## Ontological Mathematics & the Theory of Everything 4: Monads and the Singularity

Where do monads exist? What is the singularity? And what is the basic process by which they create a universe, and why would they create a universe is the first place? We review what we’ve learned so far and introduce a few new concepts to lay the groundwork for future videos!

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### 180 Responses to Ontological Mathematics & the Theory of Everything 4: Monads and the Singularity

1. LOL@Klimate Katastrophe Kooks says:

So these monad ‘gods’ only created our universe because they were bored, so they put us through death and destruction and horror and pain and retardation to entertain themselves?

I’m sorry, this is essentially morbid polytheistic anthropomorphization of inanimate objects, in my view, with no evidentiary basis. If you can show me a monad or the mathematics of all this, I’ll reconsider, but until then, I’ll continue to subscribe to “we simply do not know, nor can we know, exactly how the universe began, simply because the genesis of the universe occurred outside space-time”.

“One of the Fundamental Secrets of the Universe: All waveforms, no matter what you scribble or observe in the universe, are actually just the sum of simple sinusoids of different frequencies.”

Yes, and a sinusoid is a circular function:

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

That doesn’t imply, however, that photons (a circular path consisting of the electric and magnetic interactions oscillating in quadrature about a common axis, extended into a spiral via the photon’s movement through space-time, viewed from certain frames as a sinusoid) are in any way intelligent.

Thus, these monads, a much more basic construct, can not possibly be in any way intelligent. They may be “programmed” with certain innate simple behaviors (just as all of the universe is, via the fundamental physical laws, which we have discovered and which describe and explain how things interact in the universe), but that doesn’t necessarily lend intelligence to them.

I remain skeptical but open-minded, and await your further explanation.

2. immortal600 says:

It is very simple. GOD created the Universe. What is God? Beyond any human comprehension. It is really that simple.

3. boomie789 says:

5. CD Marshall says:

God, angels, demons if (my personal opinion not inserted) they exist or something that exists has been described as such are by all human account aliens, for they are not of this world. Something as simple as spectrum proves humans have an incredibly limited scope of comprehension based on physical observation.

Light musings…
The idea of “jumping” through space instead of traveling in space came from somewhere. What if “aliens” discarded space travels a long time ago, advanced enough to travel to point A to B with no distance to impede the process.

6. LOL@Klimate Katastrophe Kooks says:

CD Marshall wrote:
“Light musings…
The idea of “jumping” through space instead of traveling in space came from somewhere. What if “aliens” discarded space travels a long time ago, advanced enough to travel to point A to B with no distance to impede the process.”

That gets complicated with relativity taken into account. Photons, while traveling great distances at c, experience no time (thus they can travel infinite distance if nothing’s in their path). Traveling directly from point A to B with no distance between would seem to be a trade-off of space for time, so it’d take them a long time to do so. Can’t imagine why they’d make that trade-off, rather than attempting to get as near c as possible to slow time, except for the energy consumption getting to that speed, and slowing down again at the end of the journey.

7. CD Marshall says:

It’s all beyond me maybe “dimension jumping” is possible under the right technology? You’d have to know those dimensions exist and how to get to them (and out of them).

8. You just change your coordinates through or via the singularity. Voila space jumps.

9. CD Marshall says:

If you could access a black hole and use it to jump back into our space theoretically you could travel anywhere in time or other existence (assuming that’s how a black hole even works). A species that could do that would be near as gods (little g) as you could get.

10. J Cuttance says:

…a tough gang of astrophysicists…the bottom cartoon
https://www.thefarside.com/comic-collections/532/doomed-scallywags-april2022

11. Richard says:

There was no big bang. It is ridiculous to sum up all of the universe and let it come out of a hypothetical point billions of years ago. There are so many errors involved in these concepts that it is becoming comical to watch the state of astro-physics.
Serious philosophical arguments are completely missing in these presentations.

12. boomie789 says:

Very relevant but long. I’m only half way through.

13. Joseph E Postma says:

Platonism is updated with ontological mathematics. The “domain of forms” is the frequency domain. I’ll get into this soon. Good background!

14. LOL@Klimate Katastrophe Kooks says:

One should also keep in mind that the innate simple behaviors which combine to make up the seemingly complex behavior we see in the universe aren’t really ‘in’ the entities, it’s the interaction between entities (‘entities’ here being anything… bosons, leptons, quarks, energy, etc.)

Imagine space which is completely empty… no matter, no energy… a ‘perfect’ vacuum (and neglect for the moment that this is an impossibility because time and space would collapse; neglect as well for the moment that invariant-mass matter would unwind back into energy in such an environment).

{ ASIDE: It may be that the genesis of the universe came about because these monads were a form of energy or matter which doesn’t interact to expand space and thus slow time (dark matter?), and they somehow transformed to what we have now, which does expand space and slow time. Alas, we’ll never know for sure… our light-cone is divorced from and at universal genesis. }

Now place a single atom in that ‘perfect’ vacuum. Without a ruler of some sort (some means of comparison (measurement) and hence an interaction) present in the universe, we cannot know the size of the atom. Likewise, we cannot know if the atom is moving nor in what direction with nothing to compare it to.

Every single behavior we empirically observe manifests via an entity’s interaction with other objects. It is these fundamental interactions via the various fields comprising the quantum vacuum (because it is all fields, even invariant-mass matter is merely a persistent perturbation in the Higgs field) which dictate the behavior of everything. That’s why we have the electroweak interaction, the EM interaction, the strong interaction, etc… each entity interacts with these fields in certain ways, which manifests their macroscopic behavior. And that behavior is heavily dependent upon field energy density.

That’s why, for instance, one can completely stop electron capture decay by ionizing the matter in question by raising bound electron orbital radius to the point that the bound electrons are no longer bound (by increasing energy density)… no electrons to capture, no electron capture decay. That is also why one can lower bound electron orbital radius via creation of an artificially lower field energy density (as in a Casimir cavity, for one example… I’ve shown previously two such methods of doing so), even to the point of increasing electron capture decay to the point that invariant-mass matter unwinds up the Periodic Table back into the energy from whence it came (because E^2 = p^2 c^2 + m^2 c^4).

15. CD Marshall says:

If only this was even a fraction true. Then again if it were why bother coming back here?
https://www.bluestarenterprise.com/

16. CD Marshall says:

[video src="https://bluestarenterprise.com/category/videos/Project_Camelot_interviews_Ralph_Ring.mp4" /]

17. CD Marshall says:

18. CD Marshall says:
19. CD Marshall says:

This is some weird stuff.

20. CD Marshall says:

Yet he admitted it didn’t work.

21. CD Marshall says:

SO this is a little better video.

22. boomie789 says:

23. boomie789 says:

24. LOL@Klimate Katastrophe Kooks says:

As I’ve often said, an EV is akin to a car which needs to have its fuel tank replaced every x years… at a cost of thousands of dollars. If we had that situation for ICE vehicles, auto manufacturer heads would roll. If the auto manufacturers then said “Mheh, we’re not going to make or sell gas tanks for your vehicle because it’s 5 years old. Buy a newer vehicle.”, auto manufacturer heads would absolutely roll.

What we need is a battery tray for EVs that allows one to pop out the old batteries (they are just laptop cells, for the most part) and pop in new ones, much as one would do for a flashlight. I envision a long channel to hold the batteries, with a removable port at one end. You disconnect the battery from the car, open that port, slide the channel out, pop out all the batteries, pop new batteries in, slide the channel back into the pack, close that port and do the same for the other channels.

Of course, there’ll be idiots who put a cell in backwards and smoke their battery pack, so that’ll never be a reality.

Teh Stoopid are why we can’t have nice things. The problem is, idiot-proofing everything just ensures we get bigger idiots. LOL

25. CD Marshall says:

SO, the Germans were attempting, or rather the Nazis were forcing the Germans, to research zero point energy. Yes like all heat is energy but not all energy is heat all Nazis were Germans but not all Germans were Nazis. Much like today, we have one radical branch of government who controls the countries. They do not represent all people of those countries. All of these regimes rule through one thing: Fear.

26. boomie789 says:

I really don’t see the appeal of EVs.

I think I read somewhere it takes 500,000 pounds of earth to be processed to make 1 battery for 1 car. This is probably why the battery is so expensive.

27. LOL@Klimate Katastrophe Kooks says:

We need something like a stack of alternating layers of lanthanum oxide and aluminum oxide as the ‘dielectric’ layer in supercapacitors. The lanthanum layers have a slight positive charge, the aluminum layers a slight negative charge, resulting in a series of electric fields that aggregate in the same direction, which creates an electric potential between top and bottom of the stack.

Now apply a voltage in opposition to this electric potential. That allows us to pack a higher voltage into the supercap before we get ‘dielectric’ (in quotes because it’s not really a dielectric and supercaps don’t use dielectric material, they use Helmholtz double-layers) breakdown.

Because capacitor energy density is equal to the square of voltage, this allows higher energy density without sacrificing the narrow plate spacing of a supercap and thus we maintain the high capacity of supercaps (upwards of 10,000 Farad), but with higher voltages.

That allows us to replace or supplement LiIon batteries with supercaps.

That increases the number of charge/discharge cycles from the few hundred of a LiIon battery to hundreds of thousands for a supercap.

And because lanthanum oxide and aluminum oxide are usually excellent insulators (although they will channel electrons if made thick enough… in effect becoming a power source in their own right… making them thicker in a supercap would increase the potential resisting the applied voltage, but would increase plate spacing), we have lower cap leakage.

28. Note that I have answered what these guys are looking for. Should let them know.

29. LOL@Klimate Katastrophe Kooks says:

A side benefit of such a supercap would be that if it were discharged below the potential of the lanthanum oxide / aluminum oxide stack, the cap would ‘recharge’ by itself… although I suspect it’d be extremely slow.

Perhaps a (much thicker) stack of lanthanum oxide / aluminum oxide used to slowly recharge a bank of regular super caps? Leave your car sitting for a week and the charge goes up, not down.

30. We need you LOL on solving the tech…someone like you would have done it, but would have also needed ontological mathematics to understand it:

Who knows.

31. LOL@Klimate Katastrophe Kooks says:

I just got done watching ‘Red Nightmare’ (aka ‘Freedom and You’, aka ‘A Stranger In Town’) from 1957… starring Jack Webb (of Dragnet fame) as narrator, Jean Cooper (from The Young and The Restless), Robert Conrad (from Baa Baa Black Sheep) and a whole host of other notables.

A man finds his town has gone Communist, Big Brother listens to your phone calls, monitors your correspondence, requires your private passwords, runs invasive background checks on you, imposes mandatory ‘voluntary’ social obligations in support of The Party, surveils you, destroys your life if you speak against The Party… a very worrying coincidence to what is happening today.

Our government has become what we once warned about… and it is the liberals / progressives / socialists / democratic socialists / communists / whatever they call themselves to hide their true nature in our midst who have done this… we need to clean house and get our freedoms back.

Many don’t even realize that they’ve lost those freedoms, being incrementalized like a frog in a pot of water on a burner with the temperature slowly raised. Others, born after they were lost, don’t even realize the freedoms we once had. Most are too weak, too cowardly, too fat and lazy and stupid to even fight for those freedoms. Those types, if the commies are successful in tearing down the greatest nation in the history of the world, deserve to starve in the streets… and they will if the commies are successful.

32. LOL@Klimate Katastrophe Kooks says:

An even better movie… ‘Don’t be a sucker’ from 1943. The same sort of ‘divide and conquer’ tactics used by the cancel culture mobs of today, the same sort of ‘othering’ dehumanization used by the leftists of today, and with the same intent as the Nazis in Germany used… to use the useful idiots who believe the tripe they spew to gain power, to destroy their ideological enemies via lies, to tear down civilization and rebuild it in their twisted image of what society should be (with them at the top and everyone else a serf), but which can never be because it is anti-progress, anti-civilization, anti-human.

