The Fallacy Of Bell Curves
When a medium not gaining energy, the energy (v^2) will quickly be distribute among all the mass by collisions between the molecules, but the masses of objects will remain the same.
In a medium where the masses of molecules are all the same all the molecules will have identical kinetic energy (1/2mv^2).
In a medium containing matter with different masses, like the atmosphere, the gas molecules will have the same velocity (law of conservation of momentum) but different kinetic energy.
Nitrogen molecules will have kinetic energy of 14v^2, oxygen molecules 16v^2, and argon 20v^2.
Since the distribution of the molecules is uniform throughout the air mass, when no energy is added, the temperature reading will be the result of three different units of kinetic energy transferring energy to the thermometer.
This is not a bell curve.
When energy is added, by radiation, the different types of matter, absorbs different wavelengths of energy which results in different kinetic energy for different molecules. The nitrogen and oxygen molecules absorb ultraviolet energy from the sun gaining enough kinetic energy to permeate the atmosphere.
Argon, being a single atom absorbs shorter wavelengths is the x-ray and gamma ray spectrum. Because an argon atom is not much bigger than nitrogen or oxygen atoms, these wavelengths are almost completely absorbed by these elements in the upper atmosphere.
The result is even though argon starts with greater kinetic energy it cannot gain enough energy to get to the higher altitudes and is restricted to the troposphere.
In water where the molecules are all the same, it would be expected that the bell curve would consist of a single line and the wavelength of energy absorbed would only consist of three narrow wavelengths, similar to CO2 with its three atoms.
This is not the case.
Water absorbs a broad range of infrared wavelengths and the explanation for this is that the structure of water changes as it absorbs energy, creating different molecule types of structures that are able to absorb other wavelengths.
It requires 720 calories of energy to convert a gram of 0 C solid water into a gram of 100 C water gas.
Of these 720 calories only 100 appear as increased temperature so if the bell curve of water centered on a temperature of 373 K were accurate, the energy of the upper limit molecules converting to a gas would be equivalent to a temperature of 913 K.
At the other end of the bell curve the temperature of those water molecules would be equivalent to a temperature of -167 K.
There is no negative energy so there is no bell curve.
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LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“In a medium containing matter with different masses, like the atmosphere, the gas molecules will have the same velocity (law of conservation of momentum) but different kinetic energy.”
That is in direct contradiction to the Molecular Kinetic Theory, the Equipartition Theorem and the Virial Theorem. Perhaps if you stopped fleshing out your odd hobby theory with what sounds intuitive, but which is unphysical, and instead buckled down and cracked a book to study, you’d be able to understand the concept, Herb.
In point of fact, at the same kinetic temperature, lighter atoms or molecules will have higher velocity; and heavier atoms or molecules will have lower velocity, but at any given kinetic temperature (note that kinetic temperature is different than the layperson ‘temperature’ usually referred to), all atoms and molecules will have the same kinetic energy regardless of their mass.
The root mean square of the distribution of kinetic temperatures make up the layperson ‘temperature’ usually referred to.
The Equipartition Theorem equation:
KE_avg = 3/2 kT
Now, I’m going to blow all of your blather out of the water yet again, Herb. LOL
H2 at 273 K:
M = 2.01568 amu = 1.660539066605e−27 kg amu-1 * 2.01568 amu * 6.02214076e23 particle mol−1 = 0.0020156799993094936060614464 kg mol-1
Molar kinetic energy density:
(3/2) * 1.380649e−23 J K−1 * 273 K = 5.653757655e-21 J particle-1 * 6.02214076e23 particle mol−1 = 3404.77244213375178 J mol-1
v = √((3 kT) / m) = √((3 * 1.380649e−23 * 273) / 3.3471153857743664e-27) = 1838.0116092566284089214803816125227469883595064233 m sec-1
Checking our math:
1/2 m v^2 = (1/2) * 0.0020156799993094936060614464 kg mol-1 * (1838.0116092566284089214803816125227469883595064233 m sec-1)^2 = 3404.77244213375178 J mol-1
One can use the Maxwell-Boltzmann Speed Distribution Function to see that the particle speed above corresponds to the Root Mean Square speed. This speed is directly proportional to the square root of temperature, and inversely proportional to the square root of the mass.
Rotational mode energy density:
(2/2) * 1.380649e−23 J K−1 * 273 K = 3.76917177e-21 J particle-1 * 6.02214076e23 particle mol−1 = 2269.84829475583452 J mol-1
Radiation energy density:
σ = (2 π^5 KB^4) / (15 h^3 c^2) = 5.67037441918442945397099673188923087584012297029130e-8 W m-2 K-4
c = light speed (299792458 m sec-1)
e = T^4 4σ/c = 5554571841 K^4 * 7.5657332502800046477099790574307e-16 J m-3 K-4 = 4.2024408868522719181518974807104e-6 J m-3
Invariant-mass matter is a repository of energy. Not all of the KE of the atoms or molecules is going to be converted to radiation. e = aT^4 is the conversion factor between temperature and radiation energy density.
He at 280 K:
M = 4.002602 amu = 1.660539066605e−27 kg amu-1 * 4.002602 amu * 6.02214076e23 particle mol−1 = 0.00400260199862883876736821196 kg mol-1
Molar kinetic energy density:
(3/2) * 1.380649e−23 J K−1 * 280 K = 5.7987258e-21 J particle-1 * 6.02214076e23 particle mol−1 = 3492.0742996243608 J mol-1
v = √((3 kT) / m) = √((3 * 1.380649e−23 * 280) / 6.64647698907130621e-27) = 1320.947421966419638889650296868646911440259961053 m sec-1
Checking our math:
1/2 m v^2 = (1/2) * 0.00400260199862883876736821196 kg mol-1 * (1320.947421966419638889650296868646911440259961053 m sec-1)^2 = 3492.0742996243608 J mol-1
One can use the Maxwell-Boltzmann Speed Distribution Function to see that the particle speed above corresponds to the Root Mean Square speed. This speed is directly proportional to the square root of temperature, and inversely proportional to the square root of the mass.
Radiation energy density:
σ = (2 π^5 KB^4) / (15 h^3 c^2) = 5.67037441918442945397099673188923087584012297029130e-8 W m-2 K-4
c = light speed (299792458 m sec-1)
e = T^4 4σ/c = 6146560000 K^4 * 7.5657332502800046477099790574307e-16 J m-3 K-4 = 4.6503233366841065367428248875241e-6 J m-3
Invariant-mass matter is a repository of energy. Not all of the KE of the atoms or molecules is going to be converted to radiation. e = aT^4 is the conversion factor between temperature and radiation energy density.
3492.0742996243608 J mol-1 (He, 280 K) – 3404.77244213375178 J mol-1 (H2, 273 K) = 87.30185749060902 J mol-1
He at a higher temperature has a higher molar kinetic energy density than H2 at a lower temperature.
4.6503233366841065367428248875241e-6 J m-3 (He, 280 K) – 4.2024408868522719181518974807104e-6 J m-3 (H2, 273 K) = 4.478824498318346185909274068137e-7 J m-3
He at a higher temperature has a higher radiation energy density (a higher radiation pressure… remember that energy density and pressure have the same physical units: [M1 L-1 T-2]) than H2 at a lower temperature.
ALL INVARIANT-MASS OBJECTS (regardless of their mass) of a higher temperature will have higher radiation energy density than objects of a lower temperature (regardless of their mass), at all wavelengths.
https://i.imgur.com/wb9KwS0.png
ALL INVARIANT-MASS OBJECTS (regardless of their mass) of a higher kinetic temperature will have higher kinetic energy than objects of a lower kinetic temperature (regardless of their mass)… because that’s how energy works, regardless of the form of the energy.
I’m sure you can do the calculations now to figure out the kinetic energy of two different atoms or molecules at the same kinetic temperature, to prove to yourself that they’ll have the same kinetic energy regardless of their mass, yes? The only thing that changes between atoms or molecules of differing mass is the velocity of the atom or molecule at any given kinetic temperature, not the kinetic energy, regardless of the mass of the atom or molecule.
Or, to put it another way:
All atoms or molecules of the same kinetic temperature will have the same kinetic energy, regardless of mass Lighter atoms or molecules will have higher velocity, heavier atoms or molecules will have lower velocity, but at any given kinetic temperature, they will all have the same kinetic energy.
JaKo
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Hi LOL@KKK,
No pun intended, I just like brevity… 😉
While your interpretation of the theorems may be sound, your displays of calculations are rather silly. Why would you claim the results of any computation, involving a few numbers to be of a “greater accuracy” than the least accurate involved?
Cheers, JaKo
LOL@Klimate Katastrophe Kooks
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The highest precision of any calculation is to the full decimal points as calculated. The more decimal points used, the more precise the calculation.
Which is more precise:
3.14159000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
000
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364
All one is doing when using fewer decimal points is zeroing out trailing digits. That’s not more precise.
That said, I’ve had kooks in the past declare that if the end result or any of the intermediate calculation steps are rounded by even the slightest, the end result is wrong and therefore they crow that they are right and CAGW stands unmolested. So I use the highest precision numbers I can find for input values and I use the precision of my calculator.
That has allowed me to arrive at a precision of 3.8 parts per 100 trillion between the stripped-down Stefan-Boltzmann equation:
Radiant Exitance = (ε c (e_h – e_c)) / 4
… which shows that the S-B equation is supposed to be properly used by subtracting the energy density of the cooler object from the energy density of the warmer object, the EM field energy density gradient between the two objects determining the radiant exitance of the warmer object…
… to the full S-B equation:
= ε σ (T^4_h – T^4_c) A_h
… which the climastrologists use by subtracting a wholly fictive ‘cooler to warmer’ energy flow from the real (but too high because it was calculated for emission to 0 K) ‘warmer to cooler’ energy flow.
That’s where their ‘backradiation’ came from, from a misuse of the S-B equation. It’s a mathematical artifact due to misuse of the S-B equation, it doesn’t exist.
“But they’ve measured it!”, some may cry. Yeah, no.
https://claesjohnson.blogspot.com/2011/08/how-to-fool-yourself-with-pyrgeometer.html
Herb Rose (and you) now try to claim that using higher precision somehow makes it ‘wrong’. It’s not wrong, it’s as precise as the inputs and the calculator allow.
LOL@Klimate Katastrophe Kooks
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In fact, we can analogize thermodynamics to electrodynamics… the equations for each are two forms of the same thing, just for different forms of energy.
Both sets of equations can be derived from the same foundation of classical electromagnetism, showcasing that they are essentially two different forms of energy descriptions, just in different contexts and at different scales.
So if you have someone who cannot think in terms of energy density, radiation pressure, EM field energy density gradient, radiant exitance, etc., perhaps they might more easily be able to think in terms of voltage, amperage, resistance, etc.
