to  ferd berple
April 27, 2021 8:47 pm

Nothing in my essay is about GHG theory.

I’m simply pointing out that M&W base their model on applying the equation j* = σT⁴ to calculating air temperatures from convective energy flows, in a way that is not valid.

Tom Abbott
Reply to  ferd berple
April 28, 2021 7:38 am

I agree with your comments, ferd.

Reply to  Nick Schroeder
April 27, 2021 5:19 pm

Where does that extra Q come from? It’s obviously from “geo”-thermal. The gradient just continues from the core based on composition, and then joins with the sun higher up.

It’s not top down as Wilde and N&K would suggest.

It’s bottom up.

Gay-Lussac Law states that Temperature -> Pressure, not vice versa.

Gravitational pressure can not create its opposite: Atmospheric Pressure, which runs in the opposite direction.

The reason Venus is hot is … Venus is hot. The reason for high atmo pressure is … Venus is hot.

It’s very obvious if you just think about it. You’re welcome, boys.

Bob Wentworth
Reply to  Nick Schroeder
April 27, 2021 6:16 pm

The equation Q = U * A * ΔT explains a temperature difference between the surface and the upper atmosphere if and only if:

  1. the primary heat conduction mechanism for heat to leave the surface is conduction; and
  2. there is a heat sink at the top of the atmosphere to radiate the heat away into space.

In the absence of greenhouse gases in the atmosphere, #2 is not true.

#1 is not true in the Earth’s atmosphere; the primary mechanism for heat to be transferred into the atmosphere is convection of sensible and latent heat. However, this heat transfer will likely obey a law roughly like Q = U * A * ΔT. (It might involve (ΔT)ᵏ for some scaling parameter k.) So, none of this really takes away from your thesis too much.

But, #2 depends on greenhouse gases being present, so that’s very important to your argument.

Fred Souder
Reply to  Bob Wentworth
April 27, 2021 8:08 pm

I was under the impression that Saturn’s atmosphere is 96% He and H, much more at the top, of course. Yet its atmosphere seems to agree with Q=UAT. Temp gets quite hot as you descend. Ditto Jupiter, but only 80% H and He. But probes on Jupiter measured earth-range temps down at around 3 atm I believe. If greenhouse gasses have anything to do with Saturn and Jupiter’s temperature profiles, surely it is trivial.

Bob Wentworth
Reply to  Fred Souder
April 27, 2021 8:35 pm

According to the first source I checked, Jupiter and Saturn have clouds of ammonia, ammonium hydrosulfide, and water. I expect that these compounds all absorb longwave radiation and act as greenhouse gases. I don’t know how large a role they play in the thermodynamics of the gas giant planets. Though, it looks like the atmospheres have distinct troposphere and stratosphere layers. On Earth, that’s the result of the sunlight absorbing ozone layer. I’m not sure what, if anything, plays that role on the gas giants.

There are two distinct mechanisms that could give rise to temperatures declining with altitude:

  1. Heat flow to a heat sink (radiating greenhouse gases) in the upper atmosphere, leading to a temperature drop related to something like Q=UAT.
  2. Convection, driven by temperature differences between the equator and the poles, leading to an adiabatic lapse rate.

I’d expect #2 independent of whether greenhouse gases are important, but I’d expect #1 only if greenhouse gases were acting as a heat sink in a significant way.

So, I think you’d see a lapse rate either way.

Where greenhouse gases make a difference, however, is in deciding on the absolute temperature of the planet. A warmer and colder planet could still have similar rates of temperature declining with altitude. The lapse rate doesn’t determine the absolute temperature. It just determines the temperature difference between lower and upper parts of the atmosphere.

Greenhouse gases help determine the altitude at which “radiative balance” happens. Without greenhouse gases, radiative balance happens on the surface (or low in the atmosphere on a planet like Jupiter). With greenhouse gases, radiative balance happens where the greenhouse gases get thin enough to be transparent to radiation, typically at a much higher altitude.

The “radiative balance” temperature might be the same in both cases. But, if the balance point is higher, everything will be warmer than it would be if the “radiative balance” point was lower.

April 27, 2021 4:33 pm

Bob,
The only thing that we set in the model is the surface temperature at 737 Kelvin.
The minus 46 C is a consequence of the model and not a control.
Air has specific heat (Cp) and that it is the Cp of the air that stores the energy. The lapse rate -g/Cp determines the physical separation between the surface temperature and the TOA temperature.

Last edited 3 months ago by Philip Mulholland
Bob Wentworth
Reply to  Philip Mulholland
April 27, 2021 5:05 pm

The minus 46 C is a consequence of the model and not a control.

I agree that -46℃ is a “consequence of the model.” I’m not objecting to that.

But, I am pointing out that -46℃ is what your model yields as the temperature of the surface of Venus. That’s what it means for the “radiation loss” to be associated with a temperature of -46℃.

If a lapse rate were working properly, the surface would be warmer than the surface, not colder.

Reply to  Bob Wentworth
April 27, 2021 5:33 pm

The surface is warmer. The TOA emission is after convection within a hugely dense atmosphere has taken its slice.
If TOA were to exceed -46C the atmosphere would end up collapsing to the surface because all its PE would leak out to space.
You pay lip service to the existence of non radiative energy transfers but then fail to allocate any energy to them.

Bob Wentworth
Reply to  Stephen Wilde
April 27, 2021 5:45 pm

Could you clarify something: When modeling Venus, are you assuming an atmosphere with no greenhouse gases?

Bob Wentworth
Reply to  Philip Mulholland
April 27, 2021 9:03 pm

Air has specific heat (Cp) and that it is the Cp of the air that stores the energy. The lapse rate -g/Cp determines the physical separation between the surface temperature and the TOA temperature.

I zipped past this point on my first pass.

I agree that there is a lapse rate, and that the surface temperature and TOA temperature will be different.

The issue is, where does that -46℃ temperature apply?

In an atmosphere with greenhouse gases, the relevant temperature would happen around the layer where the atmosphere became thin enough that the greenhouse gases became transparent to thermal radiation. That could be high in the atmosphere.

In that case, the -46℃ temperature would apply high in the atmosphere, and the surface could be warmer. But, that would mean the temperature would depend on the level of greenhouse gases, which is something you want to argue shouldn’t matter.

In an atmosphere with no greenhouse gases (which is what I thought the DAET model assumed), then only the surface can radiate, and all the radiation passes unchanged through the atmosphere. In that case, the thermal radiant loss to space must correspond to the temperature of the surface.

So, I’m a little puzzled by how you’re thinking about this issue.

Reply to  Bob Wentworth
April 27, 2021 11:07 pm

“So, I’m a little puzzled by how you’re thinking about this issue.”
Bob,
Exploration of ideas is not your strong suit is it?

Bob Wentworth
Reply to  Philip Mulholland
April 27, 2021 11:31 pm

It sounds like you’re feeling rather cranky.

I guess I can understand that. I’m sorry.

In retrospect, I wish I hadn’t even brought up the issue of the -46℃. It doesn’t matter at all, in the big scheme of things.

I’m content to drop everything except this one issue: you used the formula j* = σT⁴ to calculate atmospheric formulas from the “thermal export” and “thermal return” energy flows.

I’m telling you, as a physicist, that using this formula is definitely not a valid way to calculate a temperature from these energy flows.

Reply to  Bob Wentworth
April 28, 2021 10:57 am

Bob,
 
Observation shows us that the atmospheric structure of the Hadley cell is present on Venus, Earth, Mars and Titan. In this list we have two slow rotators, Titan and Venus and two rapid rotators, Earth and Mars.
There is also a range of surface pressures from high pressure Venus to low pressure Mars. We also have two dominant atmospheric gas types, nitrogen for both Earth and Titan and carbon dioxide for both Venus and Mars.

Meteorology shows us that the Hadley cell is the primary and perpetual mechanism by which energy is transported from the topical zone of energy surplus towards the polar zones of energy deficit. This meridional flow is a response to solar altitude and not to planetary rotation.

As a physicist and a mathematician our challenge to you is this:
Devise a meteorological model of a Hadley cell that accounts for all possible terrestrial planet scenarios including planets that are tidally locked.

Bob Wentworth
Reply to  Philip Mulholland
April 28, 2021 12:10 pm

Devise a meteorological model of a Hadley cell that accounts for all possible terrestrial planet scenarios including planets that are tidally locked.

As I imagine you know, that’s a nontrivial problem that can’t be solved analytically in closed form, and which would require numerical modeling.

Fortunately, thermodynamics allows us to draw certain conclusions about the net effect of such atmospheric circulation without needing to examine the details of the flow.

For example, we know that energy will be conserved, and heat will flow from hot to cold. We know that temperature is a consequence of energy flows, and work done (e.g., against gravity), integrated over time.

Given these fundamental principles, a lot can be determined, even without any knowledge of detailed air flows.

The arguments that I make are based on those fundamental principles.

* * *

However, none of that is needed, for me to say that applying the formula j* = σT⁴ (with j* being set to the energy flux of a convective air flow) is NOT a valid way to compute air temperature.

There are times when that formula is relevant and appropriate, and times when it is not. In the way that you are using the formula, it is not relevant or appropriate.

It’s almost as bad as using the electrical equation V = I⋅R to calculate the “voltage” across a water pipe with a water flow rate I (with a fudge factor to make the units work out). It’s the wrong equation for the job!

Even if you and Stephen were completely right in all your beliefs about atmospheric dynamics and energy flows, that gets you nowhere if, at the end of the process, you calculate temperature using an inappropriate formula.

Reply to  Bob Wentworth
April 28, 2021 3:57 pm

Even if you and Stephen were completely right in all your beliefs about atmospheric dynamics and energy flows, that gets you nowhere if, at the end of the process, you calculate temperature using an inappropriate formula.

Bob,
This is from page 205 of Kiehl, J.T. and Trenberth, K.E. 1997. Earth’s Annual Global Mean Energy Budget

For the outgoing fluxes, the surface infrared radiation of 390 W/m2 corresponds to a blackbody emission at 15°C.

So which equation are Kiehl and Trenberth using to do this calculation if not this one?

(390/5.670367E-08)^0.25 = 288 Kelvin (15 Celsius)

Bob Wentworth
Reply to  Philip Mulholland
April 28, 2021 5:59 pm

Thanks for asking.

As I said in my essay:

Calculating a temperature from an energy flow via the Stephan-Boltzmann law [i.e., j* = 𝜀σT⁴] is valid only if the energy flow refers to thermal radiation, or to a combination of energy flows that are logically known to be equal to the amount of thermal radiation. In addition, even when dealing with thermal radiation, one cannot simply add fluxes (rather than averaging them) and compute a temperature.