Somehow I thought we’d have gotten past this sort of cheap gimmickry long ago, but we’re still just hairless apes with barely the aggregate brain capacity to sustain civilization. Sometimes Teh Stoopid rises up and destroys everything the intelligent have built. And so it goes, rinse and repeat.

33. immortal600 says:

LOL, come back over to Cfact and have fun with ‘ballonboy’ aka ‘evenminded’. He’s obsessed with you and Joe. He’s spouting the usual lies.

https://www.cfact.org/2022/04/25/the-impact-of-co2-is-overstated-so-why-dismantle-society-part-1/

34. immortal600 says:

A question. A gas under pressure gets hotter if contained in the same volume. Correct? The counter to that is the higher pressure in tires or balls that has the same temperature as air outside the containment vessel.

35. boomie789 says:

36. Nice lol

37. boomie789 says:

If you go against or stand in the way of the international banking cartel you will be destroyed and turned into the devil.

38. immortal600 says:

I didn’t phrase my question correctly. If a gas is contained at a specific volume and it’s pressure is raised, will it get hotter?

39. immortal600 says:

LOL. thanks! Good info over there!

40. LOL@Klimate Katastrophe Kooks says:

Immortal600 wrote:
“LOL, come back over to Cfact and have fun with ‘ballonboy’ aka ‘evenminded’.”

Done and dusted. He fell silent. Must have ravage his fragile psyche yet again. LOL

41. Nepal says:

immortal600, if a fixed amount of gas is contained in a fixed volume container, then the only way to increase the pressure is to apply heat. Which will also increase the temperature.

42. LOL@Klimate Katastrophe Kooks says:

Immortal600 wrote:
“I didn’t phrase my question correctly. If a gas is contained at a specific volume and it’s pressure is raised, will it get hotter?”

As long as the energy gained by the gas via compression is greater than the energy lost to ambient, yes. One can compress a gas very slowly and adiabatically.

43. Nepal says:

LolKooks, he specified constant volume, so compression is not an option.

For gas at constant volume there are two ways to increase pressure. The first, as I already said, is to add heat. This always increases temperature and pressure.

The second way is to add more gas molecules. Then what happens to temperature and pressure depend on exactly how you add the gas, as well as the temperature of the gas you’re adding. Usually, you see something like pumping room temperature gas into a room temperature tire, and not allow any heat to leave the tire. Then pressure will increase a lot, and temperature will increase a bit. Over time heat will leave the tire, and the temperature will come back down, also leading to a slight decrease in pressure.

But the exact details are very complicated here because, while you are pumping gas into the tire, it is a very non equilibrium system.

44. LOL@Klimate Katastrophe Kooks says:

Nepal wrote:
“LolKooks, he specified constant volume, so compression is not an option.”

Immortal600 wrote:
“I didn’t phrase my question correctly. If a gas is contained at a specific volume and it’s pressure is raised, will it get hotter?”

Now you’re claiming that raising the pressure isn’t compression. Get a tutor, Nepal. You need it. LOL

Just because the volume of the container is fixed doesn’t mean the pressure need be.

Also, kind of strange that you’d show up at this exact moment, after all this time, just after I drop-kicked that kook ‘evenminded’ into low orbit. He has reading comprehension problems, too. LOL

45. LOL@Klimate Katastrophe Kooks says:

Wow. This is what I call a genius application.

Exum Massbox Laser Ablation-Laser Ionization Mass Spectrometer (LA-LI MS)

Anywhere from 5 minutes to a couple hours for a sample (depending upon sample size and number of passes), runs off 120 VAC 500 W, no other consumables, desktop mass spectrometry.

Me want.

46. Nepal says:

Well, unless you allow more gas to be added, the only way to raise the pressure of a fixed volume gas is by heating it. I would never call heating compression, but maybe you meant heating or adding extra gas by “compression”, in which case I agree.

47. LOL@Klimate Katastrophe Kooks says:

Yet again you display your reading comprehension problem, Nepal… all while dancing around the topic of why you’d suddenly show up just minutes after I drop-kicked the kook ‘evenminded’ into low orbit… while displaying your scientific illiteracy and reading comprehension problem and going after little ol’ me yet again… you wouldn’t be the kook ‘evenminded’, now would you? Because the similarities are striking. LOL

Immortal600 wrote:
“I didn’t phrase my question correctly. If a gas is contained at a specific volume and it’s pressure is raised, will it get hotter?”

For a fixed volume, the only way of raising the pressure (considering that he’s wondering whether temperature will rise, so he’s obviously not heating it) is to pump more gas into the volume… to compress it.

You claimed that raising the pressure wasn’t compression. You were wrong.

Now redefine ‘compress’ to attempt to prove yourself ‘not wrong’. LOL

48. Nepal says:

No hard feelings, usually compressing means decreasing the volume, for example when gas is pushed down with a piston in an engine cycle, that’s called compression. If that’s not what you meant then all good. Now the only question is whether immortal600 meant “increasing the pressure with a fixed amount of gas,” or “increasing the pressure allowing the introduction of new gas.”

“all while dancing around the topic of why you’d suddenly show up just minutes after…”

I’m not dancing around it, I’m ignoring it because it’s stupid.

49. LOL@Klimate Katastrophe Kooks says:

Nepal wrote:
“No hard feelings, usually compressing means decreasing the volume”

So an air compressor works, according to you, by shrinking the volume of the storage tank into which the air is compressed? Strange world you live in, Nepal.

Nepal wrote:
“Now the only question is whether immortal600 meant “increasing the pressure with a fixed amount of gas,” or “increasing the pressure allowing the introduction of new gas.””

Go back and re-read what Immortal600 wrote, keep re-reading it until it sinks in:

Immortal600 wrote:
“I didn’t phrase my question correctly. If a gas is contained at a specific volume and it’s pressure is raised, will it get hotter?”

For a fixed volume, the only way of raising the pressure (considering that he’s wondering whether temperature will rise, so he’s obviously not heating it) is to pump more gas into the volume… to compress it.

For a fixed volume and a fixed amount of gas, the only way of raising the pressure is raising the temperature of the gas, but obviously that’s not what Immortal600 meant because he was wondering if temperature would increase when the pressure was raised.

50. Nepal says:

I’ll admit I misunderstood what you meant, yes.

51. LOL@Klimate Katastrophe Kooks says:

Ok, Immortal600, I tenderized the ‘evenminded’ kook for you… don’t mind the bootprints all over him, nor the multiple broken bones. As addled as he is, I doubt he’s even sensate.. LOL

I’m off to dinner and a movie.

52. LOL@Klimate Katastrophe Kooks says:

Nepal wrote:
“I’ll admit I misunderstood what you meant, yes.”

What I meant isn’t at issue, Nepal… as you have admitted, it’s what Immortal600 meant that you obviously didn’t understand Stop squirming and just admit your error and move on.

53. CD Marshall says:

I want to quote this and slap it on Twitter, I know you like being anonymous, but I’d like to give them something.
“Anonymous…
“scientist”
“Chemist”
“chemical physicist”
“pool boy”

Any corrections afore I post?

” adding any gas molecule with a higher specific heat capacity (or latent heat of vaporization) than the average specific heat capacity of the atmosphere will result in more energy convectively transited from surface to upper atmosphere and thus more energy radiatively emitted, and owing to the mean free path length / air density / altitude relation, the majority of that emitted radiation will be upwelling.

So adding more homonuclear diatomics (N2, O2) or monoatomics (Ar), or decreasing the proportion of polyatomics (CO2, H2O), will increase surface temperature by reducing convective transport of energy from the surface and radiative emission out to space (which would hinder convection because the upper atmosphere would not as effectively radiatively emit and would thus warm, lending less buoyancy to convecting air parcels… the reason homonuclear diatomics and monoatomics are actually the true ‘greenhouse gases’).”

54. CD Marshall says:

55. Nepal says:

Haha, okay LolKooks, you can twist the knife when you’re right about this little thing. Meanwhile I’ve corrected you dozens of times about substantial physics mistakes, and not once have you admitted you’re wrong. In fact you’re still here peddling perpetual motion machines almost every day, ruining this site with crack pottery.

56. LOL@Klimate Katastrophe Kooks says:

Nepal:
Name these “dozens of times”. You can’t do it because your “corrections” are based upon your fundamental misunderstanding of QFT, QM, SED and QEM.

CD Marshall:
You may refer to me as “He who shall not be named, but the intellectual god who makes libtard heads explode”. LOL

57. LOL@Klimate Katastrophe Kooks says:

One of those very “perpetual motion machines” you’re bleating about, BTW, is patented, with a working laboratory model. Another is open source, with several working replications. Now don’t you feel silly, ‘evenminded’? LOL

58. Nepal says:

LolKooks, I do not like to dwell on old arguments. But you continue to push perpetual motion machines, leading everyone here astray, and you have specifically asked for when I proved you wrong. So here is a little sampling of the many many times I have corrected you, and you never acknowledge any of it. This doesn’t even count the pages of nonsense you write about using magnets for perpetual energy, which I never bothered to respond to because it’s all bogus paragraphs without any equations.

At thermodynamic equilibrium, the Helmholtz Free Energy is zero: F = U – TS

No, this is not true. In fact let’s just use the formulas you posted:

“U = a V T^4
P = ⅓ a T^4
S = (4/3) a V T^3”

Now calculate F = U – TS = -a V T^4 / 3. Not zero.

The correct statement is, at thermodynamic equilibrium Helmholtz free energy is at a minimum with respect to all non-fixed state variables.

the anisotropy of the quantum vacuum field

no, the vacuum is isotropic

If you calculate it according to how the “standard engineering formulas” calculate upon radiant exitance, you come to the conclusion that a cooler object of higher emissivity can warm a warmer object of lower emissivity

No, the usual formulas given by thermodynamics predict heat flow only from hot to cold.

If the surrounding vacuum field energy density is higher than the energy density of the excited atom, energy cannot flow from atom to surrounding vacuum field (all atoms interact with the vacuum field), therefore that atom cannot emit.

The only way to decrease spontaneous emission to zero is to decrease the density of states of the environment to zero, as described by Fermi’s golden rule https://en.wikipedia.org/wiki/Fermi%27s_golden_rule#The_rate_and_its_derivation . Thus the energy density must be zero, exactly the opposite of your claim.

Not my theory… it’s the result of the proper application of quantum theory and the fundamental physical laws…

You keep referring to your stochastic electrodynamics as “standard quantum theory.” It is not. First it is not quantum. It tries to replace quantum electrodynamics with classical electrodynamics and classical vacuum. Your paper [1] is titled “Random electrodynamics: The theory of classical electrodynamics with classical zero-point radiation.” Second it is not standard, it is fringe theory with no convincing experimental support.

[After being told that COP refers to heat pumps, not electricity generators] Pedantism is what you do best, Nepal… energy is energy, no matter the form. LOL

No, no, no. Absolutely not. The laws of thermodynamics are all about how heat and work cannot always be converted into each other. (Work includes electricity).

Their problem, however, is that their take on radiative energetic exchange necessitates that at thermodynamic equilibrium, objects are furiously emitting and absorbing radiation

But indeed two bodies at the same temperature do constantly emit and absorb radiation, according to the Stefan-Boltzmann law. It is just the same as an ideal gas at thermal equilibrium: each molecule is constantly undergoing collisions, so the molecules are “furiously” gaining and losing energy. Thermal equilibrium does not mean all energy transfer stops, both kinetic and radiative energy are transferred. It is just heat transfer stops, that is there is no net transfer of energy.

Quoting JP, “All bodies above absolute zero temperature will radiate.” Two bodies at the same temperature with both emit and absorb radiation, but HEAT will not transfer.

