Another Botched Experiment – the Tricks of Pseudoscience
The RGHE fails the most basic laws of physics, math, science, and thermodynamics
A commentator, Max, posted this comment:
I came across this recent article by Hermann Harde (https://scc.klimarealistene.com/produkt/verification-of-the-greenhouse-effect-in-the-laboratory/) who seem to have done a proper experiment that shows that greenhouse effect “works as advertised” (although he admits there’s nothing to be worried about…).
I find it interesting because Harde is a climate realist and not an alarmist by any means. I think his experiment is the first that really tries to reproduce the GH effect, and apparently his results are supporting it…
…which is making me nervous, since I was pretty much certain that the GHE was debunked
Don’t worry, Max. We’ve been through all of this before.
Here is a link to the full pdf of the above experiment. What follows are my comments about the “experiment”:
“We measure the additional warming of a pre-heated plate due to back-radiation of the greenhouse gases carbon dioxide, methane and nitrous oxide as a function of the gas concentration, and we derive from the observed warming the radiative forcing of these gases.”
It really sounds like they have the conclusion they want already pre-determined. The experiment doesn’t look like it actually performs what needs to be tested, and for some reason they write-off the Woods experiment, and that is all that they would have had to do themselves is to repeat that.
They also then say that while the Woods-style experiment:
“answers the question how far a reduced IR transmission can warm a compartment {which was zero, as they admit}, it gives no information about interaction of greenhouse gases with IR radiation.”
And that is a load of bullocks right there. Here we have the quintessential alarmist talking point and pseudoscience, that physics behaves in a special way for GHG’s, and not the same way for other scenarios where the exact same physics should occur, such as Woods’ box experiment.
All you need to do to demonstrate the RGHE is Woods’ experiment, with a plate on top of a box. And these experiments always show no RGHE…they’re heated to what the Sun can do, and that’s it…
Max, ALL of thermodynamics is based on heat flow, and the equations for heat flow, which is all about the action from warm to cold. Cold does not radiatively or conductively increase the temperature of warm.
1) They dismiss the Woods’ experiments as not being relevant when they are precisely and perfectly relevant, providing all of the backradiation one requires to test the GHE, just as ALL greenhouse buildings in existence do. The Woods experiments use a passive ceiling, which can “emit backradiation” according to prescription, and they do not demonstrate backradiation heating.
Claiming that the exact same physics will work differently with a partially-emissive gas for the ceiling which cannot “backradiate” as well as a fully-emissive solid surface is, simply, pseudoscientific sophistry. A solid surface with full emission of “backradiation” doesn’t perform the RGHE at all, whereas a partially-emissive gas does? That’s not logical, rational, or could follow the existing laws of physics of emission.
2) They actually have two power sources, indicated by the supply feed to provide the temperature of the “Earth plate”, as well as the supply feed to provide the temperature of the “atmosphere plate”. This is no a longer a system which represents the Earth and atmosphere, and thus, the experiment isn’t even the correct experiment, anymore. The Woods-type experiment is the correct one, where you have a heated surface, and a passive receiver ceiling.
Their receiver, the “atmosphere plate”, is not passive, but has a power supply attached to it via the supply feed to provide its temperature. So that’s two independent power systems, which is not what the Earth experiences, nor what the Woods experiments do.
The real atmosphere does not have an independent power system or supply feed doing something to its temperature, and neither do the Woods experiments, and hence, you find consistency between the Woods experiments and the actual Earth & atmosphere: we do not ever measure temperatures above the heating potential of sunlight on the surface of Earth.
3) The moon’s surface gets hotter than Earth’s…it gets as high as the heating potential of sunlight. The Earth’s surface stays far cooler than the heating potential of sunlight.
4) We have the explanation for why the bottom of the atmosphere is its warmest part, due to the adiabatic gradient and the definition of an average,
So, there you go.
So… basically they threw out the Woods experiment, which is a good representation of the earth-atmosphere system, and substituted another experiment, which is not a good representation of the earth-atmosphere system?
More precisely: They threw out the Woods experiment claiming that it is not a good representation of the physics in question because it uses a solid ceiling for backradiation instead of a gas for backradiation, and substituted another experiment which they claim is a better representation of the Earth-atmosphere system because the new experiment has gas in it to create backradiation instead of a ceiling.
They admit that the Woods experiment does not show a RGHE, and so in this way they ignore it, because they imply that backradiation from a gas can do what backradiation from a solid ceiling does not do. They don’t answer or explain why backradiation from gas can create a RGHE but backradiation from a solid ceiling cannot, they merely take it for granted that you will accept that because their experiment has gas in it, then it is more like the Earth-atmosphere, hence a “more valid” experiment than Woods’.
They then also proceed to have two power sources managing the two temperatures involved, which is of course, not at all like the Earth-Sun system or Earth-atmosphere system…but because their experiment has gas in it then it is more like Earth and that’s all that matters.
It’s a complete train wreck. I’ve been through all this before. Whether they’re retarded, amateur, or hostile, they have no sense of experimentation, and that what you experiment upon is the underlying physics, the mechanism of the physics, precisely, and how it should occur, and what it should do by itself. The Woods’ experiment does all that.
They made up something that isn’t the thing any longer…even though they pretend it is by cutely using gas “which is like the atmosphere” instead of a ceiling, even though this would give better backradiation than a gas! But then the experiment is botched even if they were going to use a gas instead of a ceiling, because they use more than one power source…the cooler plate isn’t passive.
You see how sophistry works? How sophistry can trick science?
The Woods experiment represents the underlying physics involved, and in a more perfect way, because you can get full transmission of solar and full blockage and “backradiation” (per prescription, ostensibly) of IR.
Woods experiments, and all greenhouses, should easily demonstrate the RGHE, as they (should) perform the underlying physics involved more perfectly than the partially-emissive atmosphere.
They don’t perform the supposed underlying physics of the RGHE involved, because it doesn’t exist.
Ah, but – do you not find this following believable?:
“The Woods experiments use a solid ceiling for backradiation, whereas the Earth-atmosphere system actually uses a gas for backradiation.
Therefore, the Woods experiment doesn’t actually represent the Earth-atmosphere system, and therefore its results do not apply to the Earth-atmosphere.”
You see what was done there?
That being said, a gas “ceiling” and a solid ceiling would actually perform the same in terms of the underlying physics of “backradiation” ostensibly having some heating effect.
And so that is to say: a Woods experiment using gas as the ceiling would show the same results as with a solid ceiling, because the physics is actually still the same, and that result would be NO observation of a RGHE heating.
So what do you do about that, then?
You botch the experiment, and create an experiment that is no longer what you actually need to test in terms of the underlying physics, but pretend it is a better experiment “because it has gas which is like the atmosphere”, and hope that most people can’t tell what you botched…which most people, even scientists, wouldn’t.
See more here: climateofsophistry
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Geraint HUghes
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They climate crisis liars have the exact same problem with gases. CO2 gases do not induce back radiant warming either, not even on a 95% IR emitting filament.
https://www.youtube.com/watch?v=FgjT_665T6U
It also does not happen to the surface of a light bulb using CO2 gas. The entire science is completely false. It is therefore not science, it is fantasy. It is a farcical pretence at science created to deceive and defraud you.
https://theblackdragonsite.wordpress.com/2019/11/09/taller-tower-test-exposes-co2-back-radiation-nonsense/
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Herb Rose
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Bad experiments all.
The atmosphere Is being heated by the radiated energy from the sun. To deny this is to deny the validity of the laws of thermodynamics, which clearly state that ALL matter (including O2, N2, and the noble gases) absorbs radiated energy. and all matter above 0 K will radiate energy.
What energy is being absorbed depends on the structure of the matter.Atoms will absorb energy from the x-ray and shower spectrums. O2 and N2 absorb energy from the uv spectrum (which is why only 5% of uv reaches the surface. (with a concentration of 10 ppm there is no way that ozone can absorb 95% of the uv coming from the sun). Ozone is created when uv energy breaks an O2 molecule into 2 O atoms which then combine with O2 to form ozone)
Visible light is absorbed by the Earth’s surface while infra red is absorbed by CO2 and H2O..
This absorption of dIfferent energy wavelengths causes a separation of types of gases, containing different amounts of energy. At the highest altitudes in the atmosphere where the energy is greatest we find the matter with the greatest energy H and He,> He and O,> N2O2 and NO,> N2, O2, and O3,> N2, O2, Argon, CO2, H2O)
This change os energy causes a change of density and the way energy is transferred between matter. At high altitudes there are few collisions between matter and energy is transferred by radiation. As the energy decreases the frequency of collisions increase and the transfer of energy shifts to convection. In the stratosphere ozone is created by collisions between oxygen atoms and oxygen molecules.
In the troposphere collisions are frequent and energy equalizes quickly instead of the slow process of radiation.. Any experiment done in the troposphere, not done in a vacuum will have energy transfer being done by collisions, not radiation.
In convection there is no transfer of matter so any object with a greater velocity will transfer energy to an object with less velocity REGARDLESS OF THEIR MASSES. A high velocity gas molecule will transfer energy to an object with more mass even though that object has more kinetic energy. With convection cold can heat hot.
The surface of the Earth is being heated by one source (the sun) but by 2 different means, visible light and collisions with gas molecules heated bu uv. This is why the Earth and atmosphere cool when there is a solar minimum and less uv comes from the sun.
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Alan
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Your second paragraph says everything we need to know. It is astonishing that students do not know this. I assume because they have been brainwashed with nonsense. They look at images on infrared thermal cameras and see that air does not register any temperature and so they believe it does not absorb or emit radiation. They don’t understand that it is a characteristic of the measuring instrument.
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Jerry Krause
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Hi Herb,
And you just wrote: “The surface of the Earth is being heated by one source (the sun) but by 2 different means, visible light and collisions with gas molecules heated bu uv.” And you, in your comment, never mentioned the Sun’s infrared radiation.
Have a good day, Jerry
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Charles Higley
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CO2 only absorbs at three narrow IR wavelengths, equivalent ro something like -80, -200 and -400 deg C. Nothing on the planet is -200 and -400, so only something at IR equivalent to -80 would be emitted and absorbed by materials colder than -80. Another words, CO2 cannot warm anything in our real world.
However, CO2 can emit IR at these equivalent temperatures to other space. CO2 is a natural coolant, constantly sending IR that cannot be absorbed by Earth’s surface but can be lost to space.
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Herb Rose
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Hi Charles,
I believe the temperatures where CO2 absorbs energy are at -80 C, 412 C and over 700 C. -400 C is less than absolute zero (-273 C) and cannot exist anywhere.
Herb
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Jerry Krause
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Hi Allan, Herb, Geraint, and PSI Readers,,
Have you ever used the instrument termed the IR Thermometer? If not, get one and point it up at the sky, day and night. Point it down at an earth surface all hours of a day and night. During the daytime. point it at a direct sunlit surface and then point it at the your shadow right beside the sunlit surface. Go back and forth between sunlit and shaded surfaces. Point it at your forehead. And I am sure you will become convinced that this instrument is measuring something which it indicates is either F or C.
Have a good day, Jerry
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Jerry Krause
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Hi Fellows and PSI Readers,
I forgot to suggest, when the daytime sky has scattered clouds, that you point an IR-thermometer alternately at a cloud and then at the blue sky.
Have a good day, Jerry.
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sunsettommy
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Even if the AGW conjecture was correct it still couldn’t warm up the planet anyway since the postulated warm forcing increase doesn’t keep pace with the increased outflow of energy from the planet.
AGW warming is IMPOSSIBLE!
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Herb Rose
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Hi Sunsettommy,
All of the greenhouse gases combined cannot overcome the cooling by water, much less just those gases produced by man.
It takes 600 calories/gram to evaporate water. This water carries that energy up in the atmosphere where it is released into space during condensation. Since the amount of water in the atmosphere is 20 times the amount the green house gases, this means that every gram of greenhouse gases must be able to block more than 12,000 calories/gram/day from being radiated into space, in order to overcome the cooling done by evaporating water.
Herb
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Jerry Krause
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Hi Herb,
You just wrote: “It takes 600 calories/gram to evaporate water. This water carries that energy up in the atmosphere where it is released into space during condensation.” In the past you have written that water molecules, one by one, do not evaporate from a liquid water surface so there are no water molecules in the atmosphere to condense to form cloud droplets.
Which way is it???
Have a good day, Jerry
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Herb Rose
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Hi Jerry,
Exactly where in the comment did I say the water molecules were evaporating one by one? You extrapolate what you believe into what you read (if your re-read my other comment you will see where I said CO2 and H2O absorb infrared radiation.
In order to convert 1 gram of 100 C water into 1 gram of steam (individual molecules) you must add 540 calories. How can 1molecule of water gain that much more energy than the neighboring water molecules without that energy equalizing with those molecules? tThis is why a phase diagram of water shows at standard pressure water cannot exist as a gas below its boiling point.) We’ve gone through this with the tea kettle. The water comes out as invisible steam then condenses into visible water droplets on cooling. On further cooling these droplets again become invisible. If they’re cooling how can they be gaining 540 calories; gram to again become a gas? Except for steam initially escaping the tea kettle the water in the atmosphere is not a gas or individual molecules, but micro droplets (actually liquid crystals) and with further cooling these micro droplets will condense into larger droplets producing dew.
Herb
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Jerry Krause
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Hi Herb,
You ask: “Exactly where in the comment did I say the water molecules were evaporating one by one? ” When you wrote: “It takes 600 calories/gram to evaporate water.”
Have a good day, Jerry
Herb Rose
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Jerry,
1 molecule of water does not weigh 1 gram.
Jerry Krause
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Hi Herb,
You just wrote: “1 molecule of water does not weigh 1 gram.” I agree with you.
But I believe 602,300,000,000,000,000,000,000 molecules of water do weight 18 grams.
You seem to reason it easier for a 10 molecule particles to ejected from the surface of liquid water than for a single molecule to be ejected. But these 10 molecule particles are still a liquid. So it would seem it should take no more than 60 calories to “evaporate” a gram of water, or even less, because you reason it requires less energy to evaporate a particle of 10 molecules than to evaporate one molecule.
Have a good day, Jerry
Herb Rose
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Hi Jerry,
I believe that when water absorbs energy it forms liquid crystals where there is an outer shell formed by a hydroxyl ion and water molecules. In the center are hydronium ions (H3O+). As IR is absorbed more water molecules become ionized and the shell of the crystal grows thicker with an increased negative charge. (see Dr Gerald Pollack’s book) It is this negative charged shell that causes the liquid crystal to separate from the body of water (evaporate or sublimate) and rise in the atmosphere.
The calorie is a lousy unit of energy.It is defined as 1/100th the energy needed to raise 1 gram of 0C water to 100C but 1 added calorie to a gram of water does not mean a 1 C rise in temperature. It all depends on the initial temperature of the water, so you see a calorie is not a constant unit.
