Separating Fact from Fiction
In recent years, climate science has continued to evolve with new insights and metrics to better understand and address climate change. However, it is crucial to distinguish between important metrics that can be measured directly and unscientific metrics that are primarily used for propaganda to push the climate crisis narrative.
First off, let’s talk about some of the new metrics that have been invented in climate science. There’s the Global Climate Risk Index, which ranks countries based on their vulnerability to climate change. Then we have the Climate Vulnerability Monitor, which measures the impact of climate change on human health and the global economy. And let’s not forget the Climate Change Performance Index, which ranks countries based on their efforts to combat climate change.
We also have things like the Earth Energy Imbalance (EEI), which is the difference between incoming solar radiation and outgoing heat. Scientists estimate it by piecing together clues like satellite observations of sunlight bouncing back and heat escaping, tracking how much the oceans warm up, and observing changes in air movement. While uncertainties remain, scientists claim that the imbalance exists and likely sits around 0.5 W/m^2 (watts per meter squared), meaning Earth is gaining energy and warming.
However, this imbalance is near the uncertainty of the instruments and calculations used. For example, determining the exact error in measurements of incoming solar radiation (TSI), the input of heat into the system, is intricate and involves various factors. Even the most high-precision instruments, like the Total and Spectral Irradiance Sensor (TSIS), have inherent uncertainties around 0.05-0.1% due to calibration techniques and subtle changes in temperature or electronics. Furthermore, analyzing raw data involves intricate algorithms and corrections for atmospheric effects and instrument artifacts, introducing potential biases and uncertainties of around 0.1-0.2%.
Additionally, TSI naturally fluctuates by about 0.1% over an approximately 11-year cycle, and sudden bursts of energy from the Sun, like flares, can temporarily increase TSI by large amounts, making it challenging to isolate long-term trends.
Thus, current estimates suggest the total error on TSI measurements likely falls within the range of 0.2-0.3%, which translates to approximately 0.2-0.3 W/m^2. This may seem small, but it is 40-60% of the entire estimated energy imbalance and it’s crucial to consider when analyzing long-term trends and accurately estimating Earth’s energy imbalance.
The only two metrics that matter in climate science are human loss of life and economic damage normalized to GDP as these are the direct measurements of the effects of climate change on human civilization.
First of all, let’s talk about human loss of life. This metric is a direct and tangible measure of the impact of climate-related events such as extreme weather, rising sea levels, and changing ecosystems. Sure, there are some risks associated with climate change, like increased heat waves and extreme weather events, but these risks are often exaggerated.
In fact, according to a recent study, the number of deaths related to climate change has actually decreased in recent years.
Source: Irrational Fear
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Kevin Doyle
| #
Dr. Matthew Wielicki,
Your article is thoughtful, and well written.
However, you and many others respond to false assertions created by folks, such as James Hansen, at NASA in America. Hansen is the father of the ‘Greenhouse Gas Theory’. He invented it in his 1967 thesis to earn his PhD at University of Kansas.
If you review all text books and Encyclopedias prior to the 1980’s, there is no mention of ‘Greenhouse Gas Theory’, nor the magical properties of CO2. Was that because the men who landed men on the Moon were too dumb to recognize the mythical and magical powers of a trace gas called CO2?
When examined under the lense of actual Thermodynamics, his ‘Theory’ is non-sense.
‘Greenhouse Gas Theory’ asserts gases, any gas, at 15,000 ft elevation, where the air is freezing, are somehow warming the surface of the Earth.
Please, explain to me and PSI readers how this could possibly occur?
Ever study ‘Thermodynamics’ or ‘Heat Transfer’?
If so, you will certainly understand this ‘Theory’ is absolute ridiculous, stupidity.
As Joseph Goebbels once famously said, “Repeat a lie long enough, and you will come to believe it yourself”…
Reply
Herb Rose
| #
Hi Kevin,
In the troposphere the transfer of energy is almost exclusively done by convection. When molecules collide with something their momentum transfers energy to an object with less energy. The greater the mass (number) of the molecules the greater the amount of energy transferred (as long as the energy of the absorbing mass is less than the energy of the mass transferring energy. The energy equalizes with all the mass of both objects.) In the atmosphere the number of molecules decreases with increasing altitude. Even if the kinetic energy of the molecules remained the same there would be a decrease in the energy transferred to a thermometer (lower temperature). Even when the energy of the gas molecules increased the temperature can still drop due to less energy being absorbed.
Have you ever wondered why the flow of energy in the atmosphere occurs in a zig zag pattern? It is because even as the energy of molecules increases (hence less dense air) the number of molecules striking the instrument deceases. If you were to divide the temperature recorded at an altitude by the density at that altitude you would get the kinetic energy of a constant number of molecules instead of the energy in a constant volume of air. This graph shows that the energy of molecules increases with altitude and since it is energy that is being transferred the freezing gas at 15,000 ft can indeed warm the surface. (It is the sun warming the atmosphere (UV) not the Earth.
