Do Los Niños cause climatic cooling?

By Andy May

We’ve seen a lot of news stories about an upcoming El Niño, that may turn into a so-called “super” El Niño over the next year. This will affect our weather for a year or two, but what is the climatic effect of this weather feature, if any? Here we examine the history of warm ENSO events.

Los Niños warm Earth’s atmosphere for a few years because they cause excess thermal energy (heat) to be expelled from the topical Pacific Ocean and the heat is then circulated around the planet via atmospheric circulation, especially in the Northern Hemisphere where most of us live. But this is warm weather, not climate. Climate is normally defined as the average weather over a period of more than 30 years. Over 30 years, Los Niños are a cooling event since nearly all the heat they transfer to the atmosphere is eventually radiated to space. Very little of the heat released from the oceans during an El Niño is returned to the oceans because downwelling infrared radiation from the atmosphere cannot penetrate the ocean surface (Wong & Minnett, 2018). Only solar radiation can penetrate to the deeper ocean and significantly warm it.

Many Los Niños are very powerful weather features and can be traced back in time with lake sediment proxies in Ecuador as has been done by Christopher Moy and colleagues at Syracuse University (Moy et al., 2002). Figure 1 shows Moy’s El Niño proxy record and Rosenthal’s Makassar Strait proxy temperature record since 0AD. Moy’s sediment record from the Laguna Pallcacocha drainage basin is well located to record warm El Niño events since these events cause anomalous sea surface temperatures off the coast of Ecuador which initiate strong and widespread convection in the area.

Figure 1. The Moy warm El Niño record in blue (left scale) and Rosenthal’s North Pacific temperature record in orange (right scale) overlain. Data sources: (Moy et al., 2002) & (Rosenthal et al., 2013).

The important point is that during the Medieval Warm Period Los Niños were rare and did not become common until the Little Ice Age began around 1200 AD and then declined as the Little Ice Age progressed and the world became colder. They have since become common again as the world has warmed, as shown in figure 2 which is a plot of the NOAA ERSST Niño 3.4 Index where Los Niños are positive and Las Niñas are negative values.

Figure 2. The NOAA ERSST v5 ENSO index from the end of the Little Ice Age (~1850) to the present. Data source: Climate Explorer. In this plot, an El Niño is positive (0.5 or greater) and a La Niña is negative (-0.5 or less).

Los Niños were extremely rare during the Holocene Climatic Optimum, only increasing in number as the Neoglacial began as shown in figure 3. The paucity of Los Niños during the Holocene Climatic Optimum is confirmed by numerous geological proxies from around the Pacific basin as discussed in Moy et al. (Moy et al., 2002).

Figure 3. The Vinther Greenland area temperature and Moy’s warm ENSO proxy (number of events each 100 years).

The paucity of Los Niños during the Holocene Climatic Optimum has been connected to Earth’s orbital cycles by Clement et al. (Clement et al., 2000). A discussion of the effects of orbital cycles on climate can be seen here. During the Holocene Climatic Optimum, Northern Hemisphere summer insolation was maximal. It appears that when this happens Los Niños are suppressed. Since the Neoglacial began, around 3800 BC, Northern Hemisphere summer insolation has declined significantly.

Figures 1 to 3 suggest that a warm stable climate is associated with very few Los Niños, but when Earth’s climate is beginning to cool, as at the beginning of the Neoglacial Period or the early cooling years of the Little Ice Age, there are more Los Niños. Los Niños were very common as we cooled into the depths of the Little Ice Age (~1750 or so) and then as we began to warm coming out of the deepest period of the Little Ice Age the number of Los Niños dropped off. We are currently at the end of Modern Solar Maximum or the Modern Warm Period, and we are seeing more Los Niños, suggesting the world is beginning to cool.

Works Cited

Clement, A., Seager, R., & Cane, M. (2000). Suppression of El Niño during the mid-Holocene by changes in Earth’s orbit. Paleooceanography, 15(6), 731-737. https://doi.org/10.1029/1999PA000466

Moy, C., Seltzer, G., & Rodbell, D. (2002). Variability of El Niño/Southern Oscillation activity at millennial timescales during the Holocene epoch. Nature, 420, 162-165. Retrieved from https://doi.org/10.1038/nature01194

Rosenthal, Y., Linsley, B., & Oppo, D. (2013, November 1). Pacific Ocean Heat Content During the Past 10,000 years. Science. Retrieved from http://science.sciencemag.org/content/342/6158/617

Vinther, B., Buchardt, S., Clausen, H., Dahl-Jensen, Johnsen, Fisher, . . . Svensson. (2009, September). Holocene thinning of the Greenland ice sheet. Nature, 461. Retrieved from https://www.nature.com/articles/nature08355

Wong, E. W., & Minnett, P. J. (2018). The Response of the Ocean Thermal Skin Layer to Variations in Incident Infrared Radiation. Journal of Geophysical Research: Oceans, 123(4). https://doi.org/10.1002/2017JC013351

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June 17, 2026 2:07 pm

As far as I know, the IPCC does a poor job of predicting
El Niño and La Niña events. Why should we believe them
about anything else?

Nick Stokes
June 17, 2026 2:16 pm

“Very little of the heat released from the oceans during an El Niño is returned to the oceans because downwelling infrared radiation from the atmosphere cannot penetrate the ocean surface”

Dumb stuff endlessly repeated. Of course a warm atmosphere can warm the ocean. But what is missing here is mention of La Niña. That is when the equatorial Pacific turns cool and, yes, cools the atmosphere. Heat flows back into the ocean. It’s just as important for total heat flow.

This is all just redistribution of heat, not a long term source or sink. Which is why we had that cycle for centuries with steady temperature.

Reply to  Nick Stokes
June 17, 2026 2:30 pm

Yes it’s an oscillatory process involving the heat transfer between the ocean and atmosphere, based on nonlinear processes.

Reply to  Andy May
June 17, 2026 3:05 pm

This is clear, and demonstrated by the AMO, PDO, etc. Not to mention the very clear 1000-year cycle that is seen in the Medieval Warm Period, Little Ice Age, and Modern Warm Period.”

Yes, in cherry picked proxies. Admitted by you:

“Either way your quote is appropriate, thanks. Bottom line, proxies are all over the place sometimes, so you can select a group of proxies to combine to form whatever reconstruction you want. Best to deal with the proxies one at a time and in light of historical records. Basically, all reconstructions, including my own, are crap.”

Reply to  Andy May
June 17, 2026 3:31 pm

That said, you are correct, proxies are all over the place and must be chosen very carefully and we have to remember that they are only valid at their physical location,”

Certain regions, like the High Arctic (where ice–albedo feedback plays a major role) are actually more representative of global temperature changes than others.

https://postimg.cc/ZW5JFBX7

https://www.pnas.org/doi/10.1073/pnas.1616287114

Reply to  Eldrosion
June 17, 2026 3:34 pm

Here is the image.

pnas.1616287114fig01
Reply to  Eldrosion
June 17, 2026 4:36 pm

GISP looks like this.

That tiny uptick at the end starts in 1900 (according to Mann’s hockey stick). And 1940’s at the peak of that spike was similar or warmer than the first decades of this century.

GISP
Reply to  bnice2000
June 17, 2026 7:27 pm

#1) That refers to Central Greenland, not the High Arctic, which is the proxy I was citing. Did you read what Andy May just said? He wrote:

“That said, you are correct, proxies are all over the place and must be chosen very carefully and we have to remember that they are only valid at their physical location

So this doesn’t debunk anything. Andy May even acknowledged the validity of the hockey stick pattern shown in the Agassiz record:

“I guess the best way to look at it is the Arctic is the trigger and early warning system”

As such, perhaps he could write a post on WUWT about the profound future impacts of rapid warming in the High Arctic.

#2) That ice core record only extends to 95 BP, which corresponds to 1855. This would be clear to readers if the chart included an x-axis.

I pointed this out to you in a previous thread, and it was not addressed, yet you are presenting the same chart again.

So the question of whether the chart’s x-axis was deliberately omitted is even more pertinent now.

Reply to  Eldrosion
June 17, 2026 11:23 pm

Others disagree

greenland-temp
Reply to  Eldrosion
June 17, 2026 11:24 pm

As do 18O isotopes

O18-Greenland
Reply to  Eldrosion
June 17, 2026 11:28 pm

and there is no sign of any major uptick in the GHCN data, in fact 1930’s were warmer than 2000-2020

Greenland-Dec-temps
Reply to  Eldrosion
June 18, 2026 4:59 am

As such, perhaps he could write a post on WUWT about the profound future impacts of rapid warming in the High Arctic.

Perhaps instead of pushing this off on someone else you could use your obvious knowledge of Arctic climate to write it yourself.

Reply to  bnice2000
June 17, 2026 11:31 pm

Another Greenland temperature series.

As anyone can see, it was cooler in the early 2000s than it was around 1930.

And there is absolutely zero relationship to CO2 emissions.

Reply to  bnice2000
June 17, 2026 11:32 pm

oops… graph didn’t attached

CO2-Emissions-and-Greenland
Reply to  Eldrosion
June 17, 2026 8:01 pm

From your PNAS link: “In the case of the melt record, this correction is applied at the location of the ice core to remove the contributions from vertical land motion and changes in ice thickness.” What is missing is an explanation for how to derive a verifiable lapse rate. It is so cold that there is little water vapor present and a small change in water vapor will cause a relatively large change in the ‘dry’ lapse rate.

I believe that your faith in the effect of the “ice-albedo feedback” is unwarranted. The sun is absent for nominally half the year, and during the Summer, the Arctic is notoriously overcast, significantly reducing solar energy arriving at the surface. During the times when the clouds are missing, a significant portion of the solar radiation is reflected forward into space and is neglected in the arguments about the “ice-albedo feedback.”
https://wattsupwiththat.com/2016/09/12/why-albedo-is-the-wrong-measure-of-reflectivity-for-modeling-climate/

Reply to  Clyde Spencer
June 17, 2026 11:34 pm

Your allegation concerns the early Holocene elevation correction associated with Innuitian Ice Sheet thinning.

At most, that affects the amplitude of the early Holocene thermal maximum.

Those corrections are negligible to non existent in the more recent time periods of the record, so it does not directly undermine the hockey stick shape or the conclusion regarding unusually rapid recent warming.

“The sun is absent for nominally half the year, and during the Summer, the Arctic is notoriously overcast, significantly reducing solar energy arriving at the surface. During the times when the clouds are missing, a significant portion of the solar radiation is reflected forward into space and is neglected in the arguments about the “ice-albedo feedback.””

When summer sea ice retreats, highly reflective ice is replaced by darker open water. This allows a greater fraction of incoming solar radiation to be absorbed by the ocean.

Much of this energy is stored in the ocean during summer and released to the atmosphere during autumn and winter, when the temperature contrast between the relatively warm ocean and the colder atmosphere is greatest.

You don’t have the mechanism correct.

Skeptics believe that researchers who specialize in Arctic climate have somehow overlooked the fact that the Arctic experiences months of darkness and frequent cloud cover.

Reply to  Eldrosion
June 18, 2026 5:08 am

When summer sea ice retreats, highly reflective ice is replaced by darker open water. This allows a greater fraction of incoming solar radiation to be absorbed by the ocean.

Much of this energy is stored in the ocean during summer and released to the atmosphere during autumn and winter, when the temperature contrast between the relatively warm ocean and the colder atmosphere is greatest.

Ok, summer ice melts causing a retreat. That leads to an inference that fall and winter causes more sea ice to occur.

Please explain how the ocean releases heat to the atmosphere through the ice that forms. Is the release much quicker than the storage? I would also like to know the temperature differentials we are talking about. What is the value of the change in temperature that occurs during this heating?

Reply to  Andy May
June 18, 2026 11:33 am

‘See (Cronin & Jansen, 2016)’

Is this the same Timothy Cronin who maintains that even a highly linear relationship between Earth’s heat loss to space and temperature is consistent with radiative energy transfer?

https://www.pnas.org/doi/10.1073/pnas.1809868115

pnas.1809868115fig01
Reply to  Andy May
June 18, 2026 2:34 pm

Thanks, Andy. I indeed became acquainted with the graphic in your excellent ‘energy-and-matter’ post. Sorry for going off topic – just wanted to drop a hint that this author has a bias towards ‘normal’ science, i.e., defending the radiative-centric AGW narrative.

Reply to  Eldrosion
June 20, 2026 8:22 pm

When summer sea ice retreats, highly reflective ice is replaced by darker open water.

The appearance of Arctic water is highly dependent on the viewing geometry. The light coming from water has two components: 1) Diffuse reflectance from suspended particles such as sediment and plankton, which is a function of the concentration of particles and the amount of light that enters the water; and 2) Specular reflection from the planar surface, which is controlled by the angle of incidence. Light that enters the water is (1-reflectivity); that is, at glancing angles, the reflectivity is 100% and NO light enter the water, so even if there are suspended particles, there is no light to be diffusely reflected. The specular reflection is a sheaf of rays leaving the water surface at the same angle as the incident rays. To see the specular reflection, one has to be looking towards the sun at approximately the same angle as the incident rays. It is you who doesn’t “have the mechanism correct.”
comment image
Note how much brighter the specular reflection is than the ice floes.

Reply to  Clyde Spencer
June 22, 2026 9:10 am

Polarization of the incident light is also a major factor.

Reply to  Clyde Spencer
June 18, 2026 5:16 am

Greenland ice mass balance was more negative in the 1930s and 40s..

Greenland-Surface-Mass-Balance-Fettweis08
Reply to  Clyde Spencer
June 18, 2026 5:18 am

And Greenland ice sheet area is only a tiny amount down from the peak of the LIA.. and far higher than most of the last 7000-8000 years

(note are in this chart includes all the regional islands.)

Greenland-Ice-Sheet-Briner
Reply to  bnice2000
June 18, 2026 7:30 am

You’re showing this fake graph again! Briner et al. does not include data for 2015 that is your mistaken addition to the graph, in fact no data for the last 75 years was included. This is their representation of the Greenland Ice Sheet and does not include ‘regional islands’, the current Greenland Ice Sheet area is ~1.7×10^12 m^2.

Reply to  Phil.
June 18, 2026 9:00 am
Reply to  Eldrosion
June 20, 2026 8:34 pm

You claim the mantle of truth, but demonstrate above that you are only repeating what you have read elsewhere and don’t really grasp the fundamentals of the issue.

Jeff Alberts
Reply to  Eldrosion
June 17, 2026 5:07 pm

Do you chastise D’Arrigo for her proud cherry picking?

Nick Stokes
Reply to  Andy May
June 17, 2026 3:56 pm

Andy,
I’ve explained the proper scientific way of analysing it, which is balancing heat flux at the surface. But there are elementary indicators that you are wrong:
_1. Our bodies are just as opaque to IR as seawater. But it is common experience that IR warms us.

That is similar to the ocean surface. Heat is emitted by the body. When IR impinges on the surface, the net heat loss is reduced. Our bodies gain heat, but nothing had to penetrate.

_2. The example of La Niña. That is when the equatorial Pacific ocean becomes colder. Then the air becomes colder. How can that be if the air’s warmth does not pass into the ocean?

Reply to  Nick Stokes
June 17, 2026 4:38 pm

There is very little sign of any La Nina affect in the UAH data, but the major El Nino events stand out as a spike + step event …

… with basically no warming between them.

UAH-global-with-near-zero-trend-sections
Reply to  bnice2000
June 17, 2026 11:55 pm

Red cherry, green cherry, purple cherry.

Reply to  Andy May
June 17, 2026 7:44 pm

And, I believe that much of the ocean warming occurs around the coasts where the water is shallow, the heat can then move out into the open ocean by convection and overturning.
I had occasion to experience this in Hawaii. The shallow water near the beach (about 1-2 metres deep) was very warm – almost too warm (perhaps 27C) – but where the coastal shelf suddenly gave way to deep water, the temperature below was icy cold comparably. (perhaps 8C) This shallow water is warmed directly by the sun, not Nick’s stupid atmosphere.

Dave Burton
Reply to  Andy May
June 24, 2026 7:12 pm

#1 is wrong, Andy, as Wong and Minnett confirmed. LW IR absorbed by the ocean has the same warming effect as sunlight absorbed by the ocean. A Joule is a Joule, regardless of whether it arrives as shortwave radiation or longwave radiation, and regardless of the depth at which it is absorbed.

#2 is debated. A correlation has been observed between the 22-year solar cycle and the onset of some La Niñas, but it is unclear whether it’s causal. Most La Niñas onsets are not correlated with the solar cycle, and neither are El Niños.
comment image

Reply to  Nick Stokes
June 17, 2026 7:35 pm

_1. Our bodies are just as opaque to IR as seawater. But it is common experience that IR warms us.

That is similar to the ocean surface. Heat is emitted by the body. When IR impinges on the surface, the net heat loss is reduced. Our bodies gain heat, but nothing had to penetrate.

Beyond stupid.

Anthony Banton
Reply to  Mike
June 18, 2026 8:54 am

“Beyond stupid.”

Not but Ditto

Dave Burton
Reply to  Mike
June 24, 2026 7:22 pm

NIck is correct about that.

The people who think that LW IR does not warm you (or does not warm the ocean), for whatever reason, e.g., because they cannot see LW IR, or because they misunderstand the 2nd Law of thermodynamics, are just plain wrong. Full stop.

The reason “space blankets” are reflective isn’t to reflect away incoming sunlight; that’s an unfortunate side-effect. Rather, they are reflective to reflect back outgoing LW IR, and thereby help keep you warm. The reflective coating is advantageous because, in the conditions under which you need a blanket, there’s more LW IR departing from your body than there is incoming visible light.
comment image

leefor
Reply to  Nick Stokes
June 17, 2026 9:11 pm

“_1. Our bodies are just as opaque to IR as seawater. But it is common experience that IR warms us.”

When does IR warm us rather that direct sunlight?

Nick Stokes
Reply to  leefor
June 18, 2026 12:19 am

Around a fire on a cold night….

leefor
Reply to  Nick Stokes
June 18, 2026 2:20 am

Oh, only on a cold night. Got it. lol

Reply to  Nick Stokes
June 18, 2026 7:36 am

When I work out on the track in the evening after a sunny day I can feel the warmth radiating from the concrete wall as a walk near it, no direct sunshine.

Anthony Banton
Reply to  Phil.
June 18, 2026 9:01 am

And when my radiators are on in winter even though the air temperature remains the same it always feels warmer than when the thermostat switches them off.

Pub garden heaters work via IR. Had one in my parents bathroom as a kid.

Same with my log-burner – kicks out IR.

This really is bizarre – not comprehending the fact that LWIR is thermalised by water and human bodies … but then again WUWT is an alternate Universe.

Reply to  Nick Stokes
June 18, 2026 3:39 am

I’ve explained the proper scientific way of analysing it, which is balancing heat flux at the surface.”

Heat flux at the surface can NEVER balance. Since heat flux-in occurs over a different time interval then heat flux-out THEY CAN NEVER BALANCE. What has to balance is joules-in and joules-out over time.

If as Andy says, El Nino is the ocean increasing heat flux-out by dumping heat accumulated in the deeper ocean then *the* joules associated with that will contribute to the joules-in/joules-out balance, NOT to flux-in/flux-out balance.

The “balance interval” needs to be at least as long as the intervals between El Ninos. It actually needs to be *much* longer than that to observe *all* of the cyclical factors affecting the joules-in/joules-out balance.

Our bodies gain heat, but nothing had to penetrate.”

Our bodies gain heat because of metabolic heat generated inside the body, the body is a HEAT SOURCE. Where is the analog heat source for the earth? Where does the heat gain come from? Slower cooling is *NOT* heat gain, it is slower heat LOSS.

Reply to  Tim Gorman
June 18, 2026 10:04 am

“Where is the analog heat source for the earth?”

The sun.

Reply to  Bellman
June 18, 2026 3:21 pm

The sun is *NOT* an internal heat source like the body’s metabolism. There is no internal heat source in the earth that can cancel out the cooling and *raise* the temperature of the earth.

The sun is the ONLY heat source in the system that needs to be considered, all geothermal sources are orders of magnitude less.

Slower cooling is STILL cooling!

Reply to  Tim Gorman
June 18, 2026 5:53 pm

It’s irrelevant if the heat source is internal or external.

You keep making this dumb “slower cooling” chestnut, without understanding that slower cooling results in warming. If the sun is warming the oceans by the same amount each day, but each night the oceans are cooling slower, what happens to the temperature?

Reply to  Andy May
June 18, 2026 7:37 pm

Yes. Do you understand the point?

Unless Tim believes that the oceans have been always cooling, are currently at the warmest they will ever be and at best will cool at a slower rate in the future, then you have to see the cooling as part of warming/cooling cycles – either day/night or summer winter. Slowing the rate of the cooling phase without also reducing the rate of the warming phase would result in an overall rise in sea temperature.

Reply to  Bellman
June 19, 2026 5:06 am

Unless Tim believes that the oceans have been always cooling”

The entire earth has been cooling over its entire life. Otherwise, we’d be living on a molten rock.

“are currently at the warmest they will ever be and at best will cool at a slower rate in the future”

Again, the cooling function is F(t) = F0 * e^-(t/τ)

What does an exponential decay curve look like? Does the slope of the curve get less over time?

then you have to see the cooling as part of warming/cooling cycles – either day/night or summer winter.”

How about over glacial/inter-glacial periods? Heat gain/loss is a continual process over ALL time, not a piece of time limited to diurnal or seasonal.

“Slowing the rate of the cooling phase without also reducing the rate of the warming phase would result in an overall rise in sea temperature.”

Bullshite! Slower cooling IS STILL COOLING. Slower cooling DOES NOT PUSH TEMPERATURE BACK UP THE GRADIENT. It just means the temperature doesn’t go down as fast, BUT IT STILL GOES DOWN!

The temperature of the earth either goes down or it goes up. If it has been going up due to CO2, which has existed in the atmosphere over all of the millennia, then it would be either a molten ball as when it started or it would have reached temperature equilibrium with the sun.

If the temperature of the earth is going down then it simply doesn’t matter how fast it is going down, it will sooner or later reach an equilibrium with the sun. Has the earth reached an equilibrium point with the sun? If so, then do you believe there will be another glacial period in our future or do you believe we will never see another glacial period?

Reply to  Tim Gorman
June 19, 2026 5:31 am

“The entire earth has been cooling over its entire life. Otherwise, we’d be living on a molten rock.”

Really? You think the earth is colder now than it was during the last ice age?

“Again, the cooling function is F(t) = F0 * e^-(t/τ)”

Just keep ignoring the sun

“The temperature of the earth either goes down or it goes up.”

So you admit temperature can go up. So why do you keep trolling out that cooling function?

“If it has been going up due to CO2, which has existed in the atmosphere over all of the millennia, then it would be either a molten ball as when it started or it would have reached temperature equilibrium with the sun.”

Stop changing the subject. This discussion is about infra red warming the oceans not co2 specifically. But apart from that, you just keep displaying your ignorance. You said yourself in the previous comment that as temperature increases so does heat loss. If you actually tried to think through this you would understand why that tends to equilibrium rather than ever increasing temperatures.

“Has the earth reached an equilibrium point with the sun?”

Nope. And that’s exactly why the concept of the greenhouse effect was discovered.

Reply to  Bellman
June 19, 2026 5:54 am

Really? You think the earth is colder now than it was during the last ice age?”

How in Pete’s name do you think the last ice age happened if the earth wasn’t cooling somehow?

Just keep ignoring the sun”

The sun doesn’t stop the earth from cooling. It only changes the rate of cooling. I.e. “τ”. It doesn’t change e^-(t/τ) to e^(t/τ).

“So you admit temperature can go up. So why do you keep trolling out that cooling function?”

Unbelievable. Does the word “either” not mean anything to you? Your reading comprehension skills are just truly sad.

Stop changing the subject.”

No changing of the subject. Either the earth is warming or it is cooling. If it’s warming then it would be warming from the very start – meaning it would *still* be a molten rock.

You just don’t want to have to live with what you are asserting so you are looking for an excuse.

You said yourself in the previous comment that as temperature increases so does heat loss.”

So what? Are you disputing that simple truth?

” If you actually tried to think through this you would understand why that tends to equilibrium rather than ever increasing temperatures.”

You don’t even understand that the system is a three-body system. The sun, the earth, and space (about 0K).

The equilibrium point is where the heat loss to space balances the heat gain from the sun.

Reply to  Tim Gorman
June 19, 2026 7:33 am

“How in Pete’s name do you think the last ice age happened if the earth wasn’t cooling somehow?”

Dodging the question, as usual. Your claim may or may not be that the world is constantly cooling. I’m just trying to determine what you actually think “slower cooling is still cooling” means in the real world.

“The sun doesn’t stop the earth from cooling.”

That’s quite a take.

“I.e. “τ”. It doesn’t change e^-(t/τ) to e^(t/τ).”

What do you think your equation means? It clearly doesn’t describe anything over a 24 hour period. Temperatures usually rise when the sun’s out.

“Does the word “either” not mean anything to you?”

You said either the temperature goes up or down. That could mean some times they go up and sometime down which would be a sane thing to say, or it could mean the they can be ever decreasing or ever increasing. I take it that’s your opinion. That if temperatures go down they can never go up. But it’s impossible to say for sure as you keep equivocating on the subject. You change the argument whenever it suites you.

“No changing of the subject. Either the earth is warming or it is cooling.”

The subject was about oceans and whether a warmer atmosphere increased ocean temperatures.

“So what? Are you disputing that simple truth?”

Nope. I’m pointing out why that simple truth disproves your assertion.

“The sun, the earth, and space (about 0K).”

Nope, it’s the ocean the sun and the atmosphere. Even if you want to change the subject to global warming, it’s the surface, the atmosphere, space and the sun.

“The equilibrium point is where the heat loss to space balances the heat gain from the sun.”

And that heat loss is slowed down by the atmosphere.

Reply to  Bellman
June 19, 2026 12:07 pm

Dodging the question, as usual.”

Why don’t you just admit you don’t know the answer? The only one dodging the question is YOU!

“Your claim may or may not be that the world is constantly cooling.”

The earth *is* constantly cooling. There simply isn’t any doubt about it. The entire earth loses heat to space 24 hours per day. Why is this so hard to understand?

“That’s quite a take.”

You mean you have absolutely no understanding of the thermodynamics of the earth’s biosphere at all. Again, the system has three bodies, the sun, the earth, and space. And space surrounds the entire earth 24 hours per day. Since the earth is warmer than space it moves heat to space 24 HOURS PER DAY.

What do you think your equation means? It clearly doesn’t describe anything over a 24 hour period. Temperatures usually rise when the sun’s out.”

The earth radiates to space 24 hours per day! The flux it emits is driven by the decay equation F(t) = F0 * e^-(t/τ). It’s not much different from the voltage decay of a capacitor through a resistor!

F0 is a function of its own. If the earth starts with flux determined by 200K then the next second the flux it emits will be 200k – e^-(1/τ). Have you bothered to look up what “τ” is? It’s a function of its own also, it’s just doesn’t have time as an independent variable.! Now, if the earth is getting energy from the sun then F0 will no longer be 200K. F0 will change based on the balance between the energy in and the energy out at 200K.

Why do you think Tmax many days happens around 3pm local while sunset is around 9pm during the summer? The exponential decay of the earth’s temperature doesn’t start at sunset, it starts when the exponential decay value is greater than the input from the sun – about 3pm!

This is why I said this is a difficult equation to calculate. F0 is not a constant.

Apparently the heat *energy* form of the exponential decay doesn’t make any sense to you either. Q(t) = Q0 * e^-(t/τ).

 That could mean some times they go up and sometime down which would be a sane thing to say,”

Your reading comprehension skills seem to be getting worse and worse. Either that or you are cherry picking again. You left off the full context of what I said (just like you do with the GUM): “ If it’s warming then it would be warming from the very start – meaning it would *still* be a molten rock.” That does *NOT* imply that sometimes the earth warms and sometimes it cools. The earth cools *all* the time, it’s just that sometimes it cools at a slower rate than at other times!

“That if temperatures go down they can never go up”

You are *still* showing a fundamental lack of knowledge of how the biosphere works. That is *NOT* what I am saying at all. If this were the case Tmax for tomorrow could never be higher than Tmax today.

Again, for the umpteenth time, it is HEAT LOSS and HEAT GAIN, not temperature. It is Energy-in vs Energy-in!

” But it’s impossible to say for sure as you keep equivocating on the subject.”

It’s impossible for you to say because 1. you simply can’t read for meaning, and 2. all you ever do is cherry pick rather than trying to actually learn the subject.

The subject was about oceans and whether a warmer atmosphere increased ocean temperatures.”

And as you’ve been told and shown with MATH that a colder atmosphere cannot warm the ocean. The ocean *always* loses energy to the cooler atmosphere. The loss of energy simply CAN NOT increase the temperature of anything! Only heat input from a heat source can increase energy. CO2 is *NOT* a heat source.

“Nope. I’m pointing out why that simple truth disproves your assertion.”

You have yet to produce ANY math associated with thermodynamics that shows a colder object can increase the energy of a warmer object. That “simple truth” of yours exists solely in your mind and no where else (except perhaps bdgwx and Nick).

Nope, it’s the ocean the sun and the atmosphere. Even if you want to change the subject to global warming, it’s the surface, the atmosphere, space and the sun.”

Show the math. Show the thermo equations between the earth, the sun, and space that determine the energy-in and energy-out of the system.

“And that heat loss is slowed down by the atmosphere.”

Then why isn’t the earth a molten ball of rock floating in space? If energy in is consistently greater than heat out then it’s no different than applying the flame of an acetylene torch to a gram of silver. That silver becomes molten.

Again, it isn’t the RATE of energy loss that is at issue. It’s the joules-in vs the joules-out.

Reply to  Bellman
June 20, 2026 8:48 pm

Temperatures usually rise when the sun’s out.

That is only the thin veneer that we call the surface and atmosphere that periodically sees an increase in solar radiation that significantly exceeds the loss of geothermal heat. However, the net result is a loss of heat. This argument goes back to the time of Lord Kelvin. You aren’t keeping up.

Reply to  Clyde Spencer
June 21, 2026 5:23 am

” However, the net result is a loss of heat. ”

So do you think the world is peremenently cooling? If so by how much, and when should we expect the world to completely freeze up?

” This argument goes back to the time of Lord Kelvin.”

Lord Kelvin thought we were about to run out of oxygen and that the earth was only a few million years old.

Reply to  Bellman
June 21, 2026 9:11 am

So do you think the world is peremenently cooling?”

F(t) = F0 * e^-(t/τ)

Once again, F0 is based on the earth’s temperature. The only way F(t) can be zero is for temperature to be 0 (zero).

The world *IS* permanently cooling.

Reply to  Tim Gorman
June 21, 2026 10:20 am

“F(t) = F0 * e^-(t/τ) ”

It’s just number are numbers with you, isn’t it. The question is, is the world actually getting colder. How cold will it be in 10 years time based on your equation.

All your equation is saying is that the flux out is constantly dropping. So what happens when it drops below the flux from the sun? Does the flux out continue to drop. If it does, won’t that means the temperature keeps increasing?

“The world *IS* permanently cooling.”

And I’m permanently falling towards the center of the Earth, and the Earth is constantly falling towards the sun.

Reply to  Bellman
June 21, 2026 10:46 am

The question is, is the world actually getting colder.

The question for the curious is WHY is the world getting either warmer or colder. That will allow accurate prediction into the future. If science don’t or won’t do the proper thermodynamic analysis, no one will know what to expect.

Reply to  Jim Gorman
June 21, 2026 10:52 am

Which is why it’s important to understand how the greenhouse effect works rather than pretending it doesn’t.

Reply to  Bellman
June 24, 2026 3:50 am

The Earth experienced GLACIATION with TEN TIMES today’s atmospheric CO2. Where was CO2’s supposed “climate driving power” then?!

Observations Trump theory (or in this case, hypothesis).

Reply to  Bellman
June 20, 2026 8:40 pm

There is little doubt that the interior of the Earth is cooler now than it was 20,000 years ago. The surface is another story.

Reply to  Clyde Spencer
June 21, 2026 5:23 am

Well it’s the surface I’m most interested in. It’s where I live.

Reply to  Tim Gorman
June 19, 2026 8:45 am

Misuse of terminology is a cause of confusion. Regarding the heat transfer, it’s heat input and heat loss, regarding the response of the surface its heating and cooling. When heat loss rate exceeds the heat input rate the surface will cool.

Reply to  Phil.
June 20, 2026 8:59 pm

You are talking about a boundary effect between a huge source of declining heat from the residual heat and a declining source of radioactive elements providing heat, as modified by the solar energy that sweeps across the surface every 24-hours, and further modulated by changes in cloudiness. Thus, the boundary layer temperature fluctuates with the seasons and the solar elevation. However, the net effect is a slow loss of heat energy to space.

Dave Burton
Reply to  Andy May
June 24, 2026 7:33 pm

Bellman wrote, “Slower cooling results in warming.”

Andy replied, Hilarious! Did you read what you wrote?”

Bellman is right.

Andy, this is a terminology problem. It sounds like you don’t understand what “warming” means in this context.

“Warming” in this context does not mean to make something warmer than it was. It means to make something warmer than it otherwise would be.

A blanket warms you even if you’re slowing freezing to death. If being wrapped in a blanket meant that your body temperature fell more slowly, and it took longer to freeze to death than it would have taken without the blanket, then the blanket “warmed” you. That’s what “warming” means in this context.

Reply to  Dave Burton
June 25, 2026 6:04 am

A blanket warms you even if you’re slowing freezing to death. If being wrapped in a blanket meant that your body temperature fell more slowly, and it took longer to freeze to death than it would have taken without the blanket, then the blanket “warmed” you. That’s what “warming” means in this context.”

It just means slower cooling. The blanket cannot inject additional heat into the system. It can only slow the heat loss by the system.

Slower cooling is not “warming”. Saying that slowing the cooling of a thermodynamic system is “warming” the system just makes no thermodynamic sense. The blanket cannot drive the temperature profile back up the temperature gradient, the temperature gradient will never change direction, the slope will not change from negative to positive at any point, being it location or time.

If the slower cooling allows the heat source of the body to inject more heat than is lost then it is the body’s metabolism that causes the warming, not the blanket. The blanket still does nothing except slow the cooling. It’s simply not a heat source.

The same thing applies to CO2. It is not a heat source. It can’t push temperature back up the gradient.

In fact, the example itself is flawed. The blanket should have a huge hole in it representing water vapor. The example should be more like the adding a 4″ leather belt around the body.

Reply to  Dave Burton
June 25, 2026 7:11 am

A blanket warms you even if you’re slowing freezing to death.

Thermodynamically speaking, warming does not occur.

If (Tₜ < Tₜ₋₁), then thermodynamic cooling has occurred.

When the issue of time is introduced, one is dealing with a gradient. As you explained, the gradient would be reduced but not to the point of changing to a warming gradient. That is, from a negative slope to a positive slope.

These are things one learns in physical science which is revealing.

Reply to  Bellman
June 18, 2026 6:14 pm

If the sun is warming the oceans by the same amount each day, but each night the oceans are cooling slower, what happens to the temperature?

Tmin will rise. I don’t know anyone in climate science that says a rising Tmin will make the earth burn up.

Do you think Tmin can warm beyond Tmax?

Reply to  Jim Gorman
June 18, 2026 7:42 pm

Tmin will rise.

Which demonstrates the “slower cooling” argument is nonsense.

But if Tmin rises, you should also expect Tmax to rise. If you are starting each day from a warmer minimum, and you get the same amount of energy from the sun, why would you not expect to reach a higher daily maximum?

“I don’t know anyone in climate science that says a rising Tmin will make the earth burn up.”

Why do you keep wanting the world to “burn up”? Has anyone claimed it will literally do that?

Reply to  Bellman
June 19, 2026 5:10 am

But if Tmin rises, you should also expect Tmax to rise”

Tmax has a boundary condition – namely T^4. It’s why Tmax doesn’t just soar without boundary even today.

” If you are starting each day from a warmer minimum, and you get the same amount of energy from the sun, why would you not expect to reach a higher daily maximum?”

Because T^4 goes up faster than T.

Why do you keep wanting the world to “burn up”? Has anyone claimed it will literally do that?”

That *IS* the result of your hypothesis. You are the one that says an increasing Tmin implies an increasing Tmax. If slower cooling pushes Tmin up continually then Tmax will go up continually – a positive feedback loop progressing to destruction.

Reply to  Tim Gorman
June 19, 2026 5:43 am

“Tmax has a boundary condition – namely T^4. ”

Yes, if everything else is constant the earch teaches a point where it’s average temperature is sufficient to output the same energy as it’s receiving on average from the sun. But if you block some of the energy from leaving the average temperature has to rise in order to reach a new equilibrium. But that doesn’t just apply to the maximum temperature.

“Because T^4 goes up faster than T. ”

So?

“If slower cooling pushes Tmin up continually…”

It doesn’t, but thanks for resorting to another strawman argument.

Reply to  Bellman
June 19, 2026 6:05 am

 But if you block some of the energy from leaving the average temperature has to rise in order to reach a new equilibrium.”

The entire mythical meme that CO2 *blocks* or “traps” heat has been debunked over and over and over again. Yet here you are, trying to revive it. It’s based on an assumption that in and out flux has to balance at any point in time. Flux in and flux out simply can *NOT* balance at any point in time.

If CO2 “trapped” energy the earth would *still* be a molten rock since CO2 has existed in the atmosphere for most of its existence. Meaning heat would have built up and up and up over time until everything melted!

So?”

joules-in is related to Q which then determines T, a linear relation. Joules-out is related to T^4. Somehow that means nothing to you?

It doesn’t, but thanks for resorting to another strawman argument.”

You can’t have it both ways. If CO2 traps heat then Tmin is going to go up. So which is it? Does CO2 TRAP heat or does Tmin not continue to go up?

Pick one and stick with it.

Reply to  Tim Gorman
June 19, 2026 8:02 am

“The entire mythical meme that CO2 *blocks* or “traps” heat has been debunked over and over and over again.”

Debunked in your imagination.

“Flux in and flux out simply can *NOT* balance at any point in time. ”

They don’t have to balance at a single moment in time. They just have to balance over a reasonable time frame.

“If CO2 “trapped” energy the earth would *still* be a molten rock since CO2 has existed in the atmosphere for most of its existence.”

That’s you delusion. It’s not what anybody with even the slightest understanding of the subject says. More energy in the system rises temperature which means more energy is released. Hence a new equilibrium.

“joules-in is related to Q which then determines T, a linear relation. Joules-out is related to T^4. Somehow that means nothing to you?”

Every time I think these arguments can’t get any dumber….

Suppose you have an object in equilibrium. It’s temperature is such that in equals out. Double the energy in flux does not double Q. In order to return to equilibrium Q just has to increase to the point where out equals in. This will happen at the point where out has doubled.

The T⁴ part doesn’t change that. It just determines what the new equilibrium temperature will be.

“If CO2 traps heat then Tmin is going to go up. ”

What do you mean by “traps”. I suspect this is another case of you reading to much into a simple metaphore. I said that greenhouse gases block some of the outgoing energy. That blocked energy eventually radiates back to the surface increasing the temperature. That will increase the energy being emitted and again some of that will be blocked. But eventually the increased energy being emitted will be sufficient to conteract the blocking effect. Temperatures do not increase indefinitely.

Reply to  Bellman
June 19, 2026 12:49 pm

They don’t have to balance at a single moment in time. They just have to balance over a reasonable time frame.”

What’s reasonable? Millennia? Centuries? Decades? Years? Days? Hours? Seconds?

“More energy in the system rises temperature which means more energy is released. Hence a new equilibrium.”

You are running away from your own assertion! If CO2 traps heat then a new equilibrium will NEVER be reached. The energy accumulated will continue to go up. It’s only when MORE heat is input to the system that a new equilibrium will be reached. But CO2 is *NOT* a heat source, it is a reflector of emitted heat!

“Every time I think these arguments can’t get any dumber….”

It’s YOUR math, not mine. T goes up linearly, heat loss goes up by T^4. You can’t seem to grasp that means a negative feedback!

“Suppose you have an object in equilibrium. It’s temperature is such that in equals out. Double the energy in flux does not double Q.”

OMG! Let the flux = 10 joules/sec-m^2. In one second 10 joules gets transmitted through every m^2. Double the flux to 20 joule/sec-m^2.

What happens to the joules transmitted every second into every m^2?

(hint: Q is measured in joules)

In order to return to equilibrium Q just has to increase to the point where out equals in.”

How did you double the flux? CO2 is *NOT* a source of heat energy.

Once again, this is an issue with climate science trying to use flux as a metric for energy. It isn’t!

Once again, if the sun is transmitting 100 joules/sec and the earth is emitting 100 joules per second and CO2 is reflecting 50 joules/sec of that 100 joules/sec being emitted, you have a net of 50 joules/sec out! You do *NOT* have 150 joules/sec going in with only 50 joules/sec out!

You keep forgetting that the 50 joules/sec that CO2 is reflecting HAS ALREADY BEEN LOST by the earth. The earth is DOWN by 100 joules/sec before CO2 can send anything back! All CO2 can do is replace part of what has been lost, not all of it let alone MORE than has already been lost!

Heat transfer is a TIME FUNCTION. You have to analyze it over time.

That blocked energy eventually radiates back to the surface increasing the temperature.”

It can’t raise the temperature. It is only replacing part of what has already been lost. It SLOWS cooling, that is all! Cooling doesn’t result in increased temperature. Reflected heat simply can’t drive the temperature of the emitter BACK UP THE GRADIENT.

READ PLANCK! The 50 joules/sec that is reflected gets re-emitted. The emitting object now emits the 50 joules/sec that was reflected plus 50 joules/sec based on its temperature (50 + 50 = 100). 50 joules/sec re-emitted plus an additional 50 joules/sec is COOLING. Cooling does not cause a temperature increase.

Think about it for a minute. If CO2 blocks outgoing IR what does it do to incoming IR? If it blocks 50 going out it blocks 50 coming in. 50 – 50 = 0. Where is the warming?

Or are you now going to say what climate sciences says? That CO2 is a one-way blocker?

Reply to  Bellman
June 20, 2026 9:11 pm

If CO2 drives warming, why does the well-mixed CO2 result in step increases.?

Reply to  Clyde Spencer
June 21, 2026 5:01 am

I’ve no idea what you mean by “drives” or what step changes you think are happening.

bdgwx
Reply to  Clyde Spencer
June 21, 2026 6:56 am

If CO2 drives warming, why does the well-mixed CO2 result in step increases.?

I created this graphic specifically to help people visualize what is happening. Basically when you superimpose a persistent long-term trend with transient short-term cyclic processes you get a pause-up-pause-up behavior.

comment image

This alone isn’t proof that CO2 is the cause of the warming. But it does falsify the hypothesis that there is no correlation between CO2 and warming and shows that the pause-up-pause-up pattern is consistent with it.

Dave Burton
Reply to  bdgwx
June 24, 2026 8:22 pm

bdgqx, that is very interesting work!!!

In fact, I think it’s worth its own WUWT article. ‍‍‍‍‍‍ ‍‍ ‍‍‍‍‍‍ ‍‍ story tip

The “1.8°C/doubling” sensitivity is what caught my eye. Assuming realized warming is halfway between ECS and TCR, and assuming ECS = 1.5× TCR, that would make TCR = 1.44, and ECS = 2.16, which is a bit higher than the estimates I’ve calculated, but a lot lower than most models assume.

But I have a few questions:

What is TSlanom?

Is ERFanthro an estimate of the effects of non-CO2 anthropogenic changes (CH4, aerosols/particulates, etc)?

What data sources did you use?

How did you find the weights?

Did you do that work in a spreadsheet? Would you care to share it? Please? Here’s my contact info.

Reply to  Bellman
June 19, 2026 5:47 am

But if Tmin rises, you should also expect Tmax to rise.

A totally unproven assertion. It is based on your faulty assumptions. Look at this graph. It is over a 12:00 am to 12:00 am period (24 hours). The ‘heating” takes place over about 12 hours. The “cooling” takes place over all 24 hours. The predominant rate of cooling is while temperature is highest, but cooling occurs until just before heating begins again.

If you can’t tell by looking, the rate of heating is high, and the rate of cooling is lower but lasts longer. You and many others dwell on rates, W/m². These are time based values. Time matters in the calculation of totals.

comment image

Why do you keep wanting the world to “burn up”? Has anyone claimed it will literally do that?

Every time someone shows 0.3°?C/decade in the context of ever growing temperatures, then the assumption is that as long as CO2 grows so will temperature. If you want to express this value with no clarification that is up to you. If you don’t believe that the increase will continue, then you need to state for how long it will continue.

Reply to  Jim Gorman
June 19, 2026 6:54 am

“A totally unproven assertion. ”

None of these assertions are proven. They are just one set of interpretations of the models against another. I’m sure it would be possible to set up an experiment to test the claims emperically.

” The ‘heating” takes place over about 12 hours. The “cooling” takes place over all 24 hours. ”

Yet it ends up warmer than it started. I’m really not sure what point you two think you are making with all this 12 Vs 24 hour nonsense. If temperatures stay the same it’s because the total energy in over 24 hours equals the total energy out over the same period. The fact that most of the energy in happens over a 12 hour period just means you are receiving energy at twice the rate over that period. Alternatively it means the average rate of energy in in has to equal the average rate if energy out.

The simple point you keep missing is that if you reduce the rate of energy out, then that means more energy is received over 24 hours and hence you get slightly warmer. If on average you are constantly getting more energy on than out temperatures keep rising, but fortunately, so does energy out. So you end up with a warmer equilibrium, not indefinite warming as you seem to image.

” The ‘heating” takes place over about 12 hours. The “cooling” takes place over all 24 hours. ”

“Every time someone shows 0.3°?C/decade in the context of ever growing temperatures, then the assumption is that as long as CO2 grows so will temperature”

And why do you think CO2 will rise indefinitely?

And how long do you think an increase of 0.3°C / decade would have to continue before the world literally burnt? What do you even mean by the world will burn?

Reply to  Bellman
June 19, 2026 10:10 am

“Yet it ends up warmer than it started. I’m really not sure what point you two think you are making with all this 12 Vs 24 hour nonsense. If temperatures stay the same it’s because the total energy in over 24 hours equals the total energy out over the same period.“

Flux is joules per second. Energy is joules. You have to multiply joules/sec by seconds to get joules!

The time intervals involved determines amount.

Since the time intervals are different the temperatures CHANGE over time. In addition you have the complication of latent vs sensible heat.

Since both the ocean and the land are heat sinks with conductive flows that cycle in time, Nyquist requires an observation period long enough to see the longest cycle.

Energy-in and energy-out involve at least century- long cycle periods. NOT radiative balance second to second or even diurnally.

You REALLY don’t do science at all, do you?

Reply to  Tim Gorman
June 19, 2026 10:30 am

“The time intervals involved determines amount.”

And as I was explaining, so to does the rate. If you have an incoming flux of 320W/m² for 12 hours, and an outgoing flux of 160W/m² for 24 hours, you have an equilibrium.

Ignoring the rest of you attempts to deflect from this point.

Reply to  Bellman
June 19, 2026 11:03 am

Ignoring the rest of you attempts to deflect from this point.

No deflection at all! Both you and others have attempted to use flux balancing to justify warming. If you have suddenly changed your mind that it is not fluxes that must balance but actual energy over time, i.e., joules, then great, our instructing has done some good.

Reply to  Jim Gorman
June 19, 2026 11:35 am

So you accept that slower cooling can warm the oceans? Good. It the total energy balance has more energy in than energy out, the oceans warm. As the surface are is constant this equates to the average flux in being greater than the average flux out.

Reply to  Bellman
June 19, 2026 1:12 pm

SLOWER COOLING IS *NOT* WARMING!

Slower cooling is just that, slower cooling. Temperature keeps on going down, “τ” just changes. e^-(t/τ) doesn’t all of a sudden become e^(t/τ).

Reply to  Tim Gorman
June 19, 2026 2:47 pm

“SLOWER COOLING IS *NOT* WARMING!”

Shouting is not an argument.

“Slower cooling is just that, slower cooling.”

It’s just that until you accept the existence of another heat source, such as a sun. You can write your continuous cooling function all you want. It’s just you blackboard math until you factor in that heat source.

Out of interest, what value do you attribute to τ? How long will it take the oceans to freeze using your equation?

Reply to  Bellman
June 20, 2026 2:14 am

It’s just that until you accept the existence of another heat source, such as a sun.”

100 joules_emitted – 50 joules_reflected = 50 joules_cooling.

Where do you think the 100 joules_out comes from? Another dimension? The issue is how does CO2 increase temperature, not whether the sun can increase temperature.

Out of interest, what value do you attribute to τ”

Here you are, lecturing us all on thermodynamics and you have no idea of what “τ” is or what it’s value is?

There are separate “τ” decay factors for conduction, convection, and radiation.

Go look it up. You’ll just ignore it if I tell you what they are. (hint: for radiation, “τ” is related to T^3. Can you even figure out why it is T^3 and not T^4?

Reply to  Tim Gorman
June 20, 2026 5:33 am

“100 joules_emitted – 50 joules_reflected = 50 joules_cooling. ”

Where’s the sun?