With water most of the energy added becomes internal energy and does not register as temperature. You add 80 calories to 1 gram of 0 C ice and there is no temperature increases. This is because that energy is breaking the internal bonds that form the ice.
Once the temperature reaches 100 C the temperature remains constant as more energy is added. When 540 calories/gram has been added the water changes state and becomes 100 C vapor. That energy doesn’t disappear, it is becoming internal energy, melting the liquid crystal structures. That causes me to believe that 540 calories/gram has been used to build those crystals and the amount of energy needed to convert 0 C water to 100 C steam is 1180 (540 making crystals, 100 radiated energy, and 540 melting crystals.)
It is because the internal crystal structure has been destroyed that boiled water (now a liquid) will freeze into ice crystals faster than unboiled water.
It is the peculiarities of water (it should be a gas according to its molecular weight) that cause me to believe that evaporated water is not a liquid but a liquid crystal and at the second melt point (liquid crystals have 2 melt points) it is converted into a liquid.
Herb
LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“It is the peculiarities of water (it should be a gas according to its molecular weight) that cause me to believe that evaporated water is not a liquid but a liquid crystal and at the second melt point (liquid crystals have 2 melt points) it is converted into a liquid.”
Interesting theory… water is such a weird molecule due to its hydrogen bonds that it may be so. But if the phase change from liquid crystal to liquid is endothermic, shouldn’t we see same when graphing the energy required to evaporate water? We should, then, see two ‘stair-steps’ rather than one in the graph of latent heat capacity.
Herb Rose
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Hi LOL
I would refer you to Dr. Gerald Pollack’s book “The Fourth Phase of Water” where they do experiments with the “Exclusion Zone” of water. The initial shell is formed with IR energy being converted to and stored as an electrical potential (negative shell, positive center). The continues growth of the crystal is an interior thickening of the shell creating more electrical charges. When the crystal melts it is from the outside in, so there is no energy release until the shell is breached.
I recommend the book to see the evidence and see if it makes sense.
Herb
Jerry Krause
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Hi Herb and PSI Readers,
At the beginning of Book III of the The Priniciplia, Newton listed four rules of reasoning in philosophy along with his commentary to illustrate his understanding of the rule. The following is translated by Andrew Mott.
Rule III. “the qualities of bodies, which admit neither intension nor remission of degrees, and which are found to belong to all bodies within the reach of our experiments, at to be esteemed the universal qualities of all bodies whatsoever.”
“For since the qualities of bodies are only known to us by experiments; we are to hold for universal all such as universally agree with experiments;” I stop here to ask: Where did “It takes 600 calories/gram to evaporate water” come from? Several times, I have quoted Louis Elzevir, the publisher of Galileo’s ‘Two New Sciences’, who, in a preface to a reader, wrote: “Intuitive knowledge keeps pace with accurate definition.” (as translated by Crew and de Salvio). So I am curious what a common definition of “evaporate” is. “verb—turn from liquid into vapor” (New Oxford American Dictionary). However, vapor can be a less than a definitive word. “vapor—noun—a substance diffused or suspended in the air, especially one normally liquid or solid”. So one can see why Herb might be confused.
For I consider a vapor might be a gas, the third phase of matter along with liquid and solid. “Gas—noun—a substance or matter in a state in which it will expand freely to fill the whole of a container, having no fixed shape (unlike a solid) and no fixed volume (unlike a liquid)”
Now, in his last comment I see that Herb has further confused our discussion by referring to the boiling of water. However, I do not need to turn to a dictionary to resolve this confusion.
Most of us have seen an uncovered pot of water boiling until the water in the pot boils to dryness. Most of us had probably seen that dew (liquid water) on a car’s surface evaporates to dryness at temperatures well below the boiling temperature of water. The latter is the experimental result we cannot ignore according to Newton.
Have a good day, Jerry.
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Steve Titcomve
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The AGW Alarmists attribute the whole 33`C to the presence of long wave infrared active gas molecules in the atmosphere and none to the naturally occuring adiabatic temperature gradient. If we were to dig out a quarry, 1km deep, the adiabatic temp gradient would continue and increase the new near surface air temperature at the bottom of the quarry by 10’C (if dry air) else 7’C (if water-saturated air). Could the AGW Alarmists conort their science to explain that this new temperature was also due to an increase in the proportion of LW infrared active gas molecules in the quarry?
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Herb Rose
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Hi Steve,
There is no adiabatic heating in the atmosphere. The atmosphere exist because the kinetic energy of molecules converts them to a gas and then, as more energy is added, the gas expands against gravity causing an increase in volume. Gas molecules losing energy contract because of gravity, they do not gain kinetic energy.
If the molecules at the bottom of quarries had more kinetic energy the gas would increase in volume, become less dense, and rise to the top.
Energy is transferred to a thermometer by collisions with air molecules. It is not measuring the kinetic energy of the molecules but the momentum (mv) being transferred to it. Because momentum is a product of mass and velocity the reading is the result of two variables and more cooler molecules can transfer more energy to the thermometer than fewer molecules with more energy, resulting in the cooler molecules recording a higher temperature.
Herb
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LOL@Klimate Katastrophe Kooks
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Herb Rose wrote:
“There is no adiabatic heating in the atmosphere.”
Incorrect. That’s what Maxwell claimed in his “Theory of Heat“, and Boltzmann sided with Maxwell… Loschmidt proved them both wrong. In an atmosphere sufficiently dense such that gravitational auto-compression can occur, a temperature lapse rate will always evolve due to adiabatic tradeoff of gravitational potential energy for translational mode (kinetic) energy of the atmospheric atoms and molecules. This increases specific energy and thus energy density and thus temperature (given that temperature is a measure of energy density, equal to the fourth root of energy density divided by Stefan’s Constant).
The reason the air in the bottom of a quarry (or any hole where air cannot easily advect) doesn’t convect is because it’s a balance of gravitational potential energy and translational mode (kinetic) energy which keeps the air there even though it’s warmer.
The Grand Canyon, though, is a feature where air can flow laterally (ie: advect) at the bottom of the canyon… it picks up moisture from the river, becomes slightly more buoyant, it wants to rise but it cannot due to the downwelling air coming over the canyon edges, so it flows upstream.
The downwelling air continues downwelling because it is losing energy via radiative emission as it descends (the temperature increase is great enough to put CO2 into its ‘net coolant’ (radiative) mode rather than its ‘net warmant’ (thermalization) mode, so it keeps packing more air into the canyon and flowing out of the canyon mouth… the Grand Canyon is the world’s largest natural cooler. The radiative emission is so great that one can see the outline of the canyon in IR from space.
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Herb Rose
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Hi LOL,
The thermometer is inaccurate in a gas like the atmosphere. It does not measure the kinetic energy of molecules but the momentum they transfer to it by convection. It was designed to measure the flow of energy from one medium to another and cannot give an accurate indication of temperature when a change in energy produces a change in mass (fewer molecules). The zigzag pattern the thermometer records in the atmosphere is not how energy flows. In order to get an accurate indication of the kinetic energy of the molecules at an altitude you must divide the temperature by the density at that altitude (kinetic energy per constant number of molecules.) This shows the kinetic energy of molecules increases with increasing altitude and it is the sun, not the surface of the Earth, that is heating the atmosphere.
There is no potential energy due to gravity. Gravity is not a product of mass (produces inertia) but of the energy (produces motion) associated with matter. This energy decreases with distance from the source (Keppler’s law) and orbiting objects are not being pulled towards an object but are in equilibrium with its radiated energy.
The speed of light is not constant but is a wave traveling in the energy fields (magnetic and gravity) and matter fields (electric) radiated by objects. The speed of light varies with the strength of these fields. The red, blue, and purple shift of light coming from distant stars proves relativity (e=mc^2) is wrong.
Herb
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LOL@Klimate Katastrophe Kooks
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Ok, we agree on some, and disagree on some.
I agree that the speed of light is not a constant. Einstein himself stated as much many, many times. It was a necessity for his first attempt, Special Relativity; but for General Relativity, it’s necessary that the speed of light varies with the refractive index of the medium through which it is transiting… and the quantum vacuum is a refractive medium. Only in vacuum and at a constant gravitational acceleration is the speed of light constant… except that’s nowhere in the universe due to the Inverse Square Law and the fact that gravity is a long-range interaction, except in a laboratory.
I agree that thermometers become ‘inaccurate’ at higher altitudes… not really ‘inaccurate’, though… at lower altitude, it’s measuring a mean of the translational mode (kinetic) energy of the atoms or molecules and the loss of energy radiatively is comparatively negligible. The mean because the thermal mass of the thermometer necessarily ‘smears out’ the energy each collision imparts to or takes away from the thermometer.
At higher altitudes, with sparser collisions, the ratio of energy obtained / lost via collisions to the ratio of energy gained / lost via radiation reduces, so the thermometer becomes less accurate for kinetic temperature simply because it’s gaining less energy via collision and losing proportionally more energy via radiation.
But you bring up a good point… the ‘temperature’ measured at altitude is compared to the temperate measured at surface, but they’re not comparable due to the fact that the above-described ratio decreases. That’s why the thermosphere has atoms and molecules with very high kinetic temperature, but you’d still freeze to death radiatively.
So we should be measuring the energy density of the atmosphere, rather than the temperature. Energy only flows down an energy density gradient, and we can still use the S-B equation in energy density form, but that may reveal something new which we didn’t know.
Temperature is a measure of energy density, equal to the fourth root of energy density divided by Stefan’s Constant… but that’s at Earth’s surface. It may be that Stefan’s Constant is yet another constant which isn’t constant.
It may be that the lapse rate evolves simply because as air density decreases with altitude, energy density decreases with altitude and thus that allows energy to radiatively flow warmer (surface) to cooler (space), the lapse rate predicated upon the magnitude of that flux. Or it may be a balance between that and gravitational potential energy / kinetic energy.
You’ll have to provide mathematics to back up your contention of the Mass-Energy Equivalency being incorrect.
E^2 = p^2 c^2 + m^2 c^4
Herb Rose
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Hi LOL
The photon (particle nature of light) has no basis in reality. The photoelectric effect (which was its genesis) is just another example of the piezo electric effect where a change in an ionic or metallic bond (due to pressure or changing electromagnetic field) releases an electron which is already disassociated from an atom.
Energy and matter are the two separate building blocks of the universe and the forces each produces (gravity, magnetic for energy and electric for matter) behave in opposite manners.
When two opposite magnetic poles get nearer the radiated magnetic field increases in size and strength (decreasing the internal force (fewer magnetic flux lines)). When two opposite charges get nearer their electric fields decrease in size and strength as the internal repelling force increases. Conversely when two similar magnetic fields get nearer the size and strength of the radiated magnetic field decrease (as the internal repelling force increases) but when two similar charges approach the size and strength of the radiated electric field increases as internal force becomes a radiated force.
This opposite behavior means matter and energy, although they interact, are not the same and also provides a mechanism for light. When energy is added to matter it causes its attraction to neighboring matter to increase pulling the energy fields closer. This results in the negative electric fields to become closer, increasing the repelling force between the objects. When that energy decreases, due to energy flow (equilibrium) the repelling electric force moves the objects (their fields not their matter) apart setting up an oscillation (light) that is transmitted to all neighboring objects.
Einstein was wrong.This eliminates the contradiction that as objects move closer, creating a stronger gravitational field, time expands which, because the speed of light is constant, results in an increase in distance between the objects and the objects moving further apart as they move closer together
Herb.
LOL@Klimate Katastrophe Kooks
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Herb,
I’m not really understanding your textual descriptions. Might you have mathematical or graphical representations of the processes you describe, to make it more intuitive and easily grasped?
I’m intrigued because what you state may be adjunctive to what we’ve uncovered over at Joe Postma’s website, namely that elementary charge has Earth’s gravitational acceleration baked-in due to the Millikan and Fletcher oil-drop experiment balancing electrostatic force and gravitational acceleration to arrive at elementary charge); thus that Faraday’s Constant (which is predicated upon elementary charge) has Earth’s gravitational acceleration baked-in and thus cannot be a constant anywhere except Earth’s surface; that the speed of light cannot be a constant except under a constant gravitational acceleration (which doesn’t really exist from one point in space-time to the next owing to the fact that gravity obeys the Inverse Square Law and is a long-range interaction); and I suspect Stefan’s Constant (ie: the radiation constant) is yet another constant that isn’t really a constant except on Earth’s surface because it presumes the conditions at Earth’s surface… the atmospheric density and thus the ratio between time to collision and thermalization vs. the time to radiative relaxation. In the thermosphere, as one example, that ratio certainly doesn’t hold, which is why the particles (atoms / molecules) have a very high kinetic temperature but the measured thermal temperature is low… because radiative emission dominates rather than thermal collisions.
Herb Rose
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Hi LOL,
Math is not my thing so I don’t know how to adjust my description so you would understand.
Let me give you the basic of my beliefs and see if you can put into math.
The universe is composed of two components, energy and matter, with associated forces. Energy has a radiated attractive force (gravity which a general force and magnetism where the force is converted to a directional force by electrons) and this force creates movement. Matter has an internal force (electric) which creates inertia.
Energy is attracted to positive matter and since it is a stronger force (probably by a factor of the golden constant) it is able to separate matter into positive and negative charges. (The neutron is a basic matter molecule but because it has an exposed electron when it moves through an energy/magnetic field, that field can separate it into a proton, electron, and energy (gamma ray). An alpha particle (another matter molecule) has its 2 electrons surrounded by 4 protons so is stable in a magnetic/energy field despite its internal repelling force.
When matter is separated into charges by energy it creates atoms where circulating negative charges create a directional magnetic force from the radiated attractive force. As long as the outer surface of a nucleus has only positive charges (a compressing force, not a binding force) the atom will be stable but if there are exposed electrons the atom will decay through beta emissions or being split into atoms with nuclei where no electrons are exposed. There is no force of gravity (radiated force surrounding the nucleus), no strong nuclear force (concentrated compression by energy), no weak nuclear force (Electric force of matter) only an electric force from matter and an attractive force (gravity/magnetic) from energy.
These forces decreases with distance which means their strength decreases as an area (distance squared) and it is the balancing of these force as things move that create the universe.
I have done an experiment with magnets (written about in PSI) where I make a model of a permanent magnet and show that the correct formula for a magnetic force is not F= M1M2/d^2 but M3= (M1 + M2)/d where d is the distance from M1 to the magnetic field (equilibrium point) of M2. You might want to construct a model and repeat the experiment to see if you get a similar interpretation of the results.
Herb
LOL@Klimate Katastrophe Kooks
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Herb,
I kind of get what you’re saying… you’re basing your hypothesis upon charge and energy interaction.
That’s not really the correct way of looking at it. Like charges repel and opposite charges attract, of course, but the neutron doesn’t have an exposed electron.
Think in terms of energy density and energy density gradient… a bare neutron has higher energy density than the quantum vacuum and thus is unstable and will beta decay into a proton and W- boson (which decays to an electron and electron antineutrino).