Herb
Reply
LOL@Klimate Katastrophe Kooks
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You’re looking at it from the perspective of one going up into the upper atmosphere with a mercury thermometer… that’s not what takes place, and in any case, you’re simply incorrect, Herb.
The temperature measured at altitude is compensated by atmospheric pressure at that altitude via the USSA lapse rate. That’s why the US Standard Atmosphere was conceived of… so aircraft don’t exceed their OEI (one engine inoperable) rating… it’s in Code of Federal Regulations, Title 14, Chapter 1, Subchapter A, Part 1, §1.1.
For Lidar measurement at various altitudes (mainly using Raman rotational scattering for altitudes < ~25 km), corrections are likewise applied.
https://acp.copernicus.org/articles/4/793/2004/acp-4-793-2004.pdf
At no point does temperature at altitude exceed surface temperature:
https://upload.wikimedia.org/wikipedia/commons/thumb/9/9d/Comparison_US_standard_atmosphere_1962.svg/800px-Comparison_US_standard_atmosphere_1962.svg.png
https://www.engineeringtoolbox.com/docs/documents/604/us_atmosphere_temperature_vs_elevation.png
https://en.wikipedia.org/wiki/U.S._Standard_Atmosphere
Reply
Herb Rose
| #
Hi LOL,
No I’m looking at it from how energy is transferred and the way the atmosphere is composed.
The only difference between a thermometer and a barometer is the reservoir of measuring liquid is totally enclosed in a thermometer while open in a barometer. They both receive the same energy from collisions with air molecules. Fewer collisions less energy transferred.
Try holding a weight above a scale and see what it weighs. It reads 0 just as the water in the air adds no weight to it. Now drop the weight and the scale will register a weight that then decreases and becomes stable. It measures the momentum of the weight then its mass. The barometer is not measuring atmospheric pressure (weight per unit area) because the kinetic energy of the molecules is suspending them above the scale. What it measures is the momentum of the molecules. A high pressure area is a result of the vertical momentum of the molecules. A denser, low pressure area does not have less mass but horizontal velocity that transfers less momentum to the instrument. How can you believe an instrument whose start point (sea level) is a variable?
Try looking at the chemical composition of the atmosphere. The energy of the molecules increase with increased altitude. In the troposphere oxygen is in O2, H20, and CO2 molecules. In the stratosphere the H20 and CO2 disappear and the oxygen is in the form of O2 and O3. It requires 495,000 joules/mole to split the O2 molecule into oxygen atoms and create O3. Above this level the oxygen is in the form of nitrogen-oxygen molecules where the oxygen atoms have combined with partially split nitrogen molecules. Since the energy in the atmosphere never reaches the 950,000 joules/mole necessary to split the triple bond of nitrogen near the top of the atmosphere the percentage of nitrogen begins to decreases and oxygen, as oxygen atoms increases. Since the energy in the upper atmosphere is able to split chemical bonds while close to the Earth it is not, it means the energy of the atmosphere is not coming from the surface of the Earth but from the sun as UV radiation.
Do you really believe that a concentration of 10 ppm ozone the ozone layer can absorb 95% of the UV? Try blackening a 1 sq cm of a 100,000 sq cm piece glass and see if it blocks 95% of the visible light.
I know it may be impossible for you to believe but most of the crap you learned in school and cite is false.
Herb
Reply
LOL@Klimate Katastrophe Kooks
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Yeah, everyone knows the energy ultimately comes from the sun, Herb. We treat the surface of the planet as the source because there’s no means in the S-B equation to account for reflected or emitted radiation (and in fact, given that singular photons are identical except for their clockwise or anticlockwise circular polarization, there’s no way to differentiate between reflected or emitted radiation).
Your take on the atmosphere would cause an energy gradient which increases with altitude, thus any energy flow up that gradient would have to be pumped via external energy or that energy could not flow… it’s right there in the 2nd Law of Thermodynamics in the Clausius Statement sense, which you not only fundamentally misunderstand, but which you claim is not valid.
There has never been an empirically-observed 2LoT violation at the macro scale, and we now know that 2LoT is even more rigorously hewed to at the quantum scale, with a whole subset of conditions which must be met for energy to flow, which make up the macroscopic 2LoT.
https://www.pnas.org/content/112/11/3275
Herb Rose
| #
Hi LOL,
Energy flows from the top of the atmosphere down to the surface. This is not only visible light but heat (IR) created as oxygen and nitrogen convert UV into IR.
“Hotter” air rises but at the floor of the Grand Canyon the temperature remains 10 F greater than at the top because the air molecules do not have greater kinetic energy but there are more of them transferring energy to the thermometer.