“Here you are, lecturing us all on thermodynamics and you have no idea of what “τ” is or what it’s value is?”

Not answering the question. What value is τ?

If you could answer that you could also say how long it will take the earth to cool to 0k. I suspect you won’t answer because it would demonstrate the importance of the sun.

Reply to  Bellman
June 20, 2026 6:22 am

Not answering the question. What value is τ?

If you could answer that you could also say how long it will take the earth to cool to 0k. I suspect you won’t answer because it would demonstrate the importance of the sun.

ROTFLMAO! Hilarious!

Nice try, but you are just using an argumentative fallacy known as “The Fallacy of Shifting the Burden of Proof”. It means you cannot answer a properly designed question, so you reflect it back to the questioner. In a debate, guess who gets the point?

Reply to  Jim Gorman
June 20, 2026 8:00 am

“Nice try, but you are just using an argumentative fallacy known as “The Fallacy of Shifting the Burden of Proof”. ”

Hilarious. It’s your claim that the world is cooling, and that cooling is described with a simple equation. It’s your burden to demonstrate that this equation makes sense. You know full well why you won’t say the value of τ. Because if you did it would be clear that it doesn’t work because the sun still shines.

Reply to  Bellman
June 19, 2026 3:16 pm

So you accept that slower cooling can warm the oceans? 

Nice try dude! That is not what I said. Don’t put words in my mouth.

Here is my math. Show us yours.

T₁ < Tₙ₋₁ Cooling
T₁ = Tₙ₋₁ Equalibrium
T₁ > Tₙ₋₁ Heating

Now take a heat equation of your choice and attempt to make it derive a heating inequality.

Reply to  Jim Gorman
June 19, 2026 4:40 pm

Here is my math. Show us yours.

You are just stating that if one temperature is cooler than another it’s cooling. You are as usual completely ignoring the point. And it’s absurd when you’ve admitted from the start that slower cooling raises Tmin.

My maths? Say at sunset the ocean temperature is 15°C and over night it cools rapidly by 5°C, it’s temperature at sunrise is 10°C. But if on a different night it cools slower, by 4°C and so reaches a temperature of 11°C. Then I claim that

11 > 10.

Hence, slower cooling causes warming.

Now I further claim that if the daytime sun causes similar level of rise each day then as the following day will start from a warmer temperature, it’s maximum will also be warmer.

Reply to  Bellman
June 20, 2026 2:20 am

And it’s absurd when you’ve admitted from the start that slower cooling raises Tmin.”

And what does a higher Tmin indicate?

Apply it to the equation E∝τ

” But if on a different night it cools slower, by 4°C and so reaches a temperature of 11°C.”

And what does the equation E∝τ tell you about the heat loss from the slower cooling?

Hence, slower cooling causes warming.”

15°C > 11°C
15°C > 10°C

So where did the warming happen? All you’ve done is shown slower cooling!

Reply to  Tim Gorman
June 20, 2026 5:42 am

“So where did the warming happen?”

During the day, when that sum you don’t believe in, appears.

But the real problem is how you are equivocating about the word “cooling”. You are trying to imply that it only relates to cooling over the night time, but I’m saying warming happens from day to day, and year to year.

Consider I get paid on the first of the month, then spend money throughout the month. My savings keep dropping, and so you say I’m getting poorer. But I’m saying if I spend less than I earn each month I’m getting richer as my savings increase each month.

When we talk about the oceans getting warmer it isn’t about the second by second change. It’s about wether it’s warmer now than it was a few years ago. And in this case it’s about whether it’s warmer than it would have been if the atmosphere were colder.

Reply to  Bellman
June 20, 2026 6:11 am

Consider I get paid on the first of the month, then spend money throughout the month. My savings keep dropping, and so you say I’m getting poorer. But I’m saying if I spend less than I earn each month I’m getting richer as my savings increase each month.

Your analogy is flawed. You are missing the fact that the sun’s insolation is first absorbed by the surface, then some is radiated and conducted to the atmosphere while some is stored. That is a three body problem while your analogy is basically a one body system.

The second body in a three body system, the earths surface, has several processes that distribute the income in various ways. The total adds up to what is input, but you can’t just look at the cash you spend. Some disappears without you ever knowing where it went, ala latent heat. Some goes into a savings account you can’t withdraw from, ala surface heat sink. In the end, it all adds up to zero.

You are trying to make a simple system of in->out, and worse using temperature as a proxy.

Reply to  Jim Gorman
June 20, 2026 7:36 am

“Your analogy is flawed.”

Your inability to understand a point is truly impressive. It isn’t an apology for how the sun works. It’s simply to illustrate how it’s possible for you to be claiming something is declines, when looked at another way it’s increasing.

Reply to  Bellman
June 20, 2026 9:05 am

Your inability to understand a point is truly impressive.”

Show the math!

“It’s simply to illustrate how it’s possible for you to be claiming something is declines, when looked at another way it’s increasing.”

Cooling is cooling, PERIOD. There isn’t any other way to look at it.

CO2 simply cannot warm the earth. It is *NOT* a source of heat. It is a reflector. I’ve shown you the math on how that works.

WHERE IS YOUR MATH SHOWING THAT ENERGY LOSS IS WARMING?

Reply to  Tim Gorman
June 20, 2026 11:43 am

“Show the math!”

What math? Surely you don’t need me to explain why spending less than you earn makes you richer.

“Cooling is cooling, PERIOD. There isn’t any other way to look at it.”

You lack imagination. That’s always been your problem. You get an idea in your head, write it in capitals dozens of times, and convince yourself that it’s impossible for there to be any other way of looking at it.

So in your world someone who spends money is always getting poorer, even as his bank account grows. The world is always getting cooler even as temperature rise.

” I’ve shown you the math on how that works. ”

You have not. All you do is show the equation for an object cooling whilst ignoring all heat sources. You math would be fine if you just put the earth in space and watched it cool. It does not describe what happens when the sun shines on it. And you still refuse to specify the actual value for the cooling constant. Why is that?

“WHERE IS YOUR MATH SHOWING THAT ENERGY LOSS IS WARMING?”

The same place it was the last time you asked. And what is wrong with your keyboard? Why should I take any advice from someone who can’t use a shift key properly?

Reply to  Bellman
June 20, 2026 5:06 pm

tpg: ““Cooling is cooling, PERIOD. There isn’t any other way to look at it.””

“You lack imagination. That’s always been your problem.”

So your imagination tells you that cooling is warming?

Unfreakingbelievable!

You have not. All you do is show the equation for an object cooling whilst ignoring all heat sources.”

Your reading comprehension skills are atrocious. I SHOWED you how F0 depends on where the heat source takes the temperature to. And the only heat source in the system is the SUN, not CO2!

Did you not read where I told you that F0 is complicated? That it is actual F0(t). Did you not bother to read the rest of the equation I gave you?

Reply to  Tim Gorman
June 20, 2026 5:34 pm

So your imagination tells you that cooling is warming?

As I’ve tried at length and in terms simple enough for even you to understand – there is more than one way way of describing cooling or warming. You seem to be fixated on the idea that if there is cooling during [art of the day, that means that all there is is cooling. I say that if the cooling leads to warmer conditions that is warming, not cooling.

And this is only relevant because you try to argue that because the world cools at night, it’s impossible for the atmosphere to cause warming, even when you admit the atmosphere causes the cooling rate to decrease.

Reply to  Bellman
June 20, 2026 5:55 pm

I say that if the cooling leads to warmer conditions that is warming, not cooling.

You have yet to show any math that show how cooling causes warming. The statement is truly insane.

And this is only relevant because you try to argue that because the world cools at night, it’s impossible for the atmosphere to cause warming, even when you admit the atmosphere causes the cooling rate to decrease.

I have shown you the math that shows what warming means:

Tₜ > Tₜ ₋ ₁

You have yet to show any mathematics that confirms this can occur other than,

Etot = Esun + Eghg.

I have shown you that Planck disagrees with this. If you want your assertion to stand you must provide a reference showing it is true. Maybe something that shows the atmosphere is warmer than the surface.

Reply to  Jim Gorman
June 20, 2026 6:41 pm

You have yet to show any math that show how cooling causes warming.

I have, multiple times. You just ignore it, or pretend not to understand. I even gave you a simple analogy involving money, which you pretended not to understand.

I have shown you the math that shows what warming means:
Tₜ > Tₜ ₋ ₁

And I’ve explained to you that if day 1 is warmer than day 0 that means you have warming, but you just won’t accept it despite your “math”.

Etot = Esun + Eghg

That tells you nothing about warming or cooling. It’s simply accepting that you are getting energy from two sources. What matters is the balance.

I have shown you that Planck disagrees with this.”

You have not. All you ever do is cut and paste long sections with no understanding.

If you want your assertion to stand you must provide a reference showing it is true.

Showing what is true? That getting energy from two different sources increases the total energy that you receive?

Maybe something that shows the atmosphere is warmer than the surface.

It isn’t. Why do you keep using these strawmen arguments?

Reply to  Bellman
June 20, 2026 7:18 pm

And I’ve explained to you that if day 1 is warmer than day 0 that means you have warming

This has nothing to do with radiative or energy balances at all. What happens when day 2 is colder than the other two.

The use of math in thermodynamics is required to do science. You need to show a more generalized solution that matches your assertions in order to have a proper solution.

Reply to  Jim Gorman
June 21, 2026 5:34 am

“This has nothing to do with radiative or energy balances at all. ”

So what caused the warming if not the balance?

” What happens when day 2 is colder than the other two”

Then there cooling occured.

” You need to show a more generalized solution that matches your assertions in order to have a proper solution.”

I do not need to do anything. I am not a scientist. I am not writing a paper on the subject, and if I was I doubt a paper explaining why more energy in than out results in warming would be considered very original.

Reply to  Bellman
June 21, 2026 8:45 am

there is more than one way way of describing cooling or warming”

That is weasel wording. Cooling is losing joules of energy. Warming is gaining joules of energy. dT/dt = warming (joules_in) – cooling(joules_out).

Energy = joules_in – joules_out.

If joules_out > joules_in then you have cooling.

If joules_in > joules_out then you have warming.

Jim has given you this math multiple times. And each time you just blow it off!

You seem to be fixated on the idea that if there is cooling during [art of the day, that means that all there is is cooling. “

NO ONE HAS *EVER* SAID THAT! Why do you think I pointed out to you that Tmax happens before sunset? Both heating and cooling happen simultaneously. Tmax is reached when F_out = F_in, sometime in the mid-afternoon, NOT at sunset!

There is ALWAYS cooling going on, ALWAYS!

” I say that if the cooling leads to warmer conditions that is warming, not cooling.”

You *STILL* haven’t figured it out! Cooling NEVER leads to warmer conditions. Cooling is joules_out! Joules_out can *NEVER* lead to higher temperatures because that would require *more* joules, not fewer.

What leads to warming is joules_in from a heat source! CO2 is *NOT* a heat source. It is a reflector. Since it is *NOT* a heat source it cannot warm the heat source which originated the joules_out flow. All it can do is change “τ”, the delay factor associated with the tempeature change due to the joules_out flow.

The equation we’ve been trying to get you to understand is

Energy = joules_in(sun) – joules_out(earth).

Joules_out(earth) is *ALWAYS* positive, it’s emission of joules.

ΔE is only positive when joules_in(sun) > joules_out(earth).

You and bdgwx can’t seem to get this into your heads at all!

You want ΔE to be positive even when joules_out > joules_in.

And this is only relevant because you try to argue that because the world cools at night”

The earth cools 24 hours per day, every day of every year of every century of every millennia. Until you understand that the math is always going to elude you.

“it’s impossible for the atmosphere to cause warming”

It’s impossible for the atmosphere to cause warming because it is not a heat source. IT IS A REFLECTOR. If it was a perfect reflector totally encasing the globe, you would still only have equilibrium. Joules_out would be zero. The joules in the system would remain constant. dT/dt = 0, not dT/dt > 0.

even when you admit the atmosphere causes the cooling rate to decrease.”

Slower cooling means EXACTLY THAT! It is *still* a loss of joules. A loss of joules can *NOT* cause temperature to go up!

Reply to  Tim Gorman
June 21, 2026 9:22 am

“That is weasel wording.”

Says someone who’s been claiming everything is cooling even as it gets hotter.

‘If joules_in > joules_out then you have warming.”

You do realise you’ve just contradicted yourself? You claimed that any thing that was emitting joules was cooling. Now you accept that if an object receives more joules than it emits then it’s warming.

“NO ONE HAS *EVER* SAID THAT! Why do you think I pointed out to you that Tmax happens before sunset? Both heating and cooling happen simultaneously. Tmax is reached when F_out = F_in, sometime in the mid-afternoon, NOT at sunset!’

This is the problem with arguing with you. You keep making these vague contradictory statements, and then when anyone trys to figure out what you mean you just yell at them.

Are you saying it only cools between tmax and tmin, or are you saying it’s cooling 24 hours a day but it’s also warming 24 hours a day?

“You *STILL* haven’t figured it out! Cooling NEVER leads to warmer conditions.”

I haven’t figured it out because you haven’t given a coherent argument. If cooling is just the loss of gross energy and something can also be warming at the same rate, why wouldn’t slowing the rate of cooling lead to s net warming.

“Cooling is joules_out! Joules_out can *NEVER* lead to higher temperatures because that would require *more* joules, not fewer.”

But you said at the start that joules_in > joules_out meant warming. If you reduce joules out to be less than joules in, why wouldn’t that lead to higher temperatures?

This all just seems like word games. You keep arguing that greenhouse gases don’t “cause” warming, just slower cooling. Fine, argue cause and effect like that. But that just pushes the question back a step. Would the world be colder if there were no greenhouse gases?

Reply to  Bellman
June 21, 2026 9:26 am

“Energy = joules_in(sun) – joules_out(earth).”

And the equation I’ve been trying to get you to understand is

Energy = joules_in(sun) – joules_out(earth) + joules_in(atmosphere)

There are not two earth’s pne of which only receives energy from the sun, and the other one only loses energy to the atmosphere.

Reply to  Bellman
June 20, 2026 9:02 am

During the day, when that sum you don’t believe in, appears.”

The sun is a SOURCE OF HEAT. CO2 is *NOT* a source of heat.

You don’t seem to understand that heat loss is greater during the day than it is at night. Cooling doesn’t stop when the sun comes up. The RATE of cooling actually increases!

You *still* haven’t shown a lick of math to back up any of your assertions. NONE!



Reply to  Tim Gorman
June 20, 2026 11:30 am

“The sun is a SOURCE OF HEAT.”

Finally you admit it. So why did you keep ignoring it?

“Cooling doesn’t stop when the sun comes up. The RATE of cooling actually increases!”

Did you actually mean to say that? I’ll be charitable and assume you mean the amount of energy the earth radiators is greater when it’s warmer. But that’s not the net energy transfer. The earth is getting more energy than it radiates during most of the day, hence it is warming.

Reply to  Bellman
June 20, 2026 5:02 pm

Finally you admit it. So why did you keep ignoring it?”

I DON’T keep ignoring it. What the sun does is totally separate from what CO2 does! Why is that so hard to get through your skull and into your brain?

CO2 is *NOT* a heat source. It cannot ADD heat to the earth. You must *ADD* heat in order to raise the temperature. If CO2 cannot *ADD* heat then it can’t raise the temperature! All it can do is slow the cooling!

When are you going to admit that CO2 is *NOT* a heat source?

Did you actually mean to say that?”

OH MY GOD!

The rate of cooling is related to T! When T is greater the rate of course cooling is greater!

In the equation F(t) = F0 * e^-(t/τ) what do you think determines F0?

What do you think F(t) is? It’s the FLUX – and flux is a RATE! And the flux is related to T^4. The larger T is the greater T^4 is and the greater F0 is! and the greater F(t) is!

Put down the bottle man! Do you *enjoy* making a fool of yourself?

Reply to  Tim Gorman
June 20, 2026 6:23 pm

What the sun does is totally separate from what CO2 does!

And there’s your problem. They are not separate, they both have an effect on the temperature of the earth. You cannot treat each in isolation.

CO2 is *NOT* a heat source. It cannot ADD heat to the earth.

But it does reduce the rate of heat transfer from the earth to the atmosphere.

If CO2 cannot *ADD* heat then it can’t raise the temperature!

Patently false. Again based on your inability to include the sun in your model.

All it can do is slow the cooling!

Which is all you need to raise the temperature, as long as you accept the sun exists.

OH MY GOD!

Do you think this hysterical school girl act helps your case? Here’s what you said:

Cooling doesn’t stop when the sun comes up. The RATE of cooling actually increases!

I’ll take it you did mean to say that, and all I can say is it makes no sense.

The rate of cooling is related to T!

Am I supposed to guess what T is in this case?

In the equation F(t) = F0 * e^-(t/τ) what do you think determines F0?

Again, that equation is not correct when the sun is shining.

F0 will be the initial heat flux, proportional to the temperature difference.

What do you think F(t) is?

The flux at time t.

But why do you keep asking me? Your the expert.

And the flux is related to T^4.

I don’t think it is in your equation. It’s just the cooling equation translated into flux – and that’s assuming flux is roughly proportional to temperature. If I’m wrong please provide a reference, preferably written by a human rather than an AI.

The larger T is the greater T^4 is and the greater F0 is! and the greater F(t) is!

And you still have yet to explain why any of this is relevant for a world where the sun shines.

Put down the bottle man! Do you *enjoy* making a fool of yourself?

Do you want me to dehydrate in this weather?

Reply to  Bellman
June 21, 2026 8:55 am

“Again, that equation is not correct when the sun is shining.”

I’ll say this one more time.

F(t) = F0 * e^-(t/τ)

is correct ALL THE TIME.

F0 will be the initial heat flux, proportional to the temperature difference.”

I’ll ask again since you failed to answer: “what do you think determines F0?

(hint: It is *NOT* proportional to the temperature difference! It is calculated from the absolute temperature, the relation involves T^4!)

(hint2: F0 is actually F0(t) when the sun is shining. )

I pointed out to you TWICE earlier that F0 is difficult to calculate when the sun is shining. As usual, you just blew that off. Unfreakingbelievable!

Reply to  Tim Gorman
June 21, 2026 9:55 am

“is correct ALL THE TIME.”

It isn’t correct any time. It’s just an approximation when temperature changes are small, as it assumes a linear relation between temperature and flux. But it definitely won’t be correct if there is more than one object.

Consider you out an object in a cool environment. The cooling equation tells you how temperature will change over time, and the flux will be roughly the same as the temperature change.

Now put a heat source on the object. Will the object cool according to cooling equation? Obviously not. That predicts the object keeps getting colder, whilst the object might actually be warming. And if the temperature is going up, so too is the flux out.

If I’m misunderstanding your point, could you please provide a reference for your equation.

Rather than giving all these hints, why do you try to explain what you think the equation is saying and preferably give a reference.

Reply to  Bellman
June 21, 2026 11:49 am

the flux will be roughly the same as the temperature change.

Guess again Ogreat one! Stefan-Boltzmann law says that
I = σT⁴. That is an exponential function. I is not roughly the same as the change in the temperature.

I =( 5.67×10⁻⁸)(273)⁴ = 314 W/m²
I =( 5.67×10⁻⁸)(275)⁴ = 324 W/m²

Funny how the temperature changes by 2K while the flux changed by 10 W/m². Not exactly roughly the same.

It would behoove you to do the math on these problems first before just making an assertion about what you think will occur.

Reply to  Jim Gorman
June 21, 2026 11:59 am

“Guess again Ogreat one! Stefan-Boltzmann law says that
I = σT⁴. ”

You are repeating my point. It’s only an approximation for small changes. Did you actually read what I said.

It’s just an approximation when temperature changes are small, as it assumes a linear relation between temperature and flux.

” That is an exponential function.”

It’s not. It’s a power function.

” Not exactly roughly the same. ”

Take it up with your brother. He’s the one claiming the equation is always correct.

“It would behoove you to do the math on these problems first before just making an assertion about what you think will occur.”

That’s why I keep asking for a reference. I can’t find anything online explicitly using that function for flux. I assumed it was just based on the cooling equation for temperature, but that it’s obviously only an approximation dur to the fourth power issue. So did Tim just invent the equation or is there an actual logic behind it?

Reply to  Bellman
June 20, 2026 9:31 pm

It is a transient warming only at the surface that is a function of the reflectivity and specific heat capacity of terrestrial materials.

Reply to  Bellman
June 20, 2026 12:43 pm

You are trying to imply that it only relates to cooling over the night time, but I’m saying warming happens from day to day, and year to year.”

Cooling happens 24 hours per day, seven days per week, 52 weeks per year, 1000 years per millennia.

Warming happens 12 hours per day, seven days per week, 52 weeks per year, 1000 years per millennia.

The only one ignoring the time functions here is you!

Reply to  Tim Gorman
June 20, 2026 2:48 pm

“Cooling happens 24 hours per day, seven days per week, 52 weeks per year, 1000 years per millennia.”

“Warming happens 12 hours per day, seven days per week, 52 weeks per year, 1000 years per millennia.”

So what happens if the totak warming is greater than the total cooling?

Reply to  Tim Gorman
June 22, 2026 9:37 am

As I mentioned above this is a result of using the term cooling to ambiguously describe two processes: the change in surface temperature and the heat loss from the surface.

A more appropriate terminology is to refer to cooling and heat loss.
Heat loss occurs over 24 hours per day, surface cooling does not.

Reply to  Bellman
June 20, 2026 12:50 pm

 It’s about wether it’s warmer now than it was a few years ago”

No, it’s about whether it’s warmer now than it was a millennia ago, or 10 millennia ago.

IT’S A TIME FUNCTION whose low frequency cycles are far longer than just a few years.

  1. Earth is not isothermal. It is not a black body
  2. Earth does not emit what it receives immediately. It is not a black body.
  3. Earth does not radiate equally from all surface area. The earth is not a black body
  4. The thermodynamic system consists of three bodies, not two. Sun, earth, space.
  5. The rate of cooling of the earth is larger at Tmax than at Tmin.
  6. Reflective bodies are not heat sources. They can only return what they receive.

You continue to ignore all these factors.

Reply to  Bellman
June 20, 2026 7:47 am

Hence, slower cooling causes warming.

No, it proves nothing about heating.

If it started at 15°C both nights, then it cooled over both nighttimes. One night it cooled by by 5°C and the second night it cooled by by 4°C. Please note, it cooled both nights.

You have shown no basis for assuming that the cold atmosphere caused one night to cool less than another. There are any number of conditions that could cause this, such as, clouds, a higher dew point, precipitation, etc. Show some data to back up your assertion.

In fact, you have inadvertently confirmed what Tim and I have been telling you. Time! Time is an extremely important variable in analyzing energy. What would you say if the 3rd night, it cooled by 6°C?

Reply to  Jim Gorman
June 20, 2026 8:33 am

” Please note, it cooled both nights.”

Astonishing. I tried to explain why you keep getting confused over this with my wages analogy. You must have read it because you’ve already commented, yet you still keep making the same mistake.

We are talking about two different things. You are talking about the cooling process that happens most nights and I’m talking about how temperature changes over a longer time frame. Temperatures usually cool over night, but they can still get warmer over time. It’s a three steps forward, two steps back situation.

I could look at the daily temperatures at a station from January to June. Nearly every night shows cooling, yet I’m pretty confident there will have been warming between January and June.

“You have shown no basis for assuming that the cold atmosphere caused one night to cool less than another. ”

You were the ones describing it as slower cooling.

Reply to  Bellman
June 20, 2026 9:08 am

You are talking about the cooling process that happens most nights”

THE COOLING HAPPENS DURING THE DAY AS WELL AS AT NIGHT!

When is the cooling rate at is greatest?

 I’m talking about how temperature changes over a longer time frame”

No, you aren’t. You haven’t even considered the amount of time between El Nino’s when heat loss from the ocean is at its greatest.

“yet I’m pretty confident there will have been warming between January and June.”

Really? You think the Southern Hemisphere warms January and June?

Slower cooling is *still* COOLING.

Reply to  Tim Gorman
June 20, 2026 11:51 am

“THE COOLING HAPPENS DURING THE DAY AS WELL AS AT NIGHT!”

Strange, I always thought it got warmer during the day. But I guess you must be correct if you wrote it in capitals.

“No, you aren’t.”

Really? What was I talking about then?

“Really? You think the Southern Hemisphere warms January and June?”

Was I talking about the southern hemisphere? I just said a station and had in mind somewher in the UK, but if it was a station in the south just reverse the dates and read it as from June to January. The point’s the same. A location can be constantly cooling by your logic yet still end up warmer.

Reply to  Bellman
June 20, 2026 5:09 pm

Strange, I always thought it got warmer during the day. But I guess you must be correct if you wrote it in capitals.”

In other words you just blew it off when I pointed out to you that the temperature goes DOWN before the sun sets!

In other words it get COOLER during the day as well as warmer. The temperature goes down during the day when the cooling rate overrides the energy input from the sun. PROOF to anyone that can read that cooling happens during the day as well as at night!

Anyone but you I guess.

Reply to  Tim Gorman
June 20, 2026 5:28 pm

The temperature goes down during the day when the cooling rate overrides the energy input from the sun”

Good. So you have finally accepted that the sun exists and that it does get warmer during the day. So just what relevance do you think there is in continuously spamming the cooling equation as if it was applicable?

Reply to  Bellman
June 20, 2026 5:36 pm

Your insane reply shows you have nothing important to add to the discussion.

Exactly where has anyone claimed that insolation doesn’t warm the surface of the earth?

Emission and absorption are two different processes. They must be analyzed separately.

Reply to  Jim Gorman
June 20, 2026 6:30 pm

Your insane reply shows you have nothing important to add to the discussion.

So you don;t know either. You accept the sun is shining but think the cooling equation means the earth is continuously cooling.

Exactly where has anyone claimed that insolation doesn’t warm the surface of the earth?

Every time you’ve claimed the world is continuously cooling, and refused to accept that cooling slower can lead to warming.

They must be analyzed separately.

The world doesn’t know how to. Both effects happen on the world at the same time, they both contribute to any temperature change. If the sun provides a constant warming effect, and the atmosphere a balancing cooling effect, and then you slow down the rate of cooling, the earth warms, becasue the sun doesn’t know it isn’t supposed to contribute.

Reply to  Bellman
June 20, 2026 7:30 pm

So you don;t know either. You accept the sun is shining but think the cooling equation means the earth is continuously cooling.

Absorption and emission are two separate processes. The earth can be absorbing energy at one rate and emitting at another. Absorption is dependent on flux, emission is dependent on temperature.

When there is emission, cooling occurs. Otherwise you are denying the accepted law that bodies above 0K radiate and cool.

Reply to  Jim Gorman
June 21, 2026 5:29 am

” The earth can be absorbing energy at one rate and emitting at another. ”

And if it is, latent heat excepted, it’s temperature is changing. But this can’t go on for ever as the hotter the earth is the more it emits. You are always going to end up in some sort of equilibrium.

“When there is emission, cooling occurs. ”

For once and for all will you just define what you mean by cooling. If you mean any emission is cooling then say that and admit that by your definition of cooling temperatures may well be rising.

Reply to  Bellman
June 21, 2026 7:52 am

For once and for all will you just define what you mean by cooling.”

Cooling is a loss of joules of energy! Cooling is *NOT* an increase of joules of energy. Emission loses joules of energy. Emission is ALWAYS cooling. Have you *ever* read anything about entropy?

“If you mean any emission is cooling then say that and admit that by your definition of cooling temperatures may well be rising.”

Cooling does *NOT* cause a rise in temperature since it is a loss of joules of energy. *IF* you have an external source, e.g. the sun, inputting joules of energy at a rate higher than the cooling is losing joules of energy then the temperature can go up.

BUT THE COOLING DOESN’T CAUSE THE WARMING!

An external source of heat, i.e. joules, does the warming.

A colder body can’t warm a hotter body! PERIOD, EXCLAMATION POINT. A hotter body, like the sun, can warm a cooler body, like the earth.

CO2 as a colder body can’t warm the earth. The *SUN* warms the earth, i.e. hotter to colder.

All CO2 can do is act as a reflector of energy back to the source, i.e. the hotter earth, which retards the exponential decay factor thus causing an INCREASE in the number of joules being emitted. Planck’s “compensation”.

Tell us how the attached image is wrong. SHOW THE MATH.

exponential_decay_fast_vs_slow_cooling
Reply to  Tim Gorman
June 21, 2026 8:21 am

“Cooling is a loss of joules of energy! ”

Gross or net? That was the question. If you lose 100 joules but gain 200 joules, are you cooling or warming?

You say that emission is always cooling which implies you don’t care about the net change, but if so I’m not sure what point you are making. Everything above 0K is always cooling, so what? That doesn’t tell you if something is decreasing in temperature.

Reply to  Bellman
June 21, 2026 7:45 am

You accept the sun is shining but think the cooling equation means the earth is continuously cooling.”

IT *IS* COOLING CONTINUOUSLY! If it wasn’t then the temperature during the day would go UP until the sun went below the horizon since it would still be dumping additional Joules of heat into the earth!

Why is this so hard for you to understand? Does Tmax always happen just before sunset? Or does it happen in the middle of the afternoon? The sun dumps more and more Joules of heat into the earth from the time it rises above the horizon until it sets below the horizon. More Joules means higher temperatures. If the earth wasn’t cooling during the day then Tmax would always be at sunset!

refused to accept that cooling slower can lead to warming.”

Once again, warming means an increase in temperature. Cooling does *NOT* increase temperature. If it did an ice cube would never melt!

Reply to  Bellman
June 21, 2026 7:39 am

Good. So you have finally accepted that the sun exists and that it does get warmer during the day.”

It gets warmer because of heat input in Joules FROM THE SUN. It does *NOT* get warmer because of heat input in Joules from CO2. CO2 is *NOT* a heat source!

“So just what relevance do you think there is in continuously spamming the cooling equation as if it was applicable?”

Because the earth *COOLS* even while receiving Joules from the sun!

You *still* haven’t figured out why the temperature goes down before the sun goes down apparently. How does the temperature of the earth go down while getting input from the sun if it isn’t cooling during the day?

Reply to  Bellman
June 20, 2026 9:39 pm

The ‘Process’ of cooling happens 24-hours per day, with the RATE of cooling greater where the suns rays are impinging. However, the cooling is overpowered by the energy of the solar insolation and the surface warms for about 12-hours. The net result is determined by how long the imbalance exists.

Reply to  Clyde Spencer
June 21, 2026 4:43 am

See, this is why I wish the Germans were capable of defining their terms. If you define cooling as emitting energy irrespective of how much you receive, then everything everywhere is always cooling. But it has nothing to do with what most people would think you mean by cooling. Saying an object is cooling when it’s temperature is rising is misleading at best.

But then the problem is, why do people object when I say that slower cooling can result in warming? As I said before, it just seems like equivocating on the word cooling.

Reply to  Bellman
June 21, 2026 8:10 am

 If you define cooling as emitting energy irrespective of how much you receive, then everything everywhere is always cooling”

OMG! YOU FINALLY FIGURED THIS OUT?

What in Pete’s name do you think all of the math we’ve been giving you shows??????

“But it has nothing to do with what most people would think you mean by cooling.”

“Most people”, including apparently climate scientists, statisticians, and mathematicians have absolutely *NO* idea of thermodynamics at all! YOU ARE A PRIME EXAMPLE. You can’t even follow the simple math of

F(t) = F0 * e^-(t/τ)

F0 is related to TEMPERATURE. The only time F(t) will be zero is if F0 is 0 (zero). When will F0 be 0 (zero)? If an object has a F0 above 0 (zero) then exactly what does that imply?

Saying an object is cooling when it’s temperature is rising is misleading at best.”

No, it isn’t misleading at all! IT IS A TRUISM. The *real* problem is that you are so adamant about coming here and proving everyone wrong that you simply can’t learn anything that is true. You have your misconceptions and BY GOD! you are going to prove them correct regardless!

You haven’t even accepted yet that a colder body can’t raise the temperature of a warmer body.

So you introduce an external heat source (sun) and try to imply that it is the colder body (CO2) doing the warming instead of the external heat source!

100 joules_out(earth) – 50 joules_in(CO2) = 50 joules_out(earth)! THAT’S COOLING. It’s a loss of heat energy!

But then the problem is, why do people object when I say that slower cooling can result in warming? As I said before, it just seems like equivocating on the word cooling.”

It isn’t the COOLING that results in warming. It is the EXTERNAL SOURCE (sun) that results in the warming.

And you can’t even buy into into the math that slower cooling represent *more* dumping of joules than faster cooling!

You can’t refute the graph I keep showing you. It’s the MATH. Where is *YOUR* math showing that slower cooling *decreases* the number of joules being dumped compared to faster cooling?

There’s no equivocating going on about saying that cooling is cooling. It’s the math! You just for some reason can’t accept the math!

Reply to  Tim Gorman
June 21, 2026 8:42 am

“What in Pete’s name do you think all of the math we’ve been giving you shows??????”

Nothing of any use if you are only looking at the negative side of the equation.

“No, it isn’t misleading at all! ”

You don’t think it’s misleading to say something is cooling when it’s temperature is increasing?

“You haven’t even accepted yet that a colder body can’t raise the temperature of a warmer body.”

Why should I accept it. It’s patently not true.

This just illustrates your inability to consider you might be wrong. If someone disagreed with you it must be that they are wrong, you can ignore all their arguments, and it’s up to them to accept they are wrong.

Now I would be quite prepared to accept I was wrong if you could provide convincing arguments or evidence – but so far that had not happened. And in this case, given virtually everyone who has studied the subject agrees that the greenhouse effect is real, a really think you need to consider the possibility you are the one who doesn’t understand it.

Reply to  Bellman
June 21, 2026 5:18 pm

Nothing of any use if you are only looking at the negative side of the equation.”

That part of the equation applies 24 hours/day. You *have* to look at it over the entire 24 hours.

“You don’t think it’s misleading to say something is cooling when it’s temperature is increasing?”

It’s the TRUTH! It only seems misleading to you because you don’t understand thermodynamics at all!

Anything with a temperature greater than 0K IS COOLING ALL THE TIME. Every second of every day of every week of every month of every year of every decade of every century of every millennia.

You are stuck just like climate science is stuck in thinking that temperature is a measure of heat energy instead of kinetic energy. Temperature (of any scale) is *NOT JOULES OF ENERGY*.

The conductive heat equation of q = kA/L * (Th – Tc) gives you JOULES/SECOND, not joules.

The temperature of an object will only go up when the input of joules from a heat source is greater than the loss of joules from an object. That loss of joules happens every day of every week of every month of every year of every decade of every century of every millennia for any object that is above 0K. It is COOLING!

Why should I accept it. It’s patently not true.”

Then why doesn’t CO2 drive the temperature of the earth up at night? If it adds joules of energy to the earth during the day then it will do the same at night.

You keep confusing the input of joules from the sun with the reflection of joules by CO2. No sun – no input of joules and no increase in the temperature of earth, only a decrease – i.e. cooling.

A colder body cannot warm a hotter body.

The second law is: q ∝ (Th – Tc).

By definition Th ≥ Tc. That makes the exchange of heat positive (or zero, i.e. equilibrium) and the net flow of joules is from Th to Tc *ALWAYS*. That means Th COOLS and Tc WARMS – ALWAYS.

copilot:
———————————-
tpg question: can a cold object *ever* increase the temperature of a hotter body
No — a colder object can never raise the temperature of a hotter object by itself.
—————————-

You can continue with your misconception that a colder object can raise the temperature of a hotter object but its a delusion of your own making – it is a non-physical belief.

Reply to  Tim Gorman
June 21, 2026 6:36 pm

That part of the equation applies 24 hours/day. You *have* to look at it over the entire 24 hours.

Even your brother disagrees.

Regardless, how does the flux keep decreasing if the temperature is the same or increasing?

And you still won’t give any figures? How much flux is lost each day? How long before it drops to near zero?

It’s the TRUTH!

Any chance of a reference for this “truth”?

You’ve said elsewhere that cooling requires output to be greater than input, and quoted Planck saying

The fact that the body A is cooled by B and heated by B′ is due entirely to the fact that B is a weaker, B′ a stronger emitter than A.

It only seems misleading to you because you don’t understand thermodynamics at all!

Says the person who claims that the term “estimated value” in statistics means the value that you will always get.

Anything with a temperature greater than 0K IS COOLING ALL THE TIME.

Assuming this is a definition they use in thermodynamics, it’s still misleading in the context of greenhouse gases. You are specifically saying that the greenhouse effect cannot cause warming, but relying on the use of the term cooling which can be applied to an object that is warming. Your statement that reducing the rate of cooling cannot cause warming, becomes doubly wrong if you accept that cooling tells you nothing about whether an object is warming.

You are stuck just like climate science is stuck in thinking that temperature is a measure of heat energy instead of kinetic energy.

Nope. Just another of your strawman arguments. Try to engage with what I say, not with what you wish I said.

“The conductive heat equation of q = kA/L * (Th – Tc) gives you JOULES/SECOND, not joules.”

Who ever said it didn’t? q is Watts. Now list all the assumptions in that equation.

The temperature of an object will only go up when the input of joules from a heat source is greater than the loss of joules from an object. That loss of joules happens every day of every week of every month of every year of every decade of every century of every millennia for any object that is above 0K. It is COOLING!

Again, a reference is required. And an explanation that although you say the object is cooling, it’s quite possible it’s temperature is rising.

Then why doesn’t CO2 drive the temperature of the earth up at night?

It does. The warming of the greenhouse effect is generally larger at night than during the day.

If it adds joules of energy to the earth during the day then it will do the same at night.

It does.

You keep confusing the input of joules from the sun with the reflection of joules by CO2.

You still don’t understand what reflection means in thermodynamics. CO2 does not reflect energy it absorbs it. If it reflected energy we’d be getting twice as much energy back.

Regardless, I am not confusing the two at all. Energy from the atmosphere is added to energy from the sun. Do you want me to go through the equations I showed you the last time we had this conversation?

No sun – no input of joules and no increase in the temperature of earth, only a decrease – i.e. cooling.

Indeed. Nobodies disputing that. What I’m disputing is the fact we are not in a situation where there is no sun.

A colder body cannot warm a hotter body.

Why do you think if people didn’t believe you the first time, that endlessly repeating your credo will make them change their minds. You would be correct if this was a two body problem with no external heat source, but we’ve known for centuries that this is not the case with the earth.

The second law is: q ∝ (Th – Tc).

I don’t think it is, but feel free to provide a reference.

That means Th COOLS and Tc WARMS – ALWAYS.

List your assumptions. Is there any that mention external heat sources?

by itself”

What do you think that means?

Reply to  Bellman
June 22, 2026 7:18 am

Even your brother disagrees.”

No, he doesn’t. It’s your lack of reading comprehension skills that makes you think so.

“Regardless, how does the flux keep decreasing if the temperature is the same or increasing?”

Which flux are you speaking of? The flux being emitted by the earth depends on its temperature. Why do you think I keep telling you that in the equation F(t) = F0 * e^-(t/τ) that F0 is actually F0(t)?

And you still won’t give any figures? How much flux is lost each day? How long before it drops to near zero?”

Do you enjoy making a fool of yourself? Flux is a RATE, not an amount. How many times have I pointed that out to you? You don’t “lose flux”, you lose JOULES.

I’ve told you to go look up how “τ” is calculated for radiation. It is a simple formula but the components are difficult to characterize completely because the earth is not homogeneous or isothermal. (hint: non-isothermal should be a clue as to what one of the factors is)

Basically the temperature function approaches T(t) = T0 * e^-(t/τ) if the final temperature is considered to be approaching 0 (zero).

Can you solve this for “t”? It’s really quite simple. Assume T0 = 250K as a nice round number. Solve it in terms of “τ”. Assume that the final temperature is T(t) = 2.5K (1%).

Reply to  Tim Gorman
June 22, 2026 10:39 am

“No, he doesn’t.”

Guess again Ogreat one! Stefan-Boltzmann law says that
I = σT⁴. That is an exponential function. I is not roughly the same as the change in the temperature.

It would behoove you to do the math on these problems first before just making an assertion about what you think will occur.

“Which flux are you speaking of? ”

It’s your equation, you should know.

“The flux being emitted by the earth depends on its temperature.”

But not in you equation. It’s only depending on time. As I say, it’s the equation for temperature change over time. Put an object in an environment at absolute zero and no heat source and that’s how temperature will change. The flux sdroends in temperature so would follow that equation, apart from the T⁴ factor. But if temperature isn’t changing in the same was as the cooling equation say because there’s a heat source it follow that your equation will be wrong.

“F0 is actually F0(t)’

Then your equation should have stated that. But making F0 a function based on time, rather than the initial flux makes the whole equation meaningless. You are just saying the flux at any point in time is some unknown value multiplied by some other value.

“Flux is a RATE, not an amount.”

That’s why I asked how much flux was lost each day. Per day is a rate. If the flux is X joules per day tgat’s how much energy you lose during the day.

“I’ve told you to go look up how “τ” is calculated for radiation.”

It’s your equation you need to provide a reference or explain how you calculated tau. But the reason you won’t do that if you did it would be clear that earth would be uninhabital within a few days.

Reply to  Bellman
June 23, 2026 5:28 am

But not in you equation. It’s only depending on time”

Unfreakingbelievable. You can’t even give the math to calculate “τ”.

I’ve told you that F0 is a function of temperature. F0 is joules/sec-m^2. I is joules/sec-m^2. Do you see any relationship between the two?

If you can’t do the math for the simplest parts of the functional relationships you’ll never understand how the thermodynamics of the system works.

But if temperature isn’t changing in the same was as the cooling equation say because there’s a heat source it follow that your equation will be wrong.”

The equation is correct! ALL THE TIME. I’ve told you that during the day F0 is F0(t). And F0(t) is a complex equation with components consisting of the insolation of the sun and the cooling rate of the earth. I’ve even given you a proportional relationship based on the divergence theorem!

It’s simply not worth continuing this with you. You simply can’t keep up with the math.

You can’t even grasp that (Th – Tc) will give the same answer for an infinite number of combinations of Th and Tc – and it is Th and Tc that determines the internal operation of the system. All you can see is the NET, not the components.

Reply to  Tim Gorman
June 23, 2026 5:48 am

“I’ve told you that F0 is a function of temperature.”

Hold on, are you saying t in that equation is temperature, not time? You really need to define your terms, and provide a reference.

If it is temperature your equation makes even less sense. Flux is proportional to temperature to the 4th power, not to any exponential function.

“If you can’t do the math for the simplest parts of the functional relationships you’ll never understand how the thermodynamics of the system works.”

Congratulations, you’ve identified your problem.

“The equation is correct! ALL THE TIME”

Then it should be easy to provide a reference to such a miraculous equation.

“It’s simply not worth continuing this with you. ”

Then stop. It’s evident you are capable of accepting your mistakes. Until you do these conversations will continue indefinitely.

Reply to  Bellman
June 24, 2026 1:14 pm

Hold on, are you saying t in that equation is temperature, not time? You really need to define your terms, and provide a reference.”

FLUX value is based on temperature. Your lack of reading comprehension skills is showing again.

F0(t) is the flux profile over time. The flux profile over time is based on the temperature profile over time.

Why do you have such a hard time understanding simple algebra?

Reply to  Tim Gorman
June 24, 2026 1:52 pm

“Your lack of reading comprehension skills is showing again.”

Don’t blame me for your sloppy writing. As I said, you have to define your terms and provide a reference. Usually lower case t means time and uppercase T means temperature.

“F0(t) is the flux profile over time.”
So why say it wdrpended in temperature?

“The flux profile over time is based on the temperature profile over time.”

Then why not just say flux is proportionate to T⁴? All your exponential time factors are meaningless. If the temperature is constant then the flux will be constant. If temperature rises at a constant rate flux will increase with the forth power of temperature.

Unless you can provide a reference for this equation, it just looks like another thing you’ve misunderstood and are now confusing yourself with.

Reply to  Tim Gorman
June 23, 2026 7:01 am

The equation is correct! ALL THE TIME. I’ve told you that during the day F0 is F0(t). And F0(t) is a complex equation with components consisting of the insolation of the sun and the cooling rate of the earth.

How in the universe can the earth be both warming and cooling? /sarc

Maybe because absorption and emission are two different things.

Reply to  Bellman
June 23, 2026 12:34 pm

That’s why I asked how much flux was lost each day. Per day is a rate. If the flux is X joules per day tgat’s how much energy you lose during the day.

The problem with that is that it is neither linear (T⁴) nor properly measured in a fashion that you can use. Most everything you show is simply a state value. That is, something at a given instant. Incoming over a day is (1360 W/m²) × sin(θ) from 0 to π and neither ist linear. Outgoing depends on a exponential time based rate of cooling. Both ocean and land extract some of the incoming energy for storage and is not linear either.
Why don’t you dig into the equations, one of which I have posted for (dT/dt)?

Reply to  Bellman
June 22, 2026 7:37 am

bellman: ““You don’t think it’s misleading to say something is cooling when it’s temperature is increasing?””

tpg: “It’s the TRUTH”

Any chance of a reference for this “truth”?”

copilot:
—————-
does any object above 0K lose joules (cooling) all the time?

Short, precise, operational answer:
Yes. Any object above 0 K is always losing energy.
But it may lose energy more slowly than it gains it — which means it can still warm up.
———————————–

chatgpt;
—————————
does any object with a temperature above 0K emit joules all the time

Yes — any object with a temperature above 0 K continuously emits electromagnetic radiation and therefore continuously emits energy (measured in joules).
———————–

wikipedia:
—————————-
Thermal radiation is electromagnetic radiation emitted by the thermal motion of particles in matter. All matter with a temperature greater than absolute zero emits thermal radiation.
—————————–

astronomy.swin.edu.au
—————————-
All objects with a temperature above absolute zero (0 K, -273.15 oC) emit energy in the form of electromagnetic radiation.
——————————

HOW MANY REFERENCES DO YOU NEED? Why can’t you do this simple research yourself? Afraid of what you will find?

Losing energy is COOLING. It happens 24 hours per day, 7 days per week, 52 weeks per year, ten years per decade, 100 years per century, etc. for every object above 0K.

Using your assertion that losing energy can result in a temperature rise (i.e. warming) the universe will never suffer a heat death where all things are in thermal equilibrium. In fact, what the universe will ultimately become is undefined, some form of plasma I suppose.

Reply to  Tim Gorman
June 22, 2026 7:47 am

AI does not count as a reference, and not a single one of your references say that emitting energy is called cooling.

But if you do insist on using statistical word generators here’s my question to copilot.

“does any object above 0K cool all the time?”

**No — an object above 0 K does *not* cool all the time.**
It only cools when it is **hotter than its surroundings**, and it stops cooling once it reaches **thermal equilibrium**.

Here’s the clean, physics‑accurate version:

### 🌡️ Core principle: cooling depends on *temperature difference*
Heat flows from **hotter → colder**, never the other way around.
So an object above 0 K can be in three different states:

– **Cooling** — if it is hotter than its environment
– **Warming** — if it is colder than its environment
– **Staying constant** — if it is the same temperature as its environment

Being above 0 K simply means it has thermal energy. It does *not* force it to lose energy.

### 🔥 But doesn’t everything above 0 K radiate energy?
Yes — every object above absolute zero emits thermal radiation.
But it also **absorbs** radiation from its surroundings.

– If emission > absorption → it cools
– If emission < absorption → it warms
– If emission = absorption → temperature stays constant

This balance point is called **radiative equilibrium**.

### 🧊 Why this matters
If you put a warm mug in a cold room, it cools.
If you put that same mug in a hotter room, it warms.
If you put it in a room at exactly the same temperature, it stays the same.

Nothing in physics forces “continuous cooling” just because the temperature is above 0 K.

### ✔️ Final takeaway
**Objects above 0 K do not cool continuously.**
They only cool when they are hotter than their surroundings.

Reply to  Bellman
June 22, 2026 7:50 am

By the way, here’s what copilot told me when I pasted your exact question

“does any object above 0K lose joules (cooling) all the time?”

**No — an object above 0 K does *not* lose joules all the time.**
It *always emits* radiation, but it does **not** always lose *net* energy.

The key is **net heat flow**, not emission alone.

## 🔥 The essential rule
An object above 0 K:

– **Emits** thermal radiation (always true)
– **Absorbs** radiation from its surroundings (also always true)

Whether it *loses* joules depends on the balance:

– **Net loss** → it cools
– **Net gain** → it warms
– **Net zero** → temperature stays constant

So the correct statement is:

> **Objects above 0 K radiate energy, but they do not necessarily lose energy.**

## 🌡️ Example to make it concrete
A rock at 20°C in a 20°C room:

– Emits infrared radiation
– Absorbs exactly the same amount from the room

**Net energy change = 0**
So it does *not* cool, even though it is above 0 K.

A rock at 20°C in a 0°C freezer:

– Emits more than it absorbs
– Net energy loss → it cools

A rock at 20°C in a 40°C sauna:

– Absorbs more than it emits
– Net energy gain → it warms

## 🧊 Why the confusion happens
People hear “everything above absolute zero emits radiation” and assume that means “everything cools.”
But emission alone doesn’t determine cooling — **net exchange** does.

