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In the "Surface Effects" section in Overview it states the following: Personalized cooling technology is another example of an application where optical spectral selectivity can be beneficial. Conventional personal cooling is typically achieved through heat conduction and convection. However, the human body is a very efficient emitter of infrared radiation, which provides an additional cooling mechanism.
Two problems
(1)I believe this is false (2)No reference is provided.
(2) doesn't need any explanation. I will now explain why I believe this to be false. Consider thermal radiation and conduction from a human body. If the human metabolizes at M=150 watts (about 3000 kcal/day) then their radiative flux is about M/A where A ~=2m^2 is the surface area which admittedly has considerable uncertainty. This heat flux then is about 75W/m^2 and has to be carried away by radiation and conduction. Radiation is by far the dominant effect. I'm going to ignore the emissivity because it is very close to 1. The radiative flux, phi_rad is given by phi_rad = sigma (Ts^4 - Tw^4) and the conductive flux, phi_cond= kcond(Ts-Tw)/L. Here Ts is the temperature of human skin, Ts=310K, and Tw is the temp of the surrounding walls, say Tw=295K, with L the distance from the person in the center of the room to the walls, say L=5m and kcond = .026W/m-K is the thermal conductivity of air. phi_rad ~= 76W/m^2 is two or three orders of magnitude larger than phi_cond ~=0.1W/m^2, so in what sense is "personal cooling" "typically achieved through heat conduction"? I think the main mechanisms by which heat leaves a resting body are (1)radiation, (2)exhalation of water vapor, and (3)sweat and evaporative cooling during exertion.
I don't see any way to get significant amounts of heat transport from conduction/convection as compared to these other mechanisms. Presumably this is discussed in a text book somewhere? — Preceding unsigned comment added by 76.113.29.12 (talk) 00:50, 13 November 2021 (UTC)
The article states that the color of an object has little effect in its thermal radiation at IR wavelengths at ordinary temperatures. This is clearly untrue. Every auto mechanic knows that black-painted auto radiators cool their contents better than light colored ones. I had a frying pan which has a dark nonstick surface on one side and shiny aluminum on the other. If it is heated to maybe 100 or 150C and held vertically at a horizontal distance from my face, no heat is felt when the shiny side is facing me, when the dark side is facing me the heat can be felt instantly from a meter away.
I came here to learn more about this effect and how to quantify it, only to read a claim that the effect does not exist at all!
Hopefully someone (I'm not a physicist) can fix this part of the article. — Preceding unsigned comment added by 199.115.12.254 (talk) 22:19, 20 September 2016 (UTC)
I've reworked the article, when I found many factual errors and other nonsense. Please post a comment and compare with the old version, as you see fit. Again, refrain from correcting as per old version - it was full of factual errors.
--PoorLeno 22:43, 7 April 2006 (UTC)
Beginning of article: I believe radiators heat via thermal conduction and convection. I find it hard to believe that EM radiation is warming me as I stand by my radiator. — Preceding unsigned comment added by 129.55.200.20 (talk) 17:16, 19 June 2006
Heron is right. And for your radiator it will be about 20 Joule per second per squared meter. — Preceding unsigned comment added by 80.61.176.81 (talk) 16:16, 25 October 2006
I'd suggest you fix the first link to rimron.co.uk, as it appears to be broken. Though if you have a legit link to a brightness vs. temperature graph, I'd rather appreciate seeing it. — Preceding unsigned comment added by 69.4.71.1 (talk) 17:31, 25 June 2007
Reverted to last garbage-free version. LouScheffer 17:45, 25 June 2007 (UTC)
I think its cool that the author of this article defined all the constants used in the equations. I wish more article would do that. Props to whoever the author was. —Preceding unsigned comment added by TheLibertarian (talk • contribs) 23:02, 24 November 2006
I was looking for information on how color changes the radiative spectrum. Can we get some examples of emissivity correction factor functions, perhaps with a chart? -- Beland (talk) 19:44, 21 November 2007 (UTC)
it could stand to be redone such that the colors of the lines corresponded to the peak wavelength. That is, going from blueish, to greenish, to reddish, to black. Hantavirus (talk) 15:25, 12 June 2009 (UTC)
whatever are the units for the Y axis? There seems to be a scaling factor involved, but the designation kJ/nm is mysterious.72.74.53.59 (talk) 16:06, 19 March 2014 (UTC)JRB
It would be nice to get an explanation of how this fits into the larger scheme of things:
I'm just layperson looking for a framework. -Craig Pemberton (talk) 19:30, 2 October 2009 (UTC)
Another layperson's request. The following sentence is intriguing: "Thermal radiation is generated when heat from the movement of charged particles within atoms is converted to electromagnetic radiation."
