Ontological Mathematics & the Theory of Everything 5: The Mind and the Brain – Consciousness and God

Is it possible to define God and explain consciousness? In an earlier episode we realized that we would have to explain God if we were to explain existence. Learn how we can now do that, through mathematical reason.

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132 Responses to Ontological Mathematics & the Theory of Everything 5: The Mind and the Brain – Consciousness and God

  1. Philip Mulholland says:

    The brain is a transceiver of thought energy.

    It therefore follows that memories are the transmission of information through time.
    In effect, our brains are time machines.

  2. Pingback: Rp. Ontological Mathematics & the Theory of Everything 5: The Mind and the Brain – Consciousness and God – Unexpected Objects

  3. CD Marshall says:

    So the bot Alex is back on Twitter or Anubis. Evenminded. Balloon Boy. Erik. Bugs Bunny. Eli Rabbet or whoever he is posing as.

  4. CD Marshall says:

    LOL@Klimate Katastrophe Kooks
    Alex says hi and he really misses you. 😁
    “Um yeah, that’s not written by a scientist. It was written by a crank that thinks he has a perpetual free energy machine.”

    I guess you didn’t know you didn’t have a science degree. Now that’s weird.

  5. LOL@Klimate Katastrophe Kooks says:

    Poor Alex… even after all this time, he finds himself utterly unable to pull his head out of liberalism’s arse so he can take a look around at reality. LOL

  6. LOL@Klimate Katastrophe Kooks says:

    Eli Rabbet is Joshua Halpern… I’m unsure if Joshua Halpern is ‘evenminded’, but he very well could be. If so, I’ve drop-kicked him across the width and breadth of the internet for more than 3 years, often causing him to melt down so badly that he just repeated a certain word (“DISMISSED!!!”) for hours on end, or found himself utterly unable to form complete sentences. LOL

  7. Jopo says:

    Hey guys need help in tidying up my terminology and perhaps numbers. Really hoping you guys could help out. My thoughts are based upon the inspiration of many..
    From Nikolov and Zeller, Connolly and Connolly, Postma, Holmes and others that escape me at this moment.
    Please advise honestly. If it is all a coincidence or it does not prove anything? Or my application of the math is just wrong.
    So I have recently noticed that the molar mass of “air” when converted to joules and watts equates to 390 Watts or 288 Kelvin when we assume the stp of 101325 pascals.

    This is based solely on the energy contained within an electron here on Earth

    Molar mass of air is approx.. 28.96 grams per mole. When I calculate how many electrons are in a “mole of air” I get 8.74E+24 electrons or 14.519 moles of electrons in mole of air
    Knowing that an electron on earth has 1.602E-19 Joules of energy one can then calculate that 1 “mole of air” has 1400859.92 joules of energy. This equates to 390 watts. Which equates to 288 Kelvin. Surely this is not a coincidence.

    VENUS
    I was then also then able to use the exact same process and calculate the temperature on Venus @ 1 BAR – 100,000 pascals with a molar mass of 43.45 grams per mole which is 1.314000E+25 electrons per Venusian mole of air.

    The observed average temp on Venus @1 Bar or 100,000 pascals on Venus is 340 Kelvin

    My calculations are bang on 340 Kelvin.

    The catch though is that I had to change the value of energy (joules) an electron here on earth to that of Venusian using Earths perspective. This is unusual but it works!

    So 1.60e-19 joules / Venusian gravity multiplied by Earths gravity multiplied by the ratio of distance with respect to the sun for Venus and Earth. Then apply the inverse square law.
    So Venus is 1.91 times closer to the Sun.
    Using the inverse square law on 1.91 the factor is now 1.91^.25 which equates to 1.175. This inspiration come from the Robert Holmes paper
    On the Apparent Relationship Between Total Solar Irradiance and the Atmospheric Temperature at 1 Bar on Three Terrestrial-type Bodies

    So 1.602e-19 / 8.87 * 9.80665 * 1.175 = 2.08e-19 Joules per electron

    Therefore 1.314000E+25 electrons per Venusian mole * 2.08e-19 joules per electron / 3600 to get Watts = 759 Watts thus 340 Kelvin

    Below is a summary table of above.

    Earth

    Venus

  8. Nepal says:

    @jopo, I’m having trouble seeing some of the steps in your post. Can you explain the bolded parts more?

    Knowing that an electron on earth has 1.602E-19 Joules of energy one can then calculate that 1 “mole of air” has 1400859.92 joules of energy. **This equates to 390 watts. *

    I’m really not sure about the energy per electron part. For the second part, it looks like you’re trying to convert Joules to Watts. The only way to do that is to divide by some amount of time… but what amount of time did you choose, and why does that make sense?

  9. LOL@Klimate Katastrophe Kooks says:

    Jopo wrote:
    “Knowing that an electron on earth has 1.602E-19 Joules of energy”

    “The catch though is that I had to change the value of energy (joules) an electron here on earth to that of Venusian using Earths perspective. This is unusual but it works!”

    The elementary charge doesn’t change, it’s always 1.602176634e−19 C (Coulomb).

    1 C = (1 / 1.602176634e−19) e

    So 1 C = 6241509074460762607.7762409809304 e.

    Or, in simple language, 6241509074460762607.7762409809304 electrons equates to 1 C.

    8.74e24 electron / mol of air (from your post):
    8740000000000000000000000 / 6241509074460762607.7762409809304 e C-1 = 1400302.378116 C

    Faraday constant (F) = 96485.3321233 C mol-1, which gives us
    1400302.378116 C / 96485.3321233 C mol-1 =
    14.513111447100926787218348661908 mol of electrons.

    Pretty on-track so far… you’ll have to describe how you got Joules from Coulombs, and how you converted that to temperature in Kelvin.

    A Joule is a unit of energy, a Coulomb is a unit of electric charge… one cannot interconvert unless one knows the voltage, given that 1 J/C = 1 V.

    That’s sort of like asking how many watts equal one amp. You don’t know unless you also know the voltage.

  10. LOL@Klimate Katastrophe Kooks says:

    Ah, I think I see what you did.

    You set the Earth J/C = 1 volt as a reference. Then you shifted the Earth’s J/C of 1 volt to 1.299731849665709205443324421869 V for Venus. Is that it? I’m not sure that’s valid, but then I can’t find anything about the J/C ratio of either planet. We know the atmospheric electrical gradient of Earth is ~120 V / m in a flat field on a clear day… do we know what it is on Venus? Because that’d give us a direct comparison that may lend corroboration to your calculations.

    (0.0000000000000000001602176634 / 8.87 m s-2) * 9.80665 m s-2 * (Earth aphelion + perihelion / 2 =(152100000 km + 147,095,000 km)/2 = 149,597,500 km / Venus aphelion + perihelion / 2 = (108,939,000 km + 107,477,000 km)/2 = 108,208,000 km) = 0.00000000000000000024489076916351516038737703543436 = 2.4489076916351516038737703543436e-19

    So whereas you get 2.08240e-19 J/e for Venus, I get 2.4489e-19… you might check to see if that’s just due to rounding in your calculations, and whether that invalidates your calculations.

  11. Jopo says:

    My numbers are correct. The only issue is that I have created a constant of 3600 or I have converted to watt hours. Which is thus difficult for me to explain why. It is a massive coincidence though that then using the SB law then equates this to 288 K. The next coincidence is that we can use gravity and the inverse square law to calc the temp on Venus at 1 Bar which is the equivalent of Earth pressure. So the sticky point is why convert to Watt hours. Yep I need help here. Calculating Watts required to heat up a gas. Is it normal to convert joules to watt hours. To a point I say yes it is. We do it to calculate size of elements to use to heat up water or even AIR

    As for the mention that the 1.602e-19 EV or joules is applied universally, I disagree. The fact that we can apply conversions using gravity from Venus and Earth tells me different.
    SI unit for joules is kg.m2/s2
    SI unit for power is kg.m2/s3

    I never said I solved this. I said I have come across and I need to tidy up. What is the significance of the constant 3600? Why does it only apply to around the 1 Bar mark. These are other questions that need to be addressed after. I suspect that is relevant to optical density. I am way over my head here and need guidance for or against.

  12. Jopo says:

    Kooks if you apply the inv sq law as outlined in Robert Holmes paper then the 2.4489e-19 divide by 1.175 you will get the 2.0824e-19

  13. LOL@Klimate Katastrophe Kooks says:

    Aside… you want to terraform Venus? You want to clear the sulfuric acid from its atmosphere and make it more Earth-like?

    Lime (CaCO3)… millions upon millions upon millions of tons of limestone ground to a very fine powder and applied at the top of the atmosphere. It’ll react with the sulfuric acid to form gypsum (CaSO4), with which one could then build buildings.

    H2SO4 (sulfuric acid) + CaCO3 (lime) → CaSO4 (gypsum) + H2O (water) + CO2 (carbon dioxide)

    The atmosphere would clear, the sulfuric acid trapping energy in the atmosphere would chemically react and no longer be in the atmosphere, so the Atmospheric Infrared Window would open wide, the planet would cool.

    The resultant water due to the chemical reaction would start acting as a literal refrigerant (in the strict refrigeration cycle sense) just as it does in Earth’s atmosphere, and the additional CO2 would act as a net atmospheric radiative coolant just as it does in Earth’s atmosphere.

    As well, the CO2 would be excellent plant food, but I suspect that one would have to grow the plants in a partially-shaded building due to the higher solar insolation of Venus. That gets O2 into the atmosphere.

    A few million years of that, and you’ve got a practically livable planet with a much less-dense atmosphere, although it’d be much hotter than Earth due to nearly double the solar insolation… and I couldn’t imagine living on a planet with a day longer than its year… the heat buildup would be brutal even after we’d terraformed it, and the night-side would now get brutally cold, so the winds between day and night side would be vicious.

    But if we’re at that point on the Kardashev Scale where we’re terraforming a planet, I’m sure we can think of ways to spin it up to a 24 hour day and move it outward in orbital radius.

  14. boomie789 says:

    “2000 Mules” (2022)
    “Documentary on the alleged criminal voter fraud and ballot stuffing during the 2020 USA Presidential Election, which may have changed the legitimate outcome.”

    (https://odysee.com/@BannedYouTubeVideos:4/2000-Mules-2022:a?r=AUHBg8UpFrFRjosg7dZhKdpHwQ2D37MM)

  15. LOL@Klimate Katastrophe Kooks says:

    Jopo wrote:
    “Kooks if you apply the inv sq law as outlined in Robert Holmes paper then the 2.4489e-19 divide by 1.175 you will get the 2.0824e-19”

    2.4489076916351516038737703543436e-19 / 1.175 = 2.0841767588384268969138471100797e-19

    Yeah, you’re right… that’s either an odd coincidence, or you’ve got something there. Keep digging, you’ll find the connections, if they’re there, such that you can fully explain it.

  16. Nepal says:

    Jopo, I still don’t understand why you say “an electron on earth has 1.602E-19 Joules of energy”. An electron has a charge of 1.6E-19 Coulumbs, but that is another thing entirely… sort of like the difference between being 6 feet tall and having 6 dollars in your pocket. Same number, but representing totally different things.

    Also, I see that you got to watts from joules by dividing by an hour. The thing is, an hour is just a random amount of time that humans made up. You’re trying to calculate something fundamental about nature, so it can’t possibly rely on some random person who decided to divide the day into 24 parts.

    All in all, this is very cool, but there’s no doubt in my mind it’s a coincidence.

  17. Joseph E Postma says:

    If it has meaning, given the alignment of the numbers, then it has meaning the same way that the Great Pyramid of Giza and its internal mathematical alignment has meaning: it indicates a design to the universe and the constants of nature.

    You guys should watch this – I’ve checked the math and it is all correct – the Great Pyramid has mathematical relations encoded in it which indicate that its builders were aware of profound mathematical properties of nature, including the speed of light, etc.:

    As Nepal points out – the way Jopo is getting the numbers doesn’t make any sense: however, if Jopo is coming up with them utilizing basic whole numbers which are accepted long-standing features of our society and world…then it indicates a hidden allusion to knowledge across disparate sources you wouldn’t expect.

    If you can come up relations like these using random numbers…then of course it is meaningless…you’re just using the numbers you need to. But if you do it using numbers which are meaningful, and they’re whole numbers, or constants, and relate to human or mathematical existence, then you may have found something meaningful.

  18. boomie789 says:

  19. CD Marshall says:

    I need a competent periodic table to download or copy. What do you guys use? provided you don’t have it memorized.

  20. LOL@Klimate Katastrophe Kooks says:

    There’s this, but it’s online-only:
    https://inl.gov/periodic-table/

    Here’s one that’s online and can be downloaded as a PDF:
    https://periodic-table-hd.netlify.app/

  21. LOL@Klimate Katastrophe Kooks says:

    I created my own, but it’s not fully complete yet… it’s really difficult to draw the electron configuration for the atoms on the higher end of the Periodic Table and have them fit into the spreadsheet.

    I got the idea (and copied the simpler graphics) from an online Periodic Table, but I wanted more info. That Periodic Table at the time didn’t go past Xenon, so I had to make the graphics for the rest of them (I’m up to Platinum) myself… and as you know, I’m no artist.

    I work on it from time to time, whenever I get a roundtuit.

  22. CD Marshall says:

    Thank you, perfect.

  23. CD Marshall says:

    Gerald Kutney, an alleged PhD Chemist, actually gave me this link to explain the GHGE. 😂
    https://climatekids.nasa.gov/greenhouse-effect/#:~:text=The%20Short%20Answer%3A,a%20comfortable%20place%20to%20live

  24. LOL@Klimate Katastrophe Kooks says:

    All cells contain reverse-transcriptase, an enzyme. Some viruses and bacteria also manufacture reverse-transcriptase to allow them to inject themselves more easily into the DNA strand, what is known as reverse-transcription.

    Some diseases also make one especially prone to undergoing reverse-transcription, HIV amongst them.

    So the mRNA can be transcribed into the DNA in the liver, making it a permanent part of your DNA… that means you’re a permanent spike protein factory, your immune system is constantly attacking the spike protein production (ie: attacking your own body), which can lead to immune system exhaustion (which is what they used to call it, now they’ve got cases of immune system exhaustion so bad that they’re starting to call it ‘VAIDS’ (Vaccine Acquired Immune Deficiency Syndrome). Yes, it is functionally the same as having AIDS.

    If that reverse transcription occurs in germline cells (and we now know that can take place), the resultant offspring’s immune system won’t recognize the spike protein nor the virus as a threat… the virus would have free rein in that child’s body, making them a superspreader and variant factory.

    They really screwed the pooch when they started mucking about in DNA and RNA, something they just barely understand. We’re just now starting to see the horrors that will come about because of this essentially unregulated (because the CDC is controlled by the pharmaceutical companies) experiment in gene therapy… and that is what the mRNA ‘vaccines’ are… gene therapy. Even the Pfizer CEO has admitted this.

  25. LOL@Klimate Katastrophe Kooks says:

    Well, the CDC, NIH, FDA, etc., etc., etc.

  26. LOL@Klimate Katastrophe Kooks says:

    And that, of course, explains the increasing rate of hepatitis we’re now seeing. The gene-therapy ‘vaccine’ mRNA (with a non-organic end-cap so it can’t be broken down as easily by the liver) undergoes reverse transcription into liver cells in as little as 6 hours after first exposure.