## ✔️ Final takeaway
**Objects above 0 K do not continuously lose joules.**
They only lose energy when they are **hotter than their surroundings**.

If you want, I can also break down **blackbody radiation** or the **Stefan–Boltzmann law** to show the math behind this.

Reply to  Bellman
June 23, 2026 4:13 am

 *net* energy”

I explained this to you in simplistic terms. Net = Gross1 + Gross2 + … + GrossN.

You refuse to admit that Gross1, Gross2, …., GrossN exists. In your worldview only Net exists.

That is non-physical.

There are an infinite combinations of Gross1, Gross2, …, GrossN that can result in the same Net.

If you are doing a PHYSICAL analysis of a system you *have* to know Gross1, Gross2, …., GrossN in order to determine how the system is operating.

Gross1 can be a cooling process, Gross2 a warming process, GrossN can be anything!

The earth’s thermodynamic process is a prime example. You apparently don’t believe that the state of the earth is determined by the internal cooling processes and the external warming process.

That leads to the idiocy that incoming flux and outgoing flux has to balance at all times.

Numbers is just numbers, right?

Reply to  Tim Gorman
June 23, 2026 4:31 am

One strawman after another.

Reply to  Bellman
June 23, 2026 7:29 am

And not one lick of refutation from you. Just the argumentative fallacy of Argument by Dismissal.

Just the assertion that emitting thermal energy is not a cooling process.

Reply to  Tim Gorman
June 23, 2026 8:08 am

“Just the argumentative fallacy of Argument by Dismissal.”

We’ve been arguing tis for years and you still refuse to understand and just reply with strawmen arguments. Sometimes it,’s not worth wasting time on a sentence by sentence refutation. “One horse-laugh is worth ten thousand syllogisms”.

“Just the assertion that emitting thermal energy is not a cooling process.”

Just the assertion that despite repeatedly asking you, you have yet to provide a single piece of evidence that it is. A single comment from a reputable thermodynacs reference that says objects are continuously cooling even when they get warmer. Should be easy if this is a common usage.

Reply to  Bellman
June 25, 2026 5:00 am

Pointing out to you that thermodynamic systems have MULTIPLE COMPONENTS that define its operation and it’s state at any defined point IS NOT A STRAWMAN.

It’s obvious that you don’t even know what a strawman argument is!

strawman argument: “A strawman argument is a logical fallacy in which someone misrepresents or exaggerates an opponent’s position to make it easier to attack.”
You refuse to admit that you *have* to know all the heating and cooling processes in a thermodynamic system. All you want to speak to is NET result as a state at a cherry picked point in time. This leads to your misconception that cooling doesn’t happen during daylight hours, only heating.

It’s why you simply refuse to address why Tmax doesn’t happen at sunset instead of md-afternoon.

It’s why you seem to believe that CO2 “traps” heat, i.e. emission by the earth of reflected heat – because you can’t accept that the earth emits that reflected heat with new rays as laid out by Planck. So you just leave out the emittance of CO2 reflected heat in your sequence of the heating and cooling by earth.

No one trying to explain to you how the thermodynamic system known as the earth actually works is misrepresenting or exaggerating your positions.

  1. The earth cools more during the day than at night.
  2. cooling and heating are time functions
  3. joules in/out over time are the appropriate balance components
  4. flux-in and flux-out will never balance
  5. the earth is neither homogenous nor isothermal
  6. reflected heat is re-emitted by new rays
  7. new rays represent the difference between slower/faster cooling
  8. T^4 is a negative feedback
  9. T^4 flux sets a boundary condition on daytime Tmax
  10. T^4 flux represents *cooling* of an object
  11. cooling is a decaying exponential function
  12. a “state” value at any point in time does not define whether an object is cooling or heating over time. (see No. 2 and No. 3)

You’ve denied each of these during this discussion. Every piece of math you’ve been presented is trying to show how each of these are true as refutation of your denials.

Now, come back and tell us that you agree with all these points but we are still wrong with our math and with the conclusions the math imply.

Reply to  Tim Gorman
June 25, 2026 5:35 am

“Pointing out to you that thermodynamic systems have MULTIPLE COMPONENTS that define its operation and it’s state at any defined point IS NOT A STRAWMAN. ”

The strawman is you claiming I’m not saying that. There are multiple components, the atmosphere, the surface and the sun for starters. I’m the one pointing out you have to consider them both together. You keep insisting in ignoring one or the other.

But let’s go through what you actually said.

You refuse to admit that Gross1, Gross2, …., GrossN exists. In your worldview only Net exists

That’s a strawman. I’ve continually explained the greenhouse effect using gross values. I do not think that only net exists.

You apparently don’t believe that the state of the earth is determined by the internal cooling processes and the external warming process.

A blatant strawman. The entire argument for te greenhouse effect is based on balancing internal and external processes.

“That leads to the idiocy that incoming flux and outgoing flux has to balance at all times. ”

Complete strawman. If they balanced all the time there would be no warming. The point of talking about an equalibrium is to figure out what the state will be when you teach that new equalibrium, not to claim that you are always in an equalibrium.

And second strawman nobody says that fluxes are always equal at any point in time. There will always be the daily and annual cycles and many smaller fluctuations. That’s why the earth is not at a static temperature all the time. The equalibrium happens over a longer period of time.

“Numbers is just numbers, right?”

Such an idiotic strawman it’s not even worth arguing. Especially with someone who claimed his own meaningless equation is right all the time.

Reply to  Bellman
June 25, 2026 5:44 am

“It’s why you simply refuse to address why Tmax doesn’t happen at sunset instead of md-afternoon.”

Oh look, another strawman even as you deny using strawman arguments. I’ve already told you why it happens. I am not refusing to address it, it’s just irrelevant to the argument.

“It’s why you seem to believe that CO2 “traps” heat, i.e. emission by the earth of reflected heat”

And there’s another. You are the one who thinks it’s reflected heat , not I.

“So you just leave out the emittance of CO2 reflected heat in your sequence of the heating and cooling by earth.”

Not even sure if that’s a strawman or just plane ignorance. The entire argument for the greenhouse effect is based on not leaving out the reemiting of energy from ghgs.

Reply to  Bellman
June 25, 2026 5:52 am

“T^4 flux sets a boundary condition on daytime Tmax”

Could you actually calculate that boundary condition? What is the maximum Tmax possible given just the energy from the sun? And us the actual observed maximums warmer than this boundary?

“T^4 flux represents *cooling* of an object”

When all else fails just ignore the evidence and repeat your misunderstanding if what the word “cooling” means.

“cooling is a decaying exponential function”

State the assumptions. Do they mention anything about additional heat sources?

“You’ve denied each of these during this discussion. ”

Exact quotes please.

Reply to  Bellman
June 25, 2026 8:01 am

Could you actually calculate that boundary condition? What is the maximum Tmax possible given just the energy from the sun? And us the actual observed maximums warmer than this boundary?”

I *could* calculate it. I don’t need to. See the graph of the earth’s temperature during the day. Temperature depends on the amount of joules inserted into the system. The sun will insert joules into the system from sunrise to sunset. Meaning if the earth was not also cooling at the same time during the day, Tmax would occur when the maximum amount of joules was inserted, NOT THE RATE OF INSERTION BUT THE TOTAL AMOUNT INSERTED. It wouldn’t matter if the sun’s insolation is less at sunrise and sunset compared to directly overhead, it would still be inserting heat into the system the whole time. So the temperature would rise all day, just at a faster rate at some points – but it would *NEVER* have a negative slope until sunset.

Look at the attached graph showing the temperature here at my location for 6-24-2026. The temperature curve stops rising about 6pm, not 9pm when the sun sets. At 6pm the cooling rate of heat loss of the earth overtakes the heating rate of insertion from the sun. The cooling rate set a boundary condition on Tmax. F_out = F_in.

The earth doesn’t know when the sun sets so it can start cooling. It just cools ALL THE TIME.

If you want to assume that the 6′ air temp is equal to the earth’s temp then *YOU* do the calculation for LWIR F_out at 6pm. *I* can do it. But I’ll bet you can’t. If you want to assume a 0.9 emissivity go right ahead.

But, as I say, you don’t need to know the absolute values to see the effect on the temperature.

When all else fails just ignore the evidence and repeat your misunderstanding if what the word “cooling” means.”

I KINOW what cooling means in a thermodynamic system. You just continue your focus on NET and not on the gross components. It’s just like you saying you don’t consider all measurement uncertainty to be random, Gaussian, and cancels yet it shows up in every assertion you make on metrology.

It’s *ONLY* by looking at the gross cooling and heating that you can tell why the temperature curve looks as it does. If all you focus on is the NET value you can’t tell if the temperature curve turned over at 6pm because the sun went dark or not! Again, the earth doesn’t know when the sun stops inputting joules into the system so it can start cooling. It just cools ALL THE TIME.

Until you accept the thermodynamic definition of cooling as the loss of joules of heat you will *NEVER* be able to understand any of this.

sunrise_sunset_6_24_2026
Reply to  Tim Gorman
June 25, 2026 8:35 am

“I *could* calculate it.”

Ha ha. “I could do it , but I choose not to.” Yes, that’s completely convincing.

You won’t do it because you know it would destroy your argument. The earth is far warmer than the maximum possible using just the sun. That’s what people understood 200 years ago, and I’d why the greenhouse effect was discovered.

Reply to  Bellman
June 25, 2026 1:20 pm

I’ve given you the equations. I started the derivation for you. Do you think it is *that* hard to plug in values to evaluate the equation?

It isn’t the plugging that it difficult. It is coming up with the values to use. Not IMPOSSIBLE, just difficult. And there is no reason to have to do so.

The results of the slower and faster curves are self-evident except to someone like you that is locked into a non-physical dogma.

The greenhouse effect was not discovered. It was created to make simplified assumptions, that were not physical work, for climate science. The greenhouse effect REQUIRES acceptance of the dogma that the earth will retain reflected heat instead of re-emitting it like Planck laid out.

The greenhouse effect is behind every single projection, forecast, or prediction of climate science that has failed to come to pass over the past 50 years. Be it crop failures, species extinction, mass migration, starvation, whatever. But all that reflected heat can do is change Tmin because of SLOWER COOLING. Tmax is limited by T^4, i.e. heating is limited by cooling.

Climate science exacerbates these misconceptions by using mid-range diurnal temperatures WHICH ARE NOT A METRIC FOR CLIMATE. Mid-range temperatures can’t distinguish climates let alone Tmax from Tmin changes. And then to top *that*, climate science refuses to recognize that heating and cooling processes are TIME FUNCTIONS that happen at the same time while S-B temperature is a STATE FUNCTION.

And here you are trying to justify all these non-physical assumptions of climate science.

Reply to  Tim Gorman
June 25, 2026 2:29 pm

“I’ve given you the equations”
Yes, many times. What I’m asking for is a reference so we can check you understand what it means, or that you haven’t just get it from a chatbot hallucination. And then I’m asking you to demonstrate how to use it in an actual example to illustrate why the greenhouse effect is wrong.

“The greenhouse effect was not discovered. It was created to make simplified assumptions, that were not physical work, for climate science. ”

You really like to show of your ignorance. The greenhouse effect was developed long before there was such a thing as climate science. It was developed by scientists who I suspect understood the subject a little better than you – if you can imagine such a thing.

It really is the height of hypocrisy to suggest I’m not alowed to correct your mistakes because you are authorities on thermodynamics, and then with the next breath claim that practically every scientist who has ever studied the subject is wrong.

Reply to  Bellman
June 26, 2026 7:20 am

What I’m asking for is a reference so we can check you understand what it means, or that you haven’t just get it from a chatbot hallucination. “

One more time – I GAVE YOU THE MATH!

I started with a recognized and accepted equation, the S-B equation.

From there I derived the power law expression for the change in temperature over time. I approximated that power law decay using an easier-to-understand approximation consisting of exponential decay that is familiar to anyone with even a modicum of training in the physical sciences and in calculus.

I gave you a graph of how the flux graph changes for different slopes of decay, faster cooling vs slower cooling using an exponential decay example. And I showed you how integrating those curves gives more area under the curve for slower cooling.

You don’t *need* a reference confirming my derivation. You have my derivation. Show me where it is wrong. Since you can’t do the differential equations and calculus FIND SOMEONE ELSE THAT CAN. Have them SHOW MY DERIVATION TO BE WRONG.

Find someone else that can confirm your assertion that emitting joules of energy is not a cooling process.

Otherwise just accept that you are doing nothing but carping about something you don’t understand and will never understand. Then go away!

Reply to  Bellman
June 25, 2026 7:29 am

I am not refusing to address it, it’s just irrelevant to the argument.”

No, it *IS* the argument. If the earth wasn’t cooling 24/7 then Tmax would occur at a different time than it actually does.

*YOU* want to ignore that fact so you call it “irrelevant”. That’s the Argument by Dismissal fallacy.

And there’s another. You are the one who thinks it’s reflected heat , not I.”

Planck: “For example, if we let the rays emitted by the body fall
back on it, say by suitable refection, the body, while again absorbing these rays, will necessarily be at the same time emitting new rays, and this is the compensation required by the second principle.” (bolding mine, tpg)

Planck doesn’t require optical reflection, just “suitable reflection”. And an object absorbing heat from a source and sending part of that absorbed heat back to the source IS SUITABLE REFLECTION.

The entire argument for the greenhouse effect is based on not leaving out the reemiting of energy from ghgs.”

Your lack of reading comprehension skills is showing again. I didn’t leave out the re-emitting of energy from ghg’s. *YOU LEFT OUT THE RE-EMITTANCE OF THAT REFLECTED ENERGY.*

By not reflecting the earth re-emitting the reflected energy from CO2 YOU MAKE THAT REFLECTED HEAT INTO TRAPPED HEAT. If it never gets re-emitted by the earth then it *is* trapped*, retained, or however you want to characterize it.

See the shaded part of the attached image. If that shaded area doesn’t represent the re-emittance of reflected energy then what does it represent? Where does it come from? Where does it go? Why do you continue to ignore that it exists?

slower_faster_2
Reply to  Tim Gorman
June 25, 2026 8:25 am

“If the earth wasn’t cooling 24/7 then Tmax would occur at a different time than it actually does.”

If it was cooling 24/7 then the maximum temperature would be the starting temperature and everything else would be colder.

Even using your definition of cooling, nothing you say makes any sense. The reason cooling starts when it does is because that’s when energy out is greater than energy in. This is due to a combination of factors. Both the temperature rising so increasing energy out, and energy from the sun declining, along with over heating processes such as heat from the ground etc.

“*YOU* want to ignore that fact so you call it “irrelevant”. ”

I’d like you to explain why you think it’s relevant.

“say by suitable refection”

What do you think the word “say” means? You seem to think that paragraph means the only way an object gets energy back is via reflection.

“And an object absorbing heat from a source and sending part of that absorbed heat back to the source IS SUITABLE REFLECTION.”

Have you tried asking your chatbot to explain it to you? That seems to be the only authority you accept. Let me try it for you:

Copilot “In thermodynamics what is the difference between reflection and absorption?’

Short answer

In thermodynamics, absorption means a material takes in incoming energy (such as light or heat) and converts it—usually into thermal energy. Reflection means the energy is bounced off the surface without being converted. The key difference is whether the incoming energy is retained and transformed (absorption) or redirected unchanged (reflection).

Reply to  Bellman
June 25, 2026 8:29 am

“If it never gets re-emitted by the earth then it *is* trapped*, retained, or however you want to characterize it.”

The heat is re-emitted. That’s the whole reason the E increases. Your inability to even try to understand this is why these arguments become so pathetic.

Reply to  Bellman
June 25, 2026 10:01 am

If the redirected heat from CO2 gets RE-EMITTED by the earth, then how can it ADD to the heat energy in the earth?

Reply to  Tim Gorman
June 25, 2026 10:46 am

I’ve gone over the equations multiple times. I’m not wasting more time explaining it again to someone who is determined not to see the woods for the trees.

Reply to  Bellman
June 25, 2026 12:46 pm

I’ve gone over the equations multiple times. I’m not wasting more time explaining it again to someone who is determined not to see the woods for the trees.”

In other words you can’t explain how re-emitted heat can heat the earth. There aren’t any woods and no trees.

There’s just the fact that you got caught trying to say that CO2 traps heat and raises the temperature of the earth while that heat that causes the temperature rise is actually re-emitted thus lowering the temperature of the earth.

That’s what happens every time you come on WUWT and start lecturing people on how reality works. You get caught and all you have to offer is ABBA! ABBA! ABBA!



Reply to  Tim Gorman
June 25, 2026 5:02 pm

In other words you can’t explain how re-emitted heat can heat the earth.

Apparently not to you. That would mean you accepting that you were wrong about something, or maybe the concept that 1 + 1 = 2 is too complicated for you.

There’s just the fact that you got caught trying to say that CO2 traps heat…

When did I say that? It’s risky trying top use any sort of metaphor with someone of your limited imagination. You take every word to it’s most ridiculously logical meaning.

A better description might be delayed, or diverted. Some of the energy doesn’t go straight into space, but is sent back to earth. It will then get another chance to get into space. Eventually you have twice as much energy hanging around, but it isn’t trapped in the sense it can never get out. It just takes a little more time.

Reply to  Bellman
June 26, 2026 8:51 am

“Apparently not to you.”

Not to anyone! You can’t even say if the earth retains that heat reflected by CO2 or it re-emits again.

“When did I say that?”

When you refuse to admit that the earth re-emits what it gets from CO2. When you leave out the retain or re-emit step in the sequence.

What happens to that reflected heat? Does the earth retain it or re-emit it?

You’ve only got two choices. Pick one and tell us which you pick.

There’s only one choice that matches what Planck wrote. Which one is it?

A better description might be delayed, or diverted.”

Diverted? Diverted to where?

Delayed? Yep, it’s a time function – A CONTINUOUS TIME FUNCTION. meaning at any point in time it is losing *MORE* heat energy due to the slower cooling. See the attached graph!

You *still* haven’t explained how losing MORE heat from slower cooling can change the slope of the tempeature curve from negative to positive in order to raise the temperature of an object!

You *still* haven’t explained what that shaded area on the graph represents even after its been given to you at least four times. You just keep carping that the math is wrong without showing where its wrong.

slower_faster_2
Reply to  Tim Gorman
June 26, 2026 9:24 am

Part of the problem with people is that they have taken to heart that CO2 radiates half up and half down. In reality, a body radiates the same in ALL directions based upon its temperature. The value of radiation is not divided by the number of sides, or a sphere would radiate a miniscule amount in every direction.

Reply to  Jim Gorman
June 26, 2026 12:07 pm

Yes, it radiates in all directions. But any sideways radiation will generally just move the radiation around the atmosphere. It’s only the radiation which gets out of the atmosphere that has an effect, and that can only go in one of two directions.

But as I keep having to remind you this is only intended to be the simplest possible model to explain the general concept. If you want a more complex model the first issue is the assumption that all energy is absorbed by the atmosphere. Then the assumption that the atmosphere doesn’t absorb anything from the sun. Then the fact that you are treating the atmosphere as a single layer, if ignoring differences across continents and ignoring the daily and seasonal cycles.

Reply to  Bellman
June 26, 2026 12:28 pm

It’s only the radiation which gets out of the atmosphere that has an effect, and that can only go in one of two directions.

You still don’t understand radiation. Here is a slide that might help you to understand.

comment image

Bodies radiate in all directions based on the body’s temperature. As the body being discussed rises in altitude, less and less radiation actually strikes the earth and more and more simply goes off into space.

It is not half up and half down. The flux reaching the earth also diminishes based upon 1/r².

Reply to  Jim Gorman
June 26, 2026 6:08 pm

The flux reaching the earth also diminishes based upon 1/r².

I really don’t think you understand this. The energy cannot reduce like that, it would break the first rule of thermodynamics. The inverse square rule is about radiating into space from a point. The energy gets spread out in all directions and any object it hits only gets a fraction of the total amount. This doesn’t happen with the earth and the atmosphere because the earth fills out all the area below the atmosphere. It doesn’t matter how spread out the energy gets, it will all still hit the earth.

Reply to  Bellman
June 27, 2026 3:07 am

I really don’t think you understand this. The energy cannot reduce like that, it would break the first rule of thermodynamics.”

You *really* just need to go away. You *really* are making a fool of yourself.

Flux has the dimension of joules/sec-m^2. What happens to m^2 as an EM wavefront expands? You have the same energy spread over a sphere with an increasing radius.

That’s not a violation of the first rule of thermodynamics, it’s called the INVERSE SQUARE LAW OF EM RADIATION.

The inverse square rule is about radiating into space from a point. “

And you think the sun isn’t considered as a point source? Perhaps it’s an infinitely large flat plate emitting a plane wave? You don’t understand enough 3D to even know why the sun’s insolation is considered to be a plane wave when it reaches the earth, do you?

This doesn’t happen with the earth and the atmosphere because the earth fills out all the area below the atmosphere”

No, it doesn’t! Like I told you before, you must never have heard the story about Columbus and the horizon!

 It doesn’t matter how spread out the energy gets, it will all still hit the earth.”

Unfreakingbelievable! The EM wave generated by CO2 is *NOT* a spherical wavefront and doesn’t obey the inverse square law. All of the radiation from CO2 in the atmosphere strikes the earth.



Reply to  Tim Gorman
June 27, 2026 7:31 am

“What happens to m^2 as an EM wavefront expands? You have the same energy spread over a sphere with an increasing radius. ”

Yes, that’s what I said.

“That’s not a violation of the first rule of thermodynamics, it’s called the INVERSE SQUARE LAW OF EM RADIATION.”

Indeed it isn’t, it’s just the further an object is from the source the less energy it receives per area. The rest if the energy continues to spread of through space.

“And you think the sun isn’t considered as a point source?”

Try to read the thread you are jumping into. This was about the atmosphere, not the sun.

“No, it doesn’t!”

You think there is somewher below the atmosphere that isn’t the earth?

“All of the radiation from CO2 in the atmosphere strikes the earth.”

Why would you think that. I’ve already told you the the approximation is that half the radiation hits earth. The other half goes into space. What doesn’t happen is the inverse square rule.

The simple approximation is that the earth and atmosphere are infinite planes and the atmosphere radiates in two directions, up and down, equally. Obviously this is not reality, as the earth is round and the atmosphere has depth. What percentage of the energy radiated by the atmosphere actually hits the earth I couldn’t say. But it is going to be close to the 1/2, and it is not going to be determined by the inverse square rule.

Reply to  Bellman
June 30, 2026 3:57 am

Why would you think that. I’ve already told you the the approximation is that half the radiation hits earth. The other half goes into space. What doesn’t happen is the inverse square rule.”

Half does *NOT* heat the earth. Not only does MORE than half NOT hit the earth, part of the radiation that does hit it does so at an angle. Only the normal part is absorbed, the rest is reflected back. It’s called the Divergence Theorem. I’ve given you the math for that already – and you couldn’t figure it out any more than you can the power-law decay derivation from S-B.

And the inverse square law *ALWAYS* happens. Why do you think it is called a LAW?

Do you know what a steradian is? Do you know how it applies in 3D spatial analysis? Do you know how it relates to the inverse square law?

Reply to  Bellman
June 26, 2026 2:16 pm

 Then the fact that you are treating the atmosphere as a single layer, if ignoring differences across continents and ignoring the daily and seasonal cycles.”

Oh malarky! You keep throwing out the excuse that there is such a thing as a global average temperature and then you come out with this excuse?

How many times have you been told that heat gain/loss is a TIME FUNCTION. I’ve probably told you that over 100 times just myself!

And the fact that as a molecule rises in elevation the earth subtends a smaller and smaller angle is a 3D FACT. it is ignoring nothing.

Reply to  Tim Gorman
June 26, 2026 5:46 pm

How many times have you been told that heat gain/loss is a TIME FUNCTION

And how many times have you been told that it doesn’t matter. All you need to do is work out what the equilibrium balance will be. But if you want to actually get around to explaining how your time based functions work, and what they mean in practical terms please don’t let me stop you.

Reply to  Bellman
June 27, 2026 2:31 am

And how many times have you been told that it doesn’t matter. All you need to do is work out what the equilibrium balance will be.”

THERE IS NO EQUILIBRIUM BALANCE! There is nothing to work out. The earth’s thermodynamic system *NEVER* reaches equilibrium. It’s always moving, changing, etc. What do you think the word “cycle” means?

Does a sine wave ever reach equilibrium in time?

The only equilibrium that will ever be reached in reality is the heat death of the universe. Everything at the same temperature. Can *YOU* work out when that will happen? Can *YOU* work out what that final temperature will be?

But if you want to actually get around to explaining how your time based functions work, and what they mean in practical terms please don’t let me stop you.”

The time based functions work. Why do you suppose so many people graph temperature on the vertical axis and time on the horizontal axis?

Newton’s Law of Cooling *is* a time based function. It isn’t *my* time based function, it is science’s time based function. And here you are, as usual, trying to claim it is wrong and how it works isn’t described by the math.



Reply to  Tim Gorman
June 27, 2026 6:24 am

What do you think the word “cycle” means?

It means you are cycling about an equilibrium point. If there was no equilibrium you would not cycle, you would just wander of randomly.

The earth’s thermodynamic system *NEVER* reaches equilibrium.

But it does cycle about it. Why else do you think temperatures have only changed by a few degrees across millions of years?

Reply to  Bellman
June 30, 2026 3:50 am

It means you are cycling about an equilibrium point. If there was no equilibrium you would not cycle, you would just wander of randomly.”

That cycling does *NOT* have to be random or symmetric. And the cycling is the VARIANCE of the measurand – i.e. it’s uncertainty.

You can’t just ignore it. And since it has a systematic component inherited from a multiplicity of factors, it can’t be considered to be random, Gaussian, and therefore cancels out.

“But it does cycle about it. Why else do you think temperatures have only changed by a few degrees across millions of years?”

Meaning there are negative feedback paths that prevents temperature from exceeding boundaries on the cool side (planet ICE) or the hot side (planet MOLTEN). And since humans have survived both excursions toward the cold side *and* the hot side, what is the impetus behind CAGW? CO2 has been higher in the past while humans survived and the planet didn’t burn up. Why should it burn up now when CO2 is not as high as in the past?

It’s all proof that climate science is looking at just a small part of the natural cycle and trying to use that small piece of the cycle to project a Planet Fire. Their observation window simply doesn’t match Nyquist requirements for the low frequency components of the cycle. Yet it is those low frequency components that are a primary identifier of the negative feedback factors that exist. The temperature models climate sciences has today (I refuse to continue calling them climate models) simply need to be expanded to cover several centuries to be considered as actual representation of the natural variation of the biosphere.

Reply to  Tim Gorman
June 30, 2026 1:17 pm

“That cycling does *NOT* have to be random or symmetric. ”

For someone who likes to insult my reading comprehension, you do like to misunderstand what I said a lot. I didn’t say anything about the cycles being random or symmetric, just that they are cycling around an equilibrium point.

“And the cycling is the VARIANCE of the measurand”

Look, a squirrel.

Reply to  Bellman
June 25, 2026 12:42 pm

The heat is re-emitted. That’s the whole reason the E increases. Your inability to even try to understand this is why these arguments become so pathetic.”

If the heat is re-emitted then how can it add energy to the earth and raise it’s temperature?

Reply to  Tim Gorman
June 25, 2026 4:53 pm

If the heat is re-emitted then how can it add energy to the earth and raise it’s temperature?

Because it will also go into the atmosphere, making it warmer. It will then in tern be re-emitted from the atmosphere to earth.

I’m really not sure why I keep trying to explain this to you. It’s obvious your own dogma won’t allow you to understand it.

Reply to  Bellman
June 26, 2026 8:42 am

Because it will also go into the atmosphere, making it warmer. It will then in tern be re-emitted from the atmosphere to earth.”

And what does the earth then do with that heat emitted by the atmosphere?

Retain it? Re-emit it back to CO2, H2O, and space?

That’s the step you always leave out.

Which is it? Retain it or Lose-it-again?



Reply to  Tim Gorman
June 26, 2026 9:32 am

Retain it? Re-emit it back to CO2, H2O, and space?

This is more complicated than one first might think. If the earth receives at 15um, it will re-emit at a Planck curve. This means a lot will go right to space through the atmospheric window, and CO2 won’t even see it again. That is part of the reason for a dip in the radiation spectrum.

Reply to  Jim Gorman
June 26, 2026 1:51 pm

That just means that the effective reflectivity over time for CO2 is even lower.

Reply to  Tim Gorman
June 26, 2026 9:51 am

“Retain it? Re-emit it back to CO2, H2O, and space?”

Yes.

Reply to  Bellman
June 26, 2026 1:52 pm

“Retain it? Re-emit it back to CO2, H2O, and space?”

So it both retains it and loses it.

That *IS* the only way it will work for you and your beliefs.

It’s called cognitive dissonance. You should look into it.

Reply to  Tim Gorman
June 26, 2026 5:32 pm

So it both retains it and loses it.

Yes. That’s the way thermodynamics works.

Does your bank retain your money or does it give it to you? The answer is both.

Does a battery save energy or release it? The answer is both.

Does an object retain energy or does it emit it? The answer is both.

Beyond your obvious dogma, I’m really not sure why you find this hard to understand.

Reply to  Bellman
June 27, 2026 2:06 am

“Yes. That’s the way thermodynamics works.

Does your bank retain your money or does it give it to you? The answer is both.”

Except the earth doesn’t “give” the energy back upon demand like the bank does. It gives it back automatically. Like you cashing your paycheck in the bank and the bank immediately giving you back the same amount in cash.

Can you get *anything* right? Stop throwing shite against the wall – none of it is sticking.

Does a battery save energy or release it? The answer is both.”

Electrical and chemical energy is not THERMAL ENERGY. You can’t even get this one right!

Does an object retain energy or does it emit it? The answer is both.”

Tell us all exactly *HOW* an object retaining heat knows when to emit it?

Tell us all how an object “tags” incoming reflected energy so it knows not to re-emit it until later.

Reply to  Tim Gorman
June 27, 2026 6:20 am

Except the earth doesn’t “give” the energy back upon demand like the bank does.

You are just incapable of understanding the simplest metaphor. The earth gives back energy in accordance with the Stefan-Boltzmann law.

Tell us all exactly *HOW* an object retaining heat knows when to emit it?

You are the one who keeps peddling the cooling equation. How do you think an object knows how to follow that equation? Unless you are asking for some deep philosophical explanation of how the universe works, you just have to accept it does. Hotter objects emit more energy – they don’t “know” they have to, they just do.

Tell us all how an object “tags” incoming reflected energy so it knows not to re-emit it until later.

It does not. It’s just energy. It has to be conserved, but it can transform from one type to another. Energy is absorbed by the earth in the form of thermal energy, and then converts it to kinetic energy. The some of this energy is converted back into thermal energy and re-emitted. It does so in accordance with the SB law.

You really should be able to understand this. It seems like basic thermodynamics which you claim to understand better than me. So what is your problem? Are you just looking for excuses, are you trolling, or are you really this ignorant?

Reply to  Bellman
June 30, 2026 3:31 am

The earth gives back energy in accordance with the Stefan-Boltzmann law.”

Which is exactly what I said! The S-B law is a metric for the intensity WITH WHICH THE ENERGY IS AUTOMATICALLY GIVEN BACK!

“You are the one who keeps peddling the cooling equation. How do you think an object knows how to follow that equation? Unless you are asking for some deep philosophical explanation of how the universe works, you just have to accept it does. Hotter objects emit more energy – they don’t “know” they have to, they just do.”

If the object is GIVING BACK energy then it is cooling. Losing energy is a cooling process. I seem to sense that you have been reduced to having to agree with me that losing energy is COOLING and you are looking for a way to express that agreement by saying it’s what *YOU* have been saying!

It does not. It’s just energy. It has to be conserved, but it can transform from one type to another. Energy is absorbed by the earth in the form of thermal energy, and then converts it to kinetic energy. The some of this energy is converted back into thermal energy and re-emitted. It does so in accordance with the SB law.”

In other words, the object loses energy – a COOLING PROCESS.

*YOU* were arguing that reflected energy from CO2 is retained by the earth and causes the temperature to go up. And you are *still* arguing that “some* of the thermal energy is RETAINED!

That re-emission you are now admitting happens, is the COMPENSATION spoken of by Planck which slows cooling by resulting in a slower decay rate – AND A SLOWER DECAY RATE IS *STILL* A COOLING PROCESS. Because ALL of that reflected energy is re-emitted it can’t push the temperature back up the gradient, it can only change the slope of the decay curve to a slower cooling rate. See the attached graph that I have given you at least six time in the past.

How does the earth know which part of the reflected energy to re-emit and which to retain? How does it distinguish reflected energy from LWIR from the sun? How does the earth retain some of the heat which would cause the temperature to go up and result in a higher rate of S-B radiation?

If the earth retains heat then why isn’t it a molten rock by now?

slower_faster_2
Reply to  Tim Gorman
June 30, 2026 1:11 pm

“Which is exactly what I said!”

Yes, that’s the problem. You keep agreeing with me, but seem to think I’m wrong.

” I seem to sense that you have been reduced to having to agree with me that losing energy is COOLING and you are looking for a way to express that agreement by saying it’s what *YOU* have been saying!”

You can call it what you want. It’s just very confusing, which I expect is your purpose. As I keep saying, it’s equivocation. Use it in one context to just mean an object is losing energy, then try to pretend that means it can’t be getting hotter.

“That re-emission you are now admitting happens…”

And this is why arguing with you is so futile. You never try to understand what people are telling you, you argue with points that exist only in your imagination, and then claim some pathetic victory when you accept what they have actually said.

Re-emission is what I’ve been saying happens throughout. Go back over everything I’ve said and try to find anywhere were I denied that objects re-emit energy. It’s fundamental to the process. Mostly I’ve been arguing with you claiming it’s reflection.

“is the COMPENSATION spoken of by Planck ”

As I kept telling you.

“Because ALL of that reflected energy is re-emitted it can’t push the temperature back up the gradient…”

And there’s your problem. This all or nothing understanding. Either the Earth re-emiis everything or it re-emitts nothing.

The problem you still can’t address is how an object emits more energy if it’s temperature has not risen. If you increase the energy going into an object and emitts more energy as compensation it is usually by the SB law because it’s temperature has increased.

“How does the earth know which part of the reflected energy to re-emit and which to retain?”

It doesn’t need to know anything. It just follows the natural laws about releasing energy according to it’s temperature.

Reply to  Bellman
June 25, 2026 10:00 am

If it was cooling 24/7 then the maximum temperature would be the starting temperature and everything else would be colder.”

frekingUnbelivable.

“The reason cooling starts when it does is because that’s when energy out is greater than energy in.”

NO! Cooling happens anytime heat loss is involved. And you can have heat loss even while the temperature of an object is going up!

This is due to a combination of factors. Both the temperature rising so increasing energy out, and energy from the sun declining, along with over heating processes such as heat from the ground etc.” (bolding mine, tpg)

You simply refuse to see that energy out is a COOLING process. It occurs in every thing with a temperature above 0K. That cooling process happens whether the temperature of the object is going up or not.

I’d like you to explain why you think it’s relevant.”

Because the cooling process is an integral part of the system thermodynamics! YOU HAVE TO KNOW IT IN ORDER TO CALCULATE THE NET.

What do you think the word “say” means? “

Your lack of reading comprehension skills is showing again! The operative words are “suitable reflection”, it doesn’t say what “suitable” has to be. *YOU* want it to be optical reflection and refuse to read and comprehend what is actually being said.

 gets energy back”

You can’t even comprehend what you write yourself!

“redirected unchanged (reflection).”

And you think CO2 “changes” received thermal energy at a specific wavelength into something else before it sends it back? What exactly does it change it into? As usual, you totally miss the operative words in a sentence “unchanged”.

All CO2 does in REDIRECT incoming energy *BACK* to the source. Nothing gets “changed”.

Reply to  Tim Gorman
June 25, 2026 10:41 am

“You simply refuse to see that energy out is a COOLING process. ”

I refuse to accept it because it’s just another one of your Gorman definitions that is not used by the test of the world. It’s quite obvious that you only use it so you can obfescate and equivocate.

Reply to  Bellman
June 25, 2026 12:52 pm

it’s just another one of your Gorman definitions that is not used by the test of the world.”

It’s used by every single engineer involved in the thermodynamic analysis of a system. You have to know the in and the out in order to properly design the NET you want. You can’t just use NET all by itself. And the in and out represent heating AND COOLING.

There is a reason why a small acetylene torch can’t heat an anvil to the melting point. The cooling processes of the anvil radiating away heat and conducting away heat happens continually while you are heating the anvil. The cooling processes don’t stop when you hit the anvil with the flame. The cooling of the earth doesn’t stop when the sun rises.

Reply to  Tim Gorman
June 25, 2026 4:49 pm

It’s used by every single engineer involved in the thermodynamic analysis of a system.

Then show me a reference. You keep giving references that say the exact opposite. If it was used all the time surely someone has written it down somewhere.

Reply to  Bellman
June 26, 2026 8:40 am

I gave you my derivation. It’s not up to me to prove myself correct in the derivation. The derivation math is correct.

You show me where my math is wrong. Or *YOU* provide a reference showing that it is wrong.

I suggest the first place you start with is Newton’s Law of Cooling. go here: http://www.geeksforgeeks.org/physics/newtons-law-of-cooling/

The problem with Newton’s Law of Cooling is that for the earth the rate of decay is not a constant. The rate changes as the temperature changes – the power law. And I gave you the power law. And I told you why I use the exponential law instead.

 You keep giving references that say the exact opposite.”

Every reference I’ve given you does the derivations the same way I did. The fact that you can’t understand the math doesn’t mean they are saying the opposite of what I am saying. Newton’s Law of Cooling shows that. The fact that you can’t understand the math enough to show what I did wrong just means you aren’t fit to judge if my derivation is correct or not. No number of references, even if they are exactly the same as mine, is going to fix that problem. It’s going to wind up just like it wound up with Possolo and the barrel – no one but you can do partial derivatives correctly.



Reply to  Tim Gorman
June 25, 2026 7:43 pm

And you think CO2 “changes” received thermal energy at a specific wavelength into something else before it sends it back?

Yes.

What exactly does it change it into?

Molecules of carbon dioxide (CO2) can absorb energy from infrared (IR) radiation. This animation shows a molecule of CO2 absorbing an incoming infrared photon (yellow arrows). The energy from the photon causes the CO2 molecule to vibrate. Some time later, the molecule gives up this extra energy by emitting another infrared photon. Once the extra energy has been removed by the emitted photon, the carbon dioxide molecule stops vibrating.

https://scied.ucar.edu/learning-zone/how-climate-works/carbon-dioxide-absorbs-and-re-emits-infrared-radiation

All CO2 does in REDIRECT incoming energy *BACK* to the source.

Nope. It can re-emit it in any direction – hence half of the energy going out to space.

Reply to  Bellman
June 26, 2026 9:36 am

“This animation shows a molecule of CO2 absorbing an incoming infrared photon (yellow arrows). The energy from the photon causes the CO2 molecule to vibrate. Some time later, the molecule gives up this extra energy by emitting another infrared photon. ” (bolding mine, tpg)

EXACTLY WHERE DID THIS THERMAL RADIATION GET CHANGED TO NON-THERMAL RADIATION?

Your reading comprehension skills are just non-existent. You just cherry pick things without even a modicum of understanding of meaning and context.

Nope. It can re-emit it in any direction – hence half of the energy going out to space.”

If you had any inkling of 3D vectors you would know that less than 1/2 goes out to space. The amount that goes back is based on the angle the earth subtends at the point of emission. As the altitude goes up, the angle subtended goes down and the less gets back to earth.

Did anyone ever explain Columbus and the horizon to you?

Reply to  Tim Gorman
June 26, 2026 9:49 am

“EXACTLY WHERE DID THIS THERMAL RADIATION GET CHANGED TO NON-THERMAL RADIATION?”

Stop shouting. It doesn’t change to non-thermal radiation. Nobody claimed it did. It’s converted into kinetic energy, hence vibration.

“Your reading comprehension skills are just non-existent. ”

Says it all.

“If you had any inkling of 3D vectors you would know that less than 1/2 goes out to space. ”

It’s a simplifying assumption. None if this is meant to be an exact 3-d model of the entire global system. It’s the simplest possible model that illustrated how the greenhouse effect can work. But as usual you try to distract from the general point by focusing on mostly irrelevant specific assumptions. All the while insisting your own one line equation is right all the time.

Reply to  Bellman
June 26, 2026 1:58 pm

Stop shouting. It doesn’t change to non-thermal radiation. Nobody claimed it did. It’s converted into kinetic energy, hence vibration.”

So now we have you in TWO cognitive dissonance situations.

  1. The earth both retains heat from CO2 and re-emits it.
  2. Thermal energy-in and thermal energy-out are both the same and different.

You just keep tying yourself in knots trying to justify your view of how things work. Does it hurt when you get up in the morning?

Reply to  Tim Gorman
June 26, 2026 5:28 pm

The earth both retains heat from CO2 and re-emits it.

Correct. How hard is that to understand?

Thermal energy-in and thermal energy-out are both the same and different

You are very confused.

Why don’t you read up on the subject rather than keep misunderstanding everything I say. You are the one tying yourself in knots to avoid understanding what is very basic science.

Reply to  Bellman
June 27, 2026 2:00 am

——————–
The earth both retains heat from CO2 and re-emits it.
Correct. How hard is that to understand?
———————

ROFL!!! You can’t even see the cognitive dissonance you are promoting!

————————-
Thermal energy-in and thermal energy-out are both the same and different

You are very confused.
————————-

It’s *YOUR* claim, not mine.

Thermal energy is thermal energy.

Thermal energy that is received and thermal energy that is sent back to the source IS THE SAME. It is a process of reflection of energy.

As usual, you didn’t even finish the steps involved in the reflection process. You keep forgetting that the earth re-emits radiation received from CO2 and leave that step out. You keep forgetting that CO2 sends out thermal energy that it receives.

Again, the mechanism of “reflection” is irrelevant to the actual phenomenon.

From wikipedia: “In classical electrodynamics, light is considered as an electromagnetic wave, which is described by Maxwell’s equations. Light waves incident on a material induce small oscillations of polarisation in the individual atoms (or oscillation of electrons, in metals), causing each particle to radiate a small secondary wave in all directions, like a dipole antenna. All these waves add up to give specular reflection and refraction, according to the Huygens–Fresnel principle.”

To hear you tell it, even a perfect mirror doesn’t reflect since the impinging EM wave gets reflected by first inducing oscillations in the atoms which then, in turn, emit secondary EM waves.

Why don’t you read up on the subject rather than keep misunderstanding everything I say. You are the one tying yourself in knots to avoid understanding what is very basic science.”

*YOU* are the one carping on supposed details that you really have no understanding of at all. You can’t even get “reflection” correct.

I’m not misunderstanding what you say, I’m merely pointing out how most of what you post is *NOT* very basic science.

Retaining and re-emitting are contradictory. It represents cognitive dissonance in “basic science”.

Thermal heat not being reflected while the same thermal energy that goes in gets sent back are contradictory. It represents cognitive dissonance in “basic science”.

You just keep carping that my math is wrong and can’t be right. But you can’t refute it. And the assertions you make while trying to show my conclusions are wrong just violate “very basic science”.

If pointing out your misconceptions bothers you then stop posting misconceptions and actually learn something about what you are lecturing everyone on.

Reply to  Tim Gorman
June 27, 2026 6:09 am

ROFL!!! You can’t even see the cognitive dissonance you are promoting!”

Stop laughing and try to understand how it works. What do you think the Stefan-Boltzmann law is saying? You keep going on about T^4, where do you think this T^4 comes from.

I really think this much easier to understand then you do. If you add energy to an object it gets hotter. Getting hotter means retaining energy. The hotter it gets the more energy it emits.

It’s *YOUR* claim, not mine.

What claim? You’re the one saying I think thermal energy in is different to thermal energy out. When did you imagine I said that?

But you are probably equivocating on the idea of “same”. You want to think everything is reflection, so it’s literally the same ray of heat bouncing of an object, as opposed to the energy being absorbed, converted into kinetic energy, and then being converted back into heat energy.

Beyond that I’ve no idea what you think is happening.

Again, the mechanism of “reflection” is irrelevant to the actual phenomenon.

But it is relevant if it is not reflection. If CO2 reflected all heat it would never get warm, and we would receive a great deal more heat back. But if it’s absorbing the energy the atmosphere warms and the amount we get back depends on the temperature. If it was all reflection there would be no energy during the might time and we would all be frozen.

From wikipedia

I can’t find that passage, but it’s irrelevant to how thermodynamics defines reflection vs absorption. Nor does it say that light is absorbed and re-emitted. It just says that light scatters of a surface.

Here are some more relevant wiki page

Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. Common examples include the reflection of light, sound and water waves. The law of reflection says that for specular reflection (for example at a mirror) the angle at which the wave is incident on the surface equals the angle at which it is reflected.

https://en.wikipedia.org/wiki/Reflection_(physics)

Diffuse reflection is the reflection of light or other waves or particles from a surface such that a ray incident on the surface is scattered at many angles rather than at just one angle as in the case of specular reflection. An ideal diffuse reflecting surface is said to exhibit Lambertian reflection, meaning that there is equal luminance when viewed from all directions lying in the half-space adjacent to the surface.

In physics, absorption of electromagnetic radiation is how matter (typically electrons bound in atoms) takes up a photon’s energy—and so transforms electromagnetic energy into internal energy of the absorber (for example, thermal energy).

In meteorology and climatology, global and local temperatures depend in part on the absorption of radiation by atmospheric gases (such as in the greenhouse effect) and land and ocean surfaces

https://en.wikipedia.org/wiki/Absorption_(electromagnetic_radiation)

Retaining and re-emitting are contradictory. It represents cognitive dissonance in “basic science”.

How. This seems to be your problem, not mine. I really don’t understand what you think the problem is. I can only guess it’s your inability to understand the difference between absorption and reflection. You think re-emit means reflect, in which case none of the energy is retained.

Honestly, though, how do you think objects get warm, and warm objects radiate heat, if you don’t think is possible for an object to retain and re-emit energy?

Thermal heat not being reflected while the same thermal energy that goes in gets sent back are contradictory.

Again, what do you mean by “the same”? If you put $10 into a bank account and a few months later withdraw it, do you think you are getting the same $10 back?

Reply to  Bellman
June 29, 2026 7:04 am

What do you think the Stefan-Boltzmann law is saying? You keep going on about T^4, where do you think this T^4 comes from.”

I KNOW where it comes from. So what? S-B gives the total THERMAL energy emitted by a body.

For a non-isothermal volume of gas you have to integrate the local emissivity and temperatures in order to determine F.

F_total = ∫ ε(x) σ T(x)^4 dV

The atmosphere is *NOT* isothermal and the emissivity varies based on things like pressure, optical depth, and path length.

But climate science, like usual (and you with them), just makes another garbage assumption by assuming that the atmosphere is a solid that is isothermal with constant emissivity. Just like with temperature where the assumption is constant pressure, humidity, terrain, and geography for the entire globe’s climate, the same thing is done for the atmosphere.

These assumptions add so much measurement uncertainty by ignoring the variance of the conditions involved that the differences climate science is trying to identify become impossible to determine.

I really think this much easier to understand then you do”

Because you simply do not know even how S-B is applied to a gas, just like you don’t know anything about enthalpy!

” Getting hotter means retaining energy.”

Getting hotter means you also cool more! T^4!

“But it is relevant if it is not reflection.”

Unbelievable! Receiving energy and then sending it back is *NOT* reflection. You want everyone to believe that an optical mirror, which absorbs energy and re-emits it, reflects light but a heat reflector which absorbs energy and re-emits it does *not* reflect heast.

Cognitive dissonance at its finest!

Your physics quote only talks about “absorption”. It doesn’t even say what happens in an optical mirror – meaning CHANGING the EM wave to changed oscillation in the reflecting material.

You even ignore the part of the quote that says: “Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated.”

Judas Priest man! An optical mirror changes the direction of a light EM wave by absorbing it and re-emitting it so that it returns to the source. A heat reflector changes the direction of a heat EM wave by absorbing it and re-emitting it so that it returns to the source!

As usual your reading comprehension skills are totally non-existent.

Reply to  Tim Gorman
June 29, 2026 11:44 am

“For a non-isothermal volume of gas ”

The earth is not a gas.

“But climate science, like usual (and you with them), just makes another garbage assumption by assuming that the atmosphere is a solid that is isothermal with constant emissivity.”

Stop lying.

“Because you simply do not know even how S-B is applied to a gas”

The earth is not a gas.

“Getting hotter means you also cool more! T^4!”

You emit more energy, yes, that’s the point

“Receiving energy and then sending it back is *NOT* reflection. ”

Yes, that’s my point.

Reply to  Bellman
June 25, 2026 5:08 pm

The key difference is whether the incoming energy is retained and transformed (absorption) or redirected unchanged (reflection).

Do you understand what this is saying. A reflection COULD be from a mirror surface and directed back to the source, OR it could be absorbed and retransmitted back to the source. Your choice, but either way the radiation is returned to the source.