So, yes, Kiehl and Trenberth used the equation j* = σT⁴ and they were correct in doing—because they were applying it to the outgoing “surface infrared radiation” flux. They were applying the formula to thermal radiation emitted by the surface, to determine the temperature of the surface.

That context of “outgoing surface infrared radiation” is the one and only context in which it is appropriate to use this formula.

You applied the formula  j* = σT⁴ to the atmospheric “thermal export” energy flow and the atmospheric “thermal return” energy flow. These energy flows are not thermal radiation.

It is inappropriate, and meaningless, to apply j* = σT⁴ to atmospheric energy flows.

(As I mentioned in my essay, it would have been closer to being valid to apply a formula like T = (𝚽/v – PE)/C to calculating atmospheric temperature T from atmospheric energy flux 𝚽. But, that would require knowing the circulation velocity, v, and heat capacity, C.)

The formula j* = σT⁴ is applicable only to outgoing thermal radiation from the surface.

That’s how K&T used the formula. It’s not how you used the formula.

There is only one place in your Table 9 where it might be applicable to use the formula j* = σT⁴. But, it depends on what you think the “radiant loss to space” energy flow means:

  • If this refers to total radiation emitted by the surface, then the formula  j* = σT⁴ is applicably.
  • If this refers to total outbound radiation at TOA, and this is different than the amount emitted by the surface, then the formula is not applicable.
Reply to  Bob Wentworth
April 28, 2021 11:55 pm

“It is inappropriate, and meaningless, to apply j* = σT⁴ to atmospheric energy flows.”

Bob,
Please observe FIG. 7. The earth’s annual global mean energy budget based study. Units are W m~2. Kiehl, J.T. and Trenberth, K.E. 1997. Earth’s Annual Global Mean Energy Budget

Bob Wentworth
Reply to  Philip Mulholland
April 29, 2021 7:06 am

Please observe FIG. 7. The earth’s annual global mean energy budget based study. Units are W m~2. Kiehl, J.T. and Trenberth, K.E. 1997

Philip,

I’m very familiar with this figure. What is it that you take from looking at this figure?

It seems to me to be entirely consistent with what I’m saying.

The figure contains many different energy fluxes (measured in W/m²), including “Thermals” and “Evapo-transpiration”, which are examples of non-radiative atmospheric energy flows.

Do K&T apply the black-body radiation law, j* = σT⁴, to convert those non-radiative atmospheric energy flows into temperatures? They do not.

Do K&T apply j* = σT⁴ to the TOA “Outgoing Longwave Radiation” flux 235 W/m² to calculate a corresponding temperature? Again, they do not.

The only energy flux that K&T apply j* = σT⁴ to in order to calculate a temperature is the “Surface Radiation” flux, 390 W/m².

In other words, K&T appear to be doing exactly what I’m saying you ought to be doing: apply j* = σT⁴ to surface thermal radiation, and only surface thermal radiation.

Is there something else you wanted me to get from looking at this figure?

* * *

PS. When I said “It is inappropriate, and meaningless, to apply j* = σT⁴ to atmospheric energy flows” I was particularly referring to non-radiative energy flows involving energy transport by convectively circulating air, as are featured in your Noonworld paper.

Last edited 3 months ago by Bob Wentworth
Bob Wentworth
Reply to  Philip Mulholland
April 29, 2021 8:51 am

Please observe FIG. 7. The earth’s annual global mean energy budget based study. Units are W m~2. Kiehl, J.T. and Trenberth, K.E. 1997.

It occurs to me now that maybe you think my objection to applying the black-body radiation law j* = σT⁴ to non-radiative energy flows related to inappropriate units?

If so, then that is not what the objection is about.

The objection relates to the kind of energy flow, not to the units used to measure it.

Vegetable oil and motor oil for your engine both technically contain energy that can be measured in calories, but vegetable oil is appropriate for use in a cooking recipe, and motor oil is not.

The Stephan-Bolzmann law is a law about radiation, not a law about convection. A radiation law is inappropriate and meaningless to apply to convection.

Reply to  Bob Wentworth
April 29, 2021 9:22 am

“A radiation law is inappropriate and meaningless to apply to convection.”
So what radiative process are K&T referring to when they include Latent Heat in their diagram?

Bob Wentworth
Reply to  Philip Mulholland
April 29, 2021 9:45 am

So what radiative process are K&T referring to when they include Latent Heat in their diagram?

They are not referring to any radiative process.

The diagram is not a diagram of radiative energy flows. It is a diagram of energy flows. It includes both radiative and non-radiative energy flows. Latent Heat is a non-radiative energy flow.

K&T don’t apply the S-B radiation law to Latent Heat.

As I said, an energy flow being included in the diagram does not mean that K&T or I believe it is appropriate to apply the S-B radiation law to that energy flow.

K&T only apply the S-B radiation law to one particular energy flow, the surface thermal radiation energy flow.

Is this clarifying matters at all?

(I’m not sure what the apparent lack of shared understanding is about.)

Last edited 3 months ago by Bob Wentworth
Reply to  Bob Wentworth
April 29, 2021 2:44 pm

“The diagram is not a diagram of radiative energy flows. It is a diagram of energy flows.”
So where in the diagram is the energy flow for the descending mass?

Last edited 3 months ago by Philip Mulholland
Bob Wentworth
Reply to  Philip Mulholland
April 29, 2021 3:59 pm

So where in the diagram is the energy flow for the descending mass?

It would have been a bit more precise for me to have said, “K&T’s diagram is an energy flow diagram in which (a) for radiative flows the energy flows in both directions are shown explicitly, and (b) for non-radiative flows only the net heat flow is shown.”

Heat flows by definition one go in only one direction. Heat only flows only from warmer to cooler.

“Heat” flow is defined as the net energy flow associated with thermal energy. If energy is actually flowing in both directions, one must subtract the opposing energy flows from one another, to come up with a “net” flow in the direction of whichever directional energy flow was larger.

For thermal radiation, K&T explicitly show an upward radiant energy flow and a downward radiant energy flow.

They don’t do this for “thermals” (sensible heat) or “evapo-transpiration” (latent heat). For these flows, they only show the net heat transfer.

K&T don’t show an energy flow for the descending mass because they don’t need to. They’ve already accounted for the net effect of both the ascending and descending masses in the heat flow that they show.

When K&T say that the “thermal” flow rate is 24 W/m², they mean that this is the net energy transfer rate from the surface to the atmosphere.

But, if you looked at the flow rate associated with total energy density of the air (internal energy plus PE), perhaps there might be an upward energy flux of 1024 W/m² and a downward energy flux of 1000 W/m². We don’t know, because K&T chose to focus on the net heat transfer (1024 – 1000) = 24 W/m² flowing from the surface into the atmosphere.

* * *

In general…

Thermodynamic energy flow diagrams can be drawn in:

  1. a “heat flow perspective” (in which only net heat flow is shown) or
  2. an “energy flow” perspective (in which opposing energy flows are shown) or
  3. in a “mixed” perspective (in which for some energy transfers opposing flows are shown, and for others, only the unidirectional net flow is shown).

K&T are using a mixed perspective.

All three perspectives are legitimate ways of analyzing a system. All three perspectives ultimately lead to the same end result, in terms of temperatures. But the diagrams look different.

It’s easy to get confused if you think someone is using one perspective, when they are actually using a different perspective.

Last edited 3 months ago by Bob Wentworth
Bob Wentworth
Reply to  Philip Mulholland
April 30, 2021 1:41 pm

So where in the diagram is the energy flow for the descending mass?

I am curious if my explanation of this made sense to you?

why do you support the K&T contention that the surface insolation is divided by 4

I’m also curious if my short (or long) explanation of this made sense to you?

Martin Mason
Reply to  Bob Wentworth
April 29, 2021 9:10 am

The derivation of lapse rate doesn’t include radiative properties. GHGS’s aren’t necessary for lapse rate calculation unless they change the bulk physical properties of the atmosphere.

Nick Schroeder
April 27, 2021 4:42 pm

‘…without any need to reference the radiative effects of greenhouse gases.”
For good reason.
Radiative forcing is as a bogus as caloric, phlogiston, spontaneous generation and luminiferous ether.

K-T Handout.jpg
Nick Schroeder
April 27, 2021 4:48 pm

“The Second Law of Thermodynamics requires that heat only flows from hot to cold, not cold to hot (unless something like a heat pump is involved).”
So, downwelling “extra” W/m^2 from the GHGs is not possible.
And if there is no “extra” downwelling there can be no “extra” upwelling.
And the GHG warming loop disappears from the balance with zero effects.

WUWT Bastardi loop.jpg
Jim Gorman
Reply to  Nick Schroeder
April 28, 2021 8:53 am

It’s not that there is no W/m^2 down welling. It is that there is less W/m^2 coming down than there is going up. That means the surface will continue getting cooler, not hotter by 33 deg, i.e., 255 + 33 = 288 K

A lot of the 72 absorbed by the ‘atmosphere’ is actually directly from the sun and is near IR absorbed by H2O. That means any energy sent to earth from this is just indirect sun energy, and shouldn’t be included in “back radiation”.

By assuming a transparent atmosphere, you end up with two pairs of bodies. The sun/planet and planet/atmosphere that can be analyzed separately. This isn’t dissimilar to the earth.

Bob Wentworth
Reply to  Jim Gorman
April 30, 2021 2:00 pm

It’s not that there is no W/m^2 down welling. It is that there is less W/m^2 coming down than there is going up.

Absolutely everyone agrees that “there is less W/m^2 coming down than there is going up.”

It’s a misconception some people have, that someone is claiming there is more longwave radiation coming down than going up.

That means the surface will continue getting cooler, not hotter by 33 deg, i.e., 255 + 33 = 288 K

What it means is that radiative heat transfer away from the surface is less efficient than it would otherwise be. In a simple system, the radiative heat transfer rate goes as 𝛷 = 𝜀𝜎(T₂⁴-T₁⁴). “Back-radiation” corresponds to that -T₁⁴ term being subtracted.

Some systems are constrained such that you can’t reduce the heat transfer rate; there is heat that has to be gotten rid of in steady-state, e.g., the heat from sunlight absorbed by the surface. For such systems, instead of 𝛷 decreasing, T₂⁴ increases to keep 𝛷 constant.

That is the phenomenon that leads to surface temperature being increased by back-radiation, even though downwelling longwave radiation is, on average, less than upward longwave radiation.

A lot of the 72 absorbed by the ‘atmosphere’ is actually directly from the sun and is near IR absorbed by H2O. That means any energy sent to earth from this is just indirect sun energy, and shouldn’t be included in “back radiation”.

It doesn’t really matter whether that power is counted as “indirect sun energy” or “back radiation.” Either way, it is power that reaches the surface and affects the energy balance of the surface.

