A WUWT “Comment Rebuke”

Guest post by Rud Istvan

A few days ago, I posted another Lindzen Bode ECS reconciliation. It should have been controversial, stimulating many comments because of the divergence to higher climate models and also to Monckton’s often here posted much lower estimates. It was only mildly so, most about my penetration f/(1-f) versus Bode 1/(1-f) one phrase in one sentence goof, which did not affect the post’s conclusions since it only used Lindzen’s correct curve posted here decade ago.

Recently, Dr. Wentworth posted a ‘mathematical proof ‘ that the GHE must exist, even though misnamed (because real greenhouses retard local convective cooling, while greenhouse gases retard radiative cooling to space). This misnomer is no different than the equivalent ocean acidification misnomer we are also forced to live with in popular discourse. Warmunists long ago picked their definitional terms of debate, and WUWT skeptics are mostly stuck with them.  Definitional quibbling may satisfy some, but probably isn’t an effective tactic.

I was very surprised at the number of negative comments at WUWT (now well over 600) to this rigorous post with an obviously verifiable conclusion. They spanned the gamut from epistemological (really a proof, or something else?), to the old Venus/Mars ‘analogies’, to the new ignores convection (true, but convention only moves heat around in the atmosphere; it cannot not make it go away like radiation to space does), to even the very old gravitational density heating canard (ignoring that since Earths atmosphere got densified (aka ‘pumped up’) by gravitational consolidation about 4.5 billions years ago, unlike a newly pressurized bicycle tire it has had a LONG time to cool back down).  I chose not to name names; this possible guest post is only a general rebuke.

The surprisingly controversial post’s conclusion is also easily personally verified by simple observation. Tyndall proved in 1859 that both CO2 and H2O were GHG, while N2 and O2 are not. So a personal experiment can be conduced in any desert (mine was a summer day in the Mohave outside Palm Springs during a boring conference). The dry desert heats up a lot from insolation during the day, and cools down a lot at night thanks to desert low specific humidity, so not much GHG except well-mixed CO2, and therefore not much GHE at night. Burrr!

There are only two even semi-rational (but still erroneous) arguments why the CO2 GHE might not exist despite Tyndall’s experimental GHG evidence.

  1. The CO2 window overlaps the H2O window. Note, this does not say the GHE does not exist; only that CO2 ‘cannot’ be a contributor so IPCC is wrong. This assertion is frequently found on the internet in graphical form, but erroneously portrayed for two reasons (both errors are present in negative WUWT comments). The usual stuff omits radiation intensity; and while some windows do overlap, others don’t. Fine IR scale matters. The actual overlaps/windows plus their radiative intensities were provided in essay Sensitive Uncertainty in ebook Blowing Smoke. The essay’s illustration is reproduced below. Forgive the unfortunate insolation extra zero K typo, which I just caught myself. Absorption is a metric of the degree of a gas GHE effect at some frequency (wavelength), from 0 (none) to 100% (full).

Reality has to do with radiation intensity and window ‘shoulders’.

  • GHG are saturated, so can have not have any further effect. This misunderstands saturation, since it depends on the effective radiative level (ERL). As CO2 increases, the ERL rises unconstrained, since CO2 is unaffected by the lapse rate, while H2O is and so decreases. This also reduces their mid troposphere’s overlapping radiative windows. More CO2 raises the ERL. As Callendar’s 1938 curve reproduced below (from Climate Audit) first showed, the GHE never saturates.

          Also note that over the region of Callendar’s curve of present interest, the curve is approximately linear, which is why in my comment to UAH’s most recent report, I did not bother to make the log correction correctly suggested in subsequent comments to my back of the envelope implied 1.7C ECS fraction that Roy’s new update nicely implicitly brackets. ‘Good enough for government work’.

To summarize without any math, the GHE exists. It experimentally must, and easily provably does. The GHE issues are how much when (ECS), not if. Any  ‘skeptical’ arguments to the contrary are fairly easily rebutted, as done here.

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June 6, 2021 6:11 am

Neither you nor Bob Wentworth have successfully rebutted a GHE from convection.
Convection does not just move energy around.
It converts kinetic to potential in uplift and converts potential to kinetic in descent.
That process takes time and is slower than radiation so the system must heat up.
The process is renewed in every successive convective overturning cycle so the ‘time to cool down’ concept does not apply.

Kevin kilty
Reply to  Stephen Wilde
June 6, 2021 7:01 am

It converts kinetic to potential in uplift and converts potential to kinetic in descent.

What you are describing is a perpetual motion machine of the first kind. Here is an example of exactly what you describe, if my embedded link works correctly. Donald Simanek (proprietor of the Museum of Unworkable Machines) calls it the “heavier on one side seduction”. In reality, these convection cells as you are describing them are always converting potential energy locally to kinetic, and the potential energy is constantly being renewed locally through heat transfer. The kinetic energy is perpetually dissipated through friction and heat transfer. It is a heat engine, and you should employ the first law of thermodynamics to offer a complete description (du=dQ-dW). You cannot just examine the dW side of this machine, and get a correct description.

Last edited 14 days ago by Kevin kilty
Reply to  Kevin kilty
June 6, 2021 7:09 am

Not correct because you ignore the underlying radiative energy throughput which is constant at hydrostatic equilibrium. In the equilibrium state new energy coming in matches old energy going out and the process continues indefinitely as a disturbance in the flow.

Trick
Reply to  Stephen Wilde
June 6, 2021 7:32 am

Stephen, hydrostatic equilibrium means the atm. is stable up/down, which is largely observed at any time but not all times in local disturbances (storms). The parcels along the hydrostatic lapse rate are equilibrated with local surroundings and cannot descend as you imagine. Your whole imaginary concept “converts kinetic to potential in uplift and converts potential to kinetic in descent”.is not actually observed. Pistons and cylinders do not inhabit the atm. to perform lab “adiabatic heating” as you imagine.

The winds at the surface hit you in the face not the top of your head since as the surface parcels warmed above the local surroundings uplift the replacement air flows in laterally at ambient (meaning no surface heating in the convection process) and at each level above in ascent until the parcels equilibrate with surroundings.
 
I’ve pointed Stephen to youtube videos showing how convection works in reality but Stephen continues to write convection works as Stephen imagines and not as in nature.

I do know Stephen will never change to face reality (because of a book he read in the 60’s but cannot recall) and this circumstance needs to be pointed out from time to time.   

Reply to  Trick
June 6, 2021 7:42 am

The downward legs of the Hadley, Ferrel and Polar cells show Trick to be incorrect.

Trick
Reply to  Stephen Wilde
June 6, 2021 8:31 am

And the upward legs of the cells exactly balance the mass movement down simply moving existing thermodynamic internal energy around within the system. The cells do not allow convective energy to escape the system to deep space for no global temperature change within the system due these cells.

Robert of Texas
Reply to  Trick
June 6, 2021 10:46 am

Maybe I am not following this correctly, but if you move hot air higher up into the atmosphere, the CO2 and water vapor molecules can then radiate any excess heat they have captured. The radiation, because it is higher in the atmosphere, is now more likely to escape the Earth’s atmosphere completely – so more likely to move up and away. The Earth’s surface is curved so at a higher altitudes the Earth’s surface becomes less of a target. Therefore heat that is trapped in an upward convection is more likely to escape – increase the convection and more heat escapes.

Lateral movement is important in that it will take heat from a warmer place to a cooler place, thus moderating heat buildup near the equators. If the heat is moved over an area where more water is available, then evaporation can increase. The higher amount of water vapor will lower the air density, and once again you get vertical air movement taking heat upwards.

These are not controversial claims, we know that they occur. It is completely ignored in any explanation of the “Greenhouse Effect” because it makes it too complicated, not because it isn’t there.

Trick
Reply to  Robert of Texas
June 6, 2021 11:59 am

“It is completely ignored in any explanation of the “Greenhouse Effect”

Robert, when observed over 4-15+ annual cycles the weather related upward convection you describe removes as much relevant energy from near the surface as downward convection adds. Convection is NOT ignored, it’s just that measurement shows convective processes (mostly resident in the troposphere) remove no meaningful net thermodynamic internal energy from surface to deep space, natural convective processes are observed to only move that energy around within the atm. 

Reply to  Trick
June 6, 2021 2:49 pm

Another thing that you can observe in the desert – if you stick around for a little while – is microbursts.

Colder and therefore denser air falling from a great height in the atmosphere. Thunderstorm NOT required!

Upward convection is warmer, less dense air. Downward convection is colder, denser air.

Warmer air becomes cooler air by losing energy. Just where do you think that energy is lost to?

gbaikie
Reply to  Robert of Texas
June 6, 2021 12:32 pm

“Maybe I am not following this correctly, but if you move hot air higher up into the atmosphere, the CO2 and water vapor molecules can then radiate any excess heat they have captured.”

It’s not like room in a house {room is too small- it doesn’t lower air density with the short room elevation. Warm air rises and becomes less density. But it’s really “all” about warm air mass movement- convection is more than a body/mass of air going up or down, rather average velocity of gas will equalize in energy/temperature, but lower density air has less mass per volume and so has less energy even as it has same average velocity of gas.

Bob Wentworth
Reply to  Robert of Texas
June 7, 2021 4:58 pm

Maybe I am not following this correctly, but if you move hot air higher up into the atmosphere, the CO2 and water vapor molecules can then radiate any excess heat they have captured.

They are radiating heat all along the way. They don’t just wait until they reach altitude to radiate. Radiation at lower altitudes involves higher radiative fluxes because the temperatures are higher and the density of radiating molecules is higher.

Once heat reaches an altitude where the atmosphere above it is transparent to radiation, much of the heat can radiate to space. However, this occurs in gas that is at a low temperature and low density.

So, radiative heat loss at high altitude is inefficient, compared to radiative heat loss from the surface.

The net effect is that that, while convection is important in carrying heat upward in the troposphere, it’s not as magically effective at promoting radiation to space as some people seem to think it is.

The Earth’s surface is curved so at a higher altitudes the Earth’s surface becomes less of a target. 

The curvature of the Earth is small on the scale of the height of the atmosphere. This effect may amount to making a 0.5% difference. Other relevant effects involved are significantly larger. Overall, losing heat from high in the atmosphere is inefficient, and this modest effect does not alter that result.

The higher amount of water vapor will lower the air density, and once again you get vertical air movement taking heat upwards.

Nobody is arguing that convection doesn’t transfer significant amounts of heat from the surface up and throughout the troposphere. It does.

That heat transfer just doesn’t necessarily have the implications that some people seem to think it does.

These are not controversial claims, we know that they occur. It is completely ignored in any explanation of the “Greenhouse Effect” because it makes it too complicated, not because it isn’t there.

Some explanations of the GHE may ignore this, because they’re just trying to explain one aspect of how things work, without going into all the details.

But other explanations of the GHE may appear to be ignoring this, but aren’t.

My recent essay is an example of the latter. I don’t talk about convection, but it is implicitly accounted for. In particular, I show that certain things must be true, regardless of any details of heat transport occurs within the atmosphere or ocean.

It’s a little like the way that one can unequivocally say “energy will be conserved” in a system without bothering to talk about the details of the particular complex processes going on within the system.

There are certain conclusions that one can validly arrive at which don’t require analyzing the details of heat transfer within the Earth’s climate system.

It turns out that the relevance of the GHE to explaining Earth’s average surface temperature is one such conclusion.

Last edited 13 days ago by Bob Wentworth
Stephen Lindsay-Yule
Reply to  Bob Wentworth
June 8, 2021 1:24 pm

“Once heat reaches an altitude where the atmosphere above it is transparent to radiation, much of the heat can radiate to space.”

Someone has the IR spectrum upside down. Radiation is transparent from the surface (8-14µm) to tropopause > 14µm (when temperature stops decreasing). Ignoring hot solar energy (106°C) 7.65µm hitting cold atmosphere 15µm (-80°) leaves 27°C at surface. Sea temperature hardly changes in the low latitudes of the pacific ocean in 24 hrs. 0.3°C.

Atmosphere > 14µm cold, surface (9,10µm) and sun hot > 8µm. Remember that.  

Bob Wentworth
Reply to  Stephen Lindsay-Yule
June 8, 2021 2:09 pm

Someone has the IR spectrum upside down.

I have no idea what you could mean by that.

Radiation is transparent from the surface (8-14µm) to tropopause > 14µm (when temperature stops decreasing).

Why in the world would you believe that?

You couldn’t possible be more wrong. The part of the atmosphere that is most absorbent to LWIR is the troposphere.

Why? Two powerful reasons:

  • The troposphere is where almost all the water vapor (and clouds) are. These are the most powerful LWIR absorbing components in the atmosphere.
  • By mass, a majority of the atmosphere is in the troposphere. That means a majority of the mass of CO₂ is in the troposphere too.

So, the majority of the materials that absorb LWIR are in the troposphere. How could the troposphere possibly be “transparent” to LWIR?

Your mention of wavelengths leads me to wonder if you are one of the people who has the mistaken idea that a particular temperature is only associated with one wavelength.

You do realize that things anywhere in the temperature range of the -80℃ to 30℃ radiated in all wavelengths over entire range from 4µm to 24 µm?

And, that water vapor and CO₂ absorb the wavelengths that they absorb, regardless of their temperature?

Jim Gorman
Reply to  Bob Wentworth
June 9, 2021 5:26 am

The troposphere is where almost all the water vapor (and clouds) are. These are the most powerful LWIR absorbing components in the atmosphere.

That is why the concentration on CO2 reduction with unreliable renewable wind/solar products and spending trillions to do so is a joke. We would be better off trying to limit water vapor and clouds.

You do realize that things anywhere in the temperature range of the -80℃ to 30℃ radiated in all wavelengths over entire range from 4µm to 24 µm?

This flies in the face of vibrational modes in molecules and atoms that allow absorption/emission at specific wavelengths. If there is radiation at all frequencies by all substances, it derives from overtone wavelengths of the fundamental vibration frequencies and the power is miniscule.

Trick
Reply to  Jim Gorman
June 9, 2021 1:42 pm

Jim, 5:26 am: “If there is radiation at all frequencies by all substances, it derives from overtone wavelengths of the fundamental vibration frequencies and the power is miniscule.

No.

In a solid, the vibrations of the constituent particles with electrons moving generate EMR at all frequencies during the vibration process and photons exit through the surface.

Consider a gas: the constituent molecules are moving very fast at room temperature & you can fairly easily compute that average speed after a 1st course meteorology course of study.

Since the molecules are moving, the frequency of the incident illumination absorbed varies from the incident light rays.  On top of that, the molecules emit while in motion so the frequency of emission observed varies with the speed. These processes makes the spectrum of a gas continuous with certain higher intensity lines of emission at quantum jumps. This is called doppler broadening for obvious reasons.

There is a third way to generate other emission frequencies in that the gas molecules are not really hard billiard balls, they have a certain “softness” & this is called collisional line broadening.

All this was an interesting and fundable research topic after quantum mechanics became understood back in the 1930s, 40s, maybe 50s. The original research is largely complete long ago no longer under study, so to find it, a visit to the library stacks is needed.      

Stephen Lindsay-Yule
Reply to  Robert of Texas
June 8, 2021 1:09 pm

Yes the heat is transported (internal energy) + absorbed heat to another area along the gravy-train of air advection. Only a small amount of heat goes up that was absorbed. From minimum to max adds 50 watts. 6 goes up the rest travels somewhere else. Reason water/sea temperatures is hotter near desert areas. Water vapor gives off latent heat, carbon dioxide doesn’t do anything. CO2 is a ideal gas at 0.04% is is part of internal energy and does not absorb excessive heat unless heat is equilibrium to negative 80 degrees Celsius. Remember TOA -80°C, solar input 107°C surface 27° (low latitudes). Land absorbs excessive heat which is transported by elsewhere after dark.

Roger Clague
Reply to  Trick
June 7, 2021 3:42 am

Trick says
“And the upward legs of the cells exactly balance the mass movement down simply moving existing thermodynamic internal energy around within the system.”
I agree.
However gravity is 0.3% less at 20 km height.
This gravity gradient causes the temperature gradient
Gravity potential energy GPE is changed to kinetic energy KE
GPE =mgh
KE = mcT
mgh = mcT

T/h = g/c

Gravity causes change of velocity
change of velocity of molecules causes change of T

Trick
Reply to  Roger Clague
June 7, 2021 10:45 am

Roger 3:42am, the .003 loss doesn’t convect out of the system control volume to deep space changing thermodynamic internal energy, it stays in the system returning exactly .003 in the cell downward leg for no change in system internal energy in the process. 

Lit
Reply to  Roger Clague
June 7, 2021 10:36 pm

The heat source creates the gradient. The heat source is the Earth surface. Look at any diagram modelling a gradient, the heat source is what determines the gradient.

ferdberple
Reply to  Trick
June 7, 2021 6:28 pm
  •  Neither conduction nor infra-red radiation are efficient at transmitting heat through the lower troposphere, so the surface and the air in touch with it get steadily warmer until such air is less dense than the air above and overturns. Whereas in water this would lead to a uniform temperature, in the atmosphere it leads to a constant temperature gradient. As air rises there is less weight of air above it, and hence a lower pressure. In response, air expands, doing work against the remaining pressure of its surroundings and hence cooling to conserve its energy.

Temperature and Altitude (wisc.edu)

  • source: Horel and Geissler
Philip Rose
Reply to  Stephen Wilde
June 7, 2021 4:45 am

Look at those cumulous clouds rising and falling. There are a lot of water droplets and Ice particles in there circulating vertically and losing heat accordingly. Convection is indeed a powerful heat engine in UK.

AndyHce
Reply to  Trick
June 6, 2021 3:44 pm

Adiabatic heating of as much as 100 degrees F has be measured. It often effects large areas (much of Europe and Australia in recent years, various areas quire regularly on an annual basis). Since energy is radiated away, the heating could not continue without solar input but the process does often warm the surface considerably.

gbaikie
Reply to  AndyHce
June 6, 2021 4:24 pm

Adiabatic heating is certainly a thing. Many times I have cited
this effect in regards to the drying out of the Mediterranean basin
{Messinian salinity crisis}:“…As winds blew across the “Mediterranean Sink“, they would heat or cool adiabatically with altitude. In the empty Mediterranean Basin, the summertime temperatures would probably have been extremely high. Using the dry adiabatic lapse rate of around 10 °C (18 °F) per kilometer, the maximum possible temperature of an area 4 km (2.5 mi) below sea level would be about 40 °C (72 °F) warmer than it would be at sea level. Under this extreme assumption, maxima would be near 80 °C (176 °F) at the lowest points of the dry abyssal plain,..”
https://en.wikipedia.org/wiki/Messinian_salinity_crisis

And every night and everywhere on Earth one gets such warming. But I refer to it as the large thermal mass of the atmosphere. But Adiabatic heating generally means increase in temperature cause higher air masses falling. Every night, it reduces the surface air cooling. Or one might have say 1 C cooling per hour of night, and having large air mass, slows this cooling {not normally making night air warmer, but that could happen in some places and time.
What is global temperature is the surface temperature of the ocean which covers 70%, and ocean is roughly, level. Is really anything to do with land and higher elevation and lower elevations- isn’t a global thing.

Stephen Lindsay-Yule
Reply to  Trick
June 8, 2021 9:43 am

Stephen isn’t talking about winds, he is talking about pressure (forced motion through compression and increased collisions) which can be horizontal as well as vertical. No wind required.

gbaikie
Reply to  Kevin kilty
June 6, 2021 12:19 pm

“What you are describing is a perpetual motion machine of the first kind. Here is an example of exactly what you describe, “
That roughly depicts, but rather than 2 d it’s 3 d.

ferdberple
Reply to  gbaikie
June 7, 2021 6:35 pm

Here is a “perpetual motion” machine that works.

What is OTEC (otecnews.org)

Perpetual motion machines run forever without using energy. How can something that is powered by the sun be a perpetual motion machines? By that reasoning, life is a perpetual motion machine of the first kind.

Michael S. Kelly
Reply to  ferdberple
June 8, 2021 1:13 am

Huh. I had no idea that OTEC was still alive. I always thought it was a viable technology, but was stupidly implemented and thus doomed. The description of how they implement it in your link is the way I would have done it, and is perfectly workable.

Plus, a couple of the girls in the splash screen picture on your link are really hot, so I like the technology even more, now. And yes, I am that shallow.

AC Osborn
Reply to  Michael S. Kelly
June 9, 2021 10:31 am

That is very interesting technology, thanks for the link.

Gary Pearse
Reply to  Kevin kilty
June 6, 2021 9:34 pm

Kevin, I think you are missing a pithy point in taking the larger bite (I agree you were offered the large bite by Steven). Consider Willis E’s view of tropical thunderstorm as a chimney that whisks enormous volumes of warm water vapor evaporated from the ocean’s surface quickly to the stratosphere where it can emit LWIR directly to space. In this case, although this shift of water vapour is rapid, it still has the effect of delaying exit of LWIR, thereby causing some degree of warming of the atmosphere. Am I making any sense here? (My wife and I are10 days into contracting Covid so that may be a factor)

Kevin kilty
Reply to  Gary Pearse
June 7, 2021 8:36 am

I am having some trouble seeing exactly what you are getting at. It seems that you are arguing about a short delay in the LWIR, which is a sort of storage of energy, and therefore must make the atmosphere warmer, as though it is a savings account in a manner of thinking. I have no data regarding this, but does this actually delay the LWIR much as opposed to competing means of getting energy to a high place where it can finally radiate to space? More to the point, does this delay accumulate and make the atmosphere warmer without end?

I hope you, et ux, feel better soon.

Last edited 13 days ago by Kevin kilty
Bob Wentworth
Reply to  Kevin kilty
June 7, 2021 5:14 pm

does this delay accumulate and make the atmosphere warmer without end?

It doesn’t. Delays have no effect at all on temperature, in steady-state.

Delays could only affect temperature in an ongoing fashion if the delays were steadily increasing, day by day and year by year. They don’t. Whatever delays are present are likely to remain roughly the same in the long run.

As an analogy, think of a river flowing from Lake A to Lake B. Does it matter how long it takes water to get from Lake A to Lake B? Does the duration of the trip affect the water level in Lake B? It does not.

Kevin kilty
Reply to  Bob Wentworth
June 8, 2021 10:27 am

Rhetorical question, Bob.

AC Osborn
Reply to  Bob Wentworth
June 9, 2021 10:34 am

Wrong, of course it affects the level in Lake B.
Try just evaporation for starters.

There is no temperature equilibrium in Earth’s climate, it never stops changing.

Bob Wentworth
Reply to  AC Osborn
June 9, 2021 3:48 pm

Wrong, of course it affects the level in Lake B.

Try just evaporation for starters.

The claim I was responding to essentially asserted that delays inherently lead to temperature increases.

Evaporation is a side-effect, not an inherent aspect of a transit delay.

Suppose we shipped the water in a pipeline, instead of a river, so that there was no evaporation. In that case, the water level in Lake B would not depend on how long the water takes in transit.

There is no temperature equilibrium in Earth’s climate, it never stops changing.

And, rivers and lakes don’t have fixed levels.

That doesn’t mean it’s not useful to talk about the mean water level, and do urban planning that relies on predicted flood levels.

Last edited 11 days ago by Bob Wentworth
Bob Wentworth
Reply to  Gary Pearse
June 7, 2021 5:09 pm

it still has the effect of delaying exit of LWIR, thereby causing some degree of warming of the atmosphere.

There is a false meme going around that “delaying the exit of LWIR” causes warming.

It doesn’t.

A transient increase in the net delay would increase temperature temporarily, but in steady-state, delays make no difference whatsoever to temperature.

ferdberple
Reply to  Kevin kilty
June 7, 2021 6:15 pm

What you are describing is a perpetual motion machine of the first kind. 
============
Nope. What is being described is an OTEC generator, using the atmosphere in place of the ocean.

ferdberple
Reply to  Kevin kilty
June 7, 2021 6:18 pm

What is OTEC (otecnews.org)

Here is an OTEC generator. Also a “perpetual motion” machine that runs “forever” using the heat gradient of the oceans, producing excess electricity in the process.

Trick
Reply to  ferdberple
June 7, 2021 7:52 pm

Well, not forever but until the sun burns out.

JCM
Reply to  Stephen Wilde
June 6, 2021 7:06 am

Please see my comment with respect to the surface energy balance components. Wentworth’s ideas do not make any reference to the various real energy exchange mechanisms between the surface skin and the atmosphere. These concepts may help consolidate the radiative components and surface properties with your views on atmospheric convective overturning. I could have written more concisely in the comment but the general concepts are there. Please let me know if you have any thoughts. https://wattsupwiththat.com/2021/06/04/mathematical-proof-of-the-greenhouse-effect/#comment-3263069

Reply to  JCM
June 6, 2021 7:22 am

HelloJCM.
The point you made is broadly correct. The thing is that S-B only applies to black bodies but neither the unevenly lit surface of a sphere nor the mobile atmosphere around it constitute a black body on their own because non radiative energy transfers are also ongoing. They should properly be described as grey bodies which are beyond the S-B parameters.
However, once in hydrostatic equilibrium the surface and atmosphere are also in thermal equilibrium and so S-B can be applied when viewing the two grey bodies together from outside the system.
Furthermore, objects within the system will also act as black bodies because they will be in thermal equilibrium with the mobile gases surrounding them.
Climate scientists usually show a flat surface for their radiative energy budgets and include upward convection as a surface cooling effect so they have to counter that with an assumed heating effect from back radiation.
They forget that we are dealing with spherical geometry so that upward convection on the lit side then comes down on the unlit side for an equal and opposite surface warming effect there. When correctly accounting for that spherical geometry one gets a balanced energy budget without any warming effect from back radiation.
One could say that they are ‘flat Earthers’.
Since Earth is a rapid rotator the ascent and descent regions get split up and spread all around the sphere.

Trick
Reply to  Stephen Wilde
June 6, 2021 7:43 am

 “S-B only applies to black bodies.. objects within the system will also act as black bodies”

There no black bodies existing Stephen. Maybe you can learn how black body radiation exists in spite of that reality but I doubt it.

“assumed heating effect from back radiation.”

Actually measured heating effect from radiation emitted by all-sky toward the surface from a hemisphere of directions.

“convection on the lit side then comes down on the unlit side”

For no change in thermodynamic internal energy of the earth/atm. system control volume thus the process described cannot alone change the global median surface temperature since only radiation can escape the control volume of interest (around earth/atm.) to deep space and no convection/conduction escapes to space.

Stephen Lindsay-Yule
Reply to  Trick
June 8, 2021 2:01 pm

Earth is a blackbody object with a uniform temperature 5.3°C 340 watts. 5.5×10^21 molecules / (4*3.1415*(6371)^2*1000*1000*1000*10780) = 0.287 per molecule. 1187g x 0.287=340 watts. This fluid motion keeps this uniform temperature between summer and winter. Summer land NH adds 3-4 degrees, winter land snow lowers this by 3 degrees. Solar radiation controls these changes, absorbed during summer and reflected during winter. TOA is warmed by solar energy in SH as NH isn’t as cold. And cools in NH summer as SH is much colder TOA.

AC Osborn
Reply to  Stephen Lindsay-Yule
June 9, 2021 10:37 am

Uniform Temperature?
I think not, it never stops changing anywhere on earth.
You need a better description, maybe “controlled”, or something similar.

JCM
Reply to  Stephen Wilde
June 6, 2021 8:34 am

I tend to think of surface energy balance with the atmosphere as complementary to total radiative energy budget of the surface skin. The surface energy balance is partitioned among various flux mechanisms described in boundary layer concepts. I make no specific mention of S-B theory. Boundary layer balances are useful and demonstrated to adequately partition net flux at various scales. It can be used for the surface budget of an individual leaf on a tree to the entire earth surface (and every scale in between). I encourage you to revisit this idea as it is meant to be complementary to your observations and completely fits within the hydrostatic equilibrium idea your propose. It is also meant encourage the radiatively minded to think beyond their knowledge horizon. The notion that different perspectives are mutually exclusive is not correct. The boundary layer concepts are meant to provide a link between these various perspectives. For convenience I will provide the key concepts HERE. Cheers.

JCM
Reply to  JCM
June 6, 2021 8:48 am

The key here in the overall climate change debate is that while the surface energy balance does rely on the concept known as GHGs for certain mechanisms, the overall surface energy balance partitioning is easily able to adapt to minor perturbation of GHG concentration to maintain the average density profile required for hydrostatic equilibrium. The main mechanism of this is adaptability is the turbulent energy flux.

gbaikie
Reply to  JCM
June 6, 2021 12:41 pm

Yeah.
But I would say this largely about the tropics which receive more sunlight, the turbulent nature of tropical ocean heat engine warms the entire world.

JCM
Reply to  gbaikie
June 7, 2021 9:31 am

Sure, in the presence of more available energy in a water rich environment turbulent flux is most certainly the dominant factor of air-ocean heat flux in the tropics with relatively large magnitude compared to other regions. However I would caution to avoid thinking in terms of isolated geographies when, in fact, all regional geographies of the atmosphere are coupled. If you’re interested in an observational case study of the surface energy balance in the pacific warm pool I recommend this paper: http://www.met.reading.ac.uk/~swrhgnrj/teaching/MT23E/grant_hignett.pdf

JCM
Reply to  JCM
June 7, 2021 11:01 am

For anyone who might point it out: yes I am aware I should be more precise in identifying turbulent flux vs the radiant components (i.e. irradiance). I get lazy and call everything heat-flux and that’s obviously not accurate.

Reply to  Stephen Wilde
June 6, 2021 9:58 am

Stephen Wilde wrote, “S-B only applies to black bodies…”

Here’s the Stefan-Boltzman relation:

   E = ε σ T⁴

where:
   sigma σ is the Stefan-Boltzman constant, 5.670374419E−8 W/m²K⁴
   Temperature T is in Kelvin
   E = radiative emission.

So, if you think that S-B relation only applies to back bodies, then what is ε ?

I’ll give you a hint. ε is always between 0 and 1, and it is actually a function of frequency, except for emissions from a perfect _____-body. (Can you fill in the blank?)

Reply to  Dave Burton
June 6, 2021 10:46 am

Stefan-Boltzmann law, statement that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature. Formulated in 1879 by Austrian physicist Josef Stefan as a result of his experimental studies, the same law was derived in 1884 by Austrian physicist Ludwig Boltzmann from thermodynamic considerations: if E is the radiant heat energy emitted from a unit area in one second (that is, the power from a unit area) and T is the absolute temperature (in kelvins), then E = σT4, the Greek letter sigma (σ) representing the constant of proportionality, called the Stefan-Boltzmann constant. This constant has the value 5.670374419 × 10−8 watt per metre2 per K4. The law applies only to blackbodies, theoretical surfaces that absorb all incident heat radiation.

From here:

https://www.britannica.com/science/Stefan-Boltzmann-law

Trick
Reply to  Stephen Wilde
June 6, 2021 12:09 pm

Since no blackbodies exist in nature (“theoretical surfaces”) that ref. is telling Stephen that the S-B law applies to no surface in existence. Yet inexpensive IR thermometers based on S-B work fine to measure real surface temperature.

There is a reason one needs to consult modern meteorological and/or science texts which will show Stephen has not done so (preferring to go with his imagination) or Stephen would have recognized the issue. 

Reply to  Stephen Wilde
June 6, 2021 12:10 pm

The thing is that since the S-B Law only applies to black bodies then it follows that the greyer a body the greater the departure from the S-B Law.
So, the more competition there is from non radiative energy transfers the greyer the body becomes. That competition arises from the mass density of an atmosphere because it is upward convection that removes surface KE before it can be radiated to space.
Thus the lit side will have an E figure of less than 1.
It then follows that the unlit side will have an E figure of more than 1 because irradiation is zero but yet energy is still being supplied to the surface from descending air.
On the face of it that balances out at equilibrium but we are left with the unlit side warmer than it otherwise would be if there were no descent warming and that less cold air circulates back to the lit side where the additional surface energy must be added to full irradiation and the temperature rises on the lit side as well.
So. what we actiually have is two separate grey bodies interacting via convection until they reach thermal equilibrium at which point the system viewed from space appears as a black body.
But the surface and the atmosphere always remain grey bodies on both the lit and unlit sides.
The S-B Law should never have been applied to such grey bodies.

Trick
Reply to  Stephen Wilde
June 6, 2021 12:35 pm

“The thing is that since the S-B Law only applies to black bodies…”

No. BBs don’t exist. Consult a relevant meteorology text Stephen. The math will be beyond Stephen but maybe not the prose.
 
Planck’s law is idealized for intensity of blackbody radiation at a temperature and frequency. S-B law converts Planck’s ideal law across the spectrum to real body radiation and computes brightness temperature with a measured factor for emissivity of the material being illuminated. Inexpensive IR thermometers work fine Stephen. 

Bob Wentworth
Reply to  Stephen Wilde
June 6, 2021 9:14 pm

So, the more competition there is from non radiative energy transfers the greyer the body becomes. 

This is an unjustified belief you have apparently invented with no empirical evidence or support from mainstream science.

Every single book on thermodynamics and heat transfer would disagree with you.

Bob Wentworth
Reply to  Stephen Wilde
June 6, 2021 9:11 pm

Citing an entry from an encyclopedia does not establish truth. Wikipedia’s entry includes grey bodies.

What is known is that:

  1. black bodies obey M = σT⁴
  2. all matter obeys M = 𝜀σT⁴ where 0 < 𝜀 ≤ 1.
  3. 𝜀 has been measured for a wide variety of materials

If you don’t consider #2 to be the Stefan-Boltzmann Law, that’s just semantic nit-picking. It’s still a valid, well-established physical law.

You just pointlessly make conversations more difficult if you refuse to call #2 the S-B Law.

Trick
Reply to  Bob Wentworth
June 7, 2021 7:34 am

Bob 9:11 pm: “all matter obeys M = 𝜀σT⁴ where 0 < 𝜀 ≤ 1.”
 
This is not quite true Bob even with semantics considered; even you can learn that you haven’t read your Planck close enough.
 
A well-known meteorology professor once had his grad. students measure the emissivity of various materials. He well knew one material sample would show an emissivity greater than 1 and wanted to thereby teach a lesson as the students struggled with their conclusion that they must have made a mistake.
 
There is a clue in Planck’s writing: “Throughout the following discussion it will be assumed that the linear dimensions of all parts of space considered, as well as the radii of curvature of all surfaces under consideration, are large compared with the wave lengths of the rays considered.”

Last edited 13 days ago by Trick
Bob Wentworth
Reply to  Trick
June 7, 2021 12:10 pm

I see that you’re right: for materials with dimensions smaller than the relevant wavelength, emissivities greater than 1 are possible.

Thanks for the information.

Stephen Philbrick
Reply to  Bob Wentworth
June 7, 2021 1:58 pm

Interesting. I’m guessing this doesn’t have any measurable impact on the conclusions though.

Bob Wentworth
Reply to  Stephen Philbrick
June 7, 2021 2:27 pm

I’m guessing this doesn’t have any measurable impact on the conclusions though.

No, it doesn’t. The conclusions depend on materials having an emissivity value. It doesn’t matter what that value is.

Trick
Reply to  Bob Wentworth
June 7, 2021 2:46 pm

Where this diffraction effect becomes important to understand is for the material particle size of the moon surface since order of 25% of the surface is estimated to be powder material with dimensions smaller than the relevant wavelength.

This circumstance affects computed lunar brightness temperatures especially over a (nonequilibrium) wide range of temperature and solar incidence angle. This issue calls into question the current estimates for global average lunar brightness temperature (or any airless celestial body) vs. unknown global equilibrium thermometer temperature.  

Bob Wentworth
Reply to  Trick
June 7, 2021 3:10 pm

Interesting. Any references?

Trick
Reply to  Bob Wentworth
June 7, 2021 4:48 pm

See ref. ‘The global surface temperatures of the Moon as measured by the Diviner Lunar Radiometer Experiment’ 2016 Williams et. al. in Icarus.

You should be able find free pdf online:

 “Brightness temperatures in Diviner’s individual infrared channels may vary depending on the distribution of sub-footprint-scale temperatures, spectral emissivities, and photometric properties….. Therefore the brightness temperatures cannot be interpreted in terms of a unique surface temperature.”

eyesonu
Reply to  Bob Wentworth
June 7, 2021 5:16 pm

Bob W and Trick,

Would that make a molecule of H2O have an emissivity of greater than 1 at the instant of condensation or possibly evaporation?

———

This comment was intended to be made with regards to Bob Wentworth’s @ 2:10 pm, June 7

Last edited 13 days ago by eyesonu
Jonas
Reply to  Stephen Wilde
June 7, 2021 2:14 am

Boltzmann did not derive any radiation law (see his original work). He used thermodynamics and Maxwells equations to show that the internal energy and the radiation pressure is proportional to the temperature raised to 4 – at thermodynamic equilibrium in a cavity.
He commented that there is a resemblance between his expression for internal energy and Stefans observed radiation law.

Thermal radiation is generated by lattice vibrations (phonons). To me it seems logical that there should be similarities between the vibration spectum (phonon dispersion relations) and the emitted radiation spectrum – since the radiation spectrum is generated by the phonon spectrum.

If so, it is very likely that there exist more terms than only one T^4.
As long as people use this correction factor – emissivity – it will always be possible to fit SB law to the observed radiation.

Main point – radiation generated from lattice vibrations is not the same as thermodynamic equilibrium in a cavity.

I have never seen any theory that derive the thermal radiation from it´s source (lattice vibrations).

Clyde
Reply to  Dave Burton
June 7, 2021 11:33 pm

That form of the S-B equation does indeed only apply to idealized blackbodies.

Using the S-B equation in that form assumes emission to equivalently 0 K, so if one uses that form of the equation for graybodies, you’re treating graybodies as essentially idealized blackbodies with ε<1.
comment image

For graybodies, one must take into account the energy gradient by using:
q = ε σ (T_h^4 – T_c^4) A_h
To do otherwise results in a violation of Stefan’s law, and results in double the energy density in a blackbody cavity at thermodynamic equilibrium.

It also results in too-high radiant exitance figures, which one must then get around by assuming a ‘cold to warm’ energy flow (a violation of 2LoT), and subtracting that ‘cold to warm’ energy flow from the real ‘warm to cold’ energy flow. This results in nonsensical numbers such as seen on the K-T diagram.

In using the S-B equation in the form you specify for the atmosphere, just as it does in a blackbody cavity at thermodynamic equilibrium, it results in too-high energy density.

Photons not only do not, they cannot, flow from cool to warm… a photon is a persistent perturbation of the ambient EM field. A photon with lower chemical potential (from a cooler object) attempting to ascend the energy gradient toward a warmer object will be subsumed by the higher energy density of that energy gradient before it ever reaches the warmer object:comment image

Energy only flows if there is an energy gradient. That energy gradient is set up by the object’s surface molecule’s magnetic dipoles generating a radiation pressure. It is the energy density differential between cooler and warmer object which determines the radiant exitance of the warmer object, for a system with only two objects. For a complex system consisting of many objects, it is the energy density differential of a multitude of objects, but the same principle holds.

F = U – TS

Where:
F = Helmholtz Free Energy
U = internal energy
T = absolute temp
S = final entropy
TS = energy the object can receive from the environment

If U > TS, F > 0… energy must flow from object to environment.
If U = TS, F = 0… no energy can flow to or from the object.
If U < TS, F < 0… energy must flow from environment to object.

U = T^4 4εσ/c
The above formula is the Stefan-Boltzmann relation between energy density and temperature.

If ΔU = 0, then (ΔU * c/4εσ) = 0, thus no energy can flow.

U has the same physical units as pressure (J m-3) and U ∝ T. That is radiation pressure, which sets up the energy gradient.

Free energy is defined as the capacity to do work. If U = TS, p_photon = u/3 = p_object, energy cannot flow because no work can be done. Helmholtz Free Energy is zero. Photon chemical potential is zero.

Trick
Reply to  Clyde
June 8, 2021 10:08 am

Photons from a lab glass of ice water do “flow” (Clyde verbiage) into my room temperature $30 Ryobi thermometer and register 32F just fine, proving Clyde is wrong about the physics of absorbed, scattered, transmitted incident radiation.

Clyde should go back and read the original experiments leading to Planck’s Law to learn the physics.

Clyde
Reply to  Trick
June 8, 2021 11:31 am

Your non-contact thermal sensor uses a thermopile (a series-connected number of thermocouples), which converts thermal energy into an electrical signal. It does not respond to absolute temperature, it generates an output voltage proportional to the differential in energy density between a local thermistor and the thermopile…. meaning it’s measuring a relative dearth of photons from your lab glass of ice as compared to the reference point thermistor, then mathematically interpreting that to arrive at a temperature output.

Perhaps if you’d stop posturing for your own aggrandizement, and actually buckled down and studied how reality works, you’d be able to figure these sort of things out… but I doubt you will.

Clyde
Reply to  Clyde
June 8, 2021 11:37 am

In other words, photons are not flowing from your lab glass of ice water to your Ryobi’s thermopile, they’re flowing from your thermopile to the lab glass of ice water, the energy gradient determining the radiant emittance, which affects thermopile output as compared to a thermistor.

All in accord with 2LoT and Stefan’s Law, whereas your take on radiative transfer of energy blatantly violates 2LoT and Stefan’s Law.

ObjectivityIsBest
Reply to  Clyde
June 10, 2021 2:33 pm

I really don’t follow your argument. IR thermometers certainly do work in cold applications. Who hasn’t checked air conditioning and freezer temps with a newly purchased IR thermometer that remains at ambient temperature? The only potential source of the data are IR photons moving from the cold surface to the thermopile and hence against the gradient you propose. It doesn’t matter if absolute or relative temp the photons still traveled from a relatively cold to a relatively warm object.

The cosmic background radiation is about 2.7 degrees K while the WMAP detector is about 90 K and photons seem to make the 14 billion year journey just fine.

Someone mentioned no greenhouse glass in the atmosphere but funny to me that commercial greenhouses supplement carbon dioxide to 1200 ppm to speed plant growth and crop yields.best

I agree with the OP-no matter how much we loathe warmistas it is just more lunacy to counter bad science with bad science. We all need to do better.

Clyde
Reply to  ObjectivityIsBest
June 10, 2021 6:47 pm

It’s really very simple… energy only flows due to an energy gradient, just as water only flows due to a pressure gradient. In the case of energy, that pressure is radiation pressure.

If the thermopile is warmer, photons will descend the energy gradient toward the cooler surface, cooling the outward-facing side of the thermopile, which changes the output voltage of the thermopile as compared to a reference thermistor which is generally affixed to the back side of the thermopile.

It is possible to detect EM waves without relying on their thermal heating potential, by using resonance instead (which is what WMAP did). Just like a radio antenna, which upon excitation by incoming EM waves, can deliver a signal which can be amplified. Resonance does not connect to internal heating and thus does not have the frequency-temperature restriction imposed by 2LoT. You’re constructing your system such that it resonantly oscillates at the frequency of the EM waves you’re looking for, and if there are EM waves of that frequency, your system will oscillate at that frequency, then you amplify that.

The WMAP detector calculated CMB temperature by referencing antenna temperature thusly:comment image
then modeled CMB temperature thusly:comment image

WMAP did not measure absolute intensity of microwave signals, it measured the difference in antenna:CMB temperature ratio between different-frequency detectors (23, 33, 41, 61 and 94GHz), in what they call a differential pseudo-correlation radiometer…

Since WMAP employs differential receivers, the zero level of each temperature map is unspecified.

… then calculated its output based upon those differentials, then combined the resultant images to remove foreground data… but for some reason, the WMAP team weighted 61 GHz. WMAP images also had a signal-to-noise ratio that barely exceeded 1, so the detectors were mostly detecting noise. That’s why they built Planck with better shielding and cooling of the detectors… because WMAP couldn’t really show CMB anisotropy.

Due to a combination of 1/f noise and observing strategy, the noise in the WMAP sky maps is correlated from pixel to pixel. This results in certain low-l modes on the sky being less well measured than others.

In the absence of a high signal-to-noise ratio, the only indicative feature of the images is reproducibility… except WMAP couldn’t take an image of the same slice of sky twice and get identical images. That’s why they published the first-year images, then skipped publishing images for years 2 and 3 (likely when they discovered their output was not reproducible), then published an average of the 3 years.

Thus, that WMAP actually measured the temperature anisotropy of the universe is in doubt.

WMAP’s successor, Planck, cooled its detectors to as low as 0.1 K via 3-stage active refrigeration, which is why Planck was able to detect signals 10 times fainter than WMAP and collect 15 times more information than WMAP.

As to ‘greenhouse’ gases in actual greenhouses, they don’t pump in CO2 to increase greenhouse temperature, they do it so that the plants don’t have to work as hard pulling CO2 from the air to produce C6H12O6. Thus their stomata remain closed more, thus the plants require less water, grow faster, are hardier and produce more fruit/vegetable/etc.

ObjectivityIsBest
Reply to  Clyde
June 11, 2021 2:07 pm

Photons certainly aren’t charged and are Boson’s that don’t interact with any appreciable probability below 1 MeV. Thermal photons at low energy are essentially transparent to their quantum brethren and so are not subject to radiation pressure that relates to photons interaction with a non quantum solid. Water flowing downhill is not in any way a good analogy to quantum mechanical interactions.

Sorry if it appeared I was suggesting that the beneficial effects of carbon dioxide in greenhouses is due to elevated temperature, I know that is not the case and in fact carbon dioxide supplementation is often used with an exogenous heater to elevate internal greenhouse temperature to further optimize growth.

Clyde
Reply to  ObjectivityIsBest
June 11, 2021 10:55 pm

Where did I ever state that photons carry net charge? Nowhere, that’s where.

You should study up on what a photon gas is. In physics, a photon gas is a gas-like collection of photons, which has many of the same properties of a conventional gas like hydrogen or neon – including pressure, temperature, and entropy.

If photons did not interact with EM fields, then the speed of light in a vacuum would essentially be infinite (just as it would be in a ‘perfect’ vacuum (no matter, no energy)… but a ‘perfect’ vacuum (no matter, no energy) cannot exist… space and time are intricately intertwined, removing all matter and all energy would effectively erase photon transit time by shrinking space to nothing)… it is the EM field component (the non-zero expectation value) of the quantum vacuum which slows light to 299,792,458 m/s in the vacuum of space (which is not a ‘perfect’ vacuum due to not only matter, but the non-zero expectation value of the quantum vacuum).

Oh, would you look at that… exactly what I’ve been stating, from none other than CERN:
https://sci-hub.se/https://iopscience.iop.org/article/10.1209/epl/i2000-00465-1
Photon-photon interaction in a photon gas
Owing to the cosmic microwave background, the velocity of light in the Universe is reduced compared to the vacuum.

Uh-oh! Does that say that photons interact with photons to such an extent that the speed of light is slowed? LOL

Oh, would you look at that… direct sampling of electric field vacuum fluctuations by measuring its effects upon a light beam… didn’t you claim that photons of low energy couldn’t be affected by radiation pressure? Yet here we have the ultra-low radiation pressure of the quantum vacuum affecting a relatively low-energy light beam at 1.18µm (which is in the IR band):

https://sci-hub.se/https://science.sciencemag.org/content/350/6259/420

Huh… it’s almost as if I’m right, and you’re wrong. LOL

Unless you’re now claiming that an electromagnetic field cannot interact with the electric and magnetic fields (oscillating in quadrature) of a photon.

A photon is a persistent perturbation of the EM field above its ambient (at ground state, the quantum vacuum)… raise that ambient field energy density, and the photons are subsumed in the ambient EM field, and are thus no longer persistent.

Therefore, at thermodynamic equilibrium, the chemical potential of photons is always and everywhere zero. The reason is, if the chemical potential somewhere was higher than zero, photons would spontaneously disappear from that area until the chemical potential went back to zero; likewise, if the chemical potential somewhere was less than zero, photons would spontaneously appear until the chemical potential went back to zero.
comment image

ObjectivityIsBest
Reply to  Clyde
June 12, 2021 8:06 am

Photons are bosons that interact in accordance with Bose-Einsten statistics. At IR energy levels they do not interact. Quantum mechanical interactions are really not classical and you can try to describe them as much as you like with fundamentally classical ideas but not how it works. Photon-photon interaction is referred to as photon scattering and becomes important at gamma ray energies and higher, certainly not at IR energies.

I never stated that you stated photons carried charge so no reason for me to look to establish you didn’t say it. I was trying to understand what sort of gradient could possibly exist that impacted thermal photons.

It’s never fruitful to discuss quantum mechanics in terms of classical thought so good luck to you Clyde.

Clyde
Reply to  ObjectivityIsBest
June 12, 2021 4:20 pm

Who’s “describing photons with fundamentally classical ideas”? Certainly not myself. Classical physics can’t even describe the photon. Had you known that, you wouldn’t have even attempted to make your statement.

I’m quoting from the latest quantum physics knowledge, and studies done at CERN, for crying out loud. Your non sequitur is noted and mocked as the weak attempt at denigrating a superior debating opponent and far superior intellect that it is. LOL

Your denial of the studies above (which prove your stance wrong) shows that you’re not objective, nor are you here to learn… you’re here to continually reiterate your incorrect stance in defense of your ‘CAGW’ / ‘backradiation’ / ‘energy can flow willy-nilly without regard to energy gradient’ narrative, a twisted mish-mash of long-debunked ideas from the Prevost Theory of Exchanges, misapplied physics, ignored reality and conflated concepts.

The photon is a component of the ambient EM field (the photon is a persistent perturbation above the ambient)… if the ambient EM field chemical potential rises above the chemical potential of the photons in a given space, those photons will be subsumed into the ambient EM field. If the ambient EM field chemical potential in a given space is higher than the chemical potential of another space or of an object in the space, photons will manifest from the ambient EM field until chemical potential returns to zero.

Therefore, at thermodynamic equilibrium, the chemical potential of photons is always and everywhere zero. The reason is, if the chemical potential somewhere was higher than zero, photons would spontaneously disappear from that area until the chemical potential went back to zero; likewise, if the chemical potential somewhere was less than zero, photons would spontaneously appear until the chemical potential went back to zero.

This means that energy (remember, photons are quanta of energy, the electric field and magnetic field oscillating in quadrature, just another part of the ambient EM field) cannot flow if chemical potential is zero… unless you’re telling everyone that you don’t know what ‘chemical potential’ and ‘Helmholtz Free Energy’ even mean. LOL

Stephen Lindsay-Yule
Reply to  Clyde
June 8, 2021 3:34 pm

You don’t see 0K in a fluid atmosphere even when incoming radiation has gone out. Fluid atmosphere has internal energy that controls temperature gradient. Only radiatively can 0K exist but not observed in fluid atmosphere.

Clyde
Reply to  Stephen Lindsay-Yule
June 8, 2021 3:53 pm

0 K can’t really exist anywhere. Idealized blackbodies are exactly that… an idealization. They don’t actually exist.

Using q = ε σ T^4 for a graybody leads to nonsensical results. Doing so essentially treats a graybody as an idealized blackbody, but with ε<1.

It essentially assumes the graybody emits as though in a 0 K ambient and absorbs all radiation incident upon it.

So between two objects, it would assume the warmer object emits as though in a 0 K ambient and absorbs all radiation incident upon it; and it would assume the cooler object emits as though in a 0 K ambient and absorbs all radiation incident upon it, then subtracts the ‘cooler to warmer’ fictional energy flow from the ‘warmer to cooler’ real energy flow. That’s not how the real world works.
comment image

But we now have two people in these comments section advocating for using q = ε σ T^4 for graybodies, then subtracting a fictional ‘cold to warm’ energy flux from the real ‘warm to cold’ energy flux… at least one of whom is a physicist, and should know better.

They were taught this in graduate school, because it’s a convenient way of accounting for energy flow, but it’s unphysical and results in a violation of Stefan’s Law and 2LoT (in the Clausius Statement sense) if interpreted as being physical.

It leads to further complications… such as assuming at thermodynamic equilibrium between two objects, both are furiously emitting and absorbing radiation, which doubles energy density in the intervening space, a violation of Stefan’s Law.

Last edited 12 days ago by Clyde
Trick
Reply to  Clyde
June 8, 2021 4:50 pm

Clyde: …which converts the photons into thermal energy from the glass of ice water into an electrical signal and my Ryobi displays the brightness temperature 32F same as the thermometer temperature immersed in the ice water.

The photons “flowing” from my Ryobi at room temperature incident on the ice water are absorbed, scattered, and transmitted by the lab glass of ice water.

No violation of 2LOT as universe entropy is increased in the Ryobi measurement process since it is a real process.

Clyde
Reply to  Trick
June 8, 2021 5:57 pm

The energy gradient between Ryobi thermopile and lab glass of ice water determines which will emit, and by how much. If the thermopile is warmer than the lab glass of ice water, the thermopile will emit. The thermistor (usually affixed to the back of the thermopile) is used to calculate the temperature differential across the thermopile (no matter if it’s warmer on the outward-facing side or the thermistor side), and the circuitry calculates the remote temperature in a something akin to a Wheatstone bridge.

You’ll get right on detailing exactly how a photon (a quanta of energy) from a cooler object (and thus with a lower chemical potential) can even reach the warmer object (it is subsumed as it ascends the energy gradient toward the warmer object), let alone incide upon it, let alone do any work upon it. Do keep in mind a warmer object has higher energy density at all wavelengths than a cooler object.

———-
A photon is a persistent perturbation of the EM field above its ambient (at ground state, the quantum vacuum)… raise that ambient field energy density, and the photons are subsumed in the ambient EM field, and are thus no longer persistent.

Therefore, at thermodynamic equilibrium, the chemical potential of photons is always and everywhere zero. The reason is, if the chemical potential somewhere was higher than zero, photons would spontaneously disappear from that area until the chemical potential went back to zero; likewise, if the chemical potential somewhere was less than zero, photons would spontaneously appear until the chemical potential went back to zero.
———-

Energy cannot flow unless work can be done. A photon with lower chemical potential than ambient is not persistent. A photon from the cooler object will have lower chemical potential than the energy gradient between warmer and cooler object, and will be subsumed, it will not be persistent.
comment image

Are you now claiming that energy transfer has nothing to do with work? Or do you not understand what ‘chemical potential’ and ‘Helmholtz Free Energy’ mean?