Let’s discuss further your step series that ended with the earth transmitting 200 and the atmosphere transmitting 100. Here is a depiction of that.

comment image

If your hypothesis is correct there should be 100 coming in and 300 going out. Hmmmm? What happened to the Law of Conservation of Energy?

Reply to  Jim Gorman
June 25, 2026 6:32 pm

Do you understand what this is saying.

Yes. Why do you assume everyone else is as clueless as you.

A reflection COULD be from a mirror surface and directed back to the source, OR it could be absorbed and retransmitted back to the source.

It’s literally telling you that the second is not reflection.

If your hypothesis is correct

I can not claim credit for it.

there should be 100 coming in and 300 going out.

I think you made a mistake somewhere. There should be 200 coming in and 200 going out.

If you ignore the vertical arrows, then your diagram is essentially correct, except there is only a one directional 100 from the atmosphere to the earth, and another 100 going from the atmosphere to space. There are 200 coming in (100 from the sun and another 100 from the atmosphere), and 200 going out.

Also I’ve no idea why you think the earth is 244K. That would be too cold even if there were no greenhouse effect.

Reply to  Bellman
June 26, 2026 7:52 am

I think you made a mistake somewhere. There should be 200 coming in and 200 going out.

And therein is your problem. In this problem, the sun is entering 100 into the system of the earth and the atmosphere.

The earth would be in equilibrium with the sun at 100. Yet you claim that the atmosphere ADDS another 100 to the earth so that the earth ends up with 200 total.

If the atmosphere traps heat, i.e., the greenhouse theory, then it must become a third body. A body radiates in all directions based on its temperature, so the atmosphere should be radiating 100 to space AND to the earth.

If the radiation from the atmosphere to the earth is added to the radiation from the sun then the earth should be radiating at 200.

That means there is 300 roaming around the system, far exceeding what is actually introduced.

Also I’ve no idea why you think the earth is 244K. That would be too cold even if there were no greenhouse effect.

Those are purely illustrative of what SB temperatures would be at those values of flux. It does indicate that the earth would become the hottest body in the system of sun, earth, and atmosphere. Is that really possible?

If you ignore the vertical arrows, then your diagram is essentially correct,

Really? That means there is no possibility of equilibrium in this system at any time. Not between the sun and the earth. Not between the earth and the atmosphere. Somehow, you have missed an effect that would lead to maximum entropy, which defines equilibrium. I wonder what that effect might be. Compensation maybe?

Reply to  Jim Gorman
June 26, 2026 8:36 am

“The earth would be in equilibrium with the sun at 100.”

And there’s your problem. The earth should not be at equilibrium with the sun but with all incoming energy.

“If the atmosphere traps heat, i.e., the greenhouse theory, then it must become a third body.”

Traps your word, but yes you can count the atmosphere as a third body. That’s what my equation is doing.

” A body radiates in all directions based on its temperature, so the atmosphere should be radiating 100 to space AND to the earth. ”

Which it is. Hence the 1/2 A in the equation E = S + 1/2 A. It resolves when the atmosphere is radiating 200 in total, 100 to the Earth and 100 to space.

“If the radiation from the atmosphere to the earth is added to the radiation from the sun then the earth should be radiating at 200. ”

Yes.

“That means there is 300 roaming around the system, far exceeding what is actually introduced.”

That’s because the energy has accumulated. It’s being stored in the earth and the atmosphere.

“It does indicate that the earth would become the hottest body in the system of sun, earth, and atmosphere. Is that really possible?”

It’s what observations show. This whole issue started with estimates of the temperature of the earth based on just the sun’s input suggest the average temperature should be around 255K.

To state the obvious, it isn’t that the earth is hotter than the actual sun it’s that it’s hotter than the heat we get from the sun at this distance, and accounting for the fact that sun is only shining on half the earth at any one time.

” That means there is no possibility of equilibrium in this system at any time.”

As I keep saying the equalibrium isn’t at any time. Temperatures are always going up during the day and in summer, cooling at night and winter. It’s roughly balanced over a year, apart from temporary changes from the likes of El Niños and volcanic activity.

“Somehow, you have missed an effect that would lead to maximum entropy, which defines equilibrium.”

I coukdn’t explain how to calculate maximum entropy, but everything I’ve seen says it is not a problem. The sun bring a nuclear furnace has a lot to do with this.

Reply to  Bellman
June 26, 2026 9:59 am

And there’s your problem. The earth should not be at equilibrium with the sun but with all incoming energy. 4891

I have shown you how that means there should be 100 incoming to the system and following your explanation, there would be 300 going out.

That is not allowed under the conservation of energy. If you want to refute bdgwx about that, be my guest.

It’s what observations show. This whole issue started with estimates of the temperature of the earth based on just the sun’s input suggest the average temperature should be around 255K.

Sorry, that is not what observations show. That is based on a hokey value for average absorbed insolation over 24 hours. It is entirely incorrect based upon physical geometry. An average insolation should be closer to 425 W/m² which gives something closer to 295K (22°C or 71°F). Compare that to 255 + 33 = 288 from the greenhouse theory. Too close to be coincidence, it’s more that the no greenhouse is incorrect.

Reply to  Jim Gorman
June 26, 2026 11:30 am

“I have shown you how that means there should be 100 incoming to the system and following your explanation, there would be 300 going out. ”

No you haven’t. All you demonstrate is your inability to balence the books.

The earth including the atmosphere receives 100 units, and emits 100 units to space.

” An average insolation should be closer to 425 W/m² which gives something closer to 295K (22°C or 71°F). ”

Can you provide a suitable reference. Everything I’ve seen on the subject suggests a lower figure.

Reply to  Bellman
June 26, 2026 1:25 pm

That’s because the energy has accumulated. It’s being stored in the earth and the atmosphere.”

Does that energy continue to accumulate? Is it stored forever?

If so, why isn’t the earth a molten rock today?

As I keep saying the equalibrium isn’t at any time. Temperatures are always going up during the day and in summer, cooling at night and winter.”

That simply doesn’t jive with your assertion that the energy accumulates and is stored in the earth and the atmosphere.

You want to have your cake and eat it to.

  1. the heat accumulates
  2. the heat doesn’t accumulate

Which is it?

Reply to  Tim Gorman
June 26, 2026 6:01 pm

If so, why isn’t the earth a molten rock today?

If only there were resources explaining how this works, or you had ever read the previous 500 comments explaining this to you, then you wouldn’t have to keep asking the same dumb question.

No the earth is not a molten rock, because the hotter it gets the more energy it emits. Have you heard of the Stefan-Boltzmann law?

That simply doesn’t jive with your assertion that the energy accumulates and is stored in the earth and the atmosphere

It jives perfectly – you are just too dense to understand it.

  1. “the heat accumulates
  2. the heat doesn’t accumulate

Which is it?”

Heat accumulates until it reaches an equilibrium point.

Seriously, how can you claim to be an expert in thermodynamics, claim to understand this better than all scientists, and then claim not to understand that things do not warm up indefinitely. If you leave a radiator on in your house does the heat accumulate? If it does does that mean you house will eventually melt?

Reply to  Bellman
June 27, 2026 2:57 am

If only there were resources explaining how this works, or you had ever read the previous 500 comments explaining this to you, then you wouldn’t have to keep asking the same dumb question.”

In other words: “I don’t have an answer”
So you just call the question dumb. One more use of the Argument by Dismissal fallacy. Is that *ALL* you ever have to offer? Argumentative fallacies?

Seriously, how can you claim to be an expert in thermodynamics, claim to understand this better than all scientists,”

So now Newton wasn’t a scientist, eh? That’s your claim? Newton’s Law of Cooling isn’t correct and Newton wasn’t a scientist.

Unfreakingbelivable.

“If you leave a radiator on in your house does the heat accumulate?”

Why don’t you ask a HVAC engineer sometime how the cooling of a house, building, etc figures into the sizing the heating and cooling systems.

BTW, is the furnace (i.e. the sun) on all the time? Does the sun never set where you live?

Reply to  Tim Gorman
June 27, 2026 7:49 am

“In other words: “I don’t have an answer””

I keep telling you what the answer is and you should be able to figure it out yourself. That’s why it’s a dumb question.

There’s only so many ways I can point you to the SB law. Hotter objects emit more energy. That stops everything in the universe that receives heat from melting.

“So now Newton wasn’t a scientist, eh?”

What did he say about the greenhouse effect? Seriously, stop using these gotcha sytke arguments and engage your brain. All scientists who understand thermodynamics know that things do not get infiniayky hot just because you keep adding energy to them. They understand the SB law and what it implies. They understand that whilst energy accumulates in objects it will reach an equilibrium point. If you think Newton said something different give a exact quote.

“Newton’s Law of Cooling isn’t correct and Newton wasn’t a scientist. ”

Good grief, what assumptions are there in Newton’s cooling equation?

“Why don’t you ask a HVAC engineer …”

Why don’t you answer the question? You just keep claiming that if energy accumulates it will end up at a infinite temperature. I’m asking you to define what you think “heat accumulates” and gave you the example of a radiator in a room as an example. If I leave a radiator on will it reach a specific temperature or will it just keep getting warmer indefinitely. I don’t have to ask a HVAC engineer to know the answer, it’s basic physics.

“BTW, is the furnace (i.e. the sun) on all the time?”

Do you love demonstrating your inability to understand a simple metaphor. It makes no difference to the question if the radiator is on allm the time or periodically turns of. The point was about accumulating heat.

Reply to  Bellman
June 30, 2026 4:22 am

I keep telling you what the answer is and you should be able to figure it out yourself. That’s why it’s a dumb question.”

You’ve not shown one jot or tittle of math in any of your answers. You can’t even derive the power-law decay formula after having most of the derivation given to you!

There’s only so many ways I can point you to the SB law. Hotter objects emit more energy. That stops everything in the universe that receives heat from melting.”

And yet you keep claiming that the earth retains reflected heat? Just one more cognitive dissonance you keep spouting. At least you’ve changed from retaining it all to just retaining “some” of it. Retained heat *would* ultimately result in melting.

“What did he say about the greenhouse effect?”

Newton’s Law of Cooling confirms that everything above 0K COOLS. Something you simply refuse to accept! It’s part of the same thing Planck discusses as compensation via new rays. New rays emitted because of slower cooling.

They understand the SB law and what it implies. They understand that whilst energy accumulates in objects it will reach an equilibrium point.”

They also understand that if a part of the received heat is permanently retained that the object doing the retaining *will* eventually melt. Again, it’s part of why Planck says what he says of reflected heat being offset by new rays – i.e. compensation.

“Good grief, what assumptions are there in Newton’s cooling equation?”

You don’t know? Basically Newton’s Law of Cooling, an exponential decay, is a special form of the power-law decay formula. The relation t^-α ≈ e^-(kt). I’ve already tried to teach you what the difference in the curves are. Have you forgotten already? My guess is that you weren’t able to understand the math.

You just keep claiming that if energy accumulates it will end up at a infinite temperature.:

*YOU* are the one that was advocating that reflected heat from CO2 is retained in the earth. I tried to tell you why it isn’t.

Now it seems that you are trying to change your position. So, is *ANY* of the reflected heat retained or is it ALL re-emitted?

Stop trying to have it both ways!

It makes no difference to the question if the radiator is on allm the time or periodically turns of. The point was about accumulating heat.”

You simply have no grasp of reality. The radiator is *NOT* the source of heat, the furnace is!

Does the room RETAIN part of the heat radiated from the radiator as you claim the earth does with reflected heat from CO2?

Again, it’s obvious what you are now trying to do. You are going to now try and say that you never said the earth retains reflected heat. Trying to find a way out of disavowing what you actually said by saying that you didn’t say that!

As usual, you got caught making up shite to throw against the wall. Now you are trying to clean it up. Keep scrubbing. It doesn’t come off easily.

Reply to  Tim Gorman
June 30, 2026 6:37 am

“You’ve not shown one jot or tittle of math in any of your answers. ”

You are such a hypocrite. You spend 5 years whinging about my maths, insisting that maths has no relation to the real world. But now when I point out why it’s obvious in the real world that objects can’t just keep getting warmer indefinitely you demand mathematical proof.

You’ve already got the proof in your Newton equation which works just as well with warming. Put a cold object in a warm environment and it will warm up to the temperature of the environment, not warmer. This is because dT/dt is proportional to T_hot – T_cold. When the two temperatures are the same dT/dt = 0.

And I’ve given you the simplest simultaneous equations for the green house effect. It has a solution. If E -> ∞ there would not be a solution.

And I gave you a tine dependent simulation for those equations the last time you were denying the greenhouse effect. It showed exactly what you would expect. Temperatures increased, but leveled up at the predicted equilibrium level.

Meanwhile, you are the one claiming that scientists have been wrong the past century or two, yet you have not provided a “jot or tittle” of a mathematical explanation of your hypothesis. All youvevervdo is keep stating variations of Newton’s cooling law. Never an attempt to show how it works in a situation where you have a constant source of energy (the sun) along with an atmosphere that is being warmed by the Earth and is also adding to the incoming heat.

If you want anyone to believe that the laws of thermodynamics make the greenhouse effect impossible or causes infinite heat, you need to explain the maths, not just handwave and equivocate about the meaning of cooling.

Reply to  Bellman
June 30, 2026 7:08 am

“And yet you keep claiming that the earth retains reflected heat? ”

These are your words and just demonstrate your equivocation. What do you, Tim Foreman mean by “retain heat”?

Apart from the fact that an object cannot retain “heat” by the current definition of heat, what do you think happens to thermal energy when it is absorbed by an object? In my view, the energy is converted into kinetic energy. It can’t just vanish as that contradicts conservation of energy. If you want to call that “retaining” the heat, fair enough. But that does not mean the heat is locked into the object. At some point the energy is converted back into thermal energy and emitted by the object, which means the energy has gone from the object.

In reality you can’t track individual units of energy. The object is constantly receiving energy and emitting energy. All you can track is the amount of energy. If the object receives more heat than it emits, the internal energy increases and usually it gets warmer. So you could say that increase in energy is the “retained” energy, and that won’t disappear unless at some point heat out is greater than heat in. Then the internal energy reduces and it gets colder.

So, as I asked, what do you mean by retaining heat? Is it about individual units of energy, or is it the level of internal energy? Or are you just equivocating?

Reply to  Bellman
June 30, 2026 7:23 am

“Newton’s Law of Cooling confirms that everything above 0K COOLS.”

It does not. It says an object in an environment will cool, or warm, to the temperature of the environment. And that only applies in the simple case of a single object in an unchanging environment with no additional heat source.

“Something you simply refuse to accept!”

Yes, because if you use your brother’s and ever other person’s definition of cooling (Ti < T(i -1)), then it is easy to demonstrate it's false. Just leave a bottle of cold water at room temperature and see if it cools.

"Again, it’s part of why Planck says what he says of reflected heat being offset by new rays – i.e. compensation."

This is just getting absurd. You keep saying I'm wrong and the keep violently agreeing with me. You are either just arguing for the sake of arguing, or are getting tangled up on your own bizarre language. I am saying exactly the same as you. As an object gets warmer it emits more energy. This ensures the Earth does not "permently retain heat" and does not melt. You can call this increased energy compensating rays if you like, you can call warming slower cooling if you want. But until you try to understand what both of us are saying, we will just keep having these same circular arguments.

Reply to  Bellman
June 30, 2026 7:36 am

“You are going to now try and say that you never said the earth retains reflected heat. Trying to find a way out of disavowing what you actually said by saying that you didn’t say that!”

You really need to try and understand these concepts rather then playing these stupid gotch word games. “Retained” was the word copilot used to explain the difference between reflection and absorption. You still refuse to accept that distinction even though it come from your favorite authority. Instead you have now gone on a futile rampage, because for some reason you think “retained” means “permenently retained”. I’ve no idea why you would think it means permenently, but a moments thought should have made the true meaning clear to you.

Reply to  Bellman
June 25, 2026 7:17 am

That’s a strawman. I’ve continually explained the greenhouse effect using gross values. I do not think that only net exists.”

No, you haven’t. You have never reflected the gross flow that represents the emission of new rays as compensation for reflected heat as laid out in Planck’s treatise.

Planck: “For example, if we let the rays emitted by the body fall back on it, say by suitable refection, the body, while again absorbing these rays, will necessarily be at the same time emitting new rays, and this is the compensation required by the second principle.”

All you’ve ever done is deny this using the Argument by Dismissal fallacy to say it is wrong. The only rationale you’ve ever offered is that CO2 doesn’t operate like an optical mirror while ignoring Planck’s words of “ suitable refection”. “Suitable reflection” isn’t restrictive to only optical frequencies and an optical mirror.

 If they balanced all the time there would be no warming.”

Meaning that the earth should have become a molten rock long ago?

“And second strawman nobody says that fluxes are always equal at any point in time.”

Of course they do! What do you think Trenberth’s radiative balance graph shows?

“The equalibrium happens over a longer period of time.”

And yet you deny this in every thing you assert. You say you consider the gross flows and then assert that whether an object is cooling or warming is only defined by the NET, ignoring that the gross flows represent both inward and outward energy so that the object is *both* cooling and heating. Thus the assertion that the earth doesn’t cool during the day!

F(t) = F0 * e^-(t/τ) is only idiotic to YOU because you do *NOT* consider the gross flows involved in a thermodynamic system. It represents the cooling of an object for every increment of time for which the object exists above T > 0K. The fact that F0 can be F0(t) is irrelevant for defining gross flow of joules out of the object. If F(t) > 0 then the object is cooling regardless of whether it is also heating at the same time.

Again, if you do *NOT* understand what is happening inside the system then the NET value is meaningless. (Th – Tc) is meaningless unless you know Th and Tc. (200K – 50K) and (300K – 150K) gives the same NET value but represent VASTLY different systems with vastly different inputs and outputs of joules. Not only do the internal components have to withstand different temperatures they have to be designed to provide the flux needed to maintain the heating and cooling rates the temperatures generate.

Bottom line: NET isn’t what defines the cooling and heating of an object, the gross cooling and heating components do.

Reply to  Tim Gorman
June 25, 2026 7:58 am

“No, you haven’t.”

Look at all the times I explain the greenhouse effect using E, S and A. Each if those values is a gross emission. You are either demonstrating you don’t understand the maths or are lying.

“You have never reflected the gross flow that represents the emission of new rays as compensation for reflected heat as laid out in Planck’s treatise.”

Why do you think E increases? It’s that increase that is the compensation for the greater number of rays coming from the atmosphere and please learn what reflection means.

“All you’ve ever done is deny this…”

Just blatant lying at this point.

“The only rationale you’ve ever offered is that CO2 doesn’t operate like an optical mirror …”

It doesn’t act like any sort of mirror. It is not s reflecting surface. But that has nothing to do with the case. It doesn’t matter if the additional heat is coming from reflection or re-emission, the earth still has to rise in temperature to compensate.

“Meaning that the earth should have become a molten rock long ago?”

Just try to think, and actually read what I’ve actually said. I’ve explained it enough times. You admit that T⁴ is a negative feedback..that means the hotter the earth gets the more it admits. Which means you are always going to get to a point where in and out balance.

“Of course they do! What do you think Trenberth’s radiative balance graph shows?”

I should have said “every point in time”. Just try to read what I said. The equalibrium is over a reasonable period of time. At any given point of time tgee will be an imbalance. More energy in during day time than at night, more during summer than winter. But over the course of the year the in and out should be very close, or else there would be very big global changes in temperature every year.

“And yet you deny this in every thing you assert.”

Please point to one of these assertions.

“You say you consider the gross flows and then assert that whether an object is cooling or warming is only defined by the NET”

Yes because that’s true. All your own sources explain that. If there is more energy out than in you have a net heat loss and cooling. If there is more energy in than out you have a net heat gain and hence warming.

I’ve explained before that there should be no difference between equations using gross flow verses one using net flow. Something that should be glaringly obvious to anyone who understands how algebra works.

Reply to  Bellman
June 25, 2026 8:11 am

“F(t) = F0 * e^-(t/τ) is only idiotic to YOU because you do *NOT* consider the gross flows involved in a thermodynamic system. ”

I know it’s futile, but again can you just provide a reference to that equation.

All it sed you have done is take the cooling equation, treating the environment as being 0K and then replaced temperature with flux. In so doing you have ignored the T⁴ rule.

But the bigger problem is you think you apply this to all objects, regardless of the initial assumptions. The equation requires there is no external heat source – that you are only dealing with a single object and an environment. As soo as you start adding heat to the system, the temperature won’t change according to the equation, and so the flux won’t either.

To get round this you start talking about the F0 constant which is supposed to tell you the initial temperature, as if it’s a function based on time or possibly temperature. At which point the whole expo ential decay becomes meaningless.

It’s telling that you keep insisting that it works all the time but keep refusing to actually apply it to the system under discussion. You just say it’s too complicated to work out how F0 changes over time.

“F(t) = F0 * e^-(t/τ) is only idiotic to YOU because you do *NOT* consider the gross flows involved in a thermodynamic system. ”

I’ve shown how your own trusted AI authorities disagree with that. You have provided zero references to support it.

Reply to  Bellman
June 25, 2026 9:47 am

I know it’s futile, but again can you just provide a reference to that equation.”

You can find it all over. I use the e^(-t/τ) as an approximation. It is actually a power law.

exponential: F(t) = F0e^-(t/τ)
power law F(t) = (1 + 3t/τ)^-(4/3)

While the shapes of the curves are different they integrate to almost the same amount of energy radiated over an interval. The power law starts off dropping faster but ends up dropping slower. The integral from 0 to infinity both wind up as “τ”.

do a google search: “radiative loss over time, exponential vs power law”

All it sed you have done is take the cooling equation, treating the environment as being 0K and then replaced temperature with flux. In so doing you have ignored the T⁴ rule.”

No, I haven’t ignored anything.

What you need is the integral of the curve to get the total energy lost.

If you do it over a period of t ≥ 3τ you get almost exactly the same value. The exponential is just easier to explain to most people. Most people readily understand it.

The equation requires there is no external heat source – that you are only dealing with a single object and an environment”

NO! The cooling equation applies ALL THE TIME. It doesn’t matter if there is an external heat source (AND CO2 IS *NOT* AN EXTERNAL HEAT SOURCE) or not.

Once again, you are totally focused on NET and not on gross components.

The only thing that changes is whether F0 is a constant or F0(t). The equation still stands.

To get round this you start talking about the F0 constant”

I keep pointing out to you that F0 is *NOT* a constant during the day. It is an initial condition. How many times do I have to tell you that it is F0(t) during the day before it finally sinks in? I started off telling you that calculating the flux during the day was complicated because F0 is actually F0(t) – and you blew it off then and you are blowing it off now!

It’s telling that you keep insisting that it works all the time but keep refusing to actually apply it to the system under discussion. You just say it’s too complicated to work out how F0 changes over time.”

So what? And I didn’t say it was *too* complicated to work out. I just said it was complicated to work out. It is a function of time, latitude, and longitude. I even gave you a first brush at the equation. As usual, your lack of reading comprehension skills just caused you to blow it off!

“I’ve shown how your own trusted AI authorities disagree with that. You have provided zero references to support it.”

No, you haven’t. All you’ve shown is that you only look at the net result. You *still* refuse to understand that systems have both joule inputs and joule outputs that form the net. And if you only look at the NET you have absolutely NO idea of how the NET came about!

Everything above 0K radiates outward. Radiation outward is energy loss – i.e. joules. PERIOD. EXCLAMATION POINT.

All you are doing now is whining that loss of joules is not a cooling process. I guess you think your car engine just keeps on heating to destruction because there can’t be a cooling process involved while it is running.

Reply to  Tim Gorman
June 25, 2026 10:23 am

“You can find it all over.”

Then you won’t have any problem.findingnone reference.

“do a google search”

Or why don’t you do it and point me to a reference you accept.

“NO! The cooling equation applies ALL THE TIME. ”

Then your reference needs to explain it. You’ve already said it’s an approximation, but you also insist it works all the time.

Reply to  Bellman
June 25, 2026 12:29 pm

I am NOT YOUR RESEARCH ASSISTANT.

If *YOU* don’t understand and haven’t learned how thermodynamics work, then you shouldn’t be on here lecturing those of us who *have* studied the subject at university that we are all wrong!

I’ll give you the start:

C(dT/dt) = εσAT(t)^4
let k = εσA/C and
dT/dt = kT(t)^4

solve the differential equation and you wind up with

T(t) = (3kt + T^-3) * 1/3

Extend that to the equation for F_out(t) and you get the exact equation I’ve already given you.

The equation F(t) = F0 * e^-(t/τ) is an approximation that is easier to understand for *most* people and gives the exact same total from 0 to ∞.

If you *still* need a reference then go here: https://hyperphysics.gsu.edu/hbase/thermo/cootime.html

You won’t understand it any better than what I’ve already given you.

Then your reference needs to explain it. You’ve already said it’s an approximation, but you also insist it works all the time.”

Do you even the smallest clue as to what it means to say the integral of my approximation and the integral of the detailed equation are the same from 0 to ∞? Only the detailed shapes of the curves are different. One starts off faster but ends slower. Still the same area under each curve.

THE COOLING WORKS ALL THE TIME. THE APPROXIMATION DOESN’T CHANGE THAT!

Reply to  Tim Gorman
June 25, 2026 6:02 pm

I am NOT YOUR RESEARCH ASSISTANT.

Back to shouting. It really demonstrates how well you handle criticism.

I’m not asking for you to be assistant – you are the last person I’d trust with that job. What I’m asking is for you to provide evidence to support your claims.

I’ll give you the start

Fine, but I’m still going to need that reference. You are just too unreliable to be used as an authority.

C(dT/dt) = εσAT(t)^4

This is just the Stefan-Boltzmann rule. But you are treating the Power emitted as a change in temperature, which assumed no external heat source.

T(t) = (3kt + T^-3) * 1/3

As I suggested you are starting with an equation for temperature change, with no external heat source. But I think your math’s wrong.

The obvious point, is your equation is just saying that temperature will change linearly with time, which is not what you would expect.

Extend that to the equation for F_out(t) and you get the exact equation I’ve already given you.

How? F_out is proportional to T^4.

If you *still* need a reference then go here: https://hyperphysics.gsu.edu/hbase/thermo/cootime.html

Finally a reference. Pity it doesn’t give anything resembling any of your equations.

Reply to  Bellman
June 26, 2026 7:07 am

Finally a reference. Pity it doesn’t give anything resembling any of your equations.”

I gave you the reference. Apparently you can’t even tell that T^-1/3 is the same as (1/T^3). All my equation does is normalize T.

You can’t even tell that the link I gave you is solving for a time “t”, and not for dT/dt. But both my derivation and the derivation in the link work the same way. Write the differential equation, solve it, then integrate to get the total time (in one case) or the change in temperature (dT/dt).

You don’t know calculus at all let alone algebra. All you can do is carp that the calculus that other *do* know and use has to always be wrong. Just like in the example Possolo did when finding the measurement uncertainty of the volume of a barrel. All you could offer was that no one but you knew how to do partial derivatives! Yet the method was right there in there in the GUM. And all can do with that method is argue that it is wrong because it doesn’t do partial derivatives correctly!

I gave you a list of the garbage you assert that is based on the meme that the cooling process doesn’t happen 24 hours per day. Need I repeat the list?

It’s not *my* problem that you can’t solve differential equations and integrate functions. It is *YOUR* problem, not mine. I started off with S-B and went from there. You have yet to show where my derivation is wrong. All you have is the Argument by Dismissal fallacy.

SHOW ME WHERE MY MATH IS WRONG!

If you can’t then stop carping that it is wrong.

Reply to  Tim Gorman
June 26, 2026 5:52 pm

SHOW ME WHERE MY MATH IS WRONG!”

As I said this equation is obviously wrong.

T(t) = (3kt + T^-3) * 1/3

I don’t need to solve the differential to see that, it’s just not giving you a sensible result. Did you test it with a real world example?

The obvious point is that the equation is saying that temperature will always increase over time (kt), but that in all likelihood the initial temperature will be irrelevant (1/T^3). Anything that uses this equation will be close to 0K after one second and then slowly gain temperature forever.

Reply to  Bellman
June 26, 2026 5:54 pm

Regardless though, even if you get the math right is beside the point – there is no way you can apply the equation when you have an external heat source. Your equation in no way explains how temperature of flux will change in that circumstance, let alone when you also have a greenhouse effect. It’s at best telling you how a single object in an environment at 0K will behave.

Reply to  Bellman
June 27, 2026 2:52 am

Regardless though, even if you get the math right is beside the point – there is no way you can apply the equation when you have an external heat source.”

The math applies ALWAYS. Objects cool 24/7 continuously. You can’t calculate the NET if you don’t know the components creating that NET and cooling is *ALWAYS* a component of the NET – ALWAYS!

“Your equation in no way explains how temperature of flux will change in that circumstance”

You keep implying that you can know the NET before you know the components creating that NET. The cooling process is PART of the NET – ALWAYS.

 let alone when you also have a greenhouse effect”

The greenhouse effect is a phantom created solely to make a garbage assumption by climate science that the earth retains reflected heat!

They don’t even display the cognitive dissonance that you do: retained heat isn’t retained – it’s re-emitted.

Their models just show an ever-increasing amount of retained heat – no re-emission!

Reply to  Tim Gorman
June 27, 2026 3:14 pm

The math applies ALWAYS.

All you math implies is that temperatures will cool down exponentially. This is only true when you have a single object in a cooler environment, and in your case only when that environment is at 0K.

This math is wrong as soon as you consider the daily and annual cycle for the Earth.

And you can;t get around this by arguing it is only about the energy out side of the equation. The temperature equation is about temperature, not energy. And replacing temperature with energy only works as long as the temperature is correct.

You can’t calculate the NET if you don’t know the components creating that NET and cooling is *ALWAYS* a component of the NET – ALWAYS!

You can’t calculate what you call cooling unless you know the temperature. The “cooling” will be proportional to T^4. If temperature stays the same, so to will the flux out. If temperature rises, so to will the flux out.

Trying to create time series without also considering the flux in is doomed to failure.

The greenhouse effect is a phantom

Remind me about the fallacy of argument by dismissal. Just saying you don;t believe in it is not a valid argument. As well as going against the policy of this website.

Their models just show an ever-increasing amount of retained heat – no re-emission!

Who’s models? I think this might be your biggest problem. The argument is about the greenhouse effect, not global warming. The greenhouse effect does not say temperatures rise indefinitely, it explains why temperature are what they are and have been for all time. Global warming only comes into this because some of the greenhouse gases are increasing – and even then no model will show temperatures increasing indefinitely. The most you can have is a completely opaque atmosphere.

Reply to  Bellman
July 1, 2026 6:45 am

All you math implies is that temperatures will cool down exponentially. This is only true when you have a single object in a cooler environment, and in your case only when that environment is at 0K.”

You and math just don’t get along, do you? I told you already, F(t) = F0(t) e^-(t/τ) when a source of heat is involved. Why are you just repeating this back to me as if it is something you haven’t been told before?

This math is wrong as soon as you consider the daily and annual cycle for the Earth.”

NO, the math is correct. *YOU* just have to understand what F(t) = F0(t) e^-(t/τ) means. I TOLD YOU ALREADY THAT THIS GETS COMPLICATED DURING THE DAY. The math works, ALWAYS.

And you can;t get around this by arguing it is only about the energy out side of the equation.”

F(t) *IS* THE OUT SIDE OF THE EQUATION. You can’t even get this correct. Why are you on here trolling, trying to lecture people on things you have absolutely no understanding of? Do you just like to see your name on the thread?

A heat source does *NOT* cancel out the exponential decay that is caused by the object radiating and losing joules. You *seem* to be trying to imply that S-B is wrong, that the radiation out of the object goes to zero if a source is also inputting joules while it is radiating. If a source is inputting joules into the object, it’s flux out GOES UP, it doesn’t go down nor does it go to zero!

You can’t calculate what you call cooling unless you know the temperature. The “cooling” will be proportional to T^4. If temperature stays the same, so to will the flux out. If temperature rises, so to will the flux out.”

Now you are, once more, just repeating what I’ve already told you before. Do you think this is some kind of a flash of inspiration that no one else knows?

The issue is that the radiation out goes up faster than the temperature! This is called “negative feedback” and it sets a boundary condition for how high the temperature can go before the joules-out balances the joules-in. It’s why the daily temperature turns over and decreases before the sun sets.

Remind me about the fallacy of argument by dismissal. Just saying you don;t believe in it is not a valid argument. As well as going against the policy of this website.”

Nah, I’ve given you the math behind why it is a phantom! You never give *ANY* math or other evidence to back up your assertions, Argument by Dismissal includes providing no reason for the assertion you are arguing against is wrong – that’s YOU.

Planck himself says it is a phantom. Reflected heat slows cooling. I’ve given you the graph showing that. And slower cooling results in compensation for the reflected heat, not in a greenhouse effect. And slower cooling is *NOT* heating. CO2 is *NOT*, let me repeat for the umpteenth time – *NOT*, a source of heat energy. It is a reflector, not a source.

The argument is about the greenhouse effect, not global warming. “

More shite on the wall? Anthropogenic increases in the greenhouse effect generated from CO2 increases greater than natural variability *IS* supposed to be causing GLOBAL WARMING. And it is a garbage assumption made by climate science. The math doesn’t work. The math doesn’t support the assumption. CO2 increases *should* be increasing Tmax at least as much as Tmin based on the greenhouse effect. The fact that Tmax is seeing almost no increase (if any at all) stands as proof that saying anthropogenic CO2 is the cause of global warming is garbage.

even then no model will show temperatures increasing indefinitely.”

Really? See attached.

ensemble_climate_models
Reply to  Tim Gorman
July 1, 2026 7:05 am

” I told you already, F(t) = F0(t) e^-(t/τ) when a source of heat is involved. ”

You can keep telling me it as much as you like. If you can’t provide a reference why should I care what you tell me?

You’ve made F0 a function of time so what does that equation say if you don’t know what the function is? If your heat source is 2000K. and your initial temperature is 200K, what will the temperature be after 1 hour?

“You *seem* to be trying to imply that S-B is wrong, that the radiation out of the object goes to zero if a source is also inputting joules while it is radiating.”

No, I’m saying the net flux will tend to zero.

“The issue is that the radiation out goes up faster than the temperature! ”

Why do you think it’s an issue. It doesn’t matter what the first function is, as long as the energy out increases monotonically with temperature you will reach an equilibrium point. The function only determines what that temperature will be. As in my simple example, earth is emitting twice as much energy as it would without a greenhouse effect. This translates into a temperature of 2^(1/4), about 1.2. So a temperature increase of 20%.

If energy out was proportional to temperature temperature would have to double to reach the equilibrium point.

Reply to  Bellman
June 27, 2026 2:45 am

As I said this equation is obviously wrong.”

But you just can’t show where it is wrong can you? Once again, that is the Argument by Dismissal fallacy. Just declare something wrong and move on!

“I don’t need to solve the differential to see that”

ROFL! Kind of like you didn’t need to solve Possolo’s derivation of measurement uncertainty for a barrel in order to know it was wrong?

I *did* make a mistake in typing the equation. Here is the corrected one.

T(t) = (3kt + T^-3) ^(-1/3)

I have it correct on my paper but I typed it in wrong. *I* COULD TELL WHAT WAS WRONG. YOU COULDN’T.

Because you couldn’t do the derivation on your own so you didn’t have a clue as to what it should be!

The power law decay and the exponential decay are very similar. They both have a negative slope throughout, it never goes positive indicating a push back up the temperature gradient. In fact it can be simply stated as

dT/dt = -kT^4

It just depends on what “k” is as to what the actual decay time is.

Cooling of objects happens 24/7 – all the time. It’s continuous.

Reply to  Tim Gorman
June 27, 2026 5:35 am

Hilarious. He yells at me for claiming is equation was wrong, and then in the next sentence admits that it was wrong.

Not engaging with he rest of this, when Tim continues to lie about me. – “Kind of like you didn’t need to solve Possolo’s derivation of measurement uncertainty for a barrel in order to know it was wrong?”

Complete and utter lie. I’ve never said the derivation was wrong. I explained why it was right.

It’s typical Tim that he has to denigrate and lie about anyone point out his own errors.

Reply to  Bellman
June 29, 2026 6:19 am

Hilarious. He yells at me for claiming is equation was wrong, and then in the next sentence admits that it was wrong.”

You couldn’t tell me WHY it was wrong! Meaning you had absolutely no idea of what a power law decay *should* look like.

““Kind of like you didn’t need to solve Possolo’s derivation of measurement uncertainty for a barrel in order to know it was wrong?””

You are the one that said the equation was wrong! You STILL won’t admit that it is covered in the GUM in Equation 12. When I tried to show you how Possolo derived the equation you told me I didn’t know how to do partial derivatives. You *STILL* haven’t admitted that I did them correctly and in doing so derived GUM Eq 12! In fact, the last thing I remember you saying about Eq 12 was that it couldn’t be right either.

Complete and utter lie. I’ve never said the derivation was wrong. I explained why it was right.”

Your memory is faulty. You didn’t even catch that he *and I* were using relative uncertainty and you couldn’t figure out why there was no R^2 term in the H partial derivative and no H term in the R^2 partial derivative when those term cancel using relative uncertainty. All you could do was keep telling me that I didn’t understand partial derivatives!

And when I pointed you to Eq 12 you simply couldn’t figure it out either.

It’s typical Tim that he has to denigrate and lie about anyone point out his own errors.”

And it’s typical of you to admit that you can’t do simple algebraic simplification let alone actual calculus. It’s why you can’t figure out how to derive the power law decay for cooling! You just “know” it’s wrong also!

Reply to  Tim Gorman
June 29, 2026 4:12 pm

To quote someone: “I am not your research assistant.”

I have no intention of correcting your homework, I just need to look at your final equation to see it is obviously wrong. If you had the least critical faculty you would check that your “always correct” equation does what you expect it to do, and not just increase linearly.

All this nonsense could have been avoided if you just did what I asked and provide a reference to the equation. And then we could get to the meat of the problem – regardless of you actually figuring out the correct equation, it is still or nonsense because you are misusing it. You are presenting a time function for how an object cools over time with a constant environment and no heat source, and claiming it explains something about how the Earth will cool.

Reply to  Bellman
June 26, 2026 9:17 am

 But you are treating the Power emitted as a change in temperature, which assumed no external heat source.”

The heat loss is to space *and* the sun. That heat loss is solely dependent on temperature to 0K for space. You can’t even get the system configuration correct. How do you expect to analyze anything correctly?

The power emitted *DOES* change the temperature of the emitting source. IT IS HEAT LOSS WHICH YOU DENY HAPPENS APPARENTLY!

“But I think your math’s wrong.”

But you just quite can’t figure out where it’s wrong, right?

ROFL!!!! Once again, that’s the Argument by Dismissal fallacy. Just declare it wrong, provide no evidence for the declaration, and move on.

The obvious point, is your equation is just saying that temperature will change linearly with time, which is not what you would expect.”

WHY IN PETE’S NAME DO YOU IGNORE THE FACT THAT I AM USING NEWTON’S LAW OF COOLING AS AN APPROXIMATION?

The rate of heat loss changes with temperature. That’s why the power law is the most correct formula. But it is more difficult to explain than the exponential law. And the exponential law is a good approximation.

HOW MANY TIMES MUST YOU BE TOLD THIS BEFORE IT SINKS IN?

And, yes, I’m shouting. Because your ears are plugged!

How? F_out is proportional to T^4.”

No shite, sherlock! Why do you think I gave you the power law equation?

Again: power law for temperature: T(t) = (3kt + T0^-1/3)^-1/3

Again: flux is: εσT(t)^4

Again: which I approximate as F(t) = F0 * e^-(t/τ)

HOW MANY TIMES MUST YOU BE TOLD THIS BEFORE IT SINKS IN?

STOP CARPING ON MATH YOU CAN’T FIGURE OUT!

Reply to  Bellman
June 25, 2026 8:19 am

the earth still has to rise in temperature to compensate.”

No, it doesn’t. It only has to cool slower.

Again, what does the shaded area on the attached graph represent?

Is it compensation for reflected heat? If not, then what is it? If not, then where does it come from? If not then what causes it?

Does the temperature curve for either decay ever change slope from negative to positive?

Is it necessary for the curve to change from negative to positive to emit more heat?

But over the course of the year the in and out should be very close, or else there would be very big global changes in temperature every year.”

There *are* big global changes in temperature every year!

“Please point to one of these assertions.”

bellman: “**Objects above 0 K do not continuously lose joules.**”

bellman: Nothing to suggest that emitting energy whilst receiving more is called “cooling”.

bellman: The question is can you call emitting energy “cooling” even when you get more energy back.

bellman: Whether you right the equations in terms of net or gross can make no difference.

bellman: This in turn is because the earth is hotter and so emits energy at a faster rate and the sun is lower in the sky and so the input rate of energy input. It does not prove that the earth was actually cooling all the time it was getting warmer.

bellman: It’s really hard to see how this equation is happening all the time.

Yes because that’s true.”

You just showed that you don’t consider the gross flows at all. One more assertion to add to the list.

I’ve explained before that there should be no difference between equations using gross flow verses one using net flow.”

You’ve apparently started on the bottle early today. The net flow is calculated from the gross flows. You don’t calculate net flow by using net flow.

And as I keep pointing out, (200K – 50K) and (300K – 150K) give the same net flow but have vastly different gross flows with vastly different system impacts.

But you want to keep claiming that it is the net flow that defines the system, 150K = 150K.

Just as I would never want to drive over a bridge you’ve designed I would never want to be around a steam engine you’ve designed.

slower_faster_2
Reply to  Tim Gorman
June 25, 2026 8:37 am

“No, it doesn’t. It only has to cool slower.”

Huh? You are saying it compensates for more energy in by reducing the amount it emits? Do you understand what compensate means?

Reply to  Bellman
June 25, 2026 10:04 am

Huh? You are saying it compensates for more energy in by reducing the amount it emits? Do you understand what compensate means?”

Unfreakingbelievable.

How does the area under the curve INCREASING imply reducing the amount it emits?

The shaded area INCREASES the amount emitted compared to the faster decay. The area under the curve gets LARGER with slower decay.

A 3rd grader would figure this out using even the “new math”. And YOU can’t?

Reply to  Tim Gorman
June 25, 2026 10:52 am

“Unfreakingbelievable”

The logical fallacy of an argument from personal incredibility.

“How does the area under the curve INCREASING imply reducing the amount it emits?”

These are time functions. Realising the same amount of energy over a longer time interval will not compensate for the increased rate of input. You just have more incoming energy to compensate for.

Reply to  Bellman
June 25, 2026 12:39 pm

These are time functions. Realising the same amount of energy over a longer time interval will not compensate for the increased rate of input.”

You can’t read a graph any better than you can read text!

The curves are TEMPERATURE CURVES. The energy emitted is the area under each curve. The time intervals on the x-axis are the same for each curve!

The AREA UNDER THE SLOWER DECAY CURVE IS GREATER than the area under the faster decay curve for any time interval you pick!

How does a larger area under the curve per time interval becomes the same energy over a longer time interval?

Slower cooling due to reflected energy (the top curve) EMITS MORE ENERGY OVER ANY INTERVAL than the faster cooling curve without reflected energy. The difference is Planck’s “compensation” for reflected energy!

Stop trying to lecture on thermodynamics to those of us who have studied and used thermodynamics in the real world.

Once again, you are living in a statistical world which has no points of congruence with reality.

slower_faster_2
Reply to  Bellman
June 25, 2026 1:09 pm

You are saying it compensates for more energy in by reducing the amount it emits? Do you understand what compensate means?

Do you understand?

Look at a graph of temperature and notice the cooling. It start cooling shortly after insolation begins to decrease. Lately there has been about 14 hours of cooling. From about 1600 to 0600. Also note that additional cooling probably wouldn’t occur due to the dew point.

Remember, the surface retains heat, it will not show up until later. The land surface can keep the atmospheric temperature higher than it would be by releasing heat over a longer period.

Your generalizations are meaningless unless you are able to show some math to support them. They are hypothetical statements that have no math supporting them. It is definitely tiresome teaching someone science who won’t learn on their own.

I’ll leave you statements from Planck’s Theory of Heat Radiation. Study them for the term simultaneously.

Any change in the energy distribution consists of a passage of energy from one monochromatic radiation into another, and, if the temperature of the first radiation is higher, the energy transformation causes an increase of the total entropy and is hence possible in nature without compensation; on the other hand, if the temperature of the second radiation is higher, the total entropy decreases and therefore the change is impossible in nature, unless compensation occurs simultaneously, just as is the case with the transfer of heat between two bodies of different temperatures.

For example, if we let the rays emitted by the body fall back on it, say by suitable reflection, the body, while again absorbing these rays, will necessarily be at the same time emitting new rays, and this is the compensation required by the second principle.

Please note that “simultaneously” and “at the same time” are equivalent.

Study these carefully. Then tell us how they are wrong. You might want to read his section on entropy also.

Reply to  Jim Gorman
June 25, 2026 2:04 pm

“Look at a graph of temperature and notice the cooling”

What cooling? You might think an object that gets hotter is cooling, but it’s just perverse to keep using the word like that when we are discussing an object releasing more energy in response to warming.

This whole “compensation” is because Planck says that if you reflect some energy back on an object it will have to generate more rays to compensate. Unless you think it has some magic way of knowing how many new rays it has to produce this can only happen because it warms up to the point where it is releasing sufficiently more rays to balance the additional rays in.

“Lately there has been about 14 hours of cooling.”

What has this to do with compensation? It’s impossible to keep up with all these argume.ts because you keep changing the topic and the definitions. Here’s what I said:

Why do you think E increases? It’s that increase that is the compensation for the greater number of rays coming from the atmosphere and please learn what reflection means.

This is about warming due to the presence of greenhouse gases. It is not about the daily cycle.

“Your generalizations are meaningless unless you are able to show some math to support them. ”

Stop pretending you haven’t seen anu of my math. It’s one thing to try to claim it’s wrong it’s another to keep lying about me not showing them. I’ll repeat.

E = S + 1/2 A
A = E

Solution

E = 2S.

” It is definitely tiresome teaching someone science who won’t learn on their own. ”

Think how tiring it is to explain very basic and well understood mainstream.science to people who are incapable of accepting they are wrong about anything.

I argue because at some point I just want you to write down an actual set of equations that demonstrate science is wrong but it just becomes more evident each time that you can’t. You have loads of ingredients that you think must lead to a compelling argument , but you just can’t put them together in a way that makes sense. If you were capable of explaining why virtually every scoentis is wrong, you would have done it by now and not keep dragging everything down to personal insults.

Reply to  Bellman
June 26, 2026 6:38 am

Unless you think it has some magic way of knowing how many new rays it has to produce this can only happen because it warms up to the point where it is releasing sufficiently more rays to balance the additional rays in.”

YOU STILL HAVEN’T ACTUALLY LOOKED AT THE GRAPH OF TEMPERATURE UNDER FASTER AND SLOWER COOLING, HAVE YOU????

You do *NOT* need to increase the temperature in order to emit more energy. All you have to do is COOL SLOWER!

Do you see a bump in either of those curves where the slope is positive? Where temperature went up instead of down?

Both curves show continuous cooling. Yet the top curve emits more energy than the bottom curve. NO INCREASE IN TEMPERATURE, only decreasing temperature, i.e. the slope of the curves is always negative!

You are so locked into the garbage meme that reflected heat is retained that you can’t even believe what a simple graph shows. A graph that a 3rd grader could interpret.

This is about warming due to the presence of greenhouse gases. It is not about the daily cycle.”

So CO2 only appears when? If it doesn’t appear in the daily cycle then when does it appear?

And, once again, you leave out the step where the earth re-emits the reflected heat by cooling slower – the garbage climate science meme that reflected heat is retained and causes temperature to increase instead of cooling slower.

it just becomes more evident each time that you can’t.”

I’ve given you the math. You can’t refute it. F(t) = F0 e^-(t/τ) happens 24 hours per day. You say that is impossible in one sentence and then say that it possible in the next. You need to make up your mind.

Yes, F0 * e^-(tτ) is an approximation. It is a damn GOOD approximation. From 0 to ∞ it integrates to exactly the same value as the more complicated power law formula. One falls faster initially and the other falls faster in the tail. And F0 * e^-(t/τ) is far simpler to explain. YOU CAN’T EVEN FIGURE OUT WHAT THE FACTORS DETERMINING τ ARE!

 you just can’t put them together in a way that makes sense.”

The math doesn’t make sense to someone so locked into the garbage climate science meme that reflected heat is retained that they can’t even read a simple graph!

You can’t even explain why the temperature curves I show are wrong. Your only refutation is the Argument by Dismissal fallacy. Just say they are wrong and go on without showing why they are wrong.

It doesn’t matter if the curves are drawn using e^-(t/τ) or (3kt + T0^-3)^-(1/3). The slower cooling curve will *always* be higher than the faster cooling curve and neither curve will have a segment where the slope is positive. Thus slower cooling *will* emit more energy than the faster cooling – the very compensation that Planck was talking about. The area under the slower cooling curve will *always* be greater than the area under the faster cooling curve.

All you have to do is show where that math is wrong.

But you can’t, can you?

slower_faster_2
Reply to  Tim Gorman
June 26, 2026 8:58 am

Calm down. I’m not addressing any lengthy passage written in bold capitals.