By assuming a transparent atmosphere, you end up with two pairs of bodies. The sun/planet and planet/atmosphere that can be analyzed separately.

It is rarely, if ever, valid to consider a three-body thermal system as if it were two two-body systems that could be analyzed independently.

The answers do not work out the same. The three-body analysis is correct, and the “independent systems” analysis is non-physical and incorrect.

The body in the middle is affected by both other bodies. Its temperature cannot be correctly computed without considering both influences.

Nick Schroeder
April 27, 2021 4:53 pm

Any HVAC engineer modeling a house for sizing furnace and AC could apply his model to the terrestrial system with few changes.

Rick
April 27, 2021 5:10 pm

If CO2 is such a good retainer of heat can anyone tell me why it is not used in double glazed windows instead of argon. Density has been explained to me to be the reason for using argon but isn’t CO2 even denser. Since there is much more argon in the atmosphere, why is it not determined to be a culprit in global warming.

Rory Forbes
Reply to  Rick
April 27, 2021 5:44 pm

If CO2 is such a good retainer of heat

I don’t think anyone claims that CO2 “retains heat”. The conjecture is that CO2 radiates a certain percentage of escaping IR back to Earth to be reabsorbed.

Rick
Reply to  Rory Forbes
April 27, 2021 5:57 pm

So does a baseball.

Rory Forbes
Reply to  Rick
April 27, 2021 8:56 pm

Nope. A baseball heats up as it absorbs IR.

Last edited 3 months ago by Rory Forbes
Rick
Reply to  Rory Forbes
April 27, 2021 5:57 pm

Why does that not work in a window?

Rory Forbes
Reply to  Rick
April 27, 2021 8:58 pm

Why doesn’t what “work in a window”?

Rick
Reply to  Rory Forbes
April 28, 2021 6:39 am

Why would CO2 not reradiate IR back into a room. Something argon can apparently not do if it is completely transparent to IR.

Last edited 3 months ago by Rick
Bob Wentworth
Reply to  Rick
April 28, 2021 1:05 pm

The gas layer in a double-paned window is not thick enough for CO₂ between the layers to absorb enough radiation to make any noticeable difference.

Argon has only half the heat capacity of nitrogen, which means that using argon reduces the effectiveness of convective heat transport between the panes.

Rick
Reply to  Rory Forbes
April 27, 2021 6:02 pm

You just described the presumed process by which CO2 retains heat.

Rory Forbes
Reply to  Rick
April 27, 2021 8:51 pm

You just described the presumed process by which CO2 retains heat.

Nope. It scatters IR at 15 μm. It retains no heat.

Macha
Reply to  Rory Forbes
April 28, 2021 5:13 am

15um has intensity equivalent of-80C so has as much warming as trying to get a tan from a full moon.

Bob Wentworth
Reply to  Macha
April 28, 2021 1:07 pm

15um has intensity equivalent of-80C

That’s complete nonsense, as I’ve written about elsewhere.

Rick
Reply to  Rory Forbes
April 28, 2021 6:33 am

Are you saying it does not retain heat in the atmosphere?

Bob Wentworth
Reply to  Rory Forbes
April 28, 2021 1:29 pm

It scatters IR at 15 μm.

CO₂ does not technically “scatter” IR at 15 μm, though the effect is in some ways similar.

CO₂ absorbs at 15 μm, warming the mixed gas it is a part of. CO₂ in a mixed gas also radiates IR at 15 μm, and radiates more the warmer the gas is.

It retains no heat.

“Retains heat” is a vague term, but there are two ways in which it is accurate to say that CO₂ retains heat.

  1. It retains heat within the gas, in the sense that if it absorbs more than it radiates, then it will increase the temperature of the gas.
  2. It retains heat within the larger system, in the sense that some radiation headed in the direction of space is instead directed back in the general direction from which it came, reducing the radiative heat flow to space.

The conjecture is that CO2 radiates a certain percentage of escaping IR back to Earth to be reabsorbed.

I imagine it will upset some people to read me say so, but… This is not a “conjecture”; it’s an observed fact. It’s not that people just guess that IR is radiated back to Earth, they measure it. See, for example, this report on downwelling radiation fluxes as a function of time of day and month of year, as measured in Oklahoma and Kansas. (See Figure 3)

Rory Forbes
Reply to  Bob Wentworth
April 28, 2021 4:14 pm

Naw … I’m utterly bored with hearing the same gibberish attempting to validate ideas that simply don’t work. Let me know when any predictions applying that conjecture pay off. “Climate” science is thoroughly broken. I don’t believe anyone. You all have some reason why your own version of “angels on the head of a pin” is the right one. It’s worse than Marxism/socialism true believers.

Derg
Reply to  Rory Forbes
April 27, 2021 7:20 pm

Does it radiate in coming?

Or is magically only working for outgoing?

Rory Forbes
Reply to  Derg
April 27, 2021 8:55 pm

It radiates (scatters) all IR at 15 μm wave length, from all directions and in all directions.

Granum Salis
Reply to  Rory Forbes
April 27, 2021 9:31 pm

Many would say that the increased energy of vibration in the O=C=O bond is transferred by collision to adjacent molecules, which are most probably N2 or O2, long before a photon is emitted. The electrons involved fall back to a lower energy state.
Thus, the air is warmed… and convection is enhanced…

And we all go to hell in a hand-basket.

Paul Bahlin
April 27, 2021 5:32 pm

The only way to increase the temperature of an object is to increase its energy content. The model is not increasing the energy content of the planetary system. It just moves energy around. Indeed, they even show exactly equal outflow and inflow due to convection to conserve energy.

All the internals of the mechanisms are interesting and invaluable to the understanding of weather, but irrelevant to raising net temperature.

Reply to  Paul Bahlin
April 28, 2021 12:29 am

The only way to increase the temperature of an object is to increase its energy content.

No. The only way to increase the temperature of an object is to do work on it.

Paul Bahlin
Reply to  Philip Mulholland
April 28, 2021 6:41 am

Work is how you increase the content isn’t it?

Bob Wentworth
Reply to  Paul Bahlin
April 30, 2021 2:20 pm

The only way to increase the temperature of an object is to increase its energy content. The model is not increasing the energy content of the planetary system. It just moves energy around. Indeed, they even show exactly equal outflow and inflow due to convection to conserve energy.

It’s true that increasing the temperature of an object involves increasing its internal energy content. But, it’s false that one can deduce that energy content just by looking at the “equal outflow and inflow.”

A good metaphor for understanding this is to consider a lake with a damn which allows a higher rate of water outflow when the water level of the water in the lake is higher. For any given rate of water flow into the lake, the water level will automatically adjust so that in steady-state the rate of water flowing out equals the rate of water flowing in.

  • The mass of water in the lake is analogous to the internal energy of an object.
  • The water level in the lake is analogous to the temperature of the object.

There are two ways one can change the water level in the lake:

  1. One could change the rate at which water flows into the lake.
  2. One could change the design of the dam, so that a higher water level is needed to achieve the same outgoing flow rate.

The second type of change can alter the water level (corresponding to temperature) without changing the water flow rate into the lake (corresponding to solar irradiance absorbed).

So, “internal mechanisms” are, in fact, relevant to “raising net temperature.”

* * *

That said, the mechanisms proposed by M&W in their DAET model are not the type of “internal mechanisms” that are capable of “raising net temperature” sufficiently to explain observations.

The DAET model obscures this by calculating “temperature” in an illegitimate way.

JCM
April 27, 2021 5:51 pm

Mr Wentworth, I appreciate the post and the opportunity to give it another look. It is certainly providing a platform for W&M. It gives us each a reason to take another look at the science from a new point of view and to think critically.

At the core, the noonworld suggests a gamma term using empirical methods. They have demonstrated this works from both a diabatic and adiabatic point of view. At the core, the discussion demonstrates that different perspectives can lead to a similar result.

A number of your conclusions appear to be based on the principle that the ideas do not “match what one would expect to see”. “Surprising” or “peculiar” idea are dismissed based on a thought experiment. The thought experiments are expressed using simple notation of the relationships with symbols. I have no doubt that the equations work out and that this leads you to believe that your internal logical framework provides a convincing argument that “their model doesn’t have much to tell us about real planets.” However, I would be cautious of such a leap. W&M offer the opportunity to think with a new perspective. I do not see the motive to dismiss it so harshly without providing evidence. I do not know if W&M are right, that’s beside the point, I do not have evidence to prove them wrong. Nor do you.

My training is in geomatics and I suspect W&M are in the geoscience field of some sort. You will find in the “geo” sphere our training is almost centrally focused on observation and empirical analysis. These modes are equally reliant on mathematical notation as physics. I would be cautious about conflating your preference to use simple physics notation with others’ perceptions on the value and use of math. Additionally, you will find that those in the “geo” sphere will almost never been convinced by a thought experiment. We are solely driven by observation. The observational empirical approach might be different than the approach of someone who can use physics notation and thought experiments to reach conclusions, but you will find that geoscientists will rarely be convinced by this method. We all know that in the field things are ALWAYS different than what we think.

Last edited 3 months ago by JCM
JCM
Reply to  JCM
April 27, 2021 6:40 pm

The discussion above by the OP has degraded to arguing semantics and trivialities in some attempt to discredit and slander W&M; for this reason I will not be engaging in any more comments. Mr Wentworth may want to consider withdrawing the article.

Last edited 3 months ago by JCM
Alexy Scherbakoff
Reply to  JCM
April 27, 2021 10:19 pm

I worked with a person who was a PhD in mathematics. He was useless in sales and was put to work in the warehouse. He was found opening sterile packs of 100 catheters to make sure there were, in fact, 100, thereby rendering the product useless. For some reason, this amusing anecdote comes back to me when I’m reading Bob’s post.

Bob Wentworth
Reply to  JCM
April 27, 2021 7:22 pm

The only important section in my essay is the one titled ‘Non-Physical “Temperature” Calculations’

Yes, there are other sections were I identify some things as “surprising.” Those are not conclusions, just comments.

The essential conclusions I draw are all related to the section ‘Non-Physical “Temperature” Calculations’

It’s as if M&W relied on math that assumed 2+2 = 73.

Reply to  JCM
April 27, 2021 11:10 pm

“The observational empirical approach might be different than the approach of someone who can use physics notation and thought experiments to reach conclusions, but you will find that geoscientists will rarely be convinced by this method. We all know that in the field things are ALWAYS different than what we think.”
JCM,
Thank you. That is exactly our position.

Bob Wentworth
Reply to  Philip Mulholland
April 30, 2021 2:49 pm

We all know that in the field things are ALWAYS different than what we think.

Stephen’s entire argument seems to be that reality must match the way he thinks it should work.