    If heads don’t literally roll and bodies swing over this debacle, then civilization is lost. We’d have become too sheepified to survive. And superspreader variant factories are how they’ll cull us.

  27. LOL@Klimate Katastrophe Kooks says:

    Anyone interested in nuclear fusion? Webinar tomorrow:
    https://www.comsol.com/events/comsol-days/comsol-day-nuclear-fusion-101502

    Topics:
    Trends in Nuclear Fusion; Testing and Development of RF Components for ITER; Modeling RF Heating of Hydrogen Plasmas for Nuclear Fusion; Magnets, Coils, and Superconductors; Magnetohydrodynamics and Heat Transfer; Designing a Nuclear Fusion Reactor Using Multiphysics; Diagnostic Models for Nuclear Fusion; From Densification to Marangoni Flow: Optics Processing In Support of High-Power Laser Systems; Charged Particle Tracing in Magnetic Fields; Magnetomechanics and Lorentz Forces

  28. CD Marshall says:

    Thank you again, clear and concise.

  29. CD Marshall says:

    Interesting Jonae Nova mentioned Australia (who is killing everything for green energy) has enormous deposits of Thorium. How’s that for sick irony. I seriously doubt that government will allow one ounce of that used.

  30. CD Marshall says:

    *Joanne Nova

  31. CD Marshall says:

    So in the simplistic version here, energy leaving the planet is not the only means of cooling the planet. The adiabatic converts kinetic energy to potential energy. I did hear Earth has more PE than KE in the atmosphere.

  32. CD Marshall says:

    This does look like a little like a doorway. Maybe China got to Mars first. 😯

    https://img-s-msn-com.akamaized.net/tenant/amp/entityid/AAXbR2L.img?h=768&w=1366&m=6&q=60&o=f&l=f

  33. LOL@Klimate Katastrophe Kooks says:

    Remember the old adage: “What goes up, must come down.”

    So while a rising parcel of air converts temperature (atomic and molecular kinetic energy) to gravitational potential energy, a falling parcel of air does the opposite, converting gravitational potential energy to atomic and molecular kinetic energy.

    That’s known as auto-compression.

    We have a real-world example of what I describe… in the Grand Canyon.

    Temperature at the canyon bottom is consistently ~20 – 25 F warmer than the North Rim or South Rim temperature, and air flows out of the canyon on the upstream end (except on days when there’s a S-SW prevailing wind).

    Keep in mind this is in a canyon nearly a mile deep… not much sunlight reaching the bottom in the North Rim / South Rim region because its orientation is roughly east-west and the walls are pretty steep.

    Note that ‘transition temperature’ in the context below does not relate to phase change, but to a change in whether the molecule acts to thermalize radiation and thus warm the atmosphere; or absorb energy into its vibrational mode quantum states via collision, emit radiation and thus cool the atmosphere.

    All radiative molecules are dual-role molecules… they can act to warm the atmosphere at atmospheric temperatures below their ‘transition temperature’ (in the case of CO2, via absorption of a 14.98352 µm photon to become vibrationally excited, then conversion of that vibrational mode energy to translational mode energy of other atmospheric molecules upon molecular collision), or cool the atmosphere at atmospheric temperatures above their ‘transition temperature’ (via collisional transfer of translational mode energy of other atmospheric molecules to the vibrational mode energy of CO2, which is then radiatively emitted at :
    {v23(3)} → {v22(2)} 668.1 cm-1 (14.96782 µm)
    {v22(2)} → {v21(1)} 667.8 cm-1 (14.97454 µm)
    {v21(1)} → {v20(0)} 667.4 cm-1 (14.98352 µm)
    {v3(1)} → {v20(2)} 1063.7 cm-1 (9.40115 µm)


    Adapted from image at: https://web.archive.org/web/20190528024004/http://www.barrettbellamyclimate.com/page15.htm

    The ‘transition temperature’ of any given radiative molecular species is dependent upon the differential between:

    1) the combined translational mode energy (ie: the air temperature) of two colliding molecules (one CO2, one any other atom or molecule),

    -and-

    2) the lowest vibrational mode quantum state energy of the radiative molecule.

    When 2) > 1), energy flows from vibrational mode to translational mode, which is a warming process.

    When 1) > 2), energy flows from translational mode to vibrational mode, which is a cooling process.

    Compression heating and radiative emission have created a ‘wind tunnel’ effect whereby the warmer compression-heated (in accord with the Ideal Gas Law) air at the canyon bottom puts CO2 into its ‘net-coolant’ mode. The CO2 is vibrationally excited by the translational mode energy of the other atmospheric molecules because more atmospheric molecules carry sufficient kinetic energy to vibrationally excite CO2.

    The CO2 emits so much radiation from the canyon bottom that one can trace out the profile of the canyon just from its radiation signature as seen from space.

    That loss of translational mode energy of the atmospheric molecules (which flows to the vibrational mode quantum states of CO2 and is then radiatively emitted) causes the air to become denser, and it flows along the canyon bottom upstream (because it absorbs humidity from the river, making it a bit more buoyant, but not buoyant enough to flow upward overcoming the downflowing air from the canyon rim, so it’s forced to flow upstream), which is why Page Airport has a wind predominantly to the N-NE, and why the town there is warmer than surrounding areas on cold days.

    The airflow out of the upstream end of the canyon pulls more air over the canyon rims and down into the canyon, heating it via compression, whereupon the process repeats.

    https://www.ventusky.com/?p=36.34;-111.81;7&l=temperature-2m

    So the Grand Canyon is a gigantic radiative cooler.

    I’ve got the calculations for CO2’s transition temperature somewhere… it’s based upon what percentage of air molecules have sufficient kinetic energy such that during a collision, the combined kinetic energy of two colliding molecules carry enough kinetic energy to excite the CO2 vibrational mode quantum state… so, the upper end of the Boltzmann Speed Distribution Curve. As temperature increases, more molecules carry sufficient kinetic energy, so more CO2 is collisionally excited. Let me see if I can find it. Suffice to say, the average near-surface atmospheric temperature is ~288 K for a reason.

  34. Philip Mulholland says:

    @LOL@Klimate Katastrophe Kooks 2022/05/13 at 1:36 AM
    Wow, great post! Thanks.
    Here is your example map time locked to 13 May 2022 0800 UTC
    https://www.ventusky.com/?p=36.34;-111.81;7&l=temperature-2m&t=20220513/0800

  35. LOL@Klimate Katastrophe Kooks says:

    If you’re looking for a chuckle, I’ve been drop-kicking a couple new climate tardlings over at NotALotOfPeopleKnowThat. Join in the fun if you’re so inclined.

    https://notalotofpeopleknowthat.wordpress.com/2022/05/09/saturation-of-ir-wavelengths/

    Thus far, the one whinging the loudest has claimed that Kirchhoff’s Law is inviolable (it’s been invalidated for all but certain laboratory settings); that thermodynamic equilibrium isn’t a quiescent state (it’s literally the definition of TE); that energy can flow without an energy density gradient and indeed can flow against an energy density gradient without external energy pumping that system energy up the energy density gradient; that “isothermal radiation exchange” can take place; and that I’m wrong (while admitting I’m correct). LOL

    Keep in mind, this is an MIT-educated guy… so he’s presented his MIT text and asked if MIT has been teaching people energy transfer incorrectly for the last 4 decades. To which I replied, of course they have… the method they teach is a handy shortcut way of accounting for energy flows, but it misleads the scientifically-illiterate into thinking that energy can flow without regard to the energy density gradient, which is exemplified in the blather that climate loon is spewing. LOL

  36. Philip Mulholland says:

    LOL@KKK
    That’s an impressive library you have at your disposal.
    You are clearly armed to the teeth and ready to fire 🙂

  37. Jopo says:

    So having a think about the conversion of work into electrical Watts.
    By default Power is rated at work / Time. (3600) Well in my world that is the case.

    Thus the difference of work done versus work being done when it comes to nominating a wattage

    With the Joules per electron situation in my earlier post. It is stated it takes 1.602e-19 JOULES to cause an electron to lift in potential by 1 VOLT. This is just telling us the how many joules is required in one second to cause that result. It could easily be rated at 4.45e-23 Joules for the electron to lift in potential by 1 VOLT per hour! e.g 1.602e-19 / 3600 = 4.45e-23 Joules per second

  38. LOL@Klimate Katastrophe Kooks says:

    You don’t really need to know how much the electron is lifted in potential, you just need a base reference to compare the planets. So set Earth as though it were lifting electron potential by 1 V as a reference, then scale it for the other planets to match the known planetary temperature, then compare the lift in potential between the planets. If those ratios align with the rest of your math, you’ve got something, and you can then figure out how much Earth’s lift in electron potential actually is, and scale the other planets accordingly.

    If you can figure this out, it’s got far more wide-ranging implications than just the planets in our solar system… this could be huge. Keep at it. You’ll get there.

  39. Jopo says:

    Are you saying you KNOW?

  40. CD Marshall says:

    Philip Mulholland
    I wish you’d start your own blog or site or whatever. Your insights are valuable I’ve learned a great deal from you.

    All of you guys are a wealth of knowledge.

    LOL@Klimate Katastrophe Kooks might be advanced AI from the future. 😂

  41. Philip Mulholland says:

    @CDM Thanks.
    One of the most powerful features in the development of humanity is the process of technology transfer. That is the application of a given procedure or idea in a novel setting. I view my role as a transmitter of ideas rather than that of an originator or source.
    I do have a place on Research Gate where I collate my geoscience work.

  42. CD Marshall says:

    @Philip Mulholland
    Michael Crow was trying to get in contact with you at Research Gate. Not sure if he did? He’s been researching water vapor for decades, has thousands of records to prove it. He needed a veteran scientist to further his research (a colab I’m thinking?).

    https://micro6500blog.wordpress.com/2015/11/18/evidence-against-warming-from-carbon-dioxide/

    He has a link at Research Gate but I can’t get to it.

  43. boomie789 says:

  44. boomie789 says:


    two sources I like.

  45. Richard says:

    Black Hole Magic
    When you look at the latest videos about the new found black hole one is under the impression that one travels in a spaceship towards the centre of our galaxy. Nothing of this is in fact the case. The team responsible for this big scam has gathered un absurd amount of data with their so-called telescope and have fabricated images by software manipulation and a priori hypotheses of the reality of what they try to show. The final image has a resolution that is surpassing the resolution of their instruments. It’s all about data manipulation.
    Anyone with a critical mind sees right hrough this nonsense.

  46. LOL@Klimate Katastrophe Kooks says:

    Well, that sews up SamH over at NotALotOfPeopleKnowThat… he’s so butthurt that he’s now denying that he denied Kirchhoff’s own words. In fact, he’s so sparsely-read that he didn’t even know they were Kirchhoff’s own words. LOL

    Of course, reality will never sink in for the climate kooks… they’re literally insane and thus ineducable. So rather than admit his mistakes and learn from them, he stomped off in a huff. LOL

  47. CD Marshall says:

    In my meanderings I came across a paper that said CO2 actually does not have an active magnetic dipole in a relaxed state. So naturally if that is true it is no different than a homonuclear diatomic molecule until it is excited.

    Contradictions to that statement are everywhere. Did they mean something else? Or is a CO2 molecule simply never truly relaxed, something that could be achieved maybe only in a lab? For one prime reason (according to my nonscientific mind) that it is such a powerful coolant in the upper atmosphere is that a “colder” CO2 molecule will re-radiative in a unidirectional pattern. Thus when solar storms strike it absorbs (I have noticed discrepancies in absorption or reflection among a lot of papers) and redirects that energy back the way it came in with the help of NO (Nitric Oxide).

    “A volume of emitting IR photons will only equally emit in all directions when the surrounding air parcel is the same temperature.

    This is confirmed by the NASA Saber instrument. NO and CO2 protect against solar storms in the thermosphere, and 95% of the absorbed radiation gets near instantly radiated back to the colder space.” -MP

    https://www.nasa.gov/mission_pages/sunearth/news/solarstorm-power.html

  48. CD Marshall says:

    LOL@Klimate Katastrophe Kooks Copied this great reference to have. Been a bit confused on these terms myself.

    LOL@Klimate Katastrophe Kooks PERMALINK
    May 16, 2022 8:12 pm
    Fair point. That doesn’t invalidate, however, the fact that Kirchhoff’s Law of Thermal Radiation can be violated, as several studies now show.
    Fundamental physical laws aren’t violated, if they were, they wouldn’t be laws.
    The words “fact”, “theory”, “hypothesis” and “law” have very specific definitions in science:
    ———-
    Hypothesis: A tentative explanation of an empirical observation that can be tested. It is merely an educated guess.
    ———-
    Working hypothesis: A conjecture which has little empirical validation. A working hypothesis is used to guide investigation of the phenomenon being investigated.
    Scientific hypothesis: In order for a working hypothesis to be a scientific hypothesis, it must be testable, falsifiable and it must be able to definitively assign cause to observed effects.
    Null hypothesis: Also known as nullus resultarum. In the case of climate science, the null hypothesis should be that CO2 does not cause global warming.
    A Type I error occurs when the null hypothesis is rejected erroneously when it is in fact true.
    A Type II error occurs if the null hypothesis is not rejected when it is in fact false.
    ———-
    Fact: An empirical observation that has been confirmed so many times that scientists can accept it as true without having to retest its validity each time they experience whatever phenomenon they’ve empirically observed.
    Law: A mathematically rigorous description of how some aspect of the natural world behaves.
    Theory: An explanation of an empirical observation which is rigorously substantiated by tested hypotheses, facts and laws.
    Laws describe how things behave in the natural world, whereas theories explain why they behave the way they do.
    For instance, we have the law of gravity which describes how an object will behave in a gravitational field, but we’re still looking for a gravitational theory which fits into quantum mechanics and the Standard Model and explains why objects behave the way they do in a gravitational field.
    ——————–
    Hence, Kirchhoff’s Law of Thermal Radiation should actually rightfully be deemed a hypothesis… it cannot be a law, a law is a mathematically rigorous description of how some aspect of the natural world behaves… and given that Kirchhoff’s Law of Thermal Radiation only really pertains to idealized blackbody objects, and given that idealized blackbody objects are idealizations which do not exist and are in fact provable contradictions, Kirchhoff’s Law of Thermal Radiation at best describes something which does not and cannot exist.

  49. boomie789 says:


    lots of war footage

  50. LOL@Klimate Katastrophe Kooks says:

    CD Marshall wrote:
    “In my meanderings I came across a paper that said CO2 actually does not have an active magnetic dipole in a relaxed state.”

    I’m not sure, but that might be referring to the {v1} vibrational mode quantum state.