The circuit simulator:
https://tinyurl.com/yzo8hak9
Images of the circuit simulator:
https://i.imgur.com/JCjwntp.png
https://i.imgur.com/0n69gYG.png
You’ll note that the top two circuits in the circuit simulator are how the climate alarmists calculate radiant exitance… (ie: they put each object into its own separate system that doesn’t interact with other objects via the EM field, and they essentially claim all objects emit to 0 K, equivalent in electrical terms to discharging to ground… sometimes with a resistance (the analogy of emissivity) but often not… the Kiehl-Trenberth ‘Earth Energy Balance’ graphic assumes the planet’s surface is emitting to 0 K and has ε = 1… the definition of an idealized blackbody object, which is the only way they could have possibly attained 390 W m-2 surface radiant exitance at 288 K).
The bottom circuit in the circuit simulator is how reality works… objects interact via the EM field. That interaction through radiation pressure, that radiation energy density gradient, determines radiant exitance of the warmer object.
So while the climate alarmists claim that there’s no way a photon could possibly ‘know’ the temperature of an object within the photon’s path, it absolutely does ‘know’ because that photon must pass through the EM field (the photon being nothing but a quantum of EM energy; per QFT, a persistent perturbation of the EM field above the average field energy density) between objects, and thus the radiation energy density gradient between objects… and if the EM field energy density gradient is such that the chemical potential of the EM field due to that radiation energy density gradient becomes higher than the chemical potential of the photon from a cooler object, that photon likely won’t even be emitted by the cooler object, and if a photon which is emitted by a cooler object happens to be in the path of a moving, warmer object, it won’t even reach the warmer object… it will be subsumed into the background EM field (there is no law of conservation for photon number).
https://i.imgur.com/5AUJUKN.png
So the climastrologists are essentially claiming, in their claim that a cooler atmosphere can warm an already-warmer surface (via radiation emitted by that surface, no less) something analogous to a claim that a capacitor at 1.5 V, electrically connected (+) to (+) and (-) to (-) to a capacitor at 12 V, will do work upon that 12 V capacitor… the 1.5 V capacitor originally being charged to 1.5 V by that 12 V capacitor.
Common sense and reality says that’s bunkum. LOL
That, combined with the following:
https://principia-scientific.com/separating-fact-from-fiction/#comment-98166
https://principia-scientific.com/separating-fact-from-fiction/#comment-98167
https://principia-scientific.com/separating-fact-from-fiction/#comment-98198
… shows CAGW to be an absolute sham, only perpetuated by dint of the horrendous state of the educational system creating a gullible population of unthinking sheeple.
James Bernard McGinn
| #
Kkook:
if the EM field energy density gradient is such that the chemical potential of the EM field due to that radiation energy density gradient becomes higher than the chemical potential of the photon from a cooler object, that photon likely won’t even be emitted by the cooler object,
JMcG:
You are confused. Cooler objects do continue to emit photons toward warmer objects and these warmer objects will continue to absorb photons that have been emitted by cooler objects. All the LOTs are saying is that the RATE of warming of the warmer object from the photons of the cooler object will not (cannot possibly be) higher than the rate of cooling that the warmer object is already experiencing. So the cooler object is heating the warmer object. However the cooler object cannot heat the warmer object at a rate that is greater than the rate of cooling that the warmer object is already experiencing.
Kkook:
and if a photon which is emitted by a cooler object happens to be in the path of a moving, warmer object, it won’t even reach the warmer object…
JMcG:
You are confused. There is no such thing as sentient photons. Photons from cooler objects are absorbed by warmer objects. As I explained above, it’s just that the rate of cooling of the warmer object will be greater than the rate of warming it receives from the cooler object so the warmer object will continue to cool however the rate of it’s cooling will be reduced due to the warming effect from the cooler object.
So, sorry to burst your bubble, but the lOT’s are not the magic wand that yourself and Postma want them to be.
James McGinn / Genius
Paradigms Often Involve Deliberate Stupidity
https://podcasters.spotify.com/pod/show/james-mcginn/episodes/Paradigms-Often-Involve-Deliberate-Stupidity-e2e2af9
LOL@Klimate Katastrophe Kooks
| #
Jimbo “The Dunce” McGinn, the self-professed “tornado genius” who’s only ever seen a small tornado from a distance when he was a kid (which I proved via a tornado tracker); and who claims that convection doesn’t exist, instead “vortices” at the tropopause suck air up from the surface; and who claims that humid air is less buoyant than dry air; and who claims that water cannot be in the gaseous phase in the atmosphere (which I disproved via the absorption spectra of ice, liquid water and atmospheric water vapor being different because liquid water has quenched rotational mode quantum states); and who denies enormously wide swaths of long-established science weighs in with his unscientific bafflegab.
Ok, Jimbo… if, as you claim, objects are emitting even at thermodynamic equilibrium (and below that level, where the energy density is higher than the emitting object, why don’t you explain exactly how that works in mathematical terms and describe for everyone why entropy doesn’t change at thermodynamic equilibrium?
Are… are you one of ‘those kind’ who do claim that a 1.5 V capacitor electrically connected (+) to (+) and (-) to (-) to a 12 V capacitor will do work upon that 12 V capacitor, Jimbo? Because that’s a whole new level of ‘speshul’ that only one other kook has outright claimed. LOL
Read, learn, then STFU:
https://principia-scientific.com/separating-fact-from-fiction/#comment-98166
https://principia-scientific.com/separating-fact-from-fiction/#comment-98167
https://principia-scientific.com/separating-fact-from-fiction/#comment-98198
Seriously, PSI mods… how do you let this guy continue spewing his idiocy here? He makes the climate-skeptical side look like unscientific half-wits.
LOL@Klimate Katastrophe Kooks
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Is this not you, Jimbo?
Jimbo “The Dunce” McGinn dribbled:
“You are confused. Energy travels from higher density to lower density.”
And a warmer object will have higher energy density at all wavelengths than a cooler object.
https://i.imgur.com/wb9KwS0.png
And now you’re back to claiming that energy can flow willy-nilly without regard to the energy density gradient… because a kook’s particular mental defect compels them to self-contradict. LOL
James McGinn
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Kkook:
Jimbo “The Dunce” McGinn, the self-professed “tornado genius” who’s only ever seen a small tornado from a distance when he was a kid (which I proved via a tornado tracker);
JMcG:
I have no idea what the whack-job is talking about
Kkook:
who claims that convection doesn’t exist,
JMcG:
Specifically, my claim is that convection plays no role in storms. More significantly, I claim that meteorology is a confused paradigm in which it’s confused practitioner’s pretend to see evidence of convection and pretend to have observed evidence that substantiates this nonsensical claim. This dishonesty is rewarded by a public that is too dullwitted to realize that H2O does not magically adjust its boiling temperature in order to fit what people want to believe.
Kkook:
instead “vortices” at the tropopause suck air up from the surface;
JMcG:
Yes. This is exactly correct. Moreover, you have already indicated that you have no evidentiary dispute with this model. Right?
Kkook:
and who claims that humid air is less buoyant than dry air;
JMcG:
Correct. And you have indicated, once again, that you have no evidentiary dispute with this model. Right?
Kkook:
and who claims that water cannot be in the gaseous phase in the atmosphere
JMcG:
Correct. And you admitted you have no dispute. Right? (Answer my question you evasive POS.)
Kkook:
(which I disproved via the absorption spectra of ice, liquid water and atmospheric water vapor being different because liquid water has quenched rotational mode quantum states);
JMcG:
There is no such thing as “rotational mode quantum states,” you loon. You proved only that you are confused.
Kkook:
and who denies enormously wide swaths of long-established science
JMcG:
I exposed you as the confused pretender you are.
Kkook:
Ok, Jimbo… if, as you claim, objects are emitting even at thermodynamic equilibrium
JMcG:
LOL. All objects emit in all directions at all times in a manner that is proportional to their temperature. It is only confused morons like yourself who envision objects have the sentient ability to turn their emissions on or off. You are just confused.
Kkook:
describe for everyone why entropy doesn’t change at thermodynamic equilibrium?
JMcG:
It’s plainly irrational to suggest entropy is a process that can be turned on or off. You are just confused.
James McGinn / Genius
December 2023
https://podcasters.spotify.com/pod/show/james-mcginn/episodes/December-2023-e2drsf9/a-aapujqc
LOL@Klimate Katastrophe Kooks
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Jimbo “The Dunce” McGinn dribbled:
“All objects emit in all directions at all times in a manner that is proportional to their temperature.”
And what is temperature a measure of, Jimbo?
Here’s a hint:
T = 4^√(e/a)
Oh, poor ol’ dumb Jimbo, the guy who has to go around telling people he’s a ‘geeneeus’ (because to look at what he writes, he’s got to be profoundly retarded or severely mentally disturbed or both) will never figure it out. LOL
So in your ‘fantasy fizix’ world, you can just say whatever you want and it’s true; you avoid comprehensively answering questions with mathematical corroboration instead preferring handwavium non-explanations you pull straight from your sun-don’t-shine; and you deny wide swaths of long-established science in favor of demented frothings from the voices in your head… and then you’re genuinely surprised that everyone considers you to be a mental defective.
Got it. LOL
You’re the butt of a bad joke, Jimbo… and you’re the only one too dense to realize it. LOL
James McGinn
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Kkook:
And what is temperature a measure of, Jimbo?
JMcG:
LOL. You got nothing, you confused nitwit.
Kkook:
you avoid comprehensively answering questions
JMcG:
It’s not my job to figure out what your point is, you confused nitwit.
Kkook:
with mathematical corroboration
JMcG:
You make no sense . . . ever. You are just confused.
Kkook:
you deny wide swaths of long-established science
JMcG:
The length of time a notion has been believed is completely irrelevant to it’s validity. Confused pretenders, like yourself, can never grasp this simple fact.
James McGinn / Genius
The ‘Missing Link’ of Meteorology’s Theory of Storms
https://www.thunderbolts.info/wp/forum/phpBB3/viewtopic.php?t=16329
in favor of demented frothings from the voices in your head… and then you’re genuinely surprised that everyone considers you to be a mental defective.
Got it. LOL
You’re the butt of a bad joke, Jimbo… and you’re the only one too dense to realize it. LOL
LOL@Klimate Katastrophe Kooks
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Jimbo “The Dunce” McGinn dribbled:
“It’s plainly irrational to suggest entropy is a process that can be turned on or off.”
Ok, you’re claiming that entropy cannot be “turned off” (your words)… so if, as you claim, energy is flowing even at thermodynamic equilibrium, then entropy must be changing at thermodynamic equilibrium, yes?
ΔS = ΔQ / T, right?