So, it seems oddly inconsistent to claim that “things are ALWAYS different than what we think” and yet rely entirely on what Stephen thinks. I’m not aware that there is any empirical support for his assertions.

The discussion in these comments is not a debate between theory and empirical observations.

It’s about a conflict between Stephen’s unverified personal theories and the 200-year-old, extremely well-tested theory of thermodynamics that is used and confirmed daily in the fields of physics, chemistry, and engineering, among other disciplines.

I know and trust that the two of you want to get to the truth about how reality works. I want you to be successful in your quest.

I think the way to do that is by exploring the “gray areas” where things are not yet known, rather than by doing battle with what is already known with certainty.

Maybe you’ve got valuable insights to offer. But those insights won’t get any traction in the larger world if they’re provably wrong, because you’ve ignored known physics.

I want your efforts to mean something.

Last edited 3 months ago by Bob Wentworth
Keith Rowe
April 27, 2021 6:00 pm

My take of Venus is that it hasn’t cooled. The slow rotation and massive blanket of CO2 without the water has been a very large comfy blanket. The Earth cooled rapidly, Venus hasn’t. The temperature of Venus atmosphere is almost uniform on both the dark side and the light side despite it’s painfully slow rotation says a lot to me.

ferd berple
April 27, 2021 6:18 pm

Without convection, the atmosphere on a planet with a 100% transparent, non GHG atmosphere is isothermal. All the same temperature.

This is counter-intuitive. One would expect the solar energy reaching the surface would heat the atmosphere the most, making the atmosphere at the surface the warmest.

However, this is not what happens. Conduction warms the molecules nearest the surface, and those molecules with the greatest kinetic energy (KE) rebound the fastest from the surface. Because of gravity, the atmosphere is less dense in an upward direction, and this preferentially sorts the most energetic molecules upwards.

As this sorting takes place, the energetic molecules gain potential energy (PE), at the expense of giving up kinetic energy (KE). As the molecules give up KE they reach the point where they are the same KE as their neighbors, where they tend to remain. If they move lower in the atmosphere they give up PE and increase in KE, moving them upwards. If they move upwards they gain PE at the expense of KE, and fall back.

Eventually this results in all molecules having the same KE, but greater PE with altitude. However, since temperature is determined by KE, even though the molecules have different total energy, they all have the same temperature.

This point to a huge problem in using energy as a proxy for temperature, because you need to determine how much is kinetic energy and how much is potential energy.

A second problem may well be the conservation of momentum. Like energy, momentum is also conserved, even more strongly than energy. Energy need not be conserved in inelastic collisions, but momentum must always be conserved. Energy balance models appear to simply ignore momentum

Bob Wentworth
Reply to  ferd berple
April 27, 2021 9:23 pm

Without convection, the atmosphere on a planet with a 100% transparent, non GHG atmosphere is isothermal… even though the molecules have different total energy, they all have the same temperature.

I agree with this narrative.

This point to a huge problem in using energy as a proxy for temperature, because you need to determine how much is kinetic energy and how much is potential energy.

Who do you think uses energy as a proxy for temperature?

For what it’s worth, I think M&W use energy flux as a proxy for temperature. The potential energy issue isn’t a problem for them, because one can just look at the near-surface temperatures, and define that to be the level at which PE is taken to be zero. However, the problem is that energy flux cares about the velocity of the air, and that’s not something M&W have any information about.

Energy need not be conserved in inelastic collisions,

Energy is strictly conserved. In inelastic macroscopic collisions, some kinetic energy is converted to heat. But, if you account for heat, energy is conserved. For collisions at a molecular level, the only way they can be inelastic is if there is a chemical reaction (which doesn’t happen at a thermodynamically significant rate in the atmosphere).

momentum must always be conserved. Energy balance models appear to simply ignore momentum

Yes, momentum is conserved… though, once air is interacting with the Earth’s surface, momentum conservation can be trickier to track.

Kinetic energy and momentum are related by E = (1/2)p²/M. So, to some extent momentum is implicitly accounted for in energy balance equations. But, other than that, I’m not clear on why you think there is anything wrong with energy balance models ignoring momentum.

ferd berple
April 27, 2021 6:30 pm

Without convection, the atmosphere on a planet with a 100% transparent, non GHG atmosphere is isothermal. All the same temperature.
=================
continued:

Once you add convection, you no longer have an isothermal atmosphere. The work done via alternating compression and expansion of the atmosphere pumps heat from the upper atmosphere back to the surface, warming the surface above what it would be if the atmosphere was isothermal.

This is not a perpetual motion machine. The energy for this process is supplied by the sun. On earth it is likely that water is a large part of what drives convection, because the H2O molecule is quite a bit lighter than N2 or O2. (20 vs 28 vs 32).

As water evaporates it will try and move upwards until the lapse rate causes it to condense. However, the mass of rising water vapor below can continue to force the now condensed water upwards to great heights, gaining potential energy as it rises.

However, once the heat source that is evaporating the water is removed (the sun sets) there is nothing to hold the water aloft. It will fall to earth. As it does, the potential energy is now converted back to kinetic energy, pumping energy from the upper atmosphere back to the surface.

This is but one way in which potential energy from the upper atmosphere can be converted to kinetic energy at the surface by convection, warming the surface.

ferd berple
Reply to  ferd berple
April 27, 2021 8:12 pm

correction: H2O molecule is quite a bit lighter than N2 or O2. (18 vs 28 vs 32).

Bob Wentworth
Reply to  ferd berple
April 27, 2021 9:43 pm

Once you add convection, you no longer have an isothermal atmosphere. The work done via alternating compression and expansion of the atmosphere pumps heat from the upper atmosphere back to the surface, warming the surface above what it would be if the atmosphere was isothermal.

Thermodynamically, convection is a heat transport mechanism. As such, all it can do is transport heat from warmer (the surface) to cooler (heat sinks in the atmosphere created by greenhouse gases radiating, or places on the surface where the temperature is colder).

This is not a perpetual motion machine. The energy for this process is supplied by the sun.

Convection isn’t a heat pump. It’s a heat transfer process.

Heat engines and heat engines don’t exist in an open atmosphere. You need things like expanding and contracting chambers, separating heat reservoirs, and moving mechanical energy, to construct these sort of mechanisms.

You don’t have that in the atmosphere. You’ve just just basic thermodynamics.

However, once the heat source that is evaporating the water is removed (the sun sets) there is nothing to hold the water aloft. It will fall to earth. As it does, the potential energy is now converted back to kinetic energy, pumping energy from the upper atmosphere back to the surface.

This is but one way in which potential energy from the upper atmosphere can be converted to kinetic energy at the surface by convection, warming the surface.

I feel a bit puzzled by the focus on potential energy, insofar as the effect of potential energy always nets out to zero, as far as the surface is concerned (assuming a level surface). Water goes up, water comes down. It has the same potential energy at the beginning and end of the process.

Depending on how you think about it, it’s possible for some energy that goes into the atmosphere to come back. But, thermodynamically, the net effect is always more energy going into the atmosphere than coming back.

The energy balance equations always consider the net energy transfer from convection, and it always (as it thermodynamically must) involves heat flowing from warmer (typically the surface) to colder (the atmosphere or other places on the surface).

Any sort of net warming effect on a given place, just from air and water going up and down, is contrary to the Second Law of Thermodynamics.

Bob Wentworth
Reply to  Bob Wentworth
April 27, 2021 10:43 pm

Actually, I need to amend that. A convective cell can be considered a heat engine, in that it generates kinetic energy.

But, I’m very skeptical of the atmosphere’s ability to move heat from cold to hot. And, even if it could, this would only be sustainable if there were a heat source in the atmosphere. At most, the atmosphere could take the energy of sunlight absorbed in the atmosphere and transport it to the ground. But, that’s already accounted for in the calculations that say the Earth is 33 K warmer than expected, in the absence of a greenhouse effect.

To account for planetary warming, an effect needs to make heat transport to space less efficient than it would be in the case without an atmosphere. In other words, it needs to make heat loss less efficient than it would be in the pure radiative case. As far as I can tell teh only ways of doing that are to (1) block radiation from leaving (as greenhouse gases do), or (2) supply additional heat (which can’t be done without having an additional heat source somewhere; a heat pump alone wouldn’t do it).

Reply to  Bob Wentworth
April 28, 2021 3:26 am

A convective cell does NOT generate kinetic energy. It converts energy to and fro between thermal KE and non thermal PE.
That process takes time which causes energy retention within the system (radiation to space becomes less efficient which you admit is what is required) and that extra energy shows up as KE at the surface which is then recycled mechanically up and down for as long as the sun keeps shining.
I’ve lost count of the number of times and the number of ways I have directed you to the mechanics of the process but, every time, you ignore the point and go off on a tangential wild goose chase.
You have previously accepted that the atmosphere needs to make non- radiative energy transfer adjustments to retain hydrostatic equilibrium in the face of radiative imbalances.
We have told you how those adjustments work but you cannot get your head around it.

Bob Wentworth
Reply to  Stephen Wilde
April 28, 2021 5:28 am

A convective cell does NOT generate kinetic energy.

You seem to be forgetting that the term “kinetic energy” is usually used to refer to bulk movement of mass. Convection creates vertical air currents, and also creates horizontal high- and low-altitude winds, as part of the circulatory cells (e.g., Hadley cells) that it creates.

Air currents are bulk motions of air, and as such, they involve kinetic energy; (1/2)ρv² is the kinetic energy per unit volume of air, where v is the speed of the air current.

So, of course convection leads to kinetic energy.

(And convection acts a heat engine, doing mechanical work to keep air moving despite the existence of dissipative forces in the air trying to slow down that air motion. This isn’t my personal theory; see the reference. This may or may not be relevant to the discussion you and I are having.)

It converts energy to and fro between thermal KE and non thermal PE.

It does that too, although “thermal KE” is conventionally called “internal energy” and isn’t only KE, strictly speaking, since it also includes things like vibrational, rotational, and electronic energy.

That process takes time which causes energy retention within the system (radiation to space becomes less efficient which you admit is what is required)

Yes, it takes time for air and its associated energy density to travel from one place to another. Yes, energy is in some sense “retained” in the air for a while.

No, this does not have any effect whatsoever on radiation.

Retention does not necessarily affect flow rate.

Consider a river with a mass flow of 10⁵ kg/s of water. What happens to the flow if you build a dam and create a reservoir? The flow of water will be slowed until the reservoir fills up. But after it is full, in steady state, 10⁵ kg/s will flow into the reservoir, and 10⁵ kg/s will flow out of the reservoir. The flow rate will be unchanged, even though the “retention time” of the water has been increased, and water has been “delayed.” It makes no difference to the net rate of water flow through the system.

What you are talking about is exactly like that. Yes, convection takes time and temporarily “retains” energy. But, it makes no difference to the net rate of energy flow.