    From my prior writings:

    The CO2 molecule (a triatomic covalently-bonded linearly symmetric molecule with an axis of symmetry along the nuclei and a plane of symmetry perpendicular to this axis) has two rotational mode quantum states, three vibrational modes (one with two degenerate states) and four fundamental vibrational mode quantum states at 3 radiation wavelength bands centered on:

    4.25677 µm ({v3}; 2349.2 cm-1 wavenumber) {v20(0)} -> {v3(1)}
    Wien Displacement Law equivalent temperature: 680.7 K, 407.6 C, 765.7 F
    Asymmetric stretch mode; this mode is very IR-active, but the dipole moment oscillates parallel to the molecule’s symmetric axis, and therefore ΔJ = 0 Q-branch transition is forbidden (photon angular momentum is transferred to electronic mode degrees of freedom instead of rotational mode degrees of freedom, and since the resonant radiation for the vibro-rotational fine structure of the electronic mode doesn’t have sufficient energy to excite the electronic mode, it cannot be absorbed), making this very narrow-band. The radiance at this narrow frequency band is also minimal, falling at the minima between the Planck curves of solar (incoming) and terrestrial (outgoing) radiation. As discussed below, however, the CO2{v3(1)} vibrational mode quantum state is the main route for v-v (vibrational-to-vibrational) transfer of energy from vibrationally-excited N2{v1(1)} to CO2{v3(1)}, as used in CO2 lasers.
    7.20357 µm ({v1}; 1388.2 cm–1 wavenumber) {v20(0)} -> {v1(1)}
    Wien Displacement Law equivalent temperature: 402.3 K, 129.12 C, 264.4 F
    Symmetric stretch mode; this mode is IR-inactive and Raman-active, it cannot absorb IR radiation since the molecule has no change in net magnetic dipole moment unless the molecule is perturbed via collision at the same time that it absorbs a resonant photon. It does, however, Raman-scatter.
    14.98352 µm ({v2}; 667.4 cm-1 wavenumber) {v20(0)} -> {v21(1)}; {v21(1)} -> {v22(2)}; {v22(2)} -> {v23(3)}
    Wien Displacement Law equivalent temperature: 193.4 K, -79.75 C, -111.55 F
    2 degenerate bending modes with 3 practically-degenerate vibrational states.

    There is a narrow absorption band centered on ~2.7 µm, but it is swamped by the {v3} (asymmetric stretch) fundamental of H2O centered at 3755 cm-1, the {v1} (symmetric stretch) fundamental of H2O centered at 3652 cm-1 and the {v2 + v3} band of H2O centered at ~5000 cm-1, and thus has little radiance available to it except in extremely low humidity locales:
    – 2.76785 µm ({v20(0) -> v22(2) + v3(1)}; 3612.91 cm-1 wavenumber)
    – 2.69209 µm ({v20(0) -> v1(1) + v3(1)}; 3714.59 cm-1 wavenumber)

    …the only one of those vibrational modes which has any appreciable radiance available to it, and which is IR-active is 14.98352 µm, and hence this wavelength band is the largest contributor to CO2 vibrational mode quantum state energy from IR absorption.

    At 287.64 K (the latest stated average temperature of Earth) and an emissivity of 0.93643 (calculated from NASA’s ISCCP program from data collected 1983-2004), at a photon wavelength of 14.98352 µm (the primary spectral absorption wavelength of CO2), the spectral radiance is only 5.43523 W / m^2 / sr / µm (integrated radiance from 13.98352 µm – 15.98352 µm of 10.8773 W/sr-m^2 to fully take into account the absorption shoulders of CO2).

    That means that the maximum that CO2 could absorb would be 10.8773 W/sr-m^2, if all CO2 were in the CO2{v20(0)} vibrational mode quantum state.

    While the Boltzmann Factor calculates that 10.816% of CO2 will be excited in one of its {v2} vibrational mode quantum states at 288 K, the Maxwell-Boltzmann Speed Distribution Function shows that ~24.9% will be excited. This is higher than the Boltzmann Factor calculated for CO2 because faster molecules collide more often, weighting the reaction cross-section more toward the higher end.

    Thus that drops to 8.1688523 W/sr-m^2 able to be absorbed. Remember, molecules which are already vibrationally excited can not absorb radiation with energy equivalent to the vibrational mode quantum state energy at which they are already excited, unless a degenerate vibrational mode quantum state exists to absorb that energy. That radiation passes the vibrationally excited molecule by.

    That’s for all CO2, natural and anthropogenic… anthropogenic CO2 accounts for ~3.63% (per IPCC AR4) of total CO2 flux, thus anthropogenic CO2 can only absorb 0.29652933849 W/sr-m^2.

    CO2 absorbs ~50% within 1 meter, thus anthropogenic CO2 will absorb 0.148264669245 W/m^2 in the first meter, and the remainder 0.148264669245 W/m^2 within the next ~9 meters.

    The net effect of an increasing CO2 atmospheric concentration is not a ‘trapping’ of energy in the atmosphere, it is a reduction in the extinction depth at the given wavelength. The radiation which would be absorbed at a higher atmospheric CO2 concentration is already absorbed long before it reaches space, and always has been… it’s just absorbed in a shorter distance with increasing CO2 atmospheric concentration.

    That energy thermalized increases Convective Available Potential Energy, which increases convection, which carries the energy stored in the specific heat capacity (and latent heat capacity in the case of water) of the atmospheric molecules high enough in the atmosphere where collisional processes no longer dominate, radiative processes do. This is why CO2 is the most prevalent atmospheric coolant at and above TOA (TOA being where the atmospheric density reduces sufficiently that the atmosphere effectively loses its opacity for the given wavelength of radiation). A higher convection rate carries more energy to the upper atmosphere, and a higher concentration of molecules capable of emitting that radiation increases the photon emission flux (remember, the homonuclear diatomic molecules such as O2 and N2 have no net magnetic dipole and thus cannot effectively emit nor absorb radiation unless perturbed by a collision), thus increasing the radiation emitted to space, which is by definition a cooling process.

    Molecules capable of emitting [1] / Molecules incapable of emitting [2]

    If [1]/[2] ↑, total emission to space ↑.
    If [1]/[2] ↓, total emission to space ↓.

  51. CD Marshall says:

    Thank you @LOL@Klimate Katastrophe Kooks adding it to my library.

    If you could explain this in more detail I’d appreciate it, didn’t quite connect the dots: The arrows are throwing me off.
    Molecules capable of emitting [1] / Molecules incapable of emitting [2]

    If [1]/[2] ↑, total emission to space ↑.
    If [1]/[2] ↓, total emission to space ↓.

  52. CD Marshall says:

    So the wording of the Prevost theory is completely wrong using heat incorrectly to explain thermodynamic energy transfer.

  53. CD Marshall says:

    Or did someone sneak an incorrect interpretation of the Prevost theory to support climate science?

  54. LOL@Klimate Katastrophe Kooks says:

    https://archive.org/details/bub_gb_3KwRAAAAYAAJ/page/n9/mode/2up?view=theater

    You can tell he’s going to be wrong right from the first sentence, when he calls ‘heat’ a fluid. LOL

    That’s because the Prevost Theory of Exchanges (and Prevost’s Principle, it’s core tenet) assumes ‘heat’ to be a rarefied material substance with material properties such as elasticity, because it’s predicated upon Caloric Theory. It was 231 years ago, after all.

    It is certain that free radiant heat is a very rare fluid, the particles of which almost never collide with one another and do not disturb sensibly their mutual movements.

    He goes on from that on the very first page to begin discussing “the reflection of cold” being explained by this fluid he calls ‘heat’.

    Strange that he outright states that his material fluid ‘heat’ “has all the properties of light”… yet he couldn’t make the connection to the fact that it is light. Photons in transit, an energy flux… the definition of ‘heat’.

    He then goes on to discuss Caloric… something else he considers to be a fluid, another hypothesis chucked on the midden heap of scientific discovery.

    Yeah, the climate alarmists still cling to these long-debunked notions put forth by Prevost, as means of advancing their CAGW communist agenda, no matter how silly it seems in light of modern advances in science. Science be damned, they’ll cling to whatever life raft they must to keep their scam afloat. LOL

  55. Jopo says:

    @ Kook

    When it comes to Electric charge in Electrochemistry the:
    Faraday is measured in Amperes per Hour.
    Coulombs is measured in Amperes per second.

    1 Faraday of Charge is required to raise 1 mole of electrons to 1 volt.

    Converting Faradays into Coulombs into is a simple as multiplying by 3600
    The deception has been uncovered. Dividing coulombs per electron by 3600 Is how we do it.

    1 Faraday equates to 26.80148 Ampere Hours or Joules per Hour to raise 1 mole of electrons to 1 volt.. OR I.E 96485.33 Amperes per second to raise 1 mole of electrons to 1 volt.

    Knowing this. I can now further simplify the work earlier into Number of Electrons divided by Moles of electrons multiplied by Faraday ampere hour charge

    Multiply the 14.5189 moles of electrons in the Molar mass of Air by the 26.80148 Ampere Hours or Joules per Hour and we get 389.12 Ampere Hours or Joules per hour. I.e WATTS 287.8 Kelvin

    This is just a roughie. I will tidy it up tomorrow and put it to you guys again. It’s after 1AM here.

    This is out of Wiki. Pretty well sealed the deal for me.
    “one coulomb is equal to approximately 1.036×10−5 mol × NA elementary charges.”

  56. LOL@Klimate Katastrophe Kooks says:

    CD Marshall wrote:
    “If you could explain this in more detail I’d appreciate it, didn’t quite connect the dots: The arrows are throwing me off.
    Molecules capable of emitting [1] / Molecules incapable of emitting [2]

    If [1]/[2] ↑, total emission to space ↑.
    If [1]/[2] ↓, total emission to space ↓.”

    If CO2 atmospheric concentration increases, not only does that increase the aggregate specific heat capacity of the overall atmosphere (thus a parcel of air is able to convect and / or advect more energy per parcel), but that higher concentration effectively increases the emitting ‘surface area’.

    If there were only 1 CO2 molecule in the atmosphere, the effective emitting ‘surface area’ for CO2 would be that one molecule. Two molecules doubles that, four quadruples it, etc.

    Think about it this way… we all know the air warms up during the daytime. Conduction of energy (that energy from the sun, absorbed by the planet’s surface) when air contacts the planet’s surface is the major reason air warms up.

    Yet the major constituents of the atmosphere (N2 and O2) are homonuclear diatomics and thus cannot effectively radiate energy to cool down (unless their net-zero magnetic dipole is perturbed via collision). They constitute ~99% of the atmosphere. How does that ~99% cool down?

    Convection moves energy around in the atmosphere, but it cannot shed energy to space. Conduction depends upon thermal contact with other matter and since space is essentially a vacuum, conduction cannot shed energy to space… this leaves only radiative emission. The only way our planet can shed energy is via radiative emission to space.

    But N2 and O2 cannot effectively radiatively emit because, being homonuclear diatomic molecules, they have no net magnetic dipole. Only when the molecule simultaneously collides with something else (perturbing its net-zero magnetic dipole) and a resonant photon incides upon the molecule does it have any chance of absorbing radiation, and even then it doesn’t happen every single time. Same goes for radiative emission, it requires collision which perturbs the molecule’s net-zero magnetic dipole. That’s why homonuclear diatomic vibrational mode quantum states are meta-stable and relatively long-lived.

    Thus, common sense dictates that the thermal energy of the ~99% of the atmosphere which cannot effectively radiatively emit must be transferred to the so-called ‘greenhouse gases’ (CO2 being a lesser contributor in the lower atmosphere and the largest contributor in the upper atmosphere, water vapor being the main contributor in the lower atmosphere) which can radiatively emit and thus shed that energy to space.

    So, far from being ‘greenhouse gases’ which ‘trap heat’ in the atmosphere, those radiative gases actually shed energy from the atmosphere to space. They are net atmospheric coolants.

    It is the monoatomics and homonuclear diatomics which are the actual ‘greenhouse gases’.

    Monoatomics (Ar) have no vibrational mode quantum states, and thus cannot emit (nor absorb) IR. Homonuclear diatomics (O2, N2) have no net magnetic dipole and thus cannot emit (nor absorb) IR unless that net-zero magnetic dipole is perturbed via collision.

    In an atmosphere consisting of solely monoatomics and diatomics, the atoms / molecules could pick up energy via conduction by contacting the surface, just as the polyatomics do; they could convect just as the polyatomics do… but once in the upper atmosphere, they could not as effectively radiatively emit that energy, the upper atmosphere would warm, lending less buoyancy to convecting air, thus hindering convection… and that’s how an actual greenhouse works, by hindering convection.

    The chance of any N2 or O2 molecule colliding with water vapor is ~3% on average in the troposphere, and for CO2 it’s only ~0.042%. Logic dictates that as atmospheric concentration of CO2 increases, the likelihood of N2 or O2 colliding with it also increases, and thus increases the chance that N2 or O2 can transfer its translational and / or vibrational mode energy to the vibrational mode energy of CO2, which can then shed that energy to space via radiative emission. (And yes, t-v and v-v collisional processes do occur from N2 to CO2… if you doubt me, I can post the maths and studies which prove it.)

    Thus, common sense dictates that the thermal energy of the ~95.94 – 99.74% (depending upon humidity) of the atmosphere which cannot effectively radiatively emit (N2, O2, Ar) must be transferred to the so-called ‘greenhouse gases’ (CO2 being a lesser contributor below the tropopause and the largest contributor above the tropopause, water vapor being the main contributor below the tropopause) which can radiatively emit and thus shed that energy to space.

    So can anyone explain how increasing the concentration of the major radiative coolant gases (H2O, CO2) in the atmosphere (and thus increasing the likelihood that Ar, N2 and O2 will transfer their energy to those radiative coolant gases and then out to space via radiative emission) will result in more ‘heat trapping’, causing global warming? I thought not.

    CAGW is based upon a flawed understanding of physics, first promulgated by Svante Arrhenius back in 1896, before we even had a firm grasp on molecular physics and quantum mechanics. It was before the discovery of the photon, before the discovery of the electron, before the discovery of the proton, before the discovery of the neutron, before the discovery of atomic nuclei, before the Planck blackbody formula, before Special and General Relativity, obviously long before we discovered exactly why atoms and molecules emit at specific spectra, and a full decade before Einstein fully explained the discrepancies up to that date between Equipartition Theorem theory and empirical observation, ushering in a new quantum theory of matter. Arrhenius didn’t even know enough about molecular physics to call it “CO2”, he called it “carbonic acid”! Further, his experiment failed to account for water vapor, used 9.7 µm radiation (while CO2 absorbs mainly at 14.98352 µm) and he over-estimated the absorption coefficient of CO2 by 253%, forcing him to later revise his estimate (which was still wrong) of temperature forcing from CO2. Yet his flawed experiment is still touted by the climate alarmists as their basis for alarm and hence their basis for tearing down our modern society, deindustrialization and depopulation of the planet.

    Arrhenius didn’t have the technical knowledge at the time to reason his way through the problem properly, as we did immediately above.

  57. LOL@Klimate Katastrophe Kooks says:

    So, if the climate alarmists were actually serious about cooling the planet, rather than their commie machinations, I’ve provided them 3 options:

    1) Increase atmospheric CO2 concentration to increase the specific heat capacity of the bulk air and to provide more emitters in the upper atmosphere to increase convection and dump more energy to space.

    2) Dig deep E-W furrows in the ground, deep enough that gravitational auto-compression puts a high enough percentage of CO2 molecules into the net-cooling mode via t-v collisions… especially so for {v3(1)} → {v20(2)} 1063.7 cm-1 (9.40115 µm) emission. That’s higher energy than 14.98352 µm emission and it’s in the Infrared Atmospheric Window, so it has a nearly unfettered path out to space. It requires a higher molecular kinetic energy and thus more gravitational auto-compression and thus a deeper furrow, though.

    3) Water, water everywhere. Water literally acts as a refrigerant in the troposphere, in the strict ‘refrigeration cycle’ sense.

    A liquid evaporates at the heat source (the surface) [in the evaporator], it is transported (convected) [via an A/C compressor], it emits radiation to the heat sink and undergoes phase change (emits radiation in the upper atmosphere, the majority of which is upwelling owing to the mean free path length / altitude / air density relation) [in the condenser], it is transported (falls as rain or snow) [via that A/C compressor], and the cycle repeats.

    We live in the equivalent of the evaporator of a gigantic world-sized A/C unit. Water is the refrigerant, the working fluid. The upper atmosphere is the condenser. Convection stands in as the A/C compressor to move the working fluid through the A/C unit.

    2) and 3) would work especially well… want to cool an area off? Either douse it with water or uncover your furrows to allow radiation to escape to space. Want to warm an area up? Stop dousing it with water or cover your furrows so radiation cannot escape to space.

  58. Climate Heretic says:

    LOL@KKK you have mentioned several times that Kirchoff’s radiation law has been invalidated. The only two sources that I have been able to ascertain are Dr Pierre-Marie Robitaille and Stephen Crothers. If you know of other sources could you please provide links if possible.

    Regards
    Climate Heretic

  59. LOL@Klimate Katastrophe Kooks says:

    Climate Heretic wrote:
    “LOL@KKK you have mentioned several times that Kirchoff’s radiation law has been invalidated. The only two sources that I have been able to ascertain are Dr Pierre-Marie Robitaille and Stephen Crothers. If you know of other sources could you please provide links if possible.”

    As Einstein stated when it was reported to him that 100 authors had attacked his Relativity theory (paraphrased):
    “If it were wrong, then only one author would have been enough!”

    Kirchhoff, in attempting to expand Kirchhoff’s Law of Thermal Radiation from an idealized blackbody case to all materials, cheated (whether he knew he was doing so or not is moot, he was doing so)… in a perfectly-reflecting cavity, he included a small piece of graphite or carbon as a thermalizer (Planck did the same, calling it a “catalyst”)… a perfectly-reflecting cavity cannot exhibit a Planckian blackbody curve except for that thermalizer being present, because the radiation cannot equilibrate. So in fact, Kirchhoff was working with the idealized blackbody case all along! Thus, Kirchhoff’s Law of Thermal Radiation doesn’t extend beyond the idealized blackbody case, and only applies at thermodynamic equilibrium, and given that a close approximation (for a certain range of wavelengths) of an idealized blackbody cavity can only be obtained in a laboratory, Kirchhoff’s Law of Thermal Radiation only applies in that case.

    The fact that you can find paper after paper discussing violation of Kirchhoff’s Law of Thermal Radiation means it cannot be a law… it is at best a hypothesis, and given that it only really describes idealized blackbody objects, and given that idealized blackbody objects are idealizations which cannot exist and which are provable contradictions, this is akin to claiming that a ‘physical law’ which states that rainbow glitter-farting unicorns only fart rainbow glitter when flying should be taken seriously.

    And in other news…

    Finally found it… as I promised earlier, here’s what I was talking about in a prior post, the ‘transition temperature’ of CO2.

    The data below peals the death knell for CAGW. You’re welcome.

    In dealing with solely translational mode energy and neglecting vibrational mode and rotational mode energy for the moment, the Equipartition Theorem states that molecules in thermal equilibrium have the same average energy associated with each of three independent degrees of freedom, equal to:
    3/2 kT per molecule, where k = Boltzmann’s Constant
    3/2 RT per mole, where R = gas constant

    Thus the Equipartition Theorem equation:
    KE_avg = 3/2 kT
    … serves well in the definition of kinetic energy (which we sense as temperature).

    It does not do as well at defining the specific heat capacity of polyatomic gases, simply because it does not take into account the increase of internal molecular energy via vibrational mode and rotational mode excitation (specific heat capacity). Energy imparted to the molecule via either photon absorption or collisional energetic exchange can excite those vibrational mode or rotational mode quantum states, increasing the total molecular energy E_tot, but not affecting temperature at all. Since we’re only looking at translational mode energy at the moment (and not specific heat capacity); and internal molecular energy is not accounted for in measuring temperature (which is a measure of translational mode energy only), this long-known and well-proven equation fits our purpose.

    Our thermometers are an instantaneous average of molecular kinetic energy. If they could respond fast enough to register every single molecule impinging upon the thermometer probe, we’d see temperature wildly jumping up and down, with a distribution equal to the Maxwell-Boltzmann Speed Distribution Function. In other words, at any given measured temperature, some molecules will be moving faster (higher temperature) and some slower (lower temperature), with an equilibrium distribution (Planckian) curve.

    The Equipartition Theorem states that in Local Thermodynamic Equilibrium conditions all molecules, regardless of molecular weight, will have the same kinetic energy and therefore the same temperature. For higher atomic mass molecules, they’ll be moving slower; for lower atomic mass molecules, they’ll be moving faster; but their kinetic (translational mode) energy will all be the same at the same temperature.

    Therefore, utilizing the equation above, at a temperature of 288 K, the average thermal energy of a molecule is 0.03722663337910374 eV. Again, this is the average… there is actually an equilibrium distribution of energies and thereby molecular speeds.

    For CO2, with a molecular weight of 44.0095 amu, at 288 K the molecule will have:
    Most Probable Speed {(2kT/m)^1/2} = 329.8802984961799 m/s
    Mean Speed {(8kT/pm)^1/2} = 372.23005645833854 m/s
    Effective (rms) Speed {(3kT/m)^1/2} = 404.0195258297897 m/s

    For N2, with a molecular weight of 28.014 amu, at 288 K the molecule will have:
    Most Probable Speed {(2kT/m)^1/2} = 413.46812435139907 m/s
    Mean Speed {(8kT/pm)^1/2} = 466.5488177761755 m/s
    Effective (rms) speed {(3kT/m)^1/2} = 506.3929647832758 m/s

    But if those molecules are at the exact same temperature, they’ll have exactly the same translational mode energy.

    This energy at exactly 288 K is equivalent to the energy of a 33.3050 µm photon.

    If two molecules collide, their translational energy is cumulative, dependent upon angle of collision. In mathematically describing the kinematics of a binary molecular collision, one can consider the relative motion of the molecules in a spatially-fixed 6N-dimensional phase space frame of reference (lab frame) which consists of 3N spatial components and 3N velocity components, to avoid the vagaries of interpreting energy transfer considered from other reference frames.

    Simplistically, for a head-on collision between only two molecules, this is described by the equation:
    KE = (1/2 mv^2) [molecule 1] + (1/2 mv^2) [molecule 2]

    The Maxwell-Boltzmann Speed Distribution Function, taking into account 3N spatial components and 3N velocity components:

    You may surmise, “But at 288 K, the combined kinetic energy of two molecules in a head-on collision isn’t sufficient to excite CO2’s lowest vibrational mode quantum state! It requires the energy equivalent to a 14.98352 µm photon to vibrationally excite CO2, and the combined translational mode energy of two molecules colliding head-on at 288 K is only equivalent to the energy of a 16.6525 µm photon!”

    True, but you’ve not taken into account some mitigating factors…
    1) We’re not talking about just translational mode energy, we’re talking about E_tot, the total molecular energy, including translational mode, rotational mode, vibrational mode and electronic mode. At 288 K, nearly all CO2 molecules will be excited in the rotational mode quantum state, increasing CO2’s E_tot. The higher a molecule’s E_tot, the less total energy necessary to excite any of its other modes.

    2) Further, the Boltzmann Factor shows that at 288 K, ~10.26671% of N2 molecules are in the N2{v1(1)} vibrationally excited state.

    N2{v1(1)} (stretch) mode at 2345 cm-1 (4.26439 µm), correcting for anharmonicity, centrifugal distortion and vibro-rotational interaction
    1 cm-1 = 11.9624 J mol-1
    2345 cm-1 = 2345 * 11.9624 / 1000 = 28.051828 kJ mol-1
    The Boltzmann factor at 288 K has the value 1 / (2805.1828 / 288R) = 0.10266710 which means that 10.26671% of N2 molecules are in the N2{v1(1)} vibrational mode quantum state.

    Given that CO2 constitutes 0.041% of the atmosphere (410 ppm), and N2 constitutes 78.08% of the atmosphere (780800 ppm), this means that 80162.3936 ppm of N2 is vibrationally excited via t-v (translational-vibrational) processes at 288 K. You’ll note this equates to 195 times more vibrationally excited N2 molecules than all CO2 molecules (vibrationally excited or not).

    Thus energy will flow from the higher-energy (and higher concentration) N2{v1(1)} molecules to CO2 molecules excited to their CO2{v20(2)} vibrational mode quantum state (and the CO2 molecule got to that vibrational mode quantum state from the CO2{v20(0)} ground state by a prior collision), exciting the CO2 to its {v3(1)} vibrational mode, whereupon it can drop to its {v1(1)} or {v20(2)} vibrational modes by emission of 9.4 µm or 10.4 µm radiation (wavelength dependent upon isotopic composition of the CO2 molecules).

    The energy flow from translational modes of molecules to N2 vibrational mode quantum states, then to CO2 vibrational mode quantum states, then to radiation constitutes a cooling process.

    3) The Maxwell-Boltzmann Speed Distribution Function gives a wide translational mode equilibrium distribution. In order for CO2 to be vibrationally excited, it requires a minimum of the energy equivalent to a 14.98352 µm photon, equating to a CO2 speed of 425.92936688660114 m/s or an N2 speed of 533.8549080851558 m/s.

    Remember I wrote above:

    For CO2, with a molecular weight of 44.0095 amu, at 288 K the molecule will have:
    Most Probable Speed {(2kT/m)^1/2} = 329.8802984961799 m/s
    Mean Speed {(8kT/pm)^1/2} = 372.23005645833854 m/s
    Effective (rms) Speed {(3kT/m)^1/2} = 404.0195258297897 m/s

    For N2, with a molecular weight of 28.014 amu, at 288 K the molecule will have:
    Most Probable Speed {(2kT/m)^1/2} = 413.46812435139907 m/s
    Mean Speed {(8kT/pm)^1/2} = 466.5488177761755 m/s
    Effective (rms) speed {(3kT/m)^1/2} = 506.3929647832758 m/s

    For CO2, the Boltzmann Factor probability of one of its molecules being at a speed of 425.92936688660114 m/s; and for N2, the Boltzmann Factor probability of one of its molecules being at a speed of 533.8549080851558 m/s is 0.8461 at 288 K. In other words, for every 100 molecules which are at the Most Probable Speed, another ~84 molecules will be at the speed necessary to vibrationally excite CO2.

    Thus at ~288 K and higher temperature, the combined translational mode energy of colliding atmospheric molecules begins to significantly vibrationally excite CO2, increasing the time duration during which CO2 is vibrationally excited and therefore the probability that the CO2 will radiatively emit. The conversion of translational mode to vibrational mode energy is, by definition, a cooling process. The emission of the resultant radiation to space is, by definition, a cooling process.

    As CO2 concentration increases, the population of molecules able to become vibrationally excited increases, thus increasing the number of CO2 molecules able to radiatively emit, thus increasing photon flux, thus increasing energy emission to space.

    As temperature increases, the population of vibrationally excited CO2 molecules increases, thus increasing the number of CO2 molecules able to radiatively emit, thus increasing photon flux, thus increasing energy emission to space.

    This is why I state that CO2 becomes a net atmospheric coolant at approximately 288 K… the exact solution is near to impossible to calculate, given the nearly infinite number of angles of molecular collision, the equilibrium distribution of molecular speed, and the fact that atmospheric molecular composition varies spatially and temporally with altitude and water vapor concentration variations.

    In an atmosphere sufficiently dense such that collisional energy transfer can significantly occur, all radiative molecules play the part of atmospheric coolants at and above the temperature at which the combined translational mode energy of two colliding molecules exceeds the lowest vibrational mode quantum state energy of the radiative molecule. Below this temperature, they act to warm the atmosphere via the mechanism the climate alarmists claim happens all the time, but if that warming mechanism occurs below the tropopause, the net result is an increase of Convective Available Potential Energy, which increases convection, which is a net cooling process.

  60. LOL@Klimate Katastrophe Kooks says:

    A better graphic:

    You can get the Maxwell Speed Distribution graph yourself by downloading the Wolfram Player, then downloading the TheMaxwellSpeedDistribution.cdf file.

  61. Joseph E Postma says:

    In undergrad I wrote a program to collide N spheres trapped in a box using Newtonian Mechanics, with the velocity distribution of the gas being initially unitary, but randomly directed vectors. After a few seconds the sphere’s would all bounce in to each other and redistribute their velocities according whatever Newtonian Mechanics necessitated. It was a graphical program and you could watch the little spheres bounce around…the idea being that this was modelling a gas trapped in a box.

    After a few seconds with a few hundred spheres colliding into each other a few million times, analysis of the new velocity distribution showed a PERFECT Maxwellian Speed Distribution. I wrote a least-squares solver to the distribution, and the errors on the fitting parameters were like 10^-9 (one billionth).

    TLDR: An ideal gas, and the Maxwellian Speed Distribution, is simply Newtonian collision mechanics with no elasticity.

    You could add in to that program elasticity in the form of a small amount of radiation emitted every time two particles (spheres) collide, depending on their emissivity or something. Then the gas would “cool”.

    Man I bet I could get way more particles into that program with today’s computers and parallelized code on my 48-core system.

    I had actually added in gravity to that program too, and if you started the particles off with zero velocity, then they would fall and collide with and bounce off of the ground, and then each other given that some some particles would always be “above” each other at random offsets, thus allowing lateral transfer of energy and thus mixing of momentum, etc.

  62. Nepal says:

    What a great idea to write that program, I might have to copy you. We all know from thermodynamics education that the Boltzmann / Maxwell-Boltzmann distributions are universal and don’t care about the microscopic details of the collisions, or materials, or frequencies. But I want to see it happen concretely in a program I write.

  63. Climate Heretic says:

    LOL@KKK

    You said:

    “As Einstein stated when it was reported to him that 100 authors had attacked his Relativity theory (paraphrased):
    “If it were wrong, then only one author would have been enough!””

    Irrelevant to the question asked.

    You said:

    “Kirchhoff, in attempting to expand Kirchhoff’s Law of Thermal Radiation from an idealized blackbody case to all materials, cheated (whether he knew he was doing so or not is moot, he was doing so)… in a perfectly-reflecting cavity, he included a small piece of graphite or carbon as a thermalizer (Planck did the same, calling it a “catalyst”)… a perfectly-reflecting cavity cannot exhibit a Planckian blackbody curve except for that thermalizer being present, because the radiation cannot equilibrate. So in fact, Kirchhoff was working with the idealized blackbody case all along! Thus, Kirchhoff’s Law of Thermal Radiation doesn’t extend beyond the idealized blackbody case, and only applies at thermodynamic equilibrium, and given that a close approximation (for a certain range of wavelengths) of an idealized blackbody cavity can only be obtained in a laboratory, Kirchhoff’s Law of Thermal Radiation only applies in that case.”

    This is just you discussing Kirchoff’s Law of Thermal Radiation saying it’s wrong. Have you written a paper saying it’s wrong or do you know of other people writing a paper saying it’s wrong?