An idealized blackbody object assumes: 1) Emission to 0 K

Actually, an idealized blackbody at temperature T always emits the same amount, namely the Planck function at each wavelength. This is true regardless of the surrounding temperature.

This means an idealized blackbody object must always be at 0 K (because it must absorb all radiation incident upon it, thus it must have an energy density of 0 J m-3)

The energy density is not zero. There is still incoming and emitted radiation, even if the incoming is absorbed.

but it must also be at ∞ K (because it must emit all radiation incident upon it, so it must have an infinite energy density)

Nonsense. A blackbody at temperature T is at temperature T, simple. It emits the Planck function for a temperature T object, which is entirely consistent and expected. It is not at infinity temperature.

A real-world graybody object assumes… Emission and absorption which is dependent upon energy density gradient

No, this is not the case. The only place I have ever seen this claim this is Charles Anderson’s blog. Absorption depends on the incoming radiation flux, not energy density, because it is incoming radiation flux that is absorbed. Emission results from excited electrons, phonons, etc. in the material, which then emit radiation. This depends on the object’s temperature, not the external energy density.

Thermodynamic equilibrium implies there can be no change of state.

Thermodynamic equilibrium means there is no macroscopic change of state. When a gas reaches thermodynamic equilibrium the molecules do not stop moving. They continue to fly about, gain and lose energy, etc. Photons are still absorbed and emitted. The point of thermodynamics is you can forget about these trillion trillion small molecules and photons, and only look at a few macroscopic variables like temperature and pressure. Thermodynamic equilibrium means these macroscopic state variables do not change.

[Referring to Griffiths Introduction to Quantum Mechanics] And in equation 9.47, ρ(ω_0) is the energy density in the fields, per unit frequency. If the fields are equipotential, then ρ(ω_0) = 0

The equation clearly says ρ(ω_0) is *not* zero. Again, this is from the most used book on quantum mechanics, it is accurate.

Calculate the energy density in a cavity at thermodynamic equilibrium, then do so with your assumption that the cavity walls emit and absorb radiation at thermodynamic equilibrium. What do you see? That your take on radiative energetic exchange doubles the energy density of the cavity space

No, this is exactly why emissivity must equal absorptivity. In a cavity at thermodynamic equilibrium, the cavity walls emit exactly as much as they absorb, so the radiation field (including energy density) is not changed.

59. Nepal says:

Which working lab demonstration of perpetual motion are you talking about? This Excel graph with an AOL email address you posted? ( jnaudin.free.fr/images/meg21iof.gif ) Real believable, I guess all humanity’s problems are solved. Oh wait, that graph is from 22 years ago, why don’t we have perpetual motion yet?

Or do you mean your own perpetual motion machine? The one that you promise provides infinite energy, but you have to plug it into the wall… not to take energy, but because you make so much energy you have to give some back, of course. Lol.

“ Mine is plugged into the wall because I use a microinverter to feed power back to the grid, just as solar panels do.”

There are no perpetual motion machines, infinite energy devices, etc., now or ever.

60. CD Marshall says:

Pure hubris.
Jim Brown-Green
@JimBlack48
·
4h
Don’t need 1000 – just provide ONE repeatable lab experiment showing CO₂ + natural LWIR 15 μm emissions warming something (CO₂ back radiation forcing).

Rob Honeycutt | 🇺🇦
@robhon_
·
4h
Jim… The lab experiment is all around you every single day. It is the greenhouse effect that keeps our planet 33°C warmer than the sun alone can warm it. That is repeatable. It’s been repeated every day for the past 4.5 billion years.

Jim D
@jimdtweet
@robhon_

@JimBlack48
and 47 others
Yes, they have a lack of situational awareness when it comes to the climate around them.

61. immortal600 says:

LOL, try sending me another email. As ‘evenminded’ has indicated, he follows this site all the time and is aware what we post. I haven’t checked in a while but will start looking again today for new messages.

62. boomie789 says:

63. boomie789 says:

64. LOL@Klimate Katastrophe Kooks says:

Nepal wrote:
“At thermodynamic equilibrium, the Helmholtz Free Energy is zero: F = U – TS

No, this is not true. In fact let’s just use the formulas you posted:

“U = a V T^4
P = ⅓ a T^4
S = (4/3) a V T^3”

Now calculate F = U – TS = -a V T^4 / 3. Not zero.”

That’s not the Helmholtz Free Energy equation. You’re attempting to conflate two concepts to prove whatever point you’re attempting (but failing) to prove.

F = U − TS

Where,
F → The Helmholtz free energy in Joules
U → System’s internal energy in Joules.
T → Absolute temperature of the environment in Kelvin.
S → system’s entropy in joules per Kelvin.
TS → energy the system can absorb from the environment.

Now, even a grade schooler can ascertain that when U = TS, F = 0. Simple math doesn’t seem, however, to be your forte. LOL

Nepal wrote:
“the anisotropy of the quantum vacuum field

no, the vacuum is isotropic”

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

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

And that underlies your entire fundamental misunderstanding of the process by which quantum vacuum energy density can be artificially lowered, allowing extraction of energy from the quantum vacuum.

Nepal wrote:
“No, the usual formulas given by thermodynamics predict heat flow only from hot to cold.”

The climastrologists don’t use “the usual formulas given by thermodynamics. If you had even an iota of reading comprehension, you’d have realized that.

The climate alarmists misuse the S-B equation, using the form meant for idealized blackbody objects upon graybody objects:
q = σ T^4
… and slapping ε onto that (sometimes) …
q = ε σ T^4

Their misuse of the S-B equation inflates radiant exitance far above what it actually is for all graybody objects, necessitating that they carry that error forward through their calculations and cancel it on the back end, essentially subtracting a wholly-fictive ‘cooler to warmer’ energy flow from the real (but calculated incorrectly and thus far too high) ‘warmer to cooler’ energy flow… which leads especially scientifically-illiterate climate alarmists to conclude that energy actually can flow ‘cooler to warmer’ (a violation of 2LoT and Stefan’s Law).

Even worse is that they then further misuse the S-B equation by converting it into what they call a “forcing formula”:
4 ε σ T3

… which builds-in a warming trend:

For 288 K, a 1 K negative temperature change, ε=0.93643 (ref: NASA ISCCP program):
4 ε σ T3 = 5.07369679087621 W m-2

For 288 K, a 1.00525093764635 K negative temperature change, ε=0.93643 (ref: NASA ISCCP program):

q = ε σ T4 = 5.07369679087621 W m-2

For 288 K, a 1 K positive temperature change, ε=0.93643 (ref: NASA ISCCP program):
4 ε σ T3 = 5.07369679087621 W m-2

For 288 K, a 0.99474906235365 K positive temperature change, ε=0.93643 (ref: NASA ISCCP program):

q = ε σ T4 = 5.07369679087621 W m-2

So their bastardized equation gives the result for a warming of 0.99474906 K while claiming it’s a warming of 1 K, and gives the result for a cooling of 1.00525093 K while claiming it’s a cooling of 1 K.

And that’s likely why they bastardized the S-B equation… to hide the fact that they continue to treat real-world graybody objects as idealized blackbody objects, and to build-in whatever warming trend they possibly could in order to help sustain their alarmist narrative.

Nepal wrote:
“The only way to decrease spontaneous emission to zero is to decrease the density of states of the environment to zero, as described by Fermi’s golden rule https://en.wikipedia.org/wiki/Fermi%27s_golden_rule#The_rate_and_its_derivation . Thus the energy density must be zero, exactly the opposite of your claim.”

You’ve got that exactly backward, as you so often get things. If the field energy density is zero, energy density gradient is at a maximum, and thus the object would be akin to an idealized blackbody object (just with emissivity < 1)… it would emit as much as its emissivity would allow it to.

Nepal wrote:
“An idealized blackbody object assumes: 1) Emission to 0 K

Actually, an idealized blackbody at temperature T always emits the same amount, namely the Planck function at each wavelength. This is true regardless of the surrounding temperature.”

Simple math yet again eludes you. Check the graphic above.

That’s sufficient to demonstrate yet again that you, ‘evenminded’, are so far off that you’re diametrically opposite to reality. That’s not surprising… the mental illness of liberalism necessitates that the sufferer be diametrically opposite to reality.

65. immortal600 says:

LOL, thanks. I got it and have attempted to respond directly.

66. Nepal says:

LolKooks, I’m not going to go through all these again. You didn’t get it the first time, clearly nothings changed. But let’s just look at the first one. It is the simplest, fifth grade math, I just don’t see how you’re disagreeing with me.

You claimed: “U-TS=0” for any system at equilibrium. Then you helpfully provided the numbers for a photon gas at equilibrium:

“U = a V T^4
S = (4/3) a V T^3”

So let’s check it. This is literally just subtraction. Grade school. We have U, we have TS, let’s subtract them.

U-TS = a V T^4 – (4/3) a V T^4
U-TS=-(1/3) a V T^4

That’s not zero.

It is so, so clear that you’re wrong. It’s literally proven by fifth grade subtraction. U-TS is not zero. How are you still in denial.

67. LOL@Klimate Katastrophe Kooks says:

You yet again fail at basic math.

F = U − TS

Where,
F → The Helmholtz free energy (J)
U → System’s internal energy (J)
T → Absolute temperature of the environment (K)
S → system’s entropy (J K-1)
TS → energy the system can absorb from the environment

Let’s set U = 100 J and TS = 100 J.

F = 100 J – 100 J = 0 J.

Now, I know that this highly advanced mathematics is far beyond your ability to comprehend, but do try your very, very best, Nepal. LOL

Now sure how you got the idea that the equation you’re using is for “numbers for a photon gas at equilibrium”, I wrote:

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

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

Photons are a very special kind of boson. When they are in radiative thermal equilibrium in a volume V at a constant temperature T, their chemical potential is zero. The number of photons in the cavity is strictly determined by the temperature of the walls.

Applying Bose-Einstein statistics for a boson gas with a chemical potential of zero and integrating over the photon standing wave states available in the cavity in radiative thermal equilibrium, one finds that
a = π^2 k^4 / 15 ɦ^3 c^3 = 7.5657e-16 J m-3 K-4,

where ɦ is Planck’s constant h/2π, k is the Boltzmann constant, and c is the speed of light. Note that a, Stefan’s Constant, is not the same as the Stefan-Boltzmann Constant.”

That’s the energy density of the cavity. But you understand little, so you getting that bit wrong is only natural for you. You yet again display your reading comprehension problem, ‘evenminded’. LOL

68. LOL@Klimate Katastrophe Kooks says:

The total energy density of a photon gas at temperature T is:

69. LOL@Klimate Katastrophe Kooks says:

Just watched “The Conqueror” (1956), directed by Howard Hughes, starring John Wayne, Susan Hayward and Agnes Moorhead… a movie so bad that Howard Hughes spent \$12 million in 1957 to purchase all the prints of it so it couldn’t be seen for 17 years… a movie so bad they literally nuked the actors. LOL

John Wayne played Genghis Khan (you can begin face-palming now), Susan Hayward played his Mongolian paramour. It was filmed downwind of the Nevada National Security Site, a nuclear test site. Two tests, code-named “Simon” and “Harry”, both larger than the bomb dropped on Hiroshima, caused massive fallout which settled over the Snow Canyon area of St. George, UT, where they filmed “The Conquerer” the next year. To top it all off, Howard Hughes had 60 tons of Snow Canyon soil trucked to Hollywood.

The director Bill Powell, John Wayne, Susan Hayward, Agnes Moorhead and 42 other cast and crew had died of cancer by 1980. A total of 90 of the cast and crew developed cancer.

70. Nepal says:

Come on man. You’re being a child. It’s a simple question of fifth grade subtraction, you’re transparently wrong. And still you double, triple, quadruple down.

You claimed U-TS=0. You gave the expressions for U and S. Simple subtraction shows that U-TS is not zero.

** Let’s set U = 100 J and TS = 100 J.

F = 100 J – 100 J = 0 J.**

Sure, if U and TS were both 100 J that would be right, but that’s not what they are lol. For a photon gas, they are (quoting you):

** U = a V T^4
S = (4/3) a V T^3**

So U-TS=-(1/3) a V T^4. not zero.

Finally, since you don’t seem to realize, photon gas is a synonym for a cavity filled with black body radiation.

71. CD Marshall says:

Elon knows how to walk the fine line with words.

72. boomie789 says:

Interesting. I remember his tune was different on Joe Rogan.

73. CD Marshall says:

The thing about Elon is you never know if he’s being serious or not. Heck of a poker face.

74. LOL@Klimate Katastrophe Kooks says:

You’ve been shown the equation for the energy density of a photon gas, you’ve been informed that what you’re misusing is the energy density of the cavity, now you’re denying simple math to try to prove whatever point you’re attempting to prove… let me guess… that ‘backradiation’ exists and thus CAGW is real, right? No matter which sock you slip on, you can’t hide your true nature. LOL

75. LOL@Klimate Katastrophe Kooks says:

No, a photon gas is not “a synonym for a cavity filled with black body radiation”… a photon gas is considered one which does not interact with invariant-mass matter and thus photon number is conserved… in a cavity at thermodynamic equilibrium, µ = 0 and N is indeterminate… the photon number is arbitrary because photon chemical potential is zero.

That’s why the boundary conditions for a photon gas in a cavity are set such that the components of E tangential to the container walls tend to zero at the walls of the container… why a “photon gas” in a cavity must be considered to be a distribution of standing waves (which you’ve also denied).

Remember? I stated that in a cavity at thermodynamic equilibrium, the photons in the cavity space set up standing waves with the wave nodes at the surface (thus no energy can be transferred to or from the walls), and should one wall change temperature, that standing wave becomes a traveling wave with the group velocity proportional to the energy density differential and in the direction of the cooler wall… you denied that. You denied standard blackbody cavity knowledge… so you could claim that photons were absorbed and emitted by the walls at thermodynamic equilibrium (with photon chemical potential at zero, with Helmholtz Free Energy at zero).

You’re attempting to conflate the energy density of the cavity itself with the energy density of the photons in the cavity space.

76. CD Marshall says:

How does a photon gas remain in equilibrium? I tried looking up photon gas but the equations were a page long,

“Chemical potential for the photon gas is equal to zero (μ = 0), we see that the free energy F of the photon gas is equal to its Helmholtz thermodynamic potential Ω.”

77. LOL@Klimate Katastrophe Kooks says:

It’s not easy… thermodynamic equilibrium is generally a state that is passed through, not remained at… the one constant throughout the universe is change.

It’s like a U-shaped curve for Helmholtz Free Energy… with the bottom of the curve being 0 J.

F = U − TS

Where,
F → The Helmholtz free energy (J)
U → System’s internal energy (J)
T → Absolute temperature of the environment (K)
S → system’s entropy (J K-1)
TS → energy the system can absorb from the environment

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

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

∴ T = 4^√(e/a)

In other words, temperature is equal to the fourth root of energy density divided by Stefan’s constant. It is a measure of energy density.

Keep in mind that Stefan’s constant above equals 4σ/c (which is sometimes known as the radiation constant).

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

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

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

This agrees with Planck’s Law: ρ(T) = aT^4 = T^4 4σ/c.

The S-B equation integrates Planck’s Radiation Formula (which calculates the energy density for a given wavelength) over all wavelengths.

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

So in the real world, the energy density gradient determines radiant exitance, energy does not flow willy-nilly without regard to energy density gradient and 2LoT applies always and everywhere.
https://i.imgur.com/IgxATSg.png

78. Nepal says:

@CD, a photon gas stays in equilibrium by a simple process. The walls containing the gas always emit a constant amount of photons, because their electrons are excited by random thermal collisions, then decay to emit photons. If there are aren’t enough photons in the box, the constant emission by the walls will increase the number of photons, toward equilibrium.

Each second, the walls also absorb a constant fraction of the photons currently in the box. If there are too many photons in the box, this absorption will happen a lot, and the number of photons will decrease, toward equilibrium.

The first process acts to replenish photons if there are too few. The second removes photons if there are too many. When there are the exact right number of photons per volume, the two processes balance. Equilibrium.

The one magic part is that every sort of matter obeys one role: good emitters are good absorbers, and bad emitters are bad absorbers. This rule means that the equilibrium number of photons is the same regardless of the wall material. If the walls absorb a lot, you might think the equilibrium number would be smaller — but it’s not, because the walls also emit a lot.

79. Nepal says:

LolKooks, come on man, we already have the equation for U and S. Anyone can check that U-TS is not zero. Nothing you say can change that, unless you change the definition of subtraction.

If you can’t admit your mistake here when it this blindingly obvious, no wonder you make bigger, whopping mistake like believing in perpetual motion machine. You are not trustworthy because you value your ego more than physics.

Helmholtz energy is not zero at equilibrium. You are wrong. After all, it’s not supposed to be zero — it’s derivative is zero at equilibrium.

80. LOL@Klimate Katastrophe Kooks says:

Nepal wrote:
“The walls containing the gas always emit a constant amount of photons, because their electrons are excited by random thermal collisions, then decay to emit photons.”

He says, denying the definition of thermodynamic equilibrium. Now calculate the entropy of your continual energy exchange, keeping in mind that radiative energetic transfer is an irreversible entropic process. Why does entropy not change at thermodynamic equilibrium in your kooky world? You must claim that radiative energetic transfer is an idealized reversible process for your blather to be true… it’s not, which utterly destroys your blather. LOL

You can’t get around the fundamental physical laws, no matter how hard you try, no matter how much you deny reality. LOL

81. CD Marshall says:

An electron in its relaxed state does not emit a photon.

82. LOL@Klimate Katastrophe Kooks says:

Do you even understand what Helmholtz Free Energy is, Nepal? It is the amount of work the system can do upon the environment, or the environment upon the system… at thermodynamic equilibrium, neither can do work upon the other, F = 0. F is called the work function. ‘A’ was originally used (and still is outside of physics), short for Arbeit, the German word for work. In fact, it always tends toward zero… that’s why it’s known as a “falling function”. Entropy increases, exergy decreases.

But then, you’ve claimed that energy can flow with no work needing be done (so you must conversely also claim that work can be done without any energy needing to flow… yet here you are bleating about ‘perpetual motions machines’ which aren’t perpetual motion machines… you simply don’t understand much of anything. LOL)… so you’ll get right on mathematically describing exactly how that work-free energy flow takes place. Your failure to do so will stand as your tacit admission that you’re so far off the rails that your fantasy blather can’t be backed up mathematically. LOL

83. LOL@Klimate Katastrophe Kooks says:

CD Marshall wrote:
“An electron in its relaxed state does not emit a photon.”

A bound electron in its ground state does not emit a photon unless it can be induced to acquire an even lower orbital radius. There’s nothing special about the ground state… the bound electron acquires its energy which determines its orbital radius from the quantum vacuum at all times, no matter the energy density, no matter the orbital radius. When we excite a bound electron to a higher orbital, we’re bumping up local quantum vacuum energy density… remember that a photon is merely a persistent perturbation of the EM field above the ambient EM field, said EM field having its ground state at the quantum vacuum ground state (ie: the ground state of the EM component of the quantum vacuum).

Boyer, Forward, Haisch, Cole, Szanecki, De Lorenci, Calloni, Moddel, Rueda, Tiplea, Davis, Simaciu, Puthoff, Dmitriyeva, Nickisch, Milea, Schiopu, Ribeiro, Pinto, Bhattacharya, Modanese, Broda, Ibison, Srivastava, Zhou and even NASA acknowledge that it is the energy density of the quantum vacuum (the EM ground state component of the quantum vacuum) which sustains a bound electron at its ground state… artificially suppressing that energy density below the usual ground state by suppressing the resonant quantum vacuum wavemodes sustaining the bound electron at its usual orbital radius (via a Casimir cavity) induces the bound electron to acquire an even lower orbital radius, and it is that quantum jump which causes a photon to be emitted.

This has been experimentally confirmed (DARPA SPAWAR Grant No. N66001-06-1-2026)… they empirically observed a very clear IR emission for N2, Ar, Xe and He, and attempted to explain it away via frictional heating, the Joule-Thomson Effect, turbulence, the radiation coming from pumping energy and surface adsorption / absorption… none of them fit with the emission observed.

The problem currently is that we cannot produce cavities with the size consistent enough (it varies from 0.1 µm to 0.4 µm), nor with a consistent shape. For instance, for Xe, you need a consistent 0.2 µm cavity size. Obviously, at that size, you need lots of them so you can blow an appreciable amount of gas through… so scaling is an issue, as well.

As you well know, the Casimir effect means that quantum vacuum energy density is geometry-dependent… change that geometry and you change the quantum vacuum energy density. Change the quantum vacuum energy density and you change the bound electron orbital radius.

84. CD Marshall says:

So basically, what you are saying is unless it reaches absolute zero it can change (reduce) ground state. However, that is saying at current ground state it is not emitting a photon unless it changes the current ground state. Correct?

85. CD Marshall says:

Or are you saying it is always trying to change its ground state by emitting energy?

86. LOL@Klimate Katastrophe Kooks says:

Even at absolute zero there is the quantum vacuum. Zitterbewegung should still exist at absolute zero because the quantum vacuum still exists at absolute zero (hence why it’s named ‘Zero Point Energy’).

I’m fairly certain the bound electron would undergo electron capture by the oppositely-charged nucleal proton(s) long before quantum vacuum energy density was locally reduced to zero (which isn’t even possible… it’s “DC to daylight” in scope, almost impossible to block all of it). Once you get down to a single integer de Broglie wave in the electron’s orbital and go to a lower quantum vacuum energy than that, the electron in-falls, guaranteed.

87. CD Marshall says:

Joe what’s your expert opinion on this paper, assuming you haven’t already seen it.
https://www.aanda.org/articles/aa/full_html/2021/12/aa41516-21/aa41516-21.html

88. CD Marshall says:

LOL@Klimate Katastrophe Kooks
Interesting. A bit over my head, but still interesting.

89. LOL@Klimate Katastrophe Kooks says:

Yes, the bound electron is always ‘trying’ to emit a photon and settle to a lower orbital radius. What prevents it from doing so, at whatever orbital radius the bound electron is at (even the ground state) is the quantum vacuum energy density. In an excited state, that excitation generally comes from a photon (a persistent perturbation of the EM field above the ambient, the EM field having its ground state at the ground state of the EM component of the quantum vacuum) or from energy applied (heating)…. remove that field energy density such that the bound electron has a higher energy density than the field (in the case of the photon, by absorbing it; in the case of heating, by cooling), and the bound electron then has an energy density gradient via which it can emit a photon to de-excite.

Energy does not flow if there is no energy density gradient, nor does it flow up an energy density gradient (unless external energy does work to push that system energy up the gradient)… 2LoT in the Clausius Statement sense in a nutshell.

90. Nepal says:

“[Paragraphs of …F = 0 … [more paragraphs]”

Interesting explanation, only problem is F is not zero, as anyone see by simple subtraction, so you’re wrong.

It the math differs from your opinion of what the math should be… something is wrong with your opinion.