Conversely, a bound neutron is ‘shielded’, the local quantum vacuum energy density in the vicinity of the nucleus is less than the bulk quantum vacuum energy density and thus the neutron is stable in a nucleus.
Remember that energy can only flow if there is an energy density gradient, and it is all about energy flow.
So a bound electron, for example, in an excited state is in that state because the local quantum vacuum energy density has been increased. The bound electron is always trying to emit a photon to de-excite, but it cannot because of that locally higher quantum vacuum energy density. When the local quantum vacuum energy density decreases (ie: the energy which was exciting that bound electron) is removed, the bound electron can then emit a photon because there is an energy density gradient. The bound electron will fall to a state which has an integer number of de Broglie waves in its orbital and which is such that the next lower state is lower than current local quantum vacuum energy density. At ground state, energy flows from the quantum vacuum to sustain the bound electron at its orbital radius as the bound electron emits Larmor radiation (a point charge undergoing acceleration (angular acceleration, in this case) in relation to its electric field will emit Larmor radiation in the form of virtual photons). Virtual photons mediate the magnetic interaction, which is why all elements exhibit some form of magnetism (usually diamagnetism, although certain valence electron configurations allow ferromagnetism to override the underlying diamagnetism).
Thus it is the energy density of the ground-state quantum vacuum which sustains the bound electron in its ground state, preventing it from ‘spiraling-in’ to the oppositely-charged nucleal proton(s).
[1] https://sci-hub.se/10.1103/physrevd.11.790
[2] https://web.archive.org/web/20190713220130/https://arxiv.org/ftp/quant-ph/papers/0106/0106097.pdf
[3] https://web.archive.org/web/20190713225420/https://www.researchgate.net/publication/13330878_Ground_state_of_hydrogen_as_a_zero-point-fluctuation-determined_state
“We show here that, within the stochastic electrodynamic formulation and at the level of Bohr theory, the ground state of the hydrogen atom can be precisely defined as resulting from a dynamic equilibrium between radiation emitted due to acceleration of the electron in its ground-state orbit and radiation absorbed from zero-point fluctuations of the background vacuum electromagnetic field, thereby resolving the issue of radiative collapse of the Bohr atom.”
[4] https://web.archive.org/web/20180719194558/https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150006842.pdf
“The energy level of the electron is a function of its potential energy and kinetic energy. Does this mean that the energy of the quantum vacuum integral needs to be added to the treatment of the captured electron as another potential function, or is the energy of the quantum vacuum somehow responsible for establishing the energy level of the ‘orbiting’ electron? The only view to take that adheres to the observations would be the latter perspective, as the former perspective would make predictions that do not agree with observation.”
Once one realizes that no action can take place without an impetus, that energy density is equivalent to pressure (and is in fact radiation pressure… with the same units as pressure), that the impetus is a pressure gradient (energy density gradient), then everything becomes much clearer.
LOL@Klimate Katastrophe Kooks
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Mheh…
“Conversely, a bound neutron is ‘shielded’, the local quantum vacuum energy density in the vicinity of the nucleus is less than the bulk quantum vacuum energy density and thus the neutron is stable in a nucleus.”
-should be-
“Conversely, a bound neutron is ‘shielded’, the local quantum vacuum energy density in the vicinity of the nucleus is greater than the bulk quantum vacuum energy density and thus the neutron is stable in a nucleus.”
Herb Rose
| #
Hi LOL,
I am one who believes in Occam’s razor and that energy and matter cannot be created or destroyed.
When an electron and proton come together it creates a neutron and releases energy. A neutron not surround by the magnetic field of an atom will split into an electron and proton and release energy (gamma radiation) in ten minutes. To me this means that the continual forming and destruction of neutrons is creating energy and violating the laws of thermodynamics.
My solution to this is that the neutron is a molecule composed of an electron and a proton (no quarks with their Baskin Robbins list of flavors.) and the opposite charges do not cancel each other out, but only mask each other. Because of the small distance between them the two fields become undetectable at any distance from the molecule but both the negative and positive fields remain.
As energy gives the molecule motion through an energy field it creates two currents going in opposite directions. The right hand rule dictates that these two currents would be forced in opposite directions by the magnetic field, creating a shearing force between the negative and positive fields which would eventually split the molecule and release the gamma ray. There is no creation of energy only absorption of energy from the energy field which breaks the molecule into its components.
Most people agree that I am simple minded.
Herb
LOL@Klimate Katastrophe Kooks
| #
Steve, you’ve got a naturally-occurring feature which perfectly fits your scenario… the Grand Canyon. That, of course, proves that CO2 is not a ‘heat trapping, global warming’ gas, but is instead a net atmospheric radiative coolant above its ‘transition temperature’.
TLDR: In an atmosphere sufficiently dense such that collisional energy transfer can significantly occur, all radiative molecules play the part of atmospheric coolants at and above the temperature at which the combined translational mode energy of two colliding molecules exceeds the lowest vibrational mode quantum state energy of the radiative molecule. Below this temperature, they act to warm the atmosphere via the mechanism the climate alarmists claim happens all the time (thermalization), but if that warming mechanism occurs below the tropopause, the net result is an increase of Convective Available Potential Energy, which increases convection, which is a net cooling process.
Note that the term ‘transition temperature’ here used isn’t related to phase change, but is related instead to when a radiative molecule changes from a ‘net warming’ (thermalization) to a ‘net cooling’ (radiative emission) role.
Temperature at the canyon bottom is consistently ~15 – 25 F warmer than the North Rim or South Rim temperature, and air flows out of the canyon on the upstream end (except on days when there’s a S-SW prevailing wind).
Keep in mind this is in a canyon nearly a mile deep… not much sunlight reaching the bottom in the North Rim / South Rim region because its orientation is roughly east-west and the walls are pretty steep.
When the climate alarmists describe thermalization and claim it’s what causes CAGW, they only tell half the story. It’s a half-truth to hide their bigger lie of CAGW.
All radiative molecules are dual-role molecules… they can act to warm the atmosphere at atmospheric temperatures below their ‘transition temperature’ (in the case of CO2, via absorption of a 14.98352 um photon to become vibrationally excited, then conversion of that vibrational mode energy to translational mode energy of other atmospheric molecules upon molecular collision), or cool the atmosphere at atmospheric temperatures above their ‘transition temperature’ (via collisional transfer of translational mode energy of other atmospheric molecules to the vibrational mode energy of CO2, which is then radiatively emitted).
The ‘transition temperature’ of any given radiative molecular species is dependent upon the differential between:
1) the combined translational mode energy of two colliding molecules,
-and-
2) the lowest vibrational mode quantum state energy of the radiative molecule.
When 2) > 1), energy flows from vibrational mode to translational mode, which is a warming process.
When 1) > 2), energy flows from translational mode to vibrational mode, which is a cooling process.
Compression heating and radiative emission have created a ‘wind tunnel’ effect whereby the warmer compression-heated (in accord with the Ideal Gas Law) air at the canyon bottom puts CO2 into its ‘net-coolant’ mode. The CO2 is vibrationally excited by the translational mode energy of the other atmospheric molecules because more atmospheric molecules carry sufficient kinetic energy to vibrationally excite CO2.
The CO2 emits so much radiation from the canyon bottom that one can trace out the profile of the canyon just from its radiation signature as seen from space.
That loss of translational mode energy of the atmospheric molecules (which flows to the vibrational mode quantum states of CO2) causes the air to become denser, and it flows along the canyon bottom upstream, which is why Page Airport has a wind predominantly to the N-NE, and why the town there is warmer than surrounding areas.
The airflow out of the upstream end of the canyon pulls more air over the canyon rims and down into the canyon, heating it via compression, whereupon the process repeats.
https://i.imgur.com/CxVTcro.png
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LOL@Klimate Katastrophe Kooks
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I wrote the below over on Joe Postma’s website in response to ‘Rosco”s comment about Pictet’s “cold transmission” experiment which proved that energy only flows from higher to lower energy density.
Rosco wrote:
“Most physicists, on seeing it demonstrated for the first time, find it surprising and even puzzling.”
They find it “surprising and even puzzling” because the education they’ve received hasn’t instilled in them the necessary foundational knowledge such that they can instantly recognize it for what it is… energy flowing from the warmer temperature measurement device at the focal point of one mirror to the cooler object at the focal point of the other mirror (and in losing energy, the temperature measurement device registers a lower temperature), said mirrors creating a view factor wherein object and temperature measurement device can only ‘see’ each other and nothing else.
Pictet proved as much 222 years ago, and the warmists have yet to pick up on that reality… because embracing ancient and outdated information (such as the long-debunked Prevost’s Principle from 1791 which claims that energy can flow willy-nilly without regard to the energy density gradient because only an object’s internal state determines radiant exitance, none of which is true) is fundamental to advancing CAGW… without that, the warmists have nothing.
In so proving, Pictet proved that energy only flows down an energy density gradient unless external energy does work upon that system energy to push it up the gradient (as happens in, for example, your refrigerator).
Most people cannot think in terms of energy, energy density and energy density gradient. We need to analogize to something they’re familiar with.
Just as I’ve been stating, energy density and pressure have the same units, energy density gradient and pressure gradient have the same units. Thus, just as, for instance, water only spontaneously flows down a pressure gradient, energy only spontaneously flows down an energy density gradient.
That’s 2LoT, in a nutshell.
So one tack to take is to ask people if water can ever spontaneously flow uphill. Of course they’ll say, “No, water cannot flow uphill on its own.”
Then show them dimensional analysis:
Force: [M1 L1 T-2]
Area: [M0 L2 T0]
Pressure: [M1 L-1 T-2]
Length: [M0 L1 T0]
Pressure Gradient: [M1 L-2 T-2]
Explain to them that pressure is the result of Force / Area. That Pressure Gradient is Pressure / Length.
Remind them that water only spontaneously flows down a pressure gradient.
Then introduce energy. Tell them that energy is much like water. It requires an impetus to flow, just as water requires an impetus (pressure gradient) to flow. In the case of energy, that impetus is an energy density gradient, which is analogous to (and in fact, literally is) a radiation pressure gradient.
Energy: [M1 L2 T−2]
Volume: [M0 L3 T0]
Energy Density: [M1 L-1 T-2]
Length: [M0 L1 T0]
Energy Density Gradient: [M1 L-2 T-2]
Explain to them that Energy Density is Energy / Volume, and Energy Density Gradient is Energy Density / Length.
Highlight the fact that Pressure and Energy Density have the same units (bolded above). Also highlight the fact that Pressure Gradient and Energy Density Gradient have the same units (bolded above).
So we’re talking about the same concept as water only spontaneously flowing down a pressure gradient (ie: downhill) when we talk of energy only spontaneously flowing down an energy density gradient.
And since a warmer object will have higher energy density at all wavelengths than a cooler object:
https://i.imgur.com/kS20QG1.png
… ‘backradiation’ can do nothing to warm the surface because energy cannot spontaneously flow from lower to higher energy density, and thus CAGW is nothing more than a complex mathematical scam perpetrated to obtain multiple billions of dollars in funding for trough-grubbing line-toeing ‘scientists’ and to push a Marxist One World Government “Build Back Better” agenda.
Watch heads implode as people realize they’ve been lied to, that CAGW is a tyrannous cake baked of distortions layered upon fabrications and topped with biggity prevarication. Watch cognitive dissonance in action for those who refuse to believe their government would lie to them. Watch the infantilized who truly believe their government has the best interests of the people closest to heart attempt to reconcile their brainwashing-from-birth with reality. And, perhaps most importantly, watch climastrologists and warmist physicists scramble for the exits.
The base physical quantities in the MLT dimensional analysis system are:
length (L)
mass (M)
time (T)
electric current (A)
thermodynamic temperature (K or θ)
amount of substance (N)
luminous intensity (J)
From there, you can derive quantities:
https://i.imgur.com/1LPkeMB.png
In fact, there are a lot of derived units:
https://climateofsophistry.com/2022/07/19/another-botched-experiment-the-tricks-of-pseudoscience/#comment-107931
The above is the short list… I cannot find the extensive list that used to be online which listed every single derived quantity known to man.
If the formula you’re using divides, then you subtract the bracketed dimensions:
E = m c^2
m = E / c^2
Energy: [M1 L2 T-2 A0 K0 N0 J0]
Velocity: [M0 L1 T-1 A0 K0 N0 J0]
Velocity: [M0 L1 T-1 A0 K0 N0 J0]
And that gives:
M1 – M0 – M0 = M1
L2 – L1 – L1 = L0
T-2 – T-1 – T-1 = T0
Mass: [M1 L0 T0 A0 K0 N0 J0]
If the equation you’re using multiplies, you add the bracketed dimensions:
E = p c
That’s momentum:
[M1 L1 T-1 A0 K0 N0 J0]
And velocity:
[M0 L1 T-1 A0 K0 N0 J0]
[M1 L1 T-1 A0 K0 N0 J0]
[M0 L1 T-1 A0 K0 N0 J0]
That gives:
Energy: [M1 L2 T-2 A0 K0 N0 J0]
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MattH
| #
By this time, he had assisted Saussure with an experiment that demonstrated the existence of what would later be called infra-red radiation.
https://en.wikipedia.org/wiki/Marc-Auguste_Pictet
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LOL@Klimate Katastrophe Kooks
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You want to see something entirely off the hook?
Start reading here, and continue to the end:
https://climateofsophistry.com/2022/05/07/ontological-mathematics-the-theory-of-everything-5-the-mind-and-the-brain-consciousness-and-god/#comment-103349
Lots of correlations uncovered, some invalidation of physical laws and principles, some corroboration of Einstein’s theory of Relativity, but mostly those odd correlations.
For instance:
Radiant Exitance =
= ε σ (T^4_h – T^4_c) Ah
= (ε c (e_h – e_c)) / 4
= Faraday’s Constant (Ah) * (Molar Mass of Atmosphere / Ratio of Atoms Per Particle)
= Faraday’s Constant (Ah) * Moles of electrons per Mole of atmospheric particles
= (((Faraday’s Constant (J electron-1) * Number of Electrons per Mole of Atmosphere) / 3600) / Stefan-Boltzmann Constant)^0.25
= (((Faraday’s Constant (J mol-1) * Moles of Electrons per Mole of Atmosphere) / 3600) / Stefan-Boltzmann Constant)^0.25
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LOL@Klimate Katastrophe Kooks
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Those aren’t approximate correlations, either… for the calculations I did, they were exact to 23 decimal places.
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LOL@Klimate Katastrophe Kooks
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Also remember that temperature is a measure of energy density, equal to the fourth root of energy density divided by Stefan’s Constant:
e = T^4 a
a = 4σ/c
e = T^4 4σ/c
T^4 = e/(4σ/c)
T = 4√(e/(4σ/c))
T = 4√(e/a)
Where:
e = energy density
a = Stefan’s Constant
T = temperature
σ = Stefan-Boltzmann Constant
c = speed of light
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LOL@Klimate Katastrophe Kooks
| #
T = 4√(e/(4σ/c))
T = 4√(e/a)
should be –
T = 4^√(e/(4σ/c))
T = 4^√(e/a)
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Nepal
| #
Why does anyone put any stock whatsoever into what Postma claims? He can’t even calculate the steady state temperatures for two blackbody plates with one plate heated by a distant constant flux point source correctly.