When a small car with greater velocity runs into the rear of a slower truck (with more kinetic energy because of its mass) the car does not gain velocity/energy. It slows down and the truck speeds up. The law of conservation of momentum clearly states that mass does not matter and that law has been demonstrated by experimentation instead of proof of “never been observed”.
The way to distinguish between reflected and emitted radiation is by the direction of the radiation. Emitted radiation goes in all directions while reflected goes in one direction. There is a difference between a light bulb and a mirror.
There are no photons, light is a wave. In metals and crystals, where the photoelectric effect occurs electrons are separated from atoms and held in place by ionic bonds. All that is necessary to have electrons flow is a disturbance of these bonds, whether done by a changing electromagnetic wave of the right size or mechanical pressure (piezo electric effect) it does not require the creation of a particle or quantum nature of energy. Both Planck and Einstein are wrong.
Herb
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
““Hotter” air rises but at the floor of the Grand Canyon the temperature remains 10 F greater than at the top because the air molecules do not have greater kinetic energy but there are more of them transferring energy to the thermometer.”
You fundamentally misunderstand the thermodynamic equations, Herb. Given that temperature in this sense is a measure of the kinetic energy of the atoms and molecules comprising the atmosphere, you’re arguing against yourself. LOL
https://principia-scientific.com/new-paper-eviscerates-un-ipccs-climate-term-radiative-forcing/#comment-98045
https://principia-scientific.com/new-paper-eviscerates-un-ipccs-climate-term-radiative-forcing/#comment-98046
Review the equations and mathematically describe for us exactly how a higher temperature results in a lower kinetic energy, Herb.
Herb Rose
| #
Hi LOL,
Temperature may be the measurement of kinetic energy but that is not what a thermometer measures. It measures the momentum of the molecules striking it. It has no means of determine the mean or average energy of those molecules.
If you use the “Search” button on PSI you can find my article “How Cold Heats Hot” that explains how energy flows to mass and it flows from an object with higher energy/unit mass to an object with a lower energy/unit mass regardless if that mass has more energy.
Herb
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“The way to distinguish between reflected and emitted radiation is by the direction of the radiation. Emitted radiation goes in all directions while reflected goes in one direction. There is a difference between a light bulb and a mirror.”
Reflection need not be specular, Herb. Again, you haven’t the depth of knowledge necessary to be arguing these topics. You need to buckle down and study. You’re showing the shallowness of your knowledge.
MattH
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Hi Herb. If you increase your vitamin D intake to at least 5,000 international units per day you can reduce the speed of onset dementia.
Your comments on ozone blocking UV light, the termination and accumulation point of much of the suns energy, earth surface, not warming the atmosphere, and the unfortunate comments you made in response to my question on the albedo effect of clouds are all quite concerning.
There is an old saying about putting ones brain into gear before opening ones mouth. Please do an audit of your ideas before commenting.
Best wishes Matt.
P.S. You can take up to 10.000 units vitamin D per day but research vitamin K2 with that.
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“Temperature may be the measurement of kinetic energy but that is not what a thermometer measures. It measures the momentum of the molecules striking it. It has no means of determine the mean or average energy of those molecules.”
A mercury thermometer, by dint of its heat capacity, automatically ‘smears out’ the individual variations in kinetic energy of the atoms and molecules striking the thermometer, Herb. It automatically averages, arriving at the average temperature. Were that not so, we’d see the thermometer jumping up and down with each strike from an atmospheric atom or molecule.
Again, you don’t have the depth of knowledge necessary to be arguing these topics, instead fleshing out your odd hobby theory with what sounds intuitive to you, but which is unphysical.
Buckle down, study. Continuing as you are only shows your lack of depth of knowledge.
Herb Rose
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Hi LOL,
Here’s an experiment you can do to give you empirical proof of what you believe. Boil a large pot of water. Submerge the bulb of a thermometer in the water. It will read 100 C which you say is the mean kinetic energy of the molecules. Turn the heat off and submerge the thermometer to where the water line is just below the 100 C line. This should not change the mean kinetic energy of the molecule. Now look at what temperature the thermometer reads.
Herb
LOL@Klimate Katastrophe Kooks
| #
A mercury thermometer for liquids is intended to be used by submerging its bulb into the liquid, not the entire thermometer, Herb.
If you believe a mis-use of a thermometer ‘proves’ your point, your depth of knowledge is even shallower than I’d previously surmised. If you believe any of us would be bamboozled by your weak attempt at ‘proving’ yourself ‘not wrong’, you have a lower opinion of us than any of us have previously surmised.
Again, stop fleshing out your odd hobby theory with what sounds intuitive but which is unphysical, and buckled down, crack a book and study.
Herb Rose
| #
Hi Matt,
If you dig into the Earth (not in Arctic or Antarctic) the temperature will decline but then begin to increase. This is where you cross from earth being heated by the sun to being heated by geothermal heat. The equilibrium point.