91. LOL@Klimate Katastrophe Kooks says:

You get the incorrect answer yet again because you don’t understand the equations you’re working with… that’s a pattern with you. LOL

Another pattern with you is that you persist in your incorrectitude even as the glaringly obvious answer stares you in the face, doubling down on stupid until I corner you, wallop you about the brainpan with yet another lesson that likely won’t sink in, whereupon you skitter away, only to return and repeat. LOL

Well, would you look at that… you are yet again wrong. LOL

92. Nepal says:

LolKooks, not sure why you’re acting like you made a point. Your original equation was F=0. The equation you’ve boxed red is Delta F <= 0. Do you see that these are not the same? Do you see that your original equation was wrong?

Why can’t you admit you were wrong? Instead you double down and try to cover your mistake with more bs. Why do you care so little about the truth?

So we don’t forget, you said F=U-TS=0. You also said U = a V T^4, S = (4/3) a V T^3.

Therefore U-TS=-(1/3) a V T^4. not zero. Admit your mistake and move on. Otherwise you’re lying to everyone here.

93. Nepal says:

As analogy: let’s say F stands for money.

Delta F <= 0: you started with \$1,000,000 and now you have \$999,990

F=0: you have \$0.

Not the same.

94. LOL@Klimate Katastrophe Kooks says:

I see simple math and even simpler logic continues to elude you, Nepal. LOL

Helmholtz Free Energy: A = −β^−1 lnZ

Assume the system in toto consists of one single particle (one energy state, no degeneracy).

Z = e^− βE_1 and A = E_1… the maximum energy is available to do work.

E = 0 is the minimum level of energy we can get to, right? Are you claiming there’s negative energy? ΔA is predicated upon E reaching a minimum of 0, right? The claim that −A measures the maximum amount of useful work that may be extracted from the system with Helmholtz Free Energy of A holds assuming that E = 0 is the minimum energy after extraction, right?

Remember me telling you this is a “falling function”? That’s because the Free Energy tends to zero as thermodynamic equilibrium is approached. Deny that and you deny pretty much everything we know about thermodynamics. LOL

-_

One can think of Helmholtz Free Energy as the energy necessary to create a system in the absence of changes in temperature or volume… with at least some (and at most all) of that system energy (U) coming from the ambient (TS).

If a higher entropy final state is created than the ambient, less work is required to create the system because TS energy flowed to the system. The Helmholtz free energy is then a measure of the amount of energy you have to put in to create a system once the spontaneous energy transfer to the system from the environment is accounted for.

Now obviously, TS can’t pump more energy U (via the energy density gradient between ambient and system) into the system than the ambient itself has (unless external energy does work to pump that system energy against the energy density gradient). Deny that and you’ll have yet again denied 2LoT in the Clausius Statement sense.

Thus if one merely allows TS to set the initial conditions of U by allowing ambient energy to flow to the system, without you putting any more energy into the system, U=TS therefore F = 0.

And that is the definition of thermodynamic equilibrium.

Therefore, just as I stated and which you’ve now repeatedly denied, at thermodynamic equilibrium, U = TS therefore F = 0.

Now deny reality again, nutter. LOL

95. Joseph E Postma says:

GUYS…honestly…shut up. We don’t give a shit about the tit for tat anymore. And stop making it worse than it needs to be.

Stop now. Ban next.

96. LOL@Klimate Katastrophe Kooks says:

If the math differs from your opinion of what the math should be… something is wrong with your math… or your brain. LOL

97. LOL@Klimate Katastrophe Kooks says:

This guy’s a sock of ‘evenminded’, I’m sure of it. Same tactics, same blather, just toned down here so he doesn’t get insta-booted, but he just can’t help but attack long-empirically-established physics to bolster his CAGW blather. Even when he’s proven wrong, he’ll just slink away and come back again with the same idiotic arguments all over again under a new sock.

98. Nepal says:

Joe, this guy is making a mockery of physics with perpetual motion machine nonsense. Worse, he just insists he’s right even when fifth grade math proves him wrong.

You’re a physicist, I know you know he’s wrong. On this one little point maybe it’s not a big deal, even though you and I both know Helmholtz free energy is not zero at equilibrium (I learned this from wikipedia in five minutes), but when he’s telling everyone he has a perpetual motion machine on his desk, it is a big deal.

So if you want to just tell us both to shut up, okay. But I wish you would defend proper physics and tell this liar he’s wrong instead.

99. LOL@Klimate Katastrophe Kooks says:

You’re the one conflating two completely different concepts and misusing the equations for same to attempt to ‘prove’ that energy can flow without an energy density gradient, that energy can flow without work being done and hence (tacitly) that work can be done without an energy flow… and you have the temerity to claim anyone else is “making a mockery of physics”. LOL

You learned nothing of the sort from Wikipedia… if you’d read correctly, you’d have realized:

ΔU_bath + ΔU + W = 0

At thermodynamic equilibrium, none of those can change because there is no energy density gradient by which energy can flow and thus no ability for work to be done… all are zero.

∴ W ≤ – ΔF

Thus if no work can be extracted from the system to ambient (either because the system has lower energy density than ambient (F < 0) or because both system and ambient are at thermodynamic equilbrium, both have the same energy density, there is no energy gradient by which energy can flow (F = 0)), ΔF ≤ 0.

Remember that:
F = U − TS

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

100. CD Marshall says:

Dr. Pat®️Paleoclimatologist

I asked Mann a question about his proxy data, some of which was mine, and he blocked me. He now refuses to speak to me.

101. CD Marshall says:

Joe will give you two a timeout he has done if before. Knock it off already. I enjoy friendly open debate (which this has gone beyond that) but when Joe says enough is enough, stop it. By all means open your own site and have at it.

Joe’s site, Joe’s rules. Respect that. I don’t want to see either one of you blocked.

102. CD Marshall says:

This how brainwashed a grad student is…
CD Marshall
@MarshallCd
@brandondaly2018

You still can’t accept the obvious flaw in “CO2 forcing” doesn’t mandate surface temperatures. You won’t accept that is a flaw. No evidence supports it so why is it being taught as if it is confirmed physics?

Brandon Daly 🇺🇦🧊
@brandondaly2018
Because evidence supports it and it is confirmed physics.

103. CD Marshall says:

No evidence supports it and no physics confirms it. That’s why climate science made up it’s own version of physics that cannot be applied anywhere else in the realm of physics. Backscattering? ffs.

I’ll buy a “backscattering” heating system that heats the house to a higher T than the original heating element with all the doors and windows open in an ambient environment below zero where the alleged forcing is taking place.

104. CD Marshall says:

105. immortal600 says:

LOL, I sent you a message direct (from my real email address) to the address attached to your message. Did you get it?

106. LOL@Klimate Katastrophe Kooks says:

No, I didn’t get it. You’re using the email address with your moniker in it?

107. boomie789 says:

I’ve only listened to half so far but that black hole discussion didn’t last long. Lol.

Got interrupted by Better Call Saul.

108. CD Marshall says:

That was a good discussion, other than a little arrogant and condescending on one side but the rest were open for a good debate on varying points of views.

It seems hard for a scientist to simply say, “I don’t know.”

We have never seen the end of the Universe; we don’t even know if such a thing exists in what we understand in linear progression.

If you remove time and space from that equation.

Assume relativity is a restriction we exist in but the universe doesn’t.

Then concepts like “beginning” and “ending” are actually a static constant not a linear constant.

Thus if we can travel outside of our time and space dimension, we can view the universe as it exists, not as we perceive it exists. You are not actually time traveling at that point you are simply entering the universe at a different constant than the one you left.

So then the thoughts are:
Once you leave this dimension are you stuck outside of it not being able to re-enter into this “constant” or another constant. Forever outside of time, maybe able to glance or peer into the constant but never actually being able to re-enter.

If aliens exist do they have this technology already? More nefarious, have they deliberately prevented mankind from achieving it.

Perhaps humans can’t leave the constant they exist in but perhaps they can glance into another constant with the correct technology.

Who knows?

109. CD Marshall says:

Look at this from “Science of Doom” we should visit his site. GIve LOL@Klimate Katastrophe Kooks
some room to vent safely.

Radiation Basics and the Imaginary Second Law of Thermodynamics
Conclusion:

When two bodies have an energy source which has created a constant surface temperature and they are subsequently brought into proximity with each other, there will be an increase in each other’s temperature. But no thermal runaway takes place, they just reach a new equilibrium.

Basic thermodynamics explains that bodies emit thermal radiation according to temperature (to the fourth power) and according to emissivity. Not according to the temperature of a different body that might happen to absorb this radiation.

And basic thermodynamics also explains that bodies absorb thermal radiation according to their absorptivity at the wavelengths (and directions) of the incident radiation. Not according to the temperature (or any other properties) of the originating body.

Therefore, there is no room in this theory for the crazy idea that colder bodies have no effect on hotter bodies. To demonstrate the opposite, the interested student would have to find a flaw in one of the two basic elements of thermodynamics described above. And just a note, there’s no point reciting a mantra (e.g., “The second law says this doesn’t happen”) upon reading this. Instead, be constructive. Explain what happens to the emitting body and the absorbing body with reference to these elementary thermodynamics theories.

Update – now that one advocate has given some explanation, a new article: Intelligent Materials and the Imaginary Second Law of Thermodynamics

Notes

I said earlier: “If this star absorbs thermal radiation from elsewhere, it must emit more radiation or its temperature will rise. If its temperature rises then it will radiate more energy.” Strictly speaking when radiation is absorbed it might go into other forms of energy. For example, if ice receives incident radiation it may melt, and all of the heat is absorbed into changing the state of the ice to water, not to increasing the temperature.
Incident radiation can also be transmitted, e.g. through a thin layer of glass, or through a given concentration of CO2, but this won’t be the case with radiation into a body like a star. The total of reflected energy plus absorbed energy plus transmitted energy has to equal the value of the incident radiation.
The Stefan-Boltzmann equation is the integral of the Planck function across all wavelengths and directions:

Spectral Intensity, Planck

Spectral Intensity, Planck

Where h, c0 and k are constants, T is temperature and λ is the wavelength.

The Planck function describes how spectral intensity changes with wavelength (or frequency) for a blackbody. If the emissivity as a function of wavelength is known it can be used in conjunction with the Planck function to determine the actual flux.

110. CD Marshall says:
111. CD Marshall says:

Yes Joe this is the site I offered the “Author” to debate you live a few years ago. He refused. Wouldn’t even come over here, it had to be on “his” site only.

112. boomie789 says:

Woke up to some kinda big news.

Meanwhile, in California…

“In retaliation against Roe v Wade being struck down, extremists in California have introduced Bill 2223, which would legalize murdering your baby up to ONE MONTH AFTER BIRTH.”

https://leginfo.legislature.ca.gov/faces/billNavClient.xhtml?bill_id=202120220AB2223

Kind of mixed on this. While it is a grotesque act where they sell the baby parts and other rumors I’ve heard, this will have negative consequences as well.

Abortion is like drugs and prostitution. The problem is upstream.

113. boomie789 says:

Lol.

114. CD Marshall says:

What that should simply entail is the decision falls back to the states.

Steven Crowder
If Roe is overturned, the decision simply returns to the states. This is the essence of our federalist system. Leftists don’t understand this, though, and they are melting down, as expected. Kinda fun to watch, actually.

115. immortal600 says:

LOL, no my name wasn’t in it. Just sent it from my real email address to the one that you sent me through the other site.

116. Nepal says:

Crowder put it well. In a federalist government, perhaps the most sacred, inviolable right a state has is to choose which rights it grants or does not grant to different people.

117. LOL@Klimate Katastrophe Kooks says:

CD Marshall wrote:
“Thus if we can travel outside of our time and space dimension, we can view the universe as it exists, not as we perceive it exists.”