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Nepal
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Should we review some of Postma’s greatest hits? Let’s see, he didn’t know that acceleration is the TIME derivative of velocity, and he claims that objects cannot emit heat to space. For the two plate problem he claims that both plates are at the same temperature which would preclude any heat flux through the system. He’s had this error pointed out to him, yet he is not competent enough to understand it.
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Nepal
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Now Postma and LOL@Klimate Katastrophe Kooks are having conniptions over on Postma’s site because they can’t refute anything of the facts that I have stated. Postma can’t calculate the proper solution to the plate problem, he couldn’t calculate a simple acceleration problem, and he claims that objects cannot emit heat to space. He’s deathly afraid to even attempt to defend himself on a neutral site because he can’t back up his blather with rigorous mathematics. Heads are exploding over on his site.
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Moffin
| #
conniption
kə-nĭp′shən
noun
A fit of violent emotion, such as anger or panic.
An attack of hysteria; a fit of rage or vexation.
An attack of hysteria; a fit of rage or vexation.
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Nepal
| #
Indeed.
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CD Marshall
| #
Alex,
You’ve been debunked on multiple sites, at this point it is appearing you are a bot. Anyone who thinks a photon heats space should have their degree revoked (if you even have one).
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Nepal
| #
I don’t recall saying that a photon heats space. Let’s see if you are able to grasp simple concepts CD. The first law of thermodynamics states, dU = W + Q. We all agree that an object at say 300K left out in deep space will cool. That means that dU for the object is negative. Next space is not doing any work on the object and so W=0. All that leaves is Q. Since dU is negative that means the Q must be negative. A negative Q means that the object is transferring heat to space. It’s really not a difficult concept CD. Are you able to get it?
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Nepal
| #
Here CD, I will educate you. Note that all calculations are for steady state. That means that the temperatures do not change and so dU=0 for all plates. Postma ignorantly believes that a system that is passing heat through it can be in equilibrium. He is of course wrong. Heat transfer is a non-equilibrium process, be it by conduction, convection, or radiation. Since Postma incorrectly claims that no heat can be transferred to space, he must modify the first law to allow for objects to cool in space. His version of the first law is:
dU = W + Q – sigma x n x A x T^4
This is fine for thin plates where it is reasonable to assume that the temperature in the plate is uniform. Note also that A in the formula is the area enclosed by the perimeter of the sides of the plate, e.g. a thin 1m x 1m plate has an area of 1 m^2). Then, n is the number of sides of the plate that are emitting to space, and will be 0, 1, or 2. An isolated plate emits to space from 2 sides. Two plates facing one another each emit to space from 1 side, and for three plates the middle plate would not be emitting to space (under the simplifying assumption that the plates are close together and the view factor of the middle plate to space is approximately zero).
So let’s apply this to the steady state analysis of a single isolated plate being heated by a point source to its right. Let’s also neglect edge effects to make the analysis easier. Note that no work is done on the plate and so
dU = 0 = 0 + q A dt – 2 sigma A T^4 dt
Here q is the heat flux supplied by the source at the location of the plate. Its SI units would be W/m^2. A is then the area of the side of the plate being irradiated by the source and dt is the time interval (which is irrelevant at steady state).
Solving for T gives, T = [q/(2 sigma)]^(1/4)
This is consistent with what Postma posts on his blog, so at least he was able to get that right. Unfortunately, he was unable to apply the same principles to the two plate system and got that one wrong. Here is how to solve that correctly. Let’s call plate 1 at T1 the plate on the right being irradiated on its right side by the distant point source. Plate 2 at T2 will be on the left. Again, we are concerned with steady state and so dU1=0 and dU2=0.
For plate 1:
dU1 = 0 = 0 + q A dt – sigma (T1^4 – T2^4) A dt – sigma T1^4 A dt
the term sigma (T1^4 – T2^4) A dt is the heat transferred from plate 1 to plate 2. The last term sigma T1^4 A dt is the energy emitted by the right side of the plate to space. For plate 2 we have:
dU2 = 0 = 0 + sigma (T1^4 – T2^4) A dt – sigma T2^4 A dt
Again, the term sigma (T1^4 – T2^4) A dt is the heat transferred from plate 1 to plate 2. The term sigma T2^4 A dt is the energy emitted to space from the left side of plate 2. These are the first law equations according to Postma’s modification to account for energy emitted to space. Now we can solve for T1 and T2.
Anyone that can perform simple algebraic manipulations can see that the solution is:
T1 = [2 q/(3 sigma)]^(1/4)
T2 = [q/(3 sigma)]^(1/4)
Postma incorrectly claims that T1 and T2 must be equal. How he ever earned a master’s degree in physics is beyond me. He can’t perform simple heat transfer calculations, he couldn’t do a simple acceleration problem, and he barely understands dimensional analysis. You should take stock of who you try to learn physics from CD. Let me guess, you will bury your head in the sand and call me a bot.
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Joseph E Postma
| #
That’s already been covered, and your insane anti-mathematical ramblings debunked long ago:
Click the link on my name.
Now go cope and seethe more…lol. Cope, and seethe! haha
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Nepal
| #
Sorry Joseph, but it has not been covered. Your word salad does not override the math.
If you want to refute what I have written above then you have to show with that I have made an algebra error, which I haven’t, or you have to explain which of my equations is wrong and WHY.
You claiming that it has already been covered is the epitome of weakness. You can’t handle the fact that I have proven your nonsense to be wrong.
So either deal with refuting my rigorously presented comment above like an actual scientist or admit your error.
Stop hiding behind your blog where you control what can and cannot be posted.
Immortal600
| #
So you are using another moniker stolen from someone else. You are fooling no one, ‘evenminded’. You keep proving to be a dishonest idiot. Joe Postma and LOL@KKK have both destroyed you on numerous occasions and it’s fun to watch.
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LOL@Klimate Katastrophe Kooks
| #
Now, you two… don’t be too hard on Not-Professor Unevenminded… the poor ineducable kook who claims to be a professor at a midwestern university teaching thermodynamics and mechanical theory, but who’s been wrong on every single topic to date, racking up an impressive “80-some-odd” (his words) errors, all of which I’ve archived for posterity.
He can’t help it, he’s not a well man… obviously that tanker truck backed up to his lean-to hovel, pumping psychotropic medication through a fire hose straight into his jugular, isn’t helping him at all. Nor has it, for years.
Leave the poor fool to squat in the squalor of his run-down, rented single-wide as his slow flaming spiral toward inevitable babbling insanity proceeds.
Just laugh at him like you would a hare-lipped water-brained half-wit, pat him on his lumpy misshapen head, boot him a bit on the backside to send him on his way, and saunter away chuckling at the thought of such a pathetic being thinking (FSVO ‘thinking’) he has any grasp upon reality whatsoever. LOL
Nepal
| #
Postma and Kooks have not destroyed anyone. I have provided the rigorous mathematical derivation above that proves they are wrong. Your whining is not surprising, but like them you are not able to actually refute anything that I have written.
LOL@Klimate Katastrophe Kooks
| #
Not-Professor Unevenminded dribbled:
“Postma and Kooks have n”{TARDSMACK}
Poor little kook. Run along now, before the intellectual giant lumbers over, pinches your misshapen soft-shelled skull between thumb and forefinger, and squeezes. Every time I do that to you, you get a bit dumber, but a bit more hilarious in your frantic deluded rantings. LOL
Remember when I had you so perturbed that all you could type for hours was “DISMISSED!!!!“, and near the end (before you ran away to nurse your savaged psyche) you were so addlepated that you were misspelling even that? LOL
Nepal
| #
Everyone can see once again that neither Kooks nor Postma has refuted the calculation provided. They are afraid of math. They cannot refute the calculation and so they resort to insults and ad hominem.
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“Heat transfer is a non-equilibrium process, be it by conduction, convection, or radiation.”
LOL@Klimate Katastrophe Kooks wrote:
“no energy flows at thermodynamic equilibrium”
Professor BalloonKnot dribbled:
“This is nonsense.”
Deluded kooks often self-contradict… it is their most predictable attribute. LOL
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Nepal
| #
“Heat transfer is a non-equilibrium process”
This is correct. Heat transfer is a non-equilibrium process.
“no energy flows at thermodynamic equilibrium”
This is incorrect. At equilibrium there is energy flowing in all directions. The example of the pressure on the walls in your rooms proves this is physically true. Thermodynamic equilibrium does not state that there are no energy flows, it says that there is no NET energy flow. It is the NET radiation transfer that is the radiative HEAT transfer. That you are not intelligent enough to understand this is astounding.
“This is nonsense.”
Indeed, what you are writing is nonsense. You don’t seem to be able to grasp what NET means.
There is no self-contradiction in anything that I have stated.
Joseph Postma: “a cool object emits toward a hotter source”.
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
““Heat transfer is a non-equilibrium process”
This is correct. Heat transfer is a non-equilibrium process.
“no energy flows at thermodynamic equilibrium”
This is incorrect. At equilibrium there is energy flowing in all directions.”
Yes, folks, he really is that stupid. LOL
‘Heat’ is defined as energy in motion, a flow of energy, a flux of energy… not the kook’s redefinition of a ‘net flow’ of energy, as that violates the Principle of Least Action and the Entropy Maximization Principle.
So the kook’s just essentially stated:
“Energy flow is bidirectional, energy flow is unidirectional.”
And he does so completely unironically, he truly believes the pabulum he spews… because he’s just simply not that bright. LOL
He’s attempted to divorce energy and work to make his kooky “continual 2LoT violations cause CAGW” hobby theory to work, and in so doing, he claims that energy flow “has nothing to do with work” (his words)… which means he must also claim that work can be done without energy having to flow, which is full-blown retardation of the lowest order. LOL
(SNIPPED out the excessive personal attack) SUNMOD
Nepal
| #
“‘Heat’ is defined as energy in motion, a flow of energy, a flux of energy”
Wrong. Heat is a NET flux of energy. The fact that there is pressure on the opposite sides of a box containing a gas at equilibrium proves that there are energy flows in both direction but no NET energy flow and thus no heat flow.
““Energy flow is bidirectional, energy flow is unidirectional.””
Not quite. Energy flow is bidirectional and NET energy flow, i.e. heat flow, is unidirectional.
These are not difficult concepts.
Joseph E Postma
| #
That’s already been covered, and your insane anti-mathematical ramblings debunked long ago:
https://climateofsophistry.com/2021/05/19/green-plate-analyzed-and-demolished/
Now go cope and seethe more…lol. Cope, and seethe! haha
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Nepal
| #
You have not debunked anything I have written. To do so all that you would have to do is to show which of my equations is wrong. The problem for you is that I have used YOUR form of the first law for the plates. The math proves you wrong Joe. You’re lucky you didn’t have to understand simple problems like these to get your master’s degree.
Anyone who follows your link will see that you ignorantly claim that the two plates are at equal temperature. That means that the irradiated plate cannot pass any heat to the second plate. Yet you claim that the second plate can pass energy out to space. The first law is violated for the second plate in your solution. You are wrong. Deal with it.
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Joseph E Postma
| #
That’s already been covered, and your insane anti-mathematical ramblings debunked long ago:
Click the link on my name.
Now go cope and seethe more…lol. Cope, and seethe! haha
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Nepal
| #
You’re stuck on repeat Joe. Hit the reset button. CD might think you are a bot.
The simple fact is that you are inferior at both physics and mathematics. If you could support your blather then you would do so with equations. You can’t do that and so you venture to dupe your followers with BS.
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Nepal
| #
“Click the link on my name.”
OK Joseph, I did that. Let’s focus on the case of the distant point source since that was implied by the context of Eli Rabbet’s original problem.
We agree on the solution for the single plate. So that’s good that there is a starting point where we have a meeting of the minds.
One point we disagree on is where you claim, “No energy is lost between the two plates, and the energy lost to space is emitted from the first plate on the hemisphere facing the point source, and from the second plate on the hemisphere facing away from the point source. Thus, you should be able to see in your mind’s eye that this scenario actually reduces to the single-plate case”.
If energy is emitted from the second plate to space. then it must be RECEIVING energy in from somewhere else. Have you asked yourself how a plate can emit energy at steady state if it is not receiving any energy? The second plate can ONLY receive energy from the first plate because it is blocked from view from the distant source by the first plate. The ONLY way that the first plate can transfer energy to the second plate is if the first plate is HOTTER than the second plate.
Hence your claim that the second plate is at the same temperature as the first is wrong, and the first law is violated for the second plate for your solution.
Cope with that Joseph.
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MattH
| #
Hi LOL@Klimate Katastrophe Kooks and Nepal.
Question please. If you have a plate @ 300 degrees celsius in intergalactic space how do mathematically (algebraically) represent the receiving environment of the energy transferred from the plate to space.
The energy inherent in the plate is measurable and quantifiable but once that energy flows to ‘space’ it is no longer measurable and quantifiable because space is infinite and the energy is fundamentally lost.
So how do you represent infinity in the equation?
Have a nice collaborative day. Matt
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Nepal
| #
Matt – The energy emitted by the plate is in the form of photons traveling through space. No energy is ever lost. You can equivalently think of that energy as the energy of the electromagnetic waves that are traveling through space. Again, energy is never lost. There is no need to represent infinity in the equations. One can apply the first law to the system as a whole or to any sub-system. Does that answer your question?
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MattH
| #
Hi Nepal. Thank you for your reply. I do and did understand all aspects of your reply.
My question would have been more appropriately termed as energy dispersed into infinity rather than lost into infinity.
So , how is a quantifiable and qualifiable energy then represented in an equation when that energy is dispersed into an infinity, and taken further, into an infinity that is ‘bombarded’ by an infinite number of constantly varying energy sources. (stars)
So , how does mathematics account for infinity?
Anyway, I was pondering the imponderable.
Cheers. Matt
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Nepal
| #
Yes, it is better to think of energy being dispersed into infinity.
“how is a quantifiable and qualifiable energy then represented in an equation when that energy is dispersed into an infinity”
Perhaps if you could provide an example where that would need to be done, then I could provide a better answer.
For example, to solve the two plate problem such an accounting is not necessary. All that we need to do is define a system, e.g. one of the plate, and to apply the first law we need the heat into/out of it and the work done on/by it.
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LOL@Klimate Katastrophe Kooks
| #
Space is effectively an infinite heat sink. Even were we to pump more energy to space, it would respond by either concretizing invariant-mass matter or by expanding. In the earlier stages of the universe, it was energetically favorable to both expand and concretize invariant-mass matter, now it’s energetically favorable to expand.