The reason the ozone is more stable in the stratosphere than at the surface is not that it is colder but that it is unable to lose energy by radiation. When ozone absorbs UV it gains energy and becomes more stable. Just as CO2 at a concentration of 440 ppm cannot block heat from the Earth ozone at a concentration of 10 ppm cannot block 95% of the uv from reaching the Earth. The density of matter at 80 km is .000005 kg/m^3. If an amount of energy, x, was equal distribute to these molecules and the same amount of energy (no decrease because of transfer of energy to other object or increase in distance) were to be spread around the 1.2 kg of matter in the atmosphere at the surface, the energy of the molecules at 80 km would be 60,000 time the energy of the air molecules at sea level. Which way does the energy flow? Now take into account the loss of energy due to transfer and distance and also the molecules in the Earth are over more than 1000 times the number on the surface what is heating what? If you want to believe that the molecules in the atmosphere are not absorbing radiated energy from the sun then the Law of thermodynamics that says all matter absorb radiated energy is wrong and there is matter that radiates but does not absorb energy whose temperature should drop to absolute 0.
If you go into an igloo made of snow it will not be dark because all the sunlight has been reflected but lit by the light being transmitted and refracted by the snow.
Please reconsider your dose of vitamin D.
Herb
Herb Rose
| #
Mercury thermometers are calibrated using water. Ah unfortunate choice sine so much of the energy becomes internal energy and is not radiated. All other types are calibrated using a mercury thermometer.You can use many other types of liquids instead of the magical mercury to make a thermometer but they react the same to energy and record how much energy is being transferred by momentum.
What happened to your no response promise? You decided, you needed how believing in what physics told you rather than thinking obligated you to repeat the stupidity to justify all the time and money you wasted learning to believe in nonsense?
Herb
LOL@Klimate Katastrophe Kooks
| #
In statistical mechanics the following molecular equation is derived from first principles: P = n k_B T for a given volume.
Therefore T = (P / (n k_B)) for a given volume.
Where: k_B = Boltzmann Constant (1.380649e−23 J·K−1); T = absolute temperature (K); P = absolute pressure (Pa); n = number of particles
If n = 1, then T = P / k_B in units of K / m³ for a given volume.
Now, knowing that you’re a pedant, you’ll likely bleat something like “Temperature does not have units of K / m³ !!!“… note the ‘for a given volume‘ blurb, Dunce. We will cancel volume in a bit. LOL
We can relate velocity to kinetic energy via the equation:
v = √(v_x² + v_y² + v_z²) = √((DOF k_B T) / m) = √(2 KE / m)
As velocity increases, kinetic energy increases.
Kinetic theory gives the static pressure P for an ideal gas as:
P = ((1 / 3) (n / V)) m v² = (n k_B T) / V
Combining the above with the ideal gas law gives:
(1 / 3)(m v²) = k_B T
∴ T = mv² / 3 k_B for 3 DOF
∴ T = 2 KE / k_B for 1 DOF
∴ T = 2 KE / DOF k_B
Note especially that temperature (T) is a direct measure of the Kinetic Energy (KE) in the equation above, Herb. You deny all of the above and blather on about momentum.
Herb Rose wrote:
“Temperature may be the measurement of kinetic energy but that is not what a thermometer measures.”
Translation:
“Temperature may be a measurement of kinetic energy, but a thermometer (a temperature measurement device) measures something else.”
Do you even listen to yourself, Herb? LOL
Stop beclowning yourself, stop fleshing out your odd hobby theory with what sounds intuitive (but which is unphysical), buckle down, crack a book and study, Herb.
Herb Rose
| #
Hi again LOL,
The misuse of the thermometer is exposing the measuring liquid to more molecules transferring energy, giving a false reading. When the thermometer is used in the atmosphere, where all the measuring liquid is exposed to one medium, changes in the number of molecules transferring energy to it does not effect its accuracy.
Herb
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“The misuse of the thermometer is exposing the measuring liquid to more molecules transferring energy, giving a false reading. When the thermometer is used in the atmosphere, where all the measuring liquid is exposed to one medium, changes in the number of molecules transferring energy to it does not effect its accuracy.”
How do you know this to be true except by having put the upper portion of that thermometer in an absolute vacuum when using it in air?
Mercury thermometers meant for liquid and gaseous use are calibrated differently to account for the fact that the thermometer meant for gaseous use is fully immersed in the gas being measured, Herb. Again, you attempt to use a non sequitur to ‘prove’ yourself ‘not wrong’. It’s not working. LOL
Herb Rose
| #
Hi LOL,
A Pirani gauge will tell you how much heat is lost by radiation and by convection at sea level. The amount lost by radiation is .4%.
Herb
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“A Pirani gauge will tell you how much heat is lost by radiation and by convection at sea level.”
What are you even talking about? A Pirani gauge is a regulated power supply (a constant-voltage op-amp), connecting a heating element (usually a platinum wire) through a Wheatstone bridge.
A Pirani gauge is used to measure the vacuum level in a vacuum system. As the vacuum increases, the heating element gets hotter, its electrical resistance increases, unbalancing the Wheatstone bridge, giving a measure of the pressure in the system.
What does that have to do with “heat lost by radiation and by convection at sea level”? A Pirani gauge isn’t even used for that purpose.
Herb Rose wrote:
“The amount lost by radiation is .4%.”
Our planet’s surface energy is removed ~76.2% by convection, advection and latent heat, and ~23.8% by radiation. Perhaps if you used the correct equipment and the proper maths to figure that out, you’d not embarrass yourself by claiming only 0.4% of surface energy is removed by radiant exitance. LOL
https://i.imgur.com/zQHdxkF.png
That’s from Andy May… an actual physicist. Where are you getting your piffle from? LOL
Herb Rose
| #
Hi LOL,
Use the Search button PSI to look up Tom Shula :An interview. His career is in the chip and thin film industry where complete vacuums are necessary in manufacturing. He will give you all his credentials and experience. I’d suggest you watch it to see the experimental evidence showing your figures are wrong.
Herb
Herb Rose
| #
OOPs The names JohnShula not the football coach.
Herb
LOL@Klimate Katastrophe Kooks
| #
That’s a mis-use of the Pirani gauge and total junk science by someone who doesn’t know his equipment nor much about the climate.
Shula claims the Pirani gauge heating element is run at “typically 50 C to 100 C”… not even close… the manufacturer’s datasheet for the Edwards APG100 Vacuum Gauge shows it runs at 100 C over ambient, as does the datasheet for the KJLC vacuum gauge. The “50 to 100 C” is the ambient operating temperature range… the ambient temperature range at which the gauge can be run without needing special calibration for a higher or lower ambient temperature. Did Shula calibrate his Pirani gauge for a lower ambient operating temperature, assuming he didn’t run it at the minimum 50 C ambient operating temperature during his tests, but instead ran it at room temperature?
Further, Tom Shula estimates the altitude at which convection and radiation balance at 50/50 at 250,000 feet… that’s ludicrous. It balances where convection is no longer possible, at the tropopause… that’s there for a reason, after all. That is where radiative-convective equilibrium is established. That’s 5 times lower than Shula purports the tropopause to be at. That Shula didn’t even know that radiative-convective equilibrium is established at the tropopause means he shouldn’t be listened to in regards to climate. That is literally what sets tropopause height, after all.
Further, he claims the Stefan-Boltzmann law “only holds strictly at 0 K, i.e., a perfect vacuum” (his words)… as if he believes a perfect vacuum will be at 0 K regardless of the radiation within that ‘perfect vacuum’ cavity… IOW, he doesn’t have a firm grasp upon even the fundamentals. Even a perfect vacuum will have radiation energy density (ie: a temperature, remember that temperature is equal to the fourth root of energy density divided by Stefan’s Constant)… one doesn’t just achieve a perfect vacuum to achieve 0 K. One must pull that perfect vacuum’s containment to as near 0 K as is practicable to achieve 0 K.
But don’t let that stop you from latching onto every bit of misapplication of science to bolster your odd hobby theory in which you deny large swaths of scientific reality. LOL
Herb Rose
| #
Hi LOL,
We will disagree. The Stefan-Boltzmann law is based on a black body where there is no convection, only radiation, which necessitates a complete vacuum..
The division of troposphere, stratosphere, mesosphere, and thermosphere is based on temperature.and as I have said that is not how energy flows. The top of the troposphere is the atmosphere becomes almost exclusively composed of oxygen and nitrogen. No argon, no CO2, and almost no H2O. It has nothing to o with the way energy is transferred. Ozone and nitrogen-oxygen molecules are created from collision.
If he is so incompetent at what he does shy is he still employed and how can computers, cell phones, and other electronics work? The problem is that your ego can never admit to being wrong and you are willing to deny any evidence or anyone who disagrees with you. You know all the right answers that others have given you but are incapable of thought. It was people like you who proclaimed the atomic clocks on satellites proved Einstein’s theory was true but when it was pointed out that the evidence from the clocks showed that his theory was wrong the evidence was ignored and the belief in the correctness was maintained. You will only accept evidence that confirm your existing beliefs. You are not scientist but disciples of a cult.
Herb
LOL@Klimate Katastrophe Kooks
| #
And finally, the Pirani vacuum gauge is designed much like a reflective cavity… it is designed to minimize radiative energy loss by reflecting energy impinging upon the walls back into the cavity, because it is designed to measure conductive / convective energy loss as a proxy of measuring depth of vacuum. So of course the energy loss from radiation is low, it’s designed to be for this instrument’s application.