Consequently, it has no effect on the balance of energy flows, and no effect on the flow of energy that ultimately gets radiated.

that extra energy shows up as KE at the surface which is then recycled mechanically up and down for as long as the sun keeps shining.

There is no “extra” energy. The rate of energy going into the convection process and the rate of energy coming out of the convection process are identical, because energy is conserved. It’s just like the rate of water flow into and out of a reservoir being identical, in steady state.

I’ve lost count of the number of times and the number of ways I have directed you to the mechanics of the process but, every time, you ignore the point and go off on a tangential wild goose chase.

I’m sure it’s frustrating.

I understand the process you are talking about, in terms of internal energy being converted to potential energy and back again. You don’t need to keep repeating it as if I don’t understand it, or don’t know that it happens. I just don’t agree with you about what it means, that it happens.

It’ frustrating to me too, that I keep explaining to you that the process does not have the implications that you believe it to have, and you have yet to understand this.

Can you see that putting a reservoir along the course of a river does not necessarily change the rate at which water flows, downriver from the reservoir?

You have previously accepted that the atmosphere needs to make non- radiative energy transfer adjustments to retain hydrostatic equilibrium in the face of radiative imbalances.

We have told you how those adjustments work but you cannot get your head around it.

From my perspective, it’s you who “can’t get your head around it” that those adjustments do not have the implications that you believe them to have.

Last edited 3 months ago by Bob Wentworth
Reply to  Bob Wentworth
April 28, 2021 6:42 am

The rate of energy flow only reduces during the formation of an atmosphere. After that the original flow rate resumes but you have an accumulated pool of energy.
Thereafter, any flow variations from other causes can be neutralised by adjusting the amount in the reservoir.
When radiative gases seek to destabilise hydrostatic equilibrium then convection adjusts the amount of energy in the atmosphere’s pool of PE in order to neutralise any thermal effect.
When radiative gases slow down radiation to space by radiating back to the ground they affect the lapse rate slopes so as to encourage faster upward convection which sends the potential ‘surplus’ of energy back to the surface sooner under falling air from where it is radiated out to space earlier.
So GHGs may slow down radiation to space but faster convective overturning speeds up to eject it to space faster for a zero net effect but a miniscule change in the circulation pattern.
The water reservoir analogy does not involve a similar process because the Gas Laws do not apply to a liquid.

Bob Wentworth
Reply to  Stephen Wilde
April 28, 2021 7:18 am

The water reservoir analogy does not involve a similar process because the Gas Laws do not apply to a liquid.

Both energy (in convection) and mass (in a river) are conserved quantities; the analogy is relevant to understanding that retention and flow rate are different things, for a conserved quantity.

Thereafter, any flow variations from other causes can be neutralised by adjusting the amount in the reservoir.

Fluctuations might get buffered, but any ongoing changes to the rate of flow into a system will be matched in steady-state by an equal flow out of the system.

If more energy flows into or out of a system, e.g., due to radiation, then this cannot be “neutralized” in steady state. Such “neutralization” would require energy to not be conserved.

GHGs may slow down radiation to space but faster convective overturning speeds up to eject it to space faster for a zero net effect but a miniscule change in the circulation pattern.

Verbal logic does not produce reliable answers.

As I see the math, it’s impossible for convection to put a dent in the way that GHGs slow down radiation to space, no matter how much convection speeds up.

(The energy balance equations don’t permit convection to reduce back-radiation—they can only end up increasing it. The more convection brings heat into the atmosphere, the more the atmosphere will generate back-radiation.)

Reply to  Bob Wentworth
April 28, 2021 1:49 am

Your overall argument may be correct but the way you put it in this comment is not.

Particularly with respect to air density/volume relationships* which are the specific product of gravitational mass and temperature. 

Differential heating alone – produced by surface colour differences (albedo) – provide the necessary gradient for a heat exchanging engine.

You must know that warm parcels of air – particularly those warmed by ‘sensible heat flux’ at the surface, will “pop-off” as they become more buoyant than their surroundings – due to the kinetic energy of the molecules that comprise that bubble. Less dense air rushes in from the sides at ground level accompanied by associated downdrafts in the adjacent atmosphere. 

However, the air in the bubble itself does not exchange heat with the surroundings directly and therefore follows the Adiabatic Lapse Rate (ALR) constant. The surrounding air is actually governed by the Environmental Lapse Rate (ELR) which is based on the real measurement of the temperature of the surrounding air. If the ELR is greater than the ALR, rising air will be warmer than the surrounding air and therefore keep rising; the atmosphere is then said to be unstable.

Perhaps you haven’t flown a glider lifted to altitude by a ‘thermal’, your flying machine gaining ‘energy’ (GPE) from the useful work of that ‘heat engine!’ 😉

*Boyle’s Law

Reply to  Scott Bennett
April 28, 2021 1:58 am

Whoops!

Less dense air rushes in from the sides at ground level accompanied by associated downdrafts in the adjacent atmosphere. 

Bob Wentworth
Reply to  Scott Bennett
April 28, 2021 5:36 am

By the time you posted your message I had already corrected myself to say that “A convective cell can be considered a heat engine.” It can and does do work which, among other things, allows a glider or hang glider to soar higher and higher.

So, yes, you’re right.

Ferdberple
Reply to  Bob Wentworth
April 28, 2021 9:13 am

Heat engines and heat engines don’t exist in an open atmosphere
====!==!!
The Earth does not have an open atmosphere. It is bound by gravity and the planet’s surface.

April 27, 2021 9:11 pm

What happened to W&M, who are G&W and M&W? 😉

David Blenkinsop
April 27, 2021 9:28 pm

I must say, I find ii a bit hard to picture idealized atmospheres that are assumed to have zero IR properties say, yet it appears both the head poster here and the writers he’s critiquing are all keen on getting this right, anyway. General principles are brought in to help with this, as in the following quote from the head poster:

The Second Law of Thermodynamics requires that heat only flows from hot to cold, not cold to hot (unless something like a heat pump is involved).”

Just setting Venus aside, and thinking about planet Earth, we do have such a thing as a rainstorm here, right? As raindrops form, presumably the release of condensation energy makes them warmer than they might be otherwise. Then the raindrops, being heavy, fall *downward* taking whatever heat they contain down with them, regardless of whether the ground happens to be colder or warmer.

Do rainstorms violate the Second Law?

whiten
Reply to  David Blenkinsop
April 28, 2021 1:26 am

Does Gravity bend light?!!!

Or as the famous Tesla’s
answer to the question of “what’s electricity”.

Laws happen to be simply for the purpose of being broken… especially and as usually by the brightest of amongst us.
😏

Empirically proven by Mann, Cuomo and Dalai Lama too… among many many others of the same or similar stature;
🤟
forevaah…

cheers

Antero Ollila
April 27, 2021 9:55 pm

It is useless to investigate a model which denies the basic physical facts like GH gases and their ability to absorb infrared radiation. This is nonsense.

David Blenkinsop
Reply to  Antero Ollila
April 28, 2021 9:50 am

In the current discussion, it certainly is easy to get the impression that key heat transfer mechanisms might be glossed over in the interest of making a kind of ideal or Platonic model. For instance, on Earth at least, storms and/or precipitation might be carrying significant heat flow related power or energy in a down direction. But in either the standard models or the W & M model, does something like that even matter? In standard climate theories, in particular, if there’s any emphasis on storms carrying power up and down, looping energy through the system more or less in a turbulent way, why *that* certainly doesn’t get discussed much.

JamesD
Reply to  David Blenkinsop
April 28, 2021 12:03 pm

Overall rainstorms remove heat from the Earth. Clouds radiate heat to space. This results in increased condensation until you get rain. The rain is cold and certainly has a lot less enthalpy than water vapor. The overall effect however is radiation to space.

Lit
April 27, 2021 10:27 pm

Suppose an audio speaker is playing music with an average acoustic energy flux of 1 watt/m². Applying the Stefan-Boltzmann (S-B) law to that acoustic energy flux yields a temperature of 65 K (-208℃).”

Now add the temperature^4 of the speaker, and you get a emissive power + the work done by the speaker. dU=Q+W

The first law is the first thing to formulate in an energy balance. Nobody does it, not climate scientists, not anyone. Which invalidates them all.

Macha
April 27, 2021 11:16 pm

Worth a read. Antarctic pressures more influential than Hadley cell.
https://reality348.wordpress.com/2021/04/24/wet-summers-in-australia-and-the-incidence-of-la-nina/

RoHa
April 27, 2021 11:44 pm

Touble? Right here in Noonworld?

April 28, 2021 12:04 am

Bob,

Thank you for raising such a strong interest in our work.
You really are doing a superb job.

April 28, 2021 1:22 am

Bob,
Thanks to your tutelage here I am now taking lessons in sophistry.
As a beginner I am not very good at it, but here is what I have learnt so far:
 
Truth Sayer: “The sun does not shine on to the surface of the Earth at night”

Lesson 1: Take ownership of the statement by restating it as true, but predicate your confirmation with “No” to imply doubt about the use of this statement by your opponent.
Sophist:
“No. The sun does not shine on to the surface of the Earth at night”.

Lesson 2: Alter the true statement and pretend that it is the equivalent of what you have just said.
Sophist:
The sun darks onto the surface of the Earth at night.
 
Lesson 3: State your dissemblance as if it is the truth.
Sophist:
Therefore, on average the sun delivers energy to the surface of the Earth at night.
 
Bob,
As a physicist you should fully appreciate the significance of dark light.

Reply to  Philip Mulholland
April 28, 2021 1:55 am

Bob has a history at Sceptical Science:

2021 SkS Weekly Climate Change & Global Warming News Roundup #13 (skepticalscience.com)

He is therefore an ideologically committed radiative alarmist and not an open minded professional as I had previously assumed.

However, we have had great exposure via him and he has failed to land a significant blow due to his lack of knowledge of non radiative energy transfers.

To be fair he seems to have only one reference at Sceptical Science and he has recently been bashing away at others independently at Quora.

https://www.quora.com/profile/Bob-Wentworth

Maybe he is searching for truth (but only possess a radiative perspective) since the others do have flaws having failed to uncover the actual convective overturning mechanism we have now put forward.

Last edited 3 months ago by Stephen Wilde
Andrew Krause
Reply to  Stephen Wilde
April 28, 2021 5:12 am

Why the personal attacks?

Bob Wentworth
Reply to  Stephen Wilde
April 28, 2021 6:02 am

I am not interested in asserting any position about climate change. I get frustrated with bad science. I don’t think it serves anybody’s interests.

I’m not interested in advocating for a “radiative perspective”; I’m interested in advocating for a holistic, rigorous scientific perspective. Radiative physics is part of reality. It doesn’t serve anyone well to deny or promote misunderstandings of that aspect of reality.

I’m open to non-radiative phenomena being important in climate. But, when people say untrue things about radiative phenomena, that bugs me, and sets of alarms that the person I’m talking to is likely to be distorting the science.

I want correct physics. I’m not wedded to any agenda about where that should lead.

Bob Wentworth
Reply to  Philip Mulholland
April 28, 2021 5:49 am

Thanks to your tutelage here I am now taking lessons in sophistry.

I am doing my best to be an honest reporter of how I am seeing things.

I certainly don’t knowingly distort anything. If I have unknowingly distorted something, I’d appreciate learning about it. Could you give me a concrete example of a place where you believe I’ve made a false argument?

I did offer you and Stephen a chance to review my essay prior to publication to correct any errors or misrepresentations you saw, but I didn’t hear back from you.

Reply to  Bob Wentworth
April 28, 2021 6:21 am

I never saw that offer.

Bob Wentworth
Reply to  Stephen Wilde
April 28, 2021 6:32 am

Hmmm. Well…

5 days ago, in a comment I told Philip about a concern and a “serious error” I had noticed. I later decided the “concern” was misplaced, but the “error” was one of the ones that made it into the current essay. I didn’t get any response. Maybe neither of you saw the comment?

Then, some time after that, I managed to get a ResearchGate account and sent both of you an advance copy of a draft of this essay, inviting corrections. My impression was that ResearchGate usually sends out email notification of a message being received. I didn’t hear back.

If you weren’t notified, that’s frustrating. I intended to do what I could to try to avoid any unnecessary misunderstandings.

Last edited 3 months ago by Bob Wentworth
Reply to  Bob Wentworth
April 28, 2021 9:40 am

“If you weren’t notified, that’s frustrating. I intended to do what I could to try to avoid any unnecessary misunderstandings.”

Bob,

Neither Stephen nor I have received any communication from you via Research Gate.

Bob Wentworth
Reply to  Philip Mulholland
April 28, 2021 11:12 am

That’s odd. I just went to Research Gate and verified the messages to each of you in my “Sent” messages list.

Double-chacking… darn. I was completely aware that there was more than one Philip Mulholland and more than one Stephen Wilde on Research Gate, and I’m sure I would have paid attention to which one I was sending to. Yet, in each case, the message was apparently sent to the wrong person.

Very frustrating.

Reply to  Bob Wentworth
April 29, 2021 8:05 am

I never knew that either.
Such common names.

Richard M
April 28, 2021 3:01 am

The problem as I see is that M&W confuse the initial conditions with the equilibrium condition. What happens initially is the gases over areas that receive more energy on the lit side energize more and rise higher. Gravity then starts to move these particles outward from the center of the lit side. You do get movement of molecules (aka convection). However, over time the molecules spread out horizontally in the atmosphere. The atmosphere itself realigns. After the initial realignment there is no convection.

You have PV = NRT in action.

This allows the creation of a thermal gradient from the center of the lit side outward. There is no movement of molecules (convection) back across the thermal gradient. You end up with more molecules of atmosphere the further you move away from the center of the lit side. What you do get is constant conduction from the center of the lit side outward. You don’t get any convection because gravity forces the height of the atmosphere to be constant.

Because the planet is a sphere the area increases as you move outward from the center of the lit side. This means the total energy increases but not the energy per unit area. So, the temperature gets cooler. This gradient goes all the way around to the center of the dark side. How this affects the density is still a little fuzzy to me because of this increase in total area. You have more energy but it is spread out over a larger area.

Because some energy gets continually conducted to the dark side, the lit side will absorb more radiation than it radiates.

Reply to  Richard M
April 28, 2021 3:13 am

The initial condition is no atmosphere. Then the atmosphere lifts of the surface and you get a vast low pressure cell on the lit side with a vast high pressure cell on the unlit side and a full convective overturning circulation linking the two.
It never realigns to zero convection.
Then , our model describes the equilibrium condition that goes on forever or until the sun stops shining.
So, no confusion.

Richard M
Reply to  Stephen Wilde
April 28, 2021 3:52 am

Stephen, sorry but it does realign. You are confusing noonworld with Earth. Because noonworld is heated on only one side this allows the realignment. If it didn’t then PV=NRT would be violated.

The expansion of the gas on the lit side never stops. If you could force the number of molecules to be constant everywhere then your view would be correct. However, there is nothing to force that to happen. Either a gas expands when heated or it doesn’t. The forcing never stops and hence the expansion only happens initially.

Essentially you have a one time advection to realign to the ideal gas law. Your view is unphysical in multiple ways as BW has pointed out. I have explained what really happens. Your low pressure cell and high pressure cell become permanent fixtures. You end up with a constant energy gradient between them driven by conduction. No movement of the atmosphere is required.

Bob Wentworth
Reply to  Richard M
April 28, 2021 6:17 am

This is a (sadly rare) case where I (now) agree with Stephen, that Noonworld would experience ongoing convection in steady-state.

I initially thought that convection wouldn’t happen in Noonworld in steady state. But, careful thought and analysis led me to the conclusion that on the hot side the change of pressure with altitude would be different than the change of pressure with altitude on the cold side, and this would inevitably lead to a pressure differential in one direction at high altitude, and an opposite pressure differential at low altitude, leading to the formation of a circulatory cell. (I originally reasoned this out in this comment.)

Last edited 3 months ago by Bob Wentworth
Richard M
Reply to  Bob Wentworth
April 28, 2021 7:20 am

I originally thought that as well. But that was before I realized the atmosphere itself would adjust to the difference. The atmosphere is not forced into a static configuration. It aligns with the energy differences because those differences never change.

This also holds at altitudes.

Bob Wentworth
Reply to  Richard M
April 28, 2021 7:26 am

Do you have a sense of some configuration with no air moving that would be stable? I couldn’t identify one.

As best I can tell, the configuration with air circulating convectively is stable. I can’t see any dynamic that would stop it.

Richard M
Reply to  Bob Wentworth
April 28, 2021 7:58 am

Yes, the configuration would have more atmospheric mass over on the dark side. The mass would move initially and then would not be required to move again.

The system attempts to find its lowest energy state and that occurs when more mass moves over to the dark side. The energy available from the sun keeps the lit side mass more active and that is what maintains the balance.

Bob Wentworth
Reply to  Richard M
April 28, 2021 11:34 am

Unfortunately, I don’t see how that could be stable.

What vertical temperature profile do you imagine being present?

Pressure changes with altitude as dP/dz = -ρ⋅g where ρ is density. But ideal gas law says P =(ρ/M)RT where . So, dP/dz = -(gM/R)⋅P/T.

If the vertical temperature profile were isothermal, this would lead to P(z) = P(0)⋅exp[(gM/RT)⋅z]. Because T is different in the two hemispheres, this would make it impossible for the pressures to match at all altitudes. Any pressure differential, at a given altitude, will lead to horizontal wind currents. Therefore, there is no stable solution with isothermal vertical temperature profiles.

Suppose the vertical temperature profile follows an adiabatic lapse rate, so some other temperature profile. The solution in this case cannot be written in simple closed form; it’s a tricky double integral. But, at every point, |dP/dz| on the cold side will be greater than |dP/dz| on the warm stide, whenever Twarm > Tcold. So, again, it follows that there will be pressure differences at some altitudes, which will again lead to horizontal air flow.

The only way pressures at two different locations can match at all altitudes is if temperatures also match at all altitudes. But, that can’t be the case between the hot and cold sides.

Therefore, there will be air circulation in steady state.

The temperature difference will cause the atmosphere to act like a heat engine, doing work to keep the atmosphere moving, despite the presence of dissipation in the fluid flow.

Wait a minute… if temperature and pressure always were in the same ratio at the same altitude, that would seem to allow pressures to be matched at all altitudes.

But, in the presence of any heat conduction at all in the air, that temperature profile could not be sustained.

So, I think the conclusion still holds.

Last edited 3 months ago by Bob Wentworth
Richard M
Reply to  Bob Wentworth
April 28, 2021 2:26 pm

You’re making this too difficult. This is basic conservation of energy. Since nothing changes on the input side the system will seek its lowest energy configuration.

Even on the lit side the energy coming in changes as you move toward the horizon. Hence, the mass changes to minimize airflow and allow a constant flow of energy. In effect, the vertical profile changes constantly as you move.

The changes will be almost undetectable as you move meter by meter but over many kilometers they start to add up.

I assume there will an ALR but it will also be constantly changing as you move. The surface temperature will also change and be slightly below the S-B value for the incoming radiation.

Everything is driven by the conduction fed kinetic energy of the molecules since no radiation occurs in the atmosphere.

Richard M
Reply to  Richard M
April 29, 2021 2:48 pm

Bob, after thinking about it some more, it is possible the lowest energy configuration will lead to some air movement. However, I doubt it will be as Stephen imagines. There are a few non-linearities in this configuration. Having a spherical shape as the heated surface and having gravity affecting the atmosphere for starters.

If this did lead to a situation where the atmosphere could not directly adjust then you could get movement of air molecules. However, this wouldn’t be convection as we think about it on Earth. It would be a constant movement of molecules possibly at all altitudes across the temperature gradient.

Or, the nonlinearities could cancel out. One would almost have to do a simulation to see.

Bob Wentworth
Reply to  Richard M
April 29, 2021 4:47 pm

I’ve thought about the situation using a 2-dimensional “toy model.” (Technically, there is a third dimension to the box, but the thermal drivers are the same along that dimension.)

Imagine there is a perfectly insulated box (high enough that pressure varies with elevation due to gravity) with an inert gas in it. There is a heat source at temperature T1 on the “floor” on the left-hand side of the box and a heat sink on the “floor” at temperature T0 , where T1 > T0, on the right-hand side of the box. Perhaps the heat source and heat sink each take up half the floor, but one could consider variations on that configuration.

Given that there is a heat differential, allowing work to be done, there is no thermodynamic law saying that the steady-state condition can’t involve ongoing air circulation.

If there is a pressure differential across any vertical surface, then there will be air movement. (Similarly, if there is a pressure differential across any horizontal surface that doesn’t match dP/dz = -ρg, then there will be air movement.)

I continue to be unable to imagine any static configuration that wouldn’t involve a pressure differential somewhere, leading to air circulation.

I’m basing my conclusions about the likely behavior of the atmosphere of Noonworld on my thinking about this toy model.

Do you predict ongoing convective circulation in this toy model?

Last edited 3 months ago by Bob Wentworth
Richard M
Reply to  Bob Wentworth
April 29, 2021 5:45 pm

Heat can flow via conduction. It is slow and steady.

Keep in mind that on noonworld the temperature of the surface and the atmosphere above it stays the same except for the slight loss of energy via conduction towards the dark side. So, the warming of the atmosphere itself is also very slow. It just replaces the conductive energy loss.

Of course, if the nonlinearities do show up then they will force some movement of the molecules to compensate. For example, if heat conducted faster at different altitudes then that portion of the vertical air column would get more dense and something would have to give. So, yes it could be the steady state would create some air movement.

But that is not a given and unless it can be demonstrated that nonlinearities exist, I would go with a stationary equilibrium.

Bob Wentworth
Reply to  Richard M
April 29, 2021 6:15 pm

Heat can flow via conduction. It is slow and steady.

Yes, but conduction has minimal importance if convective circulation is happening.

And, convective circulation will happen, unless there is an atmospheric configuration without any horizontal pressure differentials, at any altitude. That configuration also needs to be consistent with the temperature variations that could be supported by conduction.

I don’t currently believe that any such statically stable configuration is possible in my toy model. I suspect it’s not possible on Noonworld either (even without nonlinearities).

I gather you believe differently. Perhaps neither of us will convince the other, in the absence of a verifiable concrete solution to the problem.

Richard M
Reply to  Bob Wentworth
April 29, 2021 8:23 pm

Noonworld is very nicely symmetrical. I think that may be the main reason it would find a stable configuration. The entire world would follow the ideal gas law (PV=NRT). This allows for changes in pressure if they are matched by changes in one of the other factors. Hence, the world settles into a configuration where those exact changes appear.

This would not occur in your toy model. I would expect air movement in your toy model because the changes to match the IGL would be blocked by the asymmetries involved.

This entire exercise shows the problems caused by researcher bias. I could see in Stephen’s responses to my notes that he had not even considered some of the factors I mentioned. His mind settled on the initial conditions and he just assumed those would remain. I think this same mindset exists with climate researchers.

Even if there does exist some nonlinearity that would cause some air flow it wouldn’t be what Stephen thought it was. The result would not support his view..

Reply to  Richard M
April 28, 2021 6:20 am

You would have a density gradient between them and uneven surface heating that would drive convective overturning. No way to prevent it.
It is bizarre to suggest otherwise.
There would be constant expansion over the lit side and constant contraction over the unlit side.

Richard M
Reply to  Stephen Wilde
April 28, 2021 7:31 am

Stephen, where does the uneven heating of the surface occur? It’s in exactly the same place all the time. This allows the atmosphere to adjust to those differences. The atmosphere does this initially and never needs to do it again.

You are correct that there would be a density gradient but that would match the temperature gradient. As a result there would be no changes in the energy levels at both extremes and along all concentric circles in between. This allows a slow and constant conduction of energy.

Reply to  Richard M
April 28, 2021 7:42 am

Spherical geometry results in a declining heat input as one moves away from the zenith towards the horizon with zero beyond the horizon.
There would be a temperature and density gradient in the horizontal plane all the way from zenith to horizon which would cause uplift.
The uplifted air flows around at high level to the unlit side where it sinks and then it flows back again.
How could that ever be suppressed ?

Richard M
Reply to  Stephen Wilde
April 28, 2021 8:14 am

Stephen, I don’t disagree that there are gradients. The density and temperature gradients find their lowest energy levels and stay there. They find a balance by reconfiguring the atmosphere.

What you end up with is a constant energy gradient from the lit side to the dark side produced by conduction. After equilibrium is reached there is no uplifted air. It is already in its lowest energy state.

May be easier to think about the kinetic energy. As you move along the temperature gradient, from lit to dark, you always have same probabilities of movement in all directions. The density difference counters the temperature difference. This balances the energy movement of the molecules.

What normally drives convection is an effort to find a lower energy state for the overall system. Once in the lowest energy state, there won’t be any more convection.

Reply to  Richard M
April 28, 2021 10:37 am

What normally drives convection is temperature (leading to density differentials). Due to conversion of energy to and fro between KE and PE the temperature from bottom to top can never become uniform for the overall system. There will always be sufficient temperature decline with height to ensure continuing convective overturning simply because of those energy conversion exchanges.
You can have a constant gradient from one side to the other at the top and similarly a constant gradient from one side to the other at the bottom but you can’t stop the uplift and descent between the two in the vertical plane.

Richard M
Reply to  Stephen Wilde
April 28, 2021 2:39 pm

Stephen, what would cause any uplift? You have the same energy entering the atmosphere from its contact with the surface and moving in the same manner. Always. It never changes. The atmosphere already is structured to allow this energy to flow optimally.

The top and bottom, and all points in between, of the atmosphere are also structured optimally. There is no reason for convection to occur. All you have is an unchanging gradient.

Keep in mind the only contact the atmosphere has is with the surface. All you have are kinetic transfers. No part of lit side knows there is a dark side and vice versa.

Reply to  Richard M
April 29, 2021 8:01 am

Density differentials in the horizontal plane would lead to convection. How does your scenario prevent it ?

Richard M
Reply to  Stephen Wilde
April 29, 2021 3:04 pm

Stephen, it is possible that the lowest energy configuration does lead to the movement of individual molecules. However, it is most likely not convection as we are used to on Earth.

For example, if the density gradient after achieving energy minimalization was different at different altitudes then it would force some molecules to move along the gradient. This would keep molecules constantly flowing towards the dark side at higher altitudes and others constantly moving towards the lit side at lower points. Those movements would occur at all altitudes.

However, there are no bulk transfers. This is just the result of slight variations in density.

It is also possible the density gradient is constant at all altitudes and there is no ongoing movement at all.

JamesD
Reply to  Stephen Wilde
April 28, 2021 11:57 am

I would change the term “equilibrium condition” to “steady-state condition”.

Geoff Sherrington
April 28, 2021 3:49 am

Surely some uncertainty could be reduced by actual experiments, using balls rotating at different angular velocities in a vacuum , then a simulated atmosphere, with thermometers to measure any thermal differences at different angular velocities. No need to try to mimic the real atmosphere at first, just a staged approach starting from simplest and getting as complex as might be needed.
I’d do it myself, but I’m too old to go to the billiard room and play with my balls.
Geoff S

Eben
April 28, 2021 6:30 am

It is funny that now he says his last essay Atmospheric Energy Recycling was stimulated by M&W’s work , because the minute I saw it I thought to myself If gravity principles discovery were inspired by watching a falling apple, then this energy recycling by catching back its own radiation and adding it back to itself to re-radiate even more of it surely must have been inspired by watching a dog eating its own poop.

April 28, 2021 7:33 am

This seems worth a follow up:

Bob said:

“I’m content to drop everything except this one issue: you used the formula j* = σT⁴ to calculate atmospheric formulas from the “thermal export” and “thermal return” energy flows.
I’m telling you, as a physicist, that using this formula is definitely not a valid way to calculate a temperature from these energy flows.”

That usage follows K&T so if it is wrong then the consensus view held by K&T is wrong too.

However, we do it slightly differently.

In their methodology K&T specifically aim to have an average global surface radiative only flux of 390 W/m2 this then translates into a surface temperature for the whole Earth of 15 Celsius in their model.

It is divide by 4 for the entire globe involving radiation alone that is the issue that generates the junk science. The correct science is to divide by 2 for radiation onto half the globe and then introduce the non radiative energy flows that run between the two halves. Either we and K&T are both wrong or we are both right in the matter of the application of the S-B equation to convert fluxes to a final average temperature. 

The difference between us and K&T arises in our application of our model separately to a single lit hemisphere and a single dark hemisphere. and then we bring in non radiative energy transfers between them. Consequently we have two temperature surfaces to consider and not one. 

In our model we deal with both radiative and non radiative fluxes for each surface separately and we then convert those fluxes into temperature at the very end of the modelling process. Obviously, as both surfaces are hemispheres then the temperature average of both of these hemispheres will be the global average temperature.

What we did in designing the model was to ensure that it is the flux partition between radiative and non radiative processes that generates the energy balance for each hemisphere individually and only then calculate the average temperature. I think that this may be the bit that stumps Bob. He wants to calculate the average fluxes from the two surfaces before he computes the temperature whereas this step to temperature averaging is correctly done after S-B conversion in our model. It is almost as if Bob does not realise that it is cold at night, in the winter and most emphatically so at the poles.

The proper sequence is to ascertain total fluxes both radiative and non radiative for each surface separately then convert to temperature and then do the averaging to get the global temperature.

Bob seems to want to average the fluxes both radiative and non radiative for the entire globe as a single surface and then work out the average temperature of the globe.

The defect in doing that is that there is then no place for non radiative energy flows between the lit and unlit sides and back again so that leaves a huge gap which he and K&T fill with their back radiation and of course there is then no place for the neutralisation of back radiation carried out by variation of the rate of non radiative transfers between the disparate halves of the globe.

It is only by ascertaining the true nature of those non radiative convective energy flows by treating the two hemispheres separately that allows a model to eliminate the need to propose a surface heating effect from back radiation.

The difficulty Bob has is that he cannot discredit us and at the same time support K&T so he is denigrating our approach as unphysical whereas it is in fact a more realistic and inclusive approach than that of K&T.

JamesD
Reply to  Stephen Wilde
April 28, 2021 11:48 am

It would be interesting to see what the surface atmospheric temperature on the lit and unlit side were.

Bob Wentworth
Reply to  Stephen Wilde
April 28, 2021 7:23 pm

That usage follows K&T so if it is wrong then the consensus view held by K&T is wrong too.

Your usage does NOT “follow K&T.” The apply the formula j* = σT⁴. You apply it incorrectly.

It’s as if you’re saying “I followed the same cake recipe as some famous baker”–but you’ve substituted motor oil for vegetable oil in your recipe, and can’t seem to realize that you’ve invalidated the recipe!

The formula j* = σT⁴ is only correctly applied to thermal emissions from the surface. That’s what K&T did.

You applied j* = σT⁴ to atmospheric convective energy flows. This is wildly inappropriate and wrong. It’s not what K&T did.

Either we and K&T are both wrong or we are both right in the matter of the application of the S-B equation to convert fluxes to a final average temperature. 

K&T do not apply the S-B equation to convert generic fluxes — they only apply S-B to thermal radiative fluxes from the surface.

Applying j* = σT⁴ to non-radiative energy flows, as you have done, is an extremely serious error.

K&T are right and you are wrong.

Consequently we have two temperature surfaces to consider and not one. 

In our model we deal with both radiative and non radiative fluxes for each surface separately

I understand that, and agree that that makes sense.

ascertain total fluxes both radiative and non radiative… then convert to temperature

You do not use a valid or correct formula for converting non-radiative fluxes to temperature.

I think that this may be the bit that stumps Bob. He wants to calculate the average fluxes from the two surfaces before he computes the temperature whereas this step to temperature averaging is correctly done after S-B conversion in our model. 