“You do *NOT* need to increase the temperature in order to emit more energy. All you have to do is COOL SLOWER!”

You are right but you don’t consider the implications and keep confusing yourself with your odd use of language.

Consider you have two scenarios. Scenario A, a sphere at 300K is put in an environment and allowed to cool. After a specific period of time it has cooled to 200K. Scenario B, a sphere at 300K is placed in a different environment and after the same period of time has cooled to 250K. B is cooling slower.

And you are correct, at any point in time B is emitting energy at a faster rate and over the whole period it emits more energy in total than A, as you say. But this only happens because at any point in time B is warmer than A. And that’s the point I was making at the start, slower cooling means warming. And by that I mean that B is always warmer than A at a specific point in time, not that it is warming over time.

This is my point about the greenhouse effect and why it makes no sense to worry about what cooling means. The point is that the Earth is warmer with the greenhouse effect than it would be without it.

But back to my two scenarios. There’s another obvious point. If B has lost more energy than A in total, that would imply it’s got less internal energy left, which would mean it should be colder than A. That obviously is not the case as A is 200K and B is 250K. The solution to thus paradox is that whilst B lost more energy it must also have received more energy, and the extra energy it received has to be greater than the extra energy it lost. All of which shows why you have to look at net heat transfer, and the total gross energy lost does not tell you much.

Moreover this would be true even if the two spheres were warming, e.g. they were put into an environment warmer than 300K. If A warms to 350K and B warms to 400K, then B has emitted more energy and so by your definition has cooled more.

Reply to  Bellman
June 26, 2026 9:19 am

And that’s the point I was making at the start, slower cooling means warming. 

This is not warming, it is cooling slower. Physical science requires using specific terms properly. Warming has a specific meaning in thermodynamics, (Tₜ > Tₜ₋₁). It is why the mathematics is so important in defining physical phenomena.

Reply to  Jim Gorman
June 26, 2026 9:39 am

” Physical science requires using specific terms properly. Warming has a specific meaning in thermodynamics, (Tₜ > Tₜ₋₁).”

Oh the irony. According to Tim this is cooling. You have been arguing that everything is cooling all the time even as its temperature increases.

I gave you the context of what I mean by warmer. The object is warmer than it would have been if the condition was not true. A world with an atmosphere is warmer than a world without one. Maybe it would have been clearer if I said it was warmer, rather than warming. But it amounts to the same thing. The planet is warmed by having an atmosphere.

Reply to  Bellman
June 26, 2026 1:49 pm

You are right but you don’t consider the implications and keep confusing yourself with your odd use of language.”

I am considering the implications. My math works. You can’t refute it.

And I am *NOT* using being odd with my use of language. Loss of energy is *cooling*. You simply can’t understand that something can be heating and cooling at the same time. It’s because you refuse to look at the component processes and values that make up the NET.

I pointed out to you multiple times that you can’t describe a system using the NET. (Th = 200K, Tc = 50K) and (Th = 300K, Tc = 150K) give exactly the same temperature difference, 150K. Not only do those temperatures differ in the requirements for material in the system, they give big differences in the flux-in/flux-out values. Meaning the temperature decay curves will be different.

“But this only happens because at any point in time B is warmer than A”

And that’s the point I was making at the start, slower cooling means warming”

NEITHER A OR B WARMED! They both cooled. The temperature decay curve for neither of them changed slope from negative to positive.

These were TWO DIFFERENT OBJECTS. Assuming you meant isolated as well, neither could affect the other. So they both cooled, One just cooled at a slower rate! But they both cooled, they didn’t warm.

This is the kind of connundrum you get into using your logic of only looking at NET and ignoring the components making up the NET.

The point is that the Earth is warmer with the greenhouse effect than it would be without it.”

And now we are back to retained heat and ignoring that the gain/loss functions are TIME functions. If the heat is retained forever and not re-emitted so that the earth gets warmer, then earth would be a molten rock today from accumulated heat. That hasn’t happened. That implies that heat is being lost somewhere in time and space.

Think about for one second. If you take into account that Tmax is bounded by T^4, then the accumulated heat would push Tmin up to where Tmax is. The equilibrium point would be Tmax = Tmin.

The fact that this hasn’t happened over the millennia should be a clue to both you and climate science that what is happening is that you are not expanding your observation window enough to identify all of the cycles that are at play. Nyquist doesn’t just say you have to sample fast enough to catch the high frequency cycles, you also have to have an observation window long enough to identify the lowest low frequency cycle.

If your observation window isn’t long enough then the low frequency cycles just show up as non-random noise in your data because you are just catching bits and pieces of them. They create a systematic uncertainty effect, i.e. they inflate the variance of the data set. And that seems to plague climate science to no end – they are always trying to come up with assumptions so they can ignore the variance in the data.

Those low frequency cycles represent SLOW cooling is the only conclusion that is logical when looking at the entire life of the planet.

Reply to  Tim Gorman
June 26, 2026 7:59 pm

These were TWO DIFFERENT OBJECTS. Assuming you meant isolated as well, neither could affect the other.

They were two different scenarios. They do not both exist at the same time.

One just cooled at a slower rate! But they both cooled, they didn’t warm.

Why do you find it so difficult to read what I said. One is warmer than the other. They are not warming over time, but if there is an addition input, or if the environment was warmer they could be warming. The point is one can only emit more energy over time because it is warmer.

You keep trying to have this both ways. You have no problem saying that with two objects one may be emitting more energy than the other, but won’t consider that one must be warmer than the other becasue both are cooling (and using your equivocation, even objects getting warmer are cooling).

This is the kind of connundrum you get into using your logic of only looking at NET and ignoring the components making up the NET.

There is no conundrum, just a confusion on your part of the meaning of words. 1. If one object emits more energy than another it is because it is warmer than the other, even if both are cooling. 2. In the real world, if everything else is equal, the world is not cooling. This is because, 3. The earth doesn’t just lose energy, it also receives it, from the sun and the atmosphere. 4. The net exchange of energy is what causes an object to cool or warm.

And now we are back to retained heat and ignoring that the gain/loss functions are TIME functions.

If an object is “cooling slower”, how is it not retaining heat? You obsession with “TIME” functions would be fine, if you ever actually explained how your time functions resulted in no object retaining heat. So far it’s all just had waving. Your time functions are useless unless they describe the temperature of an object when you include all the factors together, earth, sun and atmosphere.

If the heat is retained forever and not re-emitted so that the earth gets warmer

It is, so your argument fails. Really try to define your terms. What do you mean by retaining heat? Objects don’t store store heat, they store energy. But what do you mean by retaining energy forever? Do you mean the amount of energy stays level, or do you mean once inside an object the energy can never leave, like a black hole?

In reality your premise is just wrong, and your inability to grasp that point despite me constantly explaining it to you, is your problem.

But for the last time, energy increases in an object because it’s getting more energy then it gives out. But that does not mean it does not “re-emit” the energy, because emitting energy is what objects above 0K do. And once it reaches a state where the same amount of energy is re-emitted then is received, it stops warming, but continues to re-emit energy in accordance with the SB law.

That implies that heat is being lost somewhere in time and space.

Yes. Just as I keep telling you.

If you take into account that Tmax is bounded by T^4

Still no idea why you think it is. The maximum temperature depends on energy in. If there was just the sun, there would be a boundary unless the sun gets warmer. But with an atmosphere the boundary also depends on how much energy is received from the atmosphere.

The T^4 rule only tells you to what temperature the earth has to rise in order to balance the total energy in.

identify all of the cycles that are at play

And he’s off with his cycles again. Strange how your always correct temperature equation doesn’t show any any cycles, just exponential cooling.

Reply to  Bellman
June 27, 2026 3:15 am

Why do you find it so difficult to read what I said. One is warmer than the other.”

So what? THEY BOTH COOLED! Neither one went back up the temperature gradient. The temperature profile never changed from a negative slope to a positive slope!

“becasue both are cooling”

Yep, and this is what causes the negative slope. One has a higher temperature at any point in time BECAUSE IT COOLED SLOWER, not because it warmed!

Yes. Just as I keep telling you.”

Energy is never lost, it is just transformed.

“Still no idea why you think it is.”

Do you EVER get out of your basement? Why do you think the daytime temperature profile doesn’t just keep rising till sunset? Why do you think that Tmax happens mid-afternoon instead of at sunset?

It’s because heat loss overtakes heat input at some point! The earth is cooling faster than it is gaining heat energy. And heat loss is T^4!

Reply to  Tim Gorman
June 27, 2026 5:30 am

THEY BOTH COOLED!

This is just getting pathetic. Stop shouting and try to connect your thoughts. You have all the elements and there is no reason why you can’t put them together to help you understand – you just don’t want to.

Both objects are cooling because of assumptions you have made. Putting two warm objects into a colder environment, with no heat source. But the fact one is cooling slower than the other means it has to be receiving additional heat. That means is will stop cooling at a higher temperature than the other object.

Now understand that this could have worked just as well in reverse. Put two objects at 0K in a warm environment and observe that one warms slower than the other. One will always be colder than the other but “THEY BOTH WARM!”

Now, your problem is trying to explain why “they both cool” is in anyway relevant to understanding the greenhouse effect. Because when you look at the current world it is not cooling. The greenhouse effect is not making the world cool slower, it’s keeping it at a warmer temperature.

Everything you argue just comes down to word games. You concentrate on an object getting colder even when it isn’t, and then change the definition of cooling to claim that even an object increasing in temperature is still cooling. At no point do you explain how any of this refutes the greenhouse effect. It’s just you yelling “cooling” all the time.

Energy is never lost, it is just transformed.

And the equivocation continues. The question was about energy being lost to space – that is energy leaving the atmosphere into space.

Do you EVER get out of your basement?

Do you ever answer a question, rather than use a childish insult?

Why do you think the daytime temperature profile doesn’t just keep rising till sunset?

It’s been explained to you on numerous occasions. The question is why do you think it’s relevant?

It’s because heat loss overtakes heat input at some point!

Yes. That’s why I keep explaining the world will not melt due to accumulated heat. What it does not mean is that the world is continuously cooling. During the day, once it’s warmed up to a sufficient point, and the total incoming heat has reduced to a sufficient point, cooling will occur. But at another point during the day the reverse happens. Incoming heat increases, the cooling results in less outgoing heat, and the in is greater than out and the earth warms. We call this the daily cycle.

The point is that overall the temperature keeps returning to the same point – the equilibrium point. It cools every night, but it warms every day, and they cancel each other out, at least over the course of a year. The only way to get a long term change in temperature is to change the equilibrium point.

And heat loss is T^4!

The T^4 is irrelevant. It would happen even if it were T^1. What matters is the hotter something is the more it emits. T^4 just ensures it happens quicker and is more stable. Remember also T^4 applies to incoming heat from the atmosphere.

Reply to  Bellman
June 29, 2026 5:36 am

Both objects are cooling because of assumptions you have made”

It is your example. If you didn’t specify it well enough that that is *YOUR* problem, not mine.

 But the fact one is cooling slower than the other means it has to be receiving additional heat.”

No, it doesn’t. I have explained that to you ad infinitum. All it requires is a reflective body associated with it. No additional heat whatsoever in the system. NONE. Yet one will cool slower than the other if one has a reflective body and the other one does not.

That means is will stop cooling at a higher temperature than the other object.”

No, it won’t. It just gets to 0K more slowly. It still cools, just at a slower rate.

Put two objects at 0K in a warm environment and observe that one warms slower than the other. One will always be colder than the other but “THEY BOTH WARM!””

So what? 1. Is the temperature of your “warm environment” higher than either body? 2. the warm environment requires it to be a heat source or *it* will be cooling as well. 3. They both cool while they are warming. So what? You haven’t described the system well enough to determine what actually happens!

Now, your problem is trying to explain why “they both cool” is in anyway relevant to understanding the greenhouse effect.”

  1. CO2 absorbs outgoing radiation.
  2. Co2 absorbs incoming radiation.
  3. How much actual radiation reaches the earth’s surface.
  4. divide the atmosphere into layers. Each layer absorbs and emits
  5. Assume each layer receives X and emits X/2 up and X/2 down.
  6. The amount of down radiation at leach layer reduces by 1/2 since it sends half back up and only 1/2 down.
  7. This gives a geometric reduction at each layer for the down radiation.
  8. 1, 1/2, 1/4, 1/8, 1/16, …..
  9. Assume the atmosphere with CO2 is 50,000m thick and it gets split into 1000m layers. That’s 50 layers. What’s the value of the sequenc at the 50th iteration?

The greenhouse affect, as laid out by climate science is a garbage mess of assumptions.

  1. CO2 is a one-way absorber. It is transparent to everything going downward so the earth’s surface gets it all.
  2. A higher concentration of CO2 at a colder temperature emits less. The fact that there are more emitters due to a higher concentration can be ignored.
  3. You can ignore the fact that collisions between CO2 and other molecules result in CO2 heating and rising and thus cooling so it emits less energy than it absorbs.
  4. Ignores that as CO2 rises it’s emissions sent to the earth gets less because the earth subtends a smaller and smaller angle.

The greenhouse effect was pulled out of someone’s backside to make a hypothesis work without making sure it is correct physically.

Yes. That’s why I keep explaining the world will not melt due to accumulated heat.”

You keep saying that the earth doesn’t cool and heat at the same time. So how do you propose it ever reaches equilibrium? If it is losing heat it is cooling. PERIOD. EXCLAMATION POINT.

You are *still* caught in the Catch-22 of wanting the earth to accumulate heat and warm while it also cools and doesn’t warm!

We call this the daily cycle.”

The daily cycle IS NOT A SUFFICIENT OBSERVATION INTERVAL! I gave you the two Nyquist restrictions on sampling. As usual, you just ignore everything.

Long term the earth is COOLING. Joules-out greater than joules-in. Otherwise the earth would be a molten rock today. There would never have been glacial periods. You are *still* caught in the connundrum.

THERMODYNAMICS require the use of TIME FUNCTIONS. Those time functions determine the observation interval.

You can’t even admit that if CO2 ADDS heat to the system then we should see Tmax increasing at a rate equal to or greater than Tmin. Yet we aren’t. Tmax has barely increased if at all. That would *NOT* be the result from another heat source being added to the system.

Reply to  Tim Gorman
June 29, 2026 9:22 am

“It is your example. If you didn’t specify it well enough that that is *YOUR* problem, not mine.”

I was demonstrating your assertion that slowing cooler means emitting more energy.

“All it requires is a reflective body associated with it. ”

A body reflecting heat back means you have received more heat. If you mean the object in question is reflective, then yes that’s another way to slow cooling, but in that case you are not emitting more, you’ve reduced the emissions. All you have done is ensure the object has to be at a higher temperature to emit the same energy as a black body.

“No, it won’t. It just gets to 0K more slowly.”

No it doesn’t. If there is a heat source the object gets to that temperature. An object in a warm environment will cool, or warm, to the temperature of the environment. That’s what your cooling equation would show if you added the environmental temperature into the equation instead of assuming it is always 0K.

T(t) = T∞ + (T0 – T∞)e^(-kt)

Reply to  Bellman
June 29, 2026 10:07 am

“The amount of down radiation at leach layer reduces by 1/2 since it sends half back up and only 1/2 down.
This gives a geometric reduction at each layer for the down radiation.”

Nonsense. By your logic hardly any leaves the atmosphere, and the atmosphere would end up storing almost unlimited energy. Think about it. Your logic applies to radiation up as well as down. If almost no energy is emitted to the surface, almost no energy would be emitted to space.

There are two problems here. As the atmosphere is not totally opaque, it follows that each layer will have to correspondingly less opaque. If 20% of the energy from the surface goes through the single layer model without being absorbed, then each layer has to absorb a much smaller proportion. About 3% I think. A lot of the re-emitted radiation can get to the surface without ever being absorbed by a lower level.

More importantly though, you are just ignoring all the different pathways involved in a 50 layer model. You are assuming that the energy from each level has to get all the way down to the surface in a single go, when it can keep wandering up and down until it finally gets emitted either to the surface or to space.

“The greenhouse affect, as laid out by climate science is a garbage mess of assumptions.”

You keep confusing the simplest possible model used here to demonstrate how it can work with what scientists assume.

“The greenhouse effect was pulled out of someone’s backside…”

Name names. Whose backside? Fourier or Tyndell, or someone else?

Reply to  Bellman
July 1, 2026 7:18 am

I was demonstrating your assertion that slowing cooler means emitting more energy.”

Meaning you can’t read a simple graph apparently.

A body reflecting heat back means you have received more heat.”

This is where your understanding fails! The reflected heat has already been lost earlier. All the reflected heat does is change the ratio of compensation heat to object heat – i.e. it cools slower. CO2 is *NOT* a source. It can only return what has already been lost.

If you mean the object in question is reflective, then yes that’s another way to slow cooling, but in that case you are not emitting more, you’ve reduced the emissions”

Now you, once again, are trying to say that S-B is wrong. It isn’t. YOU STILL DON’T UNDERSTAND THE SIMPLE GRAPH I’VE GIVEN YOU!

All you have done is ensure the object has to be at a higher temperature to emit the same energy as a black body.”

You continue to misunderstand that joules-out is a TIME FUNCTION. An object can emit more joules by cooling slower. It does *NOT* have to be at a higher temperature. If an object is at T1 and stays there while emitting 10 joules/sec over 60 seconds it will lose more joules than if the temperature decays exponentially from T1, e.g. 10->9->8->… over that same 60 seconds.

No it doesn’t. If there is a heat source the object gets to that temperature. An object in a warm environment will cool, or warm, to the temperature of the environment. That’s what your cooling equation would show if you added the environmental temperature into the equation instead of assuming it is always 0K.”

So what? CO2 IS NOT A HEAT SOURCE. IT IS A REFLECTOR. The environment would be a *source*.

Why can’t you understand that CO2 is a reflector and not a source? It can only return what has already been lost, and only a *part* of what has been lost at that!

Reply to  Tim Gorman
July 1, 2026 7:51 am

“Meaning you can’t read a simple graph apparently.”

What graph? I was replying to your words not that sketch you keep spamming.

“The reflected heat has already been lost earlier.”

How can you emit heat you have already lost? You seem to think the Earth can just create energy out of nothing. That contradicts the conservation of energy.

“All the reflected heat does is change the ratio of compensation heat to object heat – i.e. it cools slower.”

Pointless argiung this until you accept you are wrong about CO2 reflecting heat.

And you still keep trying to ignore the sun. I really don’t know how to make this any simpler for you. The heat the earth absorbed from the atmosphere is in addition to the heat it absorbs from the sun.

“Now you, once again, are trying to say that S-B is wrong.”

Stop lying about me. It really harms your case if you have to keep arguing with a straw man, rather than what I actually said.

“If an object is at T1 and stays there while emitting 10 joules/sec over 60 seconds it will lose more joules than if the temperature decays exponentially from T1, e.g. 10->9->8->… over that same 60 seconds. ”

And at the end of that 60 seconds the one that isn’t cooling will be warmer than then the one that is cooling exponentially.

And the only way it can stay at the same temperature whilst emitting energy is because it’s receiving an equal amount of energy.

“Why can’t you understand that CO2 is a reflector and not a source?”

Because it is not a reflector. You’ve had the distinction explained to you enough times. Your persistence in continuously using the wrong and misleading term can only be due to stupidity or trolling.

As to calling it not a heat source, that’s just your usual equivocating word games. You don’t need to call it a “heat source” to understand it is additional heat. Heat that the earth would not receive if it was not there.

Reply to  Tim Gorman
June 29, 2026 10:55 am

“You keep saying that the earth doesn’t cool and heat at the same time.”

I keep saying you need to stop equivocating and specify exactly what sort of “cooling” you are talking about at any moment. Cooling usually means that an object is getting colder. Saying it can mean an object gets hotter is just confusing, especially when using it to imply something is getting colder. This is a common problem with you inventing your own terms and assuming the rest of the world has to accept your definition, but here it seems to be a deliberate technique to confuse the issue.

You could just as easily say that all objects emit and receive energy at the same time.

“So how do you propose it ever reaches equilibrium? ”

The same way it does the last 1000 times you’ve asked the question. Equilibrium occurs when energy in equals energy out. This is bound to happen because the hotter an object is the more energy it emits. And when I say reach equilibrium, that may be a points objects temperature oscillates around, as in the daily cycle of the earth.

“If it is losing heat it is cooling. PERIOD. EXCLAMATION POINT. ”

Argument by assertion again. You say that cooling just means emitting energy and that it can simultaneously be warming because it is receiving energy. What happens when it’s cooling rate equals the warming rate?

“The daily cycle IS NOT A SUFFICIENT OBSERVATION INTERVAL! ”

Stop shouting and stop changing the subject. This has got nothing to do with observation intervals.

“Long term the earth is COOLING. ”

Define what you long term, and explain what you mean by cooling.

Your argument is that everything is cooling all the time even if it getting hotter, so by your logic that’s true – just not meaningful. And in the long term the sun dies and the earth cools, and over a much longer period tee may end up with the heat death of the universe, but none of this is relevant to the greenhouse effect, which has kept the earth at a survivable temperature for millions of years.

“Otherwise the earth would be a molten rock today.”

You seem to only be capable of believing in two possibilities. Ever the earth will soon cool to 0K or will warm to infinity.

“Otherwise the earth would be a molten rock today.”

You still haven’t explained how, if he earth is cooling in accordance with the cooling equation, that we are currently a lot warmer than during the last ice age.

“THERMODYNAMICS require the use of TIME FUNCTIONS. ”

Then use the time functions to demonstrate the temperature of the earth. You keep saying you’ve done the math, it all you ever do is show the cooling equation and assert it works 24 hours a day. If that were true you would be able to show how cold he earth will be a decade or a century from now. You cannot do that because the earth is not cooling at an expontial rate.

“You can’t even admit that if CO2 ADDS heat to the system then we should see Tmax increasing at a rate equal to or greater than Tmin. ”

What’s this obsession with me having to “admit” things?

Why do you think tnax should warm faster than tmin? And why do you think scientists think it should be the opposite tmin warms faster than tmax?

“Tmax has barely increased if at all. ”

Complety wrong. If anything, global land temperatures show slightly faster tmax warming. That’s certainly been the case in the UK, as I was demonstrating in the thread about the link between increased sunshine and summer temperatures.

Reply to  Bellman
June 23, 2026 7:26 am

“**Objects above 0 K do not continuously lose joules.**”

Rofl!

Only in bellman’s world does an object above 0K not emit radiation. And if it somehow does emit radiation that is not a cooling effect but a warming effect.

Our far distant heirs need never worry about the heat death of the universe.

Reply to  Tim Gorman
June 23, 2026 8:01 am

“Rofl!”

Having just used copilot as an authority, Time now thinks it’s answer is laughably wrong. I wonder what happened to change his mind.

“Only in bellman’s world does an object above 0K not emit radiation. ”

This demonstrating he didn’t even read the reply he’s laughing at. It didn’t say they don’t emit radiation, it said they do not lose Jules. He ignores the qualifying sentence from copilot

It *always emits* radiation, but it does **not** always lose *net* energy.

The key is **net heat flow**, not emission alone.

Reply to  Bellman
June 23, 2026 4:05 am

not a single one of your references say that emitting energy is called cooling.”

All matter with a temperature greater than absolute zero emits thermal radiation.”

ROFL!! In your worldview emitting heat is *NOT* cooling!

Nothing more needs to be said.

Reply to  Tim Gorman
June 23, 2026 4:22 am

“ROFL!! In your worldview emitting heat is *NOT* cooling!”

Yes. That’s why I asked for any reference that says emitting heat is cooling. Your inability to find a reference suggests I’m right.

Did you note what your plastic pal said,

“Why the confusion happens
People hear “everything above absolute zero emits radiation” and assume that means “everything cools.”
But emission alone doesn’t determine cooling — **net exchange** does.”

Reply to  Bellman
June 23, 2026 8:18 am

Yes. That’s why I asked for any reference that says emitting heat is cooling. Your inability to find a reference suggests I’m right.

Do you ever do any research or just do you just continue to spout your opinions as fact. Here is Q/A from CoPilot.

Q: I need a reference that emitting energy is a cooling process in thermodynamics

Engineering thermodynamics texts explicitly describe emission as cooling

  • Moran & Shapiro, Fundamentals of Engineering thermodynamics, 8th ed.  In the section on energy transfer by radiation:

  • “Net radiative loss results in cooling of the surface.”

Here is another Q/A.

Q: Can you supply a calculus equation, simple if possible that relates radiation to cooling versus time

A: This is the canonical calculus equation for radiative cooling.

  • (dT/dt) =−(εσA/mc)(T−T∞)
  • If T>T∞, then T4−T∞4>0, so
  • dTdt<0 → cooling.
  • If T=T∞, cooling stops.
  • If T<T∞, the sign flips → warming.

Radiation is always a cooling process when the body is hotter than its surroundings.

You should note that the equation is not dissimilar to the SB equation except it has a gradient based on time.

Now it is your turn to find a reference that shows an energy loss and falling temperature results in heating.

Reply to  Jim Gorman
June 23, 2026 9:08 am

“Engineering thermodynamics texts explicitly describe emission as cooling”

Do you have that explicit quote or are you just trusting a random word generator to understand the context of that entire book.

Is the second comment ment to be the explicit quote? “Net radiative loss results in cooling of the surface.” because the first word of sentence is important.

“Radiation is always a cooling process when the body is hotter than its surroundings.”

Yes , the second half of that sentence is important.

Really all your quotes are agreeing with me, what matters is net heat transfer. Nothing to suggest that emitting energy whilst receiving more is called “cooling”.

“Now it is your turn to find a reference that shows an energy loss and falling temperature results in heating.”

Why? That’s not what I’m saying at all. This is the problem with arguing with you two. You just don’t understand what you are arguing, but just assume I must be saying the opposite.

Let me try to make it clear what I am saying:

1. All objects above 0K emit energy.
2. If an object emits more than it receives it will cool.
3. Cooling or warming is defined by a change in temperature. If temperature goes up, it’s warming if it goes down it is cooling.
4. The reason for point 2 is that if an object emits more than it receives it’s temperature will go down.
5. Given that point 1 does not mean that all objects are cooling all the time. If they receive more energy than they emit they will be warming.

Reply to  Jim Gorman
June 23, 2026 7:10 pm

“Can you supply a calculus equation, simple if possible that relates radiation to cooling versus time

A: This is the canonical calculus equation for radiative cooling.

  • (dT/dt) =−(εσA/mc)(T⁴−T∞⁴)
  • If T>T∞, then T4−T∞4>0, so
  • dTdt<0 → cooling.
  • If T=T∞, cooling stops.
  • If T<T∞, the sign flips → warming.

Did you actually read that before posting? You do realize it’s saying the exact opposite of what you claim?

Reply to  Bellman
June 24, 2026 12:24 pm

Did you actually read that before posting? You do realize it’s saying the exact opposite of what you claim?

I did read it. You are incorrect in your analysis of the math. Let’s go over this step x step

General Equation ->

  • (dT/dt) =−(εσA/mc)(T⁴−T∞⁴)

Cooling

  • If T>T∞, then T−T∞> 0, so dT/dt < 0
  • If dT/dt is less than 0, it is negative. That makes the slope negative and therefore T is decreasing,, i.e., cooling.

Warming

  • If T<T∞, then T⁴−T∞⁴< 0, so dT/dt > 0
  • If dT/dt is greater than 0, it is positive. That makes the slope positive and therefore T is increasing, i.e.,, warming.

This is kind of simple.

Let (εσA/mc) = Z, then the general equation is:

(dT/dt) = -Z(T⁴−T∞⁴)

Did you forget the minus sign (-) in the equation?

Reply to  Jim Gorman
June 24, 2026 1:20 pm

You are incorrect in your analysis of the math”

bellman can’t read. He doesn’t have a hope of understanding math. He still hasn’t been able to decipher relative measurement uncertainty.

Reply to  Jim Gorman
June 24, 2026 1:39 pm

“You are incorrect in your analysis of the math.”

Then you need to point out where it says that all objects are constantly cooling, and why it actually says that objects that are in a warmer environment warm rather than cool.

“This is kind of simple. ”

It is, that’s why I’m surprised you don’t get why it contradicts what you are saying. I suspect we’ll go through the usual contortions and you’ll claim you were always saying the opposite to what you originally said. So to be clear the claim I was asking references for was

That’s why I asked for any reference that says emitting heat is cooling.

“Did you forget the minus sign (-) in the equation?”

No. The minus sign is correct. It’s just saying that if an object is warmer than it’s environment it will cool, so dT/dt will be negative, and if it’s cooler it will warm and dT/dt will be positive. The relevant point is this caused by the net flow energy.

Reply to  Bellman
June 24, 2026 6:31 pm

It is, that’s why I’m surprised you don’t get why it contradicts what you are saying. 

You have not proven your refutation except by assertion. You continually use temperature as a proxy for energy. Don’t be a hypocrite. This shows that the loss of energy, using temperatures, gives a negative slope.

This is in response to your question:

That’s why I asked for any reference that says emitting heat is cooling.

Now it is your turn to find a reference that refutes Planck’s statement about a hot body being cooled by a cooler body even when there is a hotter body in the syatem.

It shouldn’t be hard if it is as you say, incorrect because a cooler body warms a hotter body if there is a hotter body in the system.

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 Bi at 1000◦ C. The fact that the body A is cooled by B and heated by Bi is due entirely to the fact that B is a weaker, Bi a stronger emitter than A.

Something for you to think about when researching. In your earlier post you had “n -> ∞” steps showing a series that won’t end with equilibrium between a hot and cold body. Think about where the extra 100 units of energy disappears to.

Reply to  Jim Gorman
June 24, 2026 7:05 pm

You have not proven your refutation except by assertion.

You made a claim that some people use the term cooling to mean any loss of energy, even when the temperature of that object is increasing. And that this means all objects above 0K are constantly cooling.

I simply asked for evidence that confirms that it usual to use the term cooling in that way. A simple reference would suffice. I am not “refuting” that the word isn’t used in this way in some disciplines, just asking for evidence. The fact that after many attempts you fail to produce any evidence confirming your claim, and many that disagree with it suggests to me that you are just making it up.

I can’t prove that nobody has ever used the word in that way, that’s why the onus is on you to provide some evidence that it is. This shouldn’t be hard, if as you claim it’s common usage.

I’m not even sure why this matters so much to you, as it has nothing to do with the wrongness of all other assertions – it’s just an argument over terminology.

This shows that the loss of energy, using temperatures, gives a negative slope.

Net energy loss. Nobody disputed that. The question is can you call emitting energy “cooling” even when you get more energy back.

Now it is your turn to find a reference that refutes Planck’s statement about a hot body being cooled by a cooler body even when there is a hotter body in the syatem.

He doesn’t say that. Try actually reading the quote you keep cutting and pasting. It literally says that a colder body will cool it, but a hotter body will heat it. Not that it’s cooling in the presence of a hotter body.

In your earlier post you had “n -> ∞” steps showing a series that won’t end with equilibrium between a hot and cold body.

No I did not. If you think I did, you need to provide an exact quote. Nothing can warm up indefinitely. My simplistic greenhouse gas demonstrations always end up with an equilibrium.

Reply to  Bellman
June 22, 2026 7:44 am

You’ve said elsewhere that cooling requires output to be greater than input, and quoted Planck saying”

No one has *ever* said that. Your lack of reading comprehension skills are showing again.

Planck: “The fact that the body A is cooled by B and heated by B′ is due entirely to the fact that B is a weaker, B′ a stronger emitter than A.”

You obviously can’t even read this simple sentence.

If B is colder than A then it will cool A because it can’t provide enough energy to A to keep A from cooling. B’ will warm A because it is at a higher temperature than A.

You don’t seem to understand that B will *cool* by losing energy to both A and B’ but it will also gain energy from A and B’. It will gain more energy than it loses – BUT IT WILL STILL EMIT RADIATION THUS LOSING ENERGY – i.e. COOLING!

Can you not tell the difference between GROSS energy flows and NET energy flows?

Reply to  Tim Gorman
June 22, 2026 10:17 am

“No one has *ever* said that.”

Energy = joules_in – joules_out.

If joules_out > joules_in then you have cooling.

If joules_in > joules_out then you have warming.

https://wattsupwiththat.com/2026/06/17/do-los-ninos-cause-climatic-cooling/#comment-4209033

“Planck: “The fact that the body A is cooled by B and heated by B′ is due entirely to the fact that B is a weaker, B′ a stronger emitter than A.””

Yes. It’s heated by B’ because it’s warmer than A. “Heated by,” not cooled.

“B’ will warm A because it is at a higher temperature than A.”

See even you agree yet you still say no one says that cooling requires energy out to be greater than energy in. Do you actually understand what you are arguing? It seems to change from sentence to sentence.

“You don’t seem to understand that B will *cool* by losing energy to both A and B’ but it will also gain energy from A and B’. ”

But you are the only one calling that cooling. Every source I can find, including Planck and you yourself, says it’s cooling if the temperature goes down and warming if the temperature goes up.

“Can you not tell the difference between GROSS energy flows and NET energy flows?”

That’s exactly what I keep asking you.

Reply to  Bellman
June 23, 2026 5:18 am

It’s your focus solely on the NET instead of the gross components that leads to your view that the earth does not cool during the day!

The proof that it does is the mismatch between the slope of temperature curve going negative before the sun sets.

Yet that can’t happen in your world.

You have yet to show *any* math supporting your assertions. And you have totally failed to refute the math I have given you.

F(t) = F0 * e^-(t/τ) happens every second of every day – except in your statistical world. F(t) = F0 * e^-(t/τ) *IS* COOLING. Period. Exclamation point.

It is a gross component of the NET. Period. Exclamation point.

Yet you can’t even give the simple function to use in calculating “τ”

Reply to  Tim Gorman
June 23, 2026 6:24 am

“It’s your focus solely on the NET instead of the gross components that leads to your view that the earth does not cool during the day!”

I was specifically refering to gross flows. But surely even you can see it makes no difference. Net is just out minus in. Whether you right the equations in terms of net or gross can make no difference.

The reason I say that it doesn’t cool as it gets hotter is because you have yet to supply an iota of evidence that anyone other then you uses such a term.

“The proof that it does is the mismatch between the slope of temperature curve going negative before the sun sets. ”

You can’t even get your own delusions straight. It cools before sunset because the energy from the sun is less than the energy from the earth. This in turn is because the earth is hotter and so emits energy at a faster rate and the sun is lower in the sky and so the input rate of energy input. It does not prove that the earth was actually cooling all the time it was getting warmer.

“Yet that can’t happen in your world. ”

Stop lying about me. Address the pints I make not your own delusions.

“You have yet to show *any* math supporting your assertions. And you have totally failed to refute the math I have given you.”

Again, stop lying.

“F(t) = F0 * e^-(t/τ) happens every second of every day”

Provide a reference and explain what you terms are. You’ve already admitted that F0 is not a constant and depends on t, but you’ve also said that this temperature It’s really hard to see how this equation is happening all the time.

“Period. Exclamation point.”

It would really help your case if you don’t think that was a compelling argument.

“Yet you can’t even give the simple function to use in calculating “τ””

It’s your equation. You are the one who has to provide that information. I can’t help you solve your equation when it makes no sense. Is t temperature or time? What is the function that defines FO? How is the function correct if it ignores the T⁴ rule? Please, just pont to your source.

Reply to  Bellman
June 22, 2026 7:49 am

Says the person who claims that the term “estimated value” in statistics means the value that you will always get”

No, that is what YOU always say. The term “best estimate” always seems to be lost on you. It’s why you’ve never figured out that the true value can be anywhere in the uncertainty interval and, in the case of an asymmetric distribution the mean will *not* be the most seen value. You *always* apply the meme that “all measurement uncertainty is random, Gaussian, and cancels” so you can use the SEM as the measurement uncertainty.

Do you remember me telling you: “Sometimes the long shot wins”?

You probably don’t remember. You just blow off everything that doesn’t match your misconceptions about the real world.

Reply to  Tim Gorman
June 22, 2026 10:41 am

“No, that is what YOU always say. ”

Nope. I was trying to explain why that wasn’t what “expected value” meant. But let’s not go down that distraction again.

Reply to  Bellman
June 23, 2026 5:30 am

Nope. I was trying to explain why that wasn’t what “expected value” meant. But let’s not go down that distraction again.”

Malarky. You wouldn’t even admit that the long shot sometimes wins. You would *always* pick the average.

Reply to  Tim Gorman
June 23, 2026 5:55 am

See here’s Tim’s distraction technique all over.I said said it will be a distraction to start discussing what expected value means, so he immediately starts lying about me in the hope I’ll be suckered into this discussion again. All this just to divert attention away him losing the argument about the greenhouse effect.

I’m not falling for this again. I’ll just state that I have never said that long shots never win, or that you should always pick the average. If Tim claims override he need to find an exact quote from me.

Reply to  Bellman
June 22, 2026 7:59 am

“Assuming this is a definition they use in thermodynamics, it’s still misleading in the context of greenhouse gases.”

It is not misleading at all! It applies even in the context of greenhouse gases!

You are specifically saying that the greenhouse effect cannot cause warming, but relying on the use of the term cooling which can be applied to an object that is warming.”

Stop putting words in my mouth. At least provide the quote you are referencing.

CO2 can *NOT* warm the earth. All it can do is slow the cooling of the earth. CO2 in the atmosphere is COLDER than the earth. Colder objects can’t warm hotter objects. Object B can’t cause the temperature of Object A to rise because B is colder than A!

The warming of the earth is *NOT* from back radiation by CO2, i.e. the greenhouse effect. It is due to joule input from the sun, an object that is at a higher temperature than the earth!

You keep trying to conflate CO2 with the sun as both of them being a source of thermal energy. CO2 is *NOT* a source of thermal energy, it is a reflector of thermal energy.

Does CO2 in the earth’s atmosphere cause the temperature of the sun to rise? CO2 in the earth’s atmosphere certainly radiates toward the sun. How long will it be until the CO2 in the earth’s atmosphere causes the sun to go nova?

Reply to  Tim Gorman
June 22, 2026 9:48 am

CO2 can cause the Earth to warm because it reduces the rate of heat loss from the Earth, the equilibrium temperature increases as long as the solar irradiance stays the same because it is the result of the balance of heat input and heat loss.

Reply to  Phil.
June 23, 2026 5:08 am

CO2 doesn’t CAUSE the Earth to warm. It does not *add* joules of heat to the ssytem. It causes the Earth to *cool* more slowly. The only thing in the earth-sun-space system that can raise the temperature of the earth is the sun.

As you say, it is the heat loss and heat input that determines the NET state of the system. But it is the cooling and heating of the system that determines the NET state. You have to know the gross components to understand the system operation.

The earth *does* cool 24 hours per day. Period. It also heats 12 hours per day. Period. The two components create a NET state.

Trying to define “cooling” as only the NET state of temperature ignores the fact that a multiplicity of components can result in the same NET.

(Th – Tc) will give the same answer whether Th = 20C and Tc = 10C as for Th = 30C and Tc = 20C. But the system internals will be vastly different for each. F(t) = F0 * e^-(t/τ) will be significantly different, i.e. the cooling rate, because F0 will be different.

This applies to *any* thermodynamic system. Be it the earth-sun-space system or the HVAC system for a berm home in a river valley versus a slab ranch on a mountain top. It is the cooling and heating components that determine the NET, the NET doesn’t determine the cooling and heating components.

It’s focusing on the NET instead of the components that leads climate science to the idiocy that incoming and outgoing flux to/from the system can balance. They will *never* come close to balancing and the meme leads to all kinds of erroneous conclusions about how the system operates.

bellman’s conclusions about the earth-sun-space are a prime example. E.g. the earth doesn’t cool during the day.

Reply to  Phil.
June 23, 2026 6:28 am

CO2 can cause the Earth to warm 

Warming is:

Tₜ > Tₜ₋₁

Can CO2 cause this inequality?

Reply to  Tim Gorman
June 22, 2026 12:34 pm

“Stop putting words in my mouth. At least provide the quote you are referencing. ”

Followed immediately by

“CO2 can *NOT* warm the earth. All it can do is slow the cooling of the earth.”

There’s your exact quote.

“It is due to joule input from the sun, an object that is at a higher temperature than the earth!”

It isn’t though. That’s the problem. The earth should be considerably cooler if it was the same temperature as the sun is at this distance.

“CO2 is *NOT* a source of thermal energy, it is a reflector of thermal energy. ”

It isn’t a reflector, but say it is if you find that easier to understand. And say it isn’t a heat source. What it is is additional energy. Energy that would otherwise be lost to space. And energy that will be absorbed by the earth in addition to all the energy from the sun. The consequence is the earth absorbs more energy than it would if there were no greenhouse gases.

“CO2 in the earth’s atmosphere certainly radiates toward the sun. ”

Only a miniscul amount. Half of it’s energy radiates into space in all directions. What proportion of that space is the sun? The other half all ends up back on the earth.

And needless to say the sum is very hot and not likely to be warmed very much by it, or any other planet.

And going nova has nothing to do with being warmed.

Reply to  Bellman
June 23, 2026 5:40 am

It isn’t though. That’s the problem. The earth should be considerably cooler if it was the same temperature as the sun is at this distance.”

Once again you focus on the NET while ignoring how the NET happened!

The temperature is different because CO2 slowed the cooling, it did *NOT* add joules of energy to the system.

You HAVE to understand the components before you can understand the NET!

Except in your statistical world.

“It isn’t a reflector, but say it is if you find that easier to understand.”

CO2 is *NOT* a source of heat energy. It is *NOT* burning in the sky!

If it isn’t a source then the only thing it can do is be transparent to thermal energy or reflect thermal energy. It is *NOT* transparent to thermal energy and it is not a source of thermal energy so that only leaves it being a reflector of received thermal energy.

What CO2 doe is the exact same thing as aluminum foil. It reflects heat.

Your view on this is a result of your inability to understand how the components of a thermodynamic system works – it comes from only considering the NET.

So tell everyone what *YOU* think CO2 is. The only options available are:

  1. source of thermal energy
  2. transparent to thermal energy
  3. reflector of received thermal energy

Is the sky burning where you are?

Reply to  Tim Gorman
June 23, 2026 6:07 am

“Once again you focus on the NET while ignoring how the NET happened!”

I see this is your latest weird obsession. No I was talking about the gross energy from the sum not being enough to warm the earth to it’s current temperature.

“The temperature is different because CO2 slowed the cooling,”

Congratulations. You’ve finally explained the greenhouse effect.

“CO2 is *NOT* a source of heat energy. It is *NOT* burning in the sky!”

Things don’t have to burn to radiate energy. As you say, anything above 0K emitts energy.

“What CO2 doe is the exact same thing as aluminum foil. It reflects heat.”

Could you provide a source that says CO2 is a reflector. Maybe ask your plastic pal.

Copilot, does co2 reflect heat.

Short answer: CO₂ does not reflect heat — it absorbs and re‑emits infrared (IR) radiation, which effectively traps heat in the atmosphere.

Why it’s not “reflection”
Reflection means bouncing radiation off a surface unchanged — like a mirror.
CO₂ doesn’t do that.

Instead, it absorbs specific IR wavelengths, becomes energized, then re‑emits new IR photons. This process is more like a sponge soaking up water and squeezing it back out, not a mirror.

“So tell everyone what *YOU* think CO2 is. The only options available are: ”

It’s an absorber and emitter of heat. Whether you think that makes it a heat source will depend on how you define heat source.

“Is the sky burning where you are?”

It certainly feels like it, at the moment.

Reply to  Bellman
June 22, 2026 8:25 am

“You are specifically saying that the greenhouse effect cannot cause warming, but relying on the use of the term cooling which can be applied to an object that is warming.”

You *need* to be able to distinguish between gross flows of thermal energy and net flows of thermal energy. Your refusal to learn this is keeping you from understanding thermodynamics.

You *have* to analyze energy inputs separately from energy outputs in order to understand what is going on. A net flow is not deterministic of a system state. Vastly different gross in-flows and out-flows of thermal energy between SystemA and SystemB can result in the same net flow.

The net flow between a cold object and a colder object can be exactly the same as between a hotter object and a hot object. Yet the conditions of each object can be vastly different because of their temperatures.

In addition you can’t even establish what the net flow of thermal energy is until you figure out what the gross components of that net flow are!

For the earth the gross components are:

-Gross_out – gross_reflected. Where Gross_out ≥ Gross_reflected ALWAYS! You can’t get more reflected than what is sent out.

-Gross_in is the sun’s insolation.

Gross_out – Gross_reflected is the COOLING COMPONENT
Gross-in is the warming component.

YOU WILL ALWAYS HAVE EACH. The warming component may be zero but it is still a component of the system.

Think about it in general terms. The same thing applies to Mars. BUT THE VALUES OF THE GROSS COMPONENTS WILL BE DIFFERENT. The net flow may wind up being same (in reality they aren’t) even if the gross components are different.

Everything above 0K cools. If it didn’t then the 2nd law equation would always be q ∝ Th.

Reply to  Tim Gorman
June 22, 2026 1:37 pm

“You *need* to be able to distinguish between gross flows of thermal energy and net flows of thermal energy”

I am distinguishing between the two. That’s why I keep saying it’s the net changes that determines if something is warming it cooling.

“Gross_out – Gross_reflected is the COOLING COMPONENT
Gross-in is the warming component.”

Aside from your continuing inability to understand what reflected means, that’s ok as long as you compare the the two and accept that if out is less than in there is warming, and that this will continue untill the two are the same. But why call them warming and cooling? Why not just input and output? Avoids a lot of confusion.

Reply to  Bellman
June 22, 2026 2:05 pm

“Gross_out – Gross_reflected is the COOLING COMPONENT
Gross-in is the warming component.”

This also illustrates why seperating out cooling / warming components is ambiguous. You can just as easily see the “gross_reflected” part as part of the warming component.

Gross_out is the cooling component.

Gross_sun + gross_atmosphere are the warming components.

Reply to  Bellman
June 22, 2026 3:35 pm

Gross_sun + gross_atmosphere are the warming components.

And you are wrong. The fluxes do not add. In order to add, gross_atmosphere would need to be larger that gross_sun. That is, the hot body.

All you have to do is look at the SB equation.

net I = σ(Thot)⁴ – σ(Tcold)⁴

From a dimensional standpoint, that is:

W/m² = hot W/m² – cold W/m²

Net flux is toward the cold, no if’s, and’s, or but’s.

It also means that the hot body is radiating more than the cold body. Therefore, cooling of the hot body takes place until the cold body becomes hotter than the hot body.

I’ve given you the necessary inequalities/equality to discern what is occurring.

Tₜ < Tₜ₋₁ Cooling
Tₜ = Tₜ₋₁ Equilibrium
Tₜ . Tₜ₋₁ Heating

Your formula of:

Gross_sun + gross_atmosphere are the warming components.

results in continuous warming over and above the equilibrium between the Gross_sun and the earth. The result? The earth should be radiating vastly more than the 1360 from the sun. That is quite a machine you have there, where energy is created out of nothing.

Reply to  Jim Gorman
June 23, 2026 3:21 am

” The fluxes do not add. ”

Why on earth not? What happens to the energy that doesn’t add?

“net I = σ(Thot)⁴ – σ(Tcold)⁴”

Which tells you the net flux reduces as the atmosphere warms. Now what happens when you add the flux from the sun? If there was no atmosphere Tcold would be zero and you have

Net I = σ(Thot⁴)

Which should equal the flux from the sun. But with a warmer atmosphere net I is smaller so you have a smaller flux between the earth and the atmosphere, then you are getting from the sun. And that means the temperature of the Earth has to warm.

“Therefore, cooling of the hot body takes place until the cold body becomes hotter than the hot body.”

You keep trying to use the logic of a two body problem. You keep ignoring the sun.

“I’ve given you the necessary inequalities/equality to discern what is occurring. ”

Why do you keep saying that as if it’s a telling point? Literally all you are saying is that if temperature increases it’s warming and if it reduces it’s cooling. The only people who don’t agree with this are those saying that all objects are cooling all the time even when temperature is rising. I wonder who they are.

“Your formula of:

Gross_sun + gross_atmosphere are the warming components.

results in continuous warming over and above the equilibrium between the Gross_sun and the earth”

No it does not. It’s astonishing you just keep making that mistake when it’s been explained to you multiple times. You should understand that as something warms it emits more energy. You said it yourself with the SB equation, yet you keep ignoring it when it suites you.

” The earth should be radiating vastly more than the 1360 from the sun.”

1360 what? If you mean W/m² the earth’s surface doesn’t get anything like that much.

But regardless yes it’s the point that with a greenhouse effect the earth’s surface radiates more than it receives from the sun. That’s why it’s warmer. In the worst case, with a completely opaque atmosphere, it would be twice as much.

” That is quite a machine you have there, where energy is created out of nothing.”

No energy is created out of nothing. All the energy comes from the sun using nuclear fusion. The earth simply holds on to it for a bit longer.

Reply to  Bellman
June 23, 2026 5:47 am

jim: “results in continuous warming over and above the equilibrium between the Gross_sun and the earth”

bellman: “No it does not. ”

When you consider Gross_sun + Gross_reflected to be the total input you *are* setting up a positive feedback that will result in ever increasing temperature.