Just because someone slapped together a compendium of junk science and confidently called it ‘The Science™’ doesn’t mean you shouldn’t do your due diligence and check out their premise, their procedures, their conclusions and their possible ulterior motives for yourself, Herb.
Did you also believe the pharmaceutical industry when it told you to get that “100% safe, 100% effective” mRNA gene-therapy ‘vaccine’ and half a dozen ‘boosters’? LOL
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“The Stefan-Boltzmann law is based on a black body where there is no convection, only radiation, which necessitates a complete vacuum..”
Not true at all. You simply don’t understand the S-B equation, and you’re parroting what Tom Shula said because you don’t know any better.
https://i.imgur.com/QErszYW.gif
In point of fact, there are two forms of the S-B equation, one for idealized blackbody objects, one for graybody objects, as the animated image above shows.
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“The top of the troposphere is the atmosphere becomes almost exclusively composed of oxygen and nitrogen. No argon, no CO2, and almost no H2O.”
Not true at all. You lose credibility when you make stuff up, Herb.
https://www.nasa.gov/centers-and-facilities/langley/twenty-years-on-saber-on-timed-still-observing-the-upper-atmosphere/
“SABER has observed that the rate of increase of carbon dioxide in the mesosphere and lower thermosphere is the same as at Earth’s surface.”
You ever hear of the term “well-mixed gas”, Herb? Yeah. LOL
In fact, CO2 is the most prevalent radiative atmospheric coolant in the upper atmosphere:
https://imgur.com/R6uHyvK.png
The image above is from a presentation given by Dr. Maria Z. Hakuba, an atmospheric research scientist at NASA JPL.
https://imgur.com/O4PkEPH.png
The image above is adapted from the Clough and Iacono study, Journal Of Geophysical Research, Vol. 100, No. D8, Pages 16,519-16,535, August 20, 1995.
Note that the Clough & Iacono study is for the atmospheric radiative cooling effect, so positive numbers at right are cooling, negative numbers are warming.
The NASA SABER project on the TIMED satellite proved that CO2 and oxides of nitrogen expelled massive amounts of energy to space during a solar flare, and act as the most prevalent upper-atmosphere coolants at all times.
Herb Rose
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Hi LOL,
I’m sure there are trace amounts of many gases in the mesosphere resulting from volcanic action and the launch of tons of materials into orbits. You continue to ignore the energetic forms of all the gases and their kinetic energy while maintaining the low temperature. How can you claim they are sending large amounts of energy into space when also arguing they have very little energy and they get this energy from the surface of the Earth, not the sun? It’s the coldest region on Earth containing very few molecules with large amounts of energy.
Herb
Herb Rose
| #
Hi again,
If you would, could you give us an explanation of why, as the temperature zig zags, the density of the atmosphere continues to decline in a smooth curve. Doesn’t colder temperature produce denser gases?
Herb
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“How can you claim they are sending large amounts of energy into space when also arguing they have very little energy and they get this energy from the surface of the Earth, not the sun? It’s the coldest region on Earth containing very few molecules with large amounts of energy.”
You yet again demonstrate your inability to grasp the fundamentals, Herb. Again, go back, review the equations, and tell us exactly how the equations arrive at a high kinetic energy but a low temperature.
https://principia-scientific.com/separating-fact-from-fiction/#comment-98204
Write it out in mathematical form as proof of your contention. If you’re unable to do so, that’s evidence that you’re wrong.
You’ll be unable to do so. LOL
LOL@Klimate Katastrophe Kooks
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The entire CAGW ‘industry’ is a sham, built upon a provable mis-use of the Stefan-Boltzmann equation to inflate radiant exitance of all objects, so they can claim a ‘backradiation’ which they then claim causes warming.
https://imgur.com/a/fCDSa0j
CO2 is a net atmospheric radiative coolant at all altitudes except for negligible warming at the tropopause:
https://imgur.com/R6uHyvK.png
The image above is from a presentation given by Dr. Maria Z. Hakuba, an atmospheric research scientist at NASA JPL.
https://imgur.com/O4PkEPH.png
The image above is adapted from the Clough and Iacono study, Journal Of Geophysical Research, Vol. 100, No. D8, Pages 16,519-16,535, August 20, 1995.
Note that the Clough & Iacono study is for the atmospheric radiative cooling effect, so positive numbers at right are cooling, negative numbers are warming.