I’ve often wondered if, rather than the CMB (Cosmic Microwave Background) being the red-shifted detritus of the Big Bang, it’s really just the red-shifted ‘fun-house mirror” of the universe outside our cosmological particle horizon… all those stars we cannot really see directly, in a universe far more vast than we could ever imagine, smeared together into a nearly homogeneous background light… sort of like looking through a fine mist smears out light sources, so too would the ‘mist’ of an expanding universe which red-shifts photons, a universe of such breadth that it gravitationally lenses photons and scatters photons… you reach a point where photons from different sources at different distances all smear together into a haze.

118. LOL@Klimate Katastrophe Kooks says:

Immortal600 wrote:
“LOL, no my name wasn’t in it.”

No, I mean the email address I set up for you to send to. It’s got ‘XYZ’ in it.

119. LOL@Klimate Katastrophe Kooks says:

CD Marshall wrote:
“While it is a grotesque act where they sell the baby parts and other rumors I’ve heard”

It’s not a rumor. Project Veritas has a Planned Parenthood principal on camera haggling over the price of baby body parts. Several other instances of same.

120. LOL@Klimate Katastrophe Kooks says:

Science of Doom wrote:
“Basic thermodynamics explains that bodies emit thermal radiation according to temperature (to the fourth power) and according to emissivity. Not according to the temperature of a different body that might happen to absorb this radiation.”

“When two bodies have an energy source which has created a constant surface temperature and they are subsequently brought into proximity with each other, there will be an increase in each other’s temperature.”

Seems a bit contradictory to me… they claim that the temperature of one body doesn’t change the temperature of another body, then they state that it does.

Methinks they’re confused. LOL

Of course the two objects brought into proximity while continuing to produce the same power output are going to increase in temperature… in order to continue to output that power, the temperature must increase… and it does so because of the proximity of (and thus the temperature of) the other body in proximity.

It’s right there in the S-B equation:
q = ε σ (T_h^4 – T_c^4) A_h

121. boomie789 says:

@LOL@kooks

O and I wrote that

122. boomie789 says:

Yea I probably linked that video here in the past. The rumors I’m referring too are like the meme I linked, or @dreno-chr0me.

Also there is this video by of Sandra Bullock on Ellen.

*I forgot you can’t say the @chr0me word on word press. Go ahead and try it will be auto deleted.

123. immortal600 says:

LOL, I can’t retrieve that message. The only one I can see is the one from 5 days ago. I would like to get to the point where you can send directly to my real email and vice versa, if possible.

124. immortal600 says:

I sent a second attempt to “nobody”@ ****m.com

125. LOL@Klimate Katastrophe Kooks says:

Immortal600:
I’ll send you another email with the email address for you to send to.

Et. al:
For two stars, it’s not so much that the two stars are absorbing radiation from each other (the cooler star would absorb radiation from the warmer star, the warmer star would not absorb radiation from the cooler star… a warmer object has higher energy density at all wavelengths than a cooler object, and it is energy density (an energy density gradient… remember that temperature is a measure of energy density, equal to the fourth root of energy density divided by Stefan’s Constant) which is the impetus for any energy flow)… it’s that if both continue producing the same amount of power, but a certain portion of each star is now ‘shaded’ from emitting to the cold of space by the other star, that slows radiant exitance of the hotter star by reducing energy density gradient in the ‘shaded’ area of that star, and absorbs energy at the cooler star… which effectively reduces the available radiating area at the prior equilibrated energy density gradient.

For two stars at exactly the same temperature, neither would absorb energy from the other (photon chemical potential is zero, photon Helmholtz Free Energy is zero)… a standing wave is set up between the two stars with the energy density equal to that of the two stars. Should one star cool, that standing wave becomes a traveling wave with the group velocity proportional to the energy density gradient and in the direction of the cooler star. That effectively halts radiant exitance from the ‘shaded’ area of each star (complicated, of course, by the fact that we’re talking about a sphere, so the photons are being emitted into an ever-expanding sphere… so really the only place that will completely stop emitting on each star is the nearest point of each star to the other, where the optical surfaces are parallel… the rest of the area of each star is emitting divergently and thus doesn’t have a view factor of 1).

To do it proper justice, one would have to subdivide each star into annular rings of a certain width, starting at the optical surface point on each star closest to the other star and going outward radially, calculating the radiant exitance of each star at the outgoing angle on the sphere for any given annular ring, the distance between that annular ring and the area of the opposite star upon which that radiation would incide and the angle at which it incides (note that ‘incide’ doesn’t necessarily mean ‘absorbed by’… photons are merely persistent perturbations of the EM field above the ambient (at ground state, the ground state of the EM component of the quantum vacuum), and there is no conservation law for photons… should a photon from a cooler object pass into a region of higher chemical potential than the photon has, that photon is subsumed into the background EM field… it can do no work.

The climate loons will often argue that a photon cannot ‘know’ apriori the energy density of the object upon which it will incide… that is false… the photon being a persistent perturbation of the EM field above the field ambient, the photon must traverse the EM field energy gradient toward the other object… so it most certainly does ‘know’… it is at that point that the climate loons will deny that temperature is a measure of energy density, that the photon is a persistent perturbation of the EM field above the field ambient, that radiation pressure and energy density gradient exist… whatever flopsweat-soaked backpedaling denialism they feel they must do to keep their religious belief in the climate catastrophe ideology alive.

But suffice to say the energy density between the stars would increase, which would reduce radiant exitance in the facing areas of each star, and since each star continues producing the same amount of power, that energy must be emitted at other, non-facing regions of each star, which reduces the area emitting at the previously-equilibrated energy density gradient, and thus increases the temperature of the star.

That’s completely different than a planet and its atmosphere… the planet’s surface can be thought of as the power ‘source’ (because it’s receiving energy from its star), but the atmosphere isn’t producing energy (as the second star is) and is in fact a coolant via conduction / convection and latent heat of evaporation and radiative emission to space. The higher the proportion of polyatomic molecules in an atmosphere, able to move a higher amount of energy due to a higher specific heat capacity or latent heat capacity (as compared to the monoatomics and homonuclear diatomics), the more energy is removed from the planet’s surface via conduction / advection / convection and evaporation, and the more energy is radiatively emitted by that atmosphere to space due to the altitude / air density / mean free path length relation (ie: most of the radiation will be upwelling… almost as if an atmosphere ‘pumps’ the radiation upward toward space).

126. immortal600 says:

LOL, I will start looking for it. Thanks!!

127. CD Marshall says:

Any thoughts on Pierrehumbert’s work? Lay it on me.

128. LOL@Klimate Katastrophe Kooks says:

Well, he’s a very clear and engaging writer, and I get the feeling he’s of the IDGAF variety as regards the woke mob.

“For present-day Earth, the only important continuum is the water vapor continuum in the window around 1000 cm-1. Carbon dioxide continua are unimportant for conditions which have prevailed on Earth during the past several billion years, but they are important for plugging the gaps in the line spectra for the dense CO2 atmospheres of Venus and early Mars. Diatomic homoatomic molecules like N2, which are transparent to IR in Earthlike conditions, have collisional continua that become important in cold, dense atmospheres. For example, the continuum makes N2 one of the most important greenhouse gases on Saturn’s largest moon, Titan.” – Infrared Radiation And Planetary Temperature, Raymond T. Pierrehumbert, Physics Today, January 2011, p. 35 https://web.archive.org/web/20190613031107/http://geosci.uchicago.edu/~rtp1/papers/PhysTodayRT2011.pdf

But you’ve got to read him carefully…

“An atmospheric greenhouse gas enables a planet to radiate at a temperature lower than the ground’s, if there is cold air aloft.”

“The emission spike at the center of the feature arises because CO2 absorbs so strongly that the radiating level is in the upper stratosphere, which is considerably warmer than the tropopause; the ozone feature exhibits a similar spike.”

“We can infer that the planet has a stratosphere in which temperature increases with height, indicating the presence of an upper-level solar absorber.”

That means as CO2 atmospheric concentration increases, so too does the effective emission height and thus the effective radiating temperature and thus the radiant flux. That’s a cooling process.

That upper-level absorber is ozone (O3), which then transfers its vibrational mode energy to N2{v1(1)}, which is nearly perfectly resonant with CO2{v3(1)} so during collision with a ground-state CO2 molecule, N2 will transfer its energy via a v-v collisional energy exchange (same as occurs in a CO2 laser… only the method of excitation differs. In the atmosphere, via solar insolation-excited O3 to N2{v1(1)} to CO2{v3(1)}; in a CO2 laser via electron impact with N2), whereupon the CO2 will radiatively emit, and due to the altitude / air density / mean free path length relation, the mean free path length for upwelling radiation is exponentially longer with increasing altitude, and for downwelling radiation it is exponentially shorter with decreasing altitude (it is re-absorbed and re-emitted, approximately half of that will be upwelling with a longer mean free path length; the re-emitted downwelling radiation will have a shorter mean free path length, will be more likely to be re-absorbed and re-emitted in an upwelling direction, etc., etc., etc.).

As well, CO2 in the upper atmosphere is absorbing solar IR. Thus more CO2 acts an an IR ‘shade’… remember that approximately 50% of radiation from the sun reaching Earth is in the infrared.

That’s exactly what the NASA TIMED / SABER project showed:

Observations of infrared radiative cooling in the thermosphere on 2 daily to multiyear timescales from the TIMED/SABER instrumenthttps://web.archive.org/web/20190331170025/https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20100011897.pdf
“Abstract:. We present observations of the infrared radiative cooling by carbon dioxide (CO2) and nitric oxide (NO) in Earth’s thermosphere.”

“Based on all the foregoing discussion, of the log-dependence of CO2 forcing (Myhre et al., GRL, 1998, vol. 25, doi: org/10.1029/98GLO1908) and its possible climate-cooling effect, I have a simpler hypothesis on the ineffectiveness of CO2 in warming the climate. I realize that this explanation is unacceptable to the IPCC and to many climate-warming advocates. I believe that the ‘gap’, now 40 years long, according to Christy, has existed throughout the Industrial Revolution — and probably during the whole of the Holocene. In other words, I consider that the ‘pause’ may be permanent.”

The Stratosphere Has Cooled:https://web.archive.org/web/20190621115328/https://www.climate.gov/sites/default/files/strattempanom1960-2011.gif
The graph shows multiple analyses of data from radiosondes that have measured stratospheric temperature for several decades. Graph adapted from Figure 2.7 in Bulletin of the American Meteorological Society, State of the Climate, 2011.

“Abstract: The writers investigated the effect of CO2 emission on the temperature of atmosphere. Computations based on the adiabatic theory of greenhouse effect show that increasing CO2 concentration in the atmosphere results in cooling rather than warming of the Earth’s atmosphere.”

How increasing CO2 leads to an increased negative greenhouse effect in Antarcticahttps://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1002/2015GL066749“>https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1002/2015GL066749

Why CO2 cools the middle atmosphere – a consolidating model perspectivehttps://web.archive.org/web/20190331154613/https://www.earth-syst-dynam.net/7/697/2016/esd-7-697-2016.pdf

A Guide to CO2 and Stratospheric Coolinghttps://web.archive.org/web/20190331083854/https://climatephys.wordpress.com/2015/05/22/a-guide-to-co2-and-stratospheric-cooling/

Cooling of the mesosphere and lower thermosphere due to doubling of CO2https://web.archive.org/web/20190702041827/https://link.springer.com/article/10.1007/s00585-998-1501-z
The sensitivity of the mesosphere and lower thermosphere (MLT) to doubling of CO2 has been studied. The thermal response in the MLT is mostly negative (cooling) and much stronger than in the lower atmosphere. An average cooling at the stratopause is about 14 K. It gradually decreases to approximately 8 K in the upper mesosphere and again increases to about 40–50 K in the thermosphere.

https://web.archive.org/web/20190209033912/https://phys.org/news/2012-11-atmospheric-co2-space-junk.html
The enhanced cooling produced by the increasing CO2 should result in a more contracted thermosphere, where many satellites, including the International Space Station, operate. The contraction of the thermosphere will reduce atmospheric drag on satellites and may have adverse consequences for the already unstable orbital debris environment, because it will slow the rate at which debris burn up in the atmosphere.

Climate “Science” on Trial; Evidence Shows CO2 COOLS the Atmospherehttps://web.archive.org/web/20190331125400/https://co2islife.wordpress.com/2017/01/29/climate-science-on-trial-evidence-shows-co2-cools-the-atmosphere/

129. LOL@Klimate Katastrophe Kooks says:

Pierrehumbert goes on to write:
“The greenhouse effect shifts the planet’s surface temperature toward the photospheric temperature by reducing the rate at which the planet loses energy at a given surface temperature. The way that works is really no different from the way adding fiberglass insulation or low-emissivity windows to your home increases its temperature without requiring more energy input from the furnace.”

Except insulation doesn’t convect and CO2 does; without CO2 and water vapor (what the climastrologists claim are the two predominant ‘greenhouse gases’) removing ~76.2% of all surface energy via convection / advection / latent heat the surface would have to be far warmer to radiate that energy away; and as CO2 atmospheric concentration increases, atmospheric effective emissivity increases (because there are a greater proportion of molecules capable of emitting).

As I said, it is the homonuclear diatomics and monoatomics which are the actual ‘greenhouse gases’… they can gain energy from the surface via conduction just as the polyatomics do, they can convect just as the polyatomics do, but once in the upper atmosphere they cannot as effectively radiatively emit because monoatomics have no vibrational mode quantum states and homonuclear diatomics have net zero magnetic dipole and thus cannot emit unless that net zero magnetic dipole is perturbed via collision. Thus the upper atmosphere would warm, lending less buoyancy to convecting parcels of air, and that is how a greenhouse actually works… by hindering convection. And again, because the polyatomics are no longer removing that ~76.2% of surface energy, the surface would have to warm to radiatively emit that energy (higher radiant exitance implies higher temperature).

So on that point, he is diametrically opposite to reality.

130. CD Marshall says:

LOL

131. LOL@Klimate Katastrophe Kooks says:

It’s the caffeine… I’ve overclocked my brain with caffeine ever since I was a nuclear engineer… I think it’s stuck on overclock now, it never shuts down. I dream in full color with sound, I can rewind and direct the dreams. I’ve dreamed in code before, woke up, sat down for 40 hours straight and typed out the code and it ran without errors first time. Yeah, I have a weird brain.

132. That mother fucker Crothers is so full of shit. Listen to me destroy his pseudoskepticism…he exists just to make skepticism look ridiculous and retarded, and make rationalism look bad by association. A truly evil player, with a purpose.

133. CD Marshall says:

Where?

134. My link to psi article way up.

135. ashemann says:

Its no coincidence the ”leak” happened to be reported on today, as today is the day the smoking gun documentary 2000 mules is released for public consumption.

136. LOL@Klimate Katastrophe Kooks says:

He is to you as Ken Wheeler was to me. I destroyed Wheeler, he didn’t even dare respond to me, and he left that forum. You can do the same to Crothers.

So, Crothers claims:

“Consider a cuboid rest-mass m_0 of sides length x. Let it move with constant rectilinear velocity v in the x-direction. Its mass is given by:
m = m_0√(1 − (v^2 c^2))
and its volume is given by:
V = x^3√(1 − v^2 c^2))
So the density D of the moving mass is:
D = m_0/(x^3 √(1 − v^2 c^2))
which is infinite when v→c but this is forbidden by Special Relativity since no material object can travel at the speed of light in vacuo. So infinite densities are forbidden by Special Relativity. Now the so-called “point-mass” has a finite mass and a zero volume, so that it is infinitely dense, which is what the singularity of the alleged black hole is supposed to be. Thus, if General Relativity permits point-masses it does so in violation of Special Relativity. Yet General Relativity is supposed to be a generalization of Special Relativity to non-uniform motion. It cannot therefore violate Special Relativity.”

Ah, see what he did there? He’s confused rest mass and relativistic mass! Rest mass is an intrinsic property, it doesn’t depend upon apparent velocity from any frame of reference… for Crothers’ claim to work, he’d have to claim that a spaceship flying by a stationary object at some large fraction of c would gauge that stationary object as more massive in their frame of reference (because of a derivation of the Equivalency Principle, without other ‘landmarks’, there’s really no way to know if the spaceship is moving at a large fraction of c, or if the object is, it’s all a matter of perspective).

Rookie mistake.

Should be:
D = m_0/x^3
Thus D→∞ as x→0.

There, fixed… no more violation of SR or GR. Well, that blows the hugest of holes in his hypothesis.

Go get ’em, Joe. LOL

137. LOL@Klimate Katastrophe Kooks says:

What changes with velocity (all velocities, but especially apparent the nearer you get to c) isn’t the rest mass but the relation between momentum and energy.

138. Richard says:

L@KKK
You are dividing by zero and think you get an infinite density. Infinite is not an adjective, also it is not a number. This is a big problem and you don’t seem to notice it.
The mistake starts with the mathematical point which is a pure nothingness and which you and Joe make a reality. You can juggle with equations all day long but it is all far removed from reality.

139. immortal600 says:

LOL, nothing yet.

140. LOL@Klimate Katastrophe Kooks says:

Richard,
D = m_0/x^3
Thus D→∞ as x→0.
… means as x approaches zero, D approaches infinity. That approach is asymptotic.

In order to occupy space-time, a point with mass (which must occupy space-time, it’s got mass and thus displaces space, rejects quantum vacuum wavemodes of a shorter wavelength than its radius, this variation in quantum vacuum energy density postulated as the factor behind gravity… remember the difference between ‘weight’ and ‘mass’ is a measure of the intensity of the gravitational field: W = m g) cannot literally be infinite, but it can get darned close.

141. boomie789 says:

“(a) US/NATO/UA agree that any attempt by russia to negotiate is a deception, stalling tactic, information gathering tactic, or stunt in service of public justification of the war. “Everything a russian says is a lie”.
(b) USA/NATO/UA are more than certain that UA will now win the war and drive russia from the territory. ( wasn’t sure of this until yesterday).
(c) USA/NATO are more than certain if putin escalates it will justify direct nato involvement in peacekeeping. (Expecting RU to declar war and move to total mobilization). 2/3 of RU forces are committed. Further committment of RU forces means their total destruction.
(d) USA finally (correctly) understands russian military is a paper tiger whose conventional forces will be slaughtered as quickly as were the iraqis.
(e) UA strategy appears to be crossing into RU and attacking the distribution points within russia ‘taking the battle to the enemy’ and then using drones to take out the bridge to crimea. (I am not certain of this but this is what I see emerging .. and it’s what I would do.)
(f) Nuclear weapons are unusable and attemtps to use it would mean the military would simply kill putin.”

-Curt Doolittle

142. immortal600 says:

LOL, got it! have sent response from real email account!

143. Richard says:

L@KKK
Using a verb to get to zero and infinity, namely ‘to approach’ is just a play on words and it means nothing. I notice this is philosophically going over your head. As I said before you’re good at juggling equations but it has nothing to do with the real world. According to Joe the singularity exists and according to you it is an infinite approachment of infinite density?

144. Joseph E Postma says:

When matter reduces to the singularity, the sine and cosine components of the basis circlings are re-orthogonalized and return to unextended mind or energy in the frequency domain.

145. CD Marshall says:

2020 just watched a video where Happer says increased ghgs will raise T by a degree. I don’t get these guys, actual physicists who think this. Now if they said, increased ghgs might increase Tmin by a degree they could have some argument room. The surface T doesn’t increase by rate of cooling.

The weird part is they just don’t get the irony in the physics. Slowed cooling equals increased energy? It’s not adding more energy to the surface from the Sun.

146. LOL@Klimate Katastrophe Kooks says:

And it’s not even slowed cooling, it’s increased cooling due to the higher concentration of higher specific heat and/or latent heat capacity of polyatomic molecules (as compared to monatomics (Ar) and homonuclear diatomics (O2, N2)) transiting more energy from the surface via evaporation and convection / advection.

Think about the Earth as if it were a giant A/C system, and you’re inside it. The atmosphere is the working fluid, the refrigerant. The surface is the evaporator. Space is the heat sink to which energy is emitted. The upper atmosphere is the condenser.

What happens if you have such an A/C system, normally filled with the extremely complex, extremely high Degrees of Freedom HFC, HCFC or CFC refrigerant, but you then reduce that amount of refrigerant and replace it with a gas such as Ar?

Well, it transits less energy from evaporator to condenser, of course!

Same goes for Earth. The more complex, high-DOF molecules you have in the atmosphere (H2O, CO2), the more energy is removed from the surface via evaporation, advection and convection, the more energy is transited to the upper atmosphere, and the more energy is emitted in the upper atmosphere to space.

Analogize the Earth to a dual-pane window… take that window and place it horizontally. The bottom pane is the surface, the gas between is the atmosphere, the top pane is space. Normally the volume between the panes is filled with Ar or other similar monoatomic gas. Why? Low DOF, low specific heat capacity, less energy transited.

Now put CO2 in there, or H2O, or any high-DOF molecule. What happens? The amount of energy transited from one pane to the other increases drastically.

That’s why they don’t use CO2 as a fill gas for dual pane windows… if it were such a terrific ‘heat trapping’ gas, they would. They don’t.

147. immortal600 says:

LOL, excellent description of what is actually happening. Well done.

148. J Cuttance says:

I listened to Crothers’ astonishingly deluded ideas on the TNT piece.

Are you sure his sounding stupid is a deliberate ploy to make skeptics all sound stupid by association? That’s some 4-D chess. It’s far more likely that he really is that dopey, and has a proverbial little dangerous knowledge.

Can you imagine our side going into warmist arenas and managing to sabotage their arguments by spouting even dumber warmist stuff?

149. CD Marshall says:

So took a few guys to hunt this down but they found it.

https://www.nasa.gov/centers/goddard/news/topstory/2007/polar_climate.html

150. CD Marshall says:

REVEALED: George Soros, Clinton and Obama staffers and European governments are behind anti-Musk campaign to force big corporations to boycott Twitter – after Elon demanded to know ‘who funds these organizations?’

``````A group of 26 activist organizations and NGOs signed a letter to companies who advertise on Twitter, warning them to reconsider if Elon Musk makes changes
Musk has pledged to lift the 'censorship' of Twitter: critics worry that he will give free rein to those trafficking in hate speech and dangerous scientific theories
The letter writers said that Musk 'will further toxify our information ecosystem and be a direct threat to public safety'
They wrote: 'Twitter risks becoming a cesspool of misinformation, with your brand attached'
Musk replied, wanting to know who was behind the 26 groups signing the letter, and commenting: 'Sunlight is the best disinfectant'
``````

https://www.dailymail.co.uk/news/article-10780583/George-Soros-Clinton-Obama-staffers-European-governments-anti-Musk-campaign.html

151. boomie789 says:

152. CD Marshall says:

“Who controls the food supply controls the people; who controls the energy can control whole continents; who controls money can control the world.” -Henry Kissinger

153. boomie789 says:

Reminds me of this one.

“Give me control of a nation’s money and I care not who makes it’s laws” — Mayer Amschel Bauer Rothschild

https://history.stackexchange.com/questions/7887/did-rothschild-say-this-famous-quote-if-yes-what-did-he-mean-by-it

154. CD Marshall says:

Straight up, Mark Zuckerberg looks like a droid. 🙄

155. Richard says:

J Cuttance,
When you have an equation with a part in it that corresponds to the matter contents of the universe and you then let these contents disappear you end up with an equation without matter and you have an empty universe, I think this is a reasonable assesment. It becomes a demonstration of stupidity thereafter when people begin talking about a situation which is called ‘outside a body’ in which this equation without matter and radiation is valid. They have inserted one mass in an empty universe. This is what can be called a magical operation. Mathematical physics is full of these kinds of manipulations and should be called mathemagixs instead.