Are you now claiming that energy can flow from lower energy density to higher energy density without external energy doing work upon the system, a violation of 2LoT in the Clausius Statement sense?

You’ve been taught a convenient means of accounting for energy flow (use q = ε σ T^4 for all objects, and assume all objects emit as though in a 0 K ambient and absorb all radiation incident upon them (essentially treating graybodies as idealized blackbodies, but with ε<1), then subtract fictional ‘cold to warm’ energy flow from real ‘warm to cold’ energy flow to account for the energy gradient), but that does not represent how reality actually works, and leads people to believe that energy can flow from cooler to warmer.
comment image

Reality works via energy gradient determining radiant exitance. Reality doesn’t allow energy to flow from lower energy density to higher energy density, unless external energy does work upon the system… that’s 2LoT (in the Clausius Statement sense) in a nutshell, which you fundamentally misunderstand.

Last edited 12 days ago by Clyde
Clyde
Reply to  Clyde
June 8, 2021 6:38 pm

As an analogy… consider two batteries, one 1.5 V and one 12 V.

We electrically connect the batteries ‘+ to +’ and ‘- to -‘… according to your take on energetic transfer, you claim the 1.5 V battery can do work upon the 12 V battery.

Now analogize the 1.5 V battery to a cooler object, and the 12 V battery to a warmer object.

In other words, photon emission can only occur if the radiation pressure of the atom or molecule which would emit is greater than the ambient field radiation pressure, and the rate decreases as the differential between the emitting atom or molecule radiation pressure and the field radiation pressure decreases.

In other words, photon absorption can only occur if the photon radiation pressure is greater than the radiation pressure of the atom or molecule upon which that photon is incident, and the rate increases as the differential between the field radiation pressure and the atom or molecule radiation pressure increases.

In other words, 2LoT holds even at the quantum scale:
https://www.pnas.org/content/112/11/3275

As Rudolf Clausius wrote in his paper entitled “Entropy”:
A transfer of heat from a hotter to a colder body always occurs in those cases in which work is done by heat…

But… but… but Clausius wrote of heat flowing from cold to hot!“, you may say… yeah, no. He wrote of that only in refuting Carnot’s assumption that the equivalent of the work done by heat is found in the mere transfer of heat from a hotter to a colder body, while the quantity of heat remains undiminished.

Do remember that “heat” is definitionally an energy flux, thus we can rewrite Clausius’ statement above:
An energy flux always (and only) occurs in those cases in which work is done by that energy flux.

Your claim that photons from the cooler lab glass of ice water can flow to the warmer thermopile implies a 2LoT violation, which implies a reversal of time (ie: the principle of irreversibility), because energy is flowing from lower to higher potential without external energy doing work upon the system, thus system entropy is spontaneously decreasing. That cannot happen unless the system experiences time reversal.

There’s a million and one ways to show that your take on energy flow is flawed, a misinterpretation brought about because your institution of higher learning misled you by teaching you a convenient means of accounting for energy flow (use q = ε σ T^4 for all objects, and assume all objects emit as though in a 0 K ambient and absorb all radiation incident upon them (essentially treating graybodies as idealized blackbodies, but with ε<1), then subtract fictional ‘cold to warm’ energy flow from real ‘warm to cold’ energy flow to account for the energy gradient), and either neglected to inform you that it was merely a means of accounting for energy flow and did not represent actual energy flow in reality, or you weren’t paying attention that day.

Bob Wentworth
Reply to  Clyde
June 8, 2021 9:12 pm

They were taught this in graduate school, because it’s a convenient way of accounting for energy flow, but it’s unphysical

I know you assert that it’s “unphysical”, etc. But, I’d like to be clear: Do you see this “way of accounting” leading to incorrect predictions about the temperatures of objects, and if so, in what situations?

Clyde
Reply to  Bob Wentworth
June 8, 2021 9:35 pm

Well, as I stated above, and assuming a two-object system at thermodynamic equilibrium, it assumes that both objects are furiously emitting and absorbing radiation (when in reality, they are emitting nor absorbing none, and a standing wave has been set up in the intervening space (with the same energy density as the objects), with the standing wave’s nodes at the surfaces of the objects (just as we find in a cavity at thermodynamic equilibrium), thus no energy can be transferred between objects, nor from the intervening space to either object), which leads to calculating double the energy density in the intervening space than that expected from Stefan’s Law.

Now extrapolate that to using q = ε σ T^4 for a graybody planet… do you suppose the atmosphere is going to have a higher calculated energy density than reality? Sure it is.

In addition, it leads to the incorrect assumption that an object’s radiant exitance is much higher than it actually is…comment image
…which then causes the follow-on assumption that we must subtract a fictional ‘cold to warm’ energy flow from the real ‘warm to cold’ energy flow… which naturally leads some people to believe that energy can actually flow ‘cold to warm’… it confuses people.

Last edited 12 days ago by Clyde
Bob Wentworth
Reply to  Clyde
June 8, 2021 9:51 pm

I get that you believe the way I and some others have “accounted for” radiation leads to statements about the radiation you don’t agree with and believe confuses people.

Eventually, I’d like to get to the bottom of those issues. (It will still be at least another week before I get my copy of Morse’s “Thermal Physics,” which I’m hoping addresses some of these issues?)

And, it would be an even more urgent issue if you told me that the accounting method I’ve used leads to incorrect predictions about net heat flows and the temperatures of objects.

Everything you’ve listed sounds like it is at the level of what is being said about radiation.

So, although you haven’t directly answered my question, unless I hear otherwise, I’ll infer that your view of the physics and how I’ve been taught to think about things don’t lead to different predictions about the thermodynamic effects on the matter involved in a system.

Last edited 12 days ago by Bob Wentworth
Clyde
Reply to  Bob Wentworth
June 8, 2021 9:55 pm

U = T^4 4εσ/c
The above formula is the Stefan-Boltzmann relation between energy density and temperature.

I’ve already told you that your means of accounting for energy flow results in double the energy density in the intervening space at thermodynamic equilibrium, and as you know, U has the same physical units as pressure (J m-3) and U ∝ T… I’ll leave you to draw your own conclusions.

Trick
Reply to  Clyde
June 9, 2021 1:22 pm

“In other words, photons are not flowing from your lab glass of ice water to your Ryobi’s thermopile”

Clyde!

I see you haven’t taken my advice. Both cooling and warming increase universe entropy. Here’s a short primer:

1.      Planck law, developed from room temperature thermopile, computes the nonzero ideal intensity of radiation emitted by my glass of 32F ice water at each and every frequency (you can plug and chug if curious to know the intensity at ANY frequency).

2.      In its field of view on my glass of ice water, my room temperature IR thermometer absorbs about 95% of these emitted photons, reflects/scatters 5% and transmits none. These incident photons did not interact at all with the photons emitted by my thermometer that, going the other way, became incident on the glass of ice water, and again about 95% of those were absorbed, about 5% scattered, transmitted, and a very small percentage of those incident photons reflected right back into my thermometer.
 
3.      S-B law instructs the IR thermometer across ALL frequencies what to read out from the absorbed photons on its temperature scale: 32F.

It’s a pretty simple process to explain but a lot of hard work (using thermopiles) went toward its development.

—-

 Clyde writes: “As Rudolf Clausius wrote in his paper entitled “Entropy”:

A transfer of heat from a hotter to a colder body always occurs in those cases in which work is done by heat…

And Clyde should look up Clausius’ definition of heat used in his sentence and find his defn. of heat is NOT an energy flux as Clyde re-writes so Clyde’s conclusions are wrong after that according to Clausius.  

Last edited 11 days ago by Trick
Clyde
Reply to  Trick
June 9, 2021 2:32 pm

{sigh}

So, are you now telling us that radiative emission and absorption has nothing to do with work? Because that’s what it sounds like you’re telling us.

First, according to your take on energetic transfer, with photons flowing from cooler lab glass ice water to warmer thermopile, the thermopile’s entropy is spontaneously decreasing, energy is spontaneously flowing from lower energy density to higher energy density. That doesn’t happen, as 2LoT in the Clausius Statement sense confirms, which you fundamentally misunderstand. The significance of the Clausius inequality is that it shows that the definition of entropy, i.e. δ S = δ q r e v T (note that entropy change is defined for a reversible process) is consistent with observed reality: the entropy of an isolated system does not decrease spontaneously. That would make radiative transfer of energy a reversible process… we know it’s irreversible because it’s time asymmetric.

Second, “at every frequency” means you misunderstand how your Ryobi non-contact thermometer works… a typical non-contact thermometer uses 8 – 14 µm radiation only.

Third, it’s not about photon-photon interaction, it’s about photon (which is merely a quanta of energy) interaction with the energy density of the ambient EM field… and between a warmer and cooler object, there is an increasing energy density (the energy gradient) as the photon from the cooler object (lower chemical potential) approaches the warmer object. Since warmer objects have higher energy density at all wavelengths than cooler objects, that means the energy gradient’s chemical potential increases as one nears the warmer object… photons from the cooler object (lower chemical potential) will be subsumed by the higher chemical potential of the energy gradient before those photons can ever incide upon the warmer object.

Fourth, Clausius’ definition of heat is the same as the scientific definition of heat… energy in motion.

Fifth, you’re attempting to resurrect the Prevost Theory Of Exchanges, a long-debunked theory of energetic exchange first promulgated in 1791. It assumed that radiation is an actual material fluid which is highly rarefied, it assumes objects emit regardless of ambient radiation pressure, it is predicated upon Caloric Theory of Heat (which is also debunked).

In fact, it was none other than James Clerk Maxwell who was instrumental in tossing the Prevost Theory of Exchanges on the midden heap of scientific history after he read Joules’ paper and convinced the scientific community to instead use the Kinetic Theory of Heat.

Radiation energy density is proportional to T^4 and is derived via the thermodynamic relation between radiation pressure p and internal energy density u, using the form of the electromagnetic stress–energy tensor: p = u/3, it represents the EM field contribution to the stress–energy tensor, describes the flow of energy in spacetime, and is a representation of the law of conservation of energy.

At thermodynamic equilibrium, the Helmholtz Free Energy is zero, thus photon chemical potential is zero, and since photons are quanta of energy, and since free energy is defined as the capacity to do work, if free energy is zero, no work can be done, thus no energy can flow.

F = U – TS
Where:
F = Helmholtz Free Energy
U = internal energy
T = absolute temperature
S = final entropy
TS = energy the object can receive from the environment

If U > TS, F > 0… energy must flow from object to environment.

If U = TS, F = 0… no energy can flow to or from the
object.

If U < TS, F < 0… energy must flow from environment to object.

If U = TS, p_photon = u/3 = p_object, energy cannot flow because no work can be done. Photon chemical potential is zero. Helmholtz Free Energy is zero.

U = T^4 4εσ/c
The above formula is the Stefan-Boltzmann relation between energy density and temperature.

If ΔU = 0, then (ΔU * c/4εσ) = 0, thus no energy can flow.

U has the same physical units as pressure (J m-3) and U ∝ T. That is radiation pressure, which sets up the energy gradient.

It requires energy (an energy gradient) to perform work, to provide the impetus for any action.

2LoT (in the Clausius Statement sense… “No process is possible whose sole result is the transfer of heat from a cooler to a hotter body”) states that energy cannot flow from a lower to a higher-energy region without external work being done upon the system… not via conduction, not via radiative means, not macroscopically, not at the quantum scale [1], not ever. Do keep in mind the definition of heat: “an energy flux”. Thus: “No process is possible whose sole result is an energy flux from a cooler to a hotter body” without external energy doing work upon the system.

[1] https://www.pnas.org/content/112/11/3275

Study quantized standing wavemodes in a cavity at thermodynamic equilibrium [2] and apply that to an object at thermodynamic equilibrium with ambient.

[2] http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html

Just as in a cavity at thermodynamic equilibrium, (if you will remember, blackbody radiation used to be called cavity radiation) between two objects at thermodynamic equilibrium, each object emits blackbody radiation until a standing wave between the two objects is set up with energy density equal to the two objects, after which no energy is exchanged, that’d violate 1LoT, 2LoT and Stefan’s Law.

Neither object can do work upon the other (nor upon the space between the objects) at thermodynamic equilibrium, thus no energy can be transferred between the two objects.

If one object changes temperature, the standing wave will begin traveling toward the object of lower temperature, with the group velocity of that traveling wave proportional to the temperature differential of the two objects, the operating principle behind it all being radiation pressure and energy gradient.

At the limit of two particles at thermodynamic equilibrium, no energy flows between them.

Now, let’s expand upon that limit of two particles in thermodynamic equilibrium… now each object will have two particles.

Now, we have two objects with two particles each, Particle 1 of Object 1 is identical to Particle 1 of Object 2; and Particle 2 of Object 1 is identical to Particle 2 of Object 2. The objects are in thermodynamic equilibrium, right? The radiation each object is giving off is identical, right? As the two objects reach the point of thermal equilibrium, the macroscopic wave (ie: the tensor product of many singular photons) emitted by each object is identical, right?

A standing wave results from two waves of the same frequency with different vectors within the same medium.

Once thermodynamic equilibrium is attained, the standing wave has energy density equal to the energy density of the objects, so the objects can neither absorb nor emit radiation… that standing wave is reflected from each object (assuming no leakage of energy outside of the space between the two objects due to misdirected photons, nor no stray photons from outside entering the space between the two objects) because the standing wave’s nodes are at the surface of each object.

I encourage you to expand the number of particles in the system to whatever arbitrary number you wish, while keeping in mind what ‘thermodynamic equilibrium’ is… how energy transfer shifts temperature distribution within and between the objects until a Planckian distribution is attained in each and thus thermodynamic equilibrium is attained between them.

The photon has a chemical potential of zero at thermodynamic equilibrium, so photons at thermodynamic equilibrium are not absorbed and re-emitted, that’s a physical impossibility and results in a violation of Stefan’s Law [3] because the standing wavemode nodes are always at the surface in question at thermodynamic equilibrium.

[3] https://objectivistindividualist.blogspot.com/2018/08/the-nested-black-body-shells-model-and.html

This is why the Stefan-Boltzmann equation for graybody objects [4] takes into account Tc when calculating radiative flux from Th: q = ε σ (T_h^4 – T_c^4) A_h

[4] http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c3

You misinterpret the S-B radiant exitance equation for real-world objects. Warmer objects don’t absorb radiation from cooler objects (a violation of 2LoT in the Clausius Statement sense and Stefan’s Law); the lower energy gradient between warmer and cooler objects (as compared to between warmer object and 0 K) slows radiant exitance of the warmer object. The differing energy density between objects manifests an energy gradient, each surface’s energy density manifesting a proportional radiation pressure.

Your take on radiative transfer would violate 1LoT, 2LoT. Stefan’s Law and a whole host of other fundamental physical laws by claiming that energy can flow from a lower to a higher energy density region.

Last edited 11 days ago by Clyde
Trick
Reply to  Clyde
June 9, 2021 4:14 pm

—–
“First, according to your take on energetic transfer, with photons flowing from cooler lab glass ice water to warmer thermopile, the thermopile’s entropy is spontaneously decreasing”

No Clyde, you must have missed my statement: Both cooling and warming increase universe entropy because they are real processes thus consistent with Clausius stated 2nd law in his 9th memoir near the end.

Second, “at every frequency” means Clyde misunderstands how a Ryobi non-contact thermometer works (or any mass works)… a typical non-contact thermometer uses 8 – 14 µm radiation but it emits and absorbs ALL frequencies. All means all, there is no exception. Ever. For any mass.
 
“Photons from the cooler object (lower chemical potential) will be subsumed by the higher chemical potential of the energy gradient before those photons can ever incide upon the warmer object.

Subsumed? Incide? That’s a bit archaic. Sorry Clyde, actually look up Planck’s Law, you just wrote an exception to the law and there are none . Ever, that’s why they call it a law. Note that there is no hedging here, Planck’s Law is never identically zero, it says all bodies exhibit radiation intensity (emit) and absorb by (Kirchhoff law) at all temperatures and all frequencies. All the time.
  
“Fourth, Clausius’ definition of heat is the same as the scientific definition of heat… energy in motion.”

No. That is the anonymous writer of the encyclopedia defn. of heat and is different than Clausius’ defn. Try again. Look it up. I am not resurrecting that radiation is an actual material fluid as you should learn by consulting Clausius, Prevost or Planck’s original work. If you want to know what our illustrious predecessors really did and said, then look up their own words. The last place to look is in a blog, or encyclopedia, which are notorious spreaders of rumors, half-truths, and outright errors.

Find a firsthand quote from those guys you name that “Warmer objects don’t absorb radiation from cooler objects (a “violation of 2LoT in the Clausius Statement sense and Stefan’s Law)”. It is simply Clyde that has erroneously used that statement because cooling is a real process that increases universe entropy.

NB: The Planck function is worthy of respect, if not awe, in that it contains not one, not two, but three fundamental constants of nature: the speed of light in a vacuum c, Planck’s constant h, and Boltzmann’s constant kB. You can’t get much more fundamental than that.

Clyde
Reply to  Trick
June 9, 2021 6:00 pm

You’ll get right on explicating exactly how a higher-entropy object can decrease the entropy of a lower-entropy object… if you can’t prove that mathematically, your entire argument collapses, and you’ll be forced to concede that your take on radiative physics is incorrect.

Clyde
Reply to  Trick
June 9, 2021 6:28 pm

Trick wrote:
“Planck’s Law is never identically zero, it says all bodies exhibit radiation intensity (emit) and absorb by (Kirchhoff law) at all temperatures and all frequencies.”

Thank you for exhibiting yet again your inability to understand Planck’s Law.

Planck’s Law does not state that all objects furiously emit and absorb all the time, as you claim (except for idealized blackbody objects).

Planck’s law describes the spectral density of electromagnetic radiation of a black body in thermal equilibrium at a given temperature T.

You’re yet again confusing idealized blackbody objects and real-world graybody objects.

Idealized blackbody objects essentially assume an infinite energy gradient (equivalent to emission to 0 K ambient, absorption from ∞ K ambient).

Real-world graybody objects emit and absorb in accord with the energy gradient, which is a manifestation of the radiation pressure.
comment image

Last edited 11 days ago by Clyde
Trick
Reply to  Clyde
June 9, 2021 7:22 pm

You are correct Clyde, Planck’s Law does not state “furiously”.

“Planck’s law describes the spectral density of electromagnetic radiation of a black body..”

You are wrong though in that Planck’s law could not possibly describe a black body as none of those exist and yet Planck’s Law was established experimentally using stuff that DOES exist and in a lab at STP (not 0K). Btw, the law states radiation intensity not density. Can Clyde explain all that?  Hint: the experimental setups used can be found for free on the internet.

Your graphic shows real-world object’s emissivity .LT. 1 yet real-world objects exist with measured emissivity .GT. 1. Can Clyde explain that miss? Hint: read through the opening paragraphs in Planck’s “Theory of Heat Radiation” which Clyde can find for free on the internet.

Clyde’s graphic “this doesn’t exist” is wrong. Hint: Tc and Th are temperatures and wherever there are temperatures there are averages around which fluctuations occur.  

Clyde
Reply to  Trick
June 9, 2021 8:21 pm

Pedantism is no way to win an argument, Trick. “Furiously” was my modifier… in reality, at thermodynamic equilibrium between two objects, neither is emitting (nor absorbing) any radiation, nor is the intervening space. There is no energy gradient, so no work can be done, no energy can flow, so there is no impetus to do anything (emission or absorption). All actions need an impetus. No energy can flow unless work can be done.

Unless you are stating that radiative energy flow has nothing to do with work. Are you?

Planck’s Law refers to spectral density. Perhaps you should brush up on your definitions.

Planck heuristically derived a formula by assuming a hypothetical electrically-charged oscillator in a blackbody cavity. It was not derived “experimentally using stuff that DOES exist and in a lab at STP (not 0K)” (your words).

STP in a lab… so you’re saying Planck worked in a lab at 273.15 K? LOL

Real-world objects can only have ε>1 if their size is smaller than the emitted wavelength and if the particle is in relative isolation from other particles… it does not apply to macroscopic objects… unless you’re calling µm sized objects “macroscopic”.

So essentially, everything you’ve stated is incorrect.

Last edited 11 days ago by Clyde
Trick
Reply to  Clyde
June 10, 2021 6:09 am

“Planck’s Law refers to spectral density. Perhaps you should brush up on your definitions.

No. Just look up the units for radiant intensity from the Planck formula.

“Planck heuristically derived a formula by assuming a hypothetical electrically-charged oscillator in a blackbody cavity. It was not derived “experimentally using stuff that DOES exist and in a lab at STP (not 0K)” (your words).

No, there was no oscillator that’s hilarious actually. He used cavity lab work initially and then found the same formula theoretically.

You are way behind looking up the original work Clyde, you need to catch up so you know what you are writing about.

“and if the particle is in relative isolation from other particles

No. You just need to catch up on the historical lab work – reading about it is ok but I recommend going there and doing it yourself (it’s called replication) if you really want to up your game & understand radiative emission and absorption so you can tell when you pick up nonsense from the internet such as you do. 

Clyde
Reply to  Trick
June 10, 2021 11:25 am

If you insist upon rewriting history, creating your own definitions and denying reality, then no one can help you.

Good luck with your alternate universe, in which energy can flow willy-nilly without regard to the energy gradient… that energy only flows when there’s an energy gradient is the reason for the meta-stability of all invariant-mass matter in the universe, after all (see below)… I expect your alternate universe will self-destruct shortly as all bound electrons undergo electron capture. LOL

Clyde
Reply to  Clyde
June 10, 2021 11:46 am

And we’re all still waiting on your answer explicating exactly how a higher-entropy object can spontaneously decrease the entropy of a lower-entropy object… if you can’t prove that mathematically, your entire argument collapses, and you’ll be forced to concede that your take on radiative physics is incorrect. I note you tap-danced away from that.

Clyde
Reply to  Trick
June 9, 2021 11:22 pm

Planck’s treatise The Theory of Heat Radiation opening paragraphs state that ‘heat rays’ are identical to ‘light rays’ of the same frequency, therefore ‘light rays’ and ‘heat rays’ are synonymous.

Now, light is composed of photons. Photons are quanta of energy, consisting of the electric and magnetic fields oscillating in quadrature (see below).

Photons also have no rest frame. Since photons have no rest mass, if a photon had a rest frame, no mass and no momentum equals nothing. Therefore photons must always be in motion.

Therefore “heat” is definitionally energy in motion. I seem to remember you denying this fundamental fact. LOL

———-
The sinusoidal ‘waves’ of photons are not actually waves… they’re spirals.comment imagecomment imagecomment image

The first image above shows the real (cosine… labeled ‘Re’ in the image) and imaginary (sine… labeled ‘Im’ in the image) components of an electromagnetic ‘wave’. When viewed in line with its direction of travel, it will appear to be a circle, and when viewed orthogonal to its direction of travel, it will appear to be a sinusoid, when in reality it’s a spiral.

This is because a sinusoid is a circular function.comment image

You’ll note the peak amplitude of the sinusoid is analogous to the radius of the circle, the peak-to-peak amplitude is analogous to the diameter of the circle, and the frequency of the sinusoid is analogous to the rotational rate of the circle. You’ll further note the circumference of the circle is equal to 2 pi radians, and the wavelength of a sinusoid is equal to 2 pi radians, so the wavelength of the sinusoid is analogous to the circumference of the circle.

Thus the magnetic field and electric field (oscillating in quadrature) of a photon is a circle geometrically transformed into a spiral by the photon’s movement through space-time. This is why all singular photons are circularly polarized either parallel or antiparallel to their direction of motion. This is a feature of their being massless and hence having no rest frame, which precludes their exhibiting the third state expected of a spin-1 particle (for a spin-1 particle at rest, it has three spin eigenstates: +1, -1, 0, along the z axis… no rest frame means no 0-spin eigenstate). A macroscopic electromagnetic wave is the tensor product of many singular photons, and thus may be linearly or elliptically polarized if all singular photons comprising the macroscopic electromagnetic wave are not circularly polarized in the same direction.

For a practical lab experiment, go outside on a sunny day and stretch out a Slinky so its shadow falls upon a surface perpendicular to the incoming sunlight… you’ll see the shadow of the spiral of the Slinky appears as a sinusoid. Now turn the Slinky so its axis is aligned parallel to the incoming light such that the light is falling through the center of it, you’ll see the shadow of the spiral of the Slinky appears as a circle. Our oscilloscopes show us a shadow of reality because they can only account for the electric field and not the magnetic field of electromagnetic radiation.

The above ties into vacuum polarization (due to the high charge density in the vicinity of the nucleus of an atom) creating a geometrical transform of resonant scalar quantum vacuum wave modes to a circular (spherical, given the DOF) orbital path of an atom’s bound electron(s) (ie: the bound electron ‘spirals’ around the nucleus, (acted upon by the Lorentz force of the EM interaction between bound electron and nucleal proton and sustained by energy from the quantum vacuum), which is why a bound electron must have an integer number of de Broglie waves in its orbit (the underlying reason for quantization of energy and hence the basis of Quantum Mechanics) or it sets up a destructive-interference orbit which lowers electron orbital radius, which is how and why electron orbital radius falls to ground state from a higher excited state when the excitation energy sustaining it in that higher orbital is removed). This is what feeds energy to a bound electron to prevent it ‘spiraling in’ to the oppositely-charged proton(s) in the nucleus. At its ground state, the energy obtained from the quantum vacuum exactly equals the energy emitted via virtual photons (magnetism… which all invariant-mass matter exhibits (usually diamagnetism, although certain electron valence configurations allow ferromagnetism to override the underlying diamagnetism)), as Boyer[1], Haisch and Ibison[2], Puthoff[3] and NASA[4] showed.

[1] https://journals.aps.org/prd/abstract/10.1103/PhysRevD.11.790
[2] https://web.archive.org/web/20190713220130/https://arxiv.org/ftp/quant-ph/papers/0106/0106097.pdf
[3] https://web.archive.org/web/20190713225420/https://www.researchgate.net/publication/13330878_Ground_state_of_hydrogen_as_a_zero-point-fluctuation-determined_state
“We show here that, within the stochastic electrodynamic formulation and at the level of Bohr theory, the ground state of the hydrogen atom can be precisely defined as resulting from a dynamic equilibrium between radiation emitted due to acceleration of the electron in its ground-state orbit and radiation absorbed from zero-point fluctuations of the background vacuum electromagnetic field, thereby resolving the issue of radiative collapse of the Bohr atom.”

[4] https://web.archive.org/web/20180719194558/https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150006842.pdf
“The energy level of the electron is a function of its potential energy and kinetic energy. Does this mean that the energy of the quantum vacuum integral needs to be added to the treatment of the captured electron as another potential function, or is the energy of the quantum vacuum somehow responsible for establishing the energy level of the ‘orbiting’ electron? The only view to take that adheres to the observations would be the latter perspective, as the former perspective would make predictions that do not agree with observation.”