So you deny 2LoT in the Clausius Statement sense, then. You claim energy can flow without regard to the energy density gradient… did I mention that you’ve already contradicted yourself? LOL
And you deny the definition of thermodynamic equilibrium as the system reaching a state of quiescence.
https://en.wikipedia.org/wiki/Thermodynamic_equilibrium
“In thermodynamic equilibrium, there are no net macroscopic flows of matter nor of energy within a system or between systems.”
Note the wording, Jimbo… “no net macroscopic flows of matter” and NO flow of energy.
“A collection of matter may be entirely isolated from its surroundings. If it has been left undisturbed for an indefinitely long time, classical thermodynamics postulates that it is in a state in which no changes occur within it, and there are no flows within it.”
Likewise for a system consisting of more than a single object:
“A system in a relation of contact equilibrium with another system may thus also be regarded as being in its own state of internal thermodynamic equilibrium.”
So in order for you to make these claims, you must also claim that radiative energy transfer is an idealized reversible process, Jimbo… except we know it isn’t.
That’s what trips up the warmists, who use the same exact arguments as you’re using. Whose side are you on, again, Jimbo?
So you’re just denying everything, contradicting yourself and in the process embarrassing yourself.
I’ll tell you the same as I told Herb Rose, Jimbo:
“STFU, buckle down, crack a book and study.”
You’ve embarrassed yourself online for more than a decade. Don’t you think that’s enough? LOL
LOL@Klimate Katastrophe Kooks
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Jimbo “The Dunce” McGinn dribbled:
“There is no such thing as “rotational mode quantum states,” you loon.”
E_total = E_rotational + E_vibrational + E_electronic + E_electronic
Now you’re denying even more reality, Jimbo… that of rotational spectroscopy.
https://en.wikipedia.org/wiki/Rotational%E2%80%93vibrational_spectroscopy
Take your psychotropic medication and take a nap, Jimbo… you’re far outclassed here. Everyone here sees you as a joke, a fraud, a self-aggrandizing charlatan.
LOL@Klimate Katastrophe Kooks
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“E_total = E_rotational + E_vibrational + E_electronic + E_electronic”
— should be —
“E_total = E_rotational + E_vibrational + E_electronic + E_translational”
Are you going to deny translational mode, too, Jimbo?
Or is it that you’re simply too maleducated to even understand any of this? LOL
James McGinn
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Kkook:
Jimbo “The Dunce” McGinn dribbled:
“It’s plainly irrational to suggest entropy is a process that can be turned on or off.” Ok, you’re claiming that entropy cannot be “turned off” (your words)…
JMcG:
Right. It’s not a process. As I indicated.
Kkook:
so if, as you claim, energy is flowing even at thermodynamic equilibrium,
JMcG;
Correct. Moreover, I’m claiming that anybody that states otherwise could only be confused.
Kkook:
then entropy must be changing at thermodynamic equilibrium, yes?
JMcG: Entropy is not quantifiable, you f#cking confused nitwit. There are no units for entropy. You are just confused.
Kkook:
So you deny 2LoT in the Clausius Statement sense, then. You claim energy can flow without regard to the energy density gradient…
JMcG: Don’t put words in my mouth you confused pretender.
Kkook:
did I mention that you’ve already contradicted yourself? LOL
JMcG:
All you have is confusion. That’s the extent of it.
Kkook:
And you deny the definition of thermodynamic equilibrium as the system reaching a state of quiescence.
JMcG:
Corrrect. In my model quienscence is only achieved at absolute zero. Equilibrium just means that the exchange (an exchange that cannot be turned off) is balanced. It doesn’t mean it has been turned off, you simpleton.
Kkook:
https://en.wikipedia.org/wiki/Thermodynamic_equilibrium
“In thermodynamic equilibrium, there are no net macroscopic flows of matter nor of energy within a system or between systems.” Note the wording, Jimbo… “no net macroscopic flows of matter” and NO flow of energy.
JMcG:
There is no NET flow you fucking moron. Neither side is gaining or losing. The exchange is balanced. But the exchange is not turned off. Energy still flows.
You have no business in a scientific discussion.
Kkook:
“A collection of matter may be entirely isolated from its surroundings. If it has been left undisturbed for an indefinitely long time, classical thermodynamics postulates that it is in a state in which no changes occur within it, and there are no flows within it.”
JMcG:
I don’t know who you are quoting here. It doesn’t matter, it is just wrong. Flow cannot be turned off.
James McGinn / Genius
Narrow Range of Factors
https://podcasters.spotify.com/pod/show/james-mcginn/episodes/Narrow-Range-of-Situational-Factors-Underlie-Emergence-of-Structural-Properties-in-the-Atmosphere-e2d4jr3/a-aanql2t
LOL@Klimate Katastrophe Kooks
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https://i.imgur.com/ZaYjj5o.png
(Liquid Water)
“Note that in the above diagram, there are still vibrational states but that the rotational states are “smeared” one into the other.”
Remember me telling you that the absorption spectra of ice, liquid water and atmospheric water vapor were different because liquid water has quenched rotational mode quantum states, Jimbo, as means of disproving your contention that only liquid water could exist in the atmosphere, your claim that gaseous water could not exist in the atmosphere?
https://upload.wikimedia.org/wikipedia/commons/9/97/Water_infrared_absorption_coefficient_large.gif
“Absorption spectrum (attenuation coefficient vs. wavelength) of liquid water (red), atmospheric water vapor (green) and ice (blue line) between 667 nm and 200 μm.”
That’s why there are three empirically-measured absorption spectra, Jimbo… one for ice, one for liquid water, one for atmospheric water vapor.
You’re just wrong, Jimbo. On everything. And you attempt the pathetic tactic of denying rotational mode quantum states as some sort of weak ‘defense’ of your indefensible position.
You’ve embarrassed yourself online long enough. It’s time for you to find a new hobby, something which doesn’t require complex cogitation… you’re clearly not up to the task. LOL
Or you can continue embarrassing yourself, and I can continue drop-kicking your stupid arse so hard that you taste the steel-toe of my boot. LOL
LOL@Klimate Katastrophe Kooks
| #
Jimbo “The Dunce” McGinn dribbled:
“Entropy is not quantifiable, you f#cking confused nitwit. There are no units for entropy. You are just confused.”
Entropy is an extensive property of a thermodynamic system, which means it depends on the amount of matter that is present. In equations, the symbol for entropy is the letter S. It has SI units of joules per kelvin (J⋅K−1) or kg⋅m2⋅s−2⋅K−1.
Seems the only one confused is you, Dunce. LOL
LOL@Klimate Katastrophe Kooks
| #
Jimbo “The Dunce” McGinn dribbled:
“In my model quienscence [SIC] is only achieved at absolute zero. Equilibrium just means that the exchange (an exchange that cannot be turned off) is balanced. It doesn’t mean it has been turned off, you simpleton.”
Ok, then reconcile your statement above with your other statement:
“You are confused. Energy travels from higher density to lower density.”
And the fact that a warmer object will have higher energy density at all wavelengths than a cooler object:
https://i.imgur.com/wb9KwS0.png
… then realize you’ve just contradicted yourself, your ‘model’ is the machination of a badly malfunctioning brain badly in need of adjustment to its psychotropic medication dosage, and bleat out more of your utter stupidity, Dunce. LOL
Do you get the feeling that you’re badly out of your depth, that you’re feeling a bit punch-drunk, that your rising frustration and compulsion to lash out at a world that is frightening and inconceivable to your puny underpowered brain is considered ‘just desserts’ by those who know you’re a self-aggrandizing charlatan, Dunce?
James McGinn
| #
Kkook:
https://i.imgur.com/ZaYjj5o.png
JMcG:
What basis does the author have to the absurd claim that there is no “translational” motion in liquid water?
(Liquid Water)
“Note that in the above diagram, there are still vibrational states but that the rotational states are “smeared” one into the other.”
JMcG:
You are so convoluted. I’ve never stated water molecules are spinning in liquid H2O. So I don’t know what you are going on about.
Kkook:
Remember me telling you that the absorption spectra of ice, liquid water and atmospheric water vapor were different because liquid water has quenched rotational mode quantum states
JMcG:
So what?
Kkook:
, Jimbo, as means of disproving your contention that only liquid water could exist in the atmosphere, your claim that gaseous water could not exist in the atmosphere?
JMcG:
You make zero sense here. I can only conclude that you are thoroughly confused.
Kkook:
https://upload.wikimedia.org/wikipedia/commons/9/97/Water_infrared_absorption_coefficient_large.gif
“Absorption spectrum (attenuation coefficient vs. wavelength) of liquid water (red), atmospheric water vapor (green) and ice (blue line) between 667 nm and 200 μm.”
JMcG:
Nothing you stated here confirms the existence of gaseous H2O in earth’s atmosphere, you confused nitwit, if this is what you are claiming.
Kkook:
That’s why there are three empirically-measured absorption spectra, Jimbo… one for ice, one for liquid water, one for atmospheric water vapor.
JMcG:
Nothing you stated here confirms the existence of gaseous H2O in earth’s atmosphere, you confused nitwit.
James McGinn / Genius
LOL@Klimate Katastrophe Kooks
| #
Jimbo “The Dunce” McGinn dribbled:
“What basis does the author have to the absurd claim that there is no “translational” motion in liquid water?”
Reading comprehension problems on top of all your other… ahem… glaringly obvious psychological deficits, Dunce? LOL
“There is little translational motion for the water molecules within the interior of the liquid unless they escape from the liquid phase.”
Poor ol’ dumb Jimbo McGinn not only can’t comprehend what he reads, but he can’t do comparative analysis!
Let’s see:
liquid phase – little translational motion
gaseous phase – moving at hundreds of feet per second
See the comparison there, Dunce? “Little” is a comparison, so one must ask “to what is it being compared?”.
Had you any reading comprehension, you would have realize it was being compared to gaseous phase. It’s right there in the sentence, FFS.
Alas, you’re simply not very bright, so that simple sentence tripped you up. LOL
Jim [slap] Jim! {SLAP}… shut your stupid gob, admit you’ve been wrong about everything, go take your psychotropic medication, and take a nap. You’ve exhausted yourself mentally and physically in defending your utterly indefensible stupidity, you need a break. LOL
James McGinn
| #
Kkook:
Let’s see:
liquid phase – little translational motion
gaseous phase – moving at hundreds of feet per second
JMcG:
Yeah, so? Remember, moron, your audience can’t see your imagination. Are you or are you not saying that this represents proof that water magically defies it’s known boiling temperature/pressure? Provide a direct response, you evasive piece of shit.
Kkook:
See the comparison there, Dunce? “Little” is a comparison, so one must ask “to what is it being compared?”.
JMcG:
LOL. You got nothing. You are just the typical confused nitiwit pretending he understands what he doesn’t.
Kkook:
Had you any reading comprehension, you would have realize it was being compared to gaseous phase. It’s right there in the sentence, FFS.
JMcG:
You not nothing!!!
James McGinn / Genius
LOL@Klimate Katastrophe Kooks
| #
Jimbo “The Dunce” McGinn dribbled:
“Yeah, so? Remember, moron, your audience can’t see your imagination. Are you or are you not saying that this represents proof that water magically defies it’s known boiling temperature/pressure? Provide a direct response, you evasive piece of shit.”
Translation:
“Yeah, you caught me displaying my lack of reading comprehension. So? Now I’ll attempt to topic-shift with a non sequitur that wasn’t even mentioned in the graphic provided so that hopefully people will forget that I’m too dumb to even read simple sentences.”
You’re pathetic, Dunce. A laughingstock; a kicktoy for the intelligent; a self-aggrandizing fraud; just a sad, sorry, pathetic mess of what can easily be deemed a brain-damaged chimp in a man suit. LOL
James McGinn
| #
Kkook:
“E_total = E_rotational + E_vibrational + E_electronic + E_electronic”
— should be —
“E_total = E_rotational + E_vibrational + E_electronic + E_translational”
Are you going to deny translational mode, too, Jimbo?
Or is it that you’re simply too maleducated to even understand any of this? LOL
JMcG:
Rotational and translational are the exact same thing, you goon.
James McGinn / Genius
LOL@Klimate Katastrophe Kooks
| #
Jimbo “The Dunce” McGinn dribbled:
“Rotational and translational are the exact same thing, you goon.”
Bwahahahaaaaa! You’ve got to be joking.
Oh. You’re. not. joking.
Well, Dunce, that pretty much scotches your image as any promulgator of scientific reality. Way to shoot yourself in the foot with a Howitzer. LOL
James McGinn
| #
Kkook:
Jimbo “The Dunce” McGinn dribbled:
“Yeah, so? Remember, moron, your audience can’t see your imagination. Are you or are you not saying that this represents proof that water magically defies it’s known boiling temperature/pressure? Provide a direct response, you evasive piece of shit.”
Translation:
“Yeah, you caught me displaying my lack of reading comprehension. So? Now I’ll attempt to topic-shift with a non sequitur that wasn’t even mentioned in the graphic provided so that hopefully people will forget that I’m too dumb to even read simple sentences.”
You’re pathetic, Dunce. A laughingstock; a kicktoy for the intelligent; a self-aggrandizing fraud; just a sad, sorry, pathetic mess of what can easily be deemed a brain-damaged chimp in a man suit. LOL
JMcG:
Answer the question, you evasive jackass. Are you or are you not saying that this represents proof that water magically defies it’s known boiling temperature/pressure?
James McGinn / Genius
LOL@Klimate Katastrophe Kooks
| #
Jimbo “The Dunce” McGinn dribbled:
“Answer the question, you evasive jackass. Are you or are you not saying that this represents proof that water magically defies it’s known boiling temperature/pressure?”
Awww, are you feeling frustrated that I won’t fall for your non sequitur attempt at topic-shifting so people will hopefully forget that you put on display for all the world to see that you can’t even read a simple sentence, Dunce? LOL
You should stomp your feet and shake your fist in the air, Dunce. Maybe that’ll help. LOL
In case you’re too dense to grok it (and you certainly are), I’m openly mocking you, in full view of the entire world. And the entire world is watching, pointing, laughing at you.
James Bernard “The Dunce” McGinn can’t even read a simple sentence! And he gets all butthurt about it! News at 11! LOL
James McGinn
| #
LOL. You got nothing!!!
James McGinn / Genius
LOL@Klimate Katastrophe Kooks
| #
James Bernard “The Dunce” McGinn dribbled:
“LOL. You got nothing!!!
James McGinn / Genius”
Translation:
“I have no retort, so I’ll use my standard ‘losing an argument badly’ mantra as I beat a retreat to lick my wounds. By the way, is everyone aware that I, James Bernard McGinn, can’t even read a simple sentence? Yes, no? Because I’m willing to humiliate myself further, if need be.
James McGinn / Dunce”
LOL
LOL@Klimate Katastrophe Kooks
| #
JaKo wrote:
“Hi LOL@KKK,
No pun intended, I just like brevity…”
The pseudonym was specifically chosen to perturb the warmists, who utilize similar tactics as that eponymous group… and being largely Democrat party (and the Democrat party being started from the Democratic-Republican party when a rift over slavery developed within that party (the Republican party being started from the Democratic-Republican party with the express purpose of ending slavery), with the express purpose of the Democrat party being to expand slavery into the Kansas-Nebraska territories, the same party which withheld women’s suffrage for more than 40 years, the party of Jim Crow, the party which used the KKK as the paramilitary arm of the Democrat party, the party which started sharecropping as a reformed type of slavery, the party which supported segregation and red-lining, the party which has historically used every dirty trick in the book, which continues to this day), those warmists find the association particularly perturbing.
I’ve had warmist kooks literally have full-blown disconnects from reality, full psychological meltdowns over my moniker, replete with incoherent text-wall ramblings, impotent threats and even a couple of them going to the police to ‘report’ my name… that’s a bonus, IMO. LOL
LOL@Klimate Katastrophe Kooks
| #
JaKo wrote:
“Hi LOL@KKK,
No pun intended, I just like brevity…”
The pseudonym was specifically chosen to perturb the warmists, who utilize similar tactics as that eponymous group… and being largely Democrat party (and the Democrat party being started from the Democratic-Republican party when a rift over slavery developed within that party (the Republican party being started from the Democratic-Republican party with the express purpose of ending slavery), with the express purpose of the Democrat party being to expand slavery into the Kansas-Nebraska territories, the same party which withheld women’s suffrage for more than 40 years, the party of Jim Crow, the party which used the KKK as the paramilitary arm of the Democrat party, the party which started sharecropping as a reformed type of slavery, the party which supported segregation and red-lining, the party which has historically used every dirty trick in the book, which continues to this day), those warmists find the association particularly perturbing.
I’ve had warmist kooks literally have full-blown disconnects from reality, full psychological meltdowns over my moniker, replete with incoherent text-wall ramblings, impotent threats and even a couple of them going to the police to ‘report’ my pseudonym… that’s a bonus, IMO. LOL
LOL@Klimate Katastrophe Kooks
| #
In fact, Herb, I’ll just go ahead and do those calculations, to spare you the embarrassment of arguing against reality again. LOL
At any given kinetic temperature, regardless of atomic or molecular weight, all atoms or molecules will have the same kinetic energy.
CO2: mass = 44.0095 amu, 7.306961098775001e-26 kg
40Ar: mass = 39.962383123824 amu, 6.63590925530233e-26 kg
Assume:
3 DOF
T = 287.64 K
k_B = Boltzmann Constant (1.380649e−23 J·K−1)
v = √((T DOF k_B) / m)
CO2: √ 287.64 K * 3 * 1.380649e−23 J·K−1 / 7.306961098775001e-26 kg = 403.7927445130133423714348056539 m/s
40Ar: √ 287.64 K * 3 * 1.380649e−23 J·K−1 / 6.63590925530233e-26 kg = 423.7177812540082999576147403721 m/s
KE = 1/2 m v^2
CO2: 1/2 * 7.306961098775001e-26 kg * 403.7927445130133423714348056539 m/s * 403.7927445130133423714348056539 m/s = 5.9569481753999999999999999999999e-21 J
40Ar: 1/2 * 6.63590925530233e-26 kg * 423.7177812540082999576147403721 m/s * 423.7177812540082999576147403721 m/s = 5.9569481753999999999999999999999e-21 J
You can do the same for any other atom or molecule, you’ll get the same result.
See how simple everything gets when you actually understand the concepts, Herb? LOL
Herb Rose
| #
Hi LOL,
So what your sayings that the law of conservation of momentum is wrong and when objets with different masses but the same velocity collide the object with morass will transfer mass to the other object to have their kinetic energy be equal. There are no elastic collisions so when an oxygen molecule and a nitrogen molecule with the same velocity the molecular weight of the Nitrogen molecule will increase by 2 and the molecular weight of the oxygen molecule will decreases by two. I assume this transfer of mass is done by an exchange of neutrons between the nucleuses of the atoms. Wouldn’t this produce radioactive air?
You had to study physics and math long tine to become this stupid and spew utter nonsense that defies reality.
“kinetic temperature” what a load of crap.
Herb
LOL@Klimate Katastrophe Kooks
| #
I’m saying that you don’t understand the Law of Conservation of Momentum, the Molecular Kinetic Theory, the Equipartition Theorem, the Virial Theorem nor much else… and you seem to just be pulling your ‘facts’ from your coal chute… did you actually believe no one would check your premise, your contentions, your conclusions? LOL
Herb Rose wrote:
““kinetic temperature” what a load of crap.”
The more you deny reality, the more you show how shallow is your knowledge, Herb.
http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/kintem.html
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/temper.html
“Clearly, temperature has to do with the kinetic energy of the molecules, and if the molecules act like independent point masses, then we could define temperature in terms of the average translational kinetic energy of the molecules, the so-called “kinetic temperature“.”
https://www.thermal-engineering.org/what-is-kinetic-temperature-definition/
Again, buckle down, crack a book and study, Herb. You’re only embarrassing yourself.
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“The result is even though argon starts with greater kinetic energy it cannot gain enough energy to get to the higher altitudes and is restricted to the troposphere.”
Yet again wrong.
https://i.imgur.com/meDwRP6.png
That’s the NASA MSIE E-90 Atmosphere Model, empirically measured by literally millions of balloons, high altitude flights, satellite drag data, mass spectrometry, incoherent scattering measurements and space flight measurements.
Note the thin purple line at top-left… that’s Argon, near 100% of its surface concentration, all the way up past the Karmen Line at 100 km (the delineation point where space starts).
Stop making stuff up, you’re misleading people with your junk science, Herb.
Herb Rose
| #
Hi LOL,
Here’s a hint for you. In reality there are no atoms with the average molecular weight. Every atom (not radioactive) is an integer multiple of the mass of a neutron, so your use of the average molecular weight in any calculations is disconnected from reality.
Herb
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“In reality there are no atoms with the average molecular weight. Every atom (not radioactive) is an integer multiple of the mass of a neutron, so your use of the average molecular weight in any calculations is disconnected from reality.”
And you yet again just make stuff up, because you don’t have the depth of knowledge necessary to be arguing any of this, Herb. What about binding energy mass, Herb? Do you deny that, as well (in the process denying all nuclear fission reactions)? LOL
I didn’t use the “average molecular weight”, I used the Atomic Mass Unit, which is the exact (to the best precision possible) measure of that particular isotope’s mass. Had you even a scintilla of knowledge on this topic, you’d know that “amu” means Atomic Mass Unit, you’d know what that meant, and you’d not have embarrassed yourself yet again. LOL
You’re going to argue molecular physics with a guy who was formally trained in nuclear and molecular physics, Herb? You’ll get crushed. LOL
Herb Rose
| #
Hi LOL,
I believe in fission but not fusion or “binding energy”.
Binding energy is what holds the nucleus of a atom together and when it is too weak the atoms decays. In beta decay an electron and gamma ray are expelled from the nucleus. The gamma ray as well as the expulsion of the electron represents an expenditure (loss) of energy by the binding force. The loss of an electron from the nucleus is also a reduction of something holding the protons together. After the expenditure of things holding the nucleus together a new stable nucleus is formed contains a greater repelling force held together with less energy. You studied long and hard to believe in nonsense.
Herb
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“I believe in fission but not fusion or “binding energy”.”
First, it’s not a matter of “believing in” it, you either accede to reality, or you do not. You do not.
Second, you’ve just contradicted yourself. Again, crack a book and study. When you’ve descended to the level of beclownment where you’re contradicting yourself and you don’t even realize it? That’s pretty bad. LOL
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“The loss of an electron from the nucleus is also a reduction of something holding the protons together.”
Utter twaddle. The bound electron(s) are not in the nucleus. You simply haven’t the first faint clue what you’re talking about.
Herb Rose
| #
Hi LOL,
You continue to try to make evidence conform to your theories. As more discoveries are made as technology advances your efforts to preserve the orthodoxy is doomed.
Here’s one you can use your math skills on. Didymos and Dimorphos are binary asteroids where they recently did the. DART experiment trying to deflect an asteroid. By using the speed of Dimorphos and its distance from Didymos you can use Newton’s formula to determine the mass of Didymos (just as the mass of planets were determined). Knowing the speed and mass of the satellite crashed into Dimorphs and its change of speed you can then calculate the mass of Dimorphs and then use that speed and the combined mass of Dimorphos plus the mass of the satellite to determine how much mass Didymos lost due to the addition of the satellite’s mass to Dimorphos.
Despite what Newton and Einstein say, gravity is not a function of mass.
Herb
Herb Rose
| #
Hi LOL,
The electrons are in the neutron which is in the nucleus.
According to physics a neutron star radiates energy produced when the electrons and protons of atoms combine to form neutrons. If a neutron is not in a nucleus it will quickly decay into an electron and a proton producing energy as a gamma ray. I guess this means the first law of thermodynamics is incorrect and energy is created as neutrons are created and destroyed. That makes fusion possible where you can split a helium nucleus producing two deuterium nuclei and use the released binding energy to fuse two deuterium nuclei into a helium nucleus, releasing more energy.
Herb
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“The electrons are in the neutron which is in the nucleus.”
No. That’s a lay simplification. Again, you’re debating someone who’s been formally trained in this field.
An unstable nucleal configuration will expel a single or multiple neutrons (which each beta decay to a proton and a W – boson (which decays to an electron and an electron antineutrino) in 879.6 ± 0.8 seconds).
You’re coming awfully close to endorsing the Cyclon Theory “neutron as proton and electron egg” crackpottery. Next you’ll be claiming the neutron can change form between neutron and proton (changing size and mass by orders of magnitude) randomly and without impetus, and that there’s only one particle which makes up all invariant-mass matter in the universe. That’s Ken Wheeler crackpottery (which he stole verbatim without attribution, natch). LOL
If you like, I can provide the electron capture decay chain from Osmium (the most-dense element) all the way up the Periodic Table to Hydrogen (I can go further down the Periodic Table, but I’ve already got it all written down from Osmium)… but you’d likely not understand it, nor would you grok why electron capture decay causes nuclear transmutation up the Period Table, nor how to force that electron capture decay even for meta-stable atoms (all atoms which are non-fissionable are meta-stable, but no one would expect you to understand why that would be).
Herb Rose wrote:
“Despite what Newton and Einstein say, gravity is not a function of mass.”
Gravity is literally invariant-mass matter expanding the geodesics of space-time, decreasing energy density in that region surround the invariant-mass matter, causing an energy density gradient which is what gravity literally is… an energy density gradient (which provides the impetus for objects to move in that gravitational field… all motion (all work) requiring an impetus). Of course it is a function of mass (or energy, energy and mass being equivalent via E^2 = m^2 c^4 + p^2 c^2)… energy and mass are two forms of the same thing.
You’re denying an awful lot of empirical science… and providing handwavium explanations that are intended for laypersons, sans mathematical corroboration, while twisting those explanations to fit your odd hobby theory. Don’t do that… everyone who’s tried has only ended up embarrassing themselves. LOL
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“I believe in fission but not fusion or “binding energy”.”
Herb Rose wrote:
“That makes fusion possible where you can split a helium nucleus producing two deuterium nuclei and use the released binding energy to fuse two deuterium nuclei into a helium nucleus, releasing more energy.”
You’ve just negated your own argument and contradicted yourself yet again. You “don’t believe” in binding energy, remember? You’ll have to come up with another explanation for fission which doesn’t include binding energy. Good luck. LOL
Or you could admit you don’t know anything about this topic, you’re furiously Googling to try to keep up, and you’re failing badly at that. LOL
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“That makes fusion possible where you can split a helium nucleus producing two deuterium nuclei and use the released binding energy to fuse two deuterium nuclei into a helium nucleus, releasing more energy.”
You’ve just violated 1LoT. Crack a book and study, Herb. LOL
Herb Rose
| #
Hi LOL,
When you say you have studied and been trained you are confessing that you’ve been brainwashed into believing in nonsense. When I say I don’t believe in something then use it, it is to point out the the absurdity and contradiction of what you believe.
A neutron is a subatomic molecule just as a alpha particle is a subatomic particle. It is made from a combination of an electron and a proton (Quarks with their flavors are a result of quacks trying to make things complicated.) and it has both a negative charge and a positive charge that are equal, making it neutral overall. When this dual charged molecule moves through an energy/magntic field it represents two equal currents flowing in opposite directions. This creates a shearing force that splits the molecule creating a hydrogen atom containing both matter and energy while producing a disturbance in the energy field (gamma ray).
Energy is attracted to positive matter and having a stronger force than the electrical force of matter it is able to split matter. When there are no exposed electrons energy will form a compression force that holds a nucleus together (no binding force), If there are too many electrons and one is exposed the compression force will destroy the nucleus. If there are too few electrons in the nucleus there will not be enough glue within the nucleus to resist the compression of the energy and it will again be destroyed.
It is simple. There are two distinct components that make up the universe: matter with its electrical force snd energy with its gravity/magnetic force. The two are not the same and behave in opposite manners. When opposite charges come together the size and strength of the radiated force degreases (neutron). When opposite poles come together the size and strength of he radiated force increases (Stronger magnet, greater gravity.). When similar charges come together the size and strength of the radiated electric force increases. When similar poles come together the size and strength of the radiated field decreases.
Energy and matter are not the same and the make believe reality you believe in is a fantasy with no connection to reality.
Herb
Herb
LOL@Klimate Katastrophe Kooks
| #
Oh, oh my… there’s just so much delusional blather, miscomprehension and outright crackpottery in your prior post that you’d do well to attempt to prove your contentions mathematically, as I’ve done.
And I’ll yet again use the closing phrase from the last thread (https://principia-scientific.com/separating-fact-from-fiction/#comment-98229 wherein I posed the same challenge to you) which you ran away from:
Your inability to do so is evidence that you are wrong, Herb.
You’ll be unable to do so. LOL
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“A neutron is a subatomic molecule just as a alpha particle is a subatomic particle. It is made from a combination of an electron and a proton (Quarks with their flavors are a result of quacks trying to make things complicated.) and it has both a negative charge and a positive charge that are equal, making it neutral overall.”
Utter quackery, completely devoid of science, divorced from reality. LOL
And given that a proton and a neutron are not fundamental particles (ie: they are further divisible, which we know and have empirically proven), what gives a proton its charge, Herb?
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“That makes fusion possible where you can split a helium nucleus producing two deuterium nuclei and use the released binding energy to fuse two deuterium nuclei into a helium nucleus, releasing more energy.”
Are you even cognizant of where you’ve gone wrong, Herb? LOL
Herb Rose wrote:
“When I say I don’t believe in something then mis-use it, it is to point out the the absurdity and contradiction of what you believe.”
Fixed it for you, Herb. LOL
LOL@Klimate Katastrophe Kooks
| #
https://image1.slideserve.com/3224064/slide22-n.jpg
Oh, but however can that be, Herb?! How can the product of splitting the Helium have a greater mass than the originating Helium?
Remember, you disclaim Mass-Energy Equivalency (E^2 = m^2 c^4 + p^2 c^2), so how, oh how, oh how did that ever happen?!
It requires energy to split the Helium nucleus, you goober. That energy is then converted to mass of the resultant.
You deny all of this.
Stop denying reality, Herb. LOL
LOL@Klimate Katastrophe Kooks
| #
Just to make it glaringly obvious to even dullards…
Herb Rose wrote:
“That makes fusion possible where you can split a helium nucleus producing two deuterium nuclei and use the released binding energy to fuse two deuterium nuclei into a helium nucleus, releasing more energy.”
LOL@Klimate Katastrophe Kooks wrote:
”
https://image1.slideserve.com/3224064/slide22-n.jpg
It requires energy to split the Helium nucleus, you goober. That energy is then converted to mass of the resultant.
You deny all of this.”
Now, Herb, are you ready to STFU, crack a book and actually study, or are you going to continue beclowning yourself? LOL
Al Shelton
| #
Where are the third party judges in this debate???
Herb Rose
| #
Hi Al,
To be a judge you need to be an expert. To be an expert you have o believoethe existing beliefs. You be the judge. Is it reason or beliefs?
Glad someone is following the discussion.
Herb
Herb Rose
| #
Hi againAl,
The point of contention is that he believes that objects with the same velocity but different masses will have the same kinetic energy (kinetic temperature) while I don’t believe that or the theoretical gyrations he uses to make that claim.
Herb
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
“The point of contention is that he believes that objects with the same velocity but different masses will have the same kinetic energy (kinetic temperature) ”
Wrong yet again… as I’ve repeatedly stated:
At any given kinetic temperature, regardless of atomic or molecular mass, all atoms or molecules will have the same kinetic energy.
Then I mathematically proved it:
https://principia-scientific.com/the-fallacy-of-bell-curves/#comment-98389
Then you denied the concept of ‘kinetic temperature’. LOL
You seem to be confused on more than a few topics, Herb. LOL
Herb Rose
| #
Hi LOL,
You are right. Since all the air molecules have the same kinetic energy there is no bell curve and the thermometer is giving the kinetic energy of every molecule not the mean.
I should have known that when the binding energy of a nucleus is released it is done as mass in the form of a mushroom.
LOL@Klimate Katastrophe Kooks
| #
Yet again you’re wrong, Herb. Again, buckle down, crack a book and study.
There is only one thing in your previous post which was correct, to wit: “You are right.” LOL
LOL@Klimate Katastrophe Kooks
| #
Herb Rose wrote:
https://principia-scientific.com/separating-fact-from-fiction/#comment-98191
“Temperature may be the measurement of kinetic energy but that is not what a thermometer measures. It measures the momentum of the molecules striking it.”
Translation:
“Temperature may be a measurement of kinetic energy, but a thermometer (a temperature measurement device) measures something else.”
Do you even listen to yourself, Herb? LOL
Herb Rose wrote:
https://principia-scientific.com/the-fallacy-of-bell-curves/#comment-98451
“Since all the air molecules have the same kinetic energy there is no bell curve and the thermometer is giving the kinetic energy of every molecule not the mean.”
Have you learned something (meaning you’re not wholly ineducable, after all), or are you simply shifting your narrative as means of bolstering your odd (and grossly incorrect) hobby theory, Herb? LOL
Why is it so difficult for you to simply admit that you don’t know what you’re talking about, and go study, Herb? LOL
Herb Rose
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Hi LOL,
Your the one insisting that all the air molecules have the same kinetic energy even when they have different masses but the same velocity.
When a gas molecule collides with a thermometer it does not add mass to it, only energy. Therefore it is only measuring the energy of the gas molecule not its kinetic energy.
You have absolutely no ability to think, just repeat what you’ve been told are the right answers. I’m sure you do well in school but you are no scientist.
Herb
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“Your the one insisting that all the air molecules have the same kinetic energy even when they have different masses but the same velocity.”
AT THE SAME KINETIC TEMPERATURE… there is a distribution of velocities that make up the layperson definition of ‘temperature’ (which you deny just as you deny so many other long-established aspects of science).
I didn’t just ‘insist’, I proved it.
https://principia-scientific.com/the-fallacy-of-bell-curves/#comment-98389
You can’t prove any of your blather because unscientific blatherings don’t have any associated maths. LOL
Herb Rose wrote:
“Therefore it is only measuring the energy of the gas molecule not its kinetic energy.”
And what would that energy be classified as, Herb? Hmmm?
E_total = E_electronic + E_vibrational + E_rotational + E_translational
Which is it, Herb?
It can only be E_translational, which is kinetic energy.
Stop denying reality, buckle down, crack a book and study. Stop embarrassing yourself. LOL
LOL@Klimate Katastrophe Kooks
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Just to rub some salt in those gaping wounds in your savaged psyche over your embarrassing yourself so badly in your insistence that translational mode energy isn’t kinetic energy, Herb:
Calculate the temperature of a 40Ar atom (mass = 6.63590925530233e-26 kg, 39.962383123824 amu) moving at 50 m/s in a single translational mode DOF, and in 3 translational mode DOF.
k_B = 1.380649e-23 J K−1
v = √(v_x^2 + v_y^2 + v_z^2)
Therefore, if v_x = 50 m/s and v_y = 0 and v_z = 0; v = 50 m/s
1) Calculate the kinetic energy of the atom:
KE = (1/2) m v^2
KE = (1/2) (6.63590925530233e-26 kg) (50 m/s)^2
KE = 8.2948865691279125e-23 J
2a)
KE = (DOF / 2) k_B T
T = (2 KE) / (DOF k_B)
(2 * 8.2948865691279125e-23 J) / (1 DOF * 1.380649e-23 J K−1) = 12.015923770817800179480809387469 K for a 40Ar atom moving in only one DOF.
(2 * 8.2948865691279125e-23 J) / (3 DOF * 1.380649e-23 J K−1) = 4.0053079236059333931602697958231 K for a 40Ar atom moving in three DOF.
Sandia National Laboratories uses this method of calculating temperature.
2b)
T = m v^2 / DOF k_B
6.63590925530233e-26 kg * 2500 / 3 DOF * 1.380649e-23 J K−1 = 4.0053079236059333931602697958231 K
6.63590925530233e-26 kg * 2500 / 1 DOF * 1.380649e-23 J K−1 = 12.015923770817800179480809387469 K for a 40Ar atom moving in only one DOF.
See what I did there, Herb? I used the kinetic energy of the atom to derive the temperature of that atom… then I corroborated that (checked my math) via another mathematical route.
I also happened to prove that it is the kinetic energy that is imparted to a temperature measurement device, not your unnamed “energy of the gas molecule” and certainly not your previously-claimed “momentum”.
And in so doing, I also happened to prove you wrong yet again. LOL
Now, Herb, can you tell us what scientific principle states that an atom’s or molecule’s temperature will increase (for the same given kinetic energy) if it is constrained in fewer than 3 DOF? Can you tell us what the name of that form of ‘temperature’ would be in that instance? I bet not. LOL
Herb Rose
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Anytime I see a value with a precision of 17 decimal places I know it’s bullshit.
You need to switch careers to statistician or accountant where they ask, What answer do you want? We will provide the math for it
Kepler’s law, dv^2 = C, .states that the energy (v^2) decreases with distance from the source. Does this mean that all satellites at the same altitude have the same kinetic energy?
LOL@Klimate Katastrophe Kooks
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The last kook I debated (read: squashed like a bug) over on CFACT (the kook going by the moniker ‘evenminded’) insisted that if it wasn’t precise to the last decimal place, it was wrong. So I use the decimal places to the precision of my calculator.
Now you’re claiming that too many decimal places means it’s wrong.
You kooks can’t have it both ways. LOL
He made many of the same claims as you make, Herb. Seems scientific illiteracy and mathematical innumeracy has tendentious tendencies. LOL
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“Does this mean that all satellites at the same altitude have the same kinetic energy?”
You mean you can’t do that simple math to figure that our for yourself, Herb?
Such a shame that, rather than buckling down, cracking a book and studying, you instead choose to lead people astray with confidently-stated incorrectitudes. How many did I catch you on in just your ‘article’ above, Herb? How many years have you been leading people astray, Herb? How much damage have you done to the climate-skeptical movement with your junk science? Why do you insist upon barfing out wrong information, rather than studying actual science, Herb? Whose side are you on, Herb? LOL
Herb Rose
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Please give us a list of the combination of stable sub atomic particles (other than protons, electrons, and neutrons) that this argon atom contains to give it this precise mass.
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“Does this mean that all satellites at the same altitude have the same kinetic energy?”
There you go, folks. Herb Rose, in a desperate attempt at ‘proving’ himself ‘not wrong’ attempts to conflate satellites with air atoms and molecules.
I’m sure the more astute readers can ascertain the glaring problem with that (which Herb either was too scientifically illiterate to catch, or disingenuously ignored to push his conflation)… the fact that air molecules collide and thereby attempt to equipartition their kinetic energy, whereas satellites do not.
Now, let’s all wait with bated breath to see what new non sequitur conflation Herb will desperately pull from his dark-n-dirty to defend his kooky unscientific views. LOL
Herb Rose
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I have the wrong formula for calculating kinetic energy. I use 1/2mV^2 but you tell me this is wrong. Since the satellites have different masses I get different kinetic energy for them but I got the same results when I multiply the masses of gasses times the same velocity squared and you tell me this is wrong. What is the correct formula?
Herb Rose
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As I’ve stated repeatedly that when objects have elastic collisions the object with the greater velocity adds velocity to the object with less velocity regardless of mass. Get yourself one of those swinging ball desk ornaments and learn about the conservation of momentum.
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“Please give us a list of the combination of stable sub atomic particles (other than protons, electrons, and neutrons) that this argon atom contains to give it this precise mass.”
It’s not my job to do the learning for you, Herb.
Start here:
https://en.wikipedia.org/wiki/Isotopes_of_argon
Learn more here:
https://ptable.com/?lang=en#Properties/Series
Learn more here:
https://www.calctool.org/physical-chemistry/atom
For 40Ar, it consists of:
18 protons with a mass of 1.00727646662153 amu each, each consisting of two Up Quarks (charge +2/3) and one Down Quark (charge -1/3) for a charge of +1 each.
22 neutrons with a mass of 1.0086649158849 amu each, each consisting of one Up Quark (charge +2/3) and two Down Quarks (charge -1/3), for a charge of 0 each.
18 electrons with mass of 0.00054857990906516 amu each.
Mass Defect of 0.3690958631931 amu per nucleus [which, everyone will remember, Herb Rose denies because he denies Mass-Energy Equivalency and thus he must deny binding energy and thus he must deny fission reactions… while at the same time contradicting himself by citing examples of binding energy and fission reactions (while getting it wrong. LOL).]
Total mass = 6.63590925530233e-26 kg, 39.962383123824 amu
Get to work, Herb. You’ve got a lot of work to do just to get to the most basic level where you can even attempt to speak coherently on these topics. LOL
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“I have the wrong formula for calculating kinetic energy. I use 1/2mV^2 but you tell me this is wrong.”
You mean you still can’t figure it out, Herb? LOL
Kinetic energy in orbit = ½ mv^2 = ½GMm/r
This is junior high school stuff, Herb. Do better, crack a book and study. LOL
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“As I’ve stated repeatedly that when objects have elastic collisions the object with the greater velocity adds velocity to the object with less velocity regardless of mass.”
You have a 1000 ton concrete block moving toward you at 1 MPH. Your car is going 150 MPH toward the concrete block.
Herb Rose claims his car will “add velocity” to that concrete block, and he’ll walk away without a scratch because he was going faster than that concrete block. LOL
You should try a practical exercise to prove that, Herb. Stupidity should be painful. LOL
Herb Rose
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Do you know what an elastic collision or velocity is?
Velocity has direction so the car is traveling at 150 mph while the concrete block is traveling at -1 mph.
When the car hits the block its velocity will increase and be greater than -1 mph. In an elastic collision there is no lose of energy by heat, crumpling metal, or other means. All the kinetic energy is preserved as motion and distributed between the two masses, changing the kinetic energy of the masses but not the total kinetic energy.
Again you are adjusting the data to give the answer you believe. What kind of charge does a fraction of an electron have? Only idiots would believe that when a proton and electron combine to form a neutron (with a mass equal to the mass of a proton plus the mass of an electron) and when that neutron decays into a proton, electron and gamma ray (no mass) that the neutron is not made from a proton and an electron.
You are an egomaniac unwilling to consider the possibility that you could be wrong and when evidence is shown that disputes your infallibility you will change the evidence or ignore it.
Use your math skills to prove that something that is a function of a variable is equal to something that is an inverse function of the same variable. Einstein did it by dividing by zero. What trick are you going to use?
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“What kind of charge does a fraction of an electron have?”
Yes, folks, Herb Rose actually believes that an electron is not a fundamental particle (ie: he believes the electron can be split further). LOL
Why don’t you just admit that you’ve constructed a fantasy fyzics world for some odd reason, you’re desperately attempting to defend it, you keep getting blasted out of the water via mathematical proof whereas you keep issuing vague hand-wavey ‘explanations’ that make no sense, because you have some block that prevents you from actually learning how the universe works, Herb. LOL
Yes, I know what an elastic collision is, Herb. For the purposes of the vehicle / cement block discussion, it should have been obvious that we were assuming an elastic collision.
And the car will not increase the cement block’s velocity, it’ll barely even slow it down as the car’s direction is reversed:
https://i.imgur.com/JsGp7d3.png
Object #1 (cement block):
Mass: 90718474 kg
Initial Velocity: 1.609344 km/h
Final Velocity: 1.1238 km/h
Object #2 (car):
Mass: 907.18474 kg
Initial Velocity: -241.4016 km/h
Final Velocity: 67.82 km/h
You’ll note the vector of the car changed sign… that means the car slowed from 241.4016 km/h, was stopped completely, then was thrown at 67.82 km/h in the opposite direction… all while the cement block only slowed down by only 0.485544 km/h.
Diametrically opposite to what you claimed: “the object with the greater velocity adds velocity to the object with less velocity regardless of mass”… in point of fact, the object with the greater velocity reduced the velocity of the object with lesser velocity, whereas the vector of velocity for the faster object was completely reversed.
But kooks are often diametrically opposite to reality, Herb… it’s a hallmark of your kookery and the easiest means of spotting a kook. LOL
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“Use your math skills to prove that something that is a function of a variable is equal to something that is an inverse function of the same variable. Einstein did it by dividing by zero.”
Boy, you’re just desperately throwing out chaff as you beat a strategic retreat, aren’t you? Like I’ve never seen that before. LOL
You’ve got to stop listening to kooks, Herb… they’re misleading you. How about you crack a book and actually study, rather than blustering out your lack of knowledge?
Einstein in his 1917 research paper proposed a model of a Static Universe but arrived at a Cosmological Constant that he had trouble getting rid of (his mathematics was correct, he just didn’t believe that the universe was expanding, as the Cosmological Constant implies).
Einstein attempted to fudge that Cosmological Constant out via a common mathematical maneuver to arrive at a static universe, a static universe being widely believed at the time.
Five years later, Alexander Freidman, a Russian cosmologist found that under certain conditions Einstein’s equation involved division by zero, which is not permissible.
Einstein acquiesced to the data, adopted the view of an expanding universe, withdrew that paper and referred to his attempt at negating the Cosmological Constant as the biggest blunder of his life… when his only blunder was that his mathematics was correct, but he didn’t like the implications of the maths, and he attempted to correct it to show what everyone believed at the time, that the universe was static.
Stop embarrassing yourself, Herb. LOL
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“Only idiots would believe that when a proton and electron combine to form a neutron (with a mass equal to the mass of a proton plus the mass of an electron) and when that neutron decays into a proton, electron and gamma ray (no mass) that the neutron is not made from a proton and an electron.”
Oh, your fragile feelings are hurt and you’re going ad hominem to compensate for your shallow depth of knowledge. Like I’ve never seen that before. LOL
There’s still a Mass Defect there that you’ve not accounted for, Herb.
1.00727646662153 amu (proton) + 0.00054857990906516 amu (electron) = 1.0078250465305 amu
1.0086649158849 amu (neutron) – 1.0078250465305 amu (proton + neutron) = 0.000839869254 amu = 782227.4219792516 eV
That’s a 0.00000158502 μm photon = 0.00158502 nm photon = 0.0158502 Å photon.
That’s a gamma ray photon, Herb. LOL
You deny Mass-Energy Equivalency, Herb, so why don’t you explain to everyone exactly how the mass of a proton plus electron is lower than the mass of a neutron, without using binding energy. LOL
Or you could, you know, just admit that you haven’t the faintest clue what you’re talking about, you’re Googling as if your life depended upon it, brow furrowed as you scratch your balding pate in frustrated puzzlement, and that’s causing you to lash out. LOL
LOL@Klimate Katastrophe Kooks
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“1.0078250465305 amu (proton + neutron)”
– should be –
“1.0078250465305 amu (proton + electron)”
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“Use your math skills to prove that something that is a function of a variable is equal to something that is an inverse function of the same variable.”
Trivial, but you won’t understand a bit of this, Herb.
https://math.stackexchange.com/questions/541978/can-the-inverse-of-a-function-be-the-same-as-the-original-function
Let “f” be a function with domain “A” and range “B”.
Define the inverse function “f^-1” as the function that has range “A” and domain “B”, and such that “f^-1(f(x)) = x” for all “x” in “A”.
Prove that “f(f^-1(x)) = x” for all “x” in “B”.
PROOF #1 (comprehensive):
Consider the function “f(x) = 2x”, which has domain “A” = (-∞, ∞) and range “B” = (-∞, ∞).
The inverse function “f^-1(x)” can be found by solving the equation “f(x) = 2x” for “x”. This results in “x = 2^-1(f(x))” or “x = f^-1(2x)”.
Since “f(x) = 2x”, we can rewrite the equation as “x = f^-1(2x)”.
Next, we need to prove “f(f^-1(x)) = x” for all “x” in “B”.
Since “f(x) = 2x” and “f^-1(x) = x/2”, we can calculate “f(f^-1(x)) = f(x/2) = 2(x/2) = x”. Thus, “f(f^-1(x)) = x” for all “x” in “B”.
As a result, we can conclude that the function “f(x) = 2x” is equal to its inverse function “f^-1(x) = x/2”, since their composition is equal to the identity function.
PROOF #2 (simple):
Graph the function y=x, you’ll see that for any given value, both the function and its inverse are identical, giving a 45° diagonal line. When you graph a function, and it mirrors itself across the x=y line, that function is an inverse of itself by definition.
LOL@Klimate Katastrophe Kooks
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In fact, Herb, I’ll even provide you a real-world example of a function being equal to its inverse function…
Given that the gradient and its derivative are dual (transpose) to (of) each other, and given that tensor rank is invariant under transposition, that means that tensor rank escalates for each gradient thusly:
……../ gradient (rank 8) [ᵀ] (rank 8) 8th derivative := drop
……./ gradient (rank 7) [ᵀ] (rank 7) 7th derivative := lock
……/ gradient (rank 6) [ᵀ] (rank 6) 6th derivative := pop
…../ gradient (rank 5) [ᵀ] (rank 5) 5th derivative := crackle
…./ gradient (rank 4) [ᵀ] (rank 4) 4th derivative := snap
…/ gradient (rank 3) [ᵀ] (rank 3) 3rd derivative := jerk
../ gradient (rank 2) [ᵀ] (rank 2) 2nd derivative := acceleration
./ gradient (rank 1) [ᵀ] (rank 1) 1st derivative := velocity
/ scalar (rank 0) (affine space position)
The gradient builds upon the tensor quantity which it is operating upon to produce the next higher-rank tensor quantity, which is the derivative. The gradient and its derivative are dual (transpose) to (of) each other, and tensor rank is invariant under transposition, so the tensor rank of the gradient is the same as the tensor rank of its transpose (its derivative) and vice versa. Given that taking the dot product of a transpose is equivalent to taking the outer product, and taking the outer product escalates resultant tensor rank, that means that for a tensor of rank k, the gradient will be rank k+1, and thus so will the resultant derivative. That in no way implies that the gradient and its associated derivative have the same physical dimensionality.
A transposition of a matrix is an involution on the set of matrices, and an involution is exactly what you were talking about (although you didn’t know it).
LOL@Klimate Katastrophe Kooks
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In other words, Herb…
If a tensor has a magnitude and no vector field (i.e., rank 0 tensor), then it is called a scalar and has 1 component in 3-D space and in 4-D space-time.
If a tensor has a magnitude and one vector field (i.e., rank 1 tensor), then it is called a vector and has 3 components in 3-D space and 4 components in 4-D space-time.
If a tensor has a magnitude and two vector fields (i.e., rank 2 tensor), then it is called a dyad and has 9 components in 3-D space and 16 components in 4-D space-time.
If a tensor has a magnitude and three vector fields (i.e.: rank 3 tensor), then it is called a triad and has 27 components in 3-D space and 64 components in 4-D space-time.
If a tensor has a magnitude and four vector fields (ie: rank 4 tensor), then it is called a tetrad and has 81 components in 3-D space and 256 components in 4-D space-time.
… so on and so forth. You start with scalar position, you increase tensor rank when you add displacement (velocity), and that gradient of affine space position is one tensor rank higher, and given that tensor rank is invariant under transposition (and velocity is a transposition of the gradient of affine space position), velocity is a Rank 1 tensor.
Then if you accelerate (increase or decrease velocity or change direction), you add another gradient (the change in velocity), and thus acceleration is a Rank 2 tensor.
So on and so forth all the way up from affine space position, to velocity, to acceleration, to jerk, to snap, to crackle, to pop, to lock, to drop. All transpositions, therefore all involutions… the original function is the inverse of its transpose function, the gradient and its derivative are dual (transpose) to (of) each other.
sunsettommy
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It is good to have a lot of discussion but please let’s tone down the attacking language in the posts to facilitate a more level discussion setting.
SUNMOD-Administrator
JaKo
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Hi Sunset Englishman,
I think it would be the best to let LOL@KKK and Herb have their email addresses and then just report the outcome instead of cluttering this website.
Cheers, JaKo
LOL@Klimate Katastrophe Kooks
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I believe we have a duty to expose scientific charlatans and question their motives for promulgating verifiably false ‘science’… it’s something an enemy attempting to discredit the climate-skeptical community would do (and has done in the past… I personally perturbed BeyondSmartLady over at CFACT to such an extent that she cracked and started spewing her true (liberal) beliefs. Edward Grant Foster (aka Tamino) was caught several times using pseudonyms and dispensing junk science so he could crow on his website about the conservative rubes believing it. ‘Evenminded’ has hijacked dozens of pseudonyms and real-life names of prominent conservative climate skeptics across dozens of sites and spews utter hogwash in an attempt to discredit our side.).
So one must ask: “Why is Herb Rose so adamant about continuing to spew unscientific bafflegab in light of mathematical proof that he is wrong? Why does he use the same non-sequituring, goalpost-moving tactics that the liberals use? Whose side is he on?”
Is it to protect ‘revenue’ from the three books he wrote (back in 2015 and 2017… very few sold and there aren’t even any reviews of any of the books)? Likely not, given the dismal sales. So what’s he protecting by continuing to spew his tripe and mislead people?
“What are you protecting?”, you may ask. Well, I’ve not written any for-sale books (I give away all of my writings for free and with the encouragement to use it in any way one wishes, in whole or in part, either with or (preferably) without attribution.
What am I protecting? I’m attempting to destroy CAGW before it destroys our way of life… using bog-standard science straight of the textbooks.
In a world that wasn’t full of ineducable sheeple, the three posts below would do the job all on their own:
https://principia-scientific.com/separating-fact-from-fiction/#comment-98166
https://principia-scientific.com/separating-fact-from-fiction/#comment-98167
https://principia-scientific.com/separating-fact-from-fiction/#comment-98198
… alas, we don’t live in that world.
Herb Rose
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High LOL,
Final comment.
I also believe in exposing falsities, which includes physics as well as GHGT.
In your example of the car and concrete block you use speed instead of velocity. Velocity has direction so since the two are moving in opposite directions one has a negative velocity. If we say the concrete block has a negative velocity and the car has a positive velocity then after the collision the concrete block has had its velocity increased (slowed) and the velocity of the car has decreased (becoming negative). Since energy is v^2 the block has gained energy the car has lost energy. The car, with less kinetic energy has added energy to the concrete block, violating the 2nd law. If this law is wrong what consequences does it have on theories that use it as a premise (Planck’s work)?
Herb
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“In your example of the car and concrete block you use speed instead of velocity.”
FFS, Herb. You do understand that velocity was used, yes? Velocity being a vector (a magnitude and a direction) whereas speed is a scalar (only a magnitude), yes?
That’s why the vector of the car changed sign, Herb… the vector of the car changed direction 180 degrees. Inherently a vector was used.
https://i.imgur.com/JsGp7d3.png
Object #1 (cement block):
Mass: 90718474 kg
Initial Velocity: 1.609344 km/h
Final Velocity: 1.1238 km/h
Object #2 (car):
Mass: 907.18474 kg
Initial Velocity: -241.4016 km/h
Final Velocity: 67.82 km/h
The negative Initial Velocity for the car denotes that it is moving to the left. The positive Final Velocity for the car denoted that it is moving to the right.
You’re just wrong, Herb. On everything. And when you’re called on it, you goalpost-move, nitpick, non sequitur and attempt to topic-shift… all the same tactics used by those attempting to discredit the climate-skeptical side by promulgating idiocy under guise of posing as one of us.
Stop it. You’re not ‘proving’ yourself ‘not wrong’, you are wrong and it’s pretty obvious to anyone with even an iota of proper schooling that you’re showing the shallowness of your knowledge.
That you continue makes me lean more and more toward believing that you’re a Judas goat. Stop it. Buckle down. Crack a book. Study.
Herb Rose
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Final final comment.
The car and concrete block are moving in the same direction when the car hits the concrete block. Does the velocity of the car increase as it gains energy from the block or does it slow down as it transfers energy to the block?
Your belief in your infallibility makes any suggestion that you are wrong inconceivable.
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“The car and concrete block are moving in the same direction when the car hits the concrete block.”
You don’t even have the depth of knowledge necessary to understand what a scalar or vector is, you don’t even have the depth of knowledge necessary to understand that the car and the block start out going toward each other, Herb.
Stop, you’re not doing anything but embarrassing yourself with your own lack of knowledge.
https://i.imgur.com/JsGp7d3.png
Object #1 (cement block):
Mass: 90718474 kg
Initial Velocity: 1.609344 km/h
Final Velocity: 1.1238 km/h
Positive velocity denotes movement to the right.
Object #2 (car):
Mass: 907.18474 kg
Initial Velocity: -241.4016 km/h
Final Velocity: 67.82 km/h
Negative velocity denotes movement to the left.
Stop embarrassing yourself, buckle down, crack a book, study.
LOL@Klimate Katastrophe Kooks
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Let us not forget Herb Rose’s initial claim:
“the object with the greater velocity adds velocity to the object with less velocity regardless of mass”
https://i.imgur.com/JsGp7d3.png
Object #1 (cement block):
Mass: 90718474 kg
Initial Velocity: 1.609344 km/h
Final Velocity: 1.1238 km/h
^^^^^^^^^^^^^^^^^^^^^^^^
HEAVIER OBJECT WITH LOWER VELOCITY; VELOCITY DECREASED
Object #2 (car):
Mass: 907.18474 kg
Initial Velocity: -241.4016 km/h
Final Velocity: 67.82 km/h
You’ll note the vector of the car changed sign… that means the car slowed from 241.4016 km/h, was stopped completely, then was thrown at 67.82 km/h in the opposite direction… all while the cement block only slowed down by only 0.485544 km/h.
Diametrically opposite to what Herb Rose claimed: “the object with the greater velocity adds velocity to the object with less velocity regardless of mass”… in point of fact, the object with the greater velocity reduced the velocity of the object with lesser velocity, whereas the vector of velocity for the faster object was completely reversed.
And yet, Herb Rose continues arguing, continues spewing confidently-stated unphysicalities. Why? What is his agenda? Why is he purposefully misleading people with verifiably incorrect statements? Why does he continue doing damage to the climate-skeptical side, painting us as scientifically-illiterate by posing as one of us? Whose side is he on?
Herb Rose
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In your analysis both objects lose energy.
I was trying to simplify the scenario so you could comprehend it, instead you try to confuse things.
In the simplified scenario both the car and the block are moving in the same direction giving them both positive velocity. The car hits the block. Do both the car and block slow down or does the car slow down and the speed of the block increase? Is the car (with less kinetic energy) adding energy to the block (more kinetic energy) or not?
Are you deliberately playing stupid or just being an ass?
LOL@Klimate Katastrophe Kooks
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And now Herb Rose is attempting to change the initial conditions to ‘prove’ himself ‘not wrong’… you specifically stated that:
“the object with the greater velocity adds velocity to the object with less velocity regardless of mass”
… if that were true, it would be true in all circumstances, Herb. Obviously, the disproof is proof that you’re spewing utter garbage. Your attempt at changing the initial conditions is a rather pitiful attempt at resurrecting your now-ravaged reputation as a purveyor of scientific truth… you’ve been exposed as a scientific charlatan… there is very little in your ‘article’ above which is actually true, and PSI should be extraordinarily cautious in ever allowing you to publish on their platform in future… because you’ve yet to explicate your agenda, why you continue to use the same exact tactics as the leftists… non sequituring, goalpost shifting, changing initial conditions, spewing confidently-stated unphysicalities as gospel truth, etc.
Remember, folks, this is Herb Rose’s “final, final, final” comment… he’s obviously got an agenda if he can’t just admit he’s wrong and go crack a book to study… instead he continues pushing whatever agenda he’s got by spewing scientific falsehoods, while attempting ad hominems (attacking the messenger rather than the message).
https://i.imgur.com/JsGp7d3.png
Object #1 (cement block):
Mass: 90718474 kg
Initial Velocity: 1.609344 km/h
Final Velocity: 1.1238 km/h
^^^^^^^^^^^^^^^^^^^^^^^^
HEAVIER OBJECT WITH LOWER VELOCITY; VELOCITY DECREASED
Object #2 (car):
Mass: 907.18474 kg
Initial Velocity: -241.4016 km/h
Final Velocity: 67.82 km/h
You’ll note the vector of the car changed sign… that means the car slowed from 241.4016 km/h, was stopped completely, then was thrown at 67.82 km/h in the opposite direction… all while the cement block only slowed down by only 0.485544 km/h.
Diametrically opposite to what Herb Rose claimed: “the object with the greater velocity adds velocity to the object with less velocity regardless of mass”… in point of fact, the object with the greater velocity reduced the velocity of the object with lesser velocity, whereas the vector of velocity for the faster object was completely reversed.
I know no one likes being told they’re wrong, Herb… but you’ve been wrong on practically everything, because your depth of knowledge is simply too shallow to even cogently converse on these topics.
Your goalpost-shifting, initial-condition-changing, ad-hominem attacking, non sequituring, topic-changing tactics aren’t going to change that, Herb.
You cracking a book and studying might. Why don’t you try that, Herb?…
… unless promulgating scientific truth isn’t your ultimate purpose. LOL
Why don’t you answer the questions, Herb?
“And yet, Herb Rose continues arguing, continues spewing confidently-stated unphysicalities. Why? What is his agenda? Why is he purposefully misleading people with verifiably incorrect statements? Why does he continue doing damage to the climate-skeptical side, painting us as scientifically-illiterate by posing as one of us? Whose side is he on?“
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“In your analysis both objects lose energy.”
And here we have a perfect example of Herb Rose confidently stating a scientific falsehood. Where did the energy go in this elastic collision in a closed system, Herb?
Knowing that it is an elastic collision, were you lying when you stated that both objects lost energy, as means of bolstering your as-yet explicated agenda?
Or do you just not believe in 1LoT, not even in a closed system during an elastic collision such as was this example? LOL
Why don’t you put that to mathematics to prove your contention, Herb?
Oh, that’s right, because you can’t. LOL
In this frictionless example of a cement block and a vehicle, it requires an energy expenditure to decelerate just as it requires an energy expenditure to accelerate. The energy of the faster object worked to decelerate the slower object (diametrically opposite to what you claimed… you claimed it would “add velocity” (your words) to the slower object.
You’ll find that the system (which is comprised of both objects) has the same exact level of momentum in the Final Condition as it had in the Initial Condition.
It is, after all, an elastic collision. And in a closed system, no less. Or do you not know what an elastic collision is, Herb? LOL
LOL@Klimate Katastrophe Kooks
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Mheh, Omni Calculator glitched, giving the incorrect Final Velocity for the second object. I discovered this as I was doing the calculations by hand just to be sure.
Here’s the correct interaction:
https://i.imgur.com/rQOAGEN.png
LOL@Klimate Katastrophe Kooks
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Ah, no it didn’t glitch, it just showed Final Velocity in m/s rather than km/h, despite my having toggled it to km/h.
Report sent to Omni Calculator to fix that bit of Javascript.
sunsettommy
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Nah it is more fun revealing their arguments here while it is my interest that the debate/discussion doesn’t run off the rails into overt bickering and personal attacks that sully’s this website.
Howdy
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Perhaps extend the recent comment list to 50 places so we can keep some semblance of track of other article replies during this, ‘my ego is bigger than yours’ fiasco?
James McGinn
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Kkook:
Ok, then reconcile your statement above with your other statement: “You are confused. Energy travels from higher density to lower density.”
JMcG:
I mispoke. I should have said net energy flow will be from higher density to lower density. Now do you get it?
Kkook:
Do you get the feeling that you’re badly out of your depth,
JMcG:
I can’t even imagine how frustrating it must be to be so sure you are right and so completely incapable of explaining how and why.
James McGinn / Genius
LOL@Klimate Katastrophe Kooks
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Jimbo “The Dunce” McGinn dribbled:
“I mispoke. I should have said net energy flow will be from higher density to lower density. Now do you get it?”
Then you would be wrong. Don’t feel bad, kooks often eschew reality and side with ‘fantasy fyzix’… you’re so prototypical you’re predictable. LOL
https://i.imgur.com/QErszYW.gif
Are you even capable of understanding that, Dunce? LOL