The Kiehl-Trenberth graphic:
https://i.imgur.com/Ghi99y0.png
… proves the climastrologists misuse the Stefan-Boltzmann equation to inflate radiant exitance of all objects, in the process inventing out of thin air a ‘backradiation’ that they then claim causes CAGW… it treats a real-world (graybody) surface as if it were an idealized blackbody object, with emission to 0 K ambient and ε = 1. That’s the only way that diagram can get to 390 W m-2 surface radiant exitance:
https://imgur.com/ml0glky.png
That’s proof-positive that they’ve misused the S-B equation to fit their narrative. Had they used the actual emissivity (0.93643, per the NASA ISCCP program), they couldn’t have arrived at 390 W m-2 (see below, left), and had they used the proper form of the S-B equation for graybody objects, they’d not have even gotten close to 390 W m-2 (see below, right).
https://imgur.com/9zIhMRs.png
The S-B equation for graybody objects isn’t meant to be used to subtract a fictive ‘cooler to warmer’ energy flow from the incorrectly-calculated and thus too high ‘warmer to cooler’ energy flow, it’s meant to be used to subtract cooler object radiation energy density (temperature is a measure of radiation energy density, the fourth root of radiation energy density divided by Stefan’s constant) from warmer object radiation energy density. Radiant exitance of the warmer object is predicated upon the radiation energy density gradient.
A proper application of the fundamental physical laws shows that ‘backradiation’ does not (can not) warm the surface.
How else can we prove this? Entropy.
The climate doomsayer’s problem, however, is that their take on radiative energy exchange necessitates that at thermodynamic equilibrium, objects are furiously emitting and absorbing radiation (this is brought about because they claim that objects emit only according to their temperature (rather than according to the radiation energy density gradient), thus for objects at the same temperature in an environment at the same temperature, all would be furiously emitting and absorbing radiation… in other words, they claim that graybody objects emit > 0 K), and they’ve forgotten about entropy… if the objects (and the environment) are furiously emitting and absorbing radiation at thermodynamic equilibrium as their incorrect take on reality must claim, why does entropy not change?
The second law states that there exists a state variable called entropy S. The change in entropy (ΔS) is equal to the energy transferred (ΔQ) divided by the temperature (T).
ΔS = ΔQ / T
Only for reversible processes does entropy remain constant. Reversible processes are idealizations. All real-world processes are irreversible.
The climastrologists claim that energy can flow from cooler to warmer because they cling to the long-debunked Prevost Principle, which states that an object’s radiant exitance is dependent only upon that object’s internal state, and thus they treat real-world graybody objects as though they’re idealized blackbody objects via: q = σ T4.
… thus the climate alarmists claim that all objects emit radiation if they are above 0 K. In reality, idealized blackbody objects emit radiation if they are above 0 K, whereas graybody objects emit radiation if their temperature is greater than 0 K above the ambient.
But their claim means that in an environment at thermodynamic equilibrium, all objects (and the ambient) would be furiously emitting and absorbing radiation, but since entropy doesn’t change at thermodynamic equilibrium, the climastrologists must claim that radiative energy transfer is a reversible process. Except radiative energy transfer is an irreversible process, which destroys their claim.
In reality, at thermodynamic equilibrium, no energy flows, the system reaches a quiescent state (the definition of thermodynamic equilibrium), which is why entropy doesn’t change. A standing wave is set up by the photons remaining in the intervening space between two objects at thermodynamic equilibrium, with the standing wave nodes at the surface of the objects (and being wave nodes, no energy can be transferred into or out of the objects). Should one object change temperature, the standing wave becomes a traveling wave, with the group velocity proportional to the radiation energy density differential, and in the direction toward the cooler object. This is standard cavity theory, applied to objects.
All idealized blackbody objects above absolute zero emit radiation, assume emission to 0 K and don’t actually exist, they’re idealizations.
Real-world graybody objects with a temperature greater than zero degrees above their ambient emit radiation. Graybody objects emit (and absorb) according to the radiation energy density gradient.
It’s right there in the S-B equation, which the climate alarmists fundamentally misunderstand:
https://i.imgur.com/QErszYW.gif
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LOL@Klimate Katastrophe Kooks
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Radiant Exitance =
= ε σ (T^4_h – T^4_c) A_h
= (ε c (e_h – e_c)) / 4
= σ / a * Δe
= Faraday’s Constant (Ah) * (Molar Mass of Atmosphere / Ratio of Atoms Per Particle)
= Faraday’s Constant (Ah) * Moles of electrons per Mole of atmospheric particles
= (((Faraday’s Constant (J electron-1) * Number of Electrons per Mole of Atmosphere) / 3600) / Stefan-Boltzmann Constant)^0.25
= (((Faraday’s Constant (J mol-1) * Moles of Electrons per Mole of Atmosphere) / 3600) / Stefan-Boltzmann Constant)^0.25
It is the second equation, the Stefan-Boltzmann equation stripped down to its roots, which shows that it should properly be used by subtracting the energy density of the cooler object from the energy density of the warmer object to arrive at the energy density gradient, which is what determines radiant exitance of the warmer object… there is an EM field gradient between the objects. This is why cooler objects not only cannot warm warmer objects, but they cannot even emit in the direction of the warmer object… because a warmer object will have higher energy density at all wavelengths than a cooler object:
https://i.imgur.com/wb9KwS0.png
Do remember that temperature (T) is a measure of radiation energy density (e), equal to the fourth root of radiation energy density divided by Stefan’s Constant.
e = T^4 a
a = 4σ/c
e = T^4 4σ/c
T^4 = e/(4σ/c)
T = 4^√(e/(4σ/c))
T = 4^√(e/a)
As Δe → 0, ΔT → 0, q → 0. As q → 0, the ratio of graybody object total emissive power to idealized blackbody object total emissive power → 0. In other words, emissivity → 0. At thermodynamic equilibrium for a graybody object, there is no radiation energy density gradient and thus no impetus for photon generation.