From a person's perspective on the Earth there may seem to be definitive day and night times, but the world may also seem flat. Mention of the earth absorbing radiation from the sun during the 'day' and radiating it outward at 'night' should be replaced with a more accurate description of that movement of energy.
Though to make a system to absorb a specific radiation can be considered work, thermal radiation has the same properties of heat, just as mechanical energy can make work with 100% efficiency, but thermally-distributed mechanical energy is heat. If we use light from the Sun (a black body at 5800 K), we can make work from it only up to a thermodinamic efficiency of (T_sun-T_earth)/(T_sun), like with any other heat. And, if we exchange thermal radiation between two bodies, we have exactly the same behavior of heat. So, I think it's not incorrect to refer to it as a method of heat transfer.--GianniG46 (talk) 11:55, 15 December 2010 (UTC)
Not only engineers use "heat" in a broader sense. Also some old physicists, such as Max Planck, which, in his book The Theory of Heat Radiation begins the first chapter, "Radiation of Heat" (p.1), with the statement: "Heat may be propagated in a stationary medium in two entirely different ways, namely, by conduction and by radiation." But Planck was German, so maybe there is some language difference.
My main point, however, is not whether thermal radiation is heat or not heat. My point is that it is not like a piston that moves and makes work. In principle I can convert piston movement in whichever energy I want with 100% efficiency, and, in the same way, I can convert a single wavelength radiation into an electrical energy or a piston movement with 100% efficiency. But not thermal radiation, which has a temperature associated with it, and is subjected to the 2nd law of thermodynamics, unless I use a Maxwell's demon. So, definitely, thermal radiation is not work, in the thermodynamic acception. --GianniG46 (talk) 19:10, 15 December 2010 (UTC)
Thermal radiation is generated by a system at temperature T. When you use it to make work on a colder system, you are subtracting thermal energy from the first system, which will cool to conserve energy, so you are using heat to produce work. If you could do this with 100% efficiency, you would violate 2nd law, in the same way that you would violate it if you used conduction between the two bodies. For e.g. a Carnot cycle there is no conceptual difference between exchanging thermal energy by conduction or by radiation. And, after all, the statistics of black body radiation is calculated in the same way used for calculating a gas, with partition functions, densities of states, Boltzmann factors and so on. So it would conceptually simpler to call heat both mechanical and electromagnetic thermal energy, and, since, when it is possible, physics must be simple... --GianniG46 (talk) 23:20, 15 December 2010 (UTC) — Preceding unsigned comment added by GianniG46 (talk • contribs)
The first part of what you say is exactly what I was trying to say with my limited English. Instead, I do not understand the second part: I have not said that thermal radiation is work, but that I can use it to make work (just like I can use heat), with an efficiency that depends on the temperatures of the source and the sink of the radiation. A monochromatic light beam (like a moving piston or a molecular beam) is energy with no entropy, and can be entirely converted into work or into any other form of energy. A thermal radiation, instead (like thermal movements of molecules), contains entropy (see H.U.Fuchs, The Dynamics of Heat, p.229), so it can be converted, or make work, only within the Carnot limits. And the temperatures to be considered are just the temperatures of the emitting and the absorbing bodies: efficiency=(T_sun-T_earth)/T_sun) in the case of sunlight. If you could obtain work with a greater efficiency, you could use this work to build a heat pump (which has the inverse of the above efficiency) to heat the Sun and cool the Earth, which is, of course, strictly forbidden by the 2nd law. So, definitely, thermal radiation is not work, but, like any other form of thermal energy, can be converted to work (e.g. by a photovoltaic cell) within the Carnot limits, see for example S.Jenks, Solar Energy, 3.3 - Entropy limitations, p.21.