    You said:

    “The fact that you can find paper after paper discussing violation of Kirchhoff’s Law of Thermal Radiation means it cannot be a law… it is at best a hypothesis, and given that it only really describes idealized blackbody objects, and given that idealized blackbody objects are idealizations which cannot exist and which are provable contradictions, this is akin to claiming that a ‘physical law’ which states that rainbow glitter-farting unicorns only fart rainbow glitter when flying should be taken seriously.”

    Then what are the names of these papers other than Robitaille and Crothers?

    You said

    “And in other news…”

    Not what I asked for. I reiterate, what are the names of these papers other Robitaille and Crothers ones?

    Regards
    Climate Heretic

  64. LOL@Klimate Katastrophe Kooks says:

    I’m not here to service your every request, Pal. Go DYOFDW. The last part of my prior post was in relation to something discussed prior. Science is done by attempting to falsify every assumption, every hypothesis, every law. You’re not doing science, you’re doing the opposite of it… defending the status quo while apparently refusing to do the research to find out for yourself.

    Had you bothered to check, Robitaille comes to his conclusion not just via gedanken means, but empirically.

    Even Planck tacitly admitted to the non-universality of Kirchhoff’s Law of Thermal Radiation. Planck used the coefficients of emission and absorption, which correspond to emissivity and absorptivity, writing: K_v = e_v/a_v. In Section 48 of his book Planck admits that his equation in Section 26 cannot be used when e_v = 0 and a_v = 0 because it results in 0/0 which is indeterminate, just as Robitaille has argued.

    Planck erred in failing to properly validate Kirchhoff’s Law (from which he derived his equation)… his attempt at validating Kirchhoff’s Law in The Theory of Heat Radiation is filled with errors… he had to redefine blackbodies, he ignored absorptivity at the interface of the blackbody, he used polarized light (when thermal radiation is never polarized), and he, like Kirchhoff, cheated a bit by using a small chunk of graphite as a thermalizer (what he called a ‘catalyst’) in a perfectly reflecting cavity (which cannot otherwise exhibit a blackbody spectrum).

  65. LOL@Klimate Katastrophe Kooks says:

    The thing is, at thermodynamic equilibrium, emissivity and absorptivity are zero by their very definitions… and Kirchhoff’s Law of Thermal Radiation only applies at thermodynamic equilibrium, so the one place that Kirchhoff’s Law of Thermal Radiation applies, its result is indeterminate, as Planck admits and as Robitaille addresses.

    If you doubt me, look up the definition of ‘thermodynamic equilbrium’… it is defined as a quiescent state. There is no energy density gradient, there is no impetus for photon generation, there is no chemical potential gradient by which photon absorption can take place.

    For two objects at thermodynamic equilibrium, no absorption or 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, no energy can be transferred in or out of the objects. Photon chemical potential is zero, they can do no work. Photon Helmholtz 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 energy density gradient and in the direction of the cooler object. This is standard cavity theory… deny it and you deny wide swaths of long-known and rigorously empirically-validated science.

  66. Nepal says:

    LolKooks, can you define exactly what you mean by “quiescent state”? When you say thermodynamic equilibrium is “defined as” a quiescent state, you’re just substituting one word for another, without giving a clear physical definition.

  67. Philip Mulholland says:

    Go DYOFDW

    Now that’s one I don’t have to look up.
    Well said.

  68. LOL@Klimate Katastrophe Kooks says:

    {sigh} You could, you know, DYOFDW, Nepal. There are many search engines via which you could do so.

    Thermodynamics postulates that all non-conserved properties averaged over a certain time scale t_TD, which we call the thermodynamic time scale, eventually stop varying and remain constant for as long as the system remains isolated. The quiescent state of an isolated system in which all properties remain constant on the t_TD time scale is called the state of thermodynamic equilibrium.

    Isolated composite systems tend toward a quiescent asymptotic state called thermodynamic equilibrium, in which the free variables assume constant values specified as solutions of an extremum problem. Otherwise known as the Entropy Maximum Principle.

    The definition of emissivity:
    The ratio of the total emissive power of a body to the total emissive power of a perfectly black body at that temperature.

    The definition of absorptivity:
    The ratio of the absorbed to the incident radiant power.

    We can ascertain that emissivity of a graybody object is zero at thermodynamic equilibrium thusly:

    While an idealized blackbody object emits (and absorbs) without regard to the energy density gradient, this is not so for a graybody object.

    As ΔT → 0, q → 0. As q → 0, the ratio of graybody total emissive power as compared to idealized blackbody object emissive power → 0. In other words, emissivity → 0. Do remember that temperature is a measure of energy density, equal to the fourth root of energy density divided by Stefan’s Constant. At thermodynamic equilibrium for a graybody object, there is no energy density gradient and thus no impetus for photon generation.

    As ΔT → 0, photon chemical potential → 0, photon Helmholtz Free Energy → 0. At zero chemical potential, zero Helmholtz 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 = absorbed / incident
    ρ = reflected / incident
    τ = transmitted / incident

    α + ρ + τ = incident = 1

    For opaque surfaces τ = 0 ∴ α + ρ = incident = 1

    If α = 0, 0 + ρ = incident = 1 ∴ ρ = 1 … 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 or 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, no energy can be transferred in or out of the objects. Photon chemical potential is zero, they can do no work. Photon Helmholtz 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 energy density gradient and in the direction of the cooler object.

    Even Planck tacitly admitted to the non-universality of Kirchhoff’s Law of Thermal Radiation. Planck used the coefficients of emission and absorption, which correspond to emissivity and absorptivity, writing: K_v = e_v/a_v. In Section 48 of his book, Planck admits that this equation (in Section 26) cannot be used when e_v = 0 and a_v = 0 because it results in 0/0 which is indeterminate.

    The thing is, at thermodynamic equilibrium, emissivity and absorptivity are zero for graybody objects by their very definitions… and Kirchhoff’s Law of Thermal Radiation only applies at thermodynamic equilibrium, so for the one condition to which Kirchhoff’s Law of Thermal Radiation applies, its result is indeterminate for graybody objects, as Planck tacitly admits.

    Idealized blackbody objects can never attain the quiescent state of thermodynamic equilibrium except at 0 K because they emit and absorb without regard to the energy density gradient (all idealized blackbody objects emit when > 0 K)… one of the contradictions which make idealized blackbody objects an impossibility, especially in light of Kirchhoff’s Law of Thermal Radiation which postulates that emissivity and absorptivity of idealized blackbody objects are equal at thermodynamic equilibrium.

    Except emissivity of an idealized blackbody object is pinned to 1 all the time by definition, and absorptivity of an idealized blackbody object is pinned to 1 all the time by definition (idealized blackbody objects maximally emit and absorb)… so Kirchhoff’s Law of Thermal Radiation describes the definition of idealized blackbody objects in only one case (at thermodynamic equilibrium), while that definition applies to idealized blackbody objects all the time. 1 does equal 1, after all.

    So this law cannot be a law… it describes objects (idealized blackbody objects) which do not and cannot exist, and for graybody objects at thermodynamic equilibrium, its result is indeterminate. It is, at best, a hypothesis. A hypothesis which has failed the most basic of tests in the case of graybody objects and which cannot apply to idealized blackbody objects at thermodynamic equilibrium because idealized blackbody objects can never attain the quiescent state of thermodynamic equilibrium.

    One might as well be making up ‘physical laws’ about rainbow glitter-farting unicorns.

  69. Philip Mulholland says:

    Science is done by attempting to falsify every assumption, every hypothesis, every law. You’re not doing science, you’re doing the opposite of it… defending the status quo while apparently refusing to do the research to find out for yourself.

    or as Richard Feynman. said:

    Study hard what interests you the most in the most undisciplined, irreverent and original manner possible.

  70. Climate Heretic says:

    LOL@KKK

    You said:

    I’m not here to service your every request, Pal. Go DYOFDW. The last part of my prior post was in relation to something discussed prior.”

    I do my own research and I only asked you for one simple thing, “can you name one other paper”. Yet you allude to paper after paper. Surely you would have been able to do that, but obviously not. Why? because you do not have any references to any other paper and if you did then it would have been very simple for you to cut and paste one reference, but you did not.

    You said:

    “Science is done by attempting to falsify every assumption, every hypothesis, every law.”

    Tell me something I do not know

    You said:

    “You’re not doing science, you’re doing the opposite of it… defending the status quo while apparently refusing to do the research to find out for yourself.”

    1) You do not know what I’m doing, so do not pretend you do.
    2) You do not know where I stand in regards to the status quo.
    3) As I said before I do my own research.
    4) I’m not your PAL.

    You are a condescending and arrogant person. If you were a truly kind and helpful person and really determined to break the status quo and advance the new paradigm. Then you would have posted a link to at least one referenced paper. I would then have thanked you for your time and effort. However, you did not.

    Regards
    Climate Heretic

  71. LOL@Klimate Katastrophe Kooks says:

    Climate Heretic wrote:
    “Surely you would have been able to do that, but obviously not.”

    Already done. Planck, who tacitly corroborated what Robitaille wrote. Just because you don’t like the answer and thereby ignore it doesn’t change reality, libtard.

  72. LOL@Klimate Katastrophe Kooks says:

    You come onto the comments section and demand that I answer your questions when you could just as easily look the answers up for yourself, and you have the unmitigated temerity to call me “condescending and arrogant”?

    Then, when I provide a detailed explanation, including how Planck tacitly corroborated what Robitaille wrote about, you completely ignore that and claim I didn’t provide you an answer because I didn’t DYOFDW and provide a link to a famous book written by a famous physicist (Planck)?

    I’m not your search engine, Pal, and I can tell you why you’re wrong, but I can’t understand it for you… that hard work of arriving at an understanding of the principles is entirely up to you, and you seem to be completely recalcitrant to even do the simple work of looking up a famous book by a famous physicist (Planck). I even provided you the sections of the book to look in, but you can’t be troubled to even lift a finger to educate yourself.

    I’ll not spoon-feed you just so you can turn around and spit it out while you claim “You’re not spoon-feeding me and how dare you not spoon-feed me!“.

    GFY. Pal. LOL

  73. Climate Heretic says:

    You said and I quote

    “The fact that you can find paper after paper discussing violation of Kirchhoff’s Law of Thermal Radiation means it cannot be a law”

    You still have not named one reference paper. Whether Planck tacitly implied that the law was not valid is not a refutation of the law.

    I’m not asking you questions about this subject. I’m only asking you for one reference to at least one paper that invalidates Kirchhoff’s Law of Thermal Radiation other than Pierre or Crothers.

    Grow up. Your ranting and raving is unbecoming. Show some decency and provide one paper that invalidates Kirchhoff’s Law of Thermal Radiation other than Pierre or Crothers.

    Regards
    Climate Heretic
    PS Sorry I should have said “Could you please provide one link to a paper that invalidates Kirchhoff’s Law of Thermal Radiation other than Pierre or Crothers. It would be very much appreciated.

  74. LOL@Klimate Katastrophe Kooks says:

    Climate Heretic wrote:
    “Whether Planck tacitly implied that the law was not valid is not a refutation of the law.”

    It is noted that you don’t understand the definition of the word “law” in this context.

    As to the “paper after paper discussing violation of Kirchhoff’s Law of Thermal Radiation”, again, DYOFDW. Are you incapable of using a search engine?

    And again, the fact that Planck tacitly admitted that Kirchhoff’s Law of Thermal Radiation arrived at an indeterminate result for absorptivity of zero and emissivity of zero means that at thermodynamic equilibrium for graybody objects the law cannot apply to graybody objects at thermodynamic equilibrium… their emissivity and absorptivity at thermodynamic equilibrium are zero as I prove above, so Kirchhoff’s Law of Thermal Radiation gets an indeterminate result in this case.

    And in the case of idealized blackbody objects, their emissivity and absorptivity are pinned at 1 all the time by the very definition of idealized blackbody objects (idealized blackbody objects maximally emit and absorb without regard to the energy density gradient), so what is Kirchhoff’a Law of Thermal Radiation even describing in this case?

    So it cannot apply to graybody objects at thermodynamic equilibrium (indeterminate result) and it does not apply to idealized blackbody objects (emissivity and absorptivity are always one, and idealized blackbody objects can never attain thermodynamic equilibrium)… so you’ll get right on detailing when, exactly, it does apply.

    Hint: It doesn’t. It’s a flawed ‘law’ that by all rights should be deemed an invalidated hypothesis. LOL

    Learn to read for comprehension and you won’t embarrass yourself quite so much. LOL

  75. Climate Heretic says:

    LOL@KKK

    Obviously you cannot produce One reference paper. That speaks volumes about you.

    Regards
    Climate Heretic
    PS Could you please provide one link to a paper that invalidates Kirchhoff’s Law of Thermal Radiation other than Pierre or Crothers. It would be very much appreciated.

  76. Jopo says:

    Hi LOL
    Mate I just looked up how the charge of 1.602E-19 Coulombs per electron was determined.
    It appears that the end result was amount of charge required to “off set gravity” and cause the very fine spray of a droplet of oil to be suspended in air. I.e to negate gravity on Earth..

    Thus the reference to adjusting the joules per electron is backed up or in reality comes from Millikan and Fletchers work when comparing to Venus.

    https://duckduckgo.com/?q=Millikan+oil+droplet+gravity+charge&t=chromentp&atb=v314-1&ia=web

  77. Richard says:

    L@KKK
    It is interesting to read you agree with Pierre-Marie Robitaille on the subject of Kirchhoff’s law.
    Have you studied more subjects he discusses in his papers or in the videos on his Sky Scholar channel?

    What does Joe think of all of this? And is Stephen Crothers still a mother fucker?

  78. Jopo says:

    Here is the revised piece on the Electrochemical aspect of our atmosphere.

  79. Nepal says:

    Robitaille thinks he has disproved black holes, big bang, Kirchhoff’s law of radiation, gaseous sun, Planck’s constant… wow, what a list of accomplishments! Either he has single-handedly revolutionized every field of physics… or he’s full of himself and wants to be the best scientist ever, so he pretends everyone else is stupid and wrong about everything.

    Seems like an easy bet to me. Robitaille hates real science, he only likes his own ego.

    .

    But let’s look at his evidence. Here is a paper that has been posted https://z.zz.ht/xSW7e.pdf . The first sentence is:

    “Kirchhoff’s Law of Thermal Emission asserts that, given sufficient dimensions to neglect diffraction, the radiation contained within arbitrary cavities must always be black, or normal, dependent only upon the frequency of observation and the temperature, while independent of the nature of the wall”

    First of all, that’s not what we call Kirchhoff’s law. Kirchhoff’s law is that a body in thermal equilibrium has equal absorptivity and emissivity at each wavelength.

    But Kirchhoff’s law does imply that, inside a closed cavity at equilibrium, the radiation inside is blackbody radiation, independent of the nature of the walls.

    Robitaille tests a different hypothesis: that any old open hole, not at equilibrium, should have perfectly blackbody radiation inside. Of course this hypothesis will fail — no one has ever said that an open hole not at equilibrium is a blackbody! But for some reason Robitaille thinks so. For convenience, let’s call this “Robitaille’s a-hole’s-a-hole” hypothesis.

    So what does he do? He drills a few holes in a few metals, waves a hot metal stick around, and takes a picture with an iPhone thermal camera. Surprise surprise, the holes have some hot radiation inside, not perfect blackbody radiation set by the temperature of the hole. Robitaille declare victory.