You guessed wrong about what I want.

Yes, it is correct to average temperatures after you calculate them. But, it is not correct to apply “S-B conversion” to non-radiative fluxes.

Bob seems to want to average the fluxes both radiative and non radiative for the entire globe as a single surface and then work out the average temperature of the globe.

Nothing could possibly be further from what I want.

The only thing I want is for you to process non-radiative fluxes correctly. You can’t do that with the S-B formula.

* * *

K&T apply S-B to radiative emissions from the surface. That is valid.

You apply S-B to non-radiative energy flows in the atmosphere. That is invalid.

Reply to  Bob Wentworth
April 29, 2021 7:58 am

Bob
What significance do you think your objection has ?
Obviously the non radiative processes contain energy and that energy is in the form of radiative thermal energy both before and after the non radiative process does its bit.
What would you do instead ?

Reply to  Stephen Wilde
April 29, 2021 9:01 am

Bob
Yet again you have not interpreted our work correctly.
With regard to the equation that you object to the fact is that we never used it to directly quantify any thermal effect from a non radiative process.
All the information required is available from the radiative fluxes at either end of the non radiative process.
K&T did just that in relation to their upward values for thermals and evapo-transpiration. One assumes they used the radiative energy at the surface to quantify those non radiative processes.
Their problem, and yours, is that they never brought it down again within the descent phase which is what we have now done with our two surface lit side and unlit side model which is far more realistic than their flawed single surface model.
Once one brings it down again then ‘hey presto’ there is a surface warming effect without any need for radiative gases and their back radiation.
Only one answer is correct. Either it is back radiation or it is bulk mass energy transfer.
Since we get the outcome of surface warming with a fully transparent atmosphere the logic is irrefutable and the radiative theory collapses.
Now, about your promise to concede all issues if we could deal with your objection over the S-B formula ?

Bob Wentworth
Reply to  Stephen Wilde
April 29, 2021 10:48 am

With regard to the equation that you object to the fact is that we never used it to directly quantify any thermal effect from a non radiative process.

You specifically applied the equation j* = σT⁴ to calculating the “air temperature” associated with the atmospheric “thermal export” and “thermal return” energy flows. These are non-radiative processes.

Are you asserting that the atmospheric “thermal export” and “thermal return” energy flows are radiative processes?

All the information required is available from the radiative fluxes at either end of the non radiative process.

If you believe that, then why don’t you accept that the planetary temperature is the value computed by applying j* = σT⁴ to the “radiative fluxes”? The only radiative fluxes you name are the ones predicting a temperature of -46℃.

(And, it’s not relevant to consider the radiative fluxes at “either end”. So, for example, solar insolation is not a flux for which it is appropriate to apply the S-B law, unless you assume this equals the surface thermal radiation flux. Only the radiative fluxes for surface thermal radiation flux are relevant to apply the S-B law to.)

K&T did just that in relation to their upward values for thermals and evapo-transpiration. One assumes they used the radiative energy at the surface to quantify those non radiative processes.

K&T developed an “energy budget” in which various processes, including thermals, evapo-transpiration, and radiation, balanced.

They arrived at a specific number for surface thermal radiation flux, 390 W/m², and applied the S-B law to that, to calculate a temperature.

You did not do anything like that.

You did the equivalent of ignoring the “surface thermal radiation flux” number you calculated, and instead applying the S-B law to the thermal and evapo-transpiration flux.

That’s not how K&T did it, and it’s not how you should be doing it, if you want to calculate a legitimate result.

You also labeled your result “air temperature.” The S-B law cannot be used to calculate “air temperature” in the way you’ve done it. Applying the S-B law is straightforward only when dealing with surfaces, not with gases.

Since we get the outcome of surface warming with a fully transparent atmosphere the logic is irrefutable and the radiative theory collapses

The “outcome” is only relevant if legitimate means have been used to get to that outcome.

You have not used legitimate means.

It’s like doing a calculation that assumes 2 + 2 = 73 and claiming you’ve proven something. It the steps along the way are not valid, then the result is not valid.

Now, about your promise to concede all issues if we could deal with your objection over the S-B formula ?

I am still waiting for you to deal with my objection over the S-B formula.

You haven’t.

You’re making vague, unsubstantiated claims that the non-radiative processes used in your calculations are somehow also thermal radiation.

That’s magical thinking. It’s not logic or math or physics.

If you think the non-radiative processes in your model somehow lead to thermal radiation, then put that into your model. It’s not there, at present.

Last edited 3 months ago by Bob Wentworth
Reply to  Bob Wentworth
April 29, 2021 11:04 am

K&T allocate energy going into thermals and evapotranspiration and base the amount by reference to radiation at the surface.
They never bring it down again for the descent phase so to force their ‘balance’ they propose a heating effect from back radiation.
We have corrected that error by returning the same amount back to the surface as went up in the first place so that back radiation becomes irrelevant.

Reply to  Stephen Wilde
April 29, 2021 11:36 am

So, a non radiative process takes thermal energy from the surface as per the K&E analysis.
By including that in their graphic they impliedly accept that there is an energy partitioning effect at the surface so our partitioning approach is clearly correct.
Then they fail to return it to the surface in the descent phase.
We have not used S-B to calculate a non radiative process any more than they have.
Once one completes the overturning cycle by adding back to the surface on the unlit side the same amount of KE as was taken from the surface on the lit side the greenhouse effect then appears even in the absence of radiative gases.
So, we are not guilty of the error that you are accusing us of.
They and you do not realise that the energy taken up by thermals and evapotranspiration cannot then go to space insofar as it is converted to PE via adiabatic ascent. PE is not heat and cannot radiate away.
Before it can be released to space it has to come back to the surface again in adiabatic descent.

Reply to  Stephen Wilde
April 29, 2021 11:43 am

The thermal export and thermal return numbers are correctly calculated from the radiation budget at the base of uplift and the base of descent. In both locations the air temperatures match the surface temperature so we are dealing with surfaces and not the bulk atmosphere.
Those temperatures therefore do represent radiative fluxes, one going upwards into PE and the other coming downwards out of PE.
We have not applied S-B to the energy content of the non radiative process itself.

Bob Wentworth
Reply to  Stephen Wilde
April 29, 2021 1:40 pm

So, a non radiative process takes thermal energy from the surface as per the K&E analysis.

It’s fine that you have non-radiative processes, as do K&T.

Where you K&T differ from you is that K&T actually calculate a quantity called “Surface Radiation” and then apply the S-B black-body radiation law,  j* = σT⁴, to that.

You don’t do that.

You imply that, in some vague, unspecified way, your “thermal export” and “thermal return” values relate to thermal radiation, and use those to calculate temperature from S-B, inappropriately.

So, we are not guilty of the error that you are accusing us of.

If you are going to calculate a temperature using S-B, you need to actually calculate a value for surface thermal radiation. Either you haven’t done that at all, or it’s the value that corresponds to -46℃. Since you assure me the latter value is not the surface temperature, that means you have not calculated surface radiation at all.

Your model needs to include an equation for surface radiation, or it is not, and cannot be, valid for you to calculate temperature from the S-B law.

Maybe the physics you describe is correct. Maybe it’s not.

But, regardless, your calculation is not valid if you are not calculating surface thermal radiation and applying the S-B law to that.

You are not currently doing that. So, your calculation is not valid.

Last edited 3 months ago by Bob Wentworth
Bob Wentworth
Reply to  Stephen Wilde
April 29, 2021 1:18 pm

K&T allocate energy going into thermals and evapotranspiration and base the amount by reference to radiation at the surface.

That characterization is false for evapo-transpiration. K&T deduce their value based on estimates of global precipitation.

K&T do deduce the value going into thermals based on an energy-balance calculation. But, they also check their result against values independently calculated using a bulk heat transfer formula.

So, K&T’s convection values are not simply “made up” to balance their equations.

They never bring it down again for the descent phase so to force their ‘balance’ they propose a heating effect from back radiation.

You are misunderstanding what K&T are doing.

It would be inappropriate for K&T to “bring it down again” in the way you describe.

Energy can be tracked in an “energy flow perspective” or a “heat flow perspective.” In an energy flow perspective, you might look at both energy up and energy down. In a heat flow perspective, you consider the net effect, when energy down is subtracted from energy up, and that’s what you talk about.

K&T are using a “heat flow perspective” for talking about convection. That is a valid perspective (typical of how people apply thermodynamics to convection), and leaves nothing out.

You prefer to use an “energy flow perspective” in which you talk about “energy up” and “energy down” separately. That’s ok, but it doesn’t mean that K&T did anything wrong. They are just using a different perspective, which lumps “up” and “down” together and offers a figure for the net effect.

K&T don’t “force their balance” by proposing a heating effect from back radiation. They supply measured values for how much back radiation is observed to exist.

Last edited 3 months ago by Bob Wentworth
Reply to  Bob Wentworth
April 29, 2021 1:35 pm

However calculated, the fact is that any KE converted to PE during adiabatic uplift has to be returned to the surface during adiabatic descent.
So where do they show it ?

Bob Wentworth
Reply to  Stephen Wilde
April 29, 2021 1:58 pm

any KE …

A minor pet peeve: I wish you would stop talking about “KE” when you mean thermal “internal energy.” It’s inaccurate to refer to internal energy as KE, though this is the way science teachers sometimes talk about the subject to high school students. Internal energy includes some components of energy that are not, technically, kinetic energy.

However calculated, the fact is that any KE converted to PE during adiabatic uplift has to be returned to the surface during adiabatic descent.

So where do they show it ?

It is true that an internal energy converted to PE will be converted back to internal energy when air returns to the surface.

That portion of the energy carried by convection is balanced during uplift and descent.

But, not all the energy carried by convection is balanced in that way.

It is not true that any internal energy put into the air (then converted to PE and back again during convection) “has to be returned to the surface.”

Some of that energy is eventually transferred to the upper atmosphere and is radiated away.

There is net heat transfer from the surface to the atmosphere. This is demanded by thermodynamics, given that the atmosphere is colder than the surface.

Therefore, the energy that goes up must be greater than the energy that comes back down. Thermodynamics demands this.

The “Thermal” flux that K&T report is the net heat flow upward, related to the difference in internal energy of the air when it starts to uplift and the amount of internal energy in the air when it returns.

There is a difference because not all the processes involved are adiabatic. The expansion and contraction are adiabatic. However, condensation of water vapor leads to some heating of the air, and radiation leads to some net cooling of the air.

There is net heat transfer from the surface to the atmosphere. The total energy flow is not precisely balanced. The numbers K&T report reflect that imbalance in energy flows.

Reply to  Bob Wentworth
April 29, 2021 2:09 pm

I don’t dispute that back radiation exists and can be measu