Sorry for my unclear English, but the references I am giving say clearly what I was trying to express. --GianniG46 (talk) 10:02, 16 December 2010 (UTC)
P.S. Note also that both of the references I have given are from physicists, as it appears from their home pages, [1] and [2], and that they speak of "thermal radiation...comparable to the ideal gas...photon gas", "radiation of heat" and "energy transfer as heat".--GianniG46 (talk) 11:14, 16 December 2010 (UTC)
a) Section 3.3.1 by Jenks begins with the words "The general efficiency limit for a photovoltaic system is the Carnot limit.", and ends with a calculated factor 1-T_ambient/T_sun ≈ 95% (=1-290K/5800K). According to you this is not the Carnot factor I was speaking of?
b) You say that "heat radiation" is an old-fashioned term. But Fuchs uses consecutively three expressions in the last few lines of section 5.4: "transfer of entropy by radiation", "radiation of heat", and "radiative transfer of heat".
c) The last phrase of this same section is "In this chapter we shall study the photon gas; radiative transfer of heat will be discussed in chapter 7 and 12". Unfortunately, neither chapter 7 nor chapter 12 are shown in the preview. That is why you find only a photon gas, with no Carnot engines (which, however are discussed in Jenks).
d) You say that thermal radiation is not work, but performs work when absorbed. More precisely you should say "may perform work". Because in thermodynamics work is not the single "atomic" work, i.e. making an electron to jump from a level to another, or changing the velocity of a single molecule by a collision, but is a coherent action made on a macroscopic quantity of matter. Moving a piston is work, generating photovoltaic currents is work. Is not work heating a body by casual collisions between molecules (conduction), or giving statistical energies to molecules by exciting their electrons with casual, thermal photons (heating by radiation).
e) And, if thermal radiation is a form of energy that may perform work, exactly the same can be told for any other form of thermal energy. So why in the article do you specify this for radiation, as contrapposed to other forms of heat (or, if you prefer, thermal energy) transfer? --GianniG46 (talk) 23:57, 16 December 2010 (UTC)
I applied myself to my favorite sport and googled for you 21,000 academic references (only sites ".edu") and 38,500 book references which make "gross errors" and "do not follow today's 'standard' physics", i.e. they use the (exact) expression "radiative heat transfer". --GianniG46 (talk) 08:37, 17 December 2010 (UTC)
Think of photons as just another particle, with a few differences from massive particles (e.g. E=pc rather than E=p^2/2m, and no number conservation). Just as massive particles do not form a thermodynamic system until they have thermally equilibrated, yielding a Maxwell distribution, so photons do not form a thermodynamic system until they have thermally equilibrated, yielding a black body (Planck) distribution. The Maxwell and Planck distributions are both isotropic. Such a "photon gas" is a thermodynamic system: it can be described by all the usual thermodynamic parameters like temperature, entropy, internal energy, etc. A monochromatic light beam is analogous to a mono-energetic stream of massive particles, it is not a thermodynamic system. Treating the sun and the earth as black body radiators is not a strictly thermodynamic problem, since there is no thermally equilibrated photon gas involved, only "beams" of photons which happen to have a black body distribution of energies, but are not isotropic. On the other hand, absorption of photons by a massive body from a thermally equilibrated photon gas can in no way to be considered to be performing work on the massive body any more than if it were absorbing energy from a Maxwell-distributed gas of massive particles. The temperature of black body radiation is a fully defined thermodynamic property of the radiation, not merely a label for the spectral distribution of the radiation. PAR (talk) 18:04, 18 December 2010 (UTC)