This ties into the 2nd Law Of Thermodynamics (2LoT)… a bound electron is always trying to emit a photon to achieve a lower energy state, but the energy sustaining the bound electron in its current state prevents the photon being emitted because energy can only flow from a higher to a lower energy density region. When that excitation energy is removed, a photon can be emitted, electron orbit no longer has an integer number of de Broglie waves, a destructive-interference orbit is thus set up, and the electron falls to a lower state in which there are an integer number of de Broglie waves in the orbit. At ground state, energy flows from the quantum vacuum to sustain the electron in its ground state orbital as it emits Larmor radiation in the form of virtual photons (a point charge undergoing acceleration (in this case angular acceleration) will emit Larmor radiation), which it does because the quantum vacuum is anisotropic (it fluctuates) under vacuum polarization in the high charge density in the vicinity of the nucleus of an atom. Thus 2LoT holds even in the quantum realm.

This ties into the very underpinnings of the meta-stability of invariant-mass matter (and hence the continued existence of the universe as we know it) and provides insight into the connection between classical and quantum theory.
———-

Trick
Reply to  Clyde
June 10, 2021 7:44 pm

“Therefore “heat” is definitionally energy in motion. I seem to remember you denying this fundamental fact. LOL”

Here’s what I wrote: That is the anonymous writer of the encyclopedia defn. of heat and is different than Clausius’ defn. Try again. Look it up.

I see Clyde still hasn’t looked it up. 

Clyde
Reply to  Trick
June 10, 2021 11:05 pm

Clausius wrote:
“Heat cannot of itself pass from a colder to a hotter body.”
You claim it can.

You’ve claimed that energy (in the form of photons) can spontaneously flow from cooler to warmer, when you claimed photons from a cooler lab glass of ice water would flow to the warmer Ryobi thermopile.

Clausius wrote:
ΔS = Q((1/T_2) – (1/T_1))
Where:
ΔS = change in entropy
Q = amount of heat
T_1, T_2 = absolute temperature

The observation that heat never flows spontaneously from cold to hot is equivalent to requiring the net entropy change to be positive for a spontaneous flow of heat.

If T_1 = T_2, then the reservoirs are in equilibrium, no energy flows (there is no energy gradient, no energy can flow), and ΔS = 0.

That destroys your take on radiative physics. Now, you can either continue denying reality, making up your own definitions and rewriting history, or you can admit your mistake and learn.

The choice is yours.

Last edited 10 days ago by Clyde
Trick
Reply to  Clyde
June 11, 2021 7:14 am

I see Clyde still hasn’t looked up Clausius’ defn. of heat and started using it correctly causing Clyde to take his demonstrated sure route to befuddlement. If heat is energy in motion (as Clyde writes) and temperature (Th,Tc) is energy in motion, then it certainly seems to follow they are the same thing.

Clyde is still caught in the state of confusion prevailing over a century or two ago when the distinction between heat (Clyde’s erroneous defn.) and temperature (Clyde’s Th and Tc pix) was vague at best. Clyde still needs to up his thermodynamics game by reading and understanding the masters of the game.

Once Clyde begins to correctly use Clausius defn. of heat I can see Clyde moving on to understand and use entropy more correctly but it will take Clyde some time to make correct use of both terms.   

Clyde
Reply to  Trick
June 11, 2021 11:46 am

You’ve been shown that your take on radiative physics causes a higher entropy object to spontaneously lower the entropy of a lower entropy object. You should know that for irreversible processes, entropy can only either remain the same or increase. You should also know that all real-world processes are irreversible.

You’ve also been shown that if there is no energy gradient, no energy can flow.

You’ve also been shown that free energy is defined as the capacity to do work, and if there is no free energy, no energy can flow.

You’ve also been shown that the modern scientific definition of ‘heat’ is energy in motion, an energy flux.

Thus your take on radiative physics has been mathematically shown to be incorrect. yet you persist in denying reality, rewriting history and making up your own definitions.

In his 1851 treatise “On the Moving Force of Heat, and the Laws regarding the Nature of Heat itself which are deducible therefrom“, Clausius wrote:
The steam-engine having furnished us with a means of converting heat into a motive power, and our thoughts being thereby led to regard a certain quantity of work as equivalent for the amount of heat expended in its production, the idea of establishing theoretically some fixed relation between a quantity of heat and the quantity of work which it can possibly produce, from which relation conclusions regarding the nature of heat itself might be deduced, naturally presents itself.

He also wrote, in refuting Carnot’s assertion that heat is never diminished:
If we assume that heat, like matter, cannot be lessened in quantity, we must also assume that it cannot be increased; but it is almost impossible to explain the ascencion of temperature brought about by friction otherwise than by assuming an actual increase of heat.

And:
It may be remarked further, that many facts have lately transpired which tend to overthrow the hypothesis that heat is itself a body, and to prove that it consists in a motion of the ultimate particles of bodies. If this be so, the general principles of mechanics may be applied to heat; this motion may be converted into work, the loss of vis viva in each particular case being proportional to the quantity of work produced.

And:
These circumstances, of which Carnot was also well aware, and the importance of which he expressly admitted, pressingly demand a comparison between heat and work, to be undertaken with reference to the divergent assumption that the production of work is not only due to an alteration in the distribution of heat, but to an actual consumption thereof; and inversely, that by the consumption of work heat may be produced.

And:
Heat can only be measured by its effects; and of the two effects mentioned, mechanical work is peculiarly applicable here, and shall therefore be chosen as a standard in the following investigation.

What is work? Is it not mechanically defined as the product of the component of the force in the direction of motion, times the distance through which the force acts?

So while Clausius never actually defined ‘heat’ in his 1851 treatise, he most certainly did equate it to work, and the mechanical definition of work is W = |F| (cosθ) |d|.

IOW, heat is energy in motion, which is the modern definition (which you deny).

IOW, if no work can be done, no energy can flow.

But then, you’re attempting a back-door means of claiming that energy flow has nothing to do with work, aren’t you? Because that’s what it seems you’re doing.

Clyde
Reply to  Trick
June 11, 2021 1:16 pm

Who said that “temperature (Th,Tc) is energy in motion“? Certainly not I… that was you.

Temperature is a measure of energy density:
U = T^4 4εσ/c
The above formula is the Stefan-Boltzmann relation between energy density and temperature. U has the same physical units as pressure (J m-3) and U ∝ T. That pressure is radiation pressure.

As for temperature for macroscopic particles such as in a gas, it is a measure of the kinetic energy (and thus the kinetic energy density per given volume).

Oh, did you confuse ‘heat’ and ‘temperature’? Sure you did… that’s a common neophyte error. Perhaps, rather than grandstanding and posturing, rewriting history and making up your own definitions, misunderstanding standard definitions and generally flop-sweating all over the place, you should damp your hubris, buckle down, brush up on your definitions and actually learn something for once. LOL

Trick
Reply to  Clyde
June 11, 2021 3:17 pm

Clyde, I see you STILL haven’t looked up Clausius’ defn. of heat and used it correctly. It is Clyde’s “energy in motion” defn. for heat that equally applies to temperature so it is Clyde that has confused heat and temperature not Clausius. Also, and again, Clyde’s picture for Th & Tc confused heat and temperature.
 
Presumably Clyde can eventually catch on to Clausius’ original writings however it is beginning to appear that may not be achieved anytime soon. Though I am always hopeful. 

Last edited 9 days ago by Trick
Clyde
Reply to  Trick
June 11, 2021 4:29 pm

Wow… so you’re now claiming that the scientific definition of heat (an energy flux) is actually the definition of temperature, despite being shown that temperature is actually a measure of energy density… you’re fundamentally confused.

I wrote above of Clausius and his equating heat and work in his 1851 treatise… but you seem to be saying that energy transfer has nothing to do with work… is that what you’re saying?

Consider two reservoirs, in contact with each other, at T_1 = T_2, which direction is the energy moving between the two reservoirs, according to your take on energetic exchange, in which you imply that energy can flow even at thermodynamic equilibrium?

How is entropy changing with your purported energy movement?

Because every time energy moves for an irreversible process, entropy will increase, and all real-world processes are irreversible.

So you’re essentially stating that the system above will continue increasing in entropy, and since temperature can be defined in terms of entropy via: 1/T ☰ (∂S/∂E)ₙ,ᵥ, you claim that the system will spontaneously cool even if it’s not losing energy to its surroundings, meaning energy must be destroyed.

Are you sure you’ve thought through your stance? LOL

Trick
Reply to  Clyde
June 11, 2021 4:46 pm

“is that what you’re saying?”

What I am evidently saying is Clyde’s “energy in motion” defn. for heat equally applies to temperature so it is Clyde that has confused heat and temperature not Clausius. And yes Clyde, Clausius does write out the definition of heat. I’m sure Clyde can find it, just dig a little harder since reading up on the subject is good practice.
 
I am consistently saying I see Clyde still hasn’t looked up Clausius’ defn. of heat, temperature and started applying physics correctly. This would be the first step for Clyde to understand for two reservoirs, in contact with each other, at T_1 = T_2, the direction energy is moving between the two reservoirs.

Clyde
Reply to  Trick
June 11, 2021 5:45 pm

So you’re now claiming that temperature is defined as an energy flux… just as heat is scientifically defined as an energy flux.

You’ll get right on presenting the mathematics of your supposition. Otherwise, you’ll be forced to admit that you’ve gotten the definitions wrong, that’s led you to false premises, which has led you to incorrect conclusions, which has compelled you to embarrass yourself online.

And we’re still waiting on your mathematical proof of exactly how a higher entropy object can spontaneously decrease the entropy of a lower entropy object. Your failure to prove this mathematically will lead you to admit the same as above. Your tap-dancing away from that mathematical proof twice now doesn’t bode well for your stance on radiative physics.

For two reservoirs, in physical contact or not, at T_1 = T_2, no energy flows. Your inability to admit to this reality doesn’t bode well for your knowledge of physics in general.

Last edited 9 days ago by Clyde
Clyde
Reply to  Clyde
June 12, 2021 12:47 am

In Clausius’ 9th memoir in The Mechanical Theory of Heat: With Its Applications to the Steam-Engine and to the Physical Properties of Bodies, he wrote:

If we denote this quantity of heat by Q, a quantity of heat given off by the body being reckoned as a negative quantity quantity of heat absorbed, then the following equation holds for the element dQ of heat absorbed during an infinitesimal change of condition,

dQ = dU + AdW

Here U denotes the magnitude which I first introduced into the theory of heat in my memoir of 1850, and defined as the sum of the free heat present in the body, and of that consumed by interior work. Since then, however, W. Thomson has proposed the term energy of the body for this magnitude, which mode of designation I have adopted as one very appropriately chosen…

heat given off
heat absorbed
Both describe energy in motion.

free heat” equates to free energy, the capability of doing work.

And what do you know, Clausius adopted Thomson’s use of the word energy to denote what he previously termed “free heat“.

So even Clausius eventually came around to the realization that heat is an energy flux.

One must wonder when Trick will achieve the same? LOL

Now, Clausius also wrote in his first memoir:
We shall forbear entering at present on the nature of the motion which may be supposed to exist within a body, and shall assume generally that a motion of the particles does exist, and that heat is the measure of their vis viva.

Vis viva‘ is the one-half the product of an object’s momentum and the square of its speed… or what we now call kinetic energy.

So Clausius did indeed confuse ‘heat‘ and ‘temperature‘ in his first memoir.

Heat‘ is definitionally an energy flux. Temperature‘ is a measure of that energy, in Clausius’ case kinetic energy.

U = T^4 4εσ/c
The above formula is the Stefan-Boltzmann relation between energy density and temperature. This is why, for instance, as one reduces the volume of a gas (compresses the gas), the ‘temperature’ increases (the kinetic energy per volume increases because the volume decreases)… the ‘heat’ (energy flow into or out of that gas) does not necessarily increase.

Trick clings to an erroneous definition from 171 years ago (because it supports his narrative that energy can flow willy-nilly without regard to energy gradient, which supports his ‘backradiation’ narrative, which supports his CAGW narrative) rather than using modern definitions… a disingenuous attempt at sustaining CAGW, which is on its deathbed.

AC Osborn
Reply to  Clyde
June 12, 2021 1:31 am

Reading this exchange shows someone who knows Physics down through the Molecular, Atomic and Quantum levels arguing the Science with someone who doesn’t.
All the someone who doesn’t understand it to the samel level can do is continually repeat the same irrelevant remark without adding anything of physics to the debate.
I think the term is “schooled”.

Clyde
Reply to  Clyde
June 12, 2021 3:23 am

Mhem…
“momentum”
-should be-
“mass”

Clyde
Reply to  Clyde
June 12, 2021 5:47 am

Now, let us examine the temperature / pressure / energy relation in an atom / molecule sense and in relation to photons:

An atom or molecule moving in a single translational mode DOF will have a higher temperature for the same kinetic energy than if it’s moving in more than a single translational mode DOF.

KE = (DOF / 2) k_B T
T = (2 KE) / (DOF k_B)

p = −1 / 3 * (the components of the stress tensor)

The 1/3 assumes that the pressure is equipartitioned in all 3 DOF (ie: static pressure). In reality, a more accurate equation for dynamic pressure would be:

p = -1 / DOF * (stress_tensor_x + stress_tensor_y + stress_tensor_z)

static pressure + dynamic pressure = total pressure
p + q = p_0
Where: p = static pressure (Pa); q = dynamic pressure (Pa); p_0 = total pressure (Pa)

Bernoulli’s Principle states that if dynamic pressure increases, static pressure must decrease. In other words, for a flowing fluid, it is trading static pressure (in 3 DOF) for dynamic pressure (in less than 3 DOF). Thus for a compressible fluid, static temperature orthogonal to the plane of flow will decrease, while stagnation temperature in the plane of flow will increase.

The equation for dynamic pressure: q = 1/2 p v²
Where: q = dynamic pressure (Pa); p = fluid density (kg/m³); v = velocity (m/s)

In statistical mechanics the following molecular equation is derived from first principles: P = n k_B T for a given volume.

Therefore T = (P / (n k_B)) for a given volume.

Where: k_B = Boltzmann Constant (1.380649e−23 J·K−1); T = absolute temperature (K); P = absolute pressure (Pa); n = number of particles

If n = 1, then T = P / k_B in units of K / m³ for a given volume.

Now, knowing that you’re a pedant, Trick, you’ll likely bleat something like “Temperature does not have units of K / m³ !!!“… note the ‘for a given volume‘ blurb. We will cancel volume in a bit.

We can relate velocity to kinetic energy via the equation:
v = √(v_x² + v_y² + v_z²) = √((DOF k_B T) / m) = √(2 KE / m)
As velocity increases, kinetic energy increases.

Kinetic theory gives the static pressure P for an ideal gas as:
P = ((1 / 3) (n / V)) m v² = (n k_B T) / V

Combining the above with the ideal gas law gives:
(1 / 3)(m v²) = k_B T

∴ T = mv² / 3 k_B for 3 DOF

∴ T = 2 KE / k_B for 1 DOF

∴ T = 2 KE / DOF k_B

See what I did there? I equated kinetic energy to pressure over that volume, thus canceling that volume, then solved for T.

———-

Now, that’s for particle kinetic energy, and we can see that temperature and pressure are intimately linked. What about photons?

The pressure P for a photon gas exerted in the x-direction on area A of the wall is summed over all i = 1 to N photons:

P = ½ ∑ 2 pix vix / V = ⅓ U/V = ⅓ e,

U is just ∑ pix vix + piy viy + piz viz for photons.

The equation for the radiation energy density is Stefan’s Law and a is Stefan’s constant.
e = aT^4

∴ T = 4^√(e/a)

In other words, temperature is equal to the fourth root of energy density divided by Stefan’s constant. It most certainly is not “energy in motion” as you claim. It is a measure of energy density.

Keep in mind that Stefan’s constant above equals 4σ/c (which is sometimes known as the radiation constant), and ε is the emissivity modifier for graybody objects.

Which is why: U = T^4 4εσ/c
The above formula is the Stefan-Boltzmann relation between energy density and temperature.

This agrees with Planck’s Law: ρ(T) = aT4 = T^4 4εσ/c, when including the graybody emissivity modifier ε.

Now that we see that all are in agreement, let us move on to chemical potential…

Among the properties of the cavity with volume V in radiative thermal equilibrium at temperature T is that:
U = a V T^4
P = ⅓ a T^4
S = (4/3) a V T^3

We can calculate the chemical potential, µ, which measures the ease with which the number n moles of photons adjusts to keep the energy density constant in the cavity in radiative thermal equilibrium:
µn = U – ST + PV
µ = 0

When photons are in radiative thermal equilibrium in a volume V at a constant temperature T, their chemical potential is zero.

Chemical potential is a measure of the ability to do work. If chemical potential equals zero, no work can be done, thus no energy can flow.

Also remember that there is no conservation law for photons. Since photons are merely persistent perturbations of the EM field above the ambient, should the ambient EM field chemical potential exceed the photon chemical potential, those photons will be subsumed by the ambient EM field.

Also remember that between two objects, one warmer and one cooler, the chemical potential increases in the intervening space as one ascends the energy gradient from cooler to warmer object. Thus photons from the cooler object will be subsumed before they ever reach the warmer object because warmer objects have higher energy density at all wavelengths than cooler objects.
comment image

See how simple it is? LOL

Trick
Reply to  Clyde
June 12, 2021 7:18 am

Very good Clyde, you made progress since you have finally found Clausius’ defn. for heat which is: the measure of (all the particles of a body) kinetic energy.

Heat is defined as a measure by Clausius! More precisely note that to Clausius heat is the measure of the total KE of the object’s particles.

So, contrary to Clyde’s statement, Clausius did NOT confuse heat with temperature which is: the local average KE of the object’s particles.

Note the difference in the words “total KE” for heat and “local average KE” for temperature. So, it is Clyde that confused heat and temperature when Clyde erroneously advocated “energy in motion” defn. which is true for both heat and temperature.

—–

“For two reservoirs, in physical contact or not, at T_1 = T_2, no energy flows.”

Now let’s use Clyde’s new enlightenment to examine Clyde’s erroneous conclusion. First let’s use one side of Clyde’s “or” for solids:

Per Clyde: “For two reservoirs, in physical contact, at T_1 = T_2, no energy flows.”

Clyde now knows to Clausius heat cannot flow between the two reservoirs because their constituent particles are not exchanged (& this would hold true even if T_1 .GT. T_2). The measure of the total KE of the particles in each reservoir certainly can change though when T_1 .GT. T_2 but per Clausius that measure of the total particle KE does NOT flow or even transfer. Just like an object does not contain work, an object does not contain heat; per Clausius the object does contain particle KE.
 
Clyde also now has learned the temperature is an average of the KE of the object’s particles. So, if a particle in one solid fluctuates to a higher speed while at same time coming into contact with a lower speed particle in the other solid then energy DOES transfer between the solids (and vice versa) even when T_1 = T_2 thus invalidating Clyde’s statement. Overall during the process net energy does not transfer.
 
This real energy exchange process also invalidates ALL of Clyde’s discussion about “energy gradients”.

There exists the birth of photons, life of photons, and death of photons. Photons are not subsumed nor incided which also invalidates Clyde’s discussion on that subject.

My IR thermometer works fine reading out ice water at 32F brightness temperature and boiling water at 212F brightness temperature from emission (birth) and absorption (death) of photons with a very short life time. 

Last edited 8 days ago by Trick
AC Osborn
Reply to  Trick
June 12, 2021 8:57 am

Where is the link to your reference?

If no energy can flow between 2 object of equal potential how can energy flow from an object with even lower potential to the same higher potential object?

Trick
Reply to  AC Osborn
June 12, 2021 9:56 am

AC 8:57am, good questions. There used to be an easy link found on the internet to Clausius’ 1st memoir for his defn. of heat is a measure but I can’t find it any more. Your own google fu is necessary if you mean linking to a Clausius ref. & if you mean another ref. please advise.

Clyde gets it wrong that no energy flows between two same temperature solids in contact so not sure if your question is to me or Clyde. Energy can transfer on contact from an object with lower temperature to a higher temperature object for the same reasons as I noted since temperature is an average of constituent particle KE & 2LOT allows this since 2LOT applies to mandate entropy increase for the universe in any real process.  

AC Osborn
Reply to  Trick
June 12, 2021 11:38 am

I am talking about Radiation transfer of energy.
All the transfer equations say that there is no flow between equally hot oblects and no temperature change in the upward direction.

Trick
Reply to  AC Osborn
June 12, 2021 1:49 pm

AC 11:38 am, then refer to Planck (not Clyde’s subsume radiation):

https://www.gutenberg.org/files/40030/40030-pdf.pdf

“A body A at 100◦ C. emits toward a body B at 0◦ C. exactly the same amount of radiation as toward an equally large and similarly situated body B0 at 1000◦ C. The fact that the body A is cooled by B and heated by B0 is due entirely to the fact that B is a weaker, B0 a stronger emitter than A.”

So apply Planck radiation statement to my IR thermometer at room temperature 72F viewing and reading out either 32F ice water or 212F boiling water and see what you get.

The transfer eqn.s are used for calculating net energy transfer. 

Clyde
Reply to  Trick
June 12, 2021 3:35 pm

The sentence immediately preceding the paragraph you quoted:
But the empirical law that the emission of any volume-element depends entirely on what takes place inside of this element holds true in all cases (Prevost’s principle).

So you are attempting to resuscitate the Prevost Theory of Exchanges, which has long been chucked on the midden-heap of scientific hypotheses. LOL

The Prevost Theory of Exchanges claimed that energy is transferred equally back and forth regardless of energy density, it’s from 1791, it assumes radiation is an actual fluid, and it’s long-debunked, first replaced by the Kinetic Theory of Heat, then by Quantum Thermodynamics.

My, oh, my, oh, my… that doesn’t bode well for your entire take on radiative exchange and energy exchange in general, your knowledge of physics, your ability to reason, and it absolutely destroys your “energy can flow willy-nilly without regard to energy gradient” narrative, which destroys your “backradiation” narrative, which destroys your “CAGW” narrative. LOL

Last edited 8 days ago by Clyde
Clyde
Reply to  Clyde
June 12, 2021 5:04 pm

Planck correctly stated:
Conduction of heat depends on the temperature of the medium in which it takes place, or more strictly speaking, on the non-uniform distribution of the temperature in space, as measured by the temperature gradient.

In other words, energy can only flow (the definition of heat) via conduction if there is a temperature (and therefore an energy density) gradient.

Where Planck erred is in his clinging to the Prevost Theory Of Exchanges in regard to radiative energy, which led him to eschew scientific reality (that energy only flows if there is an energy gradient), to wit:
But the empirical law that the emission of any volume-element depends entirely on what takes place inside of this element holds true in all cases (Prevost’s principle).

The long-debunked Prevost Theory of Exchanges (first replaced by the Kinetic Theory of Heat, then by Quantum Thermodynamics) assumed that energy flowed without regard to energy gradient. This led Planck to make the further incorrect assumption in keeping with the Prevost Theory of Exchanges:
We shall now introduce the further simplifying assumption that the physical and chemical condition of the emitting substance depends on but a single variable, namely, on its absolute temperature T.

So unless he was speaking strictly of idealized blackbody objects (and not real-world graybody objects), Planck made a mistake. Given that Planck was discussing emissivity in the very paragraph where he mentions Prevost’s principle, that means he made a mistake… he wasn’t discussing idealized blackbody objects.
comment image

The image above shows that while idealized blackbody objects do not take into account the energy gradient, real-world graybody objects most certainly do. Of course, idealized blackbody objects are just that, idealizations… they don’t actually exist.

So in the real world, the energy gradient determines radiant exitance, energy does not flow willy-nilly without regard to energy gradient and 2Lot applies always and everywhere.
comment image

So Carnot erred in assuming that heat is never consumed as work; Clausius erred in attributing ‘heat’ to the energy density of an object (‘heat’ is definitionally an energy flux, ‘temperature’ is a measure of energy density, see maths above); and Planck erred in clinging to a long-debunked radiative model, and his follow-on assumptions stemming from that led to his treating real-world objects as though they emit willy-nilly without regard to the energy gradient… even the greats of science make mistakes.

Good thing I’m around to correct them. LOL

Clyde
Reply to  Trick
June 12, 2021 2:31 pm

It wasn’t Clausius’ “definition” of heat, it was his misattribution to heat the definition of temperature… that you’ve glommed onto it because it supports your narrative, refusing to give that erroneous definition up because that destroys your ‘energy can flow willy-nilly without regard to energy gradient’ narrative, which destroys your ‘backradiation’ narrative, which destroys your ‘CAGW’ narrative.

The ‘total kinetic energy’ is merely the ‘local average kinetic energy’ times the number of atoms or molecules. You’re arguing semantics, without even knowing what you’re arguing about. LOL

Now you’re not only stating that energy can flow willy-nilly even when there’s no energy gradient, but that it cannot flow when there is an energy gradient. You’re so fundamentally confused that you’re diametrically opposite to reality. LOL

And all this while you admit that temperature is a measure of the energy of an object (just as I’ve stated… but you forget that it doesn’t necessarily have to be kinetic energy… a photon gas has temperature but no kinetic energy in the classical sense… remember, under classical physics, the photon is an impossibility… why aren’t you arguing that photons don’t even exist, since you insist upon clinging to classical physics. LOL) and is thus a measure of energy density (because whatever is being measured has a certain volume, hence energy per volume, as the maths show), and while you claim that temperature means ‘energy in motion’… so you must now claim that kinetic energy is necessarily “energy in motion” because you don’t understand that in the case of kinetic energy, it’s not the energy that’s in motion, it’s the atoms or molecules constituting the object in question.

And now you’re talking about ‘birth, life and death’ of photons, in what appears to be some attempt to anthropomorphize them, but you refuse to acknowledge that photons are a component of the EM field (a persistent perturbation above the ambient EM field), and are thus subsumed by the ambient EM field when ambient EM field chemical potential exceeds photon chemical potential; and photons are manifested from the ambient EM field when ambient EM field chemical potential of a region is above zero as compared to some other region or object.

Trick wrote:
So, if a particle in one solid fluctuates to a higher speed while at same time coming into contact with a lower speed particle in the other solid then energy DOES transfer between the solids (and vice versa)

So in your classical-physics limit, you must admit that the amount of energy exchanged particle-to-particle can only equalize the kinetic energy between the two particles (ie: atoms or molecules), and thus eventually, barring the extremely rare constructive interference which is nearly immediately smeared out over neighboring particles, all particles constituting the solid come to receive the same kinetic energy, and thus no energy can flow… so you’ve essentially stated that “thermodynamic equilibrium” means there is an energy gradient… you seem fundamentally confused. LOL

You’re arguing from the viewpoint of the classical physics-framed time-averaged statistical probability distribution treatment of an ensemble of quantum interactions of a scale sufficient to exhibit classical behavior. You can’t prove anything about instantaneous per-interaction quantum processes with that. Quantum Thermodynamics has moved beyond time-averaged statistical probabilities.

The only reason classical physics claims that molecules emit / absorb across an ensemble simultaneously is because the Kinetic Theory Of Heat is a classical physics-based time-averaged statistical equilibrium / ensemble probability theory (as Einstein admitted in his writings). It’s merely a statistical model to understand the processes, now superseded by Quantum Thermodynamics. The individual particles and the body they comprise do not operate that way in the real world.

Now let’s analogize that to photons… remember, photons are merely quanta of energy, and we’re talking of the flow of energy.

In reality, each object’s surface atoms’ or molecules’ dipole oscillations set up a surface EM field which manifests a radiation pressure. The energy density differential between cooler and warmer object sets up an energy gradient, and energy can only flow down the energy gradient from warmer to cooler. Energy from a cooler object (lower radiation pressure) does not and cannot ascend the energy gradient toward a warmer object (higher radiation pressure), because warmer objects have higher energy density at all wavelengths than cooler objects.

This app is uses the Schrodinger equation to compute energetic transfer from free space to an object:
https://physics.weber.edu/schroeder/software/BarrierScattering.html
– Set it to whatever energy level you want, as long as ‘Wavepacket Energy’ and ‘Barrier Energy’ are equal.

– Set the ‘Width’ slider to maximum. This will be akin to free space to the left, and an object to the right of the display.

– Set ‘Ramp’ to some arbitrary value (the higher the energy density of the barrier, the wider the energy gradient (ie: the ‘ramp’).

– Select the ‘Density / Phase’ option-box.

If you’re doing it right, you’ll see the free space wavemode nodes are always at the ‘ramp’, and you’ll see the wavemodes set up a standing wave (which will vary in space because ‘Wavepacket Energy’ has an uncertainty added to it). Note that even if ‘Wavepacket Energy’ is equal to ‘Barrier Energy’, no energy is imparted to the barrier (ie: the object). If the code glitches, start with ‘WavePacket Energy’ lower than ‘Barrier Energy’, then ramp it up after it’s running.

This is akin to stating that a particle (atom or molecule) in an excited quantum state cannot absorb energy which is equal to the energy which excited that quantum state, unless the particle has another quantum state available to be excited which is equal to that energy.

The electric field of a non-resonant photon, ~100,000,000 times smaller than the Coulomb field seen by the bound electron(s), slightly changes the phase of the bound electron(s), just as it changes the phase of the incident photon… no energy is exchanged if the photon is not absorbed, only the phase is shifted, altering the vector of the photon. This is how reflection from a step potential works. This is dependent upon the differential between energy density of the photon and atom / molecule. In the case of absorption, the photon constructively interferes with the available resonant quantum states of the atom or molecule. No phase shift.

As Hendrik Kramers wrote:
Another point which, this time, lies quite within the domain where our formulae are significant refers to the scattering of light by an elastically bound electron. Consider again our assembly of harmonic oscillators of frequencies k which we may imagine to take the discrete but very finely distributed values determined by [ kL – η(k) = π N (N = 1, 2, 3… ) ]. Let all of them be in the ground state (energy 3/2 ħ k) with the exception of one (frequency k’) which has the energy 5/2 ħ k’, the vibration parallel to – say – the x-axis being excited by one quantum ħ k’. In this situation light of frequency k’ coming from all directions is continuously scattered by the electron.

…the light quanta in the external field… which are affected by the presence of the electron through the appearance of the phase shift η. We might also call them phase shifted light quanta

In an ideal blackbody cavity (remember that blackbody radiation used to be called cavity radiation) at thermodynamic equilibrium, quantized standing waves are set up. The wavemode nodes are always at the walls of the cavity in this case. Because the walls and the cavity space’s EM field are of equal energy density, photons cannot be absorbed by the walls, nor can they be emitted. The photons in the standing waves are reflected from the walls. If they were absorbed / re-emitted, that would result in double the energy density in the cavity space. This is where the conventional explanation of energetic exchange falls down… it violates Stefan’s Law.

Likewise, between two objects at thermodynamic equilibrium, each object emits blackbody radiation, which sets up a standing wave between the objects. No energy is exchanged, that’d violate 1LoT, 2LoT, Stefan’s Law, etc. Neither object can do work upon the other at thermodynamic equilibrium, thus no energy can be transferred between them.

Thus, photons from one object won’t be absorbed by the other, they’ll be reflected. This increases the energy density of the standing wave until it equals that of the objects. Thus the objects cannot emit nor absorb photons, they can only reflect the photons still in the standing wave from when the objects were approaching thermodynamic equilibrium.

If one object changes temperature, the standing wave will travel toward the object of lower temperature, with the group velocity of that traveling wave proportional to the temperature differential of the two objects, the operating principle behind it all being radiation pressure.

In this circumstance, the photon has a chemical potential of zero. It can do no work. Free energy is zero. If no work can be done, no energy can flow.

Trick
Reply to  Clyde
June 12, 2021 4:19 pm

Politician Clyde writes 12:47 am last paragraph his political manifesto and repeats at 2:31 pm that I have a narrative. Sorry, Clyde, I have no narrative other than the narrative & math writings of the thermo. grandmasters.

Like the OP wrote 8:06 am, it’s not going to be fruitful discussing science hard earned by the masters in the field with a politician like Clyde, good luck and happy trails to Clyde.

—–

Some notes before I go:

NB1: “The ‘total kinetic energy’ is merely the ‘local average kinetic energy’ times the number of atoms or molecules.”

No. The particle measure of ‘total kinetic energy’ (heat) is NOT merely the ‘local average kinetic energy’ (temperature) times the number of atoms or molecules as the local temperature may not be the object’s average local temperature. The global earth near surface atm. average is ~289K which will not compute from using the temperature at the S. Pole station.

NB2: “a photon gas has temperature but no kinetic energy in the classical sense…In an ideal blackbody cavity…

No. Blackbodies do not exist Clyde.

Photons in any opaque cavity at dry ice to superheated steam equilibrium temperatures do not interact with each other, so the equilibrium distribution of a photon gas can come about only because of interactions of photons with the walls of the container. Radiation in politician Clyde’s perfectly reflecting container (which does not exist) would never evolve to the equilibrium distribution if it did not initially have this distribution. These points are made by F. E. Irons, 2004: Reappraising Einstein’s 1909 application of fluctuation theory to Planckian radiation. American Journal of Physics, Vol. 72, pp. 1059–67.

NB3: “so you’ve essentially stated that “thermodynamic equilibrium” means there is an energy gradient”

No. Clyde stated those words.

NB4: “photons are merely quanta of energy

No.  An electromagnetic wave is just as much a thing as a photon: both possess energy and momentum (linear and angular) but not, it seems, mass, thus light can exert radiation pressure, that is, transfer momentum to illuminated objects.

NB5:  “Energy from a cooler object (lower radiation pressure) does not and cannot ascend the energy gradient toward a warmer object (higher radiation pressure),”

No. See Planck link at 1:49pm: “A body A at 100◦ C. emits toward a body B at 0◦ C. exactly the same amount of radiation as toward an equally large and similarly situated body B0 at 1000◦ C. The fact that the body A is cooled by B and heated by B0 is due entirely to the fact that B is a weaker, B0 a stronger emitter than A.”

NB6: “It wasn’t Clausius’ “definition” of heat””

If Clausius defining heat can’t get through to politician Clyde, no one can.

Adios amigo. 

Clyde
Reply to  Trick
June 12, 2021 5:36 pm

‘Politician’… I’ve been called worse by better than you. LOL

You deny the definition of total kinetic energy being the average local kinetic energy times the number of particles! You deny the definitions of ‘heat’ and ‘temperature’ despite being shown mathematically that you are wrong! You deny that photons (which are a persistent perturbation of the ambient EM field above that ambient level) can interact with the ambient EM field! You deny that photons are quanta of energy! You don’t know that an electromagnetic ‘wave’ is merely the tensor product of many singular photons! You deny the precepts of 2LoT! You deny the reality that energy only flows due to an energy gradient! You cling to Planck’s incorrect statement, which is predicated upon the long-debunked Prevost Theory of Exchanges, which assumes energy flows without regard to energy gradient! You deny that the greats of science can get things wrong (Carnot assumed heat is never diminished in producing work, Clausius incorrectly attributed the definition of temperature to heat, Planck used the long-debunked Prevost Principle to assume that all objects emit energy without regard to energy gradient, whereas he correctly stated that conduction of energy can only occur due to an energy gradient)!

And you refuse to amend your incorrect knowledge… because that incorrect knowledge bolsters your ‘energy can flow willy-nilly without regard to energy gradient‘ narrative, which bolsters your ‘backradiation‘ narrative, which bolsters your ‘CAGW‘ narrative.

And now that I’ve shown you your multitude of errors and pinned you down as to why you cling to those errors, you’re running away! LOL

CAGW is well and truly dead… all that remains is unlatching the leeches still feeding off its rotting carcass. LOL

Clyde
Reply to  Trick
June 14, 2021 4:59 pm

Trick wrote:
Photons in any opaque cavity at dry ice to superheated steam equilibrium temperatures do not interact with each other,

Photons are a component of the ambient EM field (a persistent perturbation of the EM field above the ambient level)… so unless you’re claiming that the EM field cannot interact with the EM field (and you’re thus claiming that electromagnetism cannot interact with electromagnetism), you and F.E. Irons (who was a Mechanical Engineering student at School of Aerospace, Civil and Mechanical Engineering, University of New South Wales under Professor R. K. Duggins (and definitely not a quantum physicist) and who wrote a total of 8 papers) are wrong.

According to you, the weak-measurement double-slit experiment is just a hallucination (in reality, it’s a result of the photon interacting with itself and the EM component of the quantum vacuum… and if a photon can interact with itself, it most certainly can interact with other photons); the sun doesn’t gravitationally collapse because magical fairies sprinkle pixie dust on it (in reality, it is the radiation pressure brought about by photon interaction that prevents its gravitational collapse); photon gas has no pressure (a photon gas most certainly does have pressure, and given that it’s comprised of only photons, that pressure arises due to photon-photon interaction); the speed of light in vacuum is what it is because the flying spaghetti monster just likes it that way (in reality, light interaction with the CMB slows the speed of light from essentially infinity to 299,792,458 m/s). LOL

The photon interaction is with the ambient EM field and its energy density. If the ambient EM field (due to an energy gradient) has higher chemical potential than the photons in that same space, the photons will no longer be persistent perturbations above the EM field ambient, they’ll be subsumed, they’ll disappear back into the ambient EM field.

Trick wrote:
so the equilibrium distribution of a photon gas can come about only because of interactions of photons with the walls of the container.

“Interaction” doesn’t necessarily imply “absorption / emission”… electromagnetism is a force, mediated by photons and virtual photons, interaction can take place at a distance… unless you’re now claiming that electrical generators work by smacking the rotor and stator together. LOL

Trick wrote:
Radiation in politician Clyde’s perfectly reflecting container (which does not exist) would never evolve to the equilibrium distribution if it did not initially have this distribution. These points are made by F. E. Irons, 2004: Reappraising Einstein’s 1909 application of fluctuation theory to Planckian radiation. American Journal of Physics, Vol. 72, pp. 1059–67.

Of course it initially has this distribution… first, do you think thermal and radiative equilibrium just comes about in an instant without equilibration? The standing waves in the cavity space are those that are left as the cavity walls come into thermal equilibrium. Second, photons can interact at-distance, a photon need not incide upon a surface in order to interact, and that interaction need not exchange energy.

F.E. Irons makes a grave mistake… his Section IV statement implies that his closed system is comprised solely of photons, with no matter. It’s well known that radiation sans matter cannot generally equilibrate, but radiation can equilibrate via interaction with matter (and again, that ‘interaction’ doesn’t necessarily imply absorption / emission)… are you now claiming that a blackbody cavity’s walls are not comprised of matter? LOL

The electric field of a non-resonant photon, ~100,000,000 times smaller than the Coulomb field seen by the bound electron(s), slightly changes the phase of the bound electron(s), just as it changes the phase of the incident photon… no energy is exchanged if the photon is not absorbed, only the phase is shifted, altering the vector of the photon. This is how reflection from a step potential works. This is dependent upon the differential between energy density of the photon and atom / molecule. In the case of absorption, the photon constructively interferes with the available resonant quantum states of the atom or molecule. No phase shift.

As Hendrik Kramers wrote:
Another point which, this time, lies quite within the domain where our formulae are significant refers to the scattering of light by an elastically bound electron. Consider again our assembly of harmonic oscillators of frequencies k which we may imagine to take the discrete but very finely distributed values determined by [ kL – η(k) = π N (N = 1, 2, 3… ) ]. Let all of them be in the ground state (energy 3/2 ħ k) with the exception of one (frequency k’) which has the energy 5/2 ħ k’, the vibration parallel to – say – the x-axis being excited by one quantum ħ k’. In this situation light of frequency k’ coming from all directions is continuously scattered by the electron.

…the light quanta in the external field… which are affected by the presence of the electron through the appearance of the phase shift η. We might also call them phase shifted light quanta

In an ideal blackbody cavity (remember that blackbody radiation used to be called cavity radiation) at thermodynamic equilibrium, quantized standing waves are set up. The wavemode nodes are always at the walls of the cavity in this case. Because the walls and the cavity space’s field are of equal energy density, photons cannot be absorbed by the walls, nor can they be emitted (photons are nothing but energy, and energy cannot flow unless work can be done, since at thermal equilibrium, photon chemical potential is zero). The photons in the standing waves are reflected from the walls. If they were absorbed / re-emitted, that would result in double the energy density in the field. This is where the conventional explanation of energetic exchange falls down… it violates Stefan’s Law.

Likewise, between two objects at thermodynamic equilibrium, each object emits blackbody radiation, which sets up a standing wave between the objects. No energy is exchanged, that’d violate 1LoT, 2LoT, Stefan’s Law, etc. Neither object can do work upon the other at thermodynamic equilibrium, thus no energy can be transferred between them.

Clyde
Reply to  Trick
June 14, 2021 2:22 pm

You’re so fundamentally confused, your logic so twisted, that you’ve backed yourself into a corner you can’t backpedal your way out of… you’ve claimed that temperature is defined as ‘energy in motion‘, an energy flux (which is actually the definition of heat, but we’ll go with your erroneous definition of temperature for the moment to prove a point).

Now, at thermodynamic equilibrium, since no energy can flow, that would mean both of the objects which are at thermodynamic equilibrium with each other would have an absolute temperature of zero.

Right? Anytime energy cannot flow, absolute temperature must be zero according to your definition of temperature.

Awww, did you think this was my first rodeo? I’ve seen your kind so many times before that I’ve lost count. Refuting your cartoon-science contortionism is so easy for me that I can do it in my sleep.

Say, you wouldn’t have happened to have used the pseudonyms ‘Dave Burton‘ (not the skeptic Dave Burton… the kook who misappropriated his name because the skeptic Dave Burton mortally wounded the poor kook’s fragile little psyche by proving him wrong) or ‘evenminded‘ at any time in the past, would you have? Because your particular brand of blatherskite seems very familiar… or perhaps the scientifically divorced-from-reality all exhibit the same dullardry. LOL

Weekly_rise
Reply to  Stephen Wilde
June 6, 2021 7:17 am

The conversion from kinetic to potential energy in lofting air masses is balanced by the opposite conversion in falling air masses – these things are occurring simultaneously. The net kinetic energy vs potential energy in the system is not changing. The only way for the earth system to lose energy is via radiation to space, so it is the radiative balance at the top of the atmosphere that determines the planetary temperature.

Last edited 14 days ago by Weekly_rise
Reply to  Weekly_rise
June 6, 2021 7:33 am

I agree that at hydrostatic equilibrium the total amount of potential energy is constant. It also follows that total KE plus total PE is also a constant because KE becomes PE with height.
Anything that attempts to change those constants will destroy hydrostatic equilibrium and the atmosphere will be lost. That is why the radiative theory is a physical impossibility.
What happens in reality is that any destabilising influence from back radiation is neutralised by convective adjustments which are made possible by the changes in lapse rate slopes caused by those destabilising influences. Those adjustments restore hydrostatic equilibrium in the face of massive disruptions such as from volcanoes or meteorite strikes. Radiative imbalances are trivial in comparison.
Convection always changes to balance energy going out with energy coming in.

Weekly_rise
Reply to  Stephen Wilde
June 6, 2021 8:22 am

“Convection always changes to balance energy going out with energy coming in.”

It is not possible for convection to maintain earth’s energy balance with sunlight because convection does not carry energy away from the earth system.

Reply to  Weekly_rise
June 6, 2021 9:35 am

Convection controls the rate at which the kinetic energy from the initial ascent is returned to the surface in descent ready for radiation to space.
Therefore, it does control the rate at which energy flows through the system and can adjust as necessary.

philincalifornia
Reply to  Weekly_rise
June 6, 2021 9:53 am

That doesn’t seem right. That would imply that radiation to space does not vary with altitude, wouldn’t it?

Reply to  philincalifornia
June 6, 2021 10:03 am

If radiative gases are present then there must be radiation to space from within the atmosphere and it would vary with the temperature at the radiating height.
For a largely transparent atmosphere such as that of Earth the vast majority goes from the surface.

philincalifornia
Reply to  Stephen Wilde
June 6, 2021 11:08 am

I thought it would be implicit in my question that I was talking about 13 – 18 micron wavelength photons and, of course, I’m ignoring water vapor. My question related to convection and altitude, linearity versus logarithmic. I’m trying to learn more about it.

Bob Wentworth
Reply to  Stephen Wilde
June 7, 2021 5:20 pm

For a largely transparent atmosphere such as that of Earth the vast majority goes from the surface.

That’s true if you consider 17% to be the “vast majority.”

Do you ever check facts?

DocSiders
Reply to  philincalifornia
June 6, 2021 12:26 pm

Moist warm air rises convectively. At altitude (above lots of humid air and other hothouse gasses) it gives up lots and lots of its heat in a couple HOH phase changes… condensation and freezing)…lots of which is free to radiate into space way up there (actually enhanced by increased CO2 concentrations…collisions add Q to CO2 and other GHG’s which they can then release in radiation).

The air returning to the earth is way colder…and heavier because —> drier (else it wouldn’t be falling)…and the cold liquid rain and frozen solid ice falls.

So, where is there some balance between the energy rising and energy falling from convection?? Rising (and escaping) energy easily exceeds falling energy by at least an order of magnitude… No?

Last edited 14 days ago by DocSiders
Dave Fair
Reply to  DocSiders
June 6, 2021 4:42 pm

Uh, I’m not sure that “balance between the energy rising and energy falling from convection?” is a physical thing. Pure energy has many ways to go, not just falling. Show me some math.

Ron
Reply to  Weekly_rise
June 6, 2021 10:03 am

Convection dampens the day to night side amplitude thereby really “trapping” heat.

The temperature amplitude between day and night times and poles vs. equator decreases by atmospheric mass which can “store” heat due to the pressure-temperature relation given by the ideal gas law.

And that doesn’t even touch the latent heat capacity of the oceans.

gbaikie
Reply to  Weekly_rise
June 6, 2021 12:55 pm

“It is not possible for convection to maintain earth’s energy balance with sunlight because convection does not carry energy away from the earth system.”
But it carries energy within the earth system.
If topic energy balance and global temperature- such why we in an icehouse climate v a greenhouse climate. It about the average temperature of the entire ocean which averages about 3.5 C.
A greenhouse climate would would have ocean 10 C or more warmer than 3.5 C

Nicholas McGinley
Reply to  Weekly_rise
June 6, 2021 6:22 pm

I do not understand why there seems to be an underlying “one or the other” mentality.
Is that just my imagination?
Obviously both processes, and many others, are going on at the same time.

DocSiders said what I was gonna say: The energy that escapes via radiation after lifting processes (convective + orographic + any other lifting mechanisms) move air to altitude is not present when that air descends, obviously.
More moisture in the air when it is lifted by any such mechanism, increases the energy transported aloft tremendously.
The higher air is lifted, the more chance that the energy contained in the moisture as latent heat, will be released by phase changes.
Every gram of water that changes from vapor to liquid droplets releases 540 calories per gram, and each gram of liquid that freezes releases another 80 calories per gram.
This is a gigantic amount of energy compared to the amount of energy it takes to cool or heat liquid water between 0° and 100° C, which is of course 100 calories per gram.
It is also gigantic compared to the amount of thermal energy in dry air.
It is mostly dry air which is descending, after much or most of the latent heat which was in the moist ascending air has been released.

Not all of the latent heat which was released is radiated away to space in any single cycle of lifting and descent, but again, much of it has been.
And these cycles of lifting and descent are repeated over and over again.
The diurnal cycle only adds more layers of complexity to an already enormously complicated set of processes.

Dave Fair,
I think you misunderstood the part of what DocSiders said that you commented on. He was questioning that there is a balance there, as you correctly doubted yourself.

Convection accelerates the radiative process that allows the atmosphere to shed head to space. More heat at the surface adds more moisture to rising air on a global and even a local scale, and so enhances this accelerated transport of energy aloft and so further enhances radiative losses to space.

Ultimately, it is moisture that is responsible for regulating the temperature of the atmosphere.
The warmistas seem to think that there is a tight constraint on heat loss to space, and that CO2 constrains it all the more, but they ignore that at night, there is plenty of excess cooling capacity. We see this every night when cooling slows and then stops long before the following Sunrise starts a new cycle of daytime heating.
Likewise, there is a nearly unlimited amount of cooling capacity at high latitudes, especially during the long polar nights.

Moisture is what keep the Earth from becoming frigidly cold every night.

If CO2 was as important as warmistas claim, it’s effect would be particularly evident in Antarctica, and yet that place has not warmed as CO2 has risen.

Nicholas McGinley
Reply to  Nicholas McGinley
June 6, 2021 6:38 pm

I do not understand why, among some who are commenting, there seems to be an underlying “one or the other” mentality.
Is that just my imagination?
Obviously both processes, and many others, are going on at the same time.

DocSiders said what I was gonna say: The energy that escapes via radiation after lifting processes (convective + orographic + any other lifting mechanisms) move air to altitude is not present when that air descends, obviously.
More moisture in the air when it is lifted by any such mechanism, increases the energy transported aloft tremendously.
The higher air is lifted, the more chance that the energy contained in the moisture as latent heat, will be released by phase changes.
And on top of that, the higher it is lifted, the easier it will be for it to radiate away to space.
When air starts out warmer at the surface, or when it starts out more moist, the more buoyant it will tend to be and hence the higher it will tend to ascend.
So as the air near the ground grow’s warmer or more moist, for whatever reason, the more easily this accumulating energy (the energy contained in warmer air or the latent heat contained in more moist air) can be transported aloft and lost to space.

Every gram of water that changes from vapor to liquid droplets releases 540 calories per gram, and each gram of liquid that freezes releases another 80 calories per gram.
This is a gigantic amount of energy compared to the amount of energy it takes to cool or heat liquid water between 0° and 100° C, which is of course 100 calories per gram.
It is also gigantic compared to the amount of thermal energy in dry air.
It is mostly dry air which is descending, after much or most of the latent heat which was in the moist ascending air has been released.

Not all of the latent heat which was released is radiated away to space in any single cycle of lifting and descent, but again, much of it has been.
And these cycles of lifting and descent are repeated over and over again.
The diurnal cycle only adds more layers of complexity to an already enormously complicated set of processes.

Dave Fair,
I think you misunderstood the part of what DocSiders said that you commented on. He was questioning that there is a balance there, as you correctly doubted yourself.

Convection accelerates the radiative process that allows the atmosphere to shed head to space. More heat at the surface adds more moisture to rising air on a global and even a local scale, and so enhances this accelerated transport of energy aloft and so further enhances radiative losses to space.

Ultimately, it is moisture that is responsible for regulating the temperature of the atmosphere.
The warmistas seem to think that there is a tight constraint on heat loss to space, and that CO2 constrains it all the more, but they ignore that at night, there is plenty of excess cooling capacity. We see this every night when cooling slows and then stops long before the following Sunrise starts a new cycle of daytime heating.
Likewise, there is a nearly unlimited amount of cooling capacity at high latitudes, especially during the long polar nights.
And again, when there is more energy in the air, more energy will be transported to polar regions by winds and water currents.
Where it will be far easier for it to wind up being transferred to space.

Moisture is what keeps the Earth from becoming frigidly cold every night.

Weather is the set of processes that transport energy aloft and to higher latitudes within the atmosphere, and climate is the average of this weather in any given location.

If CO2 was as important as warmistas claim, it’s effect would be particularly evident in Antarctica, and yet that place has not warmed as CO2 has risen.

Last edited 14 days ago by Nicholas McGinley
Gary Pearse
Reply to  Nicholas McGinley
June 7, 2021 2:05 am

“I do not understand why, among some who are commenting, there seems to be an underlying “one or the other” mentality.
Is that just my imagination?
Obviously both processes, and many others, are going on at the same time.”

Nicholas, I, too seem to detect some kind of muddying the waters in discussions of late. Perhaps this is a new suttle onslaught against the effectiveness of WUWT? Do Wentworth’s recent papers have “strawman polarizing” characteristics, or are they being abused by those muddiers. And the use of unprecedented volume of negative dissing of comments. I think the trashing of Rud’s posting on Lindzen (an obvious target in climate jostling), is a good example.

Alasdair
Reply to  Weekly_rise
June 6, 2021 10:46 am

Forget energy balances for the moment. There are three ways energy moves up through the atmosphere. Radiation, convection and buoyancy. Sadly the latter two get conflated in the literature; but are totally different; the latter not dependent on temperature differential. The energy that arrives, by whatever means, at the Tropopause then relies on Radiation alone for balance, as you suggest. The fact that the Cirrus Clouds have growing ice crystals means that there here we have a net radiation of energy to space; this being alongside other radiation and due primarily to the buoyancy factor.
Attempting to explain matters purely in radiative terms just does not work unless the buoyancy factor is incorporated into the equations. There are large energies involved here which should not be ignored.

Reply to  Alasdair
June 6, 2021 11:38 am

The fact that the Cirrus Clouds have growing ice crystals means that there here we have a net radiation of energy to space

Alasdair,
Thank you! At last someone who recognises that solid ice crystal particles in Cirrus clouds at the tropopause are efficient and effective thermal radiators to space (both day and night), while at the same time acting as a daytime albedo control.

Trick
Reply to  Philip Mulholland
June 6, 2021 12:15 pm

Philip, those icy cirrus clouds also add radiation absorbed at the L&O surface above that of clear sky. Dr. Spencer a few years ago showed how to detect icy cirrus moving in overnight with ambient surface water thermometer temperature probes increasing water temperature over nearby ambient water not in view of the icy cirrus radiation.  

Reply to  Trick
June 6, 2021 3:56 pm

nearby ambient water not in view of the icy cirrus radiation. 

Trick,
I would need a link please.
A while back we had some observational comments from a crop grower in Florida on the subject of night frosts and cirrus veils.
He reported significant surface temperature variations associated with the advection of high level cirrus.
I would be genuinely interested to know the height and distribution of these jet stream veils and if the surface temperature effect he observed is a mirror reflection of the thermal radiation sourced from the warm surface waters of the adjacent Gulf of Mexico.

Trick
Reply to  Philip Mulholland
June 6, 2021 4:16 pm
Last edited 14 days ago by Trick
Nicholas McGinley
Reply to  Philip Mulholland
June 6, 2021 6:49 pm

Hi Phillip.
Here I am.
Yes, glad at least someone saw my recent comments regarding the possibility that the sudden increases that can be observed when cirrus moves in on cold nights, could possibly be due to reflection effects from warmer areas nearby.

It should be possible to get answers to questions such as these, although it would likely entail numerous careful observations over wide areas during conditions that occur somewhat infrequently in any given location.
Infrequently, but repeatedly, I should add.
Or at least the occurrence of such large effects as to be readily apparent may be what is infrequent: It seems to me that such effects may be more common that I observed, just not as dramatic under less optimal conditions for observing the effect.