The energy is ‘fundamentally lost’ locally only if one considers the ‘system’ to be local. Energy is never created nor destroyed, so that energy, those photons, will incide upon something somewhere else at some later time. But yes, for photons which leave the boundaries of what you consider to be your ‘system’, they are ‘fundamentally lost’ simply because they are outside your ‘system’.
The energy lost from the object is still measurable and quantifiable, because you’re calculating for that object, while holding space’s energy density essentially constant (considering it to be an infinite heat sink).
One cannot calculate upon an effective infinity except via renormalization, (and regularization by assuming new unknown physics at new scales), but there’s no need to do either here. We’re focusing on the object, not on its heat sink, space.
Photons are considered the force-carrying bosons of the EM interaction. In order to even be emitted, there must exist an energy density gradient (akin to a pressure gradient; same units for pressure and energy density; and for pressure gradient and energy density gradient), which implies there is free energy, which implies that work can be done.
What I do is forego using temperature and instead use energy density.
That makes it very clear that energy only ever spontaneously flows down an energy density gradient. No gradient, no flow (which is why thermodynamic equilibrium is considered a quiescent state… no energy flows at thermodynamic equilibrium because there is no energy density gradient to act as the impetus for photon emission nor absorption). And it absolutely will not flow against a gradient of higher chemical potential than the photons unless external energy does work to pump that system energy up the gradient (this is what happens, for example, in your refrigerator, a perfect example of 2LoT in the Clausius Statement sense).
Hence, the “all objects above 0 K emit radiation” blather only applies to idealized blackbody objects, not graybody objects. Graybody objects greater than 0 K above their ambient emit.
It’s right in the S-B equation:
https://i.imgur.com/QErszYW.gif
Idealized blackbody objects assume emission to 0 K and emissivity of 1 at all times (ie: the maximally emit and absorb). Idealized blackbody objects don’t actually exist… they’re idealizations.
Real-world graybody objects assume emission to greater than 0 K and emissivity less than 1.
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Nepal
| #
“no energy flows at thermodynamic equilibrium”
This is nonsense. No net macroscopic energy flows occur at equilibrium. Microscopic motions of particles are occurring constantly in systems at equilibrium.
“equilibrium is considered a quiescent state”
This is also nonsense. Particles are vibrating vigorously in solids at equilibrium, and are colliding with one another constantly in gases at equilibrium.
You’re just like Postma, all words and no math. You can’t solve the plate problem correctly to save your life.
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LOL@Klimate Katastrophe Kooks
| #
Your denial of long-held definitions is part and parcel of your kooky claim that continual 2LoT violations cause CAGW, Not-Professor Unevenminded. Of course, being a fundamental physical law means it’s not violated willy-nilly… not macroscopically as you’ve repeatedly attempted to claim, not at the quantum scale as you attempted to claim after your classical claim was demolished, not ever.
You’ve been proven wrong on this (with extensive maths which you were utterly unable to grasp let alone even do correctly) so many times it’s not even funny… but alas, you’re ineducable. Likely because your poor broken brain has settled into a groove worn into a rut which you find yourself utterly unable to get out of because you’re ineducable. IOW, your stupidity is recursive. LOL
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Nepal
| #
I have not denied any scientific definitions. I have denied and refuted your claims that at equilibrium there are no microscopic flows of energy.
Of course there are microscopic flows of energy. All you need to do to see how obvious this is is to consider a gas in a box. Molecules are flying around in ALL directions at equilibrium. Those are ENERGY FLOWS in all directions.
You don’t even understand what equilibrium means. You have ZERO chance of writing down the correct equations for the two plate problem.
LOL@Klimate Katastrophe Kooks
| #
Bwahahaha! You’ve learned absolutely nothing in the more than 3 years I’ve been drop-kicking your stupid arse across the width and breadth of the internet.
Now tell everyone your uber-kooky hypothesis that individual atoms and molecules have no temperature, and that atoms or molecules all moving in the same direction will have a temperature “near” absolute zero, because you’re utterly unable to understand Bernoulli’s Principle, LOL
From the last time you were utterly demolished on this topic…
That’s you claiming that graybody objects maximally emit and must absorb all radiation which falls upon it, just like idealized blackbody objects. The same topic upon which physicist Dr. Charles R. Anderson, PhD drop-kicked your stupid backside so hard that you’re still attempting, years later, to defame him in hopes of causing him reputational damage and thereby monetary loss to his materials analysis laboratory… because you’re a prototypically slimy POS libtard kook… you’re ineducable and you’re a snowflake with paper-thin skin… so anyone proving your deluded blather wrong becomes your target, Professor BalloonKnot, you creepy stalking doxing defaming kook. LOL
But Professor BalloonKnot, if the atoms or molecules comprising a body “are part of the same collection of matter at the SAME temperature” (your words), then you’ve just claimed that energy can flow from lower temperature to higher temperature, from lower potential to higher potential, without external energy doing work upon the system. In fact, you claim this “can happen quite readily” (your words).
Given this spontaneous flow of energy from lower to higher potential causes entropy to decrease, and a spontaneous decrease of entropy implies a time reversal, you’ve just claimed that time reversal “can happen quite readily” (your words). That’s the basis of your blather about time reversal only requiring a reversal of the direction of motion of the particles… you didn’t know about CPT symmetry, so you didn’t know that in order to make a system time invariant, it requires charge conjugation and parity symmetry. Not even beta decay of 60Co does that, and that’s the closest example we’ve got, given that it’s CP invariant… it still doesn’t cause time reversal.
Far from your stupid blather that 2LoT violations “can happen quite readily” (your words), a macroscopic 2LoT violation has never been empirically observed, and 2LoT is rigorously obeyed at the quantum scale… you’ve claimed that microscopically there are no restrictions on violations of 2LoT… yet again you’re diametrically opposite to reality… because you’re an insane kook. Reality confuses you. LOL
That’s you yet again reiterating your claim that objects absorb as though they’re idealized blackbody objects. That’s because you deny radiation pressure, something which has been known about for 400 years… your knowledge is only out of date by 400 years, Professor BalloonKnot… do you also believe that bloodletting can heal disease, trepanation can let the ‘bad spirits’ out, and people who can’t swim are all witches? LOL
That’s you providing comic relief by claiming that energetic transfer “has nothing to do with work” (your words)… because you don’t understand simple concepts like ‘work’, ‘potential’, ‘energy’, ‘temperature’, ‘heat’ and ‘reality’. LOL
That’s you yet again stupidly claiming that energy can spontaneously flow from a lower temperature to a higher temperature because you don’t understand that a warmer object will have higher energy density at all wavelengths than a cooler object.
That’s you detailing that this spontaneous energy flow is at the microscopic scale… between the emitting atom or molecule, the emitted photon and the atom or molecule upon which that photon is incident.
Except that doesn’t happen at the macroscopic scale, the microscopic scale, the quantum scale or any scale… you’re just a confused kook. LOL
Nepal
| #
That’s quite a lot of words and no math to back it up.
Provide your solution to the two plate problem and the equations that you solve to get your solution.
You won’t because you cannot handle mathematical physics.
Nepal
| #
“Except that doesn’t happen at the macroscopic scale, the microscopic scale, the quantum scale or any scale”
It is very easy to prove that “bidirectional” energy flow occurs.
There is pressure on the east wall of the room you are sitting in. That pressure arises from molecules that were moving eastward and then collide with the wall. There is also pressure on the west wall of the room you are sitting in. That pressure arises from molecules that were moving westward and then collide with the wall. So there are molecules traveling eastward carrying an eastward energy flux and there are molecules traveling westward carrying a westward energy flux in the room you are sitting in. At equilibrium the NET flux would be zero, i.e. the eastward flux is balanced by the westward flux.
It’s quite easy to prove your nonsense wrong.
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“Provide your solu”{TARDSMACK}
No, I do believe I’ll join the rest of the world pointing and laughing at an insane nutter who’s been denying reality for more than 3 years, and getting his stupid arse drop-kicked across the width and breadth of the internet for his stupidity. LOL
For the slow-witted, I’m referring to you, Professor BalloonKnot. LOL
For the slow-witted, I just called you slow-witted, Professor BalloonKnot. LOL
You have to tell the kook these things, otherwise he won’t get it. LOL
Nepal
| #
“No”
Not surprising. You are unable to understand and work out mathematical physics solutions.
You lose.
LOL@Klimate Katastrophe Kooks
| #
{Points, Laughs}
Already done many, many times, kook. You’ve been refuted to the point that you once lost touch with reality so badly that you couldn’t even write complete sentences. Remember? LOL
And now the voices screaming in your head are back, you’re utterly unable to stop yourself from humiliating yourself with your uber-kooky hobby theory that continual 2LoT violations cause CAGW, with all the uber-kooky baggage that entails, such as that energy flows willy-nilly without regard to the energy density gradient (because you deny the existence of radiation pressure); that atoms and molecules have no temperature but that if they’re all moving in the same direction they have a temperature “near” (your word) absolute zero (a contradiction because you don’t understand Bernoulli’s Principle, beam temperature, beam spread, nor much of anything else); that time can be reversed simply by reversing the direction of the atoms or molecules (because you don’t understand CPT symmetry), etc., etc., etc. ad hilarium.
You’re a laughingstock. Best you just crawl back into the shrubbery to nurse your already-savaged psyche, kook… it only gets worse for you from here. LOL
Nepal
| #
That’s another lie from you. You have never provided your solution to and your equations for the two plate problem.
Are you saying that Postma was wrong when he wrote, “a cool object emits toward a hotter source”?
Nepal
| #
““all objects above 0 K emit radiation”
Of course they do. This is clear to any trained physicist. You should read Einsteins seminal paper.
You’d have a difficult time solving heat transfer problems between greybody objects with the nonsense you are spewing.
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LOL@Klimate Katastrophe Kooks
| #
From the last time you were beaten to a pulp on this topic…
The above destroys your “energy can flow willy-nilly without regard to energy density gradient” blather, Professor BalloonKnot. Stop denying the fundamental physical laws, it only shows you to be a reality-denying, science-denying kook. LOL
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MattH
| #
Hi LOL@Klimate Katastrophe Kooks and Nepal.
Thank you both for your replies which I had not spotted.
I will ponder these tonight.
There is an old rule of science. Challenge the hypothesis but do not abuse the individual. That allows possibility of mutual progression or admission of error to agree on sound foundational paradigm from which to safely evolve the hypothesis.
Cheers. Matt
Nepal
| #
Einstein never stated that “adiation pressure regulates photon emission and absorption”. Why do you lie?
Obviously molecules that are emitting radiation undergo a decrease in their energy state, and those that absorb radiation undergo an increase in their energy state. At equilibrium, molecules are continuously emitting and absorbing radiation and their energy state is in a continuous flux.
Again, you don’t understand that equilibrium is at the MACROSCOPIC level. At the MICROSCOPIC level energy is flying around all over the place.
This is basic physics that you do not grasp.
LOL@Klimate Katastrophe Kooks
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MattH wrote:
“Challenge the hypothesis but do not abuse the individual.”
In this case, the individual in question is intentionally twisting scientific reality to attempt to fit it to his uber-kooky hobby theory that continual 2LoT violations cause CAGW.
When he’s proven wrong, he stalks, defames, harasses, hijacks people’s real-life names and their pseudonyms to spew his kooky blather (just as he’s doing now… that’s not the real Nepal) and attempts to disrupt forums with his blather.
He deserves every bit of denigration he gets.
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“Einstein never stated that “adiation pressure regulates photon emission and absorption”. Why do you lie?”
Displaying your crippling reading comprehension problems again? LOL
Here, read it again so you can embarrass yourself again, kook. LOL
You deny the existence of radiation pressure… your ‘knowledge’ is only about 400 years out of date, kook. LOL
Nepal
| #
“Einstein states that radiation pressure regulates photon emission and absorption”
He never made any such statement. Provide the page number where he wrote those words. Oh wait, you can’t because he never made any such statement.
You simple are unable to understand what he was talking about in the text you quoted. Nothing about radiation pressure there, only descriptions of the changes in energy states of molecules emitting and absorbing radiation.
LOL@Klimate Katastrophe Kooks
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Professor BalloonKnot dribbled:
“He never made any such statement. Provide the page number where he wrote those words. Oh wait, you can’t because he never made any such statement.”
Poor kook… you know you’re utterly defeated when you have to backpedal and deny reality so quickly because you’ve been caught in contradictions and outright misrepresentations.
http://web.ihep.su/dbserv/compas/src/einstein17/eng.pdf
Does your stupidity cause you pain, Professor BalloonKnot? Of course it does… that’s why you’ve melted down so badly in the past that you only repeated “DISMISSED!!!” for hours, and at the end you couldn’t even spell that word correctly; that’s why you’ve lost touch with reality to such an extent in the past that you couldn’t even form complete sentences; that’s why you’re forced to sock up with a new sock each time the voices screaming in your head compel you to humiliate yourself again… you’re partitioning the psychological pain of your being perpetually wrong. It’s a tactic used by psychologically weak and utterly impotent lifelong losers. LOL
Nepal
| #
Thanks for providing the proof that you lied.
Anyone can now open the file you provided the link to an search for the phrase “radiation pressure”.
That phrase does not appear in the paper.
You lose again.
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“Thanks for providing the proof that you lied.
Anyone can now open the file you provided the link to an search for the phrase “radiation pressure”.
That phrase does not appear in the paper.”
Thanks for yet again displaying your utter inability to relate energy, energy density, radiation pressure and energy density gradient… because you deny that radiation pressure exists, so you must deny that energy density exists, so you must deny that energy density gradient exists, so you must claim that energy can flow willy-nilly without regard to the energy density gradient.
https://principia-scientific.com/another-botched-experiment-the-tricks-of-pseudoscience/#comment-75414
You’re retarded. We get it. You can stop demonstrating that for the world, Professor BalloonKnot. You’ve made that fact abundantly clear. LOL
Nepal
| #
You lost Kooks. Einstein NEVER stated that “radiation pressure regulates photon emission and absorption”.
YOU ignorantly made that claim based on YOUR inability to understand Einstein’s work.
Postma makes it CRYSTAL CLEAR when he writes, “a cool object emits toward a hotter source”.
Are you saying Postma is wrong?
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“You lost Kooks. Einstein NEVER stated that “radiation pressure regulates photon emission and absorption”.”
He was discussing radiation density, which is energy density, which generates a radiation pressure, the differential between objects which sets up the energy density gradient which regulates photon emission and absorption.
You deny all of that reality so you can make your kooky “continual 2LoT violations cause CAGW” hobby theory to seem to make even a semblance of sense, and in so doing, you’ve claimed that energy flow “has nothing to do with work” (your words)… which means you must also claim that work can be done without energy having to flow, which is retardation of the lowest order… but that’s what we’ve come to expect of you.
You are, after all, the guy who claims that individual atoms and molecules don’t have a temperature, but if atoms or molecules are all moving in the same direction they have a temperature “near absolute zero” (your words) because you don’t understand Bernoulli’s Principle, beam temperature, beam spread, etc.
You’re also the guy who claims that if the direction of motion of those atoms or molecules is reversed, time reversal will take place, because you’re incapable of understanding CPT symmetry. That, of course, plays right into your claim that energy flow is an idealized reversible process, so you can get around the fact that if energy flowed at thermodynamic equilibrium, entropy would change… but since it doesn’t change at thermodynamic equilibrium, we know that energy flow is an irreversible process, which destroys every single bit of your blather.
You’re also the guy who claims that energy can flow willy-nilly without regard to the energy density gradient because you deny the existence of radiation pressure which sets up the energy density gradient which regulates photon emission and absorption… because you’re incapable of understanding much of anything.
And years of having reality jackhammered into that cement-block brain of yours has done nothing to impart clue to you… you remain just as clueless today as the day you first socked up to spew your blather.
Right, Joshua? LOL
Nepal
| #
Einstein never once stated “adiation pressure regulates photon emission and absorption”.
He did write this though, “I showed that molecules with a distribution of states in the quantum theoretical sense for temperature equilibrium are in dynamical equilibrium with the Planck radiation in this way, the Planck formula 4 was obtained in a surprisingly simple and general way It was obtained from the condition that the quantum theoretic partition of states of the internal energy of the molecules is established only by the emission and absorption of radiation.”
At equilibrium there is a DISTRIBUTION of states. Some molecules emitting and some absorbing. Equilibrium is DYNAMIC at the microscopic level. The internal energy of the molecules is established only by the emission AND absorption of radiation.
Einstein refutes your nonsense.
LOL@Klimate Katastrophe Kooks
| #
Your denial of reality hasn’t abated with time, I see. Now you’re denying even the S-B equation.
https://i.imgur.com/QErszYW.gif
That’s likely because that simple math in the S-B equation is tripping you up. LOL
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Nepal
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I have no problem understanding the SB equation. I have implemented it properly to solve the two plate problem.
The problem that neither you nor Joseph can solve properly because you are both terrified of simple math.
Nepal
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“‘Heat’ is defined as a flow of energy”
Correction, heat is the NET flow of energy. At equilibrium there are flows of energy in all directions, but the NET flow of energy is zero.
It’s not a difficult concept, but it’s not surprising that you don’t get it.
At least Postma understands that colder object DO emit energy. Joseph Postma: “a cool object emits toward a hotter source”.
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LOL@Klimate Katastrophe Kooks
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Professor BalloonKnot dribbled:
“Correction, heat is the NET flow of energy. At equilibrium there are flows of energy in all directions, but the NET flow of energy is zero.”
Professor BalloonKnot dribbled:
“Heat transfer is a non-equilibrium process, be it by conduction, convection, or radiation.”
Awww, the kook is yet again attempting to redefine reality to fit his kooky hobby theory which states that continual 2LoT violations cause CAGW… but he’s not off his rocker, right? LOL
‘Heat’ is defined as any flow of energy… and given that energy only flows from higher to lower energy density, it can only be unidirectional flow. You deny this because you deny the existence of radiation pressure… you must deny its existence for your blather to make even a semblance of sense… but anyone can look up ‘radiation pressure’ and see that you are truly insane.
Your blather is akin to stating:
“Heat transfer is two way, energy transfer is one way.”
But ‘heat’ is defined as energy in flux, an energy transfer, so you might as well be stating:
“Energy transfer is bidirectional, energy transfer is unidirectional.”
You are that retarded. LOL
Deluded kooks often self-contradict… it is their most predictable attribute. LOL
https://principia-scientific.com/another-botched-experiment-the-tricks-of-pseudoscience/#comment-75414
How many times in a single day are you going to demonstrate that you’re far too retarded to ever be right, Professor BalloonKnot? You’re so retarded that you’re fighting against yourself. LOL
I know, I know… despite the fact that you’ve repeatedly been refuted extensively on all of this in the past, the voices screaming in your head compel you to bleat… it’s a cry for help from you to the world, but no one is listening. They’re too busy laughing at you. LOL
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Nepal
| #
“‘Heat’ is defined as any flow of energy”
Wrong.
The pressure on opposite walls of a box containing a gas at equilibrium prove that there are energy flows in both directions, yet there is no heat flow.
Heat is the NET energy flow. Feynman explains it quite well.
Feynman: ”The methods of the kinetic theory that we have been using above can be used also to compute the thermal conductivity of a gas. If the gas at the top of a container is hotter than the gas at the bottom, heat will flow from the top to the bottom. (We think of the top being hotter because otherwise convection currents would be set up and the problem would no longer be one of heat conduction.) The transfer of heat from the hotter gas to the colder gas is by the diffusion of the “hot” molecules—those with more energy—downward and the diffusion of the “cold” molecules upward. To compute the flow of thermal energy we can ask about the energy carried downward across an element of area by the downward-moving molecules, and about the energy carried upward across the surface by the upward-moving molecules. The difference will give us the net downward flow of energy.”
You continue to avoid telling everyone whether Postma is right or wrong when he wrote, “a cool object emits toward a hotter source”. I don’t expect you to because you are utterly terrified of contradicting Postma.
LOL@Klimate Katastrophe Kooks
| #
The dullardry you’ve exhibited… you’ve claimed that: geometric resistance of the configuration (ie: net emissivity of the system) “does not exist” (your words); energy can flow between objects at thermodynamic equilibrium; energy can flow without work being done; energy can flow against a gradient; 2LoT violations “can happen quite readily” (your words); idealized BB objects do not maximally emit (as though in a 0 K ambient) nor absorb (as though in an ∞ K ambient); the S-B formula only applies to one real (ie: graybody) and one idealized (ie: BB) object; idealized BB objects actually exist; BB radiation is only emitted by idealized BB objects; graybody objects don’t exist; real-world (ie: graybody) objects are not graybodies; graybody objects do not emit BB radiation; all real-world objects are selective radiators; energy gradient, energy density and the resultant radiation pressure are “nonsense” (your word). You’re known as Professor BalloonKnot for a reason. LOL
You’re just a confused little laughingstock kook, ain’t ya? LOL
LOL@Klimate Katastrophe Kooks
| #
Awww, the poor deluded idiot is attempting to take quotes out of context again, despite getting drop-kicked across the width and breadth of the internet repeatedly in the past for doing so. LOL
The very next sentence after that which Professor BalloonKnot cherry-picked, Feynman stated:
If there is no “energy available to do anything” (ie: the Helmholtz free energy has dropped to zero, the photon chemical potential is zero), then no work can be done by one object upon the other… so Professor BalloonKnot is obviously forced to claim that energy can flow without work being done.
Of course, free energy is the capacity to do work, so if there’s no capacity to do work (as Feynman says above), then no energy can be transferred.
In other words, photon emission can only occur if the radiation pressure of the atom or molecule which would emit is greater than the ambient field radiation pressure, and the rate decreases as the differential between the emitting atom or molecule radiation pressure and the field radiation pressure decreases.
In other words, photon absorption can only occur if the photon radiation pressure is greater than the radiation pressure of the atom or molecule upon which that photon is incident, and the rate increases as the differential between the field radiation pressure and the atom or molecule radiation pressure increases.
But the poor deluded kook denies all of this so that his kooky hobby theory, which states that continual 2LoT violations cause CAGW, seems to make even a semblance of sense. LOL
You’re a laughingstock. You’ve been refuted many times in the past. All of those refutations are archived and available at a moment’s notice. You’ll accomplish nothing except to humiliate yourself yet again with your own stupidity. LOL
Nepal
| #
You’re flailing Kooks.
Joseph Postma: “a cool object emits toward a hotter source”.
Nepal
| #
Are you having trouble understanding Feynman?
”The transfer of heat from the hotter gas to the colder gas is by the diffusion of the “hot” molecules—those with more energy—downward and the diffusion of the “cold” molecules upward.”
Are “downward” and “upward” too technical for you?
Nepal
| #
It’s funny how everyone else can understand Feynman because he was such an outstanding teacher, but you can’t.
Try harder.
“Although when I say two things at the same temperature, which means you have balance, it doesn’t mean they have the same energy in them. It just means it’s just as easy to pick energy off of one as to pick it off the other, so as you put them next to each other, nothing apparently happens. They pass energy back and forth equally. The net result is nothing.”
Did you get that? They pass energy back and forth equally. it’s the NET result that is “nothing”.
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“Are you having trouble understanding Feynman?”
No, but you are. You’ve demonstrated that fact repeatedly, over the course of many years.
But then, you’re not really a professor teaching mechanical theory and thermodynamics… you couldn’t possibly be.
No, you’re a basement-dwelling half-wit with an odd hobby theory who is desperate to justify the lifetime spent mislearning reality, so you don’t have to admit that you’ve wasted your life on fictive ideations. LOL
Professor BalloonKnot dribbled:
“Joseph Postma: “a cool object emits toward a hotter source”.”
Awww, taking more quotes out of context? That was before he acknowledged that my take on radiative energy exchange hewed to the fundamental physical laws, to include 1LoT, 2LoT, Stefan’s Law, the Principle of Least Action and the Entropy Maximization Principle, whereas your take violates all of the above and more.
What do you hope to achieve here, kook? You’re not convincing anyone, you’re only humiliating yourself yet again by getting caught in self-contradictions, lies and assorted buffoonery. LOL
I know, I know… you can’t help yourself. Those voices screaming in your head compel you, at times, to humiliate yourself with your own stupidity. Must be hell for you. LOL
Nepal
| #
“That was before he acknowledged that my take on radiative energy exchange”
So you are saying Postma was wrong. Have you told him that he is wrong and that he should edit his posts? or are you too much of a coward to do that?
Feynman refutes your nonsense about heat flowing due to an energy gradient and about bidirectional energy flow all in one statement.
Feynman: “Although when I say two things at the same temperature, which means you have balance, it doesn’t mean they have the same energy in them. It just means it’s just as easy to pick energy off of one as to pick it off the other, so as you put them next to each other, nothing apparently happens. They pass energy back and forth equally. The net result is nothing.”
You will never admit your errors. You are too much of a coward to present your solution to the plates problem. You’re a joke.
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“Did you get that? They pass energy back and forth equally. it’s the NET result that is “nothing”.”
That’s your problem, you take these simplifications meant for lay people, and you treat them as gospel truth because they bolster your insane take on reality.
Feynman was simplifying the concept of a standing wave between two objects at thermodynamic equilibrium. Look up standard cavity theory, that’s what he was describing, outside a cavity.
But you’re the kook who claims that energy can flow without work needing to be done (and conversely, you must claim that work can be done without energy having to flow), because you’ve already stated that energy flow “has nothing to do with work” (your words).
Your fundamental misunderstandings, combined with your obviously low intelligence, combined with your oh-so-obvious mental condition, compels you to humiliate yourself from time to time by denying reality. We get it, you’re retarded and insane. You’ve made that perfectly clear. No need to keep reiterating that point, Professor BalloonKnot. LOL
(You have gone too far into the personal attacks especially when it becomes a barrage) SUNMOD
Nepal
| #
Feynman”s description is of the physics. This is physics that can be simulated directly with molecular dynamics simulations.
You don’t understand Feynman. Everyone with an IQ of at least 120 does.
You lose.
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“You don’t understand Feynman. Everyone with an IQ of at least 120 does.”
Well, that leaves you out by at least one order of magnitude, you drooling half-wit. LOL
The only one who is incapable of understanding Feynman is you… the kook who claims that energy can flow without work having to be done and conversely you must claim that work can be done without energy flowing… you’re just a bit too retarded to realize how ludicrous your statement was, that energy transfer “has nothing to do with work” (your words).
Feynman was simplifying for a lay audience. You’re obviously not even smart enough to understand that simplification, nor are you smart enough to dig deeper to figure out that he really meant a standing wave set up between two objects at thermodynamic equilibrium.
That’s the same reason you melted down when physicist Dr. Charles R. Anderson, PhD proved your idiotic blather violates Stefan’s Law… your blather necessitates that you claim that at thermodynamic equilibrium, a cavity’s walls are absorbing and emitting radiation (just as you claim objects at thermodynamic equilibrium are absorbing and emitting radiation)… which results in double the energy density than that which we calculate from Stefan’s Law. Your blather violates Stefan’s Law, and Anderson proved it.
The same applies to objects outside a cavity. At thermodynamic equilibrium, if they were absorbing and emitting radiation, the energy density in the intervening space would be double what we calculate from Stefan’s Law, and entropy would be changing… that it doesn’t change forces you to claim that either it doesn’t change by magic, or that radiative energy transfer is an idealized reversible process.
Except radiative energetic transfer is a temporal, entropic, irreversible process, which destroys every single bit of your blather.
That’s why you’re still so butthurt by Anderson that you stalk and defame him. You poor little idiot. LOL
The world is laughing at you. You’re a laughingstock. You’ve always been a laughingstock. You’ll always be a laughingstock. LOL
Nepal
| #
My IQ is clearly higher than yours . I can understand Feynman’s description of energy exchange at equilibrium. You can’t. I can provide the correct solution to the plates problem. You can’t. I am not afraid to tell Postma he is wrong. You are.
Less than two weeks ago Postma wrote, “Get them to admit energy flows both ways. This in itself is not wrong”. You are too much of a coward to tell him you think he is wrong.
I’m wiping the floor with you here showing everyone what a coward you are.
Show some courage and link to where you claim to have solved the two plate problem. Wait, you can’t because you lied about that too.
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“My IQ is clearly” zero.
Fixed it for ya, kook. LOL
You’ve been definitively refuted, you half-wit. Your blather violates Stefan’s Law. You’re drivel is fictive ideation wrought from a twisted mind which is utterly incapable of understanding reality.
Now bleat… bleat about how you’re still butthurt, after nearly 4 years, because you were definitively proven wrong by physicist Dr. Charles R. Anderson, PhD. You’re so butthurt that you’ve denied that reality for nearly 4 years.
You’re a sick joke played upon yourself. How you can even stand to live with yourself is a wonder. Obviously, you’re so retarded that you haven’t figured out that you’re a sick joke played upon yourself. LOL
(Let’s cool down the rhetorical attacks) SUNMOD
Nepal
| #
You’re a joke Kooks. Both Feynman and Einstein, in the paper that YOU linked to, describe energy flows at equilibrium.
Postma states that there are bidirectional energy flows.
Your IQ must be painfully low for you not to be able to understand this simple concept.
Go ahead and solve the plates problem. So long as you refuse, you are making my point that you are scientifically incompetent.
(Let’s back off on the personal attacks) SUNMOD
LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“You’re a joke Kooks. Both Feynman and Einstein, in the paper that YOU linked to, describe energy flows at equilibrium.”
Feynman was discussing via simplification a standing wave being set up between two objects at thermodynamic equilibrium, which refutes your deluded “energy can flow without work being done therefore work can be done without energy having to flow” drivel.
Einstein was specifically discussing radiation density, otherwise known as energy density (a photon being a quanta of energy… unless you’re going to deny that, too. LOL)… you claim energy can flow wily-nilly without regard to the energy density gradient
“Postma states that there are bidirectional energy flows.”
More quotes out of context, kook? He later stated that my take on radiative energy exchange hews to the fundamental physical laws, to include 1LoT, 2LoT, Stefan’s Law, the Principle of Least Action and the Entropy Maximization Principle, whereas your insane take on reality violates all of the above and more.
But out-of-context quotes is all you’ve got left. Your blather has definitively been refuted, it violates Stefan’s Law. It was definitively refuted nearly 4 years ago… you’ve denied reality for nearly 4 years. You are that deluded. LOL
Further, your claim that energy flow “has nothing to do with work” (your words) must mean that you also must claim that work can be done without energy having to flow… and that’s retarded.
You’ve attempted to divorce energy and work to make your kooky hobby theory work… and in so doing, you’ve beclowned yourself and made yourself the laughingstock of the entire world… you know this, which is why you use sockpuppets, rather than your real-life identity.
But rest assured, your real life identity will eventually out… and then the fun begins. LOL
Nepal
| #
Here is the solution to the plate problems.
https://principia-scientific.com/another-botched-experiment-the-tricks-of-pseudoscience/#comment-75564
Clearly Postma and I agree on the equations and their solution for the single plate. Do you agree with that solution?
If you disagree with the equations for the two-plate problem then go ahead and show which one you think is wrong and what your correction would be.
LOL@Klimate Katastrophe Kooks
| #
You’ve attempted all this before, and you only ended up humiliating yourself with your own stupidity. Whatever makes you think (FSVO ‘think’) it’ll be any different this time, kook? LOL
Remember? You claimed that energy can flow in a vacuum via other than radiative means, and that convection could occur in a 0 K vacuum? Yeah. Laughingstock. LOL
LOL@Klimate Katastrophe Kooks
| #
It’s almost as if the voices screaming in your head compel you to humiliate yourself from time to time with your own stupidity… you just can’t seem to help yourself. And that despite the fact that all of your blather was definitively refuted nearly 4 years ago by physicist Dr. Charles R. Anderson, PhD when he showed that your blather results in double the energy density in the intervening space between two objects or in a cavity space at thermodynamic equilibrium, which means your blather violates Stefan’s Law… if your blather violates even a single fundamental physical law, it’s wrong. Your violates Stefan’s Law, 1LoT, 2LoT, the Principle of Least Action, the Entropy Maximization Principle and more.
Your last attempt, during which you humiliated yourself to such an extent that you melted down so badly that you couldn’t even form cogent sentences…
Nepal
| #
Here is the solution to the plate problems:
https://principia-scientific.com/another-botched-experiment-the-tricks-of-pseudoscience/#comment-75564
If my solutions are wrong then it should be easy for you to show which of my equations you believe to be wrong and how you would correct them.
Go ahead.
Nepal
| #
“I’ve already solved a far more difficult problem”
Actually, you just provided a link where you contradict yourself.
You write:
U = a V T^4
P = ⅓ a T^4
S = (4/3) a V T^3
You also write:
“Photons are a very special kind of boson. When they are in radiative thermal equilibrium in a volume V at a constant temperature T, their chemical potential is zero.”
You also write:
µn = U – ST + PV
So let’s see what this means at equilibrium:
µn = U – ST + PV = a V T^4 – (4/3) a V T^4 + ⅓ a V T^4 = 0
Very good! That checks out. It does equal zero. But wait, you have also stated that the Helmholtz free energy is zero at equilibrium. Let’s see what you claimed the Helmholtz free energy is.
You wrote:
F = U – TS
Where:
F = Helmholtz Free Energy
So let’s look at what F is at equilibrium, which you claim is when µ = 0.
F = U – TS = a V T^4 – (4/3) a V T^4 = – (1/3) a V T^4
Oh no, that did not work out well for you. Clearly F is NOT equal to zero at equilibrium. It’s negative.
How sad for you to so easily be proven wrong with your own words.
Nepal
| #
@SUNMOD
My apologies on the rhetorical attacks. I will stick to the math and science only from now on.
(It is the numerous putdown and attacks in a comment that bothers me because it can generate a similar response that can destroy a thread in time, everyone can get excited and call someone stupid a liar or something singular in replies that isn’t going to bother me at all since a good putdown can be an effective way to expose something or make a vivid point) SUNMOD
Nepal
| #
“You’ve attempted this
sophistryin the past, kook.”You mean that I have proven that you have contradicted yourself with rigorous mathematics. How sad for you that you can’t understand basic mathematics.
You write:
U = a V T^4
P = ⅓ a T^4
S = (4/3) a V T^3
You wrote:
F = U – TS
Where:
F = Helmholtz Free Energy
You claim Helmholtz free energy is zero at equilibrium. The simple math based on what YOU wrote proves that is is NOT.
F = U – TS = a V T^4 – (4/3) a V T^4 = – (1/3) a V T^4
The MATH proves that YOU are WRONG.
Nepal
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It’s really too easy to prove you wrong. Angles can be specified from any direction.
https://uploads.disquscdn.com/images/4636bf12c29cf139cd8ab7f5418f19e35ff0e3ae1b6cfb0cdb5866b2f4f70874.png
Nepal
| #
F = U – TS = a V T^4 – (4/3) a V T^4 = – (1/3) a V T^4”
That alone proves that you are wrong when you claim that the Helmholtz free energy is zero at equilibrium. There are the equations that YOU wrote down. They PROVE that F is not equal to zero at equilibrium.
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MattH
| #
Hi LOL@Klimate Katastrophe Kooks
Thank you for the time you have given to my questions.
The concept or physics in an ‘energy pressure gradient’ makes comprehension pretty simple.
Cheers. Matt
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Nepal
| #
“The concept or physics in an ‘energy pressure gradient’ makes comprehension pretty simple.”The problem is that there is no such physical principle. Heat does not flow due to an “energy gradient”, it flows due to a temperature gradient or temperature difference. Here is a quote from Feynman. Note that he states that at equilibrium no heat flows but equilibrium does not mean that the ENERGY of the two interacting objects is the same.
Feynman – “Although when I say two things at the same temperature, which means you have balance, it doesn’t mean they have the same energy in them. It just means it’s just as easy to pick energy off of one as to pick it off the other, so as you put them next to each other, nothing apparently happens. They pass energy back and forth equally. The net result is nothing.”
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LOL@Klimate Katastrophe Kooks
| #
Professor BalloonKnot dribbled:
“”The problem is that there is no such physical principle. Heat does not flow due to an “energy gradient”, it flows due to a temperature gradient or temperature difference.”
Temperature is equal to the fourth root of energy density divided by Stefan’s Constant (aka the radiation constant).
e = T^4 a
a = 4σ/c
e = T^4 4σ/c
T^4 = e/(4σ/c)
T = 4^√(e/(4σ/c))
T = 4^√(e/a)
https://principia-scientific.com/another-botched-experiment-the-tricks-of-pseudoscience/#comment-75414
The concept of energy density creating a radiation pressure is most certainly not a physical principle in the kooky world which sockmonger Professor BalloonKnot has constructed, in which he claims that energy can flow willy-nilly without regard to the energy density gradient, that energy flow “has nothing to do with work” (his words) and hence that work can be done without energy having to flow… that’s the only way he can get his kooky hobby theory to seem to make even a semblance of sense.
But in the real world, any sane person can see that pressure and energy density have the same units because they are both a pressure. Likewise, energy density gradient and pressure gradient both have the same units because they are both a pressure gradient.
Every action requires an impetus. For an example, water can only flow if there is a pressure gradient forcing it to flow. Likewise, energy can only flow if there is a pressure gradient (energy density gradient… same units as pressure gradient) forcing it to flow.
Professor Balloonknot claims that energy can flow without any impetus, and thus he must claim that work can be done without requiring any energy. This obvious mistake doesn’t stop him from continuing to shill for his kooky theory, though… facts be damned, he’s a mission-posting kook who is desperate to justify his having spent a lifetime mislearning reality. He cannot admit that he’s wrong, thus he never learns from his mistakes, thus he continues to humiliate himself with these hilarious proclamations that energy density and energy density gradient do not exist, that energy can flow without regard to the energy density gradient, that energy flow has “nothing to do with work” (his words) and thus that work can be done without energy having to flow.
Nepal
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“Temperature is equal to the fourth root of energy density divided by Stefan’s Constant”
That’s only for a blackbody, not in general.
Feynamn’s statements are correct in general. You are wrong.
““Although when I say two things at the same temperature, which means you have balance, it doesn’t mean they have the same energy in them. It just means it’s just as easy to pick energy off of one as to pick it off the other, so as you put them next to each other, nothing apparently happens. They pass energy back and forth equally. The net result is nothing.”
Nepal
| #
I see you are bleating about getting me banned because you can’t handle the fact that I am proving you wrong at every turn with rigorous mathematics. You can’t do that with me because I back up my statements with MATH.
“That’s only for a blackbody, not in general.”
I was imprecise with my statement. e = T^4 4σ/c is the energy density of the blackbody radiation in a cavity, i.e. the energy density of the photon gas, not of the blackbody object itself.
The rest of your blather is nonsense piled upon lies. You can’t use my words to show that I have contradicted myself because I haven’t. I back up my statements with mathematics.
I solved the two plate problem and posted the solution here:
https://principia-scientific.com/another-botched-experiment-the-tricks-of-pseudoscience/#comment-75564
If you think it is wrong then point out EXACTLY where you think it is wrong.
You can’t do that. So go back to CoS and whine about me proving that you have contradicted yourself. Beg to get me banned.
Go ahead. Solve the problem if you can.
(He is BANNED suggest that you ignore him from now on unapproving his posts until he will be fully blocked) SUNMOD
Nepal
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Well, there you have it Kooks. If you could partake in a civil and rational discussion then maybe you wouldn’t be banned. Maybe if you cool down and post your solution to the plates problems with all of the governing equation and refrain from name-calling then perhaps the mod will reconsider. Otherwise, I welcome anyone else to join the discussion and walk through the steps of how this problem is solved and why Postma’s solution is wrong and self-contradictory. Any takers? Joseph? CD?
Herb Rose
| #
I thought all the sophistry arguments were suppose to happen on Joe’s site.
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Nepal
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Generally they do, but apparently Joe has brought his to this site as well. Are you able to explain the solution of the plate problem to Joe and Kooks Herb?
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Herb Rose
| #
Hi Nepal,
First I’d have to explain to you that there’s no photons (particle nature of light) and light is a disturbance traveling in the electric and magnetic fields radiated by objects. Plank’s law is wrong (it. is the structure of the object absorbing energy that determines what wavelength/frequency is absorbed, not the energy source) and it is the amplitude of electromagnetic wave that determines energy of the field and determines if energy is absorbed from the object or transferred to it.
I have discovered these are long futile discussions.
Herb.
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Nepal
| #
I understand quite well the nature of electromagnetic waves. Clearly you are unable to understand the particle description of light. That’s a shame and a gap in your understanding of modern physics.
“Plank’s law is wrong”
That’s a bold claim. Much of radiative heat transfer is based upon Planck’s law and its consequences. Engineers have successfully used these theories to successfully design countless systems and devices.
Discussions are futile if you refuse to learn. I have posted my solution to the two plate problem above. If you believe you know how to solve it differently then go ahead and show how. My guess is, like for Joseph and Kooks, math is scary for you too.
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Herb Rose
| #
Hi Nepal,
The photon was created to explain whether was no delay in the production of a current in the photoelectric effect. The argument being that a wave would need time to transfer enough energy to separated an electron from an atom and create a current. The photoelectric effect occurs in crystals and metals where electrons are already separated from atoms and all that is needed is a distortion of these bonds to cause a current. The photoelectric effect is just another version of the piezo electric effect except instead of pressure causing the distortion it is a change in the strength of the right sized electromagnetic field.
In the two plates if one plate emits energy at a red wavelength and the other only absorbs energy in the blue spectrum they will still equalize because they are both equalizing with the energy field they are in.
Math is a tool to describe reality, not reality. If you separate speed from direction it is no longer based on reality.
If the energy of light is the inverse of wavelength why are the lightwaves with the longest wavelengths/least energy able to travel so much further than the short wavelengths with more energy?
E=hfrequency
Herb
Nepal
| #
“it. is the structure of the object absorbing energy that determines what wavelength/frequency is absorbed, not the energy source”
That is correct. Planck never made any claims to the contrary. This property is called the absorptivity of the material.
“it is the amplitude of electromagnetic wave that determines energy of the field and determines if energy is absorbed from the object or transferred to it”
No Herb, the absorptivity of the object at each wavelength tells us what fraction of the incident radiation at that wavelength is absorbed.
You seem to be very confused. Perhaps if you put your claims down as equations to solve an actual problem then you will gain some clarity.
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Nepal
| #
It make no difference where the photon was “created”. It is part and parcel of modern physics. The action of individual photons can be measured in the lab. The theories that describe them are tremendously accurate.
Math is the language of science. I suggest you learn it if you want to be taken seriously.
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Herb Rose
| #
Hi Nepal,
You’re faith in “modern physics” is a religious belief based on Einstein’s infallibility. Science is about reason and evidence.
When the energy from the decay of radioactive particles didn’t agree with Einstein’s E=mc^2 the neutrino was created to make reality conform to belief.
The claim that there is no way to distinguish between acceleration and gravity in a closed container is false. If you reflect a light beam between the walls, in a gravity field it will remain at the same height while in an accelerating object it will descend as the speed increases.
The equivalence theory led to the assertion that in an increasing gravitational field time will slow. This means as an object approaches a mass and the gravity field strength increases there must be a corresponding increase in distance c=d/t). The result is as the objects get closer they get further apart.
Einstein said that if the movement of a source of light had any effect on the light then relativity was wrong. The red, blue, sand purple shift of light coming from distant stars is well established. If this shift is not due to the movement of the source (Doppler effect) the change must occur in transit, but this is impossible because at the speed of light in a vacuum there is no time and change cannot happen.
Math is used to support theory. In the general relativity theory there is a denominator (1-d/ct). Since d = ct this is dividing by zero, but that is acceptable when it supports the faith.
This is why I say the discussions are futile. Even though Einstein said one experiment could prove him wrong, in reality nothing can prove him wrong to the believers.
Herb
Nepal
| #
My faith in modern physics is based on the fact that its tenets have been confirmed and verified EXPERIMENTALLY countless times.
You have no clue what you are talking about.
(You are BANNED for hijacking a member’s name in good standing) SUNMOD
Whokoo
| #
Conniption
Immortal600
| #
To the moderators (all 3),
You are making a serious mistake in banning LOL@KKK. The poster that goes by the moniker ‘Nepal’ should be banned because he’s using someone else’s name on Joe Postma’s site to confuse people here. He is mentally deranged and has shown as much in the last 5 years on various sites (Dr. Charles Anderson, Dr. Ed Berry, Joe Postma, and Cfact. He is wrong in his science and all those sites have shown that. The only reason that clown is here is because he was banned from Joe’s site. He is a 2nd law denier and by banning LOL@KKK you have taken his side. MISTAKE. Just watch what he posts in response to this and you’ll see. He can’t deny that he has lost interactions with Drs. Anderson and Berry. He claims they are wrong and that only he is right. He is a deluded fool full of himself for no reason.
(I removed my comment thinking you are not that LOL@kook…. fella after all, but you need to stop complaining in PUBLIC about moderation actions/decisions you can send e-mail about your concerns where all three administrators can read it) SUNMOD
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Herb Rose
| #
Hi Immortal,
Nobody should be censored or banned because of what they believe, only for abuse.
Joe Postma gets emotionally attached to his beliefs and becomes insulting add ends up censoring people because they disagree with him. For someone who rails against sophistry, he frequently uses it in his arguments (not discussions). When he was claiming that cold can never heat hot, I argued that with convection, where the transfer energy is from collisions, cold can heat hot because mass is irrelevant in the conservation of momentum. His response was denial of the basics, name calling, and refusal to post my responses.
When it comes to physics we are all wrong. It is the tenacious adherence to our existing beliefs that inhibits discussions of alternate ideas and hinders the advancement of science.
The laws of thermodynamics were created when the belief was that light was energy and temperature was the expression of that energy. According to them all object with energy will radiate energy which, will decline with distance from the object. Energy can only flow from objects with more energy to objects with less energy. This means there is a point between objects where the energy from the two objects will be equal (the equilibrium point) and this will form a barrier across which energy can not flow. An object cannot transfer energy to another object unless its energy, at that distance, is greater than the source energy of the other object. Any increase in energy coming from an object will only result in a shift in the equilibrium point, not a transfer of energy.
The fact we can see light from distant stars (even during an eclipse), which passes through the greater energy radiated by the sun and the even stronger energy radiated by the Earth, means our eyes are absorbing energy radiated by that distant star in violation of the laws of thermodynamics.
Herb
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Immortal600
| #
I see my last comment disappeared. A shame.
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Herb Rose
| #
Hi Immortal,
My reply went down with the ship.I hope you had a chance to read it.
Herb
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Immortal600
| #
Herb,
No, I didn’t get a chance to see it. Could you repeat it, please.
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Moffin
| #
Hi Herb Rose.
I searched three times for your comment and immortal’s.
It appears the moderation is immoderate. I do however, support the removal of personal abuse comments.
Cheers. Moff
(Anyone can use the contact us link to post a concern about someone being a problem) SUNMOD
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Immortal600
| #
Moffin,
My comment was far from abuse. I disagreed with banning LOL@KKK. If anybody should get the boot it’s ‘Nepal’ aka ‘evenminded’ and other monikers. He’s even used MINE on Dr. Berry’s website. The mods have made a mistake and essentially that’s what I said.
(Please stop discussing moderator action/decisions in public, I am now investigating because of other voices I received properly by e-mail which YOU can do also) SUNMOD
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Immortal600
| #
Understood.
Moffin
| #
Hi Immortal600
I followed the debate. Using my other sock puppet name, MattH, I made the comment about challenging the hypothesis but not abuse the individual.
So LOL@KKK. was good on the science but lost it on the incivility. Never give power to (alleged) imposters and the fraudulent by lowering standards to the sewer or abattoir.
Curiously, in a world full of shady and corrupt behaviour concerning climate, vaccines that do not work but cripple maim and kill, and free and frank discussion being removed by corrupt tech companies, we appear to have a troll on this site inserting comments in to other peoples comments and free and frank discussion on site moderation is under attack.
Possibly a job with Pfizer, Klaus Schwab, or North Korea’s Ministry of truth would be appropriate for SunMod. but deficating in other peoples comments should be a hanging offence in an environment of striving for free speech with civility, and democracy.
Have a nice day. Matt the Moffin
(As the new Administrator I can indeed post inline comments when MODERATION requires it and discussing moderator decisions/actions in public is always unacceptable I don’t censor the debating/discussing part at all just the part about over the top and even continuous uncivil remarks which sullies debate and possibly start thread deviation from the topic. Having more than one name is normally considered a bannable offense in 99% of forums/blogs suggest you stick with just ONE name here)
(NEPAL has been banned for name hijacking) SUNMOD
Herb Rose
| #
Hi immortal,
My comment was that banning or censorship should only be done for abusive behavior, not for other reasons. The changing of a name may well be the result of censorship.
Discussions should be based on beliefs and all beliefs should be allowed. In physics we are all wrong and different points of view is the way science advances.
Joe Postma gets emotionally attached to his beliefs and, even though he derides sophistry, he is not at all hesitant to use it when someone disagrees with him. When he stated that cold can never heat hot I disagreed with him pointing out that in the transfer of energy by convection the law of conservation of momentum determines the distribution of energy and masses remain constant. It is quite possible for an object with little mass to transfer energy to a larger mass with more kinetic energy. Joe reacted with denial of basics, name calling and censoring of my comments, as I tried to explain my reasoning.
Much of what we believe comes from past theories where there were different knowledge about and understanding of reality.
An example is the laws of thermodynamics where light was thought to be energy and temperature an expression of energy (hence the misnomer “thermodynamics” rather than Energy dynamics.)
The laws of thermodynamics say that all objects with energy will radiate energy. That energy will decrease with increasing distance from the source.
Energy will flow from higher levels of energy to lower levels of energy.
Agreed?
This means the energy radiated by an object will continue to flow and decrease until it meets an energy flow from a different object of equal strength (equilibrium point). This point is a barrier preventing energy flow from both objects. If one of the objects has an increase in radiated energy this will only result in a shift in the equilibrium point. The only way an object can transfer energy to another object is if the energy level coming from it is greater than the source energy of the other object.
We are able to see light coming from distant stars (even in a total eclipse) even though that energy must pass through the greater energy emitted by the sun and the even stronger energy close to the Earth. What we see is a result of energy being radiated by the distant star and our observation is a violation of the laws of thermodynamics.
Herb
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Immortal600
| #
Herb,
Thanks for your reply. I appreciate it.
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Max Polo
| #
I will propose a different solution for the two panes problem. Instead of starting from the theory, I will start from observation (experiment). Work permitting, I plan on refining my experiment and the relevant report, but I already have got a bunch of data that seem to support my solution to the problem. This solution is different than both (fake) Nepal’s as well as Postma’s.
My experiment is briefly explained over at Zoe’s blog: https://phzoe.com/2020/03/04/dumbest-math-theory-ever/#comments
The observations say that the passive pane (2) will always have a quite lower temperature than the irradiated pane (1), depending on panes distance (i.e. depending on view factor f). If we did the test in the empty space, we’d have T2 = (f/2)^0.25T1 with T1 = (S/(2sigma))^0.25 (I will demonstrate this solution later on), where the symbols are :
Sigma: S-B constant
Blackbody behavior for both panes
T1 = temperature of pane 1 irradiated with flux S on one side (point source)
T2 = temperature of pane 2 irradiated by pane 1
f = view factor from one side of pane 1 to the facing side of pane 2 (obviously f=f12=f21)
S = solar flux on one side of pane 1
I made my experiment in ambient air in my home’s backyard, thus not in empty space, so I’ve introduced corrections for both air temperature (convection) and background radiation (from ground and sky), therefore the solution to the equations cannot be found analytically but only numerically (excel solver). The concept expressed by the above simple analytical solutions still holds true, however, the air attenuates greatly any alleged back-radiation heating (approximately from a calculated value of 4°C to a calculated value of 2°C). The experiment tells that T1 is not increased by 2°C (not even by 0.2°C) if pane 2 is added. Back-radiation is unable to heat pane 1 up. T1 remains constant with or without the second pane. The panes are 4 cm apart and pane 2 is always quite cooler than pane 1 (for instance: pane 1 = 50°C, pane 2 = 32°C).
Here below is how the relevant energy balance equations should be set up (in the simple case of empty space) in order to match measured data. Below equations represent the steady-state (temperatures not changing over time) and consider the ENERGY (not heat) emitted/received by each pane :
Pane 2, energy (not heat) balance, based on S-B law (blackbody) :
Energy flux in = energy flux out :
fsigmaT1^4 = 2sigmaT2^4
Pane 1, energy (not heat) balance, based on S-B law (blackbody) :
Energy flux in = energy flux out :
S = 2sigmaT1^4
Where S is the incoming solar flux on one side of the pane 1 (W/m2)
Note that the sun is the only energy source, pane 2 is NOT an energy source because is a passive element that in absence of the sun (that heats it via pane 1) would be at 0 K (or at the temperature of the background). Note that it would be incorrect, for pane 1, to add the energy flux from pane 2 since pane 2 is a passive element that is not introducing any new energy into the system, but rather is getting its energy from pane 1, thus its temperature will be solely dictated by the temperature of pane 1 depending on their mutual distance.
Using the above “pure S-B energy flux” approach and suitable corrections due to the real conditions (air convection, background temperature), one should be able to predict the two panes’ temperatures in a way that fits measured data, that shows no back-radiation heating whatsoever (different from (fake) Nepal’s statement), and the pane 2 temperature as being quite lower than pane 1 even if f=1 (different from Postma’s statement).
Interesting would be to make this test in the vacuum to remove some erratic fluctuations due to little air puffs that are the real problem of this experiment…maybe when I have time I’ll propose this to someone having the proper facilities.
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Herb Rose
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Hi Max,
Here’s some thoughts on the experiment.
If the experiment is not done in a vacuum the plates will be transferring energy to and absorbing energy from air molecules by convection.
If the experiment is done in a vacuum and one plate is an opaque blue that reflects a wavelength and the other is a clear (same blue) plate that will transmit and emit that wavelength, it means that neither plate can absorb or transmit energy to the other plate and they cannot equalize with each other. Equalization does occur because both plates are equalizing with the energy field they are in which absorbs and transfers all wavelengths.
Herb.
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Max Polo
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Thanks for your comment, Herb. I know that the experiment should be done in a vacuum. But I decided to simplify the execution by doing it in ambient air and correcting the results by taking into account the air convection effect by a proper equation. The full model and equations can be seen at this link: https://drive.google.com/file/d/1UXyiSyVyH-kOrso4ehMhuFGxMc86f6P2/view?usp=sharing
The equations that I wrote in my previous post were just an over-simplified version to explain the concept. Obviously, my model does take into account both the background temperature and the air temperature, to calculate what the expected back-radiation heating would be if the “consensus” approach was correct. As it can be seen, the back-radiation heating effect is calculated at 2.0°C, but 0°C is actually measured despite the very high sensitivity of the infrared thermometer, which can easily detect even 0.1°C variations.
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Rocky
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Max,
There is a problem with your solution. According to your equation for Pane 2, there is an energy flux out of sigmaT2^4 from the side of the plate facing Pane 1. However, you do not have that energy being absorbed by Pane 1. Pane 1 is a blackbody and so all radiation incident on Pane 1 must be absorbed and so it must show up in the energy balance for Pane 1.
Herb’s comment is also relevant. This experiment needs to be done in a vacuum. Additionally, if the surroundings are not at 0K then the temperature of ambient surroundings needs to be incorporated into the solution if you are going to attempt to compare to an experiment.
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Rocky
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Max,
Additionally there is a lack of symmetry in your equations. You have an energy flux into Pane 2 from Pane 1 as fsigmaT1^4, but no energy flux into Pane 1 from Pane 2 as fsigmaT2^4.
What would happen to your equations if you placed an identical source S on the other side of Pane2?
Would the Pane 2 equation become,
Energy flux in = energy flux out :
S + fsigmaT1^4 = 2sigmaT2^4
Reply
Max Polo
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Rocky, thanks for your comments. Actually, my equation for pane 2 is no different than the “consensus” equation. And you can get the same result for instance writing, for pane 2, “heat-in from pane 1 equals energy lost to space”: Q12 = sigmaAf(T1^4-T2^4) = sigmaA(1-f)T2^4 + sigmaAT2^4 from which you get T2 = T1*(f/2)^0.25. What is different from the “consensus” approach is the equation for pane 1, which deliberately neglects the presence of pane 2 because it is a passive element that cannot make a difference to pane 1. Indeed, this is what is observed in the experiment (no difference in pane 1 temperature when pane 2 is present), although – as explained in my previous post – the experiment is done in the air with proper corrections for convection heat and background radiative effect. At the link in my earlier post, there is a more detailed explanation, together with the equations used to simulate the “consensus” approach that aims to predict the expected back-radiation heating … that never happens.
Reply
Rocky
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“Actually, my equation for pane 2 is no different than the “consensus” equation.”
Agreed. However, your equation for Pane 1 is different. My question to you about Pane 2 was intended to illustrate the asymmetry in your treatments of the two panes.
So my question remains, what would happen to your equations if you placed an identical source S on the other side of Pane2?
“What is different from the “consensus” approach is the equation for pane 1, which deliberately neglects the presence of pane 2 because it is a passive element that cannot make a difference to pane 1.”
Of course it can make a difference. Pane 1 is now emitting to a finite temperature plate instead of to space. In fact, your equations are at odds with the simple Stefan-Boltzmann equation for the heat transfer between two parallel plates, q12 = sigma (T1^4 – T2^4).
NASA designs its systems based on these radiative heat transfer principles. They would have noticed if their calculations were that far off.
(You are BANNED) SUNMOD
Reply
Max Polo
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“My question to you about Pane 2 was intended to illustrate the asymmetry in your treatments of the two panes”: asymmetry comes from a simple “asymmetric” fact: Pane 1 is externally energized thus it reaches a temperature that is dictated by (a) the energy source and (b) by the background at a fixed temperature (let’s say 0K, deep space). Pane 2 is NOT externally energized and is not at a fixed temperature: it is just a passive, neutral element that gets rid of all the energy it gets, therefore it cannot alter Pane 1 temperature. Remember that this consideration comes from empirical observation and tries to interpret it, rather than neglecting reality and saying how reality should behave in theory (while it does not).
“So my question remains, what would happen to your equations if you placed an identical source S on the other side of Pane2?” My equation would predict Pane 2 to achieve an intermediate temperature between that of Pane 1 and that dictated by the additional energy source. Just like what happens with conduction, for instance: two heat sources at 1000°C, once they are put in physical contact, generate a whole body at 1000°C (not something at 2000°C, nor at 1514,00931°C…).
“Of course it can make a difference. Pane 1 is now emitting to a finite temperature plate instead of to space.” You claim something that is disproven by empirical evidence, that shows that Pane 2 does not make any difference. This is what my experiment is all about.
“In fact, your equations are at odds with the simple Stefan-Boltzmann equation for the heat transfer between two parallel plates, q12 = sigma (T1^4 – T2^4)”
Incorrect, I do use the heat transfer equation to determine the intermediate temperature achieved by Pane 2, and with the proper corrections (heat convection) it perfectly predicts measured values. My “theory” explains what HAPPENS, not what I think should happen.