This discussion was helpful. I keep seeing statements that equate "heat" with various other things. I've seen it referred to as synonymous with temperature, thermal energy, infrared radiation, thermal radiation, and transfer of thermal energy. Even in this article, I see things like "Thermal radiation is also one of the fundamental mechanisms of heat transfer" and "Thermal radiation, also known as heat", suggesting that thermal radiation is the same as heat, and also the same as the transfer of heat, meaning that heat is the same as the transfer of heat. Knowing that the definition of the term "heat" varies depending on the author, discipline, and context is very helpful. So yeah, I have no idea of whether the emission/absorption of thermal radiation is work, but now I know that heat (or transfer of heat) might be work (or convert-able) to work, depending on what definition of heat is being used.100.4.231.231 (talk) 16:23, 16 July 2019 (UTC)
It seems to me that this article is on exactly the same topic as incandescence. Should they be merged? If there's a distinction, what is it? --Steve (talk) 19:31, 16 December 2011 (UTC)
The parameters and in the equations in this section are not defined anywhere:
I'm puzzled. If it's the movement of charged particles that results in the radiation how does a neutron star radiate? Is it just the mess of ordinary matter on the surface? Would a lump of pure neutronium radiate or absorb? Thanks Number774 (talk) 12:55, 2 October 2012 (UTC)
The fourth sentence in the first paragraph reads
This motion of charges produces electromagnetic radiation in the usual way.
That is the exact issue I wanted to know, and not being a SME I don't know the usual mechanism. I presume in has to do with electrons being bounced between orbits (shells), but frankly don't know. How would this work with molecules? The page http://en.wikipedia.org/wiki/Electromagnetic_radiation on EM doesn't seem to say anything about "the usual way". This sentence prevents non-SME's from comprehending the phenomena. Please re-write or provide a link to "the usual way". Thanx Bcwilmot (talk) 08:00, 11 June 2013 (UTC)
The paragraph here concerning 'selective surfaces' uses the heating of greenhouses and cars as examples:
"A selective surface can be used when energy is being extracted from the sun. For instance, when a green house is made, most of the roof and walls are made out of glass. Glass is transparent in the visible (approximately 0.4 µm<?<0.8 µm) and near-infrared wavelengths, but opaque to mid- to far-wavelength infrared (approximately ?>3 µm).[9][10] Therefore glass lets in radiation in the visible range, allowing us to be able to see through it, but doesn’t let out radiation that is emitted from objects at or close to room temperature. This traps what we feel as heat. This is known as the greenhouse effect and can be observed by getting into a car that has been sitting in the sun."
The increased temperature inside cars and greenhouses is primarily a result of the suppression of convection, rather than the trapping of infrared radiation. The Wikipedia pages on Greenhouses and The Greenhouse Effect both make this clear, see https://en.wikipedia.org/wiki/Greenhouse_effect#Real_greenhouses. The description of greenhouses on this page is therefore inconsistent with other Wikipedia pages. — Preceding unsigned comment added by Sintilla (talk • contribs) 12:36, 2 October 2013 (UTC)
For example, fresh snow, which is highly reflective to visible light (reflectivity about 0.90), appears white due to reflecting sunlight with a peak wavelength of about 0.5 micrometers. Its emissivity, however, at a temperature of about -5 °C, peak wavelength of about 12 micrometers, is 0.99.
"reflecting sunlight with a peak wavelength of about 0.5 micrometers" -- Sunlight has this peak wavelength or the reflected light does? Or the reflectivity (a ratio) does?
"Its emissivity, however, at a temperature of about -5 °C" -- Temperature of the snow?
"peak wavelength of about 12 micrometers" -- The peak wavelength of what? The emitted light? Or of the emissivity (a ratio)? Sentence appears run-on.