As Δe → 0, ΔT → 0, photon chemical potential → 0, photon Free Energy → 0. At zero chemical potential, zero Free Energy, the photon can do no work, so there is no impetus for the photon to be absorbed. The ratio of the absorbed to the incident radiant power → 0. In other words, absorptivity → 0.
α = absorptivity = absorbed / incident radiant power
ρ = reflectivity = reflected / incident radiant power
τ = transmissivity = transmitted / incident radiant power
α + ρ + τ = 100%
For opaque surfaces τ = 0% ∴ α + ρ = 100%
If α = 0%, 0% + ρ = 100% ∴ ρ = 100% … all incident photons are reflected at thermodynamic equilibrium for graybody objects.
This coincides with standard cavity theory… applying cavity theory outside a cavity, for two graybody objects at thermodynamic equilibrium, no absorption nor emission takes place. The photons remaining in the intervening space set up a standing wave, with the wavemode nodes at the object surfaces by dint of the boundary constraints. Nodes being a zero-crossing point (and anti-nodes being the positive and negative peaks), no energy can be transferred into or out of the objects. Photon chemical potential is zero, they can do no work, photon Free Energy is zero, they can do no work. Should one object change temperature, the standing wave becomes a traveling wave with the group velocity proportional to the radiation energy density gradient and in the direction of the cooler object.
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LOL@Klimate Katastrophe Kooks
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Most people cannot think in terms of energy, energy density and energy density gradient. We need to analogize to something they’re familiar with.
Thus, just as, for instance, water only spontaneously flows down a pressure gradient (ie: downhill), energy only spontaneously flows down an energy density gradient. That’s 2LoT in the Clausius Statement sense, in a nutshell. So one tack to take is to ask people if water can ever spontaneously flow uphill. Of course they’ll say, “No, water cannot flow uphill on its own.” Then show them dimensional analysis.
mass (M), length (L), time (T), absolute temperature (K), amount of substance (N), electric charge (Q), luminous intensity (C)
We denote the dimensions like this: [Mx, Lx, Tx, Kx, Nx, Qx, Cx] where x = the number of that dimension. We typically drop out those dimensions which are not used.
Force: [M1 L1 T-2] divided by…
Area: [M0 L2 T0] equals…
Pressure: [M1 L-1 T-2] divided by…
Length: [M0 L1 T0] equals…
Pressure Gradient: [M1 L-2 T-2]
Explain to them that Pressure is Force / Area, and that Pressure Gradient is Pressure / Length. Remind them that water only spontaneously flows down a pressure gradient (ie: downhill).
Then introduce energy. Tell them that energy is much like water. It requires an impetus to flow, just as water requires an impetus (pressure gradient) to flow. In the case of radiative energy, that impetus is a radiation energy density gradient, which is analogous to (and in fact, literally is) a radiation pressure gradient.
Energy: [M1 L2 T−2] divided by…
Volume: [M0 L3 T0] equals…
Energy Density: [M1 L-1 T-2] divided by…
Length: [M0 L1 T0] equals…
Energy Density Gradient: [M1 L-2 T-2]
Explain to them that Energy Density is Energy / Volume, and Energy Density Gradient is Energy Density / Length. Highlight the fact that Pressure and Energy Density have the same units (highlighted above). Also highlight the fact that Pressure Gradient and Energy Density Gradient have the same units (bolded above).
So we’re talking about the same concept as water only spontaneously flowing down a pressure gradient (ie: downhill) when we talk of energy (of any form) only spontaneously flowing down an energy density gradient. Energy density is pressure, an energy density gradient is a pressure gradient… for energy.
It’s a bit more complicated for gases because they can convert that energy density to a change in volume (1 J m-3 = 1 Pa), for constant-pressure processes, which means the unconstrained volume of a gas will change such that its energy density (in J m-3) will tend toward being equal to pressure (in Pa). This is the underlying cause of convection.
Yes, folks, the sorry state of education is such that people have been bamboozled into believing something directly analogous to water spontaneously flowing uphill.
Use the above to ridicule the climastrologists and their enablers into silence… because their clown-world beliefs deserve nothing but denigration.
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