    All he’s really done is disprove “Robitaille’s a-hole’s-a-hole” hypothesis. The experiment has nothing to do with Kirchhoff’s law.

  80. LOL@Klimate Katastrophe Kooks says:

    Climate Heretic wrote:
    “Obviously you cannot produce One reference paper. That speaks volumes about you.”

    Already done. Planck, who corroborated what Robitaille wrote. Just because you don’t like the answer doesn’t change reality.

    That you cannot or will not use a search engine well enough to find those papers speaks volumes about you.

    That you’ve failed to explicate exactly under what conditions Kirchhoff’s Law of Thermal Radiation applies, considering that it cannot apply to graybody objects at thermodynamic equilibrium because emissivity and absorptivity go to zero for graybody objects at thermodynamic equilibrium as I prove above; and considering that it cannot apply to idealized blackbody objects because they do not and cannot exist, nor can they even attain the quiescent state of thermodynamic equilibrium except at 0 K because they emit without regard to the energy density gradient when at temperatures above 0 K, and even then absorptivity and emissivity for idealized blackbody objects is pinned to 1 by definition at all times so Kirchhoff’s Law of Thermal Radiation describes nothing new or novel about them… that speaks volumes about you as well. It says that you’re so far out of your depth that you can’t perform that explication. LOL

    Richard:
    Must I agree with all that Robitaille write if I agree with one thing he writes? No, I mustn’t. You know, small-minded people like you said the same sort of things about Einstein when he introduced Relativity. Perhaps you should look at what I’ve written, do a bit of research on your own, discover that I’m correct, then apologize to the world for wasting its oxygen lo these many years. LOL

  81. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    ““Kirchhoff’s Law of Thermal Emission asserts that, given sufficient dimensions to neglect diffraction, the radiation contained within arbitrary cavities must always be black, or normal, dependent only upon the frequency of observation and the temperature, while independent of the nature of the wall”

    First of all, that’s not what we call Kirchhoff’s law. Kirchhoff’s law is that a body in thermal equilibrium has equal absorptivity and emissivity at each wavelength.”

    Perhaps you should learn to read for comprehension, Nepal. There’s a reason Kirchhoff, in attempting to extend Kirchhoff’s Law of Thermal Radiation to all materials, had to include a graphite or carbon thermalizer… because in perfectly reflecting cavities, it cannot exhibit a blackbody curve except for that thermalizer being present. So in reality, Kirchhoff was working with the idealized blackbody case at all times… he didn’t extend Kirchhoff’s Law of Thermal Radiation to all materials, indeed he didn’t ‘extend’ it to any materials except a blackbody cavity because of that thermalizer.

    Planck did the same thing, included a thermalizer, what he called a ‘catalyst’. The two of them together have done more to confuse the weak-minded than any other scientists in history. LOL

    Kirchhoff’s Law:
    “For a body of any arbitrary material emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium, the ratio of its emissive power to its dimensionless coefficient of absorption is equal to a universal function only of radiative wavelength and temperature. That universal function describes the perfect black-body emissive power.”

  82. LOL@Klimate Katastrophe Kooks says:

    Jopo wrote:
    “Here is the revised piece on the Electrochemical aspect of our atmosphere.”

    You keep at it, Jopo, and you’ll have Nepal and Climate Heretic bashing you like they’re doing Robitaille. LOL

    I’ve found that taking a person step-by-step through the calculations, breaking it down to the smallest steps and explaining each step, makes it easier to follow. Remember, you’re going to have to convince liberals… with little logical capacity, with a 4th grade reading level and with only the most basic of math skills… if you wish to change the paradigm. So make it easy for them to perform each step. Explain each step in detail. Make them feel a sense of pride in being able to complete the steps you take to completion, “Look, Ma! I added them sums real good! I are a jeeneeooos!” LOL

  83. Nepal says:

    LolKooks, I don’t think you addressed what I wrote at all. I’m also not sure what your issue with the thermalizer is. Any closed cavity at equilibrium will have blackbody radiation inside, regardless of the nature of the walls. You can make the cavity of any material and put anything you want in it, it just has to be closed and at equilibrium. If the cavity is made out of 100% perfect reflector (a physical impossibility) then it cannot come to equilibrium, but all you need is at least some absorber, a single speck of graphite, or lettuce, eye of newt, whatever will do.

    Also, again this is related to Kirchhoff’s law, but not the same thing.

    But Robitaille tests entirely different hypothesis. He thinks OPEN HOLE (not closed cavity), NOT in equilibrium but with a super hot metal stick waving around it, should also contain perfect blackbody radiation. This is not Kirchhoff’s law, this is “Robitaille’s a-hole’s-a-hole hypothesis.” He disproves this hypothesis, good for him, but it has nothing to do with Kirchhoff’s law.

  84. boomie789 says:

  85. Nepal says:

    @boomie, yeah, why is monkeypox suddenly such a big story? It’s been around for years and never been a threat.

  86. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “Any closed cavity at equilibrium will have blackbody radiation inside, regardless of the nature of the walls.”

    Wrong. If what you claim is true, MRI could not work as we know it does. Did you not even attempt to falsify your erroneous belief before you started spouting off about the topic? LOL

    A perfectly-reflecting cavity does not allow the radiation to equilibrate to a blackbody curve… which you just admitted, so you’ve just contradicted yourself. Bit embarrassing, that. LOL

    Any graybody material at thermodynamic equilibrium is, by definition, a perfect reflector, as I show above. That pretty much scotches your babble. LOL

    Question: How exactly does one determine the spectra in a fully-closed cavity? Or did you just pull that out of your coal chute? Kirchhoff (and Planck, and Bunsen) worked with cavities which had a hole, lined up with a prism so they could see the light being emitted or a thermometer so they could measure it. LOL

    Show me where Robitaille uses a “super hot metal stick waving around an open hole NOT in equilibrium”:

    https://sci-hub.se/10.1109/TPS.2003.820958

    https://www.researchgate.net/publication/26842354_Kirchhoff%27s_Law_of_Thermal_Emission_150_Years/fulltext/57aab31408ae42ba52ac7e1d/Kirchhoffs-Law-of-Thermal-Emission-150-Years.pdf?origin=publication_detail

    https://www.researchgate.net/publication/26414792_An_Analysis_of_Universality_in_Blackbody_Radiation/fulltext/0e5f9e2ff0c41c4932dc6612/An-Analysis-of-Universality-in-Blackbody-Radiation.pdf?origin=publication_detail

  87. LOL@Klimate Katastrophe Kooks says:

    In fact, Robitaille was replicating an experiment done by De Vos in 1954 (“Evaluation of the Quality of a Blackbody” in the journal Physica), in which De Vos attempted to judge the quality of a cavity by showing that the cavities appear to be increasingly blackbody as the ratio of the length of the hole in the material to its diameter is increased. De Vos only showed the degree to which the surface of the cavity was specular or white, not whether the surface actually emitted photons at the correct temperature.

    So while De Vos’s paper is a classic in blackbody cavity research, in fact it only provides limited insight into radiation in a cavity.

    So while you’re blathering on about “a hot stick waving it outside a cavity with a hole”, you’re unknowingly bashing a classic paper in blackbody cavity research… another faux pas by you. That’s embarrassing, eh? LOL

    So Robitaille replicated De Vos’ experiment and showed that cavity radiation absolutely depended upon the cavity wall material, which means either you didn’t read Robitaille’s paper, or you were incapable of understanding it. LOL

  88. Nepal says:

    LolKooks,

    The cavity used in MRI are not closed, or in equilibrium.

    A perfectly reflecting cavity would not achieve equilibrium, luckily a perfectly reflecting material does not exist.

    A graybody at equilibrium is not a perfect reflector. There is no source saying this, anywhere. Just you and maybe Charles Anderson’s blog.

    “Question: How exactly does one determine the spectra in a fully-closed cavity?”. Indeed, to measure the cavity, you must drill a hole for radiation to escape, so it is no longer a closed cavity with perfect blackbody radiation. As the hole gets smaller and smaller, the emitted radiation gets closer and closer to perfect blackbody radiation: 90% as much radiation as a blackbody, then 99%, then 99.9%… etc. Labs measure this every day to calibrate instruments. There are also simple formulas to calculate how big the effect of the hole is. Robitaille could have used those formulas, or tried measuring a smaller hole. Instead he drilled a big gaping hole, said “See! It’s not a perfect blackbody!” and called it a day. Yeah, obviously, but that doesn’t prove anything interesting.

    You asked for Robitaille’s hot metal stick: “The heated galvanized steel rod was placed on the right near the steel hole. Note that the aluminum, copper, and brass holes all appear filled with radiation from the rod.”

  89. Nepal says:

    “So Robitaille replicated De Vos’ experiment and showed that cavity radiation absolutely depended upon the cavity wall material”

    Everyone already knew that open cavity radiation depends on wall material. Robitaille proved nothing interesting.

    A closed cavity contains perfect blackbody radiation, regardless of wall material.

    If cavity has a hole in it, the radiation is not perfect blackbody. As the size of the hole gets smaller and smaller, the cavity radiation approaches blackbody radiation: first 90%, then 99%, then 99.9%… No matter the material, if the hole is small enough, it becomes perfect blackbody radiation. This is experimental fact.

    Kirchhoff predicted that a closed cavity contains perfect blackbody radiation, and this has been confirmed experimentally. Robitaille shows that open cavity does not have perfect blackbody radiation. But Kirchhoff never said open cavity has perfect blackbody radiation.

  90. Joseph E Postma says:

    Yes, Robitaille and Crothers are a team, and they’re both entirely full of shit…I destroyed Crothers two weeks ago when he came on the PSI TNT radio with us, and he gets all of his ideas from Robitaille. They are 100% not trustworthy, and even if they ever said anything correct, it would only be by accident, or, for the purpose of placing a lie right next to it.

    If you missed it, this was Crothers/Robitaille’s “debunk” of a blackhole, paraphrasing:

    “A blackhole is supposed to be supermassive and therefore is the orbital barycenter of any non-blackhole material object orbiting around it. However, the barycenter is the center of mass of the orbital system and is only a fictional point where no mass actually is, and therefore if there is supposed to be a blackhole there which means that a mass should be there, then this contradicts itself because there’s not supposed to be any mass at the orbital barycenter.”

    You see how fucking retarded that is? If they’re willing to say something like that, you can’t trust a single world they say elsewise.

  91. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “The cavity used in MRI are not closed, or in equilibrium.

    A perfectly reflecting cavity would not achieve equilibrium, luckily a perfectly reflecting material does not exist.”

    No cavity is closed… again, how does one measure the spectra in a closed cavity? Name a few of these ‘closed-cavity’ experiments… they must be numerous, with you banging on about them, right? LOL

    A “perfectly reflecting material” absolutely does exist… graybody material at thermodynamic equilibrium is perfectly reflecting by definition, as I show above. You’ve not refuted it because you cannot. To do so, you’d have to violate the fundamental physical laws by claiming that a graybody object at thermodynamic equilibrium absorbs / emits and thus that radiative energetic exchange is an idealized reversible process and thus that entropy doesn’t change even in the face of this radiative exchange.

    There are superconducting microwave cavities that are used in particle accelerators and quantum experiments with quality factors approaching the billions.

    For frequencies below the plasma frequency, a plasma is also perfectly reflecting… look up “total reflection” and “ionospheric skip”.

    Nepal wrote:
    “If cavity has a hole in it, the radiation is not perfect blackbody. As the size of the hole gets smaller and smaller, the cavity radiation approaches blackbody radiation: first 90%, then 99%, then 99.9%… No matter the material, if the hole is small enough, it becomes perfect blackbody radiation. This is experimental fact.”

    It’s not the size of the hole, it’s the ratio of hole depth to hole diameter. And that only because it limits the radiation that can exit the cavity to a narrower and narrower angle.

    That doesn’t imply that “a closed cavity will be a perfect blackbody no matter the material” as you claim. As even you admit, a perfectly-reflecting cavity does not and cannot equilibrate the radiation to a blackbody curve, which is why Kirchhoff (and Planck) had to place a thermalizer into their cavities… and in so doing, they didn’t extend Kirchhoff’s Law of Thermal Radiation to all materials because they continued working only on the idealized blackbody case because of that thermalizer.

    For thermodynamic considerations it makes no difference if the body is hollow or not. Equilibrium is equilibrium and should exist outside and inside. A small hole is how one determines what’s happening inside the cavity… again, how is a spectra even measured in your ‘closed cavity’? Magic?

    The production of a blackbody spectrum absolutely requires the presence of a vibrational lattice and is hence intrinsically tied to the nature of the walls, contrary to Kirchhoff’s claim. That’s kind of why, after all, a perfectly-reflecting surface cannot equilibrate the radiation to a blackbody curve.

  92. Joseph E Postma says:

    Many stars emit approximately to a blackbody…one of the reasons we used to talk about was simply because they effectively are a “tiny hole” in radiative energetic terms…the amount of light they emit is insignificant compared to the amount of energy which they contain. I always thought it a neat idea. …I think there was also something about the amount of light they absorb from outside being negligible compared to the amount of light they emit. Basically a star is somewhat neat thermal equilibrium…but of course they aren’t completely and have absorption lines and Balmer discontinuities, etc., and so they also deviate considerably from a blackbody emission.

    The best “laboratory” measurement of the best Planck Curve ever recorded is that of the cosmic microwave background…smaller error bars on the fit than ANY laboratory blackbody. Really impressive.

    Crothers and Robitaille say that the cosmic microwave background is “from hydroxyl bonds in water in Earth’s oceans plus the COBE satellite had design flaws.”

    Really impressive that a design flaw and water on Earth at ~286K produces, purely by experimental accident from a satellite not looking at Earth, a perfect Planck spectrum at 2.7K just like radio telescopes say is there. I guess all the satellite does is run a C++ script to produce a Planck curve and returns THAT to the operators? lol

  93. LOL@Klimate Katastrophe Kooks says:

    Joseph E Postma wrote:
    “You see how fucking retarded that is?”

    Yeah, that’s pretty silly. The orbital barycenter of Earth / Sun is very close to the center of the sun. Definitely invariant-mass matter there. A simple look at the masses of the Earth and Sun and the distance between Earth and Sun would ascertain that.

    r_1 = a / (1 + (m_1 / m_2))
    Where:
    r_1 = distance from m_1 to the barycenter (m)
    a = distance between the center of the two bodies (m)
    m_1 & m_2 = masses of the bodies (kg)

    However, I needn’t agree with all they write. That’s the beauty of science. If it fits the facts, is testable, is falsifiable and is able to be mathematically described, that’s a pretty good indication that it represents reality.

    I don’t have to trust a thing they say, I can do my own thinking, my own research, my own gedanken experiments. If what they say makes sense, I attempt to falsify it first. If I’m unable, I have no choice but to tentatively accept it while I continue attempting to falsify it.

    Believe me, I’m trying to falsify Robitaille’s claim that the quality of a blackbody cavity depends upon cavity material… I haven’t been able to do so yet.

    Oh! I found a cavity which is closed… a radar resonator. Not a blackbody spectrum as Nepal claims. Specifically designed not to be a blackbody spectrum… it is a resonator and a bandpass filter for radar. Energy is drawn out of the cavity via a wire loop, and put into the cavity from an internal triode vacuum tube. That pretty much destroys his claim that a closed cavity will always have a blackbody spectrum regardless of cavity material.