The formula is actually Gross_sun + x(Gross_sun) where “x” is the coefficient of reflection. Gross_reflection = x * Gross_sun.

This is because CO2 is *NOT* a source of thermal energy. It can only give off what it receives, which is a portion of what the sun sends in.

Once again, you fail simple math.

Reply to  Tim Gorman
June 23, 2026 6:36 am

“When you consider Gross_sun + Gross_reflected to be the total input you *are* setting up a positive feedback that will result in ever increasing temperature. ”

When you keep saying things like that, it’s evident you have not read a thing I’ve written. Temperatures do not ever increase for the simple reason that the hotter the earth gets the more it emitts.

“The formula is actually Gross_sun + x(Gross_sun) where “x” is the coefficient of reflection. Gross_reflection = x * Gross_sun. ”

Firbtr umpteenth time try to look up what reflection means. Co tinously misusing that term detracts from your claims to understand thermodynamics.

But ignoring that, the actual equation is Gross_sun + x(Gross_earth). This is exactly what I explained to you the last time you I dusted I explained the greenhouse effect.

In the simplest example, where the atmosphere absorbes all the energy from the earth, you have

E = S + 1/2 A

Where E, S and A are the gross flux from the earth, the gross flux the earth receives from the sun, and the gross flux from the atmosphere. That last one is halved because only half the energy is radiated towards the earth. And the equal sign implies we are at equalibrium.

The other equation you need is

E = A,

And then it’s simple algebra to deduce that

E = 2S.

Reply to  Bellman
June 23, 2026 7:14 am

“ ignoring that, the actual equation is Gross_sun + x(Gross_earth).”

ROFL!

Gross_earth = Gross_sun!

The earth is not a heat SOURCE. The earth is not on fire. The earth can onle emit what it gets from the sun.

Your equation us STILL a positive feedback loop that will grow until destruction.

Everything you assert just confirms that you have no basic scientific training at all!

Only in your statistical world is the earth on fire!

Reply to  Bellman
June 23, 2026 6:56 am

Which tells you the net flux reduces as the atmosphere warms. Now what happens when you add the flux from the sun? If there was no atmosphere Tcold would be zero and you have

Where do you think the earth gets it temperature from? The sun! It is built into the equation! Tnot is derived from the sun. You don’t add it in twice!

You keep trying to use the logic of a two body problem. You keep ignoring the sun.

Keep up with the thread. The assumption is that the sun does not affect CO2. That is the simplified ideal assumption. If you want to change the system into a 3 body system where all the bodies interact, have at it, it will get complicated real quickly.

No it does not. It’s astonishing you just keep making that mistake when it’s been explained to you multiple times. You should understand that as something warms it emits more energy. You said it yourself with the SB equation, yet you keep ignoring it when it suites you.

If your statement is true, you should have no problem with showing how this inequality is reached thermodynamically for the earth’s temperature. Start with the earth as the hot body and CO2 the cold body and use standard radiative thermodynamic equations such as the Stefan-Boltzmann.

Tₜ . Tₜ₋₁ Heating

Reply to  Jim Gorman
June 23, 2026 7:18 am

You won’t get an answer. He can’t even formulate the system configuration let alone write the functional relationships in mathematical terms.

Don’t you understand that in bellman’s world the sky is on fire, the earth is on fire, and losing thermal energy is not cooling?

Reply to  Jim Gorman
June 23, 2026 7:40 am

“Where do you think the earth gets it temperature from? The sun! ”

No. It get’s all it’s energy from the sun, but it’s temperature comes from being at the right temperature to emit the same amount of energy and it receives.

” Tnot is derived from the sun. You don’t add it in twice!”

Tnot? You do add it twice if you are getting it twice. Say the sun sends 1 joule ti the earth, at a later point the earth sends that same joule ti the atmosphere and later the atmosphere sends it back to the earth. But in the mean time the sun has sent another joule. So now the earth has 2 joules.

” The assumption is that the sun does not affect CO2. ”

Yes that,’s the simplifying assumption. Greenhouses gases are transparent to rays coming from the sun but opaque to rays coming from the earth.

” If you want to change the system into a 3 body system where all the bodies interact, have at it, it will”

The thre bodies in this case don’t all interact with each other, but they do all exist at the same time, and you cannot ignore any one just because it’s inconvenient. The interactions are

S -> E
E -> A
A -> E
A -> S

I’m using A there to represent both the sun and space.

“have at it, it will get complicated real quickly. ”

It really isn’t that complicated. Even I managed it. You just have to understand that tim isn’t important if all you want to know is what the equalibrium will be. It,’s just a system of a couple of simultanious equations.

If you want to see how it evolved over time then it’s easiest to write a simulation than get into figuring out the various differential equations.

“If your statement is true”

Which statement? That warmer object emitnnmote energy? Are you suggesting the SB law is wrong?

Reply to  Bellman
June 25, 2026 4:23 am

Tnot? You do add it twice if you are getting it twice. Say the sun sends 1 joule ti the earth, at a later point the earth sends that same joule ti the atmosphere and later the atmosphere sends it back to the earth. But in the mean time the sun has sent another joule. So now the earth has 2 joules.”

Earth gets 1 joule from the sun, Total = 1
Earth emits 1 joule to atm, Total = 0
Earth gets 1 joule from Atm, total = 1
earth emits 1 joule to atm, total = 0

Cycle repeats ——-

The sum total at the end of the sequence in time is 0 joules for the earth. You skipped the portion of the sequence where the earth emits the joule it got from the atm.

This is what climate science does. It assumes the return joule from the atmosphere is “trapped”, “retained”, whatever you want to call it. It isn’t. The earth re-emits it. And since the atm is not a perfect reflector you actually wind up with a diminishing sequence that goes along with it. If the atm returns one tenth of the earth’s emitted joule then the earth re-emits that one tenth, and then the atm sends back one hundredth of the original amount, and so on …..

It is actually far more complicated than this. It suffers from climate science’s typical garbage assumptions. The absorption of heat and the emitting of heat is a time function. That absorption of the joule from the sun and the emitting of the joule from the earth involves varying flux values over varying time periods. Same with the atmosphere/earth component. It’s why you will *NEVER* see a flux balance between the insolation component of the sun and the emittance of the earth. The only way you would ever see a flux balance is if the earth and the sun were at the same temperature.

Planck: “For example, if we let the rays emitted by the body fall back on it, say by suitable refection, the body, while again absorbing these rays, will necessarily be at the same time emitting new rays, and this is the compensation required by the second principle.”

Somehow you and climate science ALWAYS ignore this part of the Planck writings. CO2 provides the “suitable reflection” and the earth responds by “emitting new rays”.

Those “new rays” are the difference between areas under a “slower” decay rate (i.e. cooling) and a “faster” decay rate (i.e. still cooling) as shown in the image I have given you at least twice. That cooling happens ALL THE TIME.

I keep telling you that you have to FULLY understand how all of the components of a system operate in order to fully understand it, both the heat input and the heat output components, i.e. heating and cooling. It’s why you and climate science keep getting the things like the input/putput sequence wrong, leading to misconceptions like “trapped” heat and flux-in/flux-out balance.

Reply to  Tim Gorman
June 25, 2026 5:06 am

“Earth gets 1 joule from the sun, Total = 1
Earth emits 1 joule to atm, Total = 0
Earth gets 1 joule from Atm, total = 1
earth emits 1 joule to atm, total = 0”

And what appened to the sun in this time line?

“Cycle repeats ——-”

And each cycle the sun has to give the earth another joule.

Reply to  Bellman
June 25, 2026 5:12 am

You are just doing exactly what I was accused of double counting. You’ve created a cycle where the earth is allowed to give up the same joule twice, whilst only getting it back once from the atmosphere.

Reply to  Bellman
June 25, 2026 1:04 pm

The earth DOES give up that joule twice. Once when it comes in from the sun and again when it comes back from CO2.

The only other option is that somehow the earth knows the joule is from CO2 and retains it instead of re-emitting.

Is the joule from CO2 tagged somehow so the earth can tell where it originated?

Reply to  Bellman
June 25, 2026 1:03 pm

What happened to the sun? It’s in the “CYCLE REPEATS”.

And each cycle the sun has to give the earth another joule.”

And the earth emits it toward the CO2. And CO2 sends it back. And the earth re-emits it. Sum = 0.

Then another joule comes in from the sun. Wash, rinse, repeat!

You keep saying the earth re-emits the heat reflected back by CO2 but then you keep leaving that re-emit step out of the sequence!

Cognitive dissonance at its finest!

Reply to  Tim Gorman
June 25, 2026 6:21 pm

What happened to the sun? It’s in the “CYCLE REPEATS”.

But you are creating a cycle where the sun only emits one joule of energy, but the earth emits two joules.

And the earth emits it toward the CO2. And CO2 sends it back. And the earth re-emits it. Sum = 0.

But now the atmosphere now has an extra joule that it hasn’t re-emitted yet.

Then another joule comes in from the sun. Wash, rinse, repeat!

You end up with a very hot atmosphere.

You keep saying the earth re-emits the heat reflected back by CO2 but then you keep leaving that re-emit step out of the sequence!

Nope, I’m not saying COL2 reflects anything. But regardless, and given this is only meant to illustrate the problem of double counting –

Cycle 1.

Earth gets 1 joule from the sun, Total = 1
Earth emits 1 joule to atm, Total = 0
Earth gets 1 joule from Atm, total = 1 (N.b. It should only get half a joule, but lets go with this toy example)

Cycle 2.

Earth gets 1 joule from the sun, Total = 2
Earth emits 2 joules to atm, Total = 0
Earth gets 2 joules from Atm, total = 2

Cycle 3.

Earth gets 1 joule from the sun, Total = 3
Earth emits 3 joules to atm, Total = 0
Earth gets 3 joules from Atm, total = 3

etc.

In a better example the atmosphere only transmits half it’s energy to the earth which is why the earth doesn’t warm up to infinity.

Cycle 1.

Earth gets 1 joule from the sun, Total = 1
Earth emits 1 joule to atm, Total = 0
Earth gets 0.5 joules from Atm, total = 0.5

Cycle 2.

Earth gets 1 joule from the sun, Total = 1.5
Earth emits 1.5 joules to atm, Total = 0
Earth gets 0.75 joules from Atm, total = 0.75

Cycle 3.

Earth gets 1 joule from the sun, Total = 1.75
Earth emits 1.75 joules to atm, Total = 0
Earth gets 0.875 joules from Atm, total = 0.875

etc.

But of course none of this happens in cycles. Sun keeps given the same joules averaged over a year, this warms up the earth with keeps emitting energy to the atmosphere in accordance with SB. This warms up the atmosphere which radiates energy in all directions in accordance with SB, half of which are in the direction of the earth which warms the earth further. You can see how this works out over time, but it’s a lot easier just top calculate what it means for a final equilibrium.

Once again, you just need two simultaneous equations.

E = S + 1/2A
A = E

=> E = 2S.

Reply to  Bellman
June 26, 2026 9:29 am

Now you want to change the goalposts! We are talking about THE surface of the EARTH, not the atmosphere.

When the earth re-emits that joule that CO2 sent back, what happens to it? (hint: space)

The bottom line is that only the sun is the SOURCE. There is no more energy in the system than what the sun puts in.

The only way the earth can go up in temperature is by retaining heat from the sun. Back radiation doesn’t ADD heat to the system, it’s just heat that’s bouncing around in the system that finally gets emitted to space.

Until you can understand that slower cooling emits more heat than faster cooling you’ll never understand why CO2 can’t raise the temperature of the earth. That increased temperature rise *has* to be based on the earth retaining more heat rather than it radiates.

So, pick one and stick with it. Does “back radiation” get retained or does it get re-emitted? And if you picked “retained” then you also need to explain why the earth doesn’t re-emit it.

And this is just a first approximation of the thermodynamic system called the biosphere. Things like dew point, evaporation, wind, etc all play a part in the system.

Reply to  Tim Gorman
June 26, 2026 11:42 am

“We are talking about THE surface of the EARTH, not the atmosphere.”

That’s what I’m talking about. Where specifically do you think I moved the goal posts?

“When the earth re-emits that joule that CO2 sent back, what happens to it? (hint: space)”

Half of it goes into space half to the surface.

“The bottom line is that only the sun is the SOURCE. There is no more energy in the system than what the sun puts in. ”

Correct. It’s just that with the greenhouse effect that energy stays around for longer.

“The only way the earth can go up in temperature is by retaining heat from the sun.”

Which is what the greenhouse effect does.

“Until you can understand that slower cooling emits more heat than faster cooling you’ll never understand why CO2 can’t raise the temperature of the earth. ”

Your problem is I do accept that slower cooling emits more energy as I explained in another comment. Using your definition of cooling. But this means the temperature of the planet with slower cooling is hotter than a planet with quicker cooling.

“That increased temperature rise *has* to be based on the earth retaining more heat rather than it radiates.”

Yes, that’s the greenhouse effect. But you need to understand that this rise in temperature only goes on till you reach a new equilibrium.

“Does “back radiation” get retained or does it get re-emitted?”

Both. I really don’t understand why you think it has to be one or the other.

Reply to  Bellman
June 26, 2026 2:12 pm

That’s what I’m talking about. Where specifically do you think I moved the goal posts?”

From the surface to the atmosphere.

“But now the atmosphere now has an extra joule that it hasn’t re-emitted yet.”
You end up with a very hot atmosphere”

Another “retained” heat excuse?

ROFL!!! That CO2 molecule loses that heat energy one way or another, through kinetic collisions or radiation. Heat energy lost through kinetic collision doesn’t get back to the earth in any significant amount. Only radiation does. And if CO2 absorbs incoming energy from the sun then it also absorbs “back” radiation from CO2 itself and converts a lot of it kinetic energy through collisions. The kinetic energy that gets transferred may not even be of sufficient amount to generate thermal radiation even if it transferred to another CO2 molecule!

Everything you are now throwing against the wall to see if it sticks just makes the amount of radiation being sent back to the earth smaller and smaller.

That means that the slower cooling exponential/power-law decay gets closer and closer to the zero-CO2 exponential/power-law decay curve. So the need for the compensation radiation energy less gets smaller and smaller as the reflected energy amount gets less and less.

This is all a prime example of why climate science is such a mess.

Reply to  Tim Gorman
June 26, 2026 5:41 pm

From the surface to the atmosphere.

Not an answer. I’ve always talked about both, you can’t talk about the greenhouse effect without both.

ROFL!!! That CO2 molecule loses that heat energy one way or another, through kinetic collisions or radiation.

Yes. Why do you find that so hard to understand. Just your assumption that CO2 is reflecting heat?

Heat energy lost through kinetic collision doesn’t get back to the earth in any significant amount

And you know that, how? Just because you don’t want it to be true?

And if CO2 absorbs incoming energy from the sun …

It doesn’t. That’s the whole point.

Everything you are now throwing against the wall to see if it sticks…

It’s all standard science. Just because you don’t understand it doesn’t make it magic.

That means that the slower cooling exponential/power-law decay gets closer and closer to the zero-CO2 exponential/power-law decay curve.

One day you are going to have to demonstrate how that works. Use actual math, maybe write an original paper. Prove all the scientists have been wrong these last century or two, because they didn’t understand the power laws. Until that happens I’ve no interest in your hand-waving claims.

Reply to  Bellman
June 26, 2026 5:51 pm

And you know that, how? Just because you don’t want it to be true?

Because hot air rises.

Reply to  Jim Gorman
June 27, 2026 2:33 am

You put it a little more succinctly than I did. bellman probably can’t figure out that hot air rises. Hot air balloons are “magic”, don’t you know?

Reply to  Jim Gorman
June 27, 2026 5:38 am

Because hot air rises.

Not an answer to the question. How do you know “Heat energy lost through kinetic collision doesn’t get back to the earth in any significant amount”.

You need to quantify how and to what extent hot air rises in the atmosphere, and what effect that has on the energy radiated back to earth.

Reply to  Bellman
June 29, 2026 6:37 am

Not an answer to the question. How do you know “Heat energy lost through kinetic collision doesn’t get back to the earth in any significant amount”.”

Of course it is! You just can’t figure it out. According to climate science cold CO2 at the top of the atmosphere emits less than warmer CO2 deeper in the atmosphere!

If CO2 gains energy IT RISES! Meaning it gets colder! So when it emits it emits less than it absorbed!

As usual, you want it *both* ways. CO2 rises and gets colder and emits less while CO2 warms but doesn’t rise and emits more than the colder CO2.

You don’t have a single thing to offer other than cognitive dissonance shite you can throw against the wall hoping something will stick!

You want CO2 to absorb radiation headed for space but *not* absorb radiation headed toward the earth! Another cognitive dissonance piece of shite you keep throwing out.

You need to quantify how and to what extent hot air rises in the atmosphere, and what effect that has on the energy radiated back to earth.”

I may have the number wrong but it’s something like 20x more likely that CO2 will lose its energy and gain its energy via collisions than be absorbing IR radiation. Every time it gains KINETIC energy it rises and if it then absorbs IR it will emit less energy because it’s colder. If it absorbs IR but has a collision *before* it emits then it is likely it doesn’t have enough energy to emit.

If CO2 blocks space-going radiation then it also blocks earth-going radiation. It’s called path loss. Since CO2 only returns PART of what it gets from the earth that PART is going to suffer blockage all the way back to the earth. So you not only have the inverse-square law working against you, you also have collisions happening that sends molecules higher in elevation before the emit again thus cutting down on the amount of energy headed toward earth.

CO2 is *NOT* a one-way filter for IR.

Reply to  Bellman
June 27, 2026 2:24 am

Not an answer. I’ve always talked about both, you can’t talk about the greenhouse effect without both.”

The subject is the surface of the earth warming. It’s what the earth does that the math applies to. The math that you can’t refute.

“Yes. Why do you find that so hard to understand. Just your assumption that CO2 is reflecting heat?”

It’s not an assumption that CO2 REFLECTS heat. CO2 is not a source of heat. It does not generate heat on its own. Part of the thermal heat CO2 receives from the earth gets sent back to the earth. The earth LOSES more heat than CO2 can send back. That’s *not* a process that pushes the earth back up the temperature gradient. It just slows down the cooling of the earth.

Warming something is pushing it back up the temperature gradient. Slowing down cooling is *not* warming.

And you know that, how? Just because you don’t want it to be true?”

What do *YOU* think that increased kinetic energy causes to happen? Most people call it convection. What do *YOU* call it?

It doesn’t. That’s the whole point.”

CO2 doesn’t absorb LWIR from the sun? How does it distinguish thermal energy from the sun from thermal energy from the earth? Is the EM wave from the sun somehow “tagged” with a different identifier?

About 1% of the sun’s insolation is in LWIR. That’s a significant amount of LWIR. And you think CO2 is transparent to it?

It’s all standard science.”

Nothing you post is “standard science”. Just look at what you’ve said here: “CO2 doesn’t absorb LWIR from the sun”. THAT IS STANDARD SCIENCE?

One day you are going to have to demonstrate how that works. Use actual math, maybe write an original paper. Prove all the scientists have been wrong these last century or two, because they didn’t understand the power laws. Until that happens I’ve no interest in your hand-waving claims.”

And here you are now, claiming that Newton’s Law of Cooling isn’t correct solely because I’ve used it. That is “standard science”? The power law decay is quite similar to the exponential decay. My guess is that you really don’t even know why they are slightly different, even after I’ve told you why.

I’ve given you the ACTUAL MATH. Straight derivation from the S-B law. And you can show no flaws in my derivation. Yet you claim that there must be – all without one single reference to any link to show a different derivation.

Tell me again who is doing the handwaving here?

Reply to  Bellman
June 22, 2026 8:29 am

Again, a reference is required. And an explanation that although you say the object is cooling, it’s quite possible it’s temperature is rising.”

I’ve just given you the references. But I also know you’ll just blow’em off.

Joules-out and Joules-in determines if temperature is going up or going down. But Joules-out IS ALWAYS > 0! And joules-out > 0 implies COOLING.

You keep wanting to say that the loss of heat energy is not cooling. Do you really understand what that implies about your understanding of thermal energy transfer?

Reply to  Tim Gorman
June 22, 2026 1:29 pm

“I’ve just given you the references. But I also know you’ll just blow’em off.”

Yes because none of them said what you claimed, and your preferred AI explicitly said it wasn’t true.

“You keep wanting to say that the loss of heat energy is not cooling.”

It’s cooling if there’s a net loss of energy. What I say is if there’s a net gain in energy and temperatures are going up, that’s warming, not cooling.

Reply to  Bellman
June 22, 2026 8:35 am

You still don’t understand what reflection means in thermodynamics. CO2 does not reflect energy it absorbs it. If it reflected energy we’d be getting twice as much energy back.”

Simply unfreakingbelievable. If CO2 absorbs heat energy from the earth what does it do with it? Is some of it re-radiated back toward the earth?

Reflection is an in-out process. As far as CO2 is concerned the “in” is from the earth and the “out” is back to the earth -REFLECTION!

“Back radiation” is REFLECTED radiation. The details of how the in-out work is irrelevant. It might be a quantum process, it might be a something else. But it is an in-out process regardless. In from the source and back out to the source.

At this point I give up. You are at exactly the same point with thermodynamics as you are with metrology. Your head is full of misconceptions and non-physical meme’s and no amount of education will change them. If you were my mule I’d be looking for the 2×4.

bdgwx
Reply to  Clyde Spencer
June 21, 2026 6:47 am

The definition of cooling is dT/dt < 0.

When you say “cooling” are you actually meaning dT/dt < 0 or do mean that the body is radiating such that the flux is F > 0?

Note that bodies can radiate more such that ΔF > 0 and still do the opposite of cooling such that dT/dt > 0.

Reply to  bdgwx
June 21, 2026 8:20 am

When you say “cooling” are you actually meaning dT/dt < 0 or do mean that the body is radiating such that the flux is F > 0?”

You are kidding right? dT/dt has to include *ALL* objects in the system.

dT/dt for an object in the system depends on inputs AND outputs of each individual object.

If an object is above 0K then F > 0 for that object! Read Planck some time.

F_out and F_in can be different. dT/dt depends on the relationship between F_out and F_in.

Note that bodies can radiate more such that ΔF > 0 and still do the opposite of cooling such that dT/dt > 0.”

This can only happen if F_out < F_in. Your statement implies that if an object is existing in a universe all of its own, all by itself, that it’s temperature can spontaneously go up, dT/dt > 0.

That’s a violation of entropy since no work has been done on the object.

All objects above 0K cool. The only way to overcome that cooling is to have an external SOURCE of joules putting in more joules than are being emitted.

In essence, dT/dt = heating – cooling. Cooling is always greater than 0 (zero). Thus heating must be greater than cooling for dT/dt to be positive. It truly *is* just that simple.

Reply to  Clyde Spencer
June 21, 2026 7:54 am

The imbalance between the sun’s input and the emission of heat from the earth shifts about mid-afternoon usually. So the surface doesn’t actually warm for 12 hours, it’s something less.

Reply to  Bellman
June 20, 2026 9:00 am

 And it’s absurd when you’ve admitted from the start that slower cooling raises Tmin.”

Slower cooling does not “RAISE” Tmin. That implies that Tmin at some point was (Tmin – X) and the reflected heat raised the temperature frrom (Tmin – X) back to Tmin.

It doesn’t do that.

The temperature COOLS from Tmax down to Tmin. No raising. If because of a slower rate of decay, Tmin(fast cooling) is lower than Tmin(slow cooling) it is *NOT* because the temperature RAISED frrom Tmin(fast cooling) UP to Tmin(slow cooling).

If Tmin(slow cooling) is at a higher point than Tmin(fast cooling) then the object is actually losing more heat per unit time because Tmin(slow cooling)^2 > Tmin(fast cooling)^4!

Reply to  Tim Gorman
June 20, 2026 11:23 am

“Slower cooling does not “RAISE” Tmin.”

It’s what Jim said.

“The temperature COOLS from Tmax down to Tmin. ”

Unless you acknowledge that I’m not talking about what happens in one day, you will never understand this, and these arguments will just be us going round in circles.

“Tmin(fast cooling) is lower than Tmin(slow cooling) ”

And so Tmin(slow cooling) is warmer than Tmin(fast cooling). Do you finally accept that point?

“Tmin(slow cooling)^2 > Tmin(fast cooling)^4”

Did you really mean to say that?

Reply to  Bellman
June 20, 2026 4:52 pm

Unless you acknowledge that I’m not talking about what happens in one day, you will never understand this, and these arguments will just be us going round in circles.”

*YOU* are the one that started talking about Tmin – cooling to 11°C one day and to only 10°C the next day!

The problem is that *YOU* can’t seem to grasp that heat loss is a TIME FUNCTION. And cooling from 15°C to 11°C on one day and from 15°C to 10°C the next day simply isn’t a long enough period to catch all the low frequency cycles involved in the thermodynamics.

And so Tmin(slow cooling) is warmer than Tmin(fast cooling). Do you finally accept that point?”

15°C to 11°C is cooling. 15°C to 10°C is cooling!

CO2 did *NOT* raise the temperature from 10°C to 11°C on the same day! It only slowed the cooling more on one day than it did on the other day.

And on the day where it went from 15°C to 11°C MORE ENERGY WAS EMITTED than on the day where it went from 15°C to 10°C!

Did you really mean to say that?”

Again, the energy emitted is the integral under the curve. And the curve with the slower decay has more area under the curve!

You can’t seem to get that into your head!

DO THE MATH!

I even did the math for you! Could you not follow it? I even post a picture showing the slow and fast decay curves! Could you not understand the graph?

Energy = ∫ F0 * e^-(t/τ)

Work it out!

Reply to  Bellman
June 20, 2026 9:21 pm

Hence, slower cooling causes warming.

I think it would be better to say that “Slower cooling allows warming,” not that it causes it.

Reply to  Clyde Spencer
June 21, 2026 4:57 am

I not sure that’s a better choice of words. If the warming wouldn’t have happened without the slower cooling, then the slower cooling caused the warning.

Maybe a clearer phrasing would be the warming happened as a result of slower cooling.

If objects fall in a river, partly blocking it and as a result the river level rises, would you say the obstruction caused the rise?

Reply to  Bellman
June 21, 2026 9:07 am

The temperature goes up because of the sun’s input. Not because of the cooling of the earth!

Reply to  Tim Gorman
June 21, 2026 10:13 am

Will temperatures be hotter or colder if there were no atmosphere?

Reply to  Bellman
June 21, 2026 10:49 am

Will temperatures be hotter or colder if there were no atmosphere?

It all depends on where you are at any given time. No different than today, the world will still enjoy a large variation of temperatures depending on your location. Trying to find an average will still be just as meaningless, maybe even less.

Reply to  Jim Gorman
June 21, 2026 12:18 pm

It’s alright, I didn’t expect you to answer. Nice work on your evasion.

So let’s ask another question. Would there be enough energy from the sun to keep the world st it’s current temperature if the greenhouse effect didn’t exist?

bdgwx
Reply to  Clyde Spencer
June 21, 2026 7:57 am

I think it would be better to say that “Slower cooling allows warming,” not that it causes it.

Similarly someone could say that closing the door allowed the kitchen oven to get warmer. It’s still the same cause and effect principal just using a different word to cheapen to the fact that closing the door resulted in the warming. I’m okay with the word choice as long as everyone knows that “allowed” means the same thing as “caused” in this context.

Reply to  Bellman
June 19, 2026 1:10 pm

And as I was explaining, so to does the rate. If you have an incoming flux of 320W/m² for 12 hours, and an outgoing flux of 160W/m² for 24 hours, you have an equilibrium.”

So now you are repeating what I’ve been telling you this whole time.

THE FLUX WILL NEVER BALANCE!

You have an equilibrium in JOULES-IN and JOULES-OUT, not in flux-in and flux-out.

And joules-in and joules-out have time intervals in centuries if not in millennia. Trenberth’s radiative balance chart simply can’t be correct. It shows rates, not amounts, and it does not show a normalized time interval!

If the flux is not balanced then neither are the temperature profiles.

Again, that’s why Tmax happens before sunset. The joules-out become greater than the joules-in!

So, now tell us one more time how a cold body can increase the joules in a warmer body so the temperature of the warmer body increases.

Reply to  Tim Gorman
June 19, 2026 2:41 pm

“So now you are repeating what I’ve been telling you this whole time.

THE FLUX WILL NEVER BALANCE!”

I’ve literally just pointed out an example of an equal input and output over a day. The balance is zero. There is no net warming or cooling. I can’t help it if you don’t understand why a net zero balence means that there will be no change from day to day. Nor can I help it if you don’t understand that temperatures can, and will, change over the course of a day, yet still end up with no overall change.

You seem to be arguing that unless the fluxes are identical from second to second there is no equalibrium, and don’t seem to understand that the equalibrium exists over longer time frames.

You are a great example of not being able to see the woods for the trees.

“So, now tell us one more time how a cold body can increase the joules in a warmer body so the temperature of the warmer body increases.”

Why? Do you think there’s some chance you’ll understand it this time?

Reply to  Bellman
June 20, 2026 2:06 am

You seem to be arguing that unless the fluxes are identical from second to second there is no equalibrium,”

No, that is what climate science is arguing.

FLUX WILL NEVER BALANCE.

“I’ve literally just pointed out an example of an equal input and output over a day. “

You pointed out that the ENERGY will balance, not the flux!

You are as bad as bdgwx. Flux is a RATE, not an amount. Energy is an amount!

It is the amount of heat that has to balance, not the rate of flow!

“Why? Do you think there’s some chance you’ll understand it this time?”

You simply can’t admit that 100 joules_out – 50 joules_reflected = 50 joules_lost, can you?

50 joules lost is COOLING, not warming.

Reply to  Bellman
June 19, 2026 10:14 am

“Alternatively it means the average rate of energy in in has to equal the average rate if energy out.”

Over what period if time? Do you have even the faintest clue about conduction, gradients, and time?

Reply to  Tim Gorman
June 19, 2026 10:33 am

“Over what period if time?”

Whatever you want. Stop trying to deflect.

Reply to  Bellman
June 19, 2026 11:02 am

Whatever you want. Stop trying to deflect.

No deflection. The time is a very important value. Take land during the spring and summer. Do you think the land emits as many joules as it revives during this time? If not, then how do those joules stored 10 – 20 inches in the soil warm the atmosphere? Can you balance fluxes or energy over this time period?

Reply to  Jim Gorman
June 19, 2026 11:31 am

Yes, deflection. The point is if you look over any period of time, if you had more energy in than out, the temperature will have increased. You were the ones who kept going on about 12 and 24 hour periods, so I assume you are interested in a single day. A year would be more appropriate to even out seasonal effects, and a longer period will even out other short term affects.

But that’s all beside the point, when you are claiming that slowing down the rate of cooling will not increase temperatures. As always when you lose that argument you start trying to deflect into the finer details in the hope that nobody notices your original misunderstanding.

Reply to  Bellman
June 20, 2026 1:42 am

But that’s all beside the point, when you are claiming that slowing down the rate of cooling will not increase temperatures.”

SHOW THE MATH!

100 joules-emitted – 50 joules-reflected = 50 joules of cooling

Increasing temperature requires ADDING HEAT ENERGY, not losing energy. Partial reflection of already emitted heat is just that, it is returning energy already emitted. That slows cooling and results in MORE energy being emitted per unit time.

E_emitted ∝ τ

τ is the decay rate. Slower cooling means a longer decay time, i.e. τ gets larger. τ getting larger means energy-emitted gets larger.

(E_emitted ∝ τ) is a pretty simple equation. SHOW THE MATH REFUTING IT!

(E_emitted ∝ τ) is not a finer detail. It’s about as simple as you can get.

Reply to  Tim Gorman
June 20, 2026 4:23 am

“100 joules-emitted – 50 joules-reflected = 50 joules of cooling”

This is just getting deranged. You just keep denying the sun exists. If the surface is emitting 100 joules it’s because it received 100 joules from the sun. With the atmosphere it gets an extra 50 joules back (not from reflection). Hence the surface gaines a net 50 joules.

100 – 100 + 50 = 50

That increases it’s temperature until balance is restored, which happens when it is emitting a value equal to what it receives from the sun plus what it receives from the atmosphere. In this simple example that happens when it emmits 200 joules, and gets back 100 from the atmosphere.

100 – 200 + 100 = 0.

In order to be emitting twice as many joules the internal energy has to have doubled. Which means temperature has risen by the fourth root of 2. About 20%. So if the temperature without an atmosphere was 250K, the temperature with this opaque atmosphere would be closer to 300K.

Everything you write just keeps ignoring the sun. Until you can accept that the earth does not just lose energy, but also gaines energy from the sun, you will never understand this.

But I really don’t get how you think this constant cooling hypothesis works. You keep asking me to do the math, yet you never do it yourself. Just how quickly do you think the earth is cooling? How long before the earth becomes uninhabital according to your hypothesis?

Reply to  Bellman
June 20, 2026 10:59 am

You just keep denying the sun exists.”

Nope. The sun is a SOURCE. It can provide additional energy. CO2 is *NOT* a heat source. It is a reflector of already emitted energy, i.e. heat loss. CO2 can *NOT* provide additional energy. It can only REFLECT back heat that has already been lost.

“If the surface is emitting 100 joules it’s because it received 100 joules from the sun. With the atmosphere it gets an extra 50 joules back (not from reflection). Hence the surface gaines a net 50 joules.”

SHOW THE MATH!

100 joules in -> 100 joules out
100 joules out -> 50 joules reflected + 50 joules lost. = 100

The surface loses a net 50 joules. It does *NOT* gain a net of 50 joules.

For the umpteenth time. Heat transfer is a TIME FUNCTION. You keep on ignoring the time involved.

That increases it’s temperature until balance is restored”

SHOW YOUR MATH! So far you haven’t even got the addition right!

You keep wanting to add the 100joules from the sun to the 50 joules reflected while ignoring that 100 joules was emitted BEFORE the 50 joules are reflected!

IT’S A TIME SEQUENCE. HEAT TRANSFER IS A TIME FUNCTION. You can’t reflect back what hasn’t yet been sent. CO2 is *NOT* a heat source, it is a reflector!

Where do you think your 50 joules reflected is coming from?

which happens when it is emitting a value equal to what it receives from the sun plus what it receives from the atmosphere.”

Judas H. Priest! For the inward flux and the outward flux to be the same BOTH OBJECTS HAVE TO BE THE SAME TEMPERATURE! For this statement to be true the earth would have to be the same temperature as the sun!

” In this simple example that happens when it emmits 200 joules, and gets back 100 from the atmosphere.

100 – 200 + 100 = 0.”

Unfreakingbelievable!

Emits 200 joules and gets back 100?

That’s 200_out – 100_back = 100_out!!!!!

You can’t even get the signs of the flux correct! It’s no wonder you never want to show the math!

Everything you write just keeps ignoring the sun.”

The sun is the ONLY SOURCE in the system! It has NOTHING to do with what CO2 is reflecting. If the sun went out tomorrow, what would the earth do? Stop emitting? What would CO2 do? Replace the sun?

Reply to  Tim Gorman
June 20, 2026 1:56 pm

“The sun is a SOURCE. ”

Yes that’s the point. So why do you keep ignoring it in your cooling equation?

“100 joules in -> 100 joules out
100 joules out -> 50 joules reflected + 50 joules lost. = 100

The surface loses a net 50 joules. It does *NOT* gain a net of 50 joules.”

You really have a hard time understanding your own maths. You are getting 100 joules in the first line and an additional 50 in the second, yet you only emit 100 in the second. Hence you are getting a net 50 joules. You are not losing a net 50 joules.

Maybe you need a simpler toy example. If Jimmy gives you 2 apples, and you give 2 apples to Karl, but Karl gives you back one of the apples, how many apples do you have?

“You keep wanting to add the 100joules from the sun to the 50 joules reflected while ignoring that 100 joules was emitted BEFORE the 50 joules are reflected!”

For the last time there is no reflection. But regardless your math is just wrong. If you get 100 of anything give 100 away and get 50 back you have 50 more than you started with, not 50 less.

This is such an obvious point that it’s really sad to see the mental contortions you go through to avoid seeing it.

“For the inward flux and the outward flux to be the same BOTH OBJECTS HAVE TO BE THE SAME TEMPERATURE! ”

Why do you keep claiming this is a two body problem. Even your brother understands there are three objects here the surface, the atmosphere, and the sun.

“That’s 200_out – 100_back = 100_out!!!!!”

And you’ve forgotten the sun again. 200 out – 100 back – 100 from the sun = 0.

“The sun is the ONLY SOURCE in the system! ”

So why do you keep ignoring it?

“If the sun went out tomorrow, what would the earth do? ”

It would get cold pretty quickly, just as your equation says. The problem for you is it hasn’t gone out, which is why your equation is wrong.

Reply to  Bellman
June 20, 2026 1:42 pm

That increases it’s temperature until balance is restored, which happens when it is emitting a value equal to what it receives from the sun plus what it receives from the atmosphere. In this simple example that happens when it emmits 200 joules, and gets back 100 from the atmosphere.

We have been over this before. Planck describes this situation as follows.

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 Bi at 1000◦ C. The fact that the body A is cooled by B and heated by Bi is due entirely to the fact that B is a weaker, Bi a stronger emitter than A.

Body A -> Earth
Body Bi -> Sun
Body B -> Atmosphere

As assumed for ideal conditions Body Bi only irradiates Body A. Body Bi and Body A will reach an equilibrium temperature, say 373K. So lets put the sun inside Body A and that Body A is at 373K and radiating at that temperature. Now lets say Body B is at 273K.

Remember Body A radiates based on its temperature and not Body B’s temperature. Planck says this.

For example, if we let the rays emitted by the body fall back on it, say by suitable reflection, the body, while again absorbing these rays, will necessarily be at the same time emitting new rays, and this is the compensation required by the second principle.

The emission at 373K is more than the emission from 273K. Stefan-Boltzmann defines what happens in this case. The net radiation will be determined from the difference between the hot body and and cold body. In this case, the net radiation is into the cold body. There is no variable used to “adjust” the temperature of the hot body because its temperature and radiation does not change. This equation changed to include time will lead you to the fact that the bodies will reach equilibrium at 373K and radiate similar flux toward each other.

Now lets look at what happens in your scenario when the Body B flux is added to Body A raising its temperature.

Step 1 Body A -> 100 Body B -> 50 Total at Body A = 150
Step 2 Body A -> 150 Body B -> 50 Total at Body A = 200
Step 3 Body A -> 200 Body B -> 50 Total at Body A = 250
Step 4 Body A -> 250 Body B -> 50 Total at Body A = 300
.
.
.
Step 10 Body A -> 550 Body B -> 50 Total at Body A = 600

Do you want to tell me where this series stops. Worse, I have shown Body B not warming at all. Think about what happens as Body B attempts to catch Body A.

This should tell you that equilibrium can never be achieved. Is that what you think?

Reply to  Jim Gorman
June 20, 2026 4:50 pm

Planck describes this situation as follows.

And you think I disagree with that?

Body A will reach an equilibrium temperature, say 373K.

That’s pretty hot.

So lets put the sun inside Body A and that Body A is at 373K and radiating at that temperature.

Why not leave it where it is?

Remember Body A radiates based on its temperature and not Body B’s temperature.

Correct.

There is no variable used to “adjust” the temperature of the hot body because its temperature and radiation does not change.

Which hot body are you talking about, the sun or the earth?

This equation changed to include time will lead you to the fact that the bodies will reach equilibrium at 373K and radiate similar flux toward each other.

Wrong. But go on.

Step 1 Body A -> 100 Body B -> 50 Total at Body A = 150
Step 2 Body A -> 150 Body B -> 50 Total at Body A = 200
Step 3 Body A -> 200 Body B -> 50 Total at Body A = 250
Step 4 Body A -> 250 Body B -> 50 Total at Body A = 300

Are those figures meant to be temperatures or emissions? It would really help your argument if you didn’t keep changing values like this.

In either case, you’ve just forgotten about body Bi, which you called the sun. But aside from that I’ve no idea what your figures are meant to mean. Why is body B (the atmosphere) staying the same temperature, whist body A (the Earth) keeps warming? You seem to just be adding the temperature of the atmosphere to the earth at each point. That’s not what happens.

Worse, I have shown Body B not warming at all.

Worse, you haven’t shown it, you’ve just assumed it.

This should tell you that equilibrium can never be achieved.

Yet here we are sitting on a planet that is still livable after billions of years. If there is no equilibrium, how come the world hasn’t melted or completely frozen by now?

Let’s think about your step diagram. I’ll assume the figures are arbitrary emission figures, and ignore the time it takes to change temperature. I’ll assume there is long enough between each step for the bodies to change to an equilibrium temperature.

Before an atmosphere –
Step 0: Sun to Earth -> 100, Earth to Space -> 100.

Add a completely opaque atmosphere. Absorbs all energy from earth, returns 50% back.

Step 1: Sun to Earth -> 100, Earth to Atmosphere -> 100, Atmosphere to Earth -> 50.

Step 2: Sun to Earth -> 100, Earth to Atmosphere -> 150, Atmosphere to Earth -> 75.

Step 2: Sun to Earth -> 100, Earth to Atmosphere -> 175, Atmosphere to Earth -> 87.5.

Step 3: Sun to Earth -> 100, Earth to Atmosphere -> 187.5, Atmosphere to Earth -> 93.75.

Step 4: Sun to Earth -> 100, Earth to Atmosphere -> 193.75, Atmosphere to Earth -> 96.875.

Step n as n tends to infinity: Sun to Earth -> 100, Earth to Atmosphere -> 200, Atmosphere to Earth -> 100.

Earth is no emitting twice as much energy as before, implying it’s temperature has risen by a factor of ⁴√2 ~ 1.19. If the initial temperature was 250K, the new temperature is 298K.

Reply to  Bellman
June 20, 2026 6:39 am

Yes, deflection. The point is if you look over any period of time, if you had more energy in than out, the temperature will have increased. 

The temperature where has increased? The atmosphere or the surface heat sink at depth? You refuse to recognize that the surface is a heat sink. Heat is trapped below the surface/atmosphere interface and is not transferred to the atmosphere until a later time. Even when it is transferred, some of it is latent heat which is not sensible.

You really need to up your game to discuss a three body system rather than a one body system. Define the 2nd bodies processes and tell us how that effects changes to the 3rd body (atmosphere).

Reply to  Jim Gorman
June 20, 2026 8:25 am

“The temperature where has increased? ”

Increased in the body of interest. The ocean in this case. Of course with the ocean there are complex interactions and hidden depths, but all of this is a distraction from the initial pint, which is your claim that a warmer atmosphere cannot warm the oceans.

” Heat is trapped below the surface/atmosphere interface and is not transferred to the atmosphere until a later time.”

Of course it’s at a later date. Everything I’ve said is about how energy in the body increases due to the balance if energy in verses out, and how that means the warming object releases more energy. It isn’t because every joul hitting the earth is reflected back.

“You really need to up your game to discuss a three body system rather than a one body system.”

Everything I’ve said has been a three body system. Ocean (or earth), atmosphere, and the sun. That’s the simplest possible model. If you want a more realistic model you need to add many more bodies, different layers of atmosphere, water, and earth for a start, and you need to consider different transport systems etc. But that doesn’t mean the simple model isn’t useful. Not least, because it demonstrates why you are wrong. It’s entirely possible for a cooler atmosphere to cause the surface to warm, and energy is not “trapped” resulting in an infinite temperature rise.

“Define the 2nd bodies processes and tell us how that effects changes to the 3rd body (atmosphere).”

I really can’t tell if you are suffering from senile dementia or just never read what I say. I went through the maths months ago describing the greenhouse effect, I’ve repeated it several times here.

Reply to  Bellman
June 20, 2026 1:00 pm

Increased in the body of interest. The ocean in this case. Of course with the ocean there are complex interactions and hidden depths, but all of this is a distraction from the initial pint, which is your claim that a warmer atmosphere cannot warm the oceans.”

A warmer atmosphere that is *still* colder than the surface CAN NOT WARM THE SURFACE. The atmosphere is *NOT* a heat source. The only heat source in the system is the sun!

Reply to  Tim Gorman
June 20, 2026 4:59 pm

A warmer atmosphere that is *still* colder than the surface CAN NOT WARM THE SURFACE.

Just keep repeating that to yourself.

The atmosphere is *NOT* a heat source.

That depends on how you define “heat source” but it isn’t relevant. You can look at the way greenhouse gases work in multiple ways. But what they are is an additional source of energy. With no atmosphere the sky is heat-less at night, and during the day the only heat is from the sun.

With an atmosphere, you can go out at night without instantly freezing, and you can survive during the day even in the shade.

Reply to  Bellman
June 19, 2026 10:51 am

Alternatively it means the average rate of energy in in has to equal the average rate if energy out.

Funny how you and your cohort criticize Dr. Frank for using a time label and yet you do the same! Why have you changed your mind about making an average have a common time label?

The simple point you keep missing is that if you reduce the rate of energy out, then that means more energy is received over 24 hours and hence you get slightly warmer. 

How does reducing energy out cause more energy to be received? Are you back to claiming that the atmosphere is warmer than the surface?

Reply to  Jim Gorman
June 19, 2026 11:22 am

“Funny how you and your cohort criticize Dr. Frank for using a time label and yet you do the same! ‘

No idea what your point is. The only thing I’ve criticized Frank for was not explaining why he doesn’t reduce uncertainty for an average, and claiming that standard deviations can be negative.

And I’m not sure why you think any criticism about Frank means you can’t point out that total energy transfer will depend on time.

“How does reducing energy out cause more energy to be received? ”

I should have said “net” energy received. It should be obvious from context.

” Are you back to claiming that the atmosphere is warmer than the surface?”

When have I ever claimed that? If the atmosphere was warmer than the ocean surface, there wouldn’t be any cooling at night.

Reply to  Jim Gorman
June 19, 2026 1:14 pm

How does reducing energy out cause more energy to be received? Are you back to claiming that the atmosphere is warmer than the surface?”

All you have to do is look up at the sky to see the flames from CO2 burning!

Reply to  Bellman
June 19, 2026 11:12 am

“And why do you think CO2 will rise indefinitely?”

It doesn’t have to rise. If CO2 “TRAPS” heat that is emitted by the earth then the amount of heat that is trapped will just keep on increasing. If every time the earth emits 100 joules, CO2 traps “x” joules of that 100 joules then for every increment of 100 joules emitted the trapped heat will go up by “x” joules. Sooner or later the amount trapped will be sufficient to scour the earth of all life.

The alternative is that CO2 does not “trap” heat. If CO2 does not trap heat then the models are all garbage. And your claim that trapped heat causes the temperature to go up is garbage as well.

You can’t seem to keep it in mind that as T goes up T^4 goes up more. That’s a negative feedback that prevents accumulation of heat. You lose more heat than you gain as temperature goes up. This is Planck’s “compensation” for reflected heat.

Now, come back and tell us that CO2 is not a reflector of heat but a source of heat.

SHOW THE MATH!

If the sun inputs 100 joules and the earth emits 100 joules and CO2 reflects back 50 joules the earth LOSES A NET 50 joules. The loss of energy is COOLING. In the next instant the earth will re-emit that reflected 50 joules PLUS the 100 joules the sun sent in. This repeats over and over and over. Reflected heat is *NOT* additional heat in the system.

Why is this so hard for you and climate science to understand? THE LOSS OF ENERGY IS COOLING, NOT WARMING.

Reply to  Tim Gorman
June 19, 2026 12:01 pm

Try reading the thread. I was commenting on Jim’s statement

then the assumption is that as long as CO2 grows so will temperature.

“If CO2 “TRAPS” heat that is emitted by the earth then the amount of heat that is trapped will just keep on increasing.”

You are either incredibly dumb, with a “b”, or deliberately trolling. I’ve explained why you are wrong multiple times.

“If every time the earth emits 100 joules, CO2 traps “x” joules of that 100 joules then for every increment of 100 joules emitted the trapped heat will go up by “x” joules”.

And so will emit more energy. Surely even you can understand that.

If 50% of the emissions are returned, then by the time temperatures have ridden enough to emit twice as much energy, the 50% that escapes will equal energy in. That’s the point when you have reached equilibrium and the temperature stops rising.

Using the T⁴ rule, a doubling of energy out occured when temperature has increased by about 20%.

The observed greenhouse effect is less than that given that the greenhouse gases are not completely opaque.

“If the sun inputs 100 joules and the earth emits 100 joules and CO2 reflects back 50 joules the earth LOSES A NET 50 joules. ”

What sort of alternative maths do they teach you? No. If you get 100 of something, lose 100, but then get back 50, you have not lost a net 50. 100 -100 + 50 does not equal -50.

“In the next instant the earth will re-emit that reflected 50 joules PLUS the 100 joules the sun sent in.”

You are really confused. First you say you’ve lost 50, now you say you’ve got an extra 50 to give away.

“This repeats over and over and over.”

Did they teach you about geometric series?

“Why is this so hard for you and climate science to understand? ”

Because it’s nonsense. And yes I have a hard time understanding why so much of the world prefers to believe in it’s own alternative math, then actually learn something

Reply to  Bellman
June 20, 2026 1:01 am

And so will emit more energy. Surely even you can understand that.”

So the earth emits everything it gets? If it gets more energy then it will emit it? No trapped heat left? It all gets emitted?

What in Pete’s name do you think we’ve been telling you? The only important issue is *when* does it all get emitted?

“If 50% of the emissions are returned, then by the time temperatures have ridden enough to emit twice as much energy,”

Are you reading anything in the thread? The temperature of the earth doesn’t have to rise in order to emit more energy, if the emission interval is longer than the receive interval. The energy emitted is F * t_x where “t_x” is the interval.

Think about it. The earth emits 100 joules and the temperature falls. It gets back 50 joules and what happens? THE EARTH STILL IS DOWN 50 JOULES! The temperature is LESS than when you start the sequence! If the temperature is LESS, then the earth has *cooled*, it hasn’t warmed!

As a reflector, CO2 cannot warm the source. It can only slow its cooling.

Nor does it have to see a temperature rise in order to emit more ENERGY. I’ve given you the formula before.

Energy ∝ τ

If you need me to, I can derive this for you by integrating the flux over time

If you make “τ”, the decay rate, longer (i.e. slower cooling) then the energy emitted over time goes UP!

I know its counter-intuitive but a slower cooling rate actually emits MORE energy per time interval!

I gave the graph showing this to you? Did you just ignore it?

Using the T⁴ rule, a doubling of energy out occured when temperature has increased by about 20%.”

Show me the math on how much energy gets emitted! You are *still* stuck in trying to make everything happen all at the same time like with a black body. Energy loss is a TIME FUNCTION in a non-black body. Where is your math showing the time intervals and the decay rates for the earth?

  1. The earth is not isothermal. It is not a black body.
  2. The earth is not homogenous. It is not a black body.
  3. The earth is a heat sink. It is not a black body

You are as bad at this as climate science. Climate science says the earth’s emissivity is not 1 – meaning it is *not* a black body. Then they turn around and assume that it is a black body so that flux-in and flux-out balances! If flux-in and flux-out balance then you have equilibrium, the temperatures of the two bodies ARE THE SAME! Is the earth the same temperature as the sun?

Have you figured out how to get (Energy ∝ τ) yet?

Reply to  Tim Gorman
June 20, 2026 4:40 am

“So the earth emits everything it gets? ”

This shouldn’t be so difficult to understand, even for you. You know all the individual parts, you just seem incapable of putting them together.

How much the earth emits depends on how hot it is. It’s proportional to the internal energy as you keep saying.

If you increase the the amount if energy the earth receives there will be an imbalance which results in the earth’s internal energy increasing. It gets hotter.

If it gets hotter the amount of energy it emmits will increase. See the first point

At some point the temperature has increased enough that the amount of energy emitted equals the amount if energy received. At that point you have a new equalibrium and the earth stops warming.

Reply to  Bellman
June 20, 2026 5:06 am

“What in Pete’s name do you think we’ve been telling you?”

I’ve lost interest in trying to fathom what you think you are telling me, because it:s always nonsense, and inconsistent.

“The temperature of the earth doesn’t have to rise in order to emit more energy, if the emission interval is longer than the receive interval. The energy emitted is F * t_x where “t_x” is the interval. ”

See, complete gibberish.

“The earth emits 100 joules and the temperature falls. It gets back 50 joules and what happens?”

The sun starts shining.

“As a reflector, CO2 cannot warm the source. It can only slow its cooling.”

Unless the sun is also shining.

“Nor does it have to see a temperature rise in order to emit more ENERGY. I’ve given you the formula before.

Energy ∝ τ”

Don’t just keep repeating this in allcaps. Explain why the earth is emitting more energy despite the internal energy not increasing.

“I know its counter-intuitive but a slower cooling rate actually emits MORE energy per time interval!”

Counter-intuitive and meaningless. The only reason you are emitting more energy is because you are receiving more energy. All you are saying is that if the sun didn’t exist the earth would quickly cool down to temperature of space, and having an atmosphere only means it takes slightly longer.

“Show me the math on how much energy gets emitted! ”

I’ve done this numerous times including in this comment section. You will just ignore it and demand I show it again.

“You are *still* stuck in trying to make everything happen all at the same time like with a black body.”

Nope. This has nothing to do with how long it takes, just that you can see what the final equilibrium will be.

” Energy loss is a TIME FUNCTION in a non-black body.”

I’m not sure you understand what a black body is. It’s a time function for any object. But as always you keep throwing up the time element as a distraction. It makes no difference to the final result. And you keep failing to provide your own calculations based on the time function, and keep ignoring the sun.

“Is the earth the same temperature as the sun?”

No, because of the greenhouse effect.

To be clear, you don’t actually think the earth will be literally the same temperature as the sun. Without the greenhouse effect it would be the same temperature as the equivalent radiation it receives from the sun. The sun is very hot but far away.

Reply to  Bellman
June 20, 2026 12:39 pm

tpg: “The energy emitted is F * t_x”

See, complete gibberish.”

Unfreakingbelievable! F IS A RATE OVER TIME! Energy is an AMOUNT.

How do you get an amount from a rate over time if you don’t multiply the rate by the time involved?

The sun starts shining.”

So what? The issue is how does CO2 cause the temperature of the earth to increase?

CO2 is *NOT* a source of heat. The sun is a source of heat. Can you not tell the difference?

tpg: ““As a reflector, CO2 cannot warm the source. It can only slow its cooling.”

Unless the sun is also shining.

That means the SUN increases the earth’s temperature, not CO2! How does CO2 as a reflector increase the temperature of the earth? How does it manage to reverse entropy when it is not a source of energy?

Does CO2 add heat to the earth when the sun is not shining?

Don’t just keep repeating this in allcaps. Explain why the earth is emitting more energy despite the internal energy not increasing.”

You don’t even understand the relationship!

(Energy ∝ τ) is the relationship for the energy the earth emits over time! The energy it emits over time is related to the decay factor. The slower the decay the more energy it emits. If CO2 slows cooling that means the decay factor gets slower. Thus the earth actually emits MORE energy per time interval!

Q(t) = Q0 * e^-(t/τ)

F(t) = dQ/dt = Q0/τ * e^-(t/τ)

Can you integrate F(t) to find the amount of energy emitted over time or must I do that simple task for you?

You asked me to explain the math – do you need to have an explanation of how to integrate this function?

Ah, hell. I may just as well do it for you.

Energy_total = ∫ F(t) dt ==> ∫ F0 * e^-(t/τ) dt, from 0 to k ==>

-F0 * τ * e^-(t/τ) evaluated from 0 to k

at 0: -F0 * τ * e^0 = -F0 * τ

at k: -F0 * τ * e^-(k/τ)

sum: F0 * [τ – τe^-k/τ), as τ ↑, F(t) ↑

This is the equation for the cooling of the earth. I’ll leave it to you to expand the equation to include the energy input from the sun (but I know you won’t do it). (hint: find F0(t))

Reply to  Tim Gorman
June 20, 2026 2:45 pm

“Unfreakingbelievable! F IS A RATE OVER TIME! Energy is an AMOUNT.”

You carefully cut out your exact quote I was calling gibberish.

The temperature of the earth doesn’t have to rise in order to emit more energy, if the emission interval is longer than the receive interval. The energy emitted is F * t_x where “t_x” is the interval.

I was questioning you claim that the temperature doesn’t have to rise in order to emit more. Your claim amounts to saying that a cold planet emits more over a century than a hot one does over a day. That might be correct but it’s irrelevant to discussing how temperature effects emissions and vice versa.

In response to me pointing out the sun shines Tim says “So what?”bwhichbexplains why he’s so confused. To him the sun shining is an irrelevance when explaining why the earth is warming rather than cooling.

“That means the SUN increases the earth’s temperature, not CO2! ”

More of Tim’s inability to consider more than one thing at a time. If the sun warms up the earth to a certain temperature it’s simply impossible that greenhouse gases could increase it further.

“(Energy ∝ τ) is the relationship for the energy the earth emits over time!”

Could you define your terms? Elsewhere you are using tau as the cooling constant now you are saying it’s the energy the earth emits I’ve time. That doesn’t make sense. The amount of energy emitted I’ve time depends on time.

“If CO2 slows cooling that means the decay factor gets slower. Thus the earth actually emits MORE energy per time interval!”

Yes because it’s hotter.

“You asked me to explain the math – do you need to have an explanation of how to integrate this function?”

What I’m asking for is a set of equations that justifies your claim. They need to be compatible with observations, whilst at the same time show how the world is constantly cooling, and greenhouse gases do not increase temperature. For extra marks they should show how the earth isn’t a frozen ice ball based only on input from the sun.

“This is the equation for the cooling of the earth.”

Now put some actual figures into the equation and tell me how cold you think the planet should now be, and how cold it will be a decade from now.

“I’ll leave it to you to expand the equation to include the energy input from the sun”

So when you say you’ve done all the math you mean apart from including the sun. Something of an important omission don’t you think?

Reply to  Bellman
June 20, 2026 5:50 am

How much the earth emits depends on how hot it is. It’s proportional to the internal energy as you keep saying.

Your system analysis is faulty. One big factor you miss is water vapor. Liquid water evaporates into water vapor using latent heat. Latent heat does not increase temperature, it increases enthalpy. Enthalpy is the total energy in a parcel. It includes “sensible heat+ latent heat”. That latent heat is released via radiation at altitude (except for dew) having never raised the temperature at the surface.

You keep wanting to define total energy by using temperature as a proxy for total heat. False assumption. Temperature is NOT a proxy for total heat except maybe over deserts and the poles. It is why climate science is going to be hung on its own petard. Models that predict only temperature instead of enthalpy miss one of the largest system components of the atmosphere, just like you.

Your simple explanations also miss the fact that the surface is a heat sink. Yes, a heat sink’s temperature will rise as it stores heat. But most of the heat on land disappears into the subsoil and only affects surface temperature at night. The oceans disperse absorbed heat through various mechanisms, but the end result is that water at depth does not radiate heat into the atmosphere.

Your lack of any math to support your assertions has gotten old. If you can’t even show basic equations and the conditions required to use them, then you are simply making stuff up.

Reply to  Jim Gorman
June 20, 2026 7:56 am

“Your system analysis is faulty. ”

This is typical Gorman distraction. Start with simplistic arguments for why something can’t happen. Then when you point out in simple terms why they are wrong, they start throwing up all manor of extra complexities andinsist you do a full system analysis. At no point will they do such an analysis, it’s all just hand waving.

“You keep wanting to define total energy by using temperature as a proxy for total heat.”

Nope.

“Your simple explanations also miss the fact that the surface is a heat sink.”

Nope. That’s the whole point.

” The oceans disperse absorbed heat through various mechanisms, but the end result is that water at depth does not radiate heat into the atmosphere. ”

I’m old enough to remember such a claim as being dismissed as “the oceans ate the warming”.

All you are saying is more heat is being stored than is observable from surface measurements. It’s the same with latent heat. Some of the extra energy caused by global warming is going into things like melting ice caps and evaporating water. I’m not sure why you think that’s a winning argument. It just suggests global warming might be worse than observed. What happens when all the ice is melted? Suddenly all that extra heat is no longer latent, but sensible.

It’s the same with heat going into the oceans. Surface temperatures don’t show this extra energy, but what happens when that extra enthalpy suddenly reaches the surface?

“Your lack of any math to support your assertions has gotten old.”

As are your snide ad homs. I am not going to do a full simulation of global climate – I’ll leave that to the experts. All I can do is explain in the simplest way possible why your simplistic claimes are wrong. And then point out the hypocrisy of you continuously demand I provide detailed models, whilst you refuse to do that yourself. You are the ones saying climate scientists are completely wrong – the onus is on you to show your workings.

bdgwx
Reply to  Bellman
June 20, 2026 10:38 am

This is typical Gorman distraction. Start with simplistic arguments for why something can’t happen. Then when you point out in simple terms why they are wrong, they start throwing up all manor of extra complexities andinsist you do a full system analysis. At no point will they do such an analysis, it’s all just hand waving.

Yep. As I say frequently if you cannot understand simple idealized scenarios you will never be able to grasp vastly more complex scenarios.

If the Gorman’s are incapable of even understanding that closing the cold door of a kitchen oven will cause the hot inside to warm further than they will not be any more capable of understanding other scenarios.

Reply to  Bellman
June 20, 2026 12:58 pm

All you are saying is more heat is being stored than is observable from surface measurements. It’s the same with latent heat”

IT’S WHY YOU WILL *NEVER* HAVE FLUX-IN EQUAL FLUX-OUT!

Radiative balance is a farce. It’s typical climate science garbage!

Reply to  Bellman
June 20, 2026 4:27 pm

 It just suggests global warming might be worse than observed. 

You just did it again. Heat and global warming as determined by temperature are not equivalent.

but what happens when that extra enthalpy suddenly reaches the surface?

Liquid water does not have “enthalpy”. It has a specific heat. As a higher temperature molecule reaches the surface it will evaporate, a phase change from liquid to gas. This requires heat to be added not released.

What happens when all the ice is melted? Suddenly all that extra heat is no longer latent, but sensible.

You really need to get out more. Ice melts from the ADDITION of heat. As it melts from added heat, ice maintains a constant temperature until the ice is gone. You should have learned this in high school chemistry. The specific term applied to the solid-to-liquid phase change is the latent heat of fusion.

From: What Is the Latent Heat of Ice? – Biology Insights

When heat energy is applied to ice at 0∘C, the energy does not increase molecular vibration or temperature. Instead, the absorbed latent heat breaks a significant portion of these hydrogen bonds. Breaking these bonds allows the molecules to slip past one another, collapsing the rigid lattice into the liquid state.



Reply to  Bellman
June 20, 2026 1:22 am

You are really confused. First you say you’ve lost 50, now you say you’ve got an extra 50 to give away.”

Huh? 100 – 50 ≠ 50?

Did they teach you about geometric series?”

It’s pretty damn obvious they haven’t taught you how to integrate a simple exponential decay. Go find an integral calculator on the internet and figure it out!

“Because it’s nonsense.”

SHOW ME THE MATH BEHIND YOUR ASSERTION OF NONSENSE! Otherwise this is just one more instance of you using the Argument by Dismissal argumentative fallacy.

(Energy ∝ τ) is *NOT* nonsense. Slow the cooling rate and you emit MORE energy over time!

And yes I have a hard time understanding why so much of the world prefers to believe in it’s own alternative math,”

WHAT alternative math? SHOW ME THE MATH.

Refute (Energy ∝ τ)! I dare you!
Refute that thermodynamic equilibrium is heat-in = heat-out and not flux-in = flux-out. I dare you!



Reply to  Tim Gorman
June 20, 2026 4:30 am

Stop asking me to explain the maths whilst refusing to do it yourself. Show how quickly you think the earth will cool, and then explain why you are ignoring the sun.

And stop with these childish strawman arguments such as, “Refute (Energy ∝ τ)! I dare you!”. No one is denying that.

Reply to  Bellman
June 20, 2026 11:43 am

Stop asking me to explain the maths whilst refusing to do it yourself.”

Judas H. Priest!

I’ve given you the math! The fact that you can’t understand it doesn’t mean I didn’t give it to you!

What in Pete’s name do you think the equation F(t) = F0 * e^-(t/τ) is?

What do you think the equation Q(t) = Q0 * e^-(t/τ) is?

What do you think F(t) = Q0/τ * e^-(t/τ) is?

what do you think (E ∝ τ) is?

I’ll give you another one: C(dT/dt) = [ S * (1-α) ] / 4 – σT^4, where α is the albedo.

Have you got a clue as to what “C” is?

I’ve told you that τ, the decay factor, is different for conduction, convection, and radiation and challenged you to figure out how to determine each. YOU FAILED!

Don’t accuse me of not explaining the math. It’s *simple* math.

e^-(t/τ) is about as simple as it gets. What explanation for it do you need?

What do you think Q0/τ * e^-(t/τ) is? It’s about as simple as it gets. It shouldn’t need any explanation!

100_out – 50_back = 50_out is about as simple as it gets. What explanation do you need for it?

And stop with these childish strawman arguments such as, “Refute (Energy ∝ τ)! I dare you!”. No one is denying that.”

YOU ARE DENYING IT!!! Every time you say a colder object can raise the temperature of a warmer object you are denying it! (Energy ∝ τ) is a direct result from Q(t) = Q0 * e^-(t/τ)!

Your only refutation is “but it’s the sun!”. When the sun doesn’t control the cooling of the earth! The emissions from the earth and the reflection of the earth’s emissions are what control the cooling of the earth!

My guess is that you don’t even understand that Q0 is *not* a constant, at least during the day. It’s Q0(t) Q_initial + ∫ S cos(θ)cos(ⱷ)dt. At night the integral = 0 and Q0(t) is the value at sunset.

You and bdgwx and Nick *still* haven’t tumbled to the fact that the thermodynamic equations controlling the heat transfer associated with the earth are TIME FUNCTIONS! And those time functions are associated with conduction, convection, and radiation.

FLUX-IN AND FLUX-OUT WILL NEVER BALANCE. They simply can’t, at least as long as the earth is not as hot as the sun!

The *only* thing that can balance are Joules-in and Joules-out and those must be evaluated over a long enough period to cover all possible cycles involved in the biosphere, including heat transport through ocean currents which can take literally *years* to move heat from the ocean to the atmosphere.

And the bottom line is that cold objects cannot add heat to warmer objects via radiation. That’s a violation of entropy unless work is involved. And there isn’t any ancient God somewhere in the solar system “fixing” that truism.

Reply to  Tim Gorman
June 20, 2026 2:15 pm

“What in Pete’s name do you think the equation F(t) = F0 * e^-(t/τ) is?”

It’s the equation for a an object cooling when against an object at 0K. It does not work if there is an additional heat source such as the sun.

It’s also meaningless unless you specify what value the cooling constant “τ”.
Unless you know that you have no idea how quickly the earth is cooling.

Lots if other similar equations ignored because they are useless unless you show how they work in the real world, which has a sun and an atmosphere.

“100_out – 50_back = 50_out is about as simple as it gets. ”

So simple it ignores the sun.

Try using this in a real world system. Say the earth is losing 50 joules per square meter a second. How long before it becomes uninhabital. Then explain why that hadn’t happened yet. Hint, it may have something to do with a hot yellow object.

“YOU ARE DENYING IT!!! ”

Nope. I’m not denying that if there was no sun the earth would cool like that. It’s just that I don’t deny the sun exists.

“Your only refutation is “but it’s the sun!”. ”

So why do you never address that cer obvious refutation of your claim?

“When the sun doesn’t control the cooling of the earth! The emissions from the earth and the reflection of the earth’s emissions are what control the cooling of the earth! ”

It’s as if you are incapable of considering two things as once. If the sun doesn’t cause the planet to cool it’s warming effects can be ignored.

Reply to  Bellman
June 20, 2026 9:07 pm

You really haven’t thought this through! There is no requirement that Tmax has to rise in a regulated system. That is why the Death Valley record hasn’t been broken in 113 years. During the Cretaceous, there were crocodiles and other reptiles living at polar latitudes. If the temperatures were the same or nearly the same over all the land, there really was little seasonality and the distinction between Tmin and Tmax was unimportant.

Reply to  Clyde Spencer
June 21, 2026 5:08 am

“You really haven’t thought this through!”

It wasn’t a thesis. Just an off the cuff remark. If you are accepting that the minimum temperature is rising, and you are adding the same amount of energy during the day time then, as a first approximation, I would expect maximums to also rise simply because the daytime warming is starting from a higher point.

“That is why the Death Valley record hasn’t been broken in 113 years.”

A single record doesn’t tell you about the trend and isn’t there some dispute about that record?

Reply to  Bellman
June 21, 2026 9:06 am

 I would expect maximums to also rise simply because the daytime warming is starting from a higher point.”

You *really* don’t get that Tmax is limited by T^4, do you?

When joules_out = joules_in you have reached Tmax. It doesn’t matter one jot, tittle, or iota what the starting temperature was.

This is no different than when your air conditioner kicks in when the house temperature reaches a certain threshold. It is a *regulated* system just like Clyde said.

The regulator in this case is joules-out.

One more time, there is a REASON why Tmax doesn’t occur at sunset!

Do you have even the faintest glimmer of understanding as to why that is?

Reply to  Tim Gorman
June 21, 2026 10:10 am

“You *really* don’t get that Tmax is limited by T^4, do you?”

I don’t get it because it’s your usual gibberish. Energy out is proportional to T⁴. That doesn’t limit anything. As temperature increases so to does the energy out and that stops when energy out equals energy in. If you increase energy in temperature will rise. Double energy in and temperature rises by 19%. There is no limit just diminishing returns.

“When joules_out = joules_in you have reached Tmax. It doesn’t matter one jot, tittle, or iota what the starting temperature was. ”

Unless the temperature of the atmosphere is contributing to joules_in.

“Do you have even the faintest glimmer of understanding as to why that is?”

Yes.

Reply to  Bellman
June 21, 2026 11:31 am

Unless the temperature of the atmosphere is contributing to joules_in.

The atmosphere does not contribute to joules_in. Only the sun provides energy into the system.

Did you not read what I said about the law of conservation of energy? It is not valid with a source in an isolated system. You must write the equations taking into account a continuing addition to the isolated system.

You keep yapping about trends over days. That is weather and is caused by the variance in all the variables involved. A simple statement that tomorrow is warmer than today has no thermodynamic significance unless you have accounted for all the different variables and their values.

Reply to  Jim Gorman
June 21, 2026 12:15 pm

“The atmosphere does not contribute to joules_in. Only the sun provides energy into the system. ”

Yet you keep quoting that Planck section that says that all objects radiate heat and don’t care about the temperature of other objects. A 100°C ball will radiate the same amount of energy to a 1000°C as it does to a 100°C one. Are you now disagreeing with Planck, or do you just not understanding what he said.

“Did you not read what I said about the law of conservation of energy?”

No. I don’t think we’ve mentioned conservation of energy and there’s so much nonsense in these comments for me to follow every one.

“You keep yapping about trends over days. ”

Only to illustrate who’s slower cooling can result in warming. This isn’t really about any current warming trend. It’s whether the surface is warmer because of the presence of greenhouse gases.

Reply to  Bellman
June 19, 2026 4:47 am

You keep making this dumb “slower cooling” chestnut, without understanding that slower cooling results in warming.”

I’m with Andy! Did you read what you wrote?

Warming requires gaining more joules, a state function. How does losing fewer joules over time result in gaining joules?

Energy is measured in JOULES, not in joules/sec-m^2.

 If the sun is warming the oceans by the same amount each day, but each night the oceans are cooling slower, what happens to the temperature?”

The ocean LOSES HEAT 24 HOURS PER DAY. It loses heat 7 days per week. It loses heat 30/31 days/month. It loses heat 12 months/year. Right on up to 1000yrs/millennia.

The ocean only gains heat 12 hours per day. 7days/wk. 30/31days/month. Right up to 1000yrs/millennia.

Do you see the difference? 24hrs vs 12 hrs?

Radiative heat loss is derived from an exponential decay. Heat loss in the integral of T^4 over the appropriate time interval.

What happens to the heat loss of the slope of the exponential decay is made less?

The heat loss equation is of the form F(t) = F0 * e^-(t/τ).

If τ, the time constant, gets larger due to slower cooling then the integral of the function INCREASES. The integral of that equation is the amount of heat lost over time “t”.

Energy loss, E, is proportional to τ: E ∝ τ

Double τ and you double the heat loss.

HAS ANYONE IN CLIMATE SCIENCE ACTUALLY STUDIED GENERAL SCIENCE let alone algebra and calculus?

Reply to  Tim Gorman
June 19, 2026 5:21 am

“Did you read what you wrote?”

Yes. Did you understand what I wrote? (Rhetorical question. Of course you didn’t.)

“How does losing fewer joules over time result in gaining joules?”

Because you are gaining more jules then you are losing.

“Energy is measured in JOULES, not in joules/sec-m^2. ”

You don’t say.

“The ocean LOSES HEAT 24 HOURS PER DAY.”

Yet, still it warms.

“The ocean only gains heat 12 hours per day.”

Simplistic but yes. That’s the point. You accept that they are not continuously cooling.

“Do you see the difference? 24hrs vs 12 hrs?”

12 hours is a different period of time than 24 hours. Another brilliant truism from you. Just what is your point?

“What happens to the heat loss of the slope of the exponential decay is made less? ”

You have less heat loss. Hence your final temperature is higher than it would have been without the slow down.

Everything you argue is just playing with words to distract from that obvious point.

“If τ, the time constant, gets larger due to slower cooling then the integral of the function INCREASES. ”

Why is t increasing? Slowing down the rate of decrease doesn’t mean you automatically get more time to reach the same destination. Unless slowing down the rate of decrease also slows the Earth’s rotation, you will still find the sun rises before you reach the same temperature.

“Double τ and you double the heat loss. ”

More words of wisdom. If only you could figure out what that implies.

Reply to  Bellman
June 19, 2026 5:45 am

Because you are gaining more jules then you are losing.”

No, because losing joules is related to T^4, not T.

“Yet, still it warms.”

Because the T^4 boundary condition hasn’t been reached yet. And actually, the ocean warming is not just a T^4 function, it also has limits set by evaporation.

Why do you think the ocean doesn’t boil every day? That would be the conclusion you are postulating.

Simplistic but yes. That’s the point. You accept that they are not continuously cooling.”

Unfreaking believable! As long as the temperature is larger than 0K, COOLING HAPPENS. Again, F(t) = F0 * e^-(t/τ).

Do you have the slightest understanding of the factors involved in what makes up”τ” ?

12 hours is a different period of time than 24 hours. Another brilliant truism from you. Just what is your point?”

What’s my point? That in the function F(t) = F0 * e^-(t/τ) that “t” is 24 hours!

Joules-in = Flux-in * 12hrs where flux-in is a constant

Joules-out = Q(t) = Q0 * e^-(t/τ) where τ is related to (1/T^3) and t is 24 hours

Joules-out is not a trivial function since it is dependent on Joules-in for part of the period.

You have less heat loss.”

No, you have MORE heat loss. The heat loss is the integral of the temperature curve. If the slope of the decay is less then you get MORE heat loss. See the attached primitive image.

 Hence your final temperature is higher than it would have been without the slow down.”

And since heat loss is related to temperature what does that imply for heat loss? You are trying to equate temperature to energy. They are *NOT* the same.

Why is t increasing?”

Read it again. “t” is not increasing, “τ” is increasing. What is “τ” dependent on?

More words of wisdom. If only you could figure out what that implies.”

ROFL! You can’t even distinguish “t” from “τ” let alone what τ” depends on.

Once again, we are back to you not having the knowledge to even discuss thermodynamics. But here you are, trying to lecture everyone that slower cooling means temperature goes up.

exponential_decay_fast_vs_slow_cooling
Robert Cutler
Reply to  Tim Gorman
June 19, 2026 8:11 am

“HAS ANYONE IN CLIMATE SCIENCE ACTUALLY STUDIED GENERAL SCIENCE let alone algebra and calculus?”

I don’t think so.

Tim, to build on many of the excellent points you’ve made, the oceans integrate solar activity, but the sea surface and atmosphere have a different response.

In this plot, the red line is the frequency response computed between temperature and sunspots. It has two distinct regions separated by a “Schwabe” notch near the middle. I also call it the Jupiter notch because both have similar 11-year periods, On the whole, the 11-year variations in TSI don’t affect climate.

The black-dashed line is the 1/freq response of an ideal integrator.

The blue-dashed line is also a frequency response computed between predicted temperature from my model (last plot, this post), and sunspots, which are the input to the model. The core model is a 99-year moving average of sunspot data. It performs four different functions, one of which is to model Earth’s integral response.

The top and middle panels have minimal averaging, but different resolutions due to the use of different windowing function.

The bottom panels have more coherent averaging which attenuates uncorrelated signals, which in this case is likely weather. I haven’t tried to model weather, so the blue-dashed line follows the ideal integrator response.

comment image

An ideal integrator also has a constant 90° phase response. For a 60-year cycle, this would imply a 15-year delay. I haven’t studied ENSO, but some of the major events appear to be separated by 20-year intervals. That’s easy to see in this plot where I plot Jupiter-Saturn conjunctions delayed by 15 years.

comment image

The 2023 spike seems out of place, which is why I favor the Hunga-Tonga explanation. However, based on the 3560-year repetition in climate, we may eventually learn that these spikes are the derivative of climate change (not the cause). Ice core data hints at the possibility of significant cool period ahead.

comment image

Sunspot data predicts slight cooling, or perhaps the optimum of our current warm period. Here, the filtered sunspot data is shifted forward 13 years.

comment image

Robert Cutler
Reply to  Robert Cutler
June 19, 2026 10:34 am

Andy, ice-core data shows a cold period around 850AD. This would be consistent with the low number of Los Niños in Figure 1.

Reply to  Robert Cutler
June 19, 2026 1:02 pm

I would have to study your graphs in detail for sure. At first glance, the matches aren’t perfect but they are very similar (the blue and black lines in the first graph).

The big question here is WHY HASN’T CLIMATE SCIENCE STUDIED THIS AND GENERATED HYPOTHESES ON IT?

The oceans will never be perfect integrators of anything, there are too many other factors at play. Clouds, winds, etc will always generate natural variability. But the integrator function doesn’t have to be perfect to be a useful metric.

Robert Cutler
Reply to  Tim Gorman
June 19, 2026 2:15 pm

This important match is the predicted vs actual (red vs blue). The blue is a moving average (boxcar or rect).

The MA filter has a known sin(freq)/(freq) response, so 1/freq plus nulls (from the numerator). The match to the black line will never be perfect because the sunspot signal doesn’t have uniform power-spectral density.

Reply to  Bellman
June 22, 2026 9:32 am

Yes, but it is an external source, not an internal source.

Reply to  Tim Gorman
June 19, 2026 8:40 am

No the heat flux at the surface will tend to balance instantaneously, when heat input is greater than heat loss the surface will warm when the reverse is true the surface will cool, if the temperature is steady then they are balanced.

Reply to  Nick Stokes
June 18, 2026 5:29 am

I’ve explained the proper scientific way of analysing it,

I don’t think so.

The basic process is that trade winds push warm surface water to the west where it accumulates. The thermocline is very deep in the western Pacific and shallow in the east. Then when trade winds decrease or reverse, the warm water essentially sloshes eastward. The western Pacific cools and the eastern Pacific warms.

Quite honestly, the atmosphere has little to do with any of this. Greenhouse theory just doesn’t apply. Any effects are weather related, and not climate change related.

The example of La Niña. That is when the equatorial Pacific ocean becomes colder. Then the air becomes colder. How can that be if the air’s warmth does not pass into the ocean?

You need to rethink this process. Somehow I think reduced radiation from the surface has something to do with this.

Reply to  Nick Stokes
June 22, 2026 6:22 am

Our bodies *generate* heat. The oceans do not. So while our bodies may get “warmed by reducing cooling,” as in the “adding blankets” analogy, just as ridiculous, the ocean does not.

As with most so-called “climate science,” you’ve got the cart before the horse, the tail wagging the dog.

Still waiting for the explanation for the GLACIATION WITH TEN TIMES TODAY’S ATMOSPHERIC CO2. Where was CO2’s “climate driving power” then?! It’s the same planet orbiting the same star and the basic physics haven’t changed.

Reply to  Andy May
June 17, 2026 4:34 pm

If you graph the last 50 years of ENSO3.4 cycles, you see very clearly that the “warming” El Nino phase greatly outweighs the La Nino phase.

El-Nino-warms-more-than-La-Nina-cools
Nick Stokes
Reply to  bnice2000
June 18, 2026 12:23 am

If you graph the last 50 years of ENSO3.4 cycles”

That graph is explicitly of El Niño. It is not designed to show La Niña.

Reply to  Andy May
June 18, 2026 7:42 am

Nick is right, you are mistaken Andy. That graph is only of the “2-year history of sea surface temperatures in the Niño-3.4 region of the tropical Pacific for all events evolving into El Niño since 1950″ (my bolding).

Nick Stokes
Reply to  Andy May
June 18, 2026 1:34 pm

Yes. But the graph just shows the index for periods fter the start of an El Niño. It does not show La Niña times, except where they are closely adjacent.

Nick Stokes
Reply to  Andy May
June 17, 2026 4:59 pm

Andy
“The data and many others disagree with you on this point.”
You don’t say who they are. But you quote just one source, Wong and Minnett. Now they are not clear thinkers on heat transfer,, but they are not saying IR can’t warm the ocean. In fact they claim, in they abstract, to have shown how it does:
“. The hypothesis is that given the heat lost through the air-sea interface
is controlled by the TSL, the TSL adjusts in response to variations in incident IR radiation to maintain the surface heat loss. This modulates the flow of heat from below and hence controls upper ocean heat content.
This hypothesis is tested using the increase in incoming longwave radiation from clouds and analyzing vertical temperature profiles in the TSL retrieved from sea-surface emission spectra. The additional energy from the absorption of increasing IR radiation adjusts the curvature of the TSL such that the upward conduction of heat from the bulk of the ocean into the TSL is reduced. The additional energy absorbed within the TSL supports more of the surface heat loss. Thus, more heat beneath the TSL is retained leading to the observed increase in upper ocean heat content.

Long-winded, but I’ve bolded the essential conclusion.

Nick Stokes
Reply to  Andy May
June 18, 2026 12:35 am

so heat flow is almost always away from the ocean and into the atmosphere.”

Yes, balancing the heat flow in via SW. But the issues is change in heat flow. An increase in down IR changes the net heat outflow. Heat from SW accumulates until the ocean is warm enough to emit the previous flux plus the added IR. Minett and Wong explain all this.

I looked at your site for references. No list in that link. an’t you just give a name of a proper scientist who agrees that warm air can’t warm the ocean?

Nick Stokes
Reply to  Andy May
June 18, 2026 1:50 pm

Fairall (2026) discusses in detail the temperature profile in the skin layer. But it does not say anywhere that the atmosphere (or down IR) cannot warm the ocean.

bdgwx
Reply to  Andy May
June 19, 2026 2:25 pm

Only the sun can warm the ocean. I don’t know how often I will have to point out the obvious to you.

If by ‘warm” you mean to cause ΔT > 0 then neither [Fairall et al. 2026] nor [Wong & Minnett 2018] say that.

Reply to  Nick Stokes
June 18, 2026 5:42 am

An increase in down IR changes the net heat outflow. 

You only have part of the process. Part of the problem s that H2O absorbs the IR into latent heat which doesn’t raise the temperature, i.e., no heat. That causes additional evaporation cooling the surface.

The other part is that hot water rises. Diffusion downward is almost impossible. Water can and does radiate IR. Heated water radiates more that cooler water. So any heating by IR at the surface is radiated quickly away with little change to the overall top several meters of water. The sun is the key here, not CO2 downward IR.

Nick Stokes
Reply to  Andy May
June 18, 2026 1:54 pm

It isn’t impossible, and they don’t say that. But it is unnecessary, because there is a strong diffusive upward flux, necessary to deliver the absorbed SW heat flux back to the surface and onwards. And other surface fluxes modulate that. Reducing it means more retained heat in the ocean – ie warming.

Reply to  Nick Stokes
June 18, 2026 5:01 pm

And other surface fluxes modulate that.

If water is a certain temperature from SW, it doesn’t “get hotter”. The depth of hot water may increase, but that is not warming. A deeper layer doesn’t heat itself to a higher temperature. You may claim that OHC increases, but not temperature. Thermodynamics is hard.

Reply to  Andy May
June 18, 2026 6:40 pm

It is frustrating teaching folks who have never spent time in a lab doing thermodynamic measurements.

Keep up the good work!

bdgwx
Reply to  Andy May
June 19, 2026 2:07 pm

The ocean is cooling, not warming.

The ocean is warming and has been for decades.

Or are you saying that the ocean is cooling at night? If so then why not discuss the what happens during the day and the net effect of both day and night?

Reply to  bdgwx
June 20, 2026 9:35 am

Of course it cools at night. Radiatively and conductivity, not to mention continuing evaporation. Both conduction and evaporation is not accounted for by strict radiation balance at the surface.

You can’t ignore that the surface is a heat sink and stores heat for a period of time.

You still haven’t shown the math determining which of these occurs.

Tₜ < Tₜ₋₁ Cooling
Tₜ = Tₜ₋₁ Equilibrium
Tₜ > Tₜ₋₁ Heating

Reply to  Nick Stokes
June 18, 2026 3:24 pm

The operative words here are: “net heat outflow”.

Net heat outflow MEANS COOLING. The cooling rate changes, that’s all. It doesn’t turn the net heat outflow into a “net heat inflow”.

Nick Stokes
Reply to  Tim Gorman
June 18, 2026 4:30 pm

No, because there is a very big inflow, which is thermalised sunlight. If heat outflow matches that, ocean temperature is steady. If more, cooling; if less, warming. When IR impinges on the surface, flux from below is diminished.

Reply to  Nick Stokes
June 18, 2026 5:05 pm

Agai, the ocean can’t warm itself. If sunlight heats water to a certain temperature, that is as warm as it will get. With evaporation and radiation it can cool, but it won’t get warmer by itself. To warm, there must be a hotter source somewhere. Where is that source?

Reply to  Andy May
June 19, 2026 6:26 am

Andy wrote:

“The first rule of holes: “When you find yourself in one, stop digging.””

That’s good advice, Andy. Now, what would you call this:

“Radiation is power.”
“Radiation by itself is not power.”

Would you call that a “hole”? I would. So would Aristotle. Would you like to try to dig yourself out of it?

Reply to  Andy May
June 20, 2026 4:25 pm

“What radiation is? This is ambiguous”

No, it’s really not.

“normally assumed to be energy”

It’s not “assumed”, Andy. That’s what it is. So why did you write the following claims?

“Radiation is a mechanism, it isn’t energy.”
and
“Radiation has power.” (please pardon my slight misquote earlier when I attributed to you the phrase “radiation is power”, which wasn’t exactly what you wrote, but definitely in the spirit)

“What hole”

This one:
“Radiation is […] energy”
versus
“It [radiation] isn’t energy”

Reply to  Andy May
June 25, 2026 7:30 am

Since you don’t appear to have any desire to dig yourself out of your self-contradictory hole, Andy, let me make it a bit more clear what I am “trying to say”, which is this:

You should probably stick to what you were taught, which is apparently some sort of petroleum engineering, and leave the radiation physics, science, and logic to the physicists. You obviously aren’t capable of grasping any of that – despite my best efforts to teach it to you.

Reply to  Nick Stokes
June 19, 2026 3:39 am

No, because there is a very big inflow, which is thermalised sunlight.”

So what is your point here exactly?

 If heat outflow matches that, ocean temperature is steady.”

But the heat outflow does *NOT* have to be the same as the heat inflow instantaneously. Different timeframes for the outflow and inflow mean the rates *can not* be the same. The total joules-in and total joules-out are what needs to balance, and this has to be totaled over the appropriate time interval.

“If heat outflow matches that, ocean temperature is steady. If more, cooling; if less, warming.”

You are *STILL* leaving out the TIME factor. I’ll fix it for you:

-If joules-out over time matches joules-in over time the ocean temperature will be steady.
-If joules-out over time is more than joules-in over time then cooling.
-if joules-out over time is less than joules-in over time then warming.

“When IR impinges on the surface, flux from below is diminished.”

So what? IR impinges on the surface 12 hours per day. Flux from below goes on for 24 hours per day. The flux does *NOT* have to match in order for there to be balance over a day. And this is just over a single day. The IR impinging on the surface changes over long time intervals. The flux from the ocean toward space changes over long time intervals. Those time intervals don’t have to be the same at any point in time. But you have to consider the joules-in and joules-out over those long time intervals in order to determine whether a heat balance exists or not.

The whole climate science idea of FLUX BALANCE is garbage. It’s the same kind of garbage that “all measurement uncertainty is random, Gaussian, and cancels” is.

And here you are, trying to say that the flux-in and flux-out should balance at any point in time.

Dave Burton
Reply to  Nick Stokes
June 24, 2026 8:39 pm

You are right, Nick. Wong and Minnett explicitly refute Andy’s contention that absorption of LW IR by the “skin layer” of the ocean does not warm the water beneath.

Reply to  Nick Stokes
June 17, 2026 7:29 pm

Of course a warm atmosphere can warm the ocean.

Dumb stuff endlessly repeated. The atmosphere is warmed by the ocean – which is warmed by the sun – not the other way round.
Or perhaps you believe that the ocean does indeed warm the atmosphere which warms the ocean which warms the atmosphere a bit more which warms the ocean a bit more until the planet eventually explodes?

Reply to  Mike
June 17, 2026 7:54 pm

He still doesn’t realize the ocean input warming from the sun is massively greater than the negligible IR effect at the ocean surface while the sun penetrates down around 600 HUNDRED feet which is nearly continuous 24/7.

LINK

Reply to  Sunsettommy
June 17, 2026 8:03 pm

Indeed. To my mind, the difference between solar warming and some immeasurable, theoretical back radiation is a beach ball to a grain of sand. But apparently it is still worth arguing about!

bdgwx
Reply to  Andy May
June 18, 2026 11:07 am

That doesn’t mean the atmosphere cannot be the cause of the ocean getting warmer. Remember the 1LOT ΔE = Ein – Eout and heat capacity ΔT = ΔE / (m*c). Therefore ΔT = (Ein – Eout) / (m*c). And when Ein > Eout then ΔT > 0. That happens when either ΔEin > 0 or ΔEout < 0 or ΔEin > ΔEout. For the microphysics of how this happens see [Wong & Minnett 2018].

Reply to  bdgwx
June 18, 2026 12:39 pm

Something you learn in engineering and most physical science lab classes is that you don’t just throw textbook equations around without satisfying all the conditions necessary to use them in unmodified form.

You define ΔE = Ein – Eout. You do realize that Ein/t can be different than Eout/t, right? Your equation is fine for an ideal world BB where what is absorbed may be instantly emitted. The real world doesn’t work that way. It is why gradients must be defined. In other words, Ein may occur in a short period of time while Eout may take much longer.

A simple example is boiling water. If you measure the energy out with a calorimeter, you can’t just remove the calorimeter at the same time you turn off the Bunsen burner. The water will continue to emit energy for some period of time after removing the heat source. The earth is no different.

Look at this graph closely. The energy input occurs over a short period of time, about 12 hours. Yet the atmosphere warms slower and emits longer than the insolation increases and continues long after insolation has decayed (about 24 hours).

There is no guarantee that what is emitted contains all the energy absorbed. The land does warm in spring and through the summer, that means it stores heat. There is no guarantee that all that heat is lost in fall and winter due to clouds and other conditions.

Why don’t you go through all the conditions necessary to use your textbook equation to model the earth. The paper you referenced should have a section that describes how that simple equation fits the physical realities.

I’ll be honest I could find no evaluation of energy in and energy out. The paper seemed to use flux as the basis for their conclusions. As I have pointed out, flux in versus flux out are not consistent over identical time periods.
comment image

Reply to  Andy May
June 18, 2026 1:23 pm

You are correct!

How many times must I show Planck’s statement about hot and cold bodies. If a cool body could push a hot body up the temperature gradient, you would end up with a never-ending increase in temperature for both bodies rather than equilibrium. This applies to conduction, convection, and radiation of heat.

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 Bi at 1000◦ C. The fact that the body A is cooled by B and heated by Bi is due entirely to the fact that B is a weaker, Bi a stronger emitter than A.

Max Planck. The Theory of Heat Radiation by Max Planck

bdgwx
Reply to  Andy May
June 18, 2026 4:53 pm

A warmer atmosphere and the IR it emits cannot cause the ocean to get warmer,

Do you think the energy just disappears?

it can only cool the ocean slower.

First…the ocean isn’t cooling. It’s warming.

Second…it’s warming because ΔEin > ΔEout. And ΔEin > ΔEout happens in part because of a change in state of the atmosphere.

The net flux is upward from the ocean to the atmosphere, the heat captured at the surface from atmospheric IR cannot move up gradient due to the cool skin effect

I don’t know how many times it has to be said. There does NOT have to be a net flux from A to B for A to cause B to warm.

The atmospheric tempeerature and radiation can modulate the solar heat leaving the ocean, but that is all.

If the atmosphere can modulate the heat leaving the ocean then atmosphere can cause the ocean to warm.

Reply to  bdgwx
June 18, 2026 5:39 pm

Do you think the energy just disappears?

Yes! From the warmer body back to the colder body. No increase in temperature of the hot body, only cooling.

Your assertion means the atmosphere is hotter than the ocean. Show us how that happens. I don’t think the atmosphere is on fire generating its own heat.

Second…it’s warming because ΔEin > ΔEout. And ΔEin > ΔEout happens in part because of a change in state of the atmosphere.

Ein is from the sun not the atmosphere. Ein is transported by the ocean to the atmosphere. Conduction, radiation, and convection.

Eout occurs because of the transport of heat from the ocean to space and the atmosphere.

Heat from a cold body simply cannot raise the temperature of hot bodies without work being done. Look up entropy.

I don’t know how many times it has to be said. There does NOT have to be a net flux from A to B for A to cause B to warm.

This is so wrong it isn’t funny. You just refuted SB, the standard heat equation Q=mc(Th -Tc),

CoPilot says:

Entropy provides a quantitative measure of energy dispersal in systems governed by the heat equation. Heat naturally flows from hot to cold regions, increasing total entropy, and the second law of thermodynamics ensures that this process is irreversible unless idealized as quasistatic. Calculating entropy changes involves integrating

δQ/T over the system, linking thermodynamic principles directly to the mathematical description of heat conduction.

Note that quasistatic refers to equilibrium.

bdgwx
Reply to  Jim Gorman
June 18, 2026 6:35 pm

Yes!

It’s ridiculous that I have to say this. Energy does NOT just disappear.

Your assertion means the atmosphere is hotter than the ocean.

No it doesn’t. This is your argument. As I keep telling you don’t expect me to defend your strawman arguments especially when they are absurd.

Ein is from the sun not the atmosphere.

Part of Ein is from the Sun. Part of it is from the atmosphere.

This is so wrong it isn’t funny.

I stand by what I said. And it’s mind numbingly easy to falsify the hypothesis that cold bodies cannot be the cause of warm bodies getting warmer.

Turn your kitchen oven on high (so that it doesn’t cycle) with the door open. Wait until the temperature inside achieves steady-state. Now close the colder door and observe the hotter inside getting warmer.

It is simple, unequivocal, and indisputable proof that cold bodies can cause hot bodies to warm whether you, Andy, or someone else understand it or not. It is such a simple and ubiquitous concept that it almost defies credulity that people here are so willing to dismiss this obvious fact.

BTW…let me nip something in the bud right now. I did NOT say that heat flows from cold to hot in an isolate system. I did NOT insinuate that heat flows from cold to hot in an isolate system. Per the 2LOT heat flows from hot to cold in an isolated system. So for those whose knee jerk reaction is to lecture me about the 2LOT make sure you lecture is regarding something I actually said otherwise I’m going to tell you what I have to tell the Gorman’s… don’t expect me to defend your strawman arguments especially when they are absurd.

You just refuted SB

Says the guy who thinks energy disappears.

Reply to  bdgwx
June 18, 2026 7:20 pm

It’s ridiculous that I have to say this. Energy does NOT just disappear.

It depends on where you look. If latent heat occurs, how do you find it and measure it? Is it hidden?

As I keep telling you don’t expect me to defend your strawman arguments especially when they are absurd.

It isn’t a strawman! You said “then atmosphere can cause the ocean to warm.” That means you are asserting the atmosphere is the hot body.

Part of Ein is from the Sun. Part of it is from the atmosphere.

The sun is the only energy source in the system. The atmosphere is not on fire, therefore it is not supplying new energy to the system.

Planck covered this. You are describing reflected heat. Reflected heat can not drive the source to a temperature that is an increase.

Your oven analogy is flawed. It is not about radiative heat it is about conduction and convection. It is actually complicated because you are changing from an open system to a closed system.

I did NOT say that heat flows from cold to hot in an isolate system.

Yes, you did. I’ll repeat from above.

You said “then atmosphere can cause the ocean to warm.”

if the atmosphere causes warming, then it must be warmer than the ocean. The 2LOT is very explicit. Heat flows from hot to cold. In a radiative system it can be no other way.

bdgwx
Reply to  Jim Gorman
June 18, 2026 8:13 pm

if the atmosphere causes warming, then it must be warmer than the ocean.

Patently False.

I’ll say it again and again. Body C does NOT have to be warmer than body H to be the cause of H getting warmer.

This happens in countless everyday scenarios. The fact that you are so indoctrinated to the contrary that you cannot even think of a single example that falsifies your claim is astonishing.

Reply to  bdgwx
June 18, 2026 9:04 pm

‘Body C does NOT have to be warmer than body H to be the cause of H getting warmer.’

What, exactly, is H getting warmer than?

bdgwx
Reply to  Frank from NoVA
June 19, 2026 8:04 am

What, exactly, is H getting warmer than?

Itself at a prior point in time. ΔT > 0.

Reply to  bdgwx
June 19, 2026 8:12 am

Sounds like a serious violation of the second law.

bdgwx
Reply to  Frank from NoVA
June 19, 2026 12:39 pm

Sounds like a serious violation of the second law.

And yet with a simple in-home experiment it can be shown that cold objects can be the cause of hot objects getting warmer. Have you considered that maybe it is your understanding of the 2LOT that is wrong?

Notice what the 2LOT does NOT say…

It does NOT say that heat (net transfer of energy) can never flow from cold to hot. Only that it cannot do so in an isolated system. There is no prohibition if the system is not isolated.

It does NOT say that cold objects cannot be the cause of hot objects getting warmer. Only that a hot object cannot warm solely as a result of heat flow (net transfer of energy) from cold to hot in an isolated system. There is no prohibition for other ways that a cold object can cause a hot object to warm futher.

The 2LOT is the most misquoted, misunderstood, misrepresented, and misapplied law by contrarians.

Reply to  bdgwx
June 19, 2026 12:53 pm

It does NOT say that heat (net transfer of energy) can never flow from cold to hot.”

You just said that entropy can be reversed without work being done.

Reply to  Frank from NoVA
June 19, 2026 12:52 pm

It is a violation.

bdgwx wants us to believe 100out – 50 back = 150 in.

Reply to  Tim Gorman
June 19, 2026 1:56 pm

deleted

Reply to  bdgwx
June 19, 2026 10:14 am

Itself at a prior point in time. ΔT > 0.

So you ARE claiming that a cold body can reverse a cooling gradient of a hot body. In other words, Tt > Tt-1.

Let’s examine the three possibiities.

Tt < Tt-1 -> cooling
Tt = Tt-1 -> equilibrium
Tt > Tt-1 -> heating

There are all kinds of equations available, the heat equation, diffusion equations, Stefan-Boltzmann, entropy, and others. Show us how these work with a cold body making the temperature of a hot body meet Tt > Tt-1.

Reply to  Jim Gorman
June 19, 2026 1:58 pm

Still waiting for your math!

Reply to  bdgwx
June 19, 2026 12:51 pm

Warmer means more joules than you started with.

Where are the extra joules coming from?

The earth emits 100 joules. CO2 reflects back 50 joules. The earth still lost 50 joules. How does that mean more joules?

CO2 is *NOT* a heat source. It is a reflector!

Reply to  Tim Gorman
June 19, 2026 2:50 pm

“The earth emits 100 joules. CO2 reflects back 50 joules. The earth still lost 50 joules. How does that mean more joules? ”

Stil denying the sun exists?

Reply to  bdgwx
June 19, 2026 3:53 am

This happens in countless everyday scenarios.”

“Body C does NOT have to be warmer than body H to be the cause of H getting warmer.”

Then YOU give us ONE example.

You simply can’t get over the fact that slower cooling is *NOT* the same thing as warming, can you?

If (Ta – Tb) is positive, i.e. Ta > Tb, give us ONE example where (Ta – Tb) can be made negative without Tb becoming greater than Ta, i.e. Tb > Ta.

As long as the temperature of ObjectA, Ta, is greater than the temperature of ObjectB, Tb, the net flow will be positive. The net flow can get larger and smaller but it will remain positive as long as Ta > Tb.

A positive net flow represents COOLING of ObjectA, not warming. It is just that simple.

bdgwx
Reply to  Tim Gorman
June 19, 2026 5:24 pm

Then YOU give us ONE example.

I already have numerous times. When I first gave you the example of the kitchen oven you claimed that closing the door would cause the inside to cool. I honestly thought it was a typo at first, but you then doubled-down and said my understanding of thermodynamics was at an elementary school level. If you cannot accept that the inside of an oven will do the exact opposite of cool when the door is closed then you certainly aren’t going to understand any other example demonstrating the same concept.

Reply to  bdgwx
June 20, 2026 1:53 am

Here is the quote you linked to:

You still haven’t figured this out, have you? If the oven door is “cooler” than the oven it should *COOL* the oven when you close the door, not warm it!”

YOU are claiming that a colder body can increase the temperature of a warmer body – a total violation of Planck and the 2nd law. It would be reversing the entropy of the warmer object.

Substitute a block of dry ice for the oven door. If you stick that block of dry ice in the door frame of the oven will it cause the temperature in the oven to go up?

you then doubled-down and said my understanding of thermodynamics was at an elementary school level”

It is *still* at that level if you think that block of dry ice will raise the temperature of the oven!

bdgwx
Reply to  Tim Gorman
June 20, 2026 7:52 am

Here is the quote you linked to:

“You still haven’t figured this out, have you? If the oven door is “cooler” than the oven it should *COOL* the oven when you close the door, not warm it!”

I am assuming you still standby those today. Yes/No?

YOU are claiming that a colder body can increase the temperature of a warmer body 

Yes.

a total violation of Planck and the 2nd law. It would be reversing the entropy of the warmer object.

And yet the hot inside of that oven warms further when you close the cold door. Have you considered that maybe it is your understanding of the laws of thermodynamics that is wrong?

Substitute a block of dry ice for the oven door. If you stick that block of dry ice in the door frame of the oven will it cause the temperature in the oven to go up?

It would be interesting experiment to perform. I have a couple of hypothesis that would be interesting to discuss sometime, but discussing them with you would be pointless since you cannot even understand the far simpler door scenario.

Reply to  bdgwx
June 20, 2026 9:25 am

I see you totally ignored the question I put to you as to what happens if you substitute a block of dry ice for the oven door.

Does the dry ice WARM the inside of the oven?

Kind of puts you in a bind, doesn’t it?

I guess in your view it is just *some* cold objects that can raise the temperature of a warmer object.

bdgwx
Reply to  Tim Gorman
June 20, 2026 12:28 pm

I see you totally ignored the question I put to you as to what happens if you substitute a block of dry ice for the oven door.

I didn’t ignore it. I said it would be pointless to discuss with you because you don’t understand the simpler door scenario. If you are incapable of understanding the door scenario you aren’t going to be any more capable of understanding more complex scenarios. I’m more than happy to discuss the scenario with someone who already understands the simpler door scenario though.

Kind of puts you in a bind, doesn’t it?

No.

I guess in your view it is just *some* cold objects that can raise the temperature of a warmer object.

Correct. Not all introductions or activations of a cold body would cause a hot body to warm further. One important requirement (easily shown via the 1LOT) is the existence of a source of energy that keeps Ein > 0. This is why closing the door of an oven when turned off will do nothing.

Reply to  bdgwx
June 20, 2026 5:12 pm

I didn’t ignore it. I said it would be pointless to discuss with you because you don’t understand the simpler door scenario.”

What’s simpler than a door plug made of dry ice?

The TRUTH is that the answer blows your assertion that cooler warms hotter right out of the water!

And you can’t admit that!

Reply to  bdgwx
June 19, 2026 6:56 am

‘Turn your kitchen oven on high (so that it doesn’t cycle) with the door open. Wait until the temperature inside achieves steady-state. Now close the colder door and observe the hotter inside getting warmer.’

The oven door, again? Not a clean experiment since you’re not controlling for convection.

bdgwx
Reply to  Frank from NoVA
June 19, 2026 7:47 am

The oven door, again?

Yep. it’s an experiment that anyone can do.

Not a clean experiment since you’re not controlling for convection.

Irrelevant. We don’t have to control for convection, conduction, and/or radiation to test the hypothesis that a cold body cannot be the cause of a hot body getting warmer. All we have to do is show that it can happen for any mode of energy transfer to falsify the hypothesis. Fun fact though…the hypothesis is false for all modes of energy transfer.

Reply to  bdgwx
June 19, 2026 8:29 am

Oh, ok. I’d be interested in knowing if Nick buys into this…

bdgwx
Reply to  Frank from NoVA
June 19, 2026 12:29 pm

Of course Nick buys into it. Why would anyone not buy the fact that closing the cold door causes the hot inside of an oven to warm further?

Nick Stokes
Reply to  bdgwx
June 19, 2026 1:06 pm

Of course I do. How do your cold clothes keep you warm?

Reply to  Nick Stokes
June 19, 2026 2:07 pm

By lessening the rate of heat loss, but that’s not what bdgwx is on about. See: https://wattsupwiththat.com/2026/06/17/do-los-ninos-cause-climatic-cooling/#comment-4208224

bdgwx
Reply to  Frank from NoVA
June 19, 2026 4:11 pm

By lessening the rate of heat loss, but that’s not what bdgwx is on about.

That’s the ΔEout < 0 case I was on about in my first post 🙂

Reply to  bdgwx
June 19, 2026 7:39 pm

Let’s review –

I asked:

‘What, exactly, is H getting warmer than?’

You replied:

‘Itself at a prior point in time.’

So we have an object, H, that is cooling over time, hence Temp_H_t+dt < Temp_H_t. By what logic would introducing a cooler object, C, whose temperature, Temp_C_t+dt < Temp_H_t+dt raise the temperature of object H above Temp_H_t+dt?

bdgwx
Reply to  Frank from NoVA
June 20, 2026 7:10 am

So we have an object, H, that is cooling over time, hence Temp_H_t+dt < Temp_H_t. By what logic would introducing a cooler object, C, whose temperature, Temp_C_t+dt < Temp_H_t+dt raise the temperature of object H above Temp_H_t+dt?

It happens when body C changes the energy flow characteristics of the system such that the amount of energy leaving the system is less than the amount entering the system.

Think of the door on the oven. By introducing it into the system you reduce the rate of energy leaving without changing the rate at which enters. The inside of the oven now has an imbalance with energy accumulating on the inside. It continues to accumulate and warm until a new balance is acheived.

Reply to  bdgwx
June 20, 2026 8:01 am

‘It happens when body C changes the energy flow characteristics of the system such that the amount of energy leaving the system is less than the amount entering the system.’

The ‘system’ in question consists only of two bodies, H and C, in an electromagnetic field. Please show me the basis in either classical or quantum electrodynamics for the existence of poly-directional (in this case bi-directional) flow of electromagnetic energy between these two bodies.

Please refer to Section 3 of the attached article:

https://ntrs.nasa.gov/api/citations/20140012672/downloads/20140012672.pdf

Reply to  Frank from NoVA
June 20, 2026 9:14 am

You are running up against the belief that 100 from the sun PLUS 50 from the cold body (atmosphere) equals a total of 150.

To admit that the thermodynamics is actually 100 MINUS 50 equals 50 of cooling destroys the GHG theory and simply can’t be true.

bdgwx
Reply to  Jim Gorman
June 20, 2026 12:17 pm

You are running up against the belief that 100 from the sun PLUS 50 from the cold body (atmosphere) equals a total of 150.

It does. That is the law of conservation of energy.

To admit that the thermodynamics is actually 100 MINUS 50 equals 50 of cooling destroys the GHG theory and simply can’t be true.

The reason you may be acting incredulous about it is because you reject the law of conservation of energy or at least are making statements that are making believe you reject it.

Reply to  bdgwx
June 20, 2026 5:10 pm

What happened to the 100 joules the earth emitted before CO2 returned part of it?

Reply to  bdgwx
June 20, 2026 7:06 pm

It does. That is the law of conservation of energy.

You are misinterpreting the law of conservation of energy. It is stated as:

The total energy of an isolated system remains constant, even though energy may change form (heat, work, internal energy).

The earth and atmosphere consists is an isolated system. If one assumes the sun has no effect on the atmosphere, then the earth becomes a sorce due to the sun’s insolation. Let’s say it radiates 100 W/m² over 1m² and the time is 1 second.

That means there is only 100 joules in the isolated system. Your assertion of

100 j + 50 j = 150 j

violates the condition of constant energy.

At this point we should be using gradients over time to show the energy exchange in the system, however we can look at what occurs at equilibrium.

What occurs at equilibrium? Both bodies are radiating 100 W/m² at each other. But that violates the conservation of energy because there is now 200 joules in the system.

Uh oh, what’s going on?

The law of conservation of energy doesn’t apply when there is a source that continuously adds energy to the isolated system.

Ask your favorite AI to show you the failure because ∆U=0 is not possible with a source in an isolated system.

Here is what CoPilot says.

if a system has a continuous source of energy, then by definition it is not isolated, so the law of conservation of energy must be written to include that energy input.”

bdgwx
Reply to  Frank from NoVA
June 20, 2026 10:32 am

We can certainly focus the conversation on radiation, but keep in mind the oven scenario has all 3 energy transfers modes in play with convection likely dominating the other two so a radiation-only analysis won’t have a lot of applicability to it. But yeah, it is still a good thing to discuss especially if we want move the conservation into radiation dominated scenarios.

Anyway the only theoretical basis you need to show that cold bodies can be the cause of hot bodies getting warmer when only radiation is in play is the 1LOT and SB law. The 0LOT and 2LOT are useful for tools in the analysis as well.

A good place to start the discussion is with the green plate effect thought experiment.

Eric Swanson and Dave Cloudman independently demonstrate the effect with experiments. And of course the JWST sunshield is another good demonstration though it is vastly more complicated.

Reply to  bdgwx
June 20, 2026 12:23 pm

‘A good place to start the discussion is with the green plate effect thought experiment.’

A better name for this would be Eli’s Blue Plate Special, which doesn’t conserve energy. Starting with his blue plate, we must have 400 W/m^2 = sigma*(T_2)^4, or T_2 =289.81K. Plug and chug for his green plate expression for T_1 and you get T_1 = 344.64K.

The irony is that, unlike the Earth, that gets fried by back radiation according to the alarmists, the heat source in Eli’s model isn’t affected by back radiation from the blue plate.

A couple of points: First, you should look at Eschenbach’s ‘Steel Greenhouse’ if you wish to at least get the geometry and concept of energy balance correct. And second, you might want to remind yourself that Earth’s atmosphere isn’t composed of condensed matter, e.g., ‘plates’, so doesn’t conform to your (or Eli’s) phenomenological physics of radiative transfer theory.

bdgwx
Reply to  Frank from NoVA
June 20, 2026 1:21 pm

A better name for this would be Eli’s Blue Plate Special, which doesn’t conserve energy.

Of course it conserves energy. In fact, having the blue plate radiating at 267 W.m-2 (262 K) and the green plate at 133 W.m-2 (220 K) is the ONLY solution that complies with the 1LOT and 2LOT and is in a stead-state.

Fsys_in = 400 W.m-2
Fsys_out = 267 W.m-2 + 133 W.m-2 = 400 W.m-2

Fbp_in = 400 W.m-2 + 133 W.m-2 = 533 W.m-2
Fbp_out = 267 W.m-2 + 267 W.m-2 = 533 W.m-2

Fgp_in = 267 W.m-2
Fgp_out = 133 W.m-2 + 133 W.m-2 = 267 W.m-2

With all 400 W.m-2 from the Sun accounted for and with both the BP and GP having energy balance the configuration complies with the 1LOT

With the heat flow being from Sun => BP => GP => Space the configuration complies with the 2LOT.

Any other configuration you try will either violate the 1LOT/2LOT or the system will not be in a steady-state and must adjust its temperature.

I created a graphic to help people visualize the energy flows.

comment image

First, you should look at Eschenbach’s ‘Steel Greenhouse’ if you wish to at least get the geometry and concept of energy balance correct.

Yes. I’m familiar with it. It’s a good thought experiment too.

And second, you might want to remind yourself that Earth’s atmosphere isn’t composed of condensed matter, e.g., ‘plates’, so doesn’t conform to your (or Eli’s) phenomenological physics of radiative transfer theory.

As I often tell people…if you can’t understand simpler idealized scenarios you will never be able to understand the vastly more complex real world. The Green Plate Effect thought experiment expresses a salient and fundamental truth that has to be understood before jumping into more complex scenarios.

Reply to  bdgwx
June 20, 2026 2:12 pm

How many times does this need to be debuncd?

I hate to tell you but at equilibrium, the blue plate should be at 290K which is (400 W/m² ÷ (5.67×10⁻⁸).²⁵ = 290K.
If the blue plate is at 290K, it radiates 400 W/m² in ALL DIRECTIONS. Check what an IR thermometer would read if pointed at one side of the blue plate in your diagram. It would read 262K, which is NOT 400K.
You have fallen into the trap of GHE. That is, CO2 radiates half up and half down thus dividing the flux in half. It doesn’t work that way. When dealing with radiation you must use a volume whereby sufficient material is contained so that the random radiation directions of all the molecules end up with a result of the radiation being in all directions. This can be verified by heating a piece of steel with a torch at 773K. At equilibrium between the torch and the steel, you don’t have to measure all six sides to determine the total temperature, only one side will do. It radiates equally in all directions at the rate specified by the temperature of the body.

I’ve gone over this time after time on X. Here is another little gem your diagram misses. That plate has 6 sides, not two. Do the other 4 sides not radiate at all? Why would that be?

bdgwx
Reply to  Jim Gorman
June 20, 2026 4:00 pm

I hate to tell you but at equilibrium, the blue plate should be at 290K which is (400 W/m² ÷ (5.67×10⁻⁸).²⁵ = 290K.

Let’s try it out.

Fbp_in = 400 W.m-2 (from Sun) + 133 W.m-2 (from GP)
Fbp_out = 400 W.m-2 (left) + 400 W.m-2 (right)
Fbp_in – Fbp_out = -267 W.m-2

Fgp_in = 400 W.m-2 (from BP)
Fgp_out = 133 W.m-2 (left) + 133 w.m-2 (right)
Fgp_in – Fgp_out = +133 W.m-2

Nope. Doesn’t work. The math using the 1LOT does not lie. You are not accounting for 267 W.m-2 from the BP and 133 W.m-2 on the GP.

Reconfigure your solution to comply with the 1LOT and resubmit for review.

Reply to  bdgwx
June 20, 2026 5:21 pm

Fbp_in = 400 W.m-2 (from Sun) + 133 W.m-2 (from GP)

Wrong!

Fbp_in is 400 W/m².

At equilibrium with the sun, which is needed to analyze without using gradients, Fbp will be at 290K.
(400 W/m² ÷ (5.67×10⁻⁸)).²⁵ = 290K.

Fbp_out will be radiating 400 W/m² IN ALL DIRECTIONS. 400 W/m² back to the sun and 400 W/m² toward the green plate.

The blue plate radiates equally in all directions. Do you disagree with that?

If you disagree, you need to provide a textbook reference that shows a body radiates based on the number of sides it has.

At this point, when the Fbp radiates 400 W/m² to the Fgp, the same thing occurs. At equilibrium, the Fgp will also be at 290K and radiating at 400 W/m² in all directions. Therefore it will be in equilibrium with the blue plate.

Since both plates will be radiating similar amounts to each other, the net radiation is zero as shown by the SB equation.

From Planck’s Theory of Heat Radiation, Section 4

We shall now consider the interior of an emitting substance assumed to be physically homogeneous, and in it we shall select any volume element dτ of not too small size.

Every point of dτ will then be the vertex of a pencil of rays diverging in all directions.

A corollary later also says that no conduction or convection can take place in which would complicate the determination of radiation effects. That means it must be isothermal, i.e., the same temperature throughout.

bdgwx
Reply to  Jim Gorman
June 20, 2026 5:49 pm

Wrong!

Fbp_in is 400 W/m².

What happened to energy hitting the BP from the GP?

At this point, when the Fbp radiates 400 W/m² to the Fgp, the same thing occurs. At equilibrium, the Fgp will also be at 290K and radiating at 400 W/m² in all directions. Therefore it will be in equilibrium with the blue plate.

Let’s try it out.

Fbp_in = 400 W.m-2 (from Sun) + 400 W.m-2 (from GP)
Fbp_out = 400 W.m-2 (left) + 400 W.m-2 (right)
Fbp_net = Fbp_in – Fbp_out = 0 W.m-2

Fgp_in = 400 W.m-2 (from BP)
Fgp_out = 400 W.m-2 (left) + 400 w.m-2 (right)
Fgp_net = Fgp_in – Fgp_out = -400 W.m-2

Fsys_in = 400 W.m-2 (from Sun)
Fsys_out = 400 W.m-2 (left) + 400 W.m-2 (right)
Fsys_net = Fsys_in – Fsys_out = -400 W.m2

You got the BP to balance but now you’re not accounting for 400 W.m-2 at the GP and 400 W.m-2 on the system as a whole.

Reconfigure your solution to comply with the 1LOT and resubmit for review.

Reply to  bdgwx
June 22, 2026 4:03 pm

Fbp_net = Fbp_in – Fbp_out = 0 W.m-2

Fgp_net = Fgp_in – Fgp_out = -400 W.m-2

You are all messed up here. The net values are between two bodies.

Sun <-> bp 400 – 400 = 0 Equalibrium
bp <-> gp 400 – 400 = 0 Equalibrium

And guess what? The sun radiates 400 to its left and the gp radiates 400 to its right. Where the sun_left and the gp_right really doesn’t matter. It could be into space or to other bodies. Who knows, who cares.

Reply to  bdgwx
June 23, 2026 9:43 am

I see you have yet to answer the questions I posed.

If you disagree, you need to provide a textbook reference that shows a body radiates based on the number of sides it has.

If the blue plate is at 290K, it radiates 400 W/m² in ALL DIRECTIONS. Check what an IR thermometer would read if pointed at one side of the blue plate in your diagram. It would read 262K, which is NOT 290K.

I’ll be really interested in this one. Your plate has 6 sides. Why do 4 of them not radiate at all? Better yet, if heat the plate to 290K, will an IR thermometer only show 262K depending on the side I point at?

Nor have you refuted the following.

You are all messed up here. The net values are between two bodies.

Sun <-> bp 400 – 400 = 0 Equalibrium

bp <-> gp 400 – 400 = 0 Equalibrium

Reply to  Jim Gorman
June 20, 2026 4:30 pm

A correction:

“which is NOT 400K”
Should be 290K

Reply to  bdgwx
June 20, 2026 2:15 pm

Look at your diagram: The blue plate has +400-267-267+133, which does not balance / conserve energy. The green plate at least balances (+267-133-133), but the whole ‘idealized scenario’ is bass-ackwards. Specifically, if the ‘Sun’ is on the left, the ‘Earth’ would schematically (logically) be on the right, hence you’re showing the lower atmosphere to be cooler than the upper atmosphere, which is the exact opposite of the so-called radiative GHE.

Maybe you and Eli should compare notes with Nicholas Schroeder.

bdgwx
Reply to  Frank from NoVA
June 20, 2026 3:49 pm

The blue plate has +400-267-267+133, which does not balance / conserve energy

Lol…check your math. It balances.

Maybe you and Eli should compare notes with Nicholas Schroeder.

See if you can figure out any other way get to the system, blue plate, and green plate to balance and comply with the 1LOT.

And by Nicholas Schroeder do mean the guy who keeps posting energy budget diagrams that do not conserve energy?

Reply to  bdgwx
June 21, 2026 6:57 am

‘Lol…check your math. It balances.’

Indeed it does. In my haste I calculated ‘left and right’ vs ‘in and out’, so my bad. So, your / Eli’s math checks out. However, I’ll stick with what I said in my second point about the science, i.e.,

‘And second, you might want to remind yourself that Earth’s atmosphere isn’t composed of condensed matter, e.g., ‘plates’, so doesn’t conform to your (or Eli’s) phenomenological physics of radiative transfer theory.’

bdgwx
Reply to  Frank from NoVA
June 21, 2026 8:05 am

The real world is vastly more complex than the Green Plate Effect thought experiment. As such there are going to numerous other considerations above and beyond what the simple thought experiment explores. No reasonable person would dismiss that.

The main point of the exercise is to get people to understand that cold bodies can be the cause of hot bodies getting warmer and that this concept applies to scenarios that are radiation dominated.

Perhaps a secondary point is that it isn’t obvious figuring out how to balance the radiation flows while also complying with the 1LOT. The Gorman’s struggles in this regard is a testament to that fact. Anyway, it gets even more interesting when you consider more than two plates like would be the case with the JWST sunshield.

Reply to  bdgwx
June 22, 2026 10:15 am

What is the ‘JWST sunshield’?

Actually, I mainly wanted to circle back to your diagram, above:

First, it’s misleading / confusing to have the ‘Sun’ on the left, so let’s replace it with a radiating ‘Earth’ that emits IR to space through two concentric shells rather than ‘plates’. Hopefully, you’ll agree with me that this set-up, with 400W radiating from the surface and 133W radiating to space, is very close (at least enough for government work) to the Trenberth-like diagrams we often see in the canonical literature.

Second, but why just two ‘plates’? Why not 10, or as many as there are in a typical RTM-centric GCM, wherein multiple ‘slabs’, each in ‘LTE’ with its neighbors, freely absorbs and emits radiation in accordance with the thermal radiative properties more correctly applicable to so-called black bodies? I think we both know the real answer to this, which is that the model would give increasingly ridiculous results for the Earth’s surface temperature, as the number of ‘plates’ increased beyond two. In other words, the model uses two plates because that number coincidentally yields results that are roughly in line with ‘observed’ IR emissions

In effect then, the only thing that models like the ‘Green Plate Effect’ succeed in showing is that conflating the thermal radiative properties of condensed matter with those of atmospheric gases is phenomenological physics.

Reply to  Frank from NoVA
June 23, 2026 9:59 am

which is that the model would give increasingly ridiculous results for the Earth’s surface temperature, as the number of ‘plates’ increased beyond two

The zeroth law of thermodynamics formalizes the idea of equivalence. Systems in thermal equilibrium can be grouped into subsets where all members share the same temperature, and no member is in equilibrium with systems outside the subset.

This is why the sun and the blue plate can be considered to be a subset and in equilibrium with each other. Likewise, the blue plate and green plate can be considered to be a subset and in equilibrium with each other. The transitive property then leads to the sun and green plate also being in equilibrium.

The sun and green plate fluxes to the blue plate do not add or there could not be equilibrium in the total system. This does have one requirement. The bodies must be ideal. Note, it would not matter if the sun was in complete view of the green plate since the zeroth law of thermodynamics would say they are all in equilibrium with each other.

One must get the fundamentals correct before moving to other, more complicated arrangements.

Reply to  bdgwx
June 25, 2026 5:43 am

 getting warmer”

It doesn’t “get” warmer. That implies injection of additional joules of heat.

It *cools* less is the proper terminology.

As long as T1 < T0 you have cooling. It doesn’t matter if T0-T1 = 10 – 5 = 5 or T0-T1 = 10 – 4 = 6.

T1 doesn’t “get” to 6 instead of 5 by injection of new heat. T1 cools just cools less.

CO2 cannot inject “new” heat. It can only replace heat that has already been lost. That results in slower cooling.

bdgwx
Reply to  bdgwx
June 21, 2026 8:42 am

And by Nicholas Schroeder do mean the guy who keeps posting energy budget diagrams that do not conserve energy?

He strikes again making energy disappear in his latest post.

Reply to  bdgwx
June 22, 2026 11:21 am

‘…making energy disappear in his latest post.’

If he added 78W/m^2 SW ‘absorbed by atmosphere’ and increased OLR by an offsetting 78W/m^2, would you (and others) be happy?

bdgwx
Reply to  Frank from NoVA
June 22, 2026 2:00 pm

My issue is that 396 W.m-2 UWIR and 333 W.m-2 DWIR is completely ignored and assumed to not even exist.

Reply to  bdgwx
June 22, 2026 3:33 pm

A good place to start the discussion is with the green plate effect thought experiment.”

Believe it or not, bdgwx, this is still the most frequently discussed topic on Roy’s blog.

And despite posts like these having been up for at least a decade, GHE denial remains persistent there.

Reply to  bdgwx
June 20, 2026 12:55 pm

It happens when body C changes the energy flow characteristics of the system”

You just changed from “heat sources” to “flow inhibitors”.

Flow inhibitors cannot *add* heat to the system. Only a heat source can *add* heat to the system. You are trying to equate insulation to the heating element in the oven.

FAIL!

Reply to  Frank from NoVA
June 20, 2026 7:12 am

Be careful here, you are using math that can confuse people who have a different understanding. Funny how you can put numbers in your math and end up with something they don’t believe!

Reply to  Jim Gorman
June 20, 2026 8:58 am

I guess that’s a step up from Nick asking Tim what he’s wearing…

Reply to  bdgwx
June 20, 2026 1:58 am

Why do you keep leaving out the heat source that is the bodies metabolism?

Will putting frozen clothes on a frozen dead body cause the body to thaw?

Reply to  Frank from NoVA
June 20, 2026 1:57 am

Put those cold clothes on a dead body. Will the temperature of the dead body go up?

Reply to  Nick Stokes
June 20, 2026 1:56 am

The clothes don’t keep you warm. They decrease the cooling rate. It’s your metabolism that keeps you warm. Put those cold clothes on a corpse. Will the temperature of the corpse go up?

Nick Stokes
Reply to  Tim Gorman
June 20, 2026 4:19 am

Do you wear clothes?

Reply to  Nick Stokes
June 20, 2026 7:10 am

Actually I do. But guess what, so do desert people. Do they need to be warmer?

Reply to  Jim Gorman
June 20, 2026 9:12 am

Lot’s of desert people wear robes from neck to toe. And they go into their tents during mid-day.

Do the robes warm them up? Does the tent warm them up?

Reply to  Nick Stokes
June 20, 2026 9:11 am

Does a dead body in a casket at a funeral service wear clothes? Do those clothes WARM the dead body?

Reply to  Tim Gorman
June 20, 2026 1:03 pm

I see our climate science/thermodynamic experts can’t seem to find an answer for whether or not clothes on a dead body warm the dead body.

Guess I’ll wait till tomorrow to see if they have an answer.

Reply to  Tim Gorman
June 20, 2026 4:54 pm

I’m neither one of those, but I can confidently say that they will not. You know why? Because a corpse doesn’t have a heat source.

This is the same reason why an atmosphere won’t warm a planet if there is no heat source such as the sun.

Reply to  bdgwx
June 19, 2026 10:05 am

We don’t have to control for convection, conduction, and/or radiation to test the hypothesis that a cold body cannot be the cause of a hot body getting warmer. 

You do have to control for the various types of heat flow if you are trying to prove that radiation from a cold body warms a hot body.

With an open door, both convection and conduction are cooling the inside of the oven. In essence, your so-called experiment is determining how much cooling takes place with an open door.

Perform your experiment in reverse. Start with the door closed and equilibrium. Then open the door and see if the cool outside warms the inside. Here is a second option. Raise the temperature in the room to 100 degrees and time how long it takes for the oven to reach equilibrium at a given temperature. Then lower the room to 50 degrees and time how much time it takes to reach the same temperature.

Lastly, let’s see your math that this experiment is destined to prove. Have you done the experiment and collected data? It should be easy to use the data to determine what occurs. You do know what Richard Feynman said, right?

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

Reply to  Jim Gorman
June 19, 2026 11:21 am

Perform your experiment in reverse. Start with the door closed and equilibrium. Then open the door and see if the cool outside warms the inside.”

100%!

Reply to  Jim Gorman
June 20, 2026 1:58 am

Put frozen clothes on a frozen dead body. Will the dead body thaw?

Reply to  bdgwx
June 19, 2026 11:20 am

“Irrelevant.”

Conductive heat loss is irrelevant? Tell it to the heat sink on the processor in your computer!!!!

We don’t have to control for convection, conduction, and/or radiation to test the hypothesis that a cold body cannot be the cause of a hot body getting warmer.”

Of course you have to control for EVERY SINGLE FACTOR. Leave it to climate science supporters to say “we can ignore reality”.

“All we have to do is show that it can happen for any mode of energy transfer to falsify the hypothesis”

Except you haven’t shown it can happen for ANY mode of energy transfer.

SHOW YOUR MATH, including *ALL* heat transfer equations.

Do you even have a clue as to what the heat loss equation is for conduction? For convection? For radiation?

bdgwx
Reply to  Andy May
June 18, 2026 7:54 pm

Sorry, the heat flow (global net) is always ocean to atmosphere

Irrelevant even if true (it’s not at least on localized scales). I’ll repeat again. A cold body can be the cause of a hot body getting warmer without heat flowing from cold to hot (which isn’t possible at least in an isolated system). This is such an obvious and intuitive concept that you should be able to rattle off real scenarios of it happening in rapid succession with little effort.

the ocean is always cooling

Patenly False. The ocean has experienced many warming periods like the one occurring today.

The best the atmosphere can do is slow the rate of cooling.

Why? Is it the law of conservation of mass ΔE = Ein – Eout, heat capacity ΔT = ΔE / (m*c), or the fact that the atmosphere can be the cause of changes in Ein and/or Eout of the ocean that you are challenging?

Reply to  bdgwx
June 19, 2026 4:13 am

A cold body can be the cause of a hot body getting warmer without heat flowing from cold to hot (which isn’t possible at least in an isolated system).”

(Th – Tc) will remain positive, indicating a cooling of Th – ALWAYS. (Th – Tc) will never go negative until Tc > Th.

“This is such an obvious and intuitive concept that you should be able to rattle off real scenarios of it happening in rapid succession with little effort.”

Then why can’t YOU provide even ONE example?

You *still* haven’t figured out that slower cooling is *NOT* the same thing as warming. The adjectives “cooling” and “warming” describe the gradient between two objects. That gradient is related to (Th-Tc). The slope of that gradient can increase and decrease but it can’t change sign unless Tc > Th. Of course when that happens the object that was Th all of a sudden becomes Tc!

“Patenly False. The ocean has experienced many warming periods like the one occurring today.”

That does *NOT* mean that the ocean all of a sudden gets cooler than the atmosphere! Tatmos does not get greater than Tocean!

or the fact that the atmosphere can be the cause of changes in Ein”

You keep being told that Ein IS THE SUN, not the atmosphere.

The atmosphere is a reflector, not a source.

At any point in time if the sun is inputting 100 joules, the earth is outputting 50 joules, and the atmosphere is reflecting 25 joules back to the earth, the earth WILL STILL BE OUTPUTTING 25 JOULES!

That is COOLING, not warming. 25 joules-out is less than it would be if the atmosphere was not reflecting heat back. But 25 joules-out IS STILL COOLING, not warming.

If you will note, the values I use are representative of values that result in joules-out and joules-in balance over time since the earth outputs 24hr/day while joules-in only occurs over 12hrs/day. The rate out (50 joules) at any time only has to be half of the joules-in (100 joules) for balance to happen.

It’s not even obvious that you recognize that the earth outputs joules DURING THE DAY as well as at night. The earth actually loses MORE heat during the day than it does at night because its temperature is larger.

Bottom line? The atmosphere can change Eout. It can *NOT* change Ein. The atmosphere is *NOT* a heat source.

Reply to  bdgwx
June 19, 2026 7:06 am

I’ll repeat again. A cold body can be the cause of a hot body getting warmer without heat flowing from cold to hot (which isn’t possible at least in an isolated system).

This cannot be. For temperature to rise, heat must be transferred. Your favorite 2LOT states this. Only added work can cause this to happen.

You have a total misunderstanding about what thermodynamic heating and temperature are.

You keep making assertions with no math to back them up. Let’s see your math that serves to support your position.

Most of us would have no problem if your statement said this.

A cold body can be the cause of a hot body remaining at a higher temperature than what would normally occur without the cold body being present.

This encompasses the fact that warming does not occur. Only cooling at a slower rate.

Tim has shown you the math that requires a reversal of the cooling gradient in order for a body to become hotter. Let us see yours.

Nick Stokes
Reply to  Andy May
June 18, 2026 5:24 pm

Andy,
it can only cool the ocean slower.”

Again, the basic arithmetic. About 160 W/m2, on average, arrives as SW and is thermalised at depth. If that 160 W/m2 then leaves via the surface, ocean temperature is stable. If 161 W/m2 leaves, ocean cools; if 159 W/m2 leaves, ocean warms.

Reply to  Nick Stokes
June 18, 2026 6:08 pm

if 159 W/m2 leaves, ocean warms.

You are out of your wheelhouse here. I hate to tell you this, but if 159 W/m² is lost, the ocean has cooled.

1 W/m² can easily be dismissed by evaporation and latent heat which is not measurable by flux. Beware equating fluxes, especially at the surface, when H2O is involved, latent heat is a real thing.

Nick Stokes
Reply to  Jim Gorman
June 20, 2026 2:11 pm

Evaporation and latent heat are measured by flux.

bdgwx
Reply to  Nick Stokes
June 18, 2026 6:39 pm

You’d think basic arithmetic (literally) and the law of conservation of energy would be so universally understood that few people would not understand that if Ein > Eout then a body’s internal energy will increase and ultimately warm.

Reply to  bdgwx
June 18, 2026 6:48 pm

You would think that basic understanding of time based variables would be understood when doing equalities.

Did you not understand what this graph shows in regards to instantaneous flux values?

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Or this one.

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Reply to  Jim Gorman
June 19, 2026 4:27 am

Neither of these two seem to understand that Ein and Eout are measured in JOULES, not in Joules/sec-m^2!

Reply to  bdgwx
June 19, 2026 4:26 am

that few people would not understand that if Ein > Eout then a body’s internal energy will increase and ultimately warm.”

E is in JOULES, not in joules/sec-m^2

You would think that those in climate science would understand that a joules/sec-m^2 is a RATE and not an amount.

A rate is a TIME FUNCTION, not a state function. An amount is a state function, not a time function.

IF Ein happens over 12 hours and Eout happens over 24 hours then the rates associated with Ein and Eout DO NOT HAVE TO BALANCE!

Reply to  Nick Stokes
June 19, 2026 4:22 am

Again, the basic arithmetic. About 160 W/m2, on average, arrives as SW and is thermalised at depth. If that 160 W/m2 then leaves via the surface, ocean temperature is stable. If 161 W/m2 leaves, ocean cools; if 159 W/m2 leaves, ocean warms.”

You are still stuck on trying to enforce FLUX BALANCE. Flux will *NEVER* balance.

You want to talk about basic arithmetic?

How many joules does 160 joules/sec-m^2 input over 12 hours?
Ans: 160 joules/sec-m^2 * 12hrs * 60min/hr * 60sec/min = 7 x 10^6 joules.

How many joules does 160 joules/sec-m^2 output over 24 hours?
Ans: 160joules/sec-m^2 * 24hrs * 60min/hr * 60sec/min = 14 x 10^6 joules.

You just turned the earth into a frozen ball!

Joules/sec-m^2 IS A RATE, not an amount. You can’t equate rates to determine balance, you have to equate amounts! A rate is a TIME FUNCTION, not a state function!

Nick Stokes
Reply to  Tim Gorman
June 19, 2026 1:04 pm

160 W/m2 is a 24hr average.

Flux has to balance in the medium term. Otherwise the sea just gets hotter and hotter or colder and colder.

Reply to  Nick Stokes
June 19, 2026 2:04 pm

Flux has to balance in the medium term. Otherwise the sea just gets hotter and hotter or colder and colder.

So you don’t believe that the ocean can store heat for long periods of time, like decades or centuries? What do you claim that a medium time is?

Do you believe that evaporative cooling by H2O doesn’t contribute to radiant flux?

Reply to  Nick Stokes
June 20, 2026 1:46 am

FLUX WILL NEVER BALANCE!

Energy is what will balance!

What do you consider “medium term”? A day? A week? A month? A year? A century? A millennia?

Reply to  Tim Gorman
June 20, 2026 7:03 am

I see there is no answer yet!

June 17, 2026 2:37 pm

Andy, my own opinion is that we should just admit we don’t know what we don’t know. That past proxies can be interpreted differently given their inherent geospatial uncertainty and lack of fine resolution fundamentally means not only do we not know, but we never can.

As a simple ‘Los Ninos’ example, we know they recur ‘regularly’—but with varying frequency and ‘strength’. Correlating that to related weather is, as the Japanese would say, ‘very difficult’ when they mean impossible. Lots of experience with that Japanese phrase in a previous ‘life’.

Izaak Walton
Reply to  Andy May
June 17, 2026 4:39 pm

That would appear to be impossible. Given that the ocean is either in one of three states (El Nino, La Nina or Neutral) the periodicity of any one of the states must correlate with those of the other two. If La Nina and Neutral are regular and periodic then there can’t be any gaps for El Nino. Alternatively if El Nino is irregular then the other two must be as well since they fit into the irregularly shaped gapes between two successive El Ninos.

Reply to  Izaak Walton
June 17, 2026 7:49 pm

WTF?

Reply to  Andy May
June 18, 2026 1:29 pm

Neither does he.

Reply to  Izaak Walton
June 18, 2026 5:47 am

You are mixing apples and oranges. The big creator of all are the trade winds in the atmosphere. They push warm water westward, then die down and let warm water slosh eastward. In order to actually determine the oscillation that causes all three conditions, one needs to know what causes trade wind variance.

Jeff Alberts
Reply to  Rud Istvan
June 17, 2026 5:10 pm

Personally, I don’t take much stock in proxies. Maybe they’re “measuring” the thing we think they are, maybe they aren’t.

June 17, 2026 2:59 pm

Andy, a separate comment. My understanding of ENSO has been formed mainly by Bob Tisdale and Willis Eschenbach, both here at WUWT. In my simple layman’s terms, ENSO is just the equatorial sloshing back and forth of the Pacific equatorial ‘surface’ warm pool, governed by the relative strength of the prevailing westerly equatorial tradewinds. Inherently not stable even tho predictably ‘regular’. Bob and Willis differ about the degree of resulting ocean heat loss, which depends on their differing views of NH near surface circulation currents. The known clockwise circulation of the very large North Pacific gyre suggests WE is more likely ‘correct’ that La Niña probably produces a net planetary heat loss while El Niño doesn’t (altho whether the associated deep thermohaline circulation surfacing ‘cools’ even locally is very unlikely). But that hypothesis observationally requires a resolution far beyond our present abilities—so again dunno.

Reply to  Andy May
June 17, 2026 4:42 pm

Their is confusion between “El Nino phase” of the ENSO cycle, and an actual El Nino event.

Mr.
Reply to  Rud Istvan
June 17, 2026 7:08 pm

Rud, if you’re a “simple layman” on this stuff, I’m the classic monkey’s uncle.

We laymen laypersons just observe that swimming in the oceans in winter is cold and not very pleasant, while swimming in the oceans in summer is refreshing and very enjoyable.

The “heat” is all on the surface in summer, and the “cold” is top to bottom in the winter.

(Sharks don’t seem to care if the oceans are “hot” or “cold” though.
It’s their larder 🙂 🙁 )

Herman Pope
June 17, 2026 2:59 pm

Gosh, we have all these correlations, one causes the other or the other way around. What about both of these are caused by something more influential and you do not have a clue as to what that is. Water is abundant, water in the ocean is abundant, water on land is abundant, ice on land is abundant, ice can be piled deep with much volume in cold places, ice can be spread thinner over large areas. It snows more in warmer times and ice accumulates and spreads and causes colder. It snows less in colder times, and accumulated ice spreads and causes colder while thawing and depleting and thinning. When the ice is almost depleted, it retreats rapidly. All the other correlations happened, but could not have caused all this.

Bob Weber
June 17, 2026 4:05 pm

“We are currently at the end of Modern Solar Maximum or the Modern Warm Period”

The Solar Modern Maximum ended 22 years ago, in 2004. The Modern Warm Period does not necessarily have to be synonymous with just the Solar Modern Maximum.

Reply to  Andy May
June 18, 2026 5:54 am

It is my opinion that the surface stores more heat over a long period of time and releases that stored heat over a long period of time. The oceans store more that land, but they both participate in the storage. Just watching the delay in atmosphere warming after land surface temperature increases in informative. It is one reason Tim has been vocal about not balancing fluxes but instead balancing joules over time.

JohnMcL
June 18, 2026 12:27 am

Three points

1. Data coverage in the Nino 3.4. region was poor in the early years. The HadSST data indicates that SST data is available from only a small number of grid cells (like 5 out of 20) in much of the late 1800s. Worse, those returning data are typically clustered at the eastern end of the Nino 3.4 region. On top of that there’s the number of observations (e.g. just one for the month, made during heavy rain might be quite different to the average of 8 measurements across a month). The uncertainty is high and no-one can do anything more than guess whether the reported temperatures (and hence anomalies) applied to the entire Nino 3.4 region.

2. Back radiation heating the ocean is rather a naive belief. Not only is the penetration into the water limited to a few tens of microns but there’s also the matter of the radiation getting down through the increasingly dense atmosphere without being absorbed by a GHG molecule that absorbs at that wavelength. Water vapor absorbs across a large part of the infra-red spectrum and I doubt that there’s any shortage of water vapor in the tropics.

3. It is very difficult to determine cause and consequence with El Nino events. Air pressure above Australia and South east Asia increases and air pressure in the eastern Pacific decreases. This means less westerly wind across the Pacific and less cold water upwelling. The higher pressure bubble also deflects what warm wind there is into the northern mid-latitudes. But are these air pressure changes the cause or a consequence?

Reply to  JohnMcL
June 18, 2026 3:42 pm

Your item 2. is very important. Climate science assumes CO2 absorbs IR that is outgoing but does not absorb IR radiated by itself! It’s just one more of the garbage assumptions climate science makes in order to make everything “balance”. Climate science also assumes that all of the IR emitted by CO2 arrives at the surface at 90° so it all gets absorbed. IR from a CO2 molecule is *NOT* a plane wave, it is a spherical wavefront meaning much of the CO2 re-emitted IR hits the surface at an angle. At best only a little more than half of the radiation emitted by a CO2 molecule in the atmosphere gets absorbed by the surface, assuming the EM wave makes it all the way to the surface!

LT3
June 18, 2026 3:40 am

Does an El-Nino really warm the globe VIA conduction? That does not compute (based on the surface area of the warm pool and the delta T of the Nino State), it would seem elevating global atmospheric water vapor levels via enhanced subtropical jet stream would be the mechanism.

jonesingforozone
June 18, 2026 6:21 am

Climate is distinguished from weather; Corellation is not causation, etc.

June 18, 2026 11:52 am

What I see in the Drroyspencer graph is that every super El Niño is followed by a new higher temperature equilibrium in the atmosphere.

Reply to  Hans Erren
June 18, 2026 3:46 pm

Beware the math behind the graph. If you are plotting averages, then El Nino effects to the average extend long after the spike it physically causes. A spike that is not accounted for is a prime cause for step changes in a plot, especially in a time series plot. It’s even worse if you are using anomalies where the spike effects remain in the baseline used to calculate the anomaly.

conrad ziefle
June 20, 2026 2:27 pm

On the surface of it, that makes no sense to me. Ninos occur frequently, only a few years apart. Climater change, by definition is something that spands (who knows) 50- several hundred years. So how is something with a short cycle causing a long term climatic drift, unless the something, is also changing, in which case, it may be an indicator of the change, not a cause of it.

Dave Burton
June 24, 2026 6:50 pm

Andy, it is true that El Niños ultimately cool the ocean, by accelerating radiative, latent and sensible heat loss, and much of that energy is ultimately radiated to space. (Arctic sea ice loss does the same thing: it accelerates heat loss from the ocean, which also raises air temperatures.) But this is wrong:

“Very little of the heat released from the oceans during an El Niño is returned to the oceans because downwelling infrared radiation from the atmosphere cannot penetrate the ocean surface (Wong & Minnett, 2018). Only solar radiation can penetrate to the deeper ocean and significantly warm it.”

The claim that downwelling LW infrared radiation from the atmosphere does not significantly warm the ocean is false. The depth at which radiation is absorbed does not matter w/r/t its warming effect, because the ocean is being constantly mixed:

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Thanks to that mixing, a joule of energy from LW IR “back radiation” absorbed at a depth of 2 µm has exactly the same warming effect as a joule of energy from sunlight absorbed at a depth of 2 meters.

In fact, incident downwelling LW IR is slightly more effective at warming the water than is sunlight, because about 6% of sunlight is reflected away, and only 2-3% of the LW IR is.

If it were true that LW IR energy absorbed by the ocean did not warm it, that would mean that the absorbed energy is immediately lost by radiation and/or evaporation & convection. But it isn’t.

In fact, the Wong & Minnett paper which you cited tested for that, and they confirmed that when LW IR is absorbed by the ocean it is not immediately lost. They took advantage of the very large changes in incident LW IR that occur when clouds pass over, and they found that there were no corresponding changes in energy emitted by the ocean. Here’s the relevant excerpt:

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They reported that there is not “a significant dependence between LWout and LWin@zenith.” In other words, absorbing more LW IR does not significantly increase the amount of LW IR emitted (other than by gradually increasing the bulk water temperature).

They also say that, “the heat from the absorbed additional IR radiation is not immediately returned to the atmosphere through the upward fluxes of LH, SH, and LWout.”

“LH” is Latent Heat (evaporative cooling), and “SH” is Sensible Heat (conductive heat loss), so they’re saying that absorbing LW IR does not accelerate energy loss by those means, either (other than by very slowly increasing the bulk water temperature, just as absorbed sunlight does).

So, we know that:

* When downwelling LW IR is absorbed in the ocean skin layer it does not affect the rate at which LW IR is emitted by the skin layer; and

* When downwelling LW IR is absorbed in the ocean skin layer it does not affect the rate of evaporation from the skin layer; and

* Even when the energy fluxes between the skin layer and the air/sky are hugely asymmetrical, the temperature of the skin layer stays within a fraction of a degree of the temperature of the water beneath.

Those facts can only be true if the skin layer is tightly thermally coupled to the water beneath. (That should not surprise you, because moving water is a marvelous conductor of heat, which is why we use it in car radiators.) So when radiation is absorbed in the top millimeter of the ocean, it merely adds to the heat content of the water, just as radiation absorbed further down does.

June 25, 2026 5:35 am