Or it may be that Florida on cold nights undergoing radiational cooling represents a rather special case, being that the peninsula is surrounded by and contains within it, bodies of very warm water, even on these cold nights.

Last edited 14 days ago by Nicholas McGinley
Nicholas McGinley
Reply to  Nicholas McGinley
June 6, 2021 7:10 pm

One way to get at something like an answer might be to quantify the effect on many occurrences, and then look for instances in which the local ground and sky conditions were close to identical, but the water temperature over the Gulf (or other adjacent areas) was substantially different.
If it is indeed a reflection effect, one might suppose that the effect is going to be greatest when the water is warmest.
The effect can be observed during months ranging from October to April, when it can be expected the Gulf of Mexico as well as lakes and rivers and bays would be at substantially different temperatures.

On at least one of the recent occasions when I was talking about this here on WUWT, I did a quick search for media accounts of this phenomenon, and was able to find several mentions of occasions where local meteorologists in various parts of the country noted that cirrus clouds had prevented or lessened the expected occurrence of k i lling frosts.
So this is not something that is unknown, or localized to rural areas of Florida.

Last edited 14 days ago by Nicholas McGinley
Reply to  Nicholas McGinley
June 7, 2021 3:14 pm

Nicholas,

Many thanks for replying.

I think that the circumstances of your observations do support the idea that this winter night time land surface warming is the result of mirror reflection from the base of the cirrus cloud veil. I checked out your location on the west side of Florida and with the warm waters of the Gulf offshore being well above freezing, then direct radiant transmission of heat from sea surface to land surface via an overlying mirror of cirrus cloud seems very plausible.

What is of particular interest is that we know that there is a direct surface to space infrared transmission window thru the atmosphere that causes night frosts. While blocking of this window by the cirrus veil is one explanation for your observations, the reflected mirror hypothesis allows for the rapid transmission of radiant energy from warm surface to cold surface by a mechanism that is in agreement with standard physics.

Nick Schroeder
Reply to  Stephen Wilde
June 6, 2021 7:25 am

Here’s an accurate graphic.
This set of equations is used by heat exchanger design engineers to design heat exchangers.
The atmosphere is just an ordinary surface heat transfer problem.

Earth Heating PPt Video 021518.jpg
Reply to  Stephen Wilde
June 6, 2021 8:54 am

Stephen,

“That process takes time and is slower than radiation so the system must heat up.”

This only matters as a change is talking place. Once a steady state has been achieved, the temperature is what it is and non radiant energy can have no further effects.

Ask yourself the following question:

What effect does convection, latent heat and other non radiant transfers of energy between the surface and the atmosphere have on the steady state temperature, once that steady state has been achieved?

Take a look at Trenberth’s obfuscated steady state energy balance that includes non radiant energy in the radiant balance. If you subtract the return of non radiant energy from the back radiation term, all that’s left are the W/m^2 required to offset the surface emissions. Whatever effect non radiant energy entering the atmosphere has is already accounted for by the steady state average temperature and its subsequent emissions.

To the extent that a Joule of energy emitted by the planet can trace its origin to latent heat entering the atmosphere, a Joule of surface emissions that would have otherwise left the planet must be returned to the surface in order to offset the lost latent heat. The same goes for convection.

Latent heat and convection have a zero sum influence on the steady state radiant behavior of the planet and the subsequent RADIANT sensitivity of the surface temperature and this is another of those physical truths obfuscated by sloppy alarmist pseudo science that calls irrational arguments settled science.

Dave Fair
Reply to  co2isnotevil
June 6, 2021 5:00 pm

Oh, Christ. The system (Earth and its atmosphere) gets energy (Sun) and releases an equal amount of energy (LW at TOA). The internal machinations of the atmosphere (dynamically and radioactively) both amplifies and reduces energy flows. Radiatively active molecules modulate energy flow within the atmosphere. Convection, conduction and clouds also modulate energy flow within the atmosphere. So called GH gasses retard outward bound LW energy and convection and conduction within the atmosphere affects the outcome.

The TOA is at a lower emissive temperature than that of the surface. Get over it. None of you keyboard cowboys know what the hell is actually going on.

Reply to  Stephen Wilde
June 6, 2021 9:31 am

Thank you Rud – I always enjoy your posts.

My previous comment on this subject is excerpted below. I remain surprised by how many commenters find the GHE controversial. I know and have corresponded over the decades with most of the leading climate skeptics in academia, and not one of them has ever raised this question! So I called it a red herring – apparently it is not, at least on this website – but I just don’t care. In my opinion, the GHE exists, and the entire discussion is a diversion.

The real climate question has always been the alleged magnitude of Climate Sensitivity, (CS). Global warming alarmists (aka “warmists”) say CS is large and thus CO2-driven warming is very-scary and dangerous; climate skeptics say CS is small and warming is small and beneficial. There is ample Earth-scale evidence that the skeptics are much more correct – the upper bound of CS is only about 1C/doubling – and the warmists are not only wrong, but know they are wrong and have known that for ~50 years.

As we correctly published in 2002, “Climate science does not support the theory of catastrophic human-made global warming – the alleged warming crisis does not exist.”
https://friendsofscience.org/assets/documents/KyotoAPEGA2002REV1.pdf

If one wants to dive into the detail, it is probable that both sides are technically wrong, and that CS does not significantly exist in physical reality – because for CS to exist temperature changes must lag atmospheric CO2 changes, and we know that in fact CO2 changes lag temperature changes at all measured time scales. Cart before horse!

I suspect it’s a scale-up issue – what might exist at a molecular scale (“CO2 is a MINOR greenhouse gas”) changes entirely when one scales up from molecular to full-Earth-scale. In engineering, we often find surprises during scale-up of processes, and most will agree that molecule-to-Earth is one very large scale-up.

My previous comment, abridged, follows:

https://wattsupwiththat.com/2021/06/04/mathematical-proof-of-the-greenhouse-effect/#comment-3261556

I know many veteran climate scientists and do not recall any of them denying the existence of the green house effect (GHE) – accordingly this paper is a bit of a “red herring”.
 
Earth’s temperature is warmed and moderated by the GHE. The primary greenhouse gas (GHG) is water vapour; atmospheric CO2 is a bit player – NOT significant – NOT a primary GHG and NOT a significant driver of global warming.

Reply to  ALLAN MACRAE
June 6, 2021 9:53 am

Thread invasion – my apologies but this needs to be published, now:

Peter Foster’s National Post take-down of “Marxist-Mark Carney”, former head of the Bank of Canada and the Bank of England, now UN climate czar.

MARK CARNEY, MAN OF DESTINY, ARISES TO REVOLUTIONIZE SOCIETY. IT WON’T BE PLEASANT
What Carney ultimately wants is a technocratic dictatorship justified by climate alarmism
by Peter Foster  Jun 05, 2021 
https://nationalpost.com/opinion/peter-foster-mark-carney-man-of-destiny-arises-to-revolutionize-society-it-wont-be-pleasant

MY PREVIOUS SITUATION ASSESSMENT
 
It’s ALL a leftist scam – false enviro-hysteria including the Climate and Green-Energy frauds, the full lockdown for Covid-19, the illogical linking of these frauds (“to solve Covid we have to solve Climate Change”), paid-and-planned terrorism by Antifa and BLM, and the mail-in ballot USA election scam – it’s all false and fraudulent.
 
The Climate-and-Covid scares are false crises, concocted by wolves to stampede the sheep.
 
The tactics used by the global warming propagandists are straight out of Lenin’s playbook.
 
The Climategate emails provided further evidence of the warmists’ deceit – they don’t debate, they shout down dissent and seek to harm those who disagree with them – straight out of Lenin.
 
The purported “science” of global warming catastrophism has been disproved numerous ways over the decades. Every one of the warmists’ very-scary predictions, some 80 or so since 1970, have failed to happen. The most objective measure of scientific competence is the ability to correctly predict – and the climate fraudsters have been 100% wrong to date.
 
There is a powerful logic that says that no rational person can be this wrong, this deliberately obtuse, for this long – that they must have a covert agenda. I made this point circa 2009, and that agenda is now fully exposed – it is the Marxist totalitarian “Great Reset” – “You will own nothing, and you’ll be happy!”
 
The wolves, proponents of both the very-scary Global Warming / Climate Change scam and the Covid-19 Lockdown scam, know they are lying. Note also how many global “leaders” quickly linked the two scams, stating ”to solve Covid we have to solve Climate Change”- utter nonsense, not even plausible enough to be specious.
 
Regarding the sheep, especially those who inhabit our universities and governments:
The sheep are well-described by Nassim Nicholas Taleb, author of the landmark text “The Black Swan”, as “Intellectual-Yet-Idiot” or IYI – IYI’s hold the warmist views as absolute truths, without ever having spent sufficient effort to investigate them. The false warmist narrative fitted their negative worldview, and they never seriously questioned it by examining the contrary evidence.
 
More, for those who can and do read and think:

CLIMATE CHANGE, COVID-19, AND THE GREAT RESET
A Climate, Energy and Covid Primer for Politicians and Media
By Allan M.R. MacRae, Published May 8, 2021 UPDATE 1e
Download the WORD file
https://thsresearch.files.wordpress.com/2021/05/climate-change-covid-19-and-the-great-reset-update-1e-readonly.docx
 
 

Last edited 14 days ago by ALLAN MACRAE
Reply to  ALLAN MACRAE
June 6, 2021 10:14 am

https://nationalpost.com/opinion/peter-foster-mark-carney-man-of-destiny-arises-to-revolutionize-society-it-wont-be-pleasant
 
Mark Carney was successful at the Bank of Canada because of events that he had no role in – the huge success of the Alberta oilsands carried the Canadian economy through his tenure and made everyone look good, regardless of their lack of contribution to Canada’s success.
 
Here is what really happened that made Canada so prosperous then, as written by an esteemed colleague, one of the most respected engineers in Canada:
 
INITIATIVES THAT DROVE MAJOR ECONOMIC GROWTH OF SYNCRUDE CANADA, THE ALBERTA OIL SANDS AND THE CANADIAN ECONOMY
 
In the 1980’s and 1990’s, Allan MacRae initiated (or co-initiated) and successfully proposed three of the four major changes that drove the successful growth of the Alberta oil sands. Changes included new income tax terms, new Crown royalty terms and a low-cost 50% production increase that reduced Syncrude unit operating costs by 30%. MacRae also recommended that Syncrude acquire new leases for growth, and technical innovations that improved performance and reduced costs.
 
MacRae incorporated these initiatives into a comprehensive strategy for Syncrude, which was implemented and was instrumental in the successful evolution and growth of Syncrude and the Alberta oil sands industry.
 
The oil sands industry became the mainstay of the Canadian economy for 15 years, with over $250 billion in new capital investments and approximately 500,000 new jobs created. Canada became the fifth-largest oil producer in the world, the largest foreign supplier of energy to the USA and the most successful economy of the G8 countries.
 
 

Reply to  ALLAN MACRAE
June 9, 2021 8:34 am

IMPORTANT VIDEO – THE BIG PICTURE:



Thanks to Yeonmi Park and Jordan Peterson for telling her courageous story. None of this is surprising to me, because of my strong education and my life experiences on six continents.

Most people do not yet realize that the North Korean model or the Chinese Communist Party (CCP) model is what western elitists like Trudeau and Biden etc want for Canada, the USA and the rest of the Western democracies – the end of freedom and the adoption of the brutal, corrupt CCP model – a few princes at the top, looking down on all the poor peasants.

Carlo, Monte
Reply to  ALLAN MACRAE
June 6, 2021 3:03 pm

Allan—as is well-known, the climate scientists talk of the ‘E’ adjective: equilibrium CS, which has always troubled me greatly, how could an equilibrium ever be attained, much less measured? This issue also extends to CMoB’s work: although he talks about ECS, it seems like it is really instantaneous (I do realize that he is using the term the same ways climate scientists do).

Any comments?

Last edited 14 days ago by Carlo, Monte
Reply to  Carlo, Monte
June 6, 2021 6:26 pm

RE Transient Climate Sensitivity or Equilibrium Climate Sensitivity (To TCS or ECS, that is the question).
 
In short, I don’t care much if at all about the alleged difference between TCS and ECS – I say it is the climate equivalent of the medieval “How many angels can dance on the head of a pin”? Why?
 
Climate Sensitivity CS is an artificial construct based on the ASSUMPTION that CO2 is a significant and measurable driver of global temperature, yet the best evidence and the cutting edge of climate science suggests that CS does not really exist – so who really cares if it is TCS or ECS?
 
Let me explain further. For clarity, I like and respect Lord Monckton, as I like and respect fellow climate skeptics John Christy, Richard McNider, Nic Lewis and Judith Curry. They have all published about CS (see below), but they are really quantifying an UPPER BOUND for CS based on the aforementioned ASSUMPTION (“that CO2 is a significant and measurable driver of global temperature”). They conclude that their calculated maximum probable CS is very low, approximately 1C/(hypothetical doubling of atmospheric CO2), such that there is no real global warming crisis, climate emergency, climate cock-up, or whatever the global warming propagandists are calling their favorite scam this week.
 
In summary, all these esteemed climate skeptics are correctly stating that “Climate science does not support the theory of catastrophic human-made global warming – the alleged warming crisis does not exist.” (Baliunas, Patterson and MacRae, 2002). This is very good of them – it’s an honest, courageous but high-risk statement in this dystopian world of rampant academic corruption, based on the 50-year-old fraud of catastrophic human-made global warming.
 
Global Warming (aka Climate Change”) alarmism is a multi-trillion-dollar political and financial fraud, and those who dispute it get vilified and sometimes get destroyed.
 
The following is the cutting edge of the science:
 
The huge decline in fossil fuel consumption during the year-plus Covid-19 lockdown had NO impact on atmospheric CO2 increase – more evidence that Ed Berry’s latest book and paper are correct – see below.
 
My friend Ed says the increase in atmospheric CO2 is primarily natural, not man-made. The smartest people on the planet think he is correct.
 
More evidence supporting Berry:

Atmospheric CO2 changes lag temperature changes at all measured time scales. (MacRae, 2008). Humlum et al (2013) confirmed this conclusion.
 
Kuo et al (1990) and Keeling (1995) made similar observations in the journal Nature, but have been studiously ignored.
 
IF CO2 is a significant driver of global temperature, CO2 changes would lead temperature changes but they do NOT – CO2 changes lag temperature changes.
 
Think about that: Kuo was correct in 1990, and for 31 years climate science has ignored that conclusion and has been going backwards!
 
Climate Sensitivity (CS) to CO2 is a fiction – so small, if it even exists, it is practically irrelevant.
 
“The future cannot cause the past.” Here is the proof, from my 2008 paper:
https://www.woodfortrees.org/plot/esrl-co2/from:1979/mean:12/derivative/plot/uah6/from:1979/scale:0.18/offset:0.17
 
In the modern data record, the lag of atmospheric CO2 changes after atmospheric temperature changes is ~9 months. This is an absolute disproof of the CAGW hypothesis, which states that increasing CO2 drives temperature. “The future cannot cause the past.”
 
In my 2019 paper below, I explained why the lag is ~9 months – it is basic calculus, the 90 degree (1/4 cycle) lag of the derivative and its integral, which is the ~3 year ENSO period.
 
My 2008 paper remains very important. My 2008 conclusion was confirmed and expanded by Humlum et al in 2013, for which I am grateful.
 
My 2008 paper has been cited by Ed Berry in his 2020-21 book and related paper, which is at the cutting edge of climate science.
“CLIMATE MIRACLE: THERE IS NO CLIMATE CRISIS – NATURE CONTROLS CLIMATE”
amazon.ca/Climate-Miracle-climate-crisis-controls-ebook/dp/B08LCD1YC3/
 
“CARBON CYCLE MODEL SHOWS NATURE CONTROLS CO2 LEVEL”
edberry.com/blog/climate/climate-physics/preprint3/
All warmists and most skeptics argue about the magnitude of climate sensitivity to increasing CO2, and whether the resulting CO2-driven global warming will be hot and dangerous or warm and beneficial. Both groups are probably wrong.
 
There is a high probability that the mainstream climate debate about the magnitude of CS is wrong – a waste of decades of vital time, tens of trillions of dollars of green energy nonsense and millions of lives. Vital energy systems have been compromised, damaged with intermittent, unreliable wind and solar generation – a debacle.
 
It is important to note that Global Cooling is happening now, even as CO2 concentrations increase – another disproof of the global warming fraud.

Cheap abundant reliable energy is the lifeblood of humanity – it IS that simple. The green sabotage of our vital energy systems, whether innocent or deliberate, has cost lives and could cost very many more.
 
Scientific details here:
“CO2, GLOBAL WARMING, CLIMATE AND ENERGY June 15, 2019”
wattsupwiththat.com/2019/06/15/co2-global-warming-climate-and-energy-2/
 
Repeating, “The future cannot cause the past.”
__________________________
 
Reference:


THE CATASTROPHIC ANTHROPOGENIC GLOBAL WARMING (CAGW) AND THE HUMANMADE CLIMATE CHANGE CRISES ARE PROVED FALSE January 10, 2020
https://thsresearch.files.wordpress.com/2020/01/the-catastrophic-anthropogenic-global-warming-cagw-and-the-humanmade-climate-change-crises-are-proved-false.pdf
[excerpt]


CATASTROPHIC: THE ALLEGED WARMING IS NOT CATASTROPHIC.
 
Based on the evidence, Earth’s climate is relatively INsensitive to increased atmospheric CO2 – climate computer models greatly exaggerate future CO2-­‐driven warming and there is no catastrophic global warming crisis.
 
Both Christy & McNider (2017) and Lewis & Curry (2018) proved that climate sensitivity to increasing CO2 is too low to cause dangerous warming. To calculate an upper-­‐bound on climate
sensitivity to CO2, both papers made the same very conservative assumption:

Both papers assumed that ALL the observed global warming is ascribed to increasing atmospheric CO2, and then calculated the maximum climate sensitivity to a hypothetical doubling of atmospheric CO2 of only about 1 degree C, which is too low to cause dangerous global warming.
 
1.     Christy and McNider (2017) analysed UAH Lower Troposphere data since 1979:
“Satellite Bulk Tropospheric Temperatures As A Metric For Climate Sensitivity” By John R. Christy and Richard T. McNider
Asia-­‐Pac. J. Atmos. Sci., 53(4), 511-­‐518, 2017
 
2.     Lewis and Curry (2018) analysed HadCRUT4v5 Surface Temperature data since 1859:
“The Impact of Recent Forcing and Ocean Heat Uptake Data on Estimates of Climate Sensitivity”
By Nicholas Lewis and Judith Curry
 
Climate computer models used by the IPCC and other global warming alarmists employ climate sensitivity values much higher than 1C/doubling, in order to create false fears of catastrophic global warming.
__________________________

Herbert
Reply to  ALLAN MACRAE
June 6, 2021 9:01 pm

Allan,
I agree with you that CO2 lags temperature on all time scales from decades to Millenia.
When faced with this conundrum, warmists try to counter with Shakun et al (2012).
The paper acknowledges that whereas temperature normally leads CO2, during the interglacial which commenced some 20,000 years ago, CO2 from the warming oceans kicked in and led the distinct warming that followed.
Put directly, the claim is “Temperature is correlated with and generally lags CO2”.
Ergo, CO2 rules the climate system.
Further the assertion is that without CO2 the normal climate forces were insufficient for the planet to escape the last Ice Age.
This paper fails to be convincing,in my opinion, because the claim that they know that the Ice Age would have continued without strong greenhouse gas accretions seems to be the rankest speculation.
Do they really know how Ice Ages terminate?

Reply to  Herbert
June 6, 2021 9:50 pm

One does not need to examine every paper by warmists and individually refute it – warmists are serial bullsh!tters, and they can fabricate new lies faster than you and I can refute them
 
Based on their dismal predictive track record. we can safely dismiss everything the warmists say and write.
 
Warmists have made 48 consecutive very-scary climate predictions to end-2020, and all have failed to happen. At 50:50 “idiot odds” for each prediction, the probability against that happening is 281 trillion to one! Serial bullsh!tters!
 
I published this new Law in early 2020. Edit: Please delete the word “Virtually”.
 
“MACRAE’S MAXIM”:
“VIRTUALLY EVERY SCARY PREDICTION BY GLOBAL WARMING ALARMISTS IS FALSE.”
 
 Best regards, Allan

stinkerp
Reply to  ALLAN MACRAE
June 6, 2021 5:11 pm

Can you explain how they arrived at an “upper bound” of climate sensitivity of 1C when CO2 is doubled? Or point to an article that explains it. Climate alarmists increase CS with positive feedback. For example they theorize that if global temperature increases 1C, reduced glacier size reduces reflected sunlight further adding some amount to warming. It’s this fudge factor that gives them latitude to make scary claims. So far their fudge factor in the climate models has resulted in predicted warming that is far greater than measured. I’d like to see experimental evidence that doubling CO2 increases warming by (no more than? ) 1C.

Last edited 14 days ago by stinkerp
Nicholas McGinley
Reply to  stinkerp
June 6, 2021 7:30 pm

Experimental evidence of what would happen to an entire planet if xyz occurs instead of abc?
How?

Reply to  stinkerp
June 6, 2021 10:08 pm

These authors both used full-Earth-scale observations to calculate their upper-bound CS. Repeating, see underlined below:

Both Christy & McNider (2017) and Lewis & Curry (2018) proved that climate sensitivity to increasing CO2 is too low to cause dangerous warming. To calculate an upper-­‐bound on climate sensitivity to CO2, both papers made the same very conservative assumption:
Both papers assumed that ALL the observed global warming is ascribed to increasing atmospheric CO2, and then calculated the maximum climate sensitivity to a hypothetical doubling of atmospheric CO2 of only about 1 degree C, which is too low to cause dangerous global warming.
 
1.     Christy and McNider (2017) analysed UAH Lower Troposphere data since 1979:
“Satellite Bulk Tropospheric Temperatures As A Metric For Climate Sensitivity” By John R. Christy and Richard T. McNider
Asia-­‐Pac. J. Atmos. Sci., 53(4), 511-­‐518, 2017
 
2.     Lewis and Curry (2018) analysed HadCRUT4v5 Surface Temperature data since 1859:
“The Impact of Recent Forcing and Ocean Heat Uptake Data on Estimates of Climate Sensitivity”
By Nicholas Lewis and Judith Curry
 
Climate computer models used by the IPCC and other global warming alarmists employ climate sensitivity values much higher than 1C/doubling, in order to create false fears of catastrophic global warming.
____________

That approach is as good as it gets – full-Earth scale so no scale-up errors, and a conservative assumption that ALL observed warming is due to increasing CO2 – so it is an upper bound solution for CS. If NO warming was due to CO2. calculated CS would be zero. Theoretically, CS could be zero or even negative, but it is highly improbable that it is more than ~1C/doubling.

More extreme/esoteric arguments are possible, but not that credible.

philincalifornia
Reply to  ALLAN MACRAE
June 6, 2021 10:18 pm

CS could be zero or even negative, but it is highly improbable that it is more than ~1C/doubling.”

That just about sums it up Allan. Well said.



Reply to  philincalifornia
June 9, 2021 7:15 am

NB the word “Theoretically”. I wrote:

“Theoretically, CS could be zero or even negative, but it is highly improbable that it is more than ~1C/doubling.”

My best guess is CS does not even exist – because the whole concept of CS relies upon “cart before horse”” – that is, CO2 changes lag temperature changes at all measured time scales – in the ice core record, and in the modern data record, as I proved in 2008.

For simplicity, consider that CS = 0.

Last edited 11 days ago by ALLAN MACRAE
Trick
Reply to  ALLAN MACRAE
June 9, 2021 1:50 pm

Allan: “CO2 changes lag temperature changes at all measured time scales

This literally would mean global temperature goes up today because I am going to drive my V-8 gas SUV tomorrow. You really may need to rethink your position on this subject. .   

Reply to  Trick
June 9, 2021 4:36 pm

Trick – take a reading comprehension course.

Then read the background papers before you comment.

Trick
Reply to  ALLAN MACRAE
June 9, 2021 5:08 pm

Something wrong with my reading comprehension Allan? Please advise. And which background papers do you recommend? I may already have them in inventory.

Reply to  Trick
June 11, 2021 10:26 am

Trickster – I doubt that you are acting in good faith.

For those that are, this paper is a good starting point.

CO2, GLOBAL WARMING, CLIMATE AND ENERGY
by Allan M.R. MacRae, B.A.Sc., M.Eng., June 15, 2019 
https://wattsupwiththat.com/2019/06/15/co2-global-warming-climate-and-energy-2/
Excel: https://wattsupwiththat.com/wp-content/uploads/2019/07/Rev_CO2-Global-Warming-Climate-and-Energy-June2019-FINAL.xlsx

Trick
Reply to  ALLAN MACRAE
June 11, 2021 11:27 am

Allen, in good faith, I’ll prefer to learn from original published papers and text books traceable to them, not blogs.

Reply to  Trick
June 11, 2021 4:22 pm

Trickster, please do – you’ll be 10-20 years behind the state of the art, mired down in the fraud of the IPCC and its minions.

Trick
Reply to  ALLAN MACRAE
June 11, 2021 4:58 pm

Allan, thermodynamic principles withstand the test of time you mention; for accuracy find them in the original research not second hand on blogs.
 
The IPCC doesn’t do original state of the art research, it’s a United Nations body not a university research effort & its objective is to provide governments at all levels with scientific information (developed by others) that they can use to develop climate policies.

Reply to  Trick
June 11, 2021 7:20 pm

More specious nonsense from Trickster.

The IPCC and its minions are global warming fraudsters, who have made 48 failed, full expired very-scary global warming predictions to end 2020, all of them false.
The probability of that being random stupidity by warmist scientists is ~281 trillion to 1.
These warmist fraudsters were not just being exceptionally stupid, they were lying to us from day 1, and they knew it.
Global warming alarmism is a fifty-year-old fraud. long past its due date.

Reply to  stinkerp
June 16, 2021 7:32 am

The IPCC’s expression of the climate sensitivity as a function of CO2 emissions is a smokescreen to isolate the sensitivity from the laws of physics. The non linear physical metric they obfuscate is W/m^2 per degree of surface temperature while the linear physical metric obfuscated by this non linear metric is W/m^2 of surface emissions per W/m^2 of forcing. There are 2 levels of indirection between the linear physical relationship that matters and how it’s been characterized and which comprised an obvious misdirection.

The lower bound of 0.4C per W/m^2 (2.2 W/m^2 of emissions per W/m^2 of forcing) was established by the IPCC at their inception as a value below which, their agenda to redistribute wealth from the developed world to the developing world using climate reparations can’t be justified.

The upper bound of 1.2 C per W/m^2 (8.8 W/m^2 of surface emissions per W/m^2 of forcing) which corresponds to the worst case projections of the alarmists was clearly pulled out of someones ass by adding enough uncertainty to make the lower bound seem unreasonably low, while the facts show that the actual physical sensitivity is less than the presumed lower limit.

The measured sensitvity to solar W/m^2 (and all W/m^2 are the same) is 1.62 W/m^2 of surface emissions per W/m^2 of non reflected solar input corresponding to 0.3C per W/m^2 and significantly below the IPCC’s lower bound. This can b calculated EXACTLY as the slop of the Stefan-Boltzmann Law given as 1/(4eoT^3) where T is the average surface temperature, o is the SB constant and e is the ratio between the emissions at TOA and the RADIANT emissions of the surface.

Nicholas McGinley
Reply to  ALLAN MACRAE
June 6, 2021 7:20 pm

Allan,
You said, “…the entire discussion is a diversion.”

And I could not agree more.
I remain conflicted on the question of what to do about it…ignore it, or keep trying to settle it.


gbaikie
Reply to  Stephen Wilde
June 6, 2021 12:15 pm

“Neither you nor Bob Wentworth have successfully rebutted a GHE from convection.”

Nor can anyone.
“Convection does not just move energy around.”
But this moving energy around is important aspect of related to global air temperature

“It converts kinetic to potential in uplift and converts potential to kinetic in descent.”

Or there lapse rate, and due to kinetic energy of gas molecule and gravity- if air warms at surface by say 5 C, the air above it, is also warmed. And if surface air cools, the air above likewise cool. A thinner atmosphere would more quickly warm and more quickly cool.
Our atmosphere has a lot thermal mass and the whole column of air warms and cools.

But I say Earth is warm because it’s water planet. The atmosphere is about 1/2 the story.

Dave Fair
Reply to  gbaikie
June 6, 2021 5:04 pm

Pleas logically construct a mathematical formulation describing your epiphany.

gbaikie
Reply to  Dave Fair
June 6, 2021 8:01 pm

Nothing I can think of which  Richard Lindzen has said, that I disagree with- he probably got some mathematical formulations for you look at.
But a lot he talks about is just commonly accepted stuff.

gbaikie
Reply to  gbaikie
June 6, 2021 8:42 pm

A WUWT blog post: Ie:
“The following description of the climate system contains nothing that is in the least controversial, and I expect that anyone with a scientific background will readily follow the description. I will also try, despite Snow’s observations, to make the description intelligible to the non-scientist.
The system we are looking at consists in two turbulent fluids (the atmosphere and oceans) interacting with each other. By ‘turbulent,’ I simply mean that it is characterized by irregular circulations like those found in a gurgling brook or boiling water, but on the planetary scale of the oceans and the atmosphere.”
https://wattsupwiththat.com/2018/10/09/richard-lindzen-lecture-at-gwpf-global-warming-for-the-two-cultures/

But anything I have said, which anyone disagree with, I am very interested in being corrected. Let make a short list:
We have cold ocean and that why we have low CO2 levels and why in
a 34 million year Ice Age.
Average ocean surface temperature is about 17 C {which is commonly said/cited]
Average land surface air according to Berkeley Earth analysis is about 10 C
3.5 C or 90% or more ocean less than 3 C, is said everywhere.
Why enter glacial and interglaciation is somehow related the Milankovitch cycles.
IPCC correctly says all models can’t predict future.
And have yet to find anyone who seriously believes in CAGW.
Most blame idea of false media news stories that ill informed and are hyping things {which is what they do with everything}

Bob Wentworth
Reply to  Stephen Wilde
June 6, 2021 8:57 pm

Neither you nor Bob Wentworth have successfully rebutted a GHE from convection.

My recent post does, in a sense, disprove “GHE from convection.” In particular, the argument rigorously establishes that, in the absence of LW-absorbing/scattering materials in the atmosphere, and given the same planetary albedo and emissivity, the average surface temperature would need to be at least 24℃ colder—no matter how much convection happens, and no matter what other processes are occurring in the atmosphere.

It’s a valid rebuttal, even if not a successful one (in the sense of persuading you).

gbaikie
Reply to  Bob Wentworth
June 6, 2021 9:16 pm

There are many problems with this.
But in terms of global temperature it is related to the average temperature of our Ocean. And entire ocean average temperature is about about 3.5 C.
And much higher global temperature occur in greenhouse global climates which have a much warmer ocean then we have have during the last 34 million years.
But within our Ice Age the ocean temperature has been about 5 C, and this has large effect upon global average temperature. All interglacial periods which are warmer have ocean temperature closer to 5 C

Robert W Turner
Reply to  Stephen Wilde
June 7, 2021 5:22 am

The pseudoscience here at WUWT is stronger than ever isn’t it?

Proclamations such as “(ignoring that since Earths atmosphere got densified (aka ‘pumped up’) by gravitational consolidation about 4.5 billions years ago, unlike a newly pressurized bicycle tire it has had a LONG time to cool back down).” show how ignorant the dogmatic defenses of the pseudoscience GHG back radiation hypothesis are.

I suppose they think the magic unicorns “pumped up” the atmosphere 4.5 billion years ago and left the system is magic thermodynamic equilibrium. It’s not the sun and the sun’s heating isn’t continuously “pumping up” the atmosphere. NA, if the sun stopped shining the perpetual motion machine of the second kind that is the GHG back radiation would keep things going.

I have shown time and time again how the hypothesis has been refuted for over 100 years and that conversation lead to Dr Wentworth coming back with:

“Are you thinking that the momentum transferred by photons to molecules will warm and cool and equal number of molecules?
If so, then I agree with that assertion.
But, momentum change is not the only effect that photon absorption has.
It also has the effect of exciting a molecule into a flexing vibration. This happens regardless of whether the molecule was moving in the -x or +x direction.
The energy in this flexing vibration can and will be transferred to other molecular modes, in that molecule or other molecules, via collisions.
Thus, photon absorption contributes to heating, even if the momentum transfers do not lead to any net heating.”

So now we see that the mental gymnastics of the GHG back radiation cult are now leading to violations of 1st Law of Thermodynamics. Dr Wentworth is now admitting that the quantum theory of radiation is correct, and that gas molecules are equally warmed and cooled when a gas is in random isotropic motion, but then goes on to claim that there will always be warming even when there is no warming. This must be the missing heat, the heat created out of the ether when incident radiation causes a gas to cool via stimulated emission rather absorption, but the gas molecule increases in temperature regardless.

Last edited 13 days ago by Robert W Turner
Bob Wentworth
Reply to  Robert W Turner
June 7, 2021 5:46 pm

So now we see that the mental gymnastics of the GHG back radiation cult are now leading to violations of 1st Law of Thermodynamics. Dr Wentworth is now admitting that the quantum theory of radiation is correct, and that gas molecules are equally warmed and cooled when a gas is in random isotropic motion,

That is not what the quantum theory of radiation says.

I agreed that momentum transfer will cause equal warming and cooling.

However, momentum transfer accounts for only one trillionth of the energy delivered to a gas by photons.

So, your analysis claiming “equal warming and cooling” is ignoring 99.9999999999% of the energy involved in photons interacting with gas molecules.

I did not agree that the net effect of photons being absorbed is that they are equally warmed and cooled. I only agreed that that was the net effect of the—negligibly important—transfer of momentum by photons.

but then goes on to claim that there will always be warming even when there is no warming.

There only appears to be a contradiction if you selectively ignore half of my words.

99.9999999999% of the energy absorbed by photons contributes to warming.

0.0000000001% does not contribute to warming.

You are focused only on a part of the phenomenon that is of negligible importance.

This must be the missing heat, the heat created out of the ether when incident radiation causes a gas to cool via stimulated emission rather absorption, but the gas molecule increases in temperature regardless.

There is always more absorption than stimulated emission, for any gas at a finite positive temperature.

It’s useless for you to read Einstein if you don’t understand what he’s saying.

The quantum theory of radiation doesn’t say what you seem to think it does.

Robert W Turner
Reply to  Bob Wentworth
June 8, 2021 6:37 am

” But for the theoretical considerations, this small effect is on an equal footing with the energy transferred by radiation because energy and momentum are very intimately related to each other; a theory may therefore be considered correct only if it can shown that the momentum transferred accordingly from the radiation to the matter leads to the kind of motion that is demanded by thermodynamics.”

It is a violation of the 1st LoT for a gas molecule to increase in temperature from absorption of a photon travelling in the opposite direction.

Bob Wentworth
Reply to  Robert W Turner
June 8, 2021 8:08 am

” But for the theoretical considerations, this small effect is on an equal footing with the energy transferred by radiation because energy and momentum are very intimately related to each other; a theory may therefore be considered correct only if it can shown that the momentum transferred accordingly from the radiation to the matter leads to the kind of motion that is demanded by thermodynamics.”

Yes, Einstein writes that. However, it is important to see what Einstein does with this idea. He writes:

In order now to show that the momenta transferred from the radiation to

the molecule according to our basic hypotheses never disturb the thermodynamic equilibrium, we need only introduce the values for ∆²/τ and R calculated in (25) and (21) respectively after the quantity

(ρ − (1/3) 𝝼 ∂ρ/∂𝝼 ) (1 − e^(− h𝝼/kT))

in (21) is replaced by

ρh𝝼/3RT

from (4). We then see that our fundamental equation (12) is satisfied identically.

We obtained, without effort, from the general quantum assumption for matter, the second Bohr rule (equation (9)) as well as Planck’s radiation formula.

In other words, Einstein concludes that as long as the radiation to which molecules are exposed satisfies Planck’s radiation formula, then “the momentum transferred accordingly from the radiation to the matter leads to the kind of motion that is demanded by thermodynamics.”

Einstein uses the idea that the velocity distribution must be of the sort demanded by thermodynamics as a constraint, to determine what must be true of the distribution of radiation to ensure that this idea is satisfied.

He uses his considerations about momentum to re-derive the Planck radiation law.

Einstein concludes that his concerns about momentum will automatically be satisfied, so long as radiation obeys the Planck radiation law.

Einstein does not use the idea that the velocity distribution must be of the sort demanded by thermodynamics to say “radiation can’t warm a gas”, as you seem to erroneously conclude.

It is a violation of the 1st LoT for a gas molecule to increase in temperature from absorption of a photon travelling in the opposite direction.

I’m deeply puzzled as to why you would claim that this conclusion follows from 1st LoT, i.e., energy conservation.

That conclusion certainly does not follow from conservation of energy.

All that conservation of energy requires is that the amount of energy transferred to a molecule must equal the amount of energy carried by the photon, h𝝼.

Of course, the momentum carried by the photon, h𝝼/c, must also be transferred.

If a molecule has mass M, and an initial velocity v0 and final velocity v1, in the direction the photon was traveling, then momentum conservation requires that:

M⋅(v2 – v1) = h𝝼/c

or

v2 = v1 + h𝝼/Mc

The change in kinetic energy experienced by the molecule will be

∆KE = (1/2)M⋅(v2² – v1²)

Applying our result for v2 to this leads to:

∆KE = (1/2)M⋅h𝝼/Mc⋅(2⋅v1 + h𝝼/Mc) = h𝝼⋅(v1/c + h𝝼/2Mc²)

Thus, energy will be conserved as long as the amount of energy used to excite the molecule’s vibrational mode is:

∆Evibration = h𝝼 – ∆KE = h𝝼⋅(1 – v1/c – h𝝼/2Mc²)

Thus, it is straightforward to arrange for energy to be conserved.

This works even if the gas molecule and photon are traveling in opposite directions, i.e., if v1 < 0.

What, exactly, is the problem you perceive in all this?

Last edited 12 days ago by Bob Wentworth
Robert W Turner
Reply to  Bob Wentworth
June 9, 2021 5:54 am

I don’t claim that a photon can’t increase the temperature of a gas molecule, e.g. increase its total partitioned kinetic energy, which happens when a photon is absorbed without subsequent stimulated emission.

The problem I see is, when travelling in opposite directions, a photon must decrease the total partitioned kinetic energy of a gas molecule through stimulated emission. A decrease in velocity of the gas molecule must occur and therefore a decrease in the internal vibrational and rotational must occur according to Kinetic Theory.

To have a gas molecule increase in velocity (increase in temperature) from the absorption of a photon travelling in the opposite direction would be akin to the velocity of a vehicle increasing as a result of smashing bugs while driving down the highway. Doing so would be creating energy.

You can’t increase/decrease the translational kinetic energy of a gas molecule without also increasing/decreasing the internal vibrational and rotational energy of that molecule. There will always be vibrational and rotational kinetic energy for all molecules above absolute zero and therefore there will always be the potential for stimulated emission and a subsequent change in velocity of gas molecules.

Bob Wentworth
Reply to  Robert W Turner
June 9, 2021 11:24 am

The states of any matter at temperature T are populated according to the Boltzmann distribution, so that the ratio of populations of two states is given by exp(-∆E/kT) where ∆E is the energy difference between the two states. For CO₂ at 0℃ and for its 15 micron vibrational transition (with a photon energy 1.3e-20 Joules), this works out to a ratio of 0.03. That means that only three percent as many molecules are in the excited state as are in the ground state.

This means that for every 3 photons that produce stimulated emission, another 100 photons will be absorbed. 97% of the time, a photon will be absorbed rather than stimulating an emission.

This is true independent of the direction in which the molecule is moving relative to the photon.

The problem I see is, when travelling in opposite directions, a photon must decrease the total partitioned kinetic energy of a gas molecule through stimulated emission.

No. As I’ve said, the probability of stimulated emission vs. absorption is unaffected by the direction in which the molecule is traveling.

No matter what direction the molecule is traveling, only a tiny minority of interactions result in stimulated emission.

A decrease in velocity of the gas molecule must occur and therefore a decrease in the internal vibrational and rotational must occur according to Kinetic Theory.

You seem to be muddling together a number of ideas.

A moment ago you were talking about stimulated emission. Now you’re talking about change in velocity, presumably due to momentum transfer?

If there were a stimulated emission that stimulated emission would actually increase the velocity of a molecule moving opposite to the photon.

In the more common case where absorption happens, there would be a slight decrease in velocity, yes. But, there are two distinct ways the energy of the molecule changes:

  1. The molecule increases its vibrational energy
  2. The translational energy of the molecule increases or decreases, depending on the relative motion of the molecule and the photon

However, effect #1 involves a trillion times as much energy as effect #2.

What typically happens is that after the photon has been absorbed, imparting vibrational energy to the molecule, the molecule with then collide with another molecule.

That collision will “thermalize” the energy, turning much of the added vibrational energy back into translational kinetic energy or rotational energy.

Overall, the net vibrational, rotational, and translational kinetic energy must increase in response to the absorption of a photon. This result is independent of the relative directions of motion of the photon and molecule.

To have a gas molecule increase in velocity (increase in temperature) from the absorption of a photon travelling in the opposite direction would be akin to the velocity of a vehicle increasing as a result of smashing bugs while driving down the highway. Doing so would be creating energy.

This is a misleading analogy.

A more quantitatively accurate analogy would be to look at in-flight refueling of an aircraft.

Suppose that a tanker aircraft produces turbulence that increases drag on the aircraft it is trying to refuel. (I don’t know if it works that way.)

Would you then say “In-flight refueling cannot possibly increase the flight range of an aircraft because the drag of the tanker forces the aircraft to burn additional fuel?” Of course not. The amount of fuel transferred by the tanker vastly exceeds any increased fuel use associated with overcoming drag during the refueling operation.

Or, in the “smashing bugs” example, it’s as if those “bugs” contained bits of enriched uranium that one could gather and add to the nuclear reactor powering our vehicle. The slowing that occurs as a result of smashing into them is insignificant in relation to the total energy content that they are carrying.

You can’t increase/decrease the translational kinetic energy of a gas molecule without also increasing/decreasing the internal vibrational and rotational energy of that molecule.

They increase/decrease together on average, but this need not be the case within an individual event.

Translational kinetic energy and vibrational/rotational energy need not change in the same direction in any one event. In the event of a photon absorbed by a molecule moving in the opposite direction, the translational energy decreases by a little while the vibrational energy increases enormously.

Subsequent collisions ensure that, on average, both translational and vibrational/rotational energy have increased.

There will always be vibrational and rotational kinetic energy for all molecules above absolute zero and therefore there will always be the potential for stimulated emission and a subsequent change in velocity of gas molecules.

I’m quite puzzled by the way you jump back and forth between the topic of photons and molecules moving in opposite directions, and the topic of stimulated emission. Do you think these topics are in some way connected? They aren’t, not in any way that I can discern.

As I calculated above, in CO₂ at 0℃, stimulated emission will be only 3% as likely as absorption. (At 20℃ the ratio is 4% and at -20℃ the ratio is 2%.) So, yes, occasionally stimulated emission happens, but most arriving photons will warm the gas rather than cooling it.

* * *

In summary:

When a photon arrives at a gas molecule it is always more likely to be absorbed than it is to stimulate an emission. This means that, on average, arriving photons will add energy to the gas, warming it.

When a photon is absorbed, the momentum transfer might increase or decrease the translational kinetic energy of the molecule depending on the relative directions of motion. However, the magnitude of this change in translational energy is insignificant compared to the increase in vibrational energy that results from the absorption. (The change in translational energy is about a trillionth of the change in vibrational energy.) Thus, absorption of a photon increases the net energy of the molecule, regardless of the direction of motion of that molecule.

* * *

Do these explanations make any sense to you?

Are there remaining concerns?

Last edited 11 days ago by Bob Wentworth
Robert W Turner
Reply to  Bob Wentworth
June 11, 2021 2:52 pm

I have two contentions.

How can the vibrational mode of a molecule be in a ground state of zero when it’s above a temperature of absolute zero? The concept of a ground state of zero is easy to understand for electronic energy transitions, but what exactly is the difference in the kinetic energy of a harmonic oscillator in the ground state 0 and in an energized state of 1?

Second, how can the kinetic theory of gases closely approximate the kinetic energy ratio partitioned between translational and vibrational/rotational kinetic energy as a function of temperature and number of atoms when the translational energy changes opposite of vibrational kinetic energy in nearly half the instances a photon is absorbed?

Bob Wentworth
Reply to  Robert W Turner
June 11, 2021 5:38 pm

The concept of a ground state of zero is easy to understand for electronic energy transitions, but what exactly is the difference in the kinetic energy of a harmonic oscillator in the ground state 0 and in an energized state of 1?

Well, one could read up on quantum harmonic oscillators. The energy levels of a quantum harmonic oscillator have energies (n + 1/2)⋅h𝝼 where 𝝼 is the resonant frequency of the oscillator.

That’s not the kinetic energy; that’s the total of kinetic+potential energy. But, I don’t know why one would want to look at kinetic energy alone.

How can the vibrational mode of a molecule be in a ground state of zero when it’s above a temperature of absolute zero?

The Boltzmann distribution is the heart of statistical mechanics and thermodynamics.

It says that the probability of any energy state at energy E being populated is proportional to exp(-E/kT) where k is Boltzmann’s constant and T is the absolute temperature.

Obeying the Boltzmann distribution for temperature T is what it means for a material to be at temperature T.

For all finite positive temperatures, lower-energy states will always have a higher probability of being populated that higher-energy states, which will be less populated.

However, at higher temperatures, the rate at which occupation probability declines with increasing energy will be reduced.

In the case of the CO₂ 15-micron vibrational transition, all you need to do is to plug the transition energy h𝝼 = 1.3e-20 Joules into the formula exp(-E/kT) and you can compute the population ratio of the states.

When I talked about the ratio of the population of the excited state to the ground state being 0.03, I was oversimplifying a bit. This is actually the ratio of the populations of each pair of successive excitation states in the oscillator.

For level n, the population probability will be exp(-(n+1/2)⋅h𝝼/kT). So, the population probabilities for n=0, 1, 2, 3… will form a sequence given by some constant times 1, 0.03, (0.03)², (0.03)³, etc.

A 15-micron photon can interact with any two adjacent energy levels in this oscillator. But, in every pair of adjacent energy levels, the ratio of populations between the more and less energetic states is about 0.03 for a temperature of 0℃ (273.15 K).

Does this make sense?

Second, how can the kinetic theory of gases closely approximate the kinetic energy ratio partitioned between translational and vibrational/rotational kinetic energy as a function of temperature and number of atoms when the translational energy changes opposite of vibrational kinetic energy in nearly half the instances a photon is absorbed?

At any given temperature, T, each degree of freedom is expected to have an average energy kT/2, whether that mode is a translational, rotational, or vibrational mode.

I guess you’re asking about how this equipartition of energy is maintained when a photon is absorbed by a molecule?

The answer is that instantaneously it’s not maintained by the photon absorption process itself (and does not need to be), but the proper ratios of energies in different modes is rapidly re-established by subsequent collisions.

This can be unpacked in a bit more detail.

When a 15 micron photon is absorbed by a CO₂, the energy in one of the molecule’s vibrational modes is increased by 1.3e-20 Joules.

The absorption of the photon also leads to a momentum transfer. Depending on the direction of motion of the molecule, this results in a change of translational kinetic energy ∆TKE somewhere in the range (-1.3e-32 J) ≤ (∆TKE) ≤ (+1.3e-32 J).

The energy change in the translational kinetic energy is very, very small compared to the change in the vibrational energy.

Instantaneously, the partition of energy between the vibrational mode and the translational/rotational modes has been brought out of balance, with respect to what would be expected in thermal equilibrium. The vibrational degree of freedom has an excess energy of 1.3e-20 J.

However, in gases at the pressures and temperatures characteristic of most of the atmosphere, molecules collide many times each second.

Each collision redistributes energy. After a few collisions, the equipartition of energy between various modes will be more or less fully restored.

The fact that the transitional kinetic energy was initially changed by a tiny amount (∆TKE) in the positive or negative direction is utterly insignificant to the end result.

Does this make sense?

* * *

You might wonder why, then, in his paper, Einstein seemed concerned to ensure that photon emission and absorption would not alter the distribution of translational velocities in a gas.

I think this makes a little more sense if one reads the entire paper and looks at what Einstein was trying to accomplish in that paper.

He was proposing a quantum theory of radiation, describing the way that radiation couples to a quantum state transition between two energy levels m and n. He proposed that this coupling could be described in a simple fashion using fixed coefficients Aₘₙ, Bₘₙ and Bₙₘ. He wanted to show that this simple model could yield predictions that corresponded to prior theory in a satisfying way.

To that end, he essentially posed the question “What must be true of thermal radiation such that momentum transfers between that radiation and the molecules of a gas will not change the statistical distributions of the velocities of gas molecules (assumed to be distributed according to a Maxwell-Boltzmann distribution with temperature T)?”

Based on that premise, Einstein used his proposed quantum theory to re-derived the Planck radiation formula for thermal radiation associated with a temperature T. In so doing, he proved that momentum transfers from radiation distributed in that way (i.e., in a Planck distribution for temperature T) would not alter the velocity distribution of gas molecules (at temperature T).

This served to demonstrate a kind of self-consistency between his new theory and prior theoretical results.

That’s what Einstein was focused on. He wasn’t thinking or writing about radiative heat transfer to gases. He was trying to show that his quantum theory was self-consistent, under the assumption that everything is at the same temperature.

Looking at the paper overall, I notice a few things…

Einstein is considering both radiation absorbed and radiation emitted in order to achieve the end result of no net change in the velocity distribution of molecules.

Einstein doesn’t talk about the energy that goes into the vibrational mode, and how this affects the partition of energy in the gas. (Or maybe he did, and I just haven’t noticed that part yet? Hmm… I ought to look for it.) Regardless, it can be shown that, if radiation obeys the Planck radiation law for temperature T, then that radiation, combined with emitted radiation, will not, on average, change the vibrational energies of a gas at temperature T. The amount of energy absorbed will be precisely balanced by the amount of energy released via spontaneous and stimulated emission.

So, basically the paper is about verifying that black-body radiation associated with temperature T neither warms nor cools a gas at temperature T. That this checks out validates Einstein’s quantum theory of radiation.

Einstein does not address at all the question of what would happen if the radiation temperature T1 is different than the gas temperature T2.

But, if you follow the logic implicit in his calculations, one can work out that a temperature difference would lead to net heating or cooling of the gas, depending on the sign of the temperature difference.

Everything Einstein wrote about it not mattering what the resonant frequencies of the vibrational modes are was written entirely in the context of assuming that T1 = T2.

Einstein didn’t assert that these don’t matter in the more general situation where T1 and T2 are different. He didn’t write about that more general situation at all in this paper.

Does any of this make sense?

* * *

Are there additional contentions/concerns?

Last edited 9 days ago by Bob Wentworth
Bob Wentworth
Reply to  Bob Wentworth
June 11, 2021 6:06 pm

Note: Upon further examination, yes, of course Einstein addressed the issue of thermal (Planck) radiation of temperature T not disturbing the population of vibrational energy levels in a gas at temperature T, when absorption and spontaneous and stimulated emission are all taken into account.

Robert W Turner
Reply to  Bob Wentworth
June 12, 2021 7:50 am

Yes it makes sense from a pure statistical explanation when considering the molecule as a quantum harmonic oscillator instead of a pure mechanical system, but there’s still the curiosity of what physical anomaly actually exists between the oscillator in the quantum ground state and non-ground state. I suppose Einstein himself was always slightly perturbed by the apparent “chance” involved in quantum systems.

Perhaps the clue is that,

It was obtained from the condition that the quantum theoretic partition of states of the internal energy of the molecules is established only by the emission and absorption of radiation

which at first seems paradoxical to the Maxwell-Boltzmann Distribution until momentum transfer from radiation is considered.

So, basically the paper is about verifying that black-body radiation associated with temperature T neither warms nor cools a gas at temperature T. That this checks out validates Einstein’s quantum theory of radiation.

But, if you follow the logic implicit in his calculations, one can work out that a temperature difference would lead to net heating or cooling of the gas, depending on the sign of the temperature difference.

Right, this seems to confirm a major contention of the violation of the 2nd LoT with the GHG hypothesis.

https://arxiv.org/pdf/0707.1161.pdf
https://www.researchgate.net/publication/276048562_Scrutinizing_the_atmospheric_greenhouse_effect_and_its_climatic_impact

Bob Wentworth
Reply to  Robert W Turner
June 12, 2021 10:59 am

there’s still the curiosity of what physical anomaly actually exists between the oscillator in the quantum ground state and non-ground state.

I’m not sure what you mean by “physical anomaly” here.

I don’t see how this is any more mysterious than the general non-intuitiveness that applies to all of quantum mechanics. Despite being non-intuitive in some ways, quantum mechanics has been one of the most successful and well-verified theories of all time.

Perhaps the clue is that,

“It was obtained from the condition that the quantum theoretic partition of states of the internal energy of the molecules is established only by the emission and absorption of radiation”

which at first seems paradoxical to the Maxwell-Boltzmann Distribution until momentum transfer from radiation is considered.

How does this seem “paradoxical to the Maxwell-Boltzmann Distribution”? I’m not perceiving anything at all paradoxical.

There was an unanswered question, perhaps, about how the various aspects of the situation related to one another—a question which Einstein ultimately answered.

But, the idea of things being “paradoxical” suggests an expectation that it “wouldn’t be expected to work out.” I suppose that’s more a matter of individual perspective, rather than relating to any objective characterization of the situation.

I guess it’s ok to differ at this level of subjective impressions. 🙂

BW: But, if you follow the logic implicit in his calculations, one can work out that a temperature difference would lead to net heating or cooling of the gas, depending on the sign of the temperature difference.

RT: Right, this seems to confirm a major contention of the violation of the 2nd LoT with the GHG hypothesis.

Why would you believe this?

Yes, thermal radiation always yields a net heat flow from warm to cool.