  94. Joseph E Postma says:

    There’s a threshold of full of shitness that tips it over the edge and splats all over. Some people ride the line very close, saying a lot of things that might make sense, but balancing with outlandish BS or sophistry. They do it in order to sow confusion and sophistry and to destroy intelligence.

    Crothers and Robitaille are entirely beyond the edge. When they say things as stupid as I’ve related above, it is clear they’re not in the slightest engaged in critical thinking, but pure unadulterated BS generation. They’re not worth discussing…only making fun of.

    “It’s not the size of the hole, it’s the ratio of hole depth to hole diameter. And that only because it limits the radiation that can exit the cavity to a narrower and narrower angle.”

    That’s what Nepal was referring to.

    Take a break, LOL.

  95. LOL@Klimate Katastrophe Kooks says:

    One more…
    Nepal wrote:
    “A perfectly reflecting cavity would not achieve equilibrium, luckily a perfectly reflecting material does not exist.”

    If Nepal has a fiber optic internet connection, his own internet connection is laughing at him. Fiber optic cables keep light in the core of the fiber via total internal reflection, making the fiber act as a waveguide.

    His own computer screen is laughing at him… the LED backlights rely upon total internal reflection to confine the LED light to the acrylic glass panel, then scatters it by etching one side of the pane, giving an approximately uniform luminous emittance.

    His own eyes are laughing at him… the view at the angle between the iris and cornea (used in diagnosing glaucoma) is blocked by total internal reflection, so a gonioscope suppresses this by replacing the air with a higher-index medium, allowing transmission at oblique incidence, typically followed by reflection in a mirror which itself may be using total internal reflection.

    Oh look… a completely closed cavity using total internal reflection which doesn’t exhibit a blackbody spectrum… useful as a laser cavity in a similar vein to the radar cavity I discussed prior:
    https://sci-hub.se/10.1364/OE.20.016033

    That proves Nepal wrong on the count of his claim that a completely closed cavity always exhibits a blackbody spectrum regardless of cavity material, and on the count of his claim that perfectly reflective materials don’t exist.

    Total internal reflection has only been known about since 1611… so it’s understandable that Nepal’s not heard of it yet. LOL

  96. Joseph E Postma says:

    It can’t be reflective…it needs to absorb. I thought that was understood.

    You’re taking a break LOL.

  97. Nepal says:

    LolKooks,

    I disagree in strongest terms with you about Kirchhoff’s law of thermal radiation. However, the last paper you posted is beyond cool, thank you for sharing it.

  98. LOL@Klimate Katastrophe Kooks says:

    It’s all there in black and white, mathematically and definitionally proved out, Nepal. I welcome your attempt at refutation. Show your work.

  99. LOL@Klimate Katastrophe Kooks says:

    I don’t subscribe to the Robitaille / Crothers working hypothesis that the CMB doesn’t exist and is solely an Earthly water microwave emission artifact, but they’re not the only ones to have seen artifacts in the CMB signal and there may be another explanation…. there’s still so much we don’t know and don’t understand.

    Ait Mansour El Houssain
    SOLEIL synchrotron
    https://www.researchgate.net/post/CMB_Anomaly_Cosmic_Microwave_Background_Symmetries_and_Inaumogienity

    Demetris Christopoulos
    National and Kapodistrian University of Athens
    https://www.researchgate.net/post/Is_CMB_radiation_just_microwave_radiation_from_the_water_of_earths_oceans

    Water absorbs (and emits) in the microwave… it may be that the signal of emission from Earth’s water is boosting the anisotropy of the signal our satellites see of the CMB, leaving an ‘imprint’ or ‘reflection’ on the CMB.

    Which means we’d have to put our CMB-measuring satellites very far out from Earth to avoid that interference.

    So it may very well be that the CMB is as I’ve postulated… a ‘fun-house mirror’ of photons from stars and planets inside and stars and planets now outside our cosmological particle horizon (emitted while still inside our cosmological particle horizon), effectively ‘smeared’ and distorted via anisotropy variations from other nearer emitters and via gravitational lensing.

  100. LOL@Klimate Katastrophe Kooks says:

    IOW…. “I could, but I won’t… because I can’t.”.

    To disprove graybody objects being perfect reflectors at thermodynamic equilibrium, you must claim that graybody objects emit and absorb at thermodynamic equilibrium (in direct contradiction to the S-B equation and the definition of ‘thermodynamic equilibrium’), you must thus claim that energy in the form of radiation can flow at thermodynamic equilibrium (in direct contradiction to 2LoT), you must thus claim that radiative transfer of energy is an idealized reversible process (in direct contradiction to reality… all real-world processes are irreversible processes, idealized reversible processes are idealizations) and hence you must claim that entropy doesn’t change when energy is transferred via radiation (again in direct contradiction to 2LoT).

    I showed you a radar resonator which is a closed cavity… why would they put a hole in a radar resonator? They extract the energy from the cavity via a wire loop. This disproves your contention that all closed cavities exhibit a blackbody spectrum, as do all resonators and band-pass filter cavities.

    And of course, the fact that Kirchhoff’s Law only applies for graybody objects at thermodynamic equilibrium, and the fact that at thermodynamic equilibrium (per 2LoT, the S-B equation and a whole host of other long-known scientific principles) that graybody objects are by definition perfect reflectors means Kirchhoff’s Law, in the one condition which it applies to graybody objects, arrives at an indeterminate result… that alone invalidates it.

    And for idealized blackbody objects, given that their emissivity and absorptivity are pinned at 1 at all times by definition (idealized blackbody objects maximally emit and absorb without regard to the energy density gradient), Kirchhoff’s Law doesn’t describe anything new about them. Nor can idealized blackbody objects actually attain the quiescent state of thermodynamic equilibrium except at 0 K (idealized blackbody objects emit > 0K and absorb all radiation incident upon them)… emission and absorption isn’t quiescence.

    But you “won’t” attempt a refutation due to your ‘high moral standards’ or somesuch.

    Those skilled in the art of kook wrangling recognize that you’re attempting to justify your inability to offer a refutation, putting that justification in the best light possible to save face as you retreat in the face of superior reasoning ability.

  101. Nepal says:

    LolKooks, Joe has repeatedly expressed concerns about this discussion already. I am showing a modicum of respect for those concerns. You can spin that however you want. If Joe & others think the discussion is worthwhile, of course I would gladly speak more to your points. But already the issue has been worn out some, and especially I have discussed your unique interpretation of equilibrium and gray bodies at excessive length in other posts.

  102. LOL@Klimate Katastrophe Kooks says:

    Whatever you have to tell yourself to assuage your feelings of inadequacy in light of having run up against superior reason ability is fine with me… just don’t expect anyone else to buy it. LOL

    My “unique interpretation” of equilibrium is the definition of thermodynamic equilibrium. As is my “unique interpretation” of graybody objects.

    You only cling to outmoded knowledge because reality challenges some dearly-held ideology which you feel you cannot part with.

  103. Kooks…would you please learn the social skill of replying more kindly and with more understanding. You don’t need to insult people as a matter of course. Nepal and others aren’t like Crothers and Robitaille and climate alarmists who obviously just bs and lie… People who are having real science discussions can just do that without the mocking and hostility. Please, try. I know that after years of debating with climate retards and total degenerate scum of the universe goblin psychopaths who are potential serial rapists and murderers like Crothers and Robitaille and Spencer can have left us scarred and jarred and defensive and disillusioned and ready to attack. But it’s not necessary here. You can work on more positive communication styles. What you have to scientifically say often makes sense, but then Nepal often makes good points too. It’s a fine line we’re at here.

  104. LOL@Klimate Katastrophe Kooks says:

    Oh, I do have those social skills, Joe…. for instance, you’ve never seen me dress you down, nor Immortal600, nor CD Marshall, nor Jopo, nor boomie789… but when a person, even in the face of a veritable avalanche of exculpatory evidence, refuses to even consider that they are incorrect and continues {and I’m following your orders here} ahem… “reiterating the same incorrectitudes”… well, I don’t suffer fools lightly.

    I do not believe that “Nepal” is who you believe him to be (a 16 year old who can cogently converse on deep topics and apparently holds graduate-level mathematical skills)… his argumentation methodology, his stated beliefs, the topics he focuses upon and his apparent dislike of certain people leads me to believe it is none other than sockmonger ‘evenminded’, forced to tone it down so he doesn’t get booted yet again, but desperately attempting to continue twisting science to further his CAGW ideology.

  105. Nepal says:

    LolKooks, the only reason you haven’t been rude to others is because they rarely correct you, as I do. Your problem is that you cannot be wrong, or even entertain notion that you could be wrong, without flying off the handle.

    As for “graduate level mathematical skills,” I’m flattered, but I’ve barely ever even busted out my calculus.

  106. LOL@Klimate Katastrophe Kooks says:

    Your “corrections” are based upon the same incorrectitudes you continue reiterating, while you absolutely refuse to consider any evidence that you might be wrong. For instance, even after being shown a closed cavity which didn’t exhibit a blackbody spectrum, you continue to make the incorrect claim that all closed cavities exhibit a blackbody spectrum regardless of the cavity material. For instance, you’ve contradicted yourself in claiming that cavity material didn’t matter in the production of a blackbody spectrum, while admitting that a perfectly reflecting cavity will not produce a blackbody spectrum… you hand-waved it away by claiming that perfectly reflecting material didn’t exist… only to be shown several examples of same.

    Immortal600 has corrected me many times, with evidence and mathematics to back it up. What he didn’t do was make an assertion sans any proof and in direct contradiction to any evidence already presented, then continued to double-down, then run away while claiming that he holds the moral high ground.

  107. LOL@Klimate Katastrophe Kooks says:

    Let us do a gedanken experiment…

    For this experiment, you need to visualize:

    A nitrogen bath at 77 K.

    A room temperature of 295 K.

    A small cavity of perfectly reflecting material [α = 0%; ε = 0%; ρ = 100%; τ = 0%]

    A large cavity of idealized blackbody material [α = 100%; ε = 100%; ρ = 0%; τ = 0%]

    Both cavities have some means of measuring their radiation spectrum.

    The small cavity will have a hinge on one wall which can open and close, and some means of remotely closing that wall of the cavity without disturbing the system.

    Now, we place the small cavity within the large cavity, with one wall of the small cavity open.

    We place the (sealed) large cavity into the liquid nitrogen bath.

    We allow the entire system to equilibrate to 77 K.

    We then close the wall of the small cavity, and remove the large cavity from the nitrogen bath.

    We then allow the entire system to equilibrate with room temperature (295 K).

    We then measure the radiation spectrum in both cavities.

    We will see that the large cavity does indeed have a blackbody spectrum corresponding to 295 K.

    We will also see that the small cavity has a blackbody spectrum still corresponding to 77 K, even though the walls of the cavity are equilibrated to 295 K. Remember, they are perfectly reflecting and thus have zero emissivity and zero absorptivity (and of course, zero transmissivity). They can do no work upon the radiation field within the cavity, nor can the photons do work upon the walls.

    And that is what generates a blackbody spectrum… the ability of the walls of the cavity to do work / have work done upon them to equilibrate the radiation injected into the cavity to a blackbody spectrum.

    If we take that large cavity and inject coherent monochromatic photons from a laser, we will still have a blackbody spectrum because the cavity walls are doing work upon the injected radiation to convert that monochromatic radiation to other wavelengths. The walls can do this because the photons can first do work upon the walls.

    If we take that small cavity and inject coherent monochromatic photons from a laser, we will have a monochromatic radiation field because the cavity walls can do no work upon the radiation field, nor the radiation field upon the walls.

    Now… at thermodynamic equilibrium, when energy density gradient is zero, when photon chemical potential is zero, when photon Helmholtz Free Energy is zero, how much work can the photons do upon the cavity walls (and the cavity walls upon the photons, regardless of cavity wall material)? ZERO.

    This is why graybody objects at thermodynamic equilibrium are considered to be perfectly reflecting.

    Do keep in mind that the definition of thermodynamic equilibrium is the minimum of Helmholtz Free Energy… a quiescent state. When Helmholtz Free Energy = 0, the state of thermodynamic equilibrium has been attained. You can’t get any more ‘minimum’ than zero.

    As ΔT → 0, q → 0. As q → 0, the ratio of graybody total emissive power as compared to idealized blackbody object emissive power → 0. In other words, emissivity → 0. Do remember that temperature is a measure of energy density, equal to the fourth root of energy density divided by Stefan’s Constant. At thermodynamic equilibrium for a graybody object, there is no energy density gradient and thus no impetus for photon generation.

    As ΔT → 0, photon chemical potential → 0, photon Helmholtz Free Energy → 0. At zero chemical potential, zero Helmholtz 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 = absorbed / incident
    ρ = reflected / incident
    τ = transmitted / incident

    α + ρ + τ = incident = 1

    For opaque surfaces τ = 0 ∴ α + ρ = incident = 1

    If α = 0, 0 + ρ = incident = 1 ∴ ρ = 1 … all incident photons are reflected at thermodynamic equilibrium for graybody objects.

    And that invalidates Kirchhoff’s Law of Thermal Radiation. The quality of a cavity absolutely depends upon the cavity material, in direct contradiction to Kirchhoff’s claims to the contrary.

  108. Nepal says:

    Ok LolKooks… to reiterate, this is my claim that you are arguing against: “A closed cavity contains perfect blackbody radiation, regardless of wall material.” You seem to have two objections.

    1) You say that a radar resonator does not contain blackbody radiation, and that “energy is drawn out of the cavity via a wire loop, and put into the cavity from an internal triode vacuum tube.”

    Obviously if you are blasting radio waves into and out of the cavity with an external power source, it is not a closed cavity, and it also won’t be in thermal equilibrium… disconnect your power source and the cavity will then be filled with blackbody radiation, according to Planck’s law, i.e. each mode contains an average number of photons predicted by Bose-Einstein distribution.

    2) You say that a perfectly reflecting cavity, with no absorption, will not equilibrate to blackbody radiation.

    But such a thing does not exist. There is no perfectly reflecting material, and even the slightest bit of absorption, like a speck of dust, and boom, the closed cavity goes all the way to the equilibrium state, which is black body radiation.

    You give the example of total internal reflection and say this has no absorption. But to have total internal reflection the light must be inside glass, and glass absorbs. Even the most perfect glass in the world, like optical fibers, have at least 0.1 dB/km loss. I calculate this means most light will be absorbed within a tenth of a millisecond.

  109. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “Obviously if you are blasting radio waves into and out of the cavity with an external power source, it is not a closed cavity”

    You’re forgetting something…

    Kirchhoff specified two conditions by which Kirchhoff’s Law of Thermal Radiation existed:
    1) thermal equilibrium must exist
    2) the entire system’s energy must be contained within the radiation field in the cavity space

    It is condition number 2 which necessitates a perfect absorber / emitter (an idealized blackbody, which must emit all that it absorbs and which must absorb all that is incident upon it) or a perfect reflector (which has absorptivity and emissivity of 0, so it cannot absorb any of the radiation), as no energy can be in the cavity walls.

    But a perfect reflector cannot equilibrate the radiation to a blackbody spectrum. In other words, the photons must be able to do work upon the cavity walls, and the walls must be able to do work upon whatever radiation spectrum we introduce to the cavity (even if that radiation is at a single wavelength) to convert it to a blackbody spectrum.

    Thus, in the case of perfectly-reflecting cavity walls, the radiation we inject into the cavity can do no work upon the cavity walls, and no work can be done upon the radiation by the walls, and thus it cannot be converted to a blackbody spectrum. Thus, because Kirchhoff’s Law of Thermal Radiation is purported to extend to all materials, this alone invalidates it. Kirchhoff’s own conditions invalidate Kirchhoff’s Law of Thermal Radiation… it requires that the walls have work done upon them by the cavity space radiation and do work upon the cavity space radiation and hold no energy.

    Only idealized blackbody walls and perfectly reflecting walls could hold no energy so all energy is in the radiation field in the cavity space. Only idealized blackbody walls can have work done upon them by the radiation field and do work upon the radiation field necessary to equilibrate the radiation to a blackbody spectrum while holding no energy so all the energy is in the radiation in the cavity space.

    So Kirchhoff’s Law only holds for idealized blackbody walls, per Kirchhoff’s own conditions… and idealized blackbody objects are idealizations… they don’t actually exist.

    Further, because idealized blackbody objects emit at temperature greater than 0 K and must absorb all radiation incident upon them, they can never attain the quiescent state of thermodynamic equilibrium except at 0 K. Absorption and emission aren’t quiescence.

    Even further, given that emissivity and absorptivity are always equal to 1 for idealized blackbody objects, Kirchhoff’s Law doesn’t describe anything about idealized blackbody objects that that isn’t already in the definition of an idealized blackbody object.

    So while you can claim that Kirchhoff’s Law of Thermal Radiation is ‘a thing’, it’s ‘a thing’ that describes something that does not and cannot exist, and by Kirchhoff’s own conditions, cannot apply to anything else.

    Now let’s hear your “Nepal’s Law of Glitter-Farting Unicorns”. LOL

  110. LOL@Klimate Katastrophe Kooks says:

    That’s kind of why, after all, Kirchhoff, after working with blackbody cavities, attempted to extend Kirchhoff’s Law of Thermal Radiation to all objects by going to the other extreme of a perfectly-reflecting cavity… except he couldn’t get it to work, which is why he had to use that small piece of graphite or carbon to equilibrate the radiation to a blackbody spectrum. I find it hard to believe that someone of such knowledge wouldn’t have realized that he was introducing a thermalizer into the cavity and thus destroying the perfect-reflector case, but it was a long time ago, so perhaps.

    And Planck (a student of Kirchhoff in 1877 and his successor at the Friedrich-Wilhelms-Universität, Berlin in 1889), in order to get his theory to work (which was predicated upon Kirchhoff’s Law of Thermal Radiation) had to do the same.

    Planck erred in failing to properly validate Kirchhoff’s Law (from which he derived his equation)… his attempt at validating Kirchhoff’s Law in The Theory of Heat Radiation is filled with errors… he had to redefine blackbodies, he ignored absorptivity at the interface of the blackbody, he used polarized light (when thermal radiation is never polarized) and thus misused Brewster’s Law, and he, like Kirchhoff, cheated a bit by using a small chunk of graphite or carbon as a thermalizer (what he called a ‘catalyst’) in a perfectly reflecting cavity (which cannot otherwise exhibit a blackbody spectrum).

  111. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “But such a thing does not exist. There is no perfectly reflecting material”

    Except graybody material at thermodynamic equilibrium, per the definition of emissivity, absorptivity, 2LoT, entropy, etc.,etc., etc. as I show above.

    See, you keep repeating these incorrectitudes without once acknowledging that I’ve already provided cases where you’re wrong. That doesn’t bode well for your credibility.

    The fact that total internal reflection reflects 100% of radiation is completely apart from the fact that straight-line transit of light through glass experiences absorption… you claimed there was no such thing as ‘perfectly reflecting material’, I showed you several such cases. Now you’re attempting to conflate things by claiming that because glass absorbs radiation, that somehow disproves the case of total internal reflection being perfectly reflecting. Again, that doesn’t bode well for your credibility.

    { Take note, Joe… not once did I call him a ‘disingenuous goal-post-moving doubling-down on stupid clue-repellent half-wit’. I was entirely polite, per your instruction. LOL }

  112. Nepal says:

    Let’s make things concrete:

    “The fact that total internal reflection reflects 100% of radiation is completely apart from the fact that straight-line transit of light through glass experiences absorption”

    Please explain how you could build a perfectly reflecting cavity using total internal reflection, where light inside is never absorbed.

    It can’t be done.

  113. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “It can’t be done.”

    Plasma, or a close approximation thereof.

    Nepal wrote:
    “It can’t be done.”

    Did you forget that below the plasma frequency, radiation incident upon the plasma is totally reflected, which is what causes the phenomenon of ionospheric skip?

    Nepal wrote:
    “It can’t be done.”

    Merely take your usual cavity, apply a high enough positive electric potential to strip valence electrons from it , ensure it remains electrically isolated after removing the electric potential. No valence electrons, no inter-molecular dipole interaction (the dominant source of blackbody radiation is transient oscillating dipoles induced by inter-molecular thermal vibrations within a material), no blackbody radiation production. Perfectly reflecting below the plasma frequency (whatever that would be for the given cavity wall material).

    Nepal wrote:
    “It can’t be done.”

    Electrical negation.

    Nepal wrote:
    “It can’t be done.”

    Or use a material such that an oscillating charge in the material, excited by an external oscillator circuit, is essentially π/2 out of phase with the cavity radiation. Thus, any emitted light is completely out of phase with the cavity radiation, and so the cavity radiation’s electric field goes to zero at the surface. Perfectly reflecting.

    Nepal wrote:
    “It can’t be done.”

    LOL

  114. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “It can’t be done.”

    Total external reflection.

    Nepal wrote:
    “It can’t be done.”

    For soft x-rays and extreme UV rays, at an angle of incidence of 0.5 – 1 degrees, due to the index of refraction being slightly less than 1 (whereas, for example, visible light has a refractive index in the rarer medium that is always greater than 1), the radiation is totally reflected. The angle of refraction can be up to 90 degrees at the critical angle. The purer the material, the higher the reflectance at higher angles of incidence, tending to 100% as the angle of incidence decreases. You’d have to configure your cavity space to ensure the radiation always has that small angle of incidence… so a torus (like a Tokamak) or somesuch.

    Nepal wrote:
    “It can’t be done.”

    Piezoelectric photorefractivity.

    Nepal wrote:
    “It can’t be done.”

    Drive a piezoelectric wall at a frequency such that it oscillates π/2 out of phase with incident radiation, the interaction of the space charge field and the optical field cancels the electric field of the cavity radiation at the surface of the wall. Total reflection.

    Nepal wrote:
    “It can’t be done.”

    LOL

  115. LOL@Klimate Katastrophe Kooks says:

    “Ok, aside from the veritable avalanche of examples you’ve provided, you can’t provide any examples of perfect reflection coupled with no absorptivity in the cavity space. It can’t be done.” – Nepal, probably

    https://link.springer.com/article/10.1007/s00340-003-1258-8
    “An active stabilization of photorefractive two-wave coupling by means of an electronic feedback loop has been used extensively during recent years in transmission geometry. It leads to 100% diffraction efficiency η and also to periodic states instead of familiar steady states.”

    Nepal wrote:
    “It can’t be done.”

    LOL

  116. Philip Mulholland says:

    I am running out of popcorn.

  117. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “It can’t be done.”

    Dark state coherent superposition population trapping.

    Nepal wrote:
    “It… it can’t be done?”

    A dark state occurs using laser light to induce transitions between energy levels, when atoms can spontaneously decay into a state that is not coupled to any other level by the laser light, preventing the atom from absorbing or emitting light from that state.

    And what happens when an atom cannot absorb or emit?

    α = absorbed / incident
    ρ = reflected / incident
    τ = transmitted / incident

    α + ρ + τ = incident = 1

    For opaque surfaces τ = 0 ∴ α + ρ = incident = 1

    If α = 0, 0 + ρ = incident = 1 ∴ ρ = 1 … all incident photons are reflected.

    Nepal wrote:
    “It {sniffle} can’t {sob} be done.”

    LOL

  118. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    “It can’t be done.”

    Cavity quantum electrodynamics with atom-like mirrors

    Nepal wrote:
    “It can’t be done.”

    https://sci-hub.se/10.1038/s41586-019-1196-1
    “It has long been recognized that atomic emission of radiation is not an immutable property of an atom, but is instead dependent on the electromagnetic environment and, in the case of ensembles, also on the collective interactions between the atoms. In an open radiative environment, the hallmark of collective interactions is enhanced spontaneous emission—super-radiance—with non-dissipative dynamics largely obscured by rapid atomic decay. Here we observe the dynamical exchange of excitations between a single artificial atom and an entangled collective state of an atomic array through the precise positioning of artificial atoms realized as superconducting qubits along a one-dimensional waveguide. This collective state is dark, trapping radiation and creating a cavity-like system with artificial atoms acting as resonant mirrors in the otherwise open waveguide.

    Nepal wrote:
    “It can’t be done.”

    Remember back when I told you that it’s all stimulated emission… that what we call spontaneous emission is merely due to the anisotropy of the EM component of the quantum vacuum, whereas what we call stimulated emission is anisotropy introduced from other processes? Yeah.

    Nepal wrote:
    “It can’t be done.”

    Remember back when I told you that emission isn’t just a function of a body’s internal state (as Prevost’s Principle and the climate alarmists claim), that the S-B equation clearly shows that it is the energy density gradient which determines radiant exitance of the warmer object, considering its (T_h^4 – T_c^4) term, and that temperature is a measure of energy density (equal to the fourth root of energy density divided by Stefan’s Constant), and thus that term in the S-B equation calculates the energy density gradient? Then I backed it up by doing the calculations using only energy densities, corroborating it with the S-B equation to a precision of 3.84 parts per 100 trillion? Yeah.

    Nepal wrote:
    “It can’t be done.”

    Remember back when you denied all that? Yeah. Rinse and repeat, eh? LOL

    Nepal wrote:
    “It can’t be done.”

    LOL

  119. Nepal says:

    First of all, you are pretending to answer my question, but actually you have made up totally new question. Question was “Please explain how you could build a perfectly reflecting cavity using total internal reflection, where light inside is never absorbed.” Instead you have just listed a bunch of other ideas for reflectors.

    More importantly, there is zero evidence that any of the mechanisms you have listed are capable of perfect reflection with exactly zero absorption.

    Remember, even the tiniest amount of absorption, like a speck of dust, will cause the radiation in a closed cavity to reach its equilibrium state, which is blackbody distribution. If your objection is seriously that a perfectly reflecting material with never thermalize, you need to provide evidence of a material that has been measured to have exactly zero absorption, ever. No such measurement exists.

    Of the two ideas you posted with actual evidence to back it up, one shows energy decaying in a few hundred nanoseconds. That’s not exactly zero absorption is it? The other is a paper that I can’t access, but it involves active feedback — which means external energy is being used, so it is not closed, not at equilibrium.

    Finally, you have slipped back into acting like a rude little child.

  120. LOL@Klimate Katastrophe Kooks says:

    Nepal wrote:
    ” Instead you have just listed a bunch of other ideas for reflectors.”

    All of which prove your contention that there is no perfectly-reflecting material wrong.

    How about that total exterior reflection? Why, that’d make for a very easy cavity, total reflection back into the cavity space. Forget that one, did ya? LOL

    Nepal wrote:
    “More importantly, there is zero evidence that any of the mechanisms you have listed are capable of perfect reflection with exactly zero absorption.”

    Except that several of them empirically measure total reflection… had you read up on any of them, you’d realize this.

    In the case of dark state coherent superposition population trapping and cavity quantum electrodynamics with atom-like mirrors, given that the atoms cannot even absorb the wavelength used, it can do nothing but reflect totally. It’s right there in the equations, which you desperately want to deny but know you cannot:

    α = absorbed / incident
    ρ = reflected / incident
    τ = transmitted / incident

    α + ρ + τ = incident = 1

    For opaque surfaces τ = 0 ∴ α + ρ = incident = 1

    If α = 0, 0 + ρ = incident = 1 ∴ ρ = 1 … all incident photons are reflected.

    Nepal wrote:
    “The other is a paper that I can’t access, but it involves active feedback — which means external energy is being used, so it is not closed, not at equilibrium.”

    Moving the goal posts again, Nepal? You claimed that (to quote you verbatim) “There is no perfectly reflecting material”.

    I showed many, many, many instances of perfectly reflecting material. You made no precondition that it need be at equilibrium, just that it exist.

    Nepal wrote:
    “Finally, you have slipped back into acting like a rude little child.”

    By being right, by proving you wrong or by pointing out that you’ve slipped back into your usual routine of denying scientific reality until you’re drubbed about the head and shoulders to the point of incoherence before you grudgingly accept scientific reality, only to apparently forget at some later date (likely when you believe others have forgotten that you’ve grudgingly accepted reality) and go right back to your original denials? You know, ‘evenminded’ does exactly that. I should know, I drop-kicked that tardling across CFACT every single day for more than 3 years. Drove him to the point that he couldn’t even form complete sentences nor count, had him so stomping mad he swore vengeance, made him cry, made him play the victim… just as you’re attempting to do now… and all because I challenged his dearly-held ideological belief of CAGW. LOL

  121. LOL@Klimate Katastrophe Kooks says:

    If you’re looking for a “perfect reflector” at thermodynamic equilibrium, then have I got a deal for you!

    https://climateofsophistry.com/2022/05/07/ontological-mathematics-the-theory-of-everything-5-the-mind-and-the-brain-consciousness-and-god/#comment-103192

    Graybody objects at thermodynamic equilibrium are by definition perfect reflectors. To claim otherwise violates so many scientific principles and fundamental physical laws that you’d have to be joking to even claim it.

    As ΔT → 0, q → 0. As q → 0, the ratio of graybody total emissive power as compared to idealized blackbody object emissive power → 0. In other words, emissivity → 0. Do remember that temperature is a measure of energy density, equal to the fourth root of energy density divided by Stefan’s Constant. At thermodynamic equilibrium for a graybody object, there is no energy density gradient and thus no impetus for photon generation.

    As ΔT → 0, photon chemical potential → 0, photon Helmholtz Free Energy → 0. At zero chemical potential, zero Helmholtz 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 = absorbed / incident
    ρ = reflected / incident
    τ = transmitted / incident

    α + ρ + τ = incident = 1

    For opaque surfaces τ = 0 ∴ α + ρ = incident = 1

    If α = 0, 0 + ρ = incident = 1 ∴ ρ = 1 … all incident photons are reflected at thermodynamic equilibrium for graybody objects.

    Did you forget about that one, Nepal? LOL

  122. boomie789 says:

  123. LOL@Klimate Katastrophe Kooks says:

    I especially liked the time when I proved ‘evenminded’ (using a different sock at the time) wrong on a key aspect of his religiously-held belief in CAGW, and he just repeated “DISMISSED!!!! for hours… I kept piling on to the point that he was so shaken that he started misspelling even that word. Then he ran away for a few months. Then he had to ditch that sock because he was too embarrassed to use it anymore, which is why he’s using ‘evenminded’ now. LOL

    Ever have anything like that happen to you, ‘Nepal’? LOL

  124. boomie789 says:

  125. LOL@Klimate Katastrophe Kooks says:

  126. Philip Mulholland says:

    I am out of popcorn.
    Give that man a beer.

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