156. CD Marshall says:

Balloon Boy over at CFACT is Evenminded. He popped up with that stupid plate diagram again I see.

157. CD Marshall says:

Wait a minute he has been pushing that diagram for well over a year and with no experiment to prove it exists? Hughes should just do the experiment and settle it once and for all.

158. LOL@Klimate Katastrophe Kooks says:

He’s been pushing that same idiotic tripe for longer than the 3 years I’ve been drop-kicking him into low orbit. I used to do it on a daily basis, but he’s so spun up now he’s on auto-froth mode. Now I just sit back and watch his hatred of reality consume him. LOL

159. immortal600 says:

It has been known for quite some time that ‘Balloon Boy’ and ‘evenminded’ are the same person. He’s admitted as much. Another dummy over there is ‘Blue’.

160. J Cuttance says:

Richard, Crothers starts off by saying hydrogen bonds are a source of energy and oceans the real cosmic background microwave source and it doesn’t improve with his explanation.

161. CD Marshall says:

You sure it’s not a bot account? Way too many aliases on way too many platforms.

162. CD Marshall says:

LOL@Klimate Katastrophe Kooks
“And it’s not even slowed cooling, it’s increased cooling due to the higher concentration of higher specific heat and/or latent heat capacity of polyatomic molecules (as compared to monatomics (Ar) and homonuclear diatomics (O2, N2)) transiting more energy from the surface via evaporation and convection / advection.”

Then they throw the “but it’s been measured at the TOA” Which is supposed to mean something, I guess to the forcing tripe.

163. CD Marshall says:

Oh yeah that’s right they claim the surface “has” to increase T to emit the E out to space, even Happer said that.

164. immortal600 says:

CD, no, that clown has two monikers. He’s used MINE on Dr. Ed Berry’s site! I had to email Dr. Berry and alert him of that fact. He’s a weak minded individual who lurks here and tries to harass me over on Cfact with his stupid math problems that have been shown to him his errors over and over. LOL does an excellent job of destroying him every time the engage. Fun to watch!

165. LOL@Klimate Katastrophe Kooks says:

CD Marshall wrote:
“Then they throw the “but it’s been measured at the TOA” Which is supposed to mean something, I guess to the forcing tripe.”

It’s far more complicated than they can grok. They’ve misinterpreted Brightness Temperature just as they’ve twisted everything else to fit their narrative.

The very graphic the climate alarmists use to claim that CO2 causes warming shows the exact opposite:

Brightness Temperature vs. Wavenumber graphic:

You’re standing on a giant ball of dirt and rock, are you not?

So an increase in altitude would increase the radius from the center of that ball, would it not?

Now, as the radius of that 3-dimensional ball increases, what happens to the surface area of that ‘ball’ of atmosphere which has that radius?

It increases, does it not?

A = 4πr^2

At the surface:
A = 4 π (6371 km)^2 = 510064471.90978827 km^2

At 5.105 km altitude:
A = 4 π (6376.105 km)^2 = 510882215.54703649 km^2

That’s an additional 817,743.63724822 km^2 area. That’s roughly equivalent to the land area of Namibia, slightly more than the land area of Mozambique, far more than the land area of Ukraine.

Now we’re getting to the heart of the matter… the area in which that radiation is emitted from CO2 increases as altitude increases.

Ok, so Brightness Temperature is the amount of energy emitted per unit surface area per unit time per unit solid angle and in a given frequency range, is it not?

So as the surface area increases for the same given amount of radiation emitted, the Brightness Temperature must decrease, yes?

Yes.

I_ν = (2 hν^3 / c)(1/(e^(hν/kT) – 1))

Where:

I_ν = Brightness, the amount of energy emitted per unit surface area per unit time per unit solid angle and in the given frequency range between ν and dν;

T = temperature of the black body

h = Planck’s constant

ν = frequency

c = speed of light

k = Boltzmann’s constant

So there’s three effects happening here to decrease Brightness Temperature with increasing altitude:

1) CO2 is convecting energy higher in the atmosphere, which increases the effective surface area (the area at the emission height is greater than the area at the surface)

and –

2) An increasing atmospheric CO2 concentration is akin to increasing the surface area of an emitting surface.

and –

3) Progressive radiative emission to space with increasing altitude.

Now, let’s really get into the details…

Brightness temperature is the temperature a black body in thermal equilibrium with its surroundings would have to be to duplicate the observed intensity of a gray body object at a frequency ν.

A brightness temperature lower than the equivalent gray body temperature implies that energy is flowing from that equivalent gray body temperature to the matter with that brightness temperature, which is shedding energy via radiative emission.

Graybodies emit according to the law: B = ε σ (T_h^4 – T_c^4) A_h
Sigma is the Stephan-Boltzmann constant, B is the brightness and T is the absolute temperature

Therefore a lower Brightness Temperature implies an absolute temperature below the temperature of the environment from which CO2 is attaining its energy. In this case, it means CO2 is radiatively cooling the atmosphere.

The image above shows the brightness temperature versus altitude. Note that as altitude increases (emission volume increases), brightness temperature decreases for the same amount of energy emitted.

That ‘dip’ in Brightness Temperature is (partly) due to progressive radiative emission of energy to space with increasing altitude, the effect of ‘surface area’ increasing with increasing altitude, and the effect of increasing atmospheric CO2 concentration increasing ‘surface area’ of that CO2 at its emission height.

All three together allow CO2 to shed energy to space more efficiently with increasing altitude and increasing atmospheric CO2 concentration.

So while the alarmists claim that increasing CO2 concentration in our atmosphere has caused that ‘dip’ in brightness temperature to deepen because CO2 is ‘trapping’ more energy in the atmosphere, in reality the ‘dip’ has deepened because the upper atmosphere has experienced a long-term cooling trend (and ironically enough, that long-term cooling trend is because of the increased CO2 concentration emitting more radiation to space). As well, an increasing atmospheric CO2 concentration more readily de-excites O2, O3 and N2 which have become vibrationally excited due to absorption of UV, making those molecules available again to absorb more UV (thus they act as a more effective ‘shade’ for downwelling radiation), thus shedding more energy to space and thus cooling the column of atmosphere.

In other words, dips in the Brightness Temperature below the profile average indicate a cooling process, whereas spikes above the profile average indicate a warming process (usually due to absorption of solar insolation). Thus the dip that the alarmists point to as causing warming is actually indicative of cooling, whereas the narrow-band upward spike in the bottom center of that dip indicates a diminution of that relatively wider-band cooling process (but under no circumstances does it indicate actual warming… it’d have to spike above the emission profile average for it to indicate that). This alone destroys the CAGW hypothesis.

Another factor accounting for that ‘dip’ is the down-conversion of energy from its absorption wavelength to a longer wavelength. That energy doesn’t just disappear, it must go somewhere… it is down-converted (mostly by collision with water vapor) and emitted at longer wavelengths.

That graphic doesn’t show CO2 ‘trapping’ energy, it shows it thermalizing and down-converting it, transferring it to primarily water vapor. That’s why the left-hand side of that graphic rises higher than it otherwise would, that energy is being emitted at longer wavelengths.

The 294K Blackbody Emission line in the graphic above is misdirection… apply a lower-temperature (287.64 K) BB curve, and take into account the fact that CO2, O3 and H2O are thermalizing and down-converting that energy, which causes the graph to the left of where those molecules are absorbing to be higher.

First, a disclaimer: I am definitely not an artist. I can’t draw a straight line to save my life, I can’t free-hand draw, I have no photo-editing skills whatsoever. I still (poorly) draw stick figures to represent people. The orange line in the image below is very likely not correct, nor to scale. The image was edited in Paint, and I pixel-by-pixel filled in the orange line… that is the extent of my graphic artistry abilities.

But when I state: “apply a lower-temperature BB curve, and take into account the fact that CO2, O3 and H2O are thermalizing and down-converting that energy, which causes the graph to the left of where those molecules are absorbing to be higher”, this is a very rough graphical approximation of what I’m talking about:

Note the absorption bands for CO2, O3 and H2O (red circled), note that the graphic rises higher to the left (longer wavelength) of those absorption bands than it would were those absorption bands non-existent.

If anyone’s got image editing skills and can fit a 287.64 K curve to that image, properly scaled, it’d be much appreciated if you could share it.

166. LOL@Klimate Katastrophe Kooks says:

One way one can determine the actual “global temperature” (or at least a rough approximation) is to draw the BB curve such that the open area below the curve is equal to the filled area above the curve.

167. That’s precisely the definition of “effective temperature”.

168. CD Marshall says:

You just blew my mind. Thank you very much I always have lots of questions and appreciate the answers. I might not be a scientist (which may be why I have more of an open mind on climate science than some), but I love every nugget of science I get.

Saving this for more study.

So that leaves to the extended question.
The troposphere is not in TE.
But is the “effective T” considered global TE? Or is it a quasi-thermal equilibrium…
So the effective T is just an average which would mean no planet actually exists in TE? As you said eluded, using Kirchhoff’s Law is just fallacious and part of “climate science” trying to fit science to a narrative not proving if the narrative fits actual science.

Why does every climate manual written (at least from 2000+) always talk about the energy balance in thermal equilibrium when such a thing simply doesn’t exist.

169. CD Marshall says:

I still can’t get over NASA having a real energy budget graph this whole time. Joe it’s pretty much your graph, incoming solar energy reaching the surface divided by 2. No GHGE in that graph.

170. CD Marshall says:

The funny thing is clouds do radiate to space that is actually taught in (some) climate, stacked decks of clouds radiate upwards and below, eventually the top deck emits directly to space.

171. J Cuttance says:

CD, the 6% radiated from the surface seems light considering it is the warmest component.

172. CD Marshall says:

That’s actually about right, between 6-12% I would put it at. Clear skies and it is higher. Latent heat is a large part of that energy as it should be from tropical oceans and the highest solar density. Remove clouds and the open window is greater.

I believe Philip Muholland did research on that from the Tibetan Plateau, where water vapor didn’t get in the way of direct IR from the surface to space. Don’t recall the results off hand?

173. CD Marshall says:

NOT that I’m an expert on it by any means 😂.

174. CD Marshall says:

LOL someone posted this as “empirical proof of global warming” 😂🤣😅

https://skepticalscience.com/empirical-evidence-for-global-warming.htm

175. CD Marshall says:

Oh and is the effective BB reduced (254 Kelvin) from lower solar intensity of increased ghg cooling?
Less solar storms always cools the upper atmosphere.

176. LOL@Klimate Katastrophe Kooks says:

CD Marshall wrote:

And that correlates to:

The image above states that ~76.2% of energy is transited from the surface via convection, advection and latent heat of evaporation.

100% – 49% = 51% (absorbed by surface)

So 51% is the amount we’re working with.

21% (radiated directly to space and radiation absorbed by atmosphere) of that 100% equates to 41.176470588235294117647058823529% of that 51%.

23% (latent heat) of that 100% equates to 45.098039215686274509803921568627% of that 51%.

7% (conduction and convection) of that 100% equates to 13.725490196078431372549019607843% of that 51%.

45.098039215686274509803921568627% + 13.725490196078431372549019607843% = 58.82352941176470588235294117647%.

So they’ve got their (radiated directly to space and radiation absorbed by atmosphere) set a bit too high.

It should be 23.8% of that 51% (radiated directly to space and radiation absorbed by atmosphere), or 12.138% of the 100%.

177. CD Marshall says:

Interest in employment at Twitter if Elon takes it over has increased by 263%. Anyone need a job?