178.38.186.135 (talk) 00:15, 9 April 2015 (UTC)
The article states "Whenever EM radiation is emitted and then absorbed, heat is transferred. This principle is used in microwave ovens, laser cutting, and RF hair removal." Microwaves are waves of energy. When this energy hits an object molecules are put into motion creating heat locally. This is what the article on microwaves states. Can someone explain this in more detail? Artsiki — Preceding unsigned comment added by Artsiki (talk • contribs) 18:04, 19 November 2016 (UTC)
This passage in the opening paragraph “All matter with a temperature greater than absolute zero emits thermal radiation.” Suggests that even matter which does not couple EM field would so how emit photons. Shouldn’t it read all matter containing charged particles? Dark matter seemingly doesn’t, and I don’t believe a neutrino cloud would emit photons. Wolfmankurd (talk) 12:09, 25 June 2018 (UTC)
I think this article would benefit from having a section connecting to the article on Photon statistics. It should include at least a brief discussion of super-Poissonian light and perhaps a discussion on the relation between the Bose-Einstein distribution and Planck's law (basically the latter is a special case of the B-E statistics, when one sets the energy 'quantum' to be ). Let me know if anybody is interested in collaborating! Mcasariego (talk) 14:46, 20 April 2020 (UTC)
This text blurb: "The net radiative heat transfer from one surface to another is the radiation leaving the first surface for the other minus that arriving from the second surface." violates 2LoT in the Clausius Statement sense. You're confusing some into thinking 2LoT can be violated nilly-willy, which a surprising proportion of them are using to claim that these continual 2LoT violations can cause catastrophic global warming. That's not scientific, does no one any good, and causes public perception of climate science to suffer.
2LoT (in the Clausius Statement sense... "No process is possible whose sole result is the transfer of heat from a cooler to a hotter body") states that energy cannot flow from a lower to a higher-energy region without external work being done upon the system... not via conduction, not via radiative means, not macroscopically, not at the quantum scale [1], not ever. Do keep in mind the definition of heat: "an energy flux". Thus: "No process is possible whose sole result is an energy flux from a cooler to a hotter body" without external energy doing work upon the system.
Radiation energy density is proportional to T^4 and is derived via the thermodynamic relation between radiation pressure p and internal energy density u, using the electromagnetic stress–energy tensor: p = u/3, it represents the EM field contribution to the stress–energy tensor, describes the flow of energy in spacetime, and is a representation of the law of conservation of energy.
At thermodynamic equilibrium, the Helmholtz Free Energy is zero, thus photon chemical potential is zero, thus no work can be done by the photon, thus no energy can flow: F = U - TS Where: F = Helmholtz Free Energy U = internal energy T = absolute temperature S = final entropy TS = energy the object can receive from the environment
If U > TS, F > 0... energy must flow from object to environment. If U = TS, F = 0... no energy can flow to or from the object. If U < TS, F < 0... energy must flow from environment to object.
If U = TS, p_photon = u/3 = p_object, energy cannot flow because no work can be done. Photon chemical potential is zero.
"Therefore, at thermodynamic equilibrium, the chemical potential of photons is always and everywhere zero. The reason is, if the chemical potential somewhere was higher than zero, photons would spontaneously disappear from that area until the chemical potential went back to zero; likewise, if the chemical potential somewhere was less than zero, photons would spontaneously appear until the chemical potential went back to zero." [2]
Thus photons from a lower energy density source are not even incident upon a higher energy density object. [3]
In a cavity (remember that blackbody radiation used to be called cavity radiation) at thermodynamic equilibrium, quantized standing-wave wavemodes are set up. The wavemode nodes are always at the cavity walls. The walls and the standing-wave wavemodes in the cavity space have the same energy density, thus photon chemical potential is zero, thus the photons cannot be absorbed and re-emitted by the cavity walls, they are reflected. The same thing occurs out here in the real world. Energy does not and cannot flow from lower energy density to higher energy density without external energy doing work upon the system.
"Another point which, this time, lies quite within the domain where our formulae are significant refers to the scattering of light by an elastically bound electron. Consider again our assembly of harmonic oscillators of frequencies k which we may imagine to take the discrete but very finely distributed values determined by [ kL - η(k) = π N (N = 1, 2, 3... ) ]. Let all of them be in the ground state (energy 3/2 ħ k) with the exception of one (frequency k') which has the energy 5/2 ħ k', the vibration parallel to - say - the x-axis being excited by one quantum ħ k'. In this situation light of frequency k' coming from all directions is continuously scattered by the electron.
