Talk:Thermal radiation

Overview

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)[reply]

Object Color and Thermal Radiation at IR Wavelengths

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)[reply]

Several years ago, a mechanic told me that every mechanic knows that's myth. — Preceding unsigned comment added by 92.67.227.181 (talk) 14:24, 4 July 2022 (UTC)[reply]

Reworked

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)[reply]

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

Try harder. Radiators heat by all three mechanisms. If you're close to the radiator then you can feel the IR radiation. You know it's radiation because you can shield yourself from it with your hand or a piece of reflective material. If you're on the opposite side of the room then most of the heat you receive will be from convection. If you touch the radiator then it's conduction. --Heron 19:54, 21 June 2006 (UTC)[reply]

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

Hm, perhaps it was a bad example. I was visualizing a radiator in vaccum. I guess a better example would be heat from a fireplace, where you feel the infrared radiation as you come close to it. But at the same time, I wanted to illustrate that the surface itself is radiating. PoorLeno 10:58, 18 December 2006 (UTC)[reply]

And what about the heat that the air itself conducts? How do you discern between this and radiation? James Callahan 21:35, 23 April 2007 (UTC)[reply]

Put a piece of thin plastic like a trash bag between you and the heat source. Tarchon (talk) 20:46, 17 August 2009 (UTC)[reply]

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)[reply]

Props

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

Non-black bodies

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)[reply]

Graph of Wien's Law

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)[reply]

agreed, as is I think this could confuse a lot of people--Xris0 (talk) 21:05, 7 August 2009 (UTC)[reply]

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[reply]

Larger context

It would be nice to get an explanation of how this fits into the larger scheme of things:

Does this govern solids only?
How does it relate to emission spectra of gasses?
What about liquids, how do they glow? Plasmas?
Objects in state transition or tri-points?
Coloured, reflective, and translucent objects?
How does it relate to an isolated object over time, as it's energy dissipates? How fast would something cool down?

I'm just layperson looking for a framework. -Craig Pemberton (talk) 19:30, 2 October 2009 (UTC)[reply]

How is movement converted to light?

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."

I am very interested in seeing the active voice version of this and think it would be a significant contribution to this article.
Isn't heat defined as the motion of particles? If so, then this explanation may be somewhat circular.
The following doesn't explain it either and is probably incorrect but I hope it will get the ball rolling and give an idea of what I'd like to see here: "Thermal radiation is generated when charged particles within atoms collide as a result of heat energy and emit electromagnetic radiation."

Jojalozzo (talk) 13:51, 12 October 2009 (UTC)[reply]

EM radiation is emitted whenever a charge undergoes some acceleration. Thermal radiation really comes from a wide variety of emission mechanisms - what they all have in common that makes them thermal radiation is that they're pumped by heat energy. Collision isn't really necessary for the process, and that doesn't really seem like a good way to explain it to me. Probably the best way to think about it is that jiggling a charge produces EM radiation and heat is jiggling. Tarchon (talk) 08:50, 17 February 2010 (UTC)[reply]

The Earth has no 'day' or 'night'

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.

Dkrocks (talk) 08:53, 7 December 2009 (UTC)[reply]

Emission/absorption of thermal radiation is work?

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)[reply]

The state, the Hamiltonian, of a system is defined by internal and external parameters that affect the system. External parameters are, for example, the volume or any kind of field that the system is located in. Thermal interactions are defined as those when the external parameters are constant. Work is when external parameters change. Heat is defined as an energy transfer by thermal interaction. In this sense thermal radiation is a change in external parameter, therefore work. The article on heat describes it also as energy transfer not by performance of work. In the thermodynamic strict usage, heat is only the exchange in form of Fourier's type of conduction. Reif, Statistical and Thermal Physics, has a whole chapter on this and he defines heat accordingly as well.

That said, however, the term heat has all kinds of usages, historically, popularly, and even in science. The usage of heat in engineering is clearly much broader, and is more or less based on the notion of its effect when applied to a system. In physics, it is usually defined from the principle of origin, its microscopic nature.

In general, it is widely acknowledged that many terms including heat in their name, are misnomers, even physically well established terms, such as heat capacity. Perhaps some languages other than English are more precise, but the ones I know are not.

Mentioning other forms of heat transfer here is not the goal of this article, so don't mention it, as then people may start to argue about just how many forms there really are, for example, mass transfer is often considered a form of 'heat transfer' in engineering too. Best to avoid the issue here, and let the heat transfer article sort it out.

Kbrose (talk) 17:59, 15 December 2010 (UTC)[reply]

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)[reply]

This is not consistent. If you claim to convert a single wavelength, than why not a spectrum of wavelengths? No matter how many photons a system absorbs, or whether a system resonates at the same frequencies, you have performed work on it. It doesn't matter what its spectral distribution. All photons are equal except for their energy. Furthermore, thermal radiation does not have a temperature of its own, the (color) temperature associated with it is the temperature of a cavity radiating it, and it is merely a label for the spectral distribution of the radiation. When radiation interacts with the sink, it is an intermediate system that is involved and whose thermal energy is modified. Therefore it attains some temperature, but it will never be the same as the source of the radiation. It is a fallacy, a mistake, to apply the Carnot theory of reversible engines to continuous radiation from a 'hot' source to a 'cold' sink. The theory of maximum thermal efficiency of a Carnot engine cannot be applied directly, if that is what you claim. Instead, it is a matter of radiative power received at the sink that must be considered. I grant you that 'heat' is used rather loosely sometimes in physics too, as I also stated historical usage. When consulting modern reference texts, one must infer whether the context is engineering or pure physics. Engineering students often find that when they take classes in the pure sciences to fulfill their prereqs, they are confronted with different, more precise definitions, much to their plight. In a 'neutral' encyclopedia one must be aware of this and disambiguate unfortunately. Kbrose (talk) 21:52, 15 December 2010 (UTC)[reply]

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)

It pretty clear you are not clear about the first and the second law. Any form of work may be converted into each other, whether it's mechanical or electronic absorption of a photon. That happens with 100% precision. That's the conservation of energy of the first law. All forms of work are thermodynamically equivalent because their conversion does not transfer entropy. They may also be converted into heat partially or completely, but the reverse is not true, one cannot convert heat to work completely, because entropy enters a system with heat, but the work does not remove it. This is where the second law applies the Carnot limit to efficiency of such an engine. Since entropy isn't removed by the work, it builds up in the system and must be removed, therefore an engine is cyclical to allow periods of entropy removal, in the form of wasted heat.

Your passage in the second sentence (When you use it to make work on a colder system,...) describes thermal radiation as work, but then you try to argue it isn't work? You state ...it will cool to conserve energy. Wrong, it cools because removing thermal energy (converting it to radiation) is synonymous with lowering its temperature; temperature is the measure of thermal energy. In E-M radiation, thermal energy is indeed converted to radiation completely, where else would the waist energy go that is not converted? As heat? The system reheating itself, i.e. increasing its thermal energy while radiating it away? Naively logical I suppose, but not a conversion actually. Thermal radiation is a form of energy, it is light or photons, that perform work on a system that absorb them. This may cause heat to be generated, indeed, and then you can try to build an engine with it, and calculate its efficiency when you know the temperature of your electronically excited states, but don't use the temperature of the Sun to do so. Kbrose (talk) 06:23, 16 December 2010 (UTC)[reply]

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)[reply]

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)[reply]

The book by Fuchs, that you are quoting, is a very modern text and it avoids the historically awkward and confusing treatment of the term 'heat' by using the entropy instead, which in many ways makes the concept of entropy very intuitive and removes its stigma of difficulty and obscurity. This is along ideas recently advertised by Herrmann in Germany. It restores some validity of the original concepts of heat as originally discussed by Black in the 1700s, except he described what we now know to be entropy, a term coined much later. The book also has very nice diagrams of thermodynamic systems illustrating the ideas.

Anyhow, as nice as this book is, you are misinterpreting it. In fact, nowhere in the text does it suggest to use the temperatures of Sun and Earth in the Carnot efficiency equation that you display above. This is an erroneous misapplication of the limit, which has been mentioned in various places in the literature, btw. The chapter you quote discusses the photon gas in a cavity and the energy exchange between the radiation field and the cavity walls, but not a Carnot engine between Sun and Earth. It simply does not support your idea.

But, you are right that thermal radiation is not work, indeed, thermal radiation is a form of energy, that performs work when absorbed. That's why I wrote in the lede ...the absorption of thermal radiation is work... Thereby, heat transfer (in the old physics/engineering meaning) by radiation is work. But thermal radiation is not heat. That's why physicists do not call it heat radiation anymore, but thermal radiation, because it originates from the thermal energy of a body. Even Planck made the distinction between heat rays (light) and heat, and he did not confuse the two. However he did call it heat radiation, an old-fashioned term.

As for your second reference, Jenks, it is talking about a different kind of efficiency, not any Carnot limit between Sun and Earth. Please read in detail.

Kbrose (talk) 19:34, 16 December 2010 (UTC)[reply]

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)[reply]

In 'standard' physics today, there is only one definition of heat, namely a thermal energy current across a system boundary. If you want to change it, then write a major paper or book (like Fuchs) and convince others, but not here first. I am not saying that today's definition is good or bad, but it certainly is very confusing to many. RE Jenks, this efficiency is the efficiency of radiative flux power of two opposing radiators, as I hinted at before. Obviously when a sink absorbs radiation, it simultaneously emits radiation based on its own temperature which cannot be used in the device. Radiative power is dependent on the source temperature, so there is a theoretical efficiency that must use the source and sink temperatures. This is a popular exercise in teaching to find the 'equilibrium' temperature of Earth, excluding all other energy sources, but this has nothing to do with thermodynamics per se. Further consider that radiative processes, and all 'heat transfer' processes, are non-equilibrium processes, and equilibrium thermodynamics does not apply in its basic form. Fuchs' book tries to unify the description of all this, which is desired by many, but not main stream yet. Kbrose (talk) 20:14, 17 December 2010 (UTC)[reply]

Briefly, you have to recognize that the book by Fuchs does not follow today's 'standard' physics in presentation. So his nomenclature is slightly different. I tried to point this out earlier. He describes thermal phenomena in terms of entropy transfer, and uses the term thermal energy current for the concept of heat as it is taught in current standard physics. This frees him up to use the term heat as it is used in everyday language of the layman. You have to read some of the introductory chapters where it becomes quickly very clear. He uses the term heat just like non-scientists use it in everyday life. In Chapter 7 this is expressed in the first sentence. The section is entitled "The Transport of Heat", and the first sentence is: "In this chapter, we shall take a closer look at the transport of entropy." So in his book, heat = entropy, which is exactly the street use of heat. These new ideas permeate the entire book, but they are actually not new, they are based on revived 18th century concepts by Black, which got misinterpreted and lost for a long time. So, it is in this definition that he uses heat radiation, as you correctly stated, but it is the radiation of entropy, not standard physics' heat. This also explains the title of the book. In chapter 4 he explains what heat is and is not. He states on page 110: "The point is this: heat is not energy."

Just googling subjects and picking out of context citations is easy, a favorite sport on Wikipedia, but can lead to gross errors, as this shows. You have to read and understand the book, and in this context, it's actually many books that need be read. Kbrose (talk) 02:32, 17 December 2010 (UTC)[reply]

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)[reply]

Google is not a judge of physics. If you want to continue to play word games, go ahead in your own time. Kbrose (talk) 20:14, 17 December 2010 (UTC)[reply]

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)[reply]

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)[reply]

Merge with Incandescence?

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)[reply]

For the time being, I edited incandescence to be a short article on the visible-light aspect of thermal radiation. I put lots of links so that readers looking for more detail will go to this article instead. --Steve (talk) 16:54, 19 December 2011 (UTC)[reply]

I see no reason why they should not be merged. --Cowlinator (talk) 02:08, 22 February 2017 (UTC)[reply]

I agree, having two seems redundant. KingSupernova (talk) 15:39, 17 May 2018 (UTC)[reply]

Incandescence is the effect that warm objects emit thermal radiation, so incandescence is the effect / mechanism, while thermal radiation is the resulting electromagnetic fields / photons themselves. Regardless, I think it's the same topic so the articles should be merged. — Preceding unsigned comment added by 92.67.227.181 (talk) 14:15, 4 July 2022 (UTC)[reply]

Radiative heat transfer equations

The parameters $E$ and $F$ in the equations in this section are not defined anywhere:

{\dot {Q}}_{1\rightarrow 2}=A_{1}E_{b1}F_{1\rightarrow 2}-A_{2}E_{b2}F_{2\rightarrow 1}

Neutron star?

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)[reply]

You're basically asking, "Why isn't a neutron star transparent?" (cf Kirchhoff's law of thermal radiation.) Phrasing it this way makes it more familiar.

The first sentence of this article is "Thermal radiation is electromagnetic radiation generated by the thermal motion of charged particles in matter." According to statistical mechanics, that claim is equivalent to "Electromagnetic radiation is absorbed when they cause charged particles in matter to move." Is it accurate? Well, that is not universally true ... electromagnetic radiation can be absorbed other ways too. In electron spin resonance, for example, radio waves are absorbed by interacting with electron magnetic moments, causing the electron to rotate (which is not "motion" as the word is normally understood ... the electron is a point particle). Neutrons have can absorb electromagnetic waves that way too, and they are not even charged. They can also absorb electromagnetic waves via their electric polarizability, which is related to their being made of quarks. (Is that related to "motion of charged particles"? Maybe...)

But I would say that the first sentence is accurate in the vast majority of cases and I don't think it really needs to be changed to be more technically accurate.

If you have specific follow-up questions about why a neutron star is not transparent, I would suggest asking at Wikipedia:Reference_desk/Science. Remember that even if one single neutron is relatively transparent, the whole neutron star (many kilometers thick and extremely dense) can still be totally opaque. --Steve (talk) 23:20, 2 October 2012 (UTC)[reply]

Thanks for that Steve. I wasn't aware of the reference desk. And I certainly hadn't considered the quarks inside the neutron! BTW you might like to consider more cross references (or even merging) betweeen this page, the Kirchoff's law one, and Black-body_radiation. Number774 (talk) 07:32, 4 October 2012 (UTC)[reply]

The usual way....What?

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)[reply]

I think the usual electrodynamical way was meant. According to the theory of electromagnetism, charged particles, such as electrons, induce an electric field which points outward, is strong near the particle and falls off in the distance. According to Maxwell's equations, the curl of the magnetic field is the change of the electric field, while the curl of the electric field in turn is the change of the magnetic field but in the opposite direction, so that the two fields induce one another. So an oscillating charged particle will induce an oscillating electromagnetic field: light. If you're interested in learning more about the subject, you could start by reading Introduction to Electrodynamics. — Preceding unsigned comment added by 92.67.227.181 (talk) 14:13, 4 July 2022 (UTC)[reply]

Greenhouse mechanism discrepancy

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)[reply]

Those arguing that it must be one or the other appear to believe that the Law of Excluded Middle applies to physics. A similar mistake is made by those who insist that global warming is either all natural or all man-made. As atmospheric scientist Craig Bohren pointed out quarter of a century ago, greenhouses that don’t show the effects of IR trapping tend to be “set in windy environments and have thin walls.” Vaughan Pratt (talk) 21:41, 12 January 2014 (UTC)[reply]

A standard old-style greenhouse only inches into the excluded middle very slightly though. The glass, air and plants are in thermal equilibrium, so the glass panes and iron beams still radiate blackbody radiation themselves. Sure, a greenhouse can be made more efficient by trying to prevent this thermal equilibrium from setting in, by using double glass maybe, but the principal mechanism by which a greenhouse operates is preventing warm air from rising up into the sky allowing cooler air to take its place.

Ambiguous, confusing

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)[reply]

12 µm is long-wavelength thermal infrared radiation emitted by the snow because it's hundreds of kelvin above absolute zero. — Preceding unsigned comment added by 92.67.227.181 (talk) 13:15, 4 July 2022 (UTC)[reply]

Microwave Radiation

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)[reply]

We know that temperature and energy density are related by Stefan's Law.

Temperature (T) is equal to the fourth root of radiation energy density (e) divided by Stefan's Constant (a) (ie: the radiation constant), 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)

... but energy doesn't flow according to temperature, it flows according to energy density gradient (see below for the explanation).

Now, a microwave uses ~2.4 GHz photons, with a wavelength of 124876 µm (in air) and a corresponding Wien Displacement Law Peak Temperature of only 0.023205194753195168 K.

How does it warm food? Well, we've discovered a means of disconnecting, via electrical means, the Stefan's Law relation between temperature and energy density, to cause a far higher energy density of the emitter.

That radiation field has a higher energy density than the food, and thus the food undergoes radiation heating. Energy flows from the radiation field to the food, which increases its kinetic energy, and since kinetic energy in this regard is a measure of temperature, the food's temperature increases... IOW, while we've found a way to disconnect the Stefan's Law relation between temperature and energy density of the microwave emitter via electrical means, the absorber (the food) still follows Stefan's Law. 76.30.103.137 (talk) 03:52, 27 October 2024 (UTC)[reply]

All bodies radiate? Don’t think so

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)[reply]

I agree with this! Mcasariego (talk) 14:35, 20 April 2020 (UTC)[reply]

I belive all matter is wrong. Gases do not emit thermal radiation due to their temperature. Liquids? 2001:14BB:695:1312:A1A5:844A:F7CF:6C05 (talk) 05:34, 4 May 2023 (UTC)[reply]

That's a good catch actually.

Section on quantum statistics?

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 $E=\hbar \omega$ ). Let me know if anybody is interested in collaborating! Mcasariego (talk) 14:46, 20 April 2020 (UTC)[reply]

Violation of 2LoT in the Clausius Statement sense

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)[reply]

did you plagiarize me?

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)[reply]

The current wording came to be in the edit on 21:44, 6 September 2021; we won't be able to tell you more without a link to your text. 92.67.227.181 (talk) 23:54, 7 July 2022 (UTC)[reply]

That text should be tightened up a bit... it confuses people into thinking blackbodies emit blackbody radiation, and graybodies emit graybody radiation. I should know, I've debated enough of them.

So first point out that there are two form of radiative emission... blackbody emission and spectral emission.

Rather than "thermal radiation", it should properly be designated as "blackbody radiation", and the explanation pointing out that blackbody radiation comes about due to inter-atomic or inter-molecular non-zero electric dipoles due to the particles (atoms or molecules) being in close proximity, and an energy density gradient that that emitted energy can spontaneously flow down.

Conversely, spectral emission comes about due to a non-zero electric dipole for that molecule (as opposed to inter-atomic or inter-molecular for blackbody radiation), and an energy density gradient that that emitted energy can spontaneously flow down.

It should be further pointed out that solids, liquids, gases of sufficient density and plasmas of sufficient density can emit blackbody radiation; but as a first approximation, our atmosphere (a 'thin gas') cannot because the atoms and molecules spend the majority of their time relatively distant from each other and thus cannot sustain the inter-atomic or inter-molecular non-zero electric dipoles necessary for blackbody emission.

And it should be noted that as gas or plasma density increases, blackbody emission increases until it eventually swamps the underlying spectral emission. Likewise for absorption.

76.30.103.137 (talk) 03:29, 27 October 2024 (UTC)[reply]

Subjective color to the eye of a black body thermal radiator

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)[reply]

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)[reply]

Thermal Radiation

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)[reply]

There is no contradiction really. Boyer is considering the inside of a cavity devoid of any matter. What he does not explicitly mention in this paragraph, is that the cavity is bounded by walls made of matter. Since photons do not interact with each other, the only way for the EM field to reach a thermal equilibrium is to interact with the walls of the cavity. The EM radiation inside the cavity is then exactly the radiation that is radiated by the walls of the cavity. Jähmefyysikko (talk) 05:20, 10 July 2023 (UTC)[reply]

Confusion of idealized blackbodies and graybodies

The text blurb:

"All matter with a temperature greater than absolute zero emits thermal radiation." is wrong. It confuses idealized blackbody objects and real-world graybody objects.

As I show here:

https://www.patriotaction.us/showthread.php?tid=2711

Idealized blackbody objects (which assume emission to 0 K and ε=1 by the very definition of idealized blackbody objects) emit radiation above 0 K. Of course, idealized blackbody objects don't actually exist... they're idealizations. The closest we can come are laboratory blackbodies which exhibit high absorptivity and emissivity in certain wavebands, but even they are not idealized blackbodies... idealized blackbodies by definition have no thