So I’ve been presented with a comment inside a thread at someone’s blog. Being the naturally curious sort, I read the article linked and found no reference to the comment at this blog. I did find it in a thread comment by “LOL@Klimate Katastrophe Kooks” – not joking about the moniker.
“Between two objects, one warmer and one cooler, the chemical potential increases in the intervening space as one ascends the energy gradient from cooler to warmer object. Thus photons from the cooler object will be subsumed before they ever reach the warmer object because warmer objects have higher energy density at all wavelengths than cooler objects.” -Hyperphysics Fails at Basic Science, Logic, Reasoning, and Math (https://climateofsophistry.com/2021/11/17/hyperphysics-fails-at-basic-science-logic-reasoning-and-math/)
There is a bit of math in the thread where the DOF appears mixed up but the math isn’t the fun bit. The fun bit is that the photons simply vanish by a process I’ve never heard of called photonic subusmation, which although I have significant experience in optics, I’ve never heard of.
I can imagine so many experiments that would prove out this subsumation were it true, but the thing which proves this concept completely moot, is the room temperature bolometer. Like this one:
Bolometers are detectors used to measure incident IR radiation. They are very sensitive to thermal radiation and are predominantly used in the IR spectrum between 10 to 5000µm (30THz to 60GHz).
The FLIR TG165 is sensitive in the range of CO2 global warming DOOM! 7.5–14 µm wavelength.
Well the sensor elements are basically imperfect blackbodies, designed to absorb in the range specified. A lens which transmits in that same range allows the light to pass through unimpeded. I believe Ruby is a good lens material in this range. The first one of these I saw was in the 90’s and it cost $10K.
What is interesting is that this is an uncooled bolometer — meaning room temperature blackbody. So can a room temperature sensor measure something colder than itself like an ice-cube or is NO energy transferred per photonic subsumation?
This device apparently can measure to -25C:
-25°C to 300°C (-13°F to 572°F)
But cannot be operated below -10 C:
-10°C to 45°C (14°F to 113°F)
This means a device can measure the emitted radiation from an object cooler than itself. The radiation is affecting the physical temperature of the micro sensor array.
So if photons emitted from a cold body are subsumed before reaching the sensor and no energy is transferred, this would NOT work. Sky thermodynamics fail.
If the photons prevent an equal photon from being emitted by the warmer object as the cold photon is subsumed, you get exactly the same energy transfer as you do with classical thermodynamics. Sky thermodynamics fail again.
I fail to imagine a third option so the concept of the photonic absorption is now rendered moot to the discussion of climate by a simple off-the-shelf bolometer device. The first argument is proven wrong by the bolometer’s existence, the second is mathematically equivalent to standard understanding of thermodynamics so CO2 warming would still exist.
The completely unfounded, unscientific concept still includes an electromagnetic rewrite of all of physics which is flatly stupid …but either way — failure.
the Air Vent 
[[I can imagine so many experiments that would prove out this subsumation were it true, but the thing which proves this concept completely moot, is the room temperature bolometer… The (Tektronics) FLIR TG165 is sensitive in the range of CO2 global warming DOOM! 7.5–14 µm wavelength.]]
Here’s the specs:
https://www.flir.com/products/tg165-x/?vertical=condition+monitoring&segment=solutions
Hmm… a killer refutation of my entire understanding of thermal physics. I’m bleeding… 🙂
[[This device apparently can measure to -25C: -25°C to 300°C (-13°F to 572°F). But cannot be operated below -10 C:-10°C to 45°C (14°F to 113°F)]]
Let’s check their figures:
Radiation of a wavelength of 7.5 microns to 14 microns has a Planck blackbody peak power (Wien) temperature of 113.2C to -66.1C, according to this handy free online calculator:
https://www.omnicalculator.com/physics/wiens-law
Here’s what Wiki says:
“Some long-wave cameras require their detector to be cryogenically cooled, typically for several minutes before use, although some moderately sensitive infrared cameras do not require this. Many thermal imagers, including some forward-looking infrared cameras (such as some LWIR enhanced vision systems (EVS)) are also uncooled… Medium-wave (MWIR) cameras operate in the 3–5 μm range. These can see almost as well, since those frequencies are less affected by water-vapor absorption, but generally require a more expensive sensor array, along with cryogenic cooling.” – https://en.wikipedia.org/wiki/Forward-looking_infrared
Back to you:
[[If the photons prevent an equal photon from being emitted by the warmer object as the cold photon is subsumed, you get exactly the same energy transfer as you do with classical thermodynamics.]]
Classical thermodynamics is the only kind. You claim there’s another kind? It’s the only branch of physics that will never be overturned. It says that a colder body can’t raise the temperature of a hotter body by any means, including radiation. I’m not sure that subsuming photons is prohibited by the need to emit an equal photon since a hotter black body is continually emitting photons at that wavelength anyway, but that’s all that reflection is, so who cares if a colder photon is reflected or absorbed and reemitted, it still can’t raise the temperature of a hotter body. Why can’t you accept that?
Wait! The FLIR meter is measuring the Planck blackbody radiation peak emitted from objects it’s looking at which are at those low temperatures, meaning that it sees the spectrum and uses software to estimate the peak. Yes, the sensor can be hotter and still register those photons striking it, although its own temperature will increase the signal-to-noise ratio. Same principle as touching dry ice or holding your hands near it and feeling the cold. It’s not sending its own photons to it, making your argument ridiculous, and since those photons can be arranged to travel mostly untouched through the lenses and clear air., even moist air, it strengthens my understanding not yours. Your diatribe actually has nothing to do with what you were trying to disprove, which is no surprise, because a killer refutation would require actual measurement of the sensor’s quantum operation, which is prohibited by the quantum measurement problem.
[[The completely unfounded, unscientific concept still includes an electromagnetic rewrite of all of physics which is flatly stupid.]]
That’s what you’re doing, rewriting the Second Law to have a quantum level exception to push a you know what concept. Funny, that’s what happened after the gigantic U.N. IPCC octopus took over and hijacked climate science to frame CO2 as causing global warming, otherwise the hoax would be too obvious to fool the unscientific and they’d have to be disbanded and find another hoax to push global Marxism.
Here’s a peer-reviewed article explaining how black body radiation maximizes entropy increase, pulling the rug under the IPCC’s feet:
https://asmedigitalcollection.asme.org/IMECE/proceedings-abstract/IMECE2006/47640/217/320872
For laffers, here’s a physics blog that has been trying to save the IPCC a little too hard, starting out with physicsts telling the 2nd Law like it is, only to be inundated by a bunch of IPCC trolls trying to muddy the waters. Where’s your comments?
https://www.physicsforums.com/threads/cooler-objects-able-to-increase-the-temperature-of-warmer-objects.932598/
BTW, you really should have sent your comments to the blog involved, Climate of Sophistry, which is run by an anti-IPCC astrophysicist who has had many battles with them that always end by them refusing to answer his objections and canceling him. This blog has a tiny readership, but it’s one of a bunch of IPCC-owned blogs trying to tell truth to power, although admitted a few are run by scientific crackpots. Here’s my cool free list, the most complete available:
http://www.historyscoper.com/climateblogs.html
In closing:
The bottom line is that all quantum effects are speculation, and that’s all this blog article is. But why fix on this tiny point in my refounding of climate science freed of IPCC hijacking? Duh, what you really want is clearly to keep your claim going that CO2’s 15 micron -80C thermal radiation can cause global warming, as if you’re a paid IPCC shill. Sorry, you didn’t defend them at all, but made me look more right by failing to directly disprove me with a direct theoretical scientific argument. Duh, and I know why: there isn’t any. When will you give up, admit I’m right and help me get the IPCC shut down, the money train stopped and refunded, and all its fake climate scientists forced to get new careers while the global Marxists search for another hoax to push? When are you going to advertise my killer disproofs on your main page and in all your margins as the best use of your readers’ time? Sorry, I’m not on Big Oil’s payroll so you’d have to go totally amateur like moi.
I linked to the peer-reviewed blog establishing black body radiation increase because the entire IPCC house of cards is built on an alleged Earth-Sun energy balance, as if there’s no such thing as entropy dispersion of energy into the heat death of the Universe when CO2 is concerned, which explains why satellite measurements of TOA Earth surface radiation shows a big notch at 15 microns. This is a macro-level law, not quantum level.
https://en.wikipedia.org/wiki/Entropy_(energy_dispersal)
You still insist that I’m a crackpot and that your IPCC gurus have the absolute truth from God on the subject of thermodynamics? Why don’t you try the acid test of telling your favorite IPCC climate science about their 15 micron -80C problem and asking for a definitive scientific refutation? Watch them block you 🙂
Let me cut to the chase: The only quality time you’ll ever spend studying real thermal physics until the IPCC falls is my cool free Internet pubs.
Start here:
https://www.quora.com/Why-do-we-say-no-net-heat-flow-between-two-objects-of-different-temperatures-but-not-no-heat-flow/answer/TL-Winslow
https://www.quora.com/Are-there-any-ways-that-people-can-help-prevent-climate-change-without-changing-their-lifestyle-or-buying-new-things/answer/TL-Winslow
http://www.historyscoper.com/climatescience101.html
—000—
Oops! I meant the detector measures a minimum peak not a maximum peak at the frigid temperature wavelength. It’s not absorbing energy and heating up. Same principle as detecting a hole in a sunlit landscape by the black spot.
HistoryScoper wrote:
“Same principle as touching dry ice or holding your hands near it and feeling the cold.”
Exactly so. Placing your finger near dry ice makes the energy density gradient between finger and dry ice steeper than it would be between, for instance, finger and room temperature wall. Photons are thus emitted by the finger to the dry ice at a faster rate than they would be to the wall, which cools the finger, which the nerves sense.
It is the same with Jeff Id’s bolometer. Pointing the sensor toward a cooler target makes the energy density gradient between sensor and target steeper than it would be between, for instance, sensor and room temperature wall. Photons are thus emitted by the sensor to the cooler target at a faster rate than they would be to the wall, which cools the sensor, which the bolometer electronics convert to a calculated temperature of the target.
Pointing the sensor at something at exactly the same temperature as the sensor would mean no photons to or from the sensor (assuming a view factor which prevents emission to or from other objects at various angles)… the electronics interpret that as the target being the same temperature as the sensor.
Directly analogous to Pictet’s experiment of 1800… and yet even today, we have some (Jeff Id apparently among them) who claim that ‘cold’ can be transferred.
Should Jeff deny this, he’s explicitly stated that photons from the cooler target strike and are absorbed by the warmer bolometer sensor (in violation of 2LoT, but let’s set that aside for now). If the bolometer sensor is already at a higher temperature than the target, and those photons purportedly warm the sensor even more, just how does he propose that the bolometer electronics register that target’s lower temperature? Hence, he can only be espousing the transmission of ‘cold’. Those photons from the cooler target must be cooling down the warmer bolometer sensor, right? Not warming it.
Or he can (finally, reluctantly, painfully given his past faux pas as regards radiative energy transfer) admit that the photon flow in this case is from bolometer sensor to cooler target, which cools the sensor, which the bolometer electronics register as the target temperature. All in accord with the fundamental physical laws, and exactly as Pictet showed 222 years ago.
Poor dumb bastard, look in the mirror and laugh at yourself, the bolometer you show operates not via thermodynamic’s but via electromagnetics, the photoelectric effect; it does not mean as you claim that the radiation from the cold is heating the sensor pixels on the CMOS/CCD chip.
The photon are being converted to electrons and then read out, what the fuck is wrong with you people.
It is the photoelectric effect.
Bolometers are not ccd’s
Fail
Still got fuck all to do with photons from a cold object or environ ”warming” the warmer sensors as you claimed you poor dumb bastard…
Infrared (IR) camera is a device that measures incident IR radiation emitted by
objects and therefore allows a remote way to measure the temperature of an object without contact. IR radiation is electromagnetic radiation, which wavelength
region, 0.75–1000 µm, lies between visible light and microwaves.
All objects with a temperature over absolute zero emit IR radiation. (Hamamatsu and Solid State
Division, 2011) The main part of the IR camera is a focal plane array (FPA) that
consists of a two-dimensional array of individual sensing elements. Those sensing
elements collect incident radiation energy from environment and convert it into an
electric signal, thus creating a thermal image of the scene. (Meola et al., 2015)
Click to access 250168454.pdf
The bolometer is a sensor that warms under radiant heat. I put the wavelength sensitivity range of this one in the post. Right there in co2 back-radiation land.
You are struggling a bit here I see.
The problem you now have is that this is proven absorption of em radiation in the co2 band of a warmer object by radiation from a cooler object.
Energy moving from cold to hot.
Net energy still moving from hot to cold.
I wrote what you’re quoting in your article. ‘Subsumed’ into the background EM field means they are no longer detectable except as the background temperature, or by a sensor of lower temperature than the background temperature.
Remember, per QFT, a photon is the electric and magnetic interaction oscillating in quadrature about a common axis, a circle geometrically transformed into a spiral by dint of the photon’s necessary movement through space-time. ‘Necessary movement’ because the photon has no invariant-mass… so if the photon had a rest frame, no mass and no relativistic momentum equals nothing. Thus all fundamental particles which have no invariant mass have no rest frame. They move at c (which, per GR, varies with a varying gravitational field, as Einstein reiterated many times… it is only with SR that light always moves at a fixed c, but SR was superceded by GR).
———-
Einstein wrote:
1913: “I arrived at the result that the velocity of light is not to be regarded as independent of the gravitational potential. Thus the principle of the constancy of the velocity of light is incompatible with the equivalence hypothesis.”
1916: “In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity.”
1920: “Second, this consequence shows that the law of the constancy of the speed of light no longer holds, according to the general theory of relativity, in spaces that have gravitational fields. As a simple geometric consideration shows, the curvature of light rays occurs only in spaces where the speed of light is spatially variable.”
This is how gravitational lensing (ie: the curvature of light rays around a gravitating body) occurs, which was what originally corroborated Einstein’s theory. Light doesn’t follow a straight line (the path of least space), it follows the path of least time. A gravitating body dilates space and slows down time, thus light attempts to bend around gravitating bodies.
———-
Thus, the photon (which we view as a sinusoid) is actually a spiral.
This is because a sinusoid is a circular function, and a circle spread over space-time is a spiral, and a spiral viewed from one dimension is a sinusoid.
(Take a Slinky outside on a sunny day and align it such that the sun falls perpendicular to the axis of the Slinky. Look at the shadow as you stretch the Slinky. It’s a sinusoid.)
You’ll note the peak amplitude of the sinusoid is analogous to the radius of the circle, the peak-to-peak amplitude is analogous to the diameter of the circle, and the frequency of the sinusoid is analogous to the rotational rate of the circle. You’ll further note the circumference of the circle is equal to 2 pi radians, and the wavelength of a sinusoid is equal to 2 pi radians, so the wavelength of the sinusoid is analogous to the circumference of the circle.
Thus the magnetic field and electric field (oscillating in quadrature about a common axis) of a photon is a circle geometrically transformed into a spiral by the photon’s movement through space-time. This is why all singular photons are circularly polarized either parallel or antiparallel to their direction of motion. This is a feature of their being massless and hence having no rest frame (if a photon had a rest frame, no rest mass and no momentum equals nothing, so massless particles must remain in motion), which precludes their exhibiting the third state expected of a spin-1 particle (for a spin-1 particle at rest, it has three spin eigenstates: +1, -1, 0, along the z axis… no rest frame means no 0-spin eigenstate). A macroscopic electromagnetic wave is the tensor product of many singular photons, and thus may be linearly or elliptically polarized if all singular photons comprising the macroscopic electromagnetic wave are not circularly polarized in the same direction.
Now, the photon is also, per QFT, a persistent perturbation of the EM field above the field ambient. If a photon encroaches upon a region of space-time that has higher chemical potential than the photon has, that photon can do no work (it has no Free Energy), it will not be detectable (except by a sensor of lower temperature, lower energy density (because remember, temperature is a measure of radiation energy density, equal to the fourth root of energy density divided by Stefan’s Constant, per Stefan’s Law)).
And that is what is meant by ‘subsumed’… the photon is no longer persistent, and thus it is no longer detectable (again, except by a sensor of lower energy density than the photon field being measured, or as the ambient).
As to your bolometer, this is analogous to Pictet’s 1800 experiment. Photons from the cooler atmosphere are not flowing from the atmosphere to the bolometer sensor, photons are flowing from the sensor to the cooler atmosphere.
The IR sensor, in an environment cooler than the sensor, is emitting photons, thus it cools. The current flow through the sensor is compared to a reference resistor’s current flow in something like a Wheatstone Bridge. The reference resistor is typically physically affixed to the back of the IR sensor.
Why is the photon no longer detectable if its chemical potential is zero? Because energy cannot flow unless there is Free Energy (ie: unless work can be done), and zero chemical potential means no Free Energy, thus no energy flow. Remember that a photon is nothing but a quantum of energy.
The S-B equation explains it:
Idealized Blackbody Object (assumes emission to 0 K and ε = 1):
q_bb = ε σ (T_h^4 – T_c^4) Ah
= 1 σ (T_h^4 – 0 K) 1 m^2
= σ T^4
Graybody Object (assumes emission to > 0 K and ε < 1):
q_gb = ε σ (T_h^4 – T_c^4) Ah
Do keep in mind that idealized blackbody objects are provable contradictions that do not and cannot exist. The closest we can come are laboratory blackbodies which are highly absorptive and emissive within a certain wavelength range… but an idealized blackbody must absorb all radiation incident upon it, it must emit all radiation it absorbs and it must have zero thermal capacity (an idealized blackbody cavity must have all its energy in the radiation field and none in the cavity walls as a pre-condition of Kirchhoff's Law of Thermal Radiation).
Remember that a warmer object will have higher energy density at all wavlengths than a cooler object:
So the climate alarmists have bamboozled the world by treating real-world graybody objects as though they're idealized blackbody objects, and using the incorrect form of the S-B equation (the form meant for idealized blackbody objects):
q_bb = σ T^4
And slapping emissivity onto that (sometimes):
q_bb = ε σ T^4
And they use that incorrect form upon real-world graybody objects when they should be using:
q_gb = ε σ (T_h^4 – T_c^4)
If you don't believe me, do the calculations for the Kiehl-Trenberth Global Energy Budget graphic or any subsequent similar graphics:
They can't get to 390 W m-2 surface radiant exitance except by treating the surface of the planet as though it's an idealized blackbody object (emission to 0 K, emissivity = 1).
So what they've done is isolated all objects into their own systems, emitting to 0 K, so the objects cannot interact through the ambient EM field, then they 'connect' the objects by subtracting a wholly-fictive 'cooler to warmer' ENERGY FLOW from the real (but too high because it was calculated for emission to 0 K) 'warmer to cooler' ENERGY FLOW.
But that's not how the S-B equation is meant to be used. It's meant to be used to subtract cooler object ENERGY DENSITY from warmer object ENERGY DENSITY (remember that temperature is a measure of radiation energy density per Stefan's Law).
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)
q = ε σ (T_h^4 – T_c^4)
∴ q = ε σ ((e_h / (4σ / c)) – (e_c / (4σ / c))) Ah
Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))
∴ q = (ε c (e_h – e_c)) / 4
Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
W m-2 = (m sec-1 (ΔJ m-3)) / 4
∴ q = σ / a * Δe
Canceling units, we get W m-2.
W m-2 = (W m-2 K-4 / J m-3 K-4) * ΔJ m-3
One can see from the immediately-above two forms of the S-B equation that the Stefan-Boltzmann equation is all about subtracting the radiation energy density of the cooler object from the radiation energy density of the warmer object.
It is the radiation energy density gradient which determines radiant exitance of the warmer object.
————————-
One can also analogize thermodynamics to electrical theory (since people think in terms of current, voltage, resistance, etc. more readily than radiant exitance, energy density, emissivity, etc.):
Circuit simulator: https://tinyurl.com/yzo8hak9
I have used the circuit simulator to arrive at the same answer as the S-B equation arrives at, to a precision of 3.8 parts per 100 trillion.
So essentially, the climate alarmists claim a battery of 1.5 V, wired '+' to '+' and '-' to '-' with a 12 V battery, will do work upon that 12 V battery when they claim that radiation from a cooler object will be absorbed by a warmer object. And that's ludicrous.
And they claim a 12 V battery with a 1 Ohm resistor hung off it will charge a 12 V battery with a 1 KOhm resistor hung off it, if the batteries are wired '+' to '+' and '-' to '-'. (with resistivity analogized to emissivity). That is equally ludicrous.
————————-
That brings up some really interesting mathematical coincidences as regards our atmosphere:
Radiant Exitance =
= ε σ (T_h^4 – T_c^4) Ah
= (ε c (e_h – e_c)) / 4
= σ / a * Δe
= Faraday’s Constant (Ah) * (Molar Mass of Atmosphere / Ratio of Atoms Per Particle)
= Faraday’s Constant (Ah) * Moles of electrons per Mole of atmospheric particles
= (((Faraday's Constant (J electron-1) * Number of Electrons per Mole of Atmosphere) / 3600) / Stefan-Boltzmann Constant)^0.25
= (((Faraday's Constant (J mol-1) * Moles of Electrons per Mole of Atmosphere) / 3600) / Stefan-Boltzmann Constant)^0.25
————————-
Mheh…
“T = 4√(e/a)”
– should be –
“T = 4^√(e/a)”
In other words, temperature is equal to the fourth root of energy density divided by Stefan’s Constant, thus temperature is a measure of radiation energy density per Stefan’s Law.
[self snip] — found some content in the middle of the LOL book.
Hey i searched your comment for bolometer and found this:
So in this instance, you do admit the bolometer is responding to the energy differential between the colder object and the warmer sensor in exactly the same magnitude as classical electromagnitism where through superposition, radiation from the cooler source strikes the sensor. This is option 2 in my post. It is odd because in your next quote — you have also chosen option 1 of my post.
This is option 1 of my post.
So option 2 means that CO2 warming works as predicted by standard physics. You are literally claiming the exact same magnitude of predicted energy transfer as a climate model.
Option 1 means the bolometer above doesn’t work.
AS UNLIKELY AS IT SEEMS (to others), YOU HAVE CHOSEN BOTH ANSWERS AND CONTRADICT YOURSELF.
Which is it LOL, is it the bolometer doesn’t work, or is it CO2 back-radiation works?
Jeff Id wrote:
“So in this instance, you do admit the bolometer is responding to the energy differential between the colder object and the warmer sensor in exactly the same magnitude as classical electromagnitism where through superposition, radiation from the cooler source strikes the sensor.”
That’s not how ‘superposition’ works. But then, you not knowing that, means you don’t understand enough of the basics to even understand my primer above.
What you’ve described in a bolometer is analogous to Pictet’s experiment of 1800. The energy in that experiment flowed from the warmer measurement device to the cooler object. Mirrors were set up such that the object and the measurement device only had a view factor which allowed them to interact with each other, and nothing else. It’s easy to see that ‘cold’ is not transmitted, that energy only flows from warmer to cooler, from higher energy density to lower.
I could mathematically demonstrate that radiation from a cooler object doesn’t even incide upon a warmer object, let alone be absorbed by it (because remember, a warmer object will have higher energy density at all wavelengths than a cooler object), but you either do not or cannot grasp the fundamentals by which you could understand it.
Mheh, why not. Let me know if you understand this, Jeff. Especially the last two paragraphs.
I took this from the website of physicist Dr. Charles R. Anderson, PhD, but it’s standard cavity theory, so you should be able to find it in any good thermodynamics tome.
Do remember that this is all based upon the concepts in the book Thermal Physics, Second Edition, written by Philip M. Morse, Professor of Physics at MIT, co-founding editor of Annals of Physics, co-founder of MIT Acoustics Laboratory, first Director of Brookhaven National Laboratory, founder of MIT Computation Center.
Read this for comprehension, you retard. LOL
>>>>>>>>>>>>>>>>>>>>>>>>>>><<<<<<<<<<<<<<<<<<<<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>><<<<<<<<<<<<<<<<<<<<<<<<<<<
Let us examine an enclosure whose walls are at a constant temperature. The radiation within this enclosure is in thermal equilibrium with the walls of the enclosure. In such a case, the energy density of the radiation, e(T), is dependent on the temperature, but independent of the volume enclosed. If the volume enclosed is V, then the total radiation energy, U, is Ve(T).
For our enclosure in which the thermal properties are described only in terms of the intensive variables T (temperature) and P, the radiation pressure on the walls, and the extensive variables V and S, the entropy, we have dU = T dS – P dV. Or, T dS = dU + P dV.
The radiation pressure on the walls of the uniform temperature enclosure is equal to twice the momentum, p, component perpendicular to the wall of photons reflected off the wall times half the perpendicular velocity component, where the perpendicular velocity component tells us how many photons strike the wall per second and only half have a perpendicular velocity component toward the wall. The pressure P for a photon gas exerted in the x-direction on area A of the wall will be summed over all i = 1 to N photons:
P = ½ ∑ 2 pix vix / V = ⅓ U/V = ⅓ e,
U is just ∑ pix vix + piy viy + piz viz for photons, so for random motion, the summation over only one component is one-third of this sum. Note that the radiation pressure on the wall does not require that one photon incident upon the wall be absorbed and another emitted. One gets the same result with a photon reflecting off the wall without absorption just as one would calculate the pressure on a wall for a perfect gas under the assumption of elastic collisions with the wall. Such a gas of point particles has a pressure of ⅔ e however, since the average kinetic energy of the perfect gas particles is ½ the dot product of the momentum and the velocity and not the dot product of the momentum and the velocity as is the case for photons.
Thus, T dS = dU + P dV
= d(Ve) + P dV = V de + e dV + ⅓ e dV
= V (de/dT) dT + (4/3) e dV
We also have
T dS = T (∂S/∂T)_V dT + T (∂S/∂V)_T dV,
So we match up the terms for dT and for dV:
(∂S/∂T)_V = (V/T) de/dT
(∂S/∂V)_T = 4e/3T
Since the order of differentiation does not matter, it is also the case that
{∂[(∂S/∂T)_V]/ ∂V}_T = {∂[(∂S/∂V)_T]/ ∂T}_V
{∂[(V/T) de/dT]/ ∂V}_T = {∂[4e/3T]/ ∂T}_V
(1/T) de/dT = (4/3T) de/dT – 4e/3T^2
de/dT = 4e/T
e = aT^4
The equation for the radiation energy density is Stefan’s Law and a is Stefan’s constant. If we open a peep hole into this cavity, the flux of energy emitted has this energy density of aT^4, though an equal amount of energy has to then be supplied to the enclosure to maintain it at temperature T as it loses energy through the peep hole. Within the cavity emitting this energy, we do not know if the photons incident upon the walls are absorbed and replaced with the emission of another of the same energy or whether those photons are simply reflected from the wall. The above results are the same in either case.
When the cavity above is at a constant temperature Tf with the walls everywhere at that temperature, we know that the emission of photon power from the interior walls equals the absorption of photon power by the walls. If the cavity was at an initial lower temperature Ti and then was heated until it came to a new higher equilibrium temperature of Tf, the emissivity of the walls had to be greater than the absorptivity of the walls during that heating process in order to increase the energy density in the cavity from aTi^4 to aTf^4. Because the energy density depends on the fourth power of the temperature, a doubling of the temperature requires a factor of 16 times greater energy density. The emissivity of the walls during such a heating process must be much greater than the absorptivity to allow that 16-fold photon energy density increase. The emissivity and the absorptivity of the walls of the cavity are not just a function of the material. They are a function of whether the material is in radiative thermal equilibrium or not as well. It is very important to realize this.
When the cavity was at Ti in a radiative equilibrium condition, the emissivity and the absorptivity were equal, but when the cavity was further heated to the new equilibrium temperature of Tf, they were not equal when Ti < T < Tf. Conversely, during a cooling from a higher equilibrium temperature to a new cooler equilibrium temperature, the absorptivity of the walls would have to be greater than the emissivity of the walls.
Among the properties of the cavity with volume V in radiative thermal equilibrium at temperature T is that:
U = a V T^4
P = ⅓ a T^4
S = (4/3) a V T^3
We can calculate the chemical potential, µ, which measures the ease with which the number n of moles of photons adjusts to keep the energy density constant in the cavity in radiative thermal equilibrium:
µn = U – ST + PV
µ = 0
Photons are a very special kind of boson. When they are in radiative thermal equilibrium in a volume V at a constant temperature T, their chemical potential is zero. The number of photons in the cavity is strictly determined by the temperature of the walls.
Applying Bose-Einstein statistics for a boson gas with a chemical potential of zero and integrating over the photon standing wave states available in the cavity in radiative thermal equilibrium, one finds that
a = π^2 k^4 / 15 ɦ^3 c^3 = 7.5657e-16 J m-3 K-4,
where ɦ is Planck’s constant h/2π, k is the Boltzmann constant, and c is the speed of light. Note that a, Stefan’s Constant, is not the same as the Stefan-Boltzmann Constant.
The cavity I have discussed here is almost always called a black-body cavity, with the implication that every photon incident upon an interior wall is absorbed and replaced by another photon of equal energy. This is usually said to be a condition in which the emissivity and the absorptivity of the walls of the cavity are both 1. Yet, at no point was such a condition ever used in the derivation of any of the properties of a cavity of volume V under a condition of radiative equilibrium at temperature T. The only requirement was that a photon incident upon a wall was either reflected or if it were absorbed, then it had to be replaced by another photon emitted from the wall at the same energy, which is easily done because the photons in the cavity at radiative thermal equilibrium have a chemical potential of zero.
When the thermal radiative properties of a cavity in radiative equilibrium are applied to a surface which is not in thermal radiative equilibrium in a closed cavity condition, we must be very aware of the differences. When such a surface is surrounded by vacuum and all of its thermal conditions are described by its radiative properties, there will be an infinitesimal volume in the vacuum arbitrarily close to the surface which has an energy density e as described above from the equilibrium cavity condition.
Let us imagine two infinite planar surfaces which are very close compared to their dimensions, with only vacuum between them and beyond them. If these two surfaces are at the same temperature T and in equilibrium, the energy density e is clearly aT^4 everywhere in the volume between the two planes.
Now let us suppose that one of the two planes is radiatively heated from the back side and reaches a new higher temperature Th, at which the temperature becomes stable. As this hotter plane radiates energy at its higher temperature, it will increase the temperature of the nearby plane. If the cooler plane has a heat capacity, it will take a while to heat up to its own new warmer equilibrium temperature. Assuming this initially cooler plane only loses heat by radiating it toward the vacuum at absolute zero, it will achieve that equilibrium new temperature such that it is equal to that of the plane heated on its backside.
Once again the energy density e is clearly aT^4 everywhere in the volume between the two planes, where T is the new higher temperature.
If this second plane has some other uniform means of losing heat than radiation over its surface area, its new equilibrium temperature will not be as high as Th. Let the lower temperature plane have the temperature Tc. Now the energy density in the vacuum arbitrarily close to either side of the hotter plane will be aTh^4 and that arbitrarily close to the cooler plane on either side will be aTc^4.
This is a condition for a plane at a temperature T with a vacuum interface that the energy density e = a T^4. Said in a different fashion, the perfectly thermally conducting plane would not be at a temperature T if the energy density of photons in the space immediately at the surface of the plane was not aT^4. In the vacuum space in between the two planes the energy density will be a continuously decreasing function from a high value of aTh^4 at the hotter plane surface to a lower value of aTc^4 at the cooler plane surface. The hotter plane emits and absorbs photons as a surface at an energy density of aTh^4. The lower temperature plane absorbs and emits photons as a surface would with an energy density of aTc^4 in the vacuum immediately adjacent to the surface. Its lower temperature compared to that of the plane radiantly heated from the backside alerts us that this second plane has some non-radiative means of heat loss.
The energy density between the two planes is not a(Th^4 + Tc^4), which is what it would have to be if it were true that the photons incident upon the warmer plane from the cooler plane was that due to an energy density of aTc^4 even as the emission energy density of photons from that warmer surface was that due to an energy density of aTh^4. There has to be a photon energy density gradient between these two planes and this is not consistent with all of the emitted photons from the cooler plane surface being incident upon the warmer plane surface as is usually hypothesized by the advocates of catastrophic man-made global warming.
They deny the existence of an electromagnetic field gradient and a gradient in the energy density with such rhetorical whimsy as “The photons emitted by the cooler surface must be absorbed by the warmer surface because the warmer surface cannot poll incident photons to see if they came from a warmer or a colder surface.” The electromagnetic field in effect does just that by controlling what photons are created by the field. We shall see that this has critically important implications for the CAGW hypothesis.
We also have to take note that there is a flow of photons from the warmer surface to the cooler surface. This case is not a true equilibrium condition of the kind found in the cavity where all surfaces were at the same temperature. The chemical potential of photons in this case is not zero. A photon incident upon the cooler surface can now be absorbed by that surface without the emission of a photon of the same energy. Indeed, the other side of the cooler plane surface is radiating energy into space, so there is a net flow of energy across this plane. More photons have to be absorbed on the side facing the hotter plane than are emitted from the side facing only vacuum as the cooler plane is initially warmed by the warmer plane. When the temperature becomes constant, then the energy flux of absorbed photons on the warmer side must equal the energy flux of photons out of the cooler plane outer surface facing the cold vacuum plus that of any other cooling mechanism. More photons have to be emitted from the cooler plane side facing only space than are absorbed on that same surface. Indeed, when space is at absolute zero, no photons would be absorbed on that side at all. The absorptivity of the surface of the plane facing the warmer plane is greater than the emissivity of that surface, which is at variance with the conditions of the cavity in radiative thermal equilibrium.
The photon radiation pressure P on the plane is proportional to aT^4 of the plane. This photon pressure is proportional to the number of photons incident upon the plane, as is the amount of energy absorbed or emitted from the plane. Therefore, we know that the power density of absorbed or emitted photons will be proportional to the energy density e and therefore to aT^4. That new proportionality with T^4 is given by the Stefan-Boltzmann constant,
σ = 5.670374419 x 10^-8 W / m^2K^4,
when every incident photon is absorbed, as it is for a black body. If the fraction of photons absorbed is less than one, we call that an absorptivity of the fraction α and if the fraction of emitted photons is less than the maximum, we call that the emissivity fraction ԑ.
So in this steady-state system, the outside surface of the cooler plane is emitting photons into vacuum in accordance with the energy density e = aTc^4 at the rate per unit area of εc σ Tc^4, where εc is the emissivity of the cooler plane outer surface. This plane is also losing energy by non-radiative means at the rate Qout. The absorption of energy by the cooler plane on the side facing the warmer plane is given by αc σ Tc^4, which is the flow of energy into the cooler plane. The origin of the energy was the emission from the warmer surface given by εh σ Th^4. Thus, we have
εh σ Th^4 = αc σ Tc4^ = Qout + εc σ Tc^4
Qout = εh σ Th^4 – εc σ Tc^4 and
Qout = (αc – εc) σ Tc^4
So the energy outflow by non-radiative means is equal to the difference of the radiated energy flux of the planes by the surfaces of each emitting toward the cooler outside environment as given by the first expression for Qout. This can readily be mistaken as the energy flow between the planes, though it is not actually so. The planes will not have a different steady state temperature unless Qout is non-zero, so the absorptivity of the cooler plane surface facing the warmer plane must be greater than the emissivity of the outside surface of that same cooler plane as given by the second expression of Qout.
The absorptivity and the emissivity of this plane will only be equal when this plane has the same temperature as the other plane. More photon energy must be absorbed than is emitted from the cooler plane. Since the photon energy emitted from the warmer plane is equal to the photon energy absorbed by the cooler plane and the photon energy density between the planes is nowhere greater than that immediately outside the warmer plane, all of the absorbed photons come from the warmer plane.
No photons emitted from the cooler plane are to be found in the vacuum between the planes. If they were, there would be a clear violation of the Law of Energy Conservation, at least given that there is reason to believe that the one lesson that one takes from the physics of a cavity in thermal equilibrium is that the internal energy density is given by e = aT^4 and that in general, surfaces at a temperature of T will have an energy density in vacuum immediately adjacent to them at the same energy density such a surface would have in the cavity at a thermal equilibrium condition.
In this case, not only do no thermally-emitted photons from the cooler plane become absorbed by the warmer plane surface, but none are even incident upon that surface.
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Sorry about the ‘retard’ comment… this was written in response to a warmist (‘evenminded’… a troll espousing exactly what you’re espousing) long ago. I neglected to remove that comment.
But as you can see, I’ve been through all this many times before. I’ve never lost, not even against warmist physicists espousing what you espouse (Bob Wentworth on WUWT was an especially hilarious example).
You have two options to chose from. They are exclusive.
Which is it? Same energy transfer as standard physics and climate models or no energy transfer and the bolometer doesn’t work?
They are not exclusive, you simply fail to grasp the basics, Jeff.
The photons from the cooler atmosphere will not even incide upon the warmer sensor (see the post I sent which you censored… absolute mathematical proof of what I state, all taken from Thermal Physics, Second Edition written by Philip M. Morse, Professor of Physics at MIT, co-founding editor of Annals of Physics, co-founder of MIT Acoustics Laboratory, first Director of Brookhaven National Laboratory, founder of MIT Computation Center.), the energy density gradient subsumes them into the background EM field before then. And one cannot distinguish individual photons of the background EM field except as the ambient temperature (or via a sensor which is cooled below ambient, giving an energy density gradient), those photons are no longer a persistent perturbation above the EM field ambient.
The sensor is emitting photons which cools it. The circuitry senses this cooling as a change in current flow through the sensor, which it compares with the current flow through a reference resistor. That comparison is done in something like a Wheatstone Bridge.
Rather than continuing to deny reality, Jeff, might I suggest that you go back and study Pictet’s 1800 experiment? This is exactly the same as your bolometer.
Place your bolometer at the focal point of one mirror, an ice cube at the focal point of the other mirror. The mirrors set up a view factor such that the ice cube and the bolometer can only ‘see’ each other, nothing else.
You claim that photons from the colder ice cube are bouncing off its mirror, going to the bolometer’s mirror, being focused into the bolometer, and changing its temperature… you’ve just espoused the idea of the transmission of ‘cold’.
In reality, the energy transfer is from the bolometer sensor, off the bolometer’s mirror, to the ice cube’s mirror, then to the ice cube. The loss of energy by the bolometer sensor causes it to cool, and the bolometer circuitry senses this.
When the bolometer and the ice cube are at exactly the same temperature (ie: thermodynamic equilibrium), there will be no photon generation whatsoever. The system has reached the quiescent state known as thermodynamic equilibrium. And that is the definition of thermodynamic equilibrium, a quiescent state.
Emission and absorption isn’t quiescence, Jeff. Your take on radiative energetic exchange (that all objects greater than 0 K emit photons at all times, and that photons from a cooler object can incide upon a warmer object) means that entropy must change, right?
2LoT states that there exists a state variable called entropy S. The change in entropy (ΔS) is equal to the energy transferred (ΔQ) divided by the temperature (T).
ΔS = ΔQ / T
Only for reversible processes does entropy remain constant. Reversible processes are idealizations. All real-world processes are irreversible.
The climastrologists claim that energy can flow from cooler to warmer because they cling to the long-debunked Prevost Principle, which states that an object’s radiant exitance is dependent only upon that object’s internal state, and thus they treat real-world graybody objects as though they’re idealized blackbody objects via:
q = σ T^4.
… thus the climate alarmists claim that all objects emit radiation if they are above 0 K. In reality, idealized blackbody objects emit radiation if they are above 0 K, whereas graybody objects emit radiation if their temperature is greater than 0 K above the ambient. Of course, idealized blackbody objects don’t actually exist… they’re idealizations.
But their claim means that in an environment at thermodynamic equilibrium, all objects (and the ambient) would be furiously emitting and absorbing radiation, but since entropy doesn’t change at thermodynamic equilibrium, the climastrologists must claim that radiative energy transfer is a reversible process. Except radiative energy transfer is an irreversible entropic temporal process, which destroys their claim.
It also destroys your claim, since you claim the same thing as the warmists… that claim is what undergirds the entire CAGW edifice.
So now you pick option 2.
So if it is option 2, do you admit that the standard gray body emissions models predict exactly the expected result from the bolometer? Or is there some difference?
Yes, isolating all objects into their own systems such that they cannot interact via the EM field, then ‘connecting’ them by subtracting a wholly-fictive ‘cooler to warmer’ ENERGY FLOW from the real (but too high because it was calculated for emission to 0 K) ‘warmer to cooler’ ENERGY FLOW arrives at the same radiant exitance of the warmer object, but it does so by inventing a ‘backradiation’ energy flow that does not exist and is therefore unphysical.
Do it correctly, by subtracting the cooler object ENERGY DENSITY from the warmer object ENERGY DENSITY (to arrive at the energy density gradient), and one can not only arrive at the correct radiant exitance of the warmer object, but one also hews to the fundamental physical laws.
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)
q = ε σ (T_h^4 – T_c^4)
∴ q = ε σ ((e_h / (4σ / c)) – (e_c / (4σ / c))) Ah
Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))
∴ q = (ε c (e_h – e_c)) / 4
Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
W m-2 = (m sec-1 (ΔJ m-3)) / 4
∴ q = σ / a * Δe
Canceling units, we get W m-2.
W m-2 = (W m-2 K-4 / J m-3 K-4) * ΔJ m-3
One can see from the immediately-above two forms of the S-B equation that the Stefan-Boltzmann equation is all about subtracting the radiation energy density of the cooler object from the radiation energy density of the warmer object.
Or weren’t you aware that the conclusions drawn from your take on radiative energy exchange violate 2LoT and Stefan’s Law?
In a cavity, if we use your take, the energy density in the cavity space is double that which we would calculate from Stefan’s Law. Thus your take on radiative energy exchange violates Stefan’s Law. In addition, as explicated in a prior post, your take on radiative energy exchange also forces you to claim that radiative energy exchange is an idealized reversible process… except it’s an irreversible temporal entropic process.
That should be enough to cause you to investigate whether what you believe is true or not, Jeff.
That’s why standard cavity theory states that at thermodynamic equilibrium, the photons remaining in the cavity space as the cavity walls and space come into thermodynamic equilibrium are perfectly reflected by the cavity wall, that a standing wave is set up in the cavity space with the wavemode nodes at the cavity wall surface by dint of boundary constraints.
Now, a wavemode node being the zero crossing point, no energy CAN be transferred either into or out of the cavity walls.
Thus, the cavity is in the quiescent state known as thermodynamic equilibrium. And that is the definition of thermodynamic equilibrium, a quiescent state.
Should one wall change temperature, that standing wave becomes a traveling wave, with the group velocity proportional to the energy density differential between the walls and in the direction of the cooler wall.
To put it in terms of standard cavity theory, in a cavity at thermodynamic equilibrium, the photons remaining in the space as the walls and cavity space come into thermodynamic equilibrium are perfectly reflected from the wall surfaces. A standing wave is set up, with the wavemode nodes at the cavity surface due to boundary constraints. A node being a zero crossing point, no energy CAN be transferred into or out of of the walls at thermodynamic equilibrium.
Should one wall change temperature, that standing wave becomes a traveling wave, with the group velocity proportional to the energy density differential between the walls and in the direction of the cooler wall.
Graybody objects become perfect reflectors at thermodynamic equilibrium.
“Aw, you can’t prove that!”, you may say. Yeah, I can. Mathematically.
The definition of emissivity: The ratio of the total emissive power of a body to the total emissive power of a perfectly black body at that temperature.
The definition of absorptivity: The ratio of the absorbed to the incident radiant power
As Δe → 0, ΔT → 0, q → 0. As q → 0, the ratio of graybody object total emissive power to idealized blackbody object total emissive power → 0. In other words, emissivity → 0. At thermodynamic equilibrium for a graybody object, there is no radiation energy density gradient and thus no impetus for photon generation.
As Δe → 0, ΔT → 0, photon chemical potential → 0, photon Free Energy → 0. At zero chemical potential, zero Free Energy, the photon can do no work, so there is no impetus for the photon to be absorbed. The ratio of the absorbed to the incident radiant power → 0. In other words, absorptivity → 0.
α = absorptivity = absorbed / incident radiant power
ρ = reflectivity = reflected / incident radiant power
τ = transmissivity = transmitted / incident radiant power
α + ρ + τ = 100%
For opaque surfaces τ = 0% ∴ α + ρ = 100%
If α = 0%, 0% + ρ = 100% ∴ ρ = 100% … all incident photons are reflected at thermodynamic equilibrium for graybody objects.
Thus, while Kirchhoff’s Law of Thermal Radiation states that absorptivity and emissivity are equal at thermodynamic equilibrium, that is only because they are both zero. Now, Kirchhoff’s Law of Thermal Radiation is a ratio. What is zero divided by zero? Indeterminate.
Even Planck tacitly admitted to the non-universality of Kirchhoff’s Law of Thermal Radiation. Planck used the coefficients of emission and absorption, which correspond to emissivity and absorptivity, writing: Kv = ev / av. In Section 48 of his book, Planck admits that this equation (in Section 26) cannot be used when ev = 0 and av = 0 (ie: at thermodynamic equilibrium) because it results in 0/0 which is indeterminate
Kirchhoff’s Law of Thermal Radiation therefore does not and cannot apply at the one condition at which it is said to apply (thermodynamic equilibrium) for graybody objects. It applies to idealized blackbody objects only (and even then, there are problems because they claim idealized blackbody objects emit at temperatures greater than 0 K and must absorb all radiation incident upon them, so idealized blackbody objects can never really reach the quiescent state of thermodynamic equilibrium except at 0 K (emission and absorption isn’t quiescence, and thermodynamic equilibrium is defined as a quiescent state). And of course, idealized blackbody objects don’t actually exist… they’re idealizations.
Non sequitur.
Not a non sequitur, you simply fail to grasp the entire picture, Jeff.
If a graybody object has emissivity of zero at thermodynamic equilibrium (at zero energy density gradient), as I’ve shown above, how exactly does it start emitting AGAINST an energy density gradient? You know… when it’s the cooler object?
Magic? Is it magic, Jeff? Describe for us exactly how that happens, Jeff. Describe for us exactly how photon generation occurs against an energy density gradient. I expect you to go all the way down to the quantum scale and to include the exact mechanism for photon generation, Jeff. If you can’t do that, you’re out of your depth.
Bon sequitur and i know very well how superposition works
Non sequitur answer, Jeff… you’ve failed to explicate exactly how energy can flow against an energy density gradient (without external energy doing work to push that system energy up the gradient), how a photon can be generated against an energy density gradient.
Here, I’ll give you a hint… a charge (either a point charge or the unequal charge distribution of a molecule) undergoing either linear or angular acceleration relative to its electric field will emit a photon if the lower energy state in question of the molecule is higher than or equal to the ambient energy density. That covers electronic mode, vibrational mode and rotational mode quantum states adequately.
Now it’s your turn, Jeff. Describe for us exactly how photon generation occurs against an energy density gradient.
It occurs the same way it always occurs and doesn’t experience any gradient.
You are the one claiming new physics.
So answer my question. Same energy transfer as standard physics or bolometer doesn’t work?
Jeff Id wrote:
“It occurs the same way it always occurs and doesn’t experience any gradient.”
Now you’re denying the existence of energy density gradient? Really, Jeff?
Jeff Id wrote:
“You are the one claiming new physics.”
I am the one espousing standard radiative and cavity theory as explicated in Thermal Physics, Second Edition written by Philip M. Morse, Professor of Physics at MIT, co-founding editor of Annals of Physics, co-founder of MIT Acoustics Laboratory, first Director of Brookhaven National Laboratory, founder of MIT Computation Center.
The absolute mathematical proof of what I state was in a post from me which you censored… absolute mathematical proof that photons from a cooler object will not even incide upon a warmer object, let alone be absorbed by the warmer object (because a warmer object will have higher energy density at all wavelengths than a cooler object). Why’d you censor that post, Jeff? If you wish, elide the offending word and post it. Go on, prove yourself wrong, Jeff.
You are the one who’s bought into the ‘new fyziks’ of the warmists to bolster their climate alarmism.
Why are you dodging the question, Jeff?
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291779
“Now it’s your turn, Jeff. Describe for us exactly how photon generation occurs against an energy density gradient.”
So is the surface energy of the earth/bolometer more or less or equal to the back radiation predicted from co2?
Jeff Id wrote:
“So is the surface energy of the earth/bolometer more or less or equal to the back radiation predicted from co2?”
‘Backradiation’ is a mathematical manifestation of the climate alarmists’ misuse of the S-B equation. It is unphysical.
They isolate each object into its own system so it cannot interact via the EM field. They treat the objects as though they’re idealized blackbody objects (emission to 0 K, and usually emissivity equal to 1, although they sometimes also include emissivity:
q = σ T^4
q = ε σ T^4
Rather than using the proper form of the S-B equation for graybody objects:
q = ε σ (T_h^4 – T_c^4)
Because they’ve isolated all objects such that they cannot interact via the EM field and treated all objects as though they’re emitting to 0 K, that inflates radiant exitance of all objects. To get their equation to balance, they subtract on the back end the wholly-fictive ‘cooler to warmer’ ENERGY FLOW from the real (but far too high because it was calculated for emission to 0 K) ‘warmer to cooler’ ENERGY FLOW.
That ‘cooler to warmer’ ENERGY FLOW is unphysical, a mathematical artifact due to their misuse of the S-B equation.
Do it correctly, by subtracting the cooler object ENERGY DENSITY from the warmer object ENERGY DENSITY (to arrive at the energy density gradient), and one can not only arrive at the correct radiant exitance of the warmer object, but one also hews to the fundamental physical laws.
As already explained to you in prior comments, your take on radiative energy exchange violates 2LoT and Stefan’s Law, and it forces you to claim that radiative energy exchange is an idealized reversible process (so you can claim that objects at thermodynamic equilibrium but greater than 0 K can emit but entropy does not change), but radiative energy exchange is a reversible temporal entropic process… so you’ve bought lock, stock and barrel into the ‘new fyziks’ which the climastrologists use to bolster their climate alarmism.
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)
q = ε σ (T_h^4 – T_c^4)
∴ q = ε σ ((e_h / (4σ / c)) – (e_c / (4σ / c))) Ah
Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))
∴ q = (ε c (e_h – e_c)) / 4
Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
W m-2 = (m sec-1 (ΔJ m-3)) / 4
∴ q = σ / a * Δe
Canceling units, we get W m-2.
W m-2 = (W m-2 K-4 / J m-3 K-4) * ΔJ m-3
One can see from the immediately-above two forms of the S-B equation that the Stefan-Boltzmann equation is all about subtracting the radiation energy density of the cooler object from the radiation energy density of the warmer object.
The climastrologists’ misuse of the S-B equation by subtracting energy flows rather than energy densities is what has led to CAGW, it undergirds the entirety of the CAW charade.
More or less or equal?
You Don’t need to copy paste a book.
Mheh…
“but entropy does not change), but radiative energy exchange is a reversible temporal entropic process”
-should be-
“but entropy does not change), but radiative energy exchange is an irreversible temporal entropic process”
Jeff Id wrote:
“More or less or equal?
You Don’t need to copy paste a book.”
Not even more or less equal, ‘backradiation’ does not exist, it is a mathematical artifact of the climastrologists’ misuse of the S-B equation. It is unphysical.
I’m not copy-n-pasting, Jeff. I wrote all of the above. Long ago. This isn’t my first rodeo with those who deny reality. Not even warmist physicists can stand against that reality. I’ve converted many to scientific reality… one formerly-warmist climastrologist even publicly apologized and wrote a book. Mikey Mann doesn’t even dare utter my pseudonym for fear that I’ll crush him again.
You can either accede to scientific reality, or you can continue denying reality. Your choice.
But one must ask… if you continue to deny scientific reality, which side are you really on?
You can do it!
Jeff Id wrote:
“You can do it!”
I can’t “do it” if you censor my replies, Jeff… it’s almost as it you don’t think people can take screenshots of those moderated and subsequently-censored posts. You know, like the one I posted with absolute mathematical proof that photons from cooler objects not only won’t be absorbed by warmer objects, but won’t even incide upon the warmer object, all in accordance with the fundamental physical laws and all taken from Thermal Physics, Second Edition, written by Philip M. Morse, Professor of Physics at MIT, co-founding editor of Annals of Physics, co-founder of MIT Acoustics Laboratory, first Director of Brookhaven National Laboratory, founder of MIT Computation Center.
How’s it going to look when it’s shown that when you’re proven wrong, you start censoring, Jeff? Do you believe people are going to flock to your website to learn from someone who not only denies scientific reality, but censors when that scientific reality becomes such a deluge that your denials become ridiculous?
Thus far, you’ve advocated for the warmist take on radiative energy exchange; you’ve advocated for the climastrologist misuse of the S-B equation (using the form meant for idealized blackbody objects upon real-world graybody objects);you’ve advocated for violation of 2LoT and Stefan’s Law; you’ve unknowingly clung to the long-debunked Prevost’s Principle; you’ve espoused ideas which force you to claim that radiative energy exchange is an idealized reversible process so you can claim that all objects greater than 0 K emit radiation and thus at thermodynamic equilibrium, those objects still emit but entropy doesn’t change… but radiative energy exchange is an irreversible entropic temporal process, so that scotches every single bit of what you’re espousing.
Who’s side are you really on, Jeff?
Or is it that educating yourself properly is simply too difficult, so you cling to your incorrectitudes, despite the fact that they bolster the climastrologist narrative?
I have censored nothing. If you went to the spam bin, i’ll have to get it in the morning.
Answer the question.
Don’t be afraid.
Your nonsensical question about an unphysicality (‘backradiation’) which is a mathematical artifact brought about by the misuse of the S-B equation, using the form of the S-B equation intended for idealized blackbody objects upon real-world graybody objects, which creates out of thin air a wholly-fictive ‘backradiation’ by calculating for emission to 0 K, thus inflating radiant exitance of all objects and necessitating the subtraction of that wholly-fictive ‘backradiation’ ENERGY FLOW from the real (but too high because it was calculated for emission to 0 K) ‘warmer to cooler’ ENERGY FLOW, when the proper way of using the S-B equation is to subtract the ENERGY DENSITY of the cooler object from the ENERGY DENSITY of the warmer object, the ENERGY DENSITY GRADIENT determining warmer object radiant exitance?
That question? Already answered more than once. Merely because you don’t like the answer because it destroys your narrative (the same narrative the warmists use, BTW) doesn’t mean it’s not been answered, Jeff.
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)
q = ε σ (T_h^4 – T_c^4)
∴ q = ε σ ((e_h / (4σ / c)) – (e_c / (4σ / c))) Ah
Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
W m-2 = W m-2 K-4 * (Δ(J m-3 / (W m-2 K-4 / m sec-1)))
∴ q = (ε c (e_h – e_c)) / 4
Canceling units, we get J sec-1 m-2, which is W m-2 (1 J sec-1 = 1 W).
W m-2 = (m sec-1 (ΔJ m-3)) / 4
∴ q = σ / a * Δe
Canceling units, we get W m-2.
W m-2 = (W m-2 K-4 / J m-3 K-4) * ΔJ m-3
One can see from the immediately-above two forms of the S-B equation that the Stefan-Boltzmann equation is all about subtracting the radiation energy density of the cooler object from the radiation energy density of the warmer object.
What is the difference in magnitude to the black body?
You cannot answer can you?
Jeff Id wrote:
“What is the difference in magnitude to the black body?
You cannot answer can you?”
The “difference in magnitude” of what? People generally ask for what they want in terms of units. ‘Magnitude’ alone doesn’t have units. Your language is imprecise.
Let’s take the K-T diagram as an example.
They treat the real-world graybody surface of Earth (emissivity of 0.93643 per NASA CCCP program) as if it were an idealized blackbody object emitting to 0 K and with emissivity of 1. That’s the only way they can get to 390 W m-2 surface radiant exitance. That is proof-positive that they’ve misused the S-B equation to bolster their narrative.
And that, of course, is a misuse of the S-B equation, using the form meant for idealized blackbody objects upon graybody objects, which inflates radiant exitance far above what it actually is.
Looking at more realistic numbers:
The left-hand image is what one would arrive at when using the proper emissivity of Earth, but still assuming emission to 0 K (which is what the climastrologists sometimes do… the typically use:
q = σ T^4
… upon real-world graybody objects (but that form of the S-B equation is meant for use on idealized blackbody objects, since it assumes emission to 0 K and emissivity of 1), and sometimes they slap emissivity onto that:
q = ε σ T^4
… which still assumes emission to 0 K).
The right-hand image is what one would arrive at by properly using the correct form of the S-B equation upon a graybody object, with assumption of emissivity less than 1 and emission to greater than 0 K.
You will note that by doing it correctly, by subtracting cooler object ENERGY DENSITY from warmer object ENERGY DENSITY, the S-B equation gives a result which completely does away with the climastrologist ‘backradiation’. IOW, ‘backradiation’ is a mathematical artifact brought about by the climastrologist misuse of the S-B equation. It is unphysical.
So one must wonder why you continue to cling to and shill for a misuse of the S-B equation and the mathematical conjuring of that ‘backradiation’.
You have heard the question. If you cannot answer, your whole theory fails. What is the big problem?
“But we can measure backradiation! Ha! I got you there!”, some less well-versed in thermodynamics may claim.
Yeah, no.
“How To Fool Yourself With A Pyrgeometer”
https://claesjohnson.blogspot.com/2011/08/how-to-fool-yourself-with-pyrgeometer.html
And that’s exactly what I’ve stated above as regards energy flow from a warmer bolometer or pyrgeometer sensor to the ambient, which cools the sensor down, which the circuitry registers as a cooler ambient.
In fact, with a near-surface mean free path length for the 13.98352 µm – 15.98352 µm waveband of only ~10.4 m (which would decrease to ~9.7 m for a doubling of atmospheric CO2 concentration), and given that atmospheric attenuation absorbs ~50% over 1/10th of the mean free path length, then 50% of the remaining radiation within the next 1/10th of the mean free path length, etc., etc…. that would mean that this ‘backradiation’ would have to nearly all come from within ~10.4 of the surface, with fully 50% of that coming from the first ~1 m of atmosphere above the surface. And that’s ludicrous.
One can arrive at the atmospheric mean free path length one of two ways… either go line by line through the waveband to determine radiant flux at the wavelength in question, and how much of that is attenuated by the atmosphere; or treat the 13.98352 µm – 15.98352 µm waveband as if it were a radio signal (convert wavelength to frequency) and use any number of websites which calculate radio signal atmospheric attenuation.
Jeff Id wrote:
“You have heard the question. If you cannot answer, your whole theory fails. What is the big problem?”
Again, Jeff, it’s been answered multiple times. That you don’t like the answer because it destroys your narrative and shows that you have a gossamer-thin grasp upon the concept of radiative energy transfer doesn’t mean it’s not been answered, merely that you refuse to accede to scientific reality.
All I’ve seen is a chicken avoiding the question.
Maybe it’s in the spam trap.
Jeff id wrote:
“All I’ve seen is a chicken avoiding the question.”
Answered here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291801
And here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291799
And here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291789
Or are you merely complaining that you cannot do the calculations to ascertain for yourself that what I state is correct, Jeff?
Why have you side-stepped the question, Jeff:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291779
“Now it’s your turn, Jeff. Describe for us exactly how photon generation occurs against an energy density gradient.”
All I’ve seen is a chicken psychologically projecting his inability to answer a question, Jeff, by stupidly claiming a question hasn’t been answered when your question has been answered multiple times and via multiple avenues of reasoning.
Just to be sure you don’t miss that question, Jeff:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291779
“Now it’s your turn, Jeff. Describe for us exactly how photon generation occurs against an energy density gradient.”
No worries, right? I mean, your inability to form a cogent basis in physics by which a photon could possibly be spontaneously emitted against an energy density gradient only destroys the underlying premise of your belief system and thus destroys your narrative, while also conveniently highlighting that you’ve spent an untold number of years believing in and shilling for the exact premise which undergirds the warmist narrative.
Chicken.
Not able to answer are you. Because you know the answer does not match your feelings.
I knew that when I wrote the post.
Jeff Id wrote:
“Chicken.
Not able to answer are you. Because you know the answer does not match your feelings.”
Psychologically-projective troll. Already anwered multiple times. That you don’t like the answer because it destroys your narrative and shows you to be a scientific neophyte doesn’t mean it’s not been answered, merely that you cannot grasp scientific reality.
Why have you side-stepped the question, Jeff:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291779
“Now it’s your turn, Jeff. Describe for us exactly how photon generation occurs against an energy density gradient.”
No worries, right? I mean, your inability to form a cogent basis in physics by which a photon could possibly be spontaneously emitted against an energy density gradient only destroys the underlying premise of your belief system and thus destroys your narrative, while also conveniently highlighting that you’ve spent an untold number of years believing in and shilling for the exact premise which undergirds the warmist narrative.
Jeff Id wrote:
“There is a bit of math in the thread where the DOF appears mixed up”
You must really enjoy being constantly wrong, eh, Jeff?
There is no mix-up. Anyone who’s worked with high pressure air realizes that a relief valve lifting can cause relief piping temperature to rise far above the static temperature of the bulk gas being relieved… study up on stagnation temperature.
https://www.chemengonline.com/temperature-effects-high-velocity-gas-flow/
“Guidelines are presented to better understand the temperature profiles of high-velocity gases In high-velocity gas flows, such as those that may occur within the discharge piping of a pressure-relief or depressurization valve, the temperature experienced at the wall of the pipe through which the gas is flowing can be much higher than the flowing stream’s “static” temperature. In fact, the wall temperature approaches the stagnation temperature, which is the temperature that would be obtained if the fluid were brought adiabatically and reversibly (isentropically) to rest.
My calculations on the cited webpage are exactly the same mathematical methodology used by Sandia National Laboratories.
https://lammps.sandia.gov/doc/compute_temp_chunk.html
An atom or molecule moving in a single DOF will have a higher calculated temperature than an identical atom or molecule moving in more DOF, for the same kinetic energy. If you understood thermodynamics, that fact would have been obvious to you, Jeff.
You have so many fundamental misconceptions that you really ought to go back to Square 1 and start your education all over again, Jeff. If you paid for it, you should by all rights get a refund. They sold you a bill of goods.
Avoidance.
Good call
Defensive non sequitur as avoidance of the fact that you’ve been demonstrably wrong on every single topic. Not such a good call… you seem to think you’re fooling anyone. LOL
I released your spam bucket this morning. In there, you gave one book and two comments.
IN NO COMMENT HAVE YOU ANSWERED MY QUESTION. — sorry for the caps, I want everyone to know that you are afraid to answer.
In your comments you said backradiation does not exist so that leaves me wondering what the gradient-subsuming nonsense is for then. It is nonsense btw.
In your second comment you claim you have never lost. My return claim is that you have never won.
—
Last chance to discuss your theory before I get grumpy. Does your theory of no back radiation mean the black body warms MORE, LESS, or the SAME as the backradiation required by standard physics.
Your nonsensical question about an unphysicality (‘backradiation’) which is a mathematical artifact brought about by the climastrologist misuse of the S-B equation, using the form of the S-B equation intended for idealized blackbody objects upon real-world graybody objects, which creates out of thin air a wholly-fictive ‘backradiation’ by calculating for emission to 0 K, thus inflating radiant exitance of all objects and necessitating the subtraction of that wholly-fictive ‘backradiation’ ENERGY FLOW from the real (but too high because it was calculated for emission to 0 K) ‘warmer to cooler’ ENERGY FLOW, when the proper way of using the S-B equation is to subtract the ENERGY DENSITY of the cooler object from the ENERGY DENSITY of the warmer object, the ENERGY DENSITY GRADIENT determining warmer object radiant exitance?
That question? Already answered more than once. Merely because you don’t like the answer because it destroys your narrative (the same narrative the warmists use, BTW) doesn’t mean it’s not been answered, Jeff.
Answered here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291786
And here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291785
And here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291789
And here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291756
And here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291794
And here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291774
And here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291783
And here:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291758
All of those answers showing that it is I who is promulgating standard thermodynamic, radiative and cavity theory as found in Thermal Physics, 2nd Edition, written by Philip M. Morse, Professor of Physics at MIT, co-founding editor of Annals of Physics, co-founder of MIT Acoustics Laboratory, first Director of Brookhaven National Laboratory, founder of MIT Computation Center.
You? You’re bleating out the idiotic hijacking of standard thermodynamic, radiative and cavity theory by the warmists to spew their alarmism to gain further funding and to push forward a Marxist agenda. And in so doing, you’re violating 2LoT, Stefan’s Law and the Principle of Maximum Entropy, among many other fundamental physical laws.
You’re simply either not intellectual enough, or you are too brainwashed, to grasp that fact, Jeff.
Now, as to questions being avoided, why have you been psychologically projecting your utter inability to answer the one question which your failure to answer destroys your entire premise, Jeff?
Just to be sure you don’t miss that question, Jeff:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291779
“Now it’s your turn, Jeff. Describe for us exactly how photon generation occurs against an energy density gradient.”
No worries, right? I mean, your inability to form a cogent basis in physics by which a photon could possibly be spontaneously emitted against an energy density gradient only destroys the underlying premise of your belief system and thus destroys your narrative, while also conveniently highlighting that you’ve spent an untold number of years believing in and shilling for the exact premise which undergirds the warmist narrative.
Your links to your books do not answer a damned thing. If you have no guts, I shall proceed to tear apart your work.
Then you’ve either not read the contents of those “links to my books”, or you’ve failed to comprehend them, Jeff.
I suggest, before you further embarrass yourself, that you make a best effort as regards comprehension of those ‘books’, Jeff.
And while you’re at it, to avoid completely humiliating yourself, I suggest you first defend the premise undergirding your blather…
Just to be sure you don’t miss that question, Jeff:
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291779
“Now it’s your turn, Jeff. Describe for us exactly how photon generation occurs against an energy density gradient.”
No worries, right? I mean, your inability to form a cogent basis in physics by which a photon could possibly be spontaneously emitted against an energy density gradient only destroys the underlying premise of your belief system and thus destroys your narrative, while also conveniently highlighting that you’ve spent an untold number of years believing in and shilling for the exact premise which undergirds the warmist narrative.
Check your moderation bin, Jeff (why are my comments always going to moderation, Jeff? You afraid of something? LOL). And do be sure to post that reply of mine… remember, screen shots. Your reputation hangs in the balance, Jeff.
For what it’s worth – I don’t think you do yourself any favors with paranoid claims about being “censored.”
Why do so many blog commenters go there – when the far more likely explanation is that blogs have wonky comment filters?
Wju do you think some of your comments go through while others don’t? Do you think Jeff reads your comments in fractions if a second and then allows only those that don’t make look bad go through?
Certainly your characterization of him, as someone who thinks he knows things when clearly he doesn’t, matches my experiences with him. But don’t undermine your own arguments with adjacent displays of bad logic.
Joshua wrote:
“For what it’s worth – I don’t think you do yourself any favors with paranoid claims about being “censored.””
It just seems strange that so many of my comments go to moderation. Had I not spoken up and reminded Jeff that screen shots had been taken and that his reputation hangs in the balance, do you believe he’d allow those comments which prove him wrong? That public ‘shot across the bow’ is, I believe, what forced him to post those comments from me.
I didn’t moderate you. The spam shall be released in an hour.
My guess is a wonky filter. Jeff’s ridiculous over-confidence shows he’s likely incapable of even accepting the possibility that he’s wrong. So then he’d have no reason to hide comments showing him wrong ’cause he thinks it can’t happen.
But that’s just speculation and like I said, some comments go through. It’s not logistically possible that he reads them that quickly and then decides which ones to filter out.
Again, look at the modus operandi.
When Jeff’s wrong he will just throw out some obviously wrong bullshit and claim it “proves” he’s right. He’s done it time and again.
Here’s from back in the day where Jeff applies the same stupid-ass, self-victimizing, paranoid logic to build up his sense of self-importance:
https://noconsensus.wordpress.com/2011/03/28/climate-blogger-censored-by-new-york-times/
Joshua,
I’ve made a hobby of admitting being wrong quite publicly here.
You’ve been here for a very short time.
The data says what it does.
Jeff –
Feel free to claim whatever you’d like.
I can only judge from what I’ve witnessed. Which is that you sometimes claim that you’re absolutely right with no possibility of being wrong, about things when you’re either obviously, or very likely wrong.
That’s what I saw the very first time I came here. It’s kind of a classic example of the phenomenon, where the pattern was even jumped on by a whole series of your readers – all of which bought into your obviously wrong, paranoid conspiracy theory (not dissimilar to LOL@KKK).
It’s such a fascinating phenomenon, where smart, knowledgeable people buy into inane theories because of their ideological “motivations.”
https://noconsensus.wordpress.com/2011/03/28/climate-blogger-censored-by-new-york-times/
Jeff –
It’s clear that LOL@KKK needs to rethink his whole approach if he thinks what he’s saying is so obviously and devastatingly true that you can’t even stand to let his comments go though. There’s something fundamentally wrong with his reasoning process.
I can’t understand the physics discussion, but when someone employs obviously implausible reasoning in one context, it suggest that they might do the same in another.
I mean we all make mistakes, and in an “emotional” moment let our thinking get biased. So then what will be interesting is to see what happens going forward, once it’s been made clear to him that his paranoid “censorship” theory is extremely unlikely. What matters most is accountability for making errors, not the errors themselves, per se.
But what’s beautiful about it, is that it’s so easy to find examples of you employing the same problematic reasoning process. And I bet that you will never admit how silly your the reasoning was behind your post from back in the day.
So here’s your test. Just be accountable. Show that you can let go of your emotional attachment. Just acknowledge how silly your “censorship” theory was.
I believe everything has been released.
Jeff Id wrote:
“You’ve been here for a very short time.”
Joshua participated in that 2011/03/28 post’s comments section, so according to you, a “very short time” (your words) is 4224 days.
Remember, you’ve “made a hobby of admitting being wrong quite publicly” (your words), Jeff. So either you’ve redefined “very short time” or you’re wrong, and you must now “quite publicly” admit to same.
Joshua wrote:
“That’s what I saw the very first time I came here. It’s kind of a classic example of the phenomenon, where the pattern was even jumped on by a whole series of your readers – all of which bought into your obviously wrong, paranoid conspiracy theory (not dissimilar to LOL@KKK).”
I like your logic, although I do believe this comes about due to the fact that the forum owner has absolute control (and zero public transparency) over who goes to and who does not go to moderation, and given past experience, it would appear that the natural inclination of most people, when presented with data which is antithetical to what they believe, have an urge to damp that data, no matter its merit. Hence, and again given past experience, it is only natural to conclude that censorship is taking place when a completely innocuous comment goes to moderation and either isn’t released until the next day (and only after it is made publicly known that screen shots have been taken of that moderated comment); or as is the case today (to which you replied), with the notice to Jeff to check his moderation bin a full 4 minutes after the initial comment which had gone into moderation, released immediately after that notice.
Joshua wrote:
“It’s clear that LOL@KKK needs to rethink his whole approach if he thinks what he’s saying is so obviously and devastatingly true that you can’t even stand to let his comments go though.”
Obviously, if what I write were so *obviously* true, it would be *obvious* even to those such as Jeff Id, who has bought lock, stock and barrel into the perversion of bog-standard radiative, thermodynamic and cavity theory (that same perversion demonstrably violating 2LoT, Stefan’s Law, the Principle of Maximum Entropy and a whole host of other fundamental physical laws); and who has demonstrated that he doesn’t really understand such simple concepts as DOF, and translational motion of atoms or molecules within a constrained DOF as regards calculated temperature (which means he doesn’t really understand kinetic temperature, nor Bernoulli’s Principle, nor stagnation temperature, nor a whole host of related concepts).
But I will admit that all moderated comments by myself have (after a time) been released from moderation. Whether the thought of reputational damage to Jeff Id as being labeled censorious did the trick, or whether Jeff Id actually doesn’t censor, time will tell.
LOL@KKK –
In all fairness, although for a short period many years ago I did write some comments here, it was many years ago and it was a short period of time – so Jeff’s “You’ve been here for a very short time” would be understandable.
> and given past experience, it would appear that the natural inclination of most people, when presented with data which is antithetical to what they believe, have an urge to damp that data, no matter its merit.
Sure. But on the other hand, sometimes people think they’re being moderated because they’re making devastating arguments that others find intolerable, when in reality those other people just think they’re assholes, or making dumb comments they don’t want to waste time with, or they’ve been moderated for just basically random or silly reasons. Again, that thread I linked from years ago being a case in point – where Jeff thought he was being moderated for what he laughably thought was a devastating comment and was sure he was being “censored” even though that view was plainly silly. And his view was then supported by a chorus of other people who bought into his vapid viewpoint.
This happens all the time in blogs. Where people write comments that they’re sure are being “censored” out when in fact, that’s not what’s happening (on top of the silly, pearl-clutching use of “censored” when at worst it would just be a blog proprietor limiting comments on a freakin’ blog).
Again, just think of the logistics. You write a comment. In a period from a fraction of a second to maybe something like three seconds, your comment either appears or doesn’t. In oder for some of your comments to pass through and show up, and some to get “censored,” a moderator would have to read all of your comments incredibly fast and then cull those that are supposedly reputationally dangerous, and let the others through. He’d have to do that basically whenever you comment, so he’d have to be just waiting around with full attention 60/60/24/7/365 to see if maybe you’re going to comment. It’s just not plausible.
If all of your comments only show after. lengthly delay, and some never show up, you’re being moderated. If some show up immediately and some only after a long delay (or not at all), then you’re likely not being moderated. If they all show up after a delay, then they’re all being held in moderation for some mechanical reason but not being filtered. Moderation filters are hardly state of the art AI language processing, and even if they were they’d likely wind up with many illogical moderations.
And on top of all of that, there’s no reputational risk for Jeff. He always thinks he’s right even when he’s obviously wrong – so he doesn’t think there’s any reputational risk. And his readers mostly think he can’t make mistakes – as was demonstrated in that thread from 10 years ago where he claimed he was being “censored” and his readers (many of them smart and knowledgeable people ) piled on in agreement even though his theory of “censorship” was laughably and obviously dumb as shit.
Nobody is in moderation.
> If you have no guts, I shall proceed to tear apart your work.
Whoa. Another internet tough guy.
If you do it again, I shall taunt you a second time.
Another slayer vanquished.
A good day’s work.
You can’t just declare “victolly!” while not having done anything, Jeff. Remember? Extraordinary claims require extraordinary evidence.
Show us this vanquishment you purport took place. Give us a link to it.
Or admit no vanquishment took place, that you’re in full-blown defensive mode, utterly unable to reconcile your incorrect take on radiative energy exchange and its rampant violations of fundamental physical laws with scientific reality.
You have been, you are, and unless you seriously buckle down and educate yourself, you will continue to be wrong. You can’t, apparently, even defend your stance (if you could have, you would have. You did not… instead you denied the existence of radiation energy density and thus energy density gradient, and demonstrated that you don’t really understand the concept of DOF, kinetic temperature, stagnation temperature, the pressure-temperature relation, Bernoulli’s Principle, etc.. LOL).
And in the process of all the above, you’ve studiously avoided answering the below question, the premise of which undergirds every single bit of your blather:
————————-
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291779
“Now it’s your turn, Jeff. Describe for us exactly how photon generation occurs against an energy density gradient.”
No worries, right? I mean, your inability to form a cogent basis in physics by which a photon could possibly be spontaneously emitted against an energy density gradient only destroys the underlying premise of your belief system and thus destroys your narrative, while also conveniently highlighting that you’ve spent an untold number of years believing in and shilling for the exact premise which undergirds the warmist narrative.
————————-
Your stance is not standard radiative, thermodynamic and cavity theory, it’s a bastardization of same. That’s why your take demonstrably violates 1Lot, 2LoT in the Clausius Statement sense, Stefan’s Law, the Entropy Maximization Principle, the Equipartition Theorem, and a whole host of other fundamental physical laws.
“But 2LoT can be freely violated at the quantum scale!”, is usually the next desperate move by those who wish to sustain the warmist premise of energy being able to flow from cooler to warmer object, and thus their claim of bidirectional energy flow with the ‘net’ being what the S-B equation calculates, and thus their climate alarmist narrative.
No macroscopic 2LoT violations have ever been empirically observed and 2LoT is rigorously obeyed at the quantum scale, with an entire family of subset 2LoT conditions which must be met before any change of state can take place.
So, having removed that avenue from your ongoing science-denialism, Jeff, what will you try next? LOL
I didn’t just declare victory. I put up an answer as the head post with a link directly above my comment — see #8. You missed it apparently. — well done.
I also checked the spam bucked and found ZERO trapped comments from you. All that bitching and it seems the ‘technical error’ is on YOUR END, despite your false accusations of censorship.
Stop being such a damned chicken brain and read kid. You suck at this.
Your ‘victolly!’ declaration is premature and, given that you find yourself utterly unable to even approach any semblance of answering that foundational question asked of you, entirely unwarranted, Jeff.
—————
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291779
“Now it’s your turn, Jeff. Describe for us exactly how photon generation occurs against an energy density gradient.”
No worries, right? I mean, your inability to form a cogent basis in physics by which a photon could possibly be spontaneously emitted against an energy density gradient only destroys the underlying premise of your belief system and thus destroys your narrative, while also conveniently highlighting that you’ve spent an untold number of years believing in and shilling for the exact premise which undergirds the warmist narrative.
—————
I had a catastrophic hard drive crash, on my main drive and on one of my external backup drives, so the O/S won’t boot, and I can’t clone a backup back onto the main drive, as I would usually do for such an occurrence. Not sure what caused it… I would suspect the ever-present power glitches as solar and wind power intermittency disrupt the grid, but my computer is on a 2000VA live UPS with inverter-generated pure sine power, fed from a large battery bank (3 days worth of battery backup power at current usage), so grid glitches shouldn’t affect them.
I’m currently scanning the drives and remapping the bad sectors. It’ll take awhile, they’re large drives. Until then. I’m left hunt-n-pecking on a phone.
I’ll be back soon enough, whereupon you can continue embarrassing yourself, Jeff.
Or you can just continue doing so in my absence, as I see you’ve been doing. LOL
First law of Thermodynamics. You lose kid.
Jeff I’d wrote:
“First law of Thermodynamics. You lose kid.”
Still attempting to push your lie that a photon subsumed into the background EM field is ‘destroyed’, Jeff, and in so doing continuing to demonstrate your utter inability to grasp QFT?
Despite the fact that you’ve had it explained to you in detail?
https://noconsensus.wordpress.com/2022/10/22/why-i-laugh-at-the-sky/#comment-291756
You’ve been repeatedly called on that lie of yours on another thread… but if lies are all you’ve got to defend the very premise which undergirds CAGW and thus which the leftist warmists equally vociferously defend with similar lies… then I guess you have no choice but to lie repeatedly in the same manner as a leftist warmist. LOL
Show everyone this purported 1LoT violation, Jeff… prove it mathematically.
Victory!
This has been my same experience every time I’ve had a discussion with sky dudes. I ask any detail and I get a refusal to answer. I get mocking and pointed to a pile of textbook style ramblings with errors distributed throughout and endless, endless conclusion. Then after many hours of patience, I explain how science actually works and poof, the sky dudes vanish.