All physicists agree with this premise.

Just as physicists generally agree that the GHE is real.

The GHE phenomenon is entirely compatible with the requirement that thermal radiation leads to a net heat flow from warm to cool. There is no violation of the 2nd LoT involved in the GHE.

* * *

It is only a straw-man misinterpretation of the GHE that involves heat flowing in any forbidden direction.

The argument of Gerlich and Tscheuschner, your first reference, relies entirely on such a straw-man misinterpretation. G&T assert that the GHE model claims something which it does not claim, then refute this mischaracterization of the GHE model, and thereby falsely assert they have shown the GHE to violate the 2nd LoT. I unpack this in a critique of G&T’s paper.

As far as I can tell, your second reference, Kramm and Dlugi, does not make any independent assertion of a violation of the 2nd LoT, only mentioning G&T’s assertion of this. (If I’ve missed such an allegation, let me know. K&D make other assertions, which I have not yet had time to analyze.)

* * *

I hope I’ve been able to put to rest the idea that Einstein’s 1917 paper offers any commentary that is directly relevant to “debunking” the physics of radiative gases which underlies the GHE.

As for the assertion that the GHE somehow violates the 2nd LoT, I’d be happy to talk this through, if you have concerns about this.

Last edited 8 days ago by Bob Wentworth
Lowell
Reply to  Stephen Wilde
June 7, 2021 8:42 am

Stephen: What your describing is the ideal gas nature of the atmosphere. Just because what you say is true about the isotropic nature of the atmosphere does not contradict the Green House Gases both retain heat and radiate heat into space. The real world of physics is much more complicated and interesting than simple convection.

Lit
Reply to  Stephen Wilde
June 7, 2021 10:32 pm

It converts kinetic to potential in uplift and converts potential to kinetic in descent.”

So, descending cold air blowing on a hot surface makes it warmer?

Then why does a fan cool you as it blows air on your skin?

dk_
June 6, 2021 6:15 am

I read all of both posts. Perhaps I’m in the minority on this one as well. Thanks.
Grumpiness about “green house” metaphors does not signify dissent.

Curious George
Reply to  dk_
June 6, 2021 7:24 am

I do believe that the greenhouse effect exists. It is just that I am not satisfied with Bob Wentworth’s “mathematical proof”.

Dave Fair
Reply to  Curious George
June 6, 2021 5:07 pm

Damnit! Provide your own “mathematical proof” that the GHE doesn’t exist! Otherwise, STFU. I’m getting tired of keyboard cowboys.

Trick
Reply to  Robert W Turner
June 7, 2021 7:51 am

Robert’s “This has been done before” is another name for a “supremely important law of the history of science” in “the Infinite Chain of Priority: Somebody Else Always Did It First” (Tony Rothman, 2003: Everything’s Relative and Other Fables from Science and Technology, John Wiley & Sons, p. xiii.

Bob Wentworth
Reply to  Robert W Turner
June 10, 2021 1:04 pm

Regarding the paper by Gerlich and Tscheuschner, it is, regrettably, the most badly argued “scientific” paper I have ever read. Please consider reading my response to that paper.

Last edited 10 days ago by Bob Wentworth
Curious George
Reply to  Dave Fair
June 7, 2021 9:01 am

Dave, please re-read the “mathematical proof”. And please re-read my comment. Why should I contradict myself?

RickWill
June 6, 2021 6:16 am

Linear analysis fails in its application to the global energy balance.

There is no “GHE”. Easily provable by looking at how emissivity and albedo respond to surface temperature.

Emissivity of water at 271K reduces from 0.96 to 0.3 after 50 days without any surface insolation because ice is an insulator.

10mm of water in the atmosphere below 273K reduces heat loss. The same water above 273K reduces energy input. Earth’s albedo over tropical oceans goes from 0.1 to 0.9 depending on where that water is located.

Linear analysis simply does not handle these complexities.

Both albedo and emissivity are complex functions of surface temperature. No linear analysis can handle these non-linear effects.

The nearest analogy for water on the surface and in the atmosphere are diodes for an electrical circuit or a non-return valve in a flow circuit. Anyone using linear analysis on such a system demonstrates their poor understanding on the system realities.

Last edited 14 days ago by RickWill
philincalifornia
Reply to  RickWill
June 6, 2021 9:55 am

Yes and sadly, although I’m most definitely not averse to being educated, I don’t feel that I have been by this thread. For example, there’s no discussion, unless I missed it about radiation with respect to altitude of the greenhouse gas molecule, and convection.

Also, I would add that linear effects of convection (i.e. not even feedbacks, positive or negative) can wipe out the logarithmic-reduced purported effects of CO2, and also remember this also has to be calculated for at least above 280ppm for it to be again purportedly man-made. There’s a lot missing here in this discussion, with the latter point being strangely and ubiquitously missing..

Dave Fair
Reply to  philincalifornia
June 6, 2021 5:09 pm

Show your work! [The math.] Phil, otherwise you are BSing.

philincalifornia
Reply to  Dave Fair
June 6, 2021 8:35 pm

I have no work. I was asking questions of people who know more than me.

philincalifornia
Reply to  philincalifornia
June 6, 2021 9:43 pm

Furthermore, what else are Sunday’s for, that and watching International cricket? That’s when I get my best bullshitting done. As a very good friend from the University of Georgia told me “The only thing you get better at as you get older is bullshitting”.

I could go one, but I’m trying to develop something more concrete regarding things that aren’t feedback, but are intrinsically linear from the get-go (correcting people on the linearity of the Calvin Cycle wrt CO2 started it).

…. and, as ever, I would like to start above 280ppm CO2.

Nicholas McGinley
Reply to  philincalifornia
June 6, 2021 11:41 pm

I am better at resting, personally.
I used to be very restless, but no more.

When I was younger, I had to contend with a very strict dad, who seriously cramped my style.
He was very sharp, too, and had a saying that “You cannot bullshit a bullshitter.”
I had a daily exercise in disproving that maxim.
But at some point I decided to focus on things that are true, and so have had not a lot of bullshitting practice time in recent decades.
But I am pretty sure my experience from younger days has given me a real nose for malarkey.
I can smell that crap from a long ways off.

Last edited 14 days ago by Nicholas McGinley
Bob Wentworth
Reply to  RickWill
June 6, 2021 9:21 pm

There is no “GHE”. Easily provable by looking at how emissivity and albedo respond to surface temperature.

You are jumping from claiming that there is a flaw in my analysis to concluding that this “proves” there is no GHE. Even if you were 100% correct about having identified a flaw in my argument (you haven’t), that wouldn’t constitute your having “proved” anything.

Identifying a flaw in a proof of “A” does not constitute a proof of “not A”.

You are totally misinterpreting the nature of the analysis when you call it “linear.”

RickWill
Reply to  Bob Wentworth
June 6, 2021 10:14 pm

Bob claimed:

You are totally misinterpreting the nature of the analysis when you call it “linear.”

When you can show me where your values for emissivity and albedo are functions of time and temperature with non-linearities then you will having something worthwhile.

I know you were surprised to see that the water surface does not get colder than 271K. You are similarly surprised that the water surface does not exceed 303K.

Any analysis has to have these hard limits. Once included you will realise there is no “GHE”. It is a myth for people unable to accept non-linearities.

None of your equations have the hard limits. This is the average temperature function for the global oceans:
TLat >-80<80 = 271.3K
For Lat -80 to +80, TLat= 271.3 + 32/80*ABS(80 – Lat)
The temperature is non-linear.

The ocean temperature only exists in the range 271.3K to 303K because hard limits prevent it from going beyond it.

Two easily demonstrated non-linearities.

If there was no atmosphere then the ocean temperature would be higher – easily verified.

Last edited 14 days ago by RickWill
Bob Wentworth
Reply to  RickWill
June 7, 2021 5:54 pm

None of your equations have the hard limits.

Nor do they need to have these hard limits.

The conclusions drawn are in a context of “if albedo and emissivity are unchanged.”

The conclusions are in the context of a “thought experiment.” It doesn’t matter if you don’t know how to change the Earth in a way that wouldn’t change the albedo and emissivity. The thought experiment remains equally valid (no less, no more).

Tom
June 6, 2021 6:23 am

There are way too many commenters here who fall into the category of people whose comprehension of math and science is fully constrained by politics and preconceived notions.

fretslider
Reply to  Tom
June 6, 2021 6:54 am

Can you share your data showing that?

Thanks.

DMacKenzie,
Reply to  fretslider
June 6, 2021 7:32 am

Just read them…

Tom
Reply to  fretslider
June 6, 2021 7:34 am

The raw data exists in the comments which I have no way to compile and analyze, but it is there for all to see.

B Clarke
Reply to  Tom
June 6, 2021 8:07 am

The vast majority of posters here,when discussing “the science ” discuss the science theres very little prejudice on the science, political opinion will of course be discussed when climate change is the topic,the two can not be separated, when politics is the main driver of climate science, when climate science is funded by governments with agendas ,when the IPCCs main function is to prove global warming. There is two sides to a debate ,there are two sides to a argument, unfortunately in climate science,( the science is settled) according to the MSM and politicians, here on WUWT no one is barred from thier opinion nothing is settled and opposing views are welcomed in open debate , you are fortunate to be able to discuss here,what ever the outcome.

Earthling2
Reply to  B Clarke
June 6, 2021 3:10 pm

“There is two sides to a debate ,there are two sides to a argument”

Actually, there are 3 sides to a debate or argument. My side, your side and the truth or facts of the matter. But humor aside, it never hurts to have more discussion on some matters, but I thought the GHE was established science here at this blog…the only issue up for real debate is what are the feedbacks, whether they be positive or negative. IMHO, it is the ECS that needs debating until the cows come home, not the existence of a GHE, which might be moot if ECS is negative. But it would still be incorrect to say there is no GHE, even if it ultimately causes cooling in the final analysis.

B Clarke
Reply to  Earthling2
June 6, 2021 3:16 pm

I don’t decide the debate, l don’t believe the science is settled,

Hence the debate,sounds like you have a lot to add to the debate.

Nicholas McGinley
Reply to  Earthling2
June 6, 2021 7:45 pm

I think there are many other issues up for debate.
-For example, what factors are responsible for all of the huge changes in GAST and/or local climate regimes, prior to the industrial revolution?

-What are the relative magnitudes of these factors?

-Do they or do they not overwhelm any possible effects from humans, including changes in land use, the effect of the actual energies we release into the environment, etc?

-Is net warming really something that will have negative consequences, regardless of the cause or causes?

-Are we or are we not reaching the end of the interglacial, and can we possibly do anything about it if we are?

-How much benefit will adding CO2 to the air have in the biosphere, crop growth, health, and yields?

-Will having more CO2 continue to cause global greening, including shrinking of marginal desert/arid regions by drastically lowering how much water plants need to grow and thrive?

  • And many more.
Last edited 14 days ago by Nicholas McGinley
Rory Forbes
Reply to  Tom
June 6, 2021 12:32 pm

The raw data exists in the comments which I have no way to compile and analyze, but it is there for all to see.

What you’re describing is merely your own unsupported opinion. If there is “raw data” observable to all, it can be compiled and analyzed by anyone with the time patience and will to do it. Hell, the entire idea of a “climate crisis” and the AGW position “is fully constrained by politics and preconceived notions”. Those is bound infiltrate any discussion of climate.

Dave Fair
Reply to  Tom
June 6, 2021 5:15 pm

FU, too. My response is about the level of your comment. Your comment is similar the level of the “racist” epithet. I have far more intellect, education and experience than many of the asshole on this Thread. The GHE is real and dipshit deniers are idiots.

Dave Fair
Reply to  Dave Fair
June 6, 2021 5:20 pm

Hey! Rereading that really felt good!

Kevin kilty
June 6, 2021 6:28 am

Two observations:

First, mechanical engineers have been designing furnaces and boilers for a very long time. Heat transfer within those devices depends on radiation transfer in an atmosphere of CO2 and water vapor, which are the residuum of combustion within them. The physics of the process is essentially the greenhouse effect. Engineers do not employ the exact physics but rather a workable approximation.

Second, I have heard no one here dispute the correctness of the transfer equation (here) and so how is it that the greenhouse effect doesn’t exist even though the transfer equation demands that it does, and enables one to calculate its effect?

There are several bits of magical thinking that pop up here from time to time and which seem to be impossible to snuff out — non-existance of the greenhouse effect and PV=nRT determines gas temperature independently of any other factors are two of them.

Last edited 14 days ago by Kevin kilty
AC Osborn
Reply to  Kevin kilty
June 6, 2021 7:12 am

Re “Second, I have heard no one here dispute the correctness of the transfer equation”.
First of all gases are not Black Bodies.
Second if you do the transfer function in the direction of Cold Object to Warm Object you get a massive NEGATIVE number which say’s that the Warmer Object has been COOLED.
How do you explain that?

Take 2 bodies one at 200K and one at 280K, transfer from hot to cold = + 4546560000
transfer form cold to hot = -4546560000

Explanation please.

Kevin kilty
Reply to  AC Osborn
June 6, 2021 7:29 am

First, Those are some pretty big numbers you have there — what are they? What units?

Second, the transfer equation does not depend on the ” total blackness” of the medium which is radiating and absorbing. Most of the time we think of the gaseous medium as being “grey”. That is, described by an emissivity.

Third, a warm body in the presence of a cool body does cool — it transfers net energy to the cool body. You cannot just use the stefan boltzmann law to calculate the transfer as you also need 1) the true angular distribution of radiated energy from both bodies (an emissivity < 1.0 is just the first order approximation), and 2) you have to employ a view factor between the two unless one body is completely enclosed by the other. If the total view of the world of either body is not solely of the other body, then you had better take into account what else is in the view or your calculations are incomplete.

Nick Schroeder
Reply to  AC Osborn
June 6, 2021 8:55 am

In his heat radiation lecture notes Planck observed that for heat radiation to interact with stuff they had to have similar dimensions.
Short wave x-rays, cosmic rays, etc. are high energy and short wave. They go together.
UV waves are shorter and cause minerals to fluoresce.
LWIR are , well, long wave and low energy and are too large too interact at the molecular level.

Nicholas McGinley
Reply to  Nick Schroeder
June 6, 2021 8:00 pm

The radio antenna on my car “interacts with” radio waves that are hundreds of meters long and have a tiny fraction of the energy of LWIR.

Bob Wentworth
Reply to  Nick Schroeder
June 6, 2021 9:26 pm

LWIR are , well, long wave and low energy and are too large too interact at the molecular level.

So, every scientist and engineer who has ever measured the absorption and emission properties of matter composed of molecules was lying? And the industrial processes that depend on these interactions don’t really work?

Gordon A. Dressler
Reply to  Nick Schroeder
June 7, 2021 8:26 am

Nick,

Please examine the particle-wave duality of EM radiation. Is a photon of LWIR “long” or “short”?

Microwave radiation has “wavelengths” longer than that of IR and LWIR, yet it clearly interacts with water at the molecular level. Why is that?

Last edited 13 days ago by Gordon A. Dressler
Gordon A. Dressler
Reply to  AC Osborn
June 6, 2021 12:06 pm

AC Osborn,

You asked for an explanation.

First, your example of two bodies, one at 200 K and the other at 280 K is nonsense for reasons of missing units, emissivity values, radiating areas, and separation distances.

Second, any physical body with a surface temperature above absolute zero will emit thermal radiation (although such radiation does not have to conform to the Stefan-Boltzmann radiation law for ideal black bodies). For a uniform body surface, such radiation will be isotropic and will be unaffected by the presence of other bodies that may intercept some of this radiation. Ask yourself, how is it possible that one body physically separated from another body would “know” that there is/are one or more external object to which radiation should be directed?

Third, the steady-state radiative power shift at one object caused by receiving radiation from a second separated object, both with temperatures above absolute zero, is simple to calculate: new power radiated by object 1 = old power radiated by object 1 + power newly received from object 2.

In the above case, the fact that object 2 also happens to be intercepting radiation from object 1 only comes into play to the extent that such radiation happens to also increase the steady-state radiating power of object 2. Yes, it is an iterative process that converges to a final steady-state equilibrium for both bodies in this given gedanken experiment.

Bob Wentworth
Reply to  Gordon A. Dressler
June 6, 2021 9:35 pm

For a uniform body surface, such radiation will be isotropic

I appreciate your comment, and I wonder about this particular claim.

Recently, it was pointed out to me that reflectivity is a function of angle of incidence, so absorptivity and emissivity would also need to be a function of angle of incidence.

Wikipedia seems to support this idea “[Directional spectral emissivity] depends upon the wavelength and upon the angle of the outgoing thermal radiation. Kirchhoff’s law actually applies exactly to this more complex emissivity: the emissivity for thermal radiation emerging in a particular direction and at a particular wavelength matches the absorptivity for incident light at the same wavelength and angle.”

angech
Reply to  Bob Wentworth
June 7, 2021 4:45 am

Bob.

“reflectivity is a function of angle of incidence, “”
Yes
Bounces off so not absorbed.

“so absorptivity and emissivity would also need to be a function of angle of incidence.”
No.

Absorption can occur from any direction.
Emission can go in any direction.

Bob Wentworth
Reply to  angech
June 7, 2021 12:18 pm

The issue is that Kirchoff’s Law requires that:

absorptivity + transmissivity + reflectivity = 1

and

absorptivity = emissivity

These rules are relevant at any given wavelength and in any given direction.

So, if reflectivity becomes large at some angle of incidence, that reduces the maximum possible emissivity at that angle.

angech
Reply to  Bob Wentworth
June 7, 2021 7:05 pm

The issue is that Kirchoff’s Law requires that:
absorptivity + transmissivity + reflectivity = 1

I would take this to mean energy in equals energy out.

That the incidence of the rays reduces absorption because more is reflected at large angles of incidence is quite logical.
Applies to the Arctic and Antarctic.

The assumption that the energy absorbed from a certain direction must be emitted as the same amount of energy is conservation of energy.
There is no directional requirement on emission.

In fact you state
“These rules are relevant at any given wavelength and in any given direction.”
Hence
“For a uniform body surface, such radiation will be isotropic”

I was concerned you were disagreeing with Gordon but apparently not. My bad.

Hence this Wikipedia statement is potentially misleading depending on how it is interpreted

“the emissivity for thermal radiation emerging in a particular direction and at a particular wavelength matches the absorptivity for incident light at the same wavelength and angle.”

In that the emissivity must match the absorption but there is no compelling reason for any particular direction for that emissivity to go.
The amount of energy absorbed at a particular angle is specified by how much is reflected and how much absorbed.
.
The emission however can be and must be in any direction it likes.
If post facto you observe the same amount of energy coming out back in the direction it came in then Kirchoff is trivially right.
Given you have already specified the angle and the amount of energy.

God does not pay dice with the universe.
If Kirchhoff means that the only way for a hot body to emit is directly back at its incoming source this would be very interesting.
And wrong for a lot of thermodynamic reasons.

Bob Wentworth
Reply to  angech
June 7, 2021 8:01 pm

The issue is that Kirchoff’s Law requires that:

absorptivity + transmissivity + reflectivity = 1

I would take this to mean energy in equals energy out.

Yes, that rule does guarantee conservation of energy.

The assumption that the energy absorbed from a certain direction must be emitted as the same amount of energy is conservation of energy.

No, that’s not what I said or meant. And, it’s not conservation of energy.

The rule is: absorptivity = emissivity

That’s not a rule about the amount of energy. The energy absorbed is (energy incident) ×(absorptivity).

Energy emitted spontaneously is (emissivity) × σT⁴

It’s perfectly possible that the amount of energy incident is ZERO while the energy is being emitted.

These are NOT the “same amount of energy” and there is no reason that they should be.

The reason “absorptivity = emissivity” is for deep thermodynamic reasons, not for reasons of energy conservation.

There is no directional requirement on emission.

That might be true if “absorptivity = emissivity” was about energy conservation, but it is not.

In fact you state

“These rules are relevant at any given wavelength and in any given direction.”

Hence

“For a uniform body surface, such radiation will be isotropic”

No. Your “hence” does not follow at all.

Because the rules are relevant “in any given direction” it follows that
ff absorptivity depends on angle, then emissivity must also depend on angle.

Note that none of this is an argument I came up with. Someone in these comment threads made the claim, then I realized it seemed to be correct, and my subsequent research has backed up the claim.

Another Wikipedia article says “the emissivity and the absorptivity, if they happen to be dependent on direction, must again be equal for any given direction.”

The emission however can be and must be in any direction it likes.

That appears to be wrong. Emission is isotropic for an ideal black-body, but not necessarily for real matter.

If post facto you observe the same amount of energy coming out back in the direction it came in then Kirchoff is trivially right.

Given you have already specified the angle and the amount of energy.

You seem to be misunderstanding what the term “emissivity” means. There is no “direction it came in.” Emissivity is about radiation that is emitted even when there is no external radiation “coming in.”

If Kirchhoff means that the only way for a hot body to emit is directly back at its incoming source this would be very interesting.

There is no “incoming source.” If you think there is, you are misunderstanding the idea of emissivity.

Last edited 13 days ago by Bob Wentworth
angech
Reply to  Bob Wentworth
June 8, 2021 8:22 am

“you are misunderstanding the idea of emissivity.”
Emissivity does not require an incoming source.
However we are discussing Kirchhoff as well as emissions.

(Gustav Kirchhoff‘s 1859 law of thermal radiation) that equates the emissivity of a surface with its absorption of incident radiation (the “absorptivity” of a surface)

Kirchoff’s law requires an incoming source, else there is no law.

“emissivities are the total hemispherical emissivities from the surfaces.a more complex “directional spectral emissivity” can also be measured. This emissivity depends upon the wavelength and upon the angle of the outgoing thermal radiation. Kirchhoff’s law actually applies exactly to this more complex emissivity: the emissivity for thermal radiation emerging in a particular direction and at a particular wavelength matches the absorptivity for incident light at the same wavelength and angle. The total hemispherical emissivity is a weighted average of this directional spectral emissivity;”

My takes from this.
Emissivity is the transfer back to space of all the energy that came in minus the reflectance.
Hence the surface temp for a blackbody at an emissivity of 1 is the same temperature for a greyer body at 0.94 if they have absorbed and emitting the same amount of energy per surface area.
The energy exiting the earth is after all what it absorbed regardless of its colour.
As you said a lower emissivity object would have to heat up more to get rid of that same amount of energy at its lower emissivity.
It does not retain energy.

Kirchoffs law is just stating that a hemisphere heats up at different rates dependent on the angle of the light on the hemisphere. Therefore it will emit at different temperatures exactly proportional to the amount of incoming energy it was able to absorb.

-Thank you for getting me to investigate it a little more and trying to sort out the problem of the outgoing radiation more clearly

Bob Wentworth
Reply to  angech
June 8, 2021 9:47 am

I’m glad you’re trying to make sense of all this.

The generally idea that there is a relationship between what a material can absorb and what it can emit, is, indeed, what Kirchoff’s law is about.

You’re right at that level, but a lot of the detailed ideas you have continue to seem a bit garbled.

Kirchoff’s law requires an incoming source, else there is no law.

No. Your apparent misinterpretation of Kirchoff’s law requires an incoming source. The law as normally understood does not.

(Gustav Kirchhoff‘s 1859 law of thermal radiation) that equates the emissivity of a surface with its absorption of incident radiation (the “absorptivity” of a surface)

The law says that emissivity (which relates to spontaneous radiation independent of any incident radiation) is numerically identical to absorptivity (which relates to incident radiation).

It does NOT say that emissivity relates, in any way, to incident radiation.

As a metaphor… Suppose there it turned out to be true that the number of pizzas delivered to colleges per day is equal to the number of dogs adopted at animal shelters. That doesn’t mean pizzas are being taken to dog shelters.

The things involved are simply numerically equal.

Kirchoff’s law is NOT an expression of energy conservation.

Emissivity is the transfer back to space of all the energy that came in minus the reflectance.

No. Look at the definition of emissivity. It’s about the ratio of power spontaneously emitted to the amount that an ideal blackbody radiator would emit. It has nothing to do with “all the energy that came in.”

You cannot change the definition of emissivity just to make Kirchoff’s law make sense to you.

Hence the surface temp for a blackbody at an emissivity of 1 is the same temperature for a greyer body at 0.94 if they have absorbed and emitting the same amount of energy per surface area.

No. That’s actually contradicted by what you quote a bit later, i.e.,

As you said a lower emissivity object would have to heat up more to get rid of that same amount of energy at its lower emissivity.

Yes.

To see this, suppose the power emitted per unit area is P. We know P = 𝜀𝜎T⁴, so T⁴ = P/𝜀𝜎.

So, an object with a lower emissivity, 𝜀, will have a higher temperature, given the same power P.

Therefore it will emit at different temperatures exactly proportional to the amount of incoming energy it was able to absorb.

Mmm… not exactly.

I know it’s not so satisfying, but I think you’d be better off accepting Kirchoff’s law (emissivity = absorptivity) as a mysterious truth, that two quantities happen to be numerically equal.

It’s nice to have an explanation, but the real explanation is very subtle, and your explanation isn’t quite right.

angech
Reply to  Bob Wentworth
June 8, 2021 3:30 pm

thanks

Jim Gorman
Reply to  angech
June 8, 2021 9:54 am

9. While the phenomenon of scattering means a continuous modification in the interior of the medium, a discontinuous change in both the direction and the intensity of a ray occurs when it reaches the boundary of a medium and meets the surface of a second medium. The latter, like the former, will be assumed to be homogeneous and isotropic. In this case, the ray is in general partly reflected and partly transmitted. The reflection and refraction may be “regular,” there being a single reflected ray according to the simple law of reflection and a single transmitted ray, according to Snell’s law of refraction, or, they may be diffuse,” which means that from the point of incidence on the surface the radiation spreads out into the two media with intensities that are different in different directions. We accordingly describe the surface of the second medium as “smooth” or “rough” respectively. Diffuse reaction occurring at a rough surface should be carefully distinguished from reflection at a smooth surface of a turbid medium. In both cases part of the incident ray goes back to the first medium as diffuse radiation. But in the first case the scattering occurs on the surface, in the second in more or less thick layers entirely inside of the second medium.

From Planck’s thesis on heat radiation.

Gordon A. Dressler
Reply to  Bob Wentworth
June 8, 2021 8:34 am

Bob Wentworth posted: “The issue is that Kirchoff’s Law requires that: “. . . and absorptivity = emissivity. These rules are relevant at any given wavelength and in any given direction.”

Well, that may be Kirchoff’s law as applied to an ideal black body, but it is certainly not generally applicable to the real world, with the understanding that radiation energy that is not absorbed is “lost” due to reflection.

Some everyday examples of the ratio of absorptivity/emissivity over the solar spectrum:
— 11.0 for nickel-oxide coated stainless steel
— 9.7 for plated black chrome metal
— 3.0 for aluminum foil
— 2.1 for dull brass, copper, galvanized steel, aluminum
— 0.68 for red brick
— 0.68 for concrete
— 0.37 for snow, ice granules
— 0.24 for light colored paints, firebrick, clay, glass
— 0.17 for anodized aluminum
— 0.16 for fresh snow having fine particles
— 0.10 for magnesium oxide paint
(source of the above listed values versus materials: http://www.redrok.com/concept.htm#emissivity )