...the light quanta in the external field... which are affected by the presence of the electron through the appearance of the phase shift η. We might also call them phase shifted light quanta..." [4]
IOW, the electric field of a non-resonant photon, ~100,000,000 times smaller than the Coulomb field seen by the bound electron(s), slightly changes the phase of the bound electron(s), as equally as it changes the phase of the incident photon... no energy is exchanged if the photon is not absorbed, only the phase is shifted, changing the photon's vector. This is how reflection from a step potential works. This is dependent upon the differential between energy density of the photon and atom / molecule. In the case of absorption, the photon constructively interferes with the available resonant quantum states of the atom or molecule. No phase shift
"This means that, at every point of the region of space that is in (pointwise) radiative equilibrium, the total, for all frequencies of radiation, interconversion of energy between thermal radiation and energy content in matter is nil (zero)." [5]
Again, photons from a lower energy density source will not even incide upon a higher energy density object. Thus they cannot be absorbed. They are reflected, which increases ambient field radiation pressure until the objects and the ambient field have the same energy density, whereupon the objects can no longer emit nor absorb, they can only reflect the photons in the quantized standing wavemodes in the ambient field.
In fact, not all photons of sufficient energy to be absorbed are absorbed due to the angular momentum selection rules... all quantum numbers (including angular momentum) must be conserved in order for a quantum transition to occur.
It's not that the object is absorbing radiation from the cooler ambient (or cooler object) thus slowing its cooling rate, it's that the radiation pressure of the cooler ambient (or cooler object) lowers radiative flux from the warmer object, thus lowering its cooling rate.
[1] https://www.pnas.org/content/112/11/3275
[2] https://en.wikipedia.org/wiki/Chemical_potential#Sub-nuclear_particles
[3] https://objectivistindividualist.blogspot.com/2017/11/solving-parallel-plane-black-body.html
[4] http://www.solvayinstitutes.be/pdf/Proceedings_Physics/1948.pdf#page253
[5] https://en.wikipedia.org/wiki/Radiative_equilibrium 71.135.33.222 (talk) 01:07, 6 February 2021 (UTC)
https://en.wikipedia.org/wiki/Thermal_radiation
Thermal radiation is electromagnetic radiation generated by the thermal motion of particles in matter. Thermal radiation is generated when heat from the movement of charges in the material (electrons and protons in common forms of matter) is converted to electromagnetic radiation. All matter with a temperature greater than absolute zero emits thermal radiation. Particle motion results in charge-acceleration or dipole oscillation which produces electromagnetic radiation.
notice there is no source for this paragraph
another thing to note is that the charge particles are not accelerating from/in heat; kinetic energy is transferred without acceleration!
2600:1010:B126:C8FD:1D19:6704:7FB3:4EDB (talk) 09:57, 5 December 2021 (UTC)
The article should explain the apparent discrepancy between the colour table in the article and Color temperature. See the talk page of that article for details. 92.67.227.181 (talk) 13:08, 4 July 2022 (UTC)
I also note that this is a different table, citing a different source, from the two tables in red heat. Either that article should be merged here, or this table should be moved to that article. I'd support merging since the read heat article is tiny, covers basically the same topic as this article, and is very unlikely to grow. — Preceding unsigned comment added by 92.67.227.181 (talk) 14:44, 4 July 2022 (UTC)
The article states right up front, "Thermal radiation is electromagnetic radiation generated by the thermal motion of particles in matter."
This seems to conflict with Timothy Boyer's statements: "Obvious ly a first step in creating a region of vacuum is to eliminate all visible matter, such as solids and liquids. Gases must also be removed. When all matter has been excluded, however, space is not empty; it remains filled with electromagnetic radiation. A part of the radiation is thermal, and it can be removed by cooling, but another component of the radiation has a subtler origin." ("The Classical Vacuum," Timothy Boyer, Scientific American, August 1985).
So here we have a vacuum with no matter which still radiates thermal radiation. Care to discuss? — Preceding unsigned comment added by Yates (talk • contribs) 01:30, 10 July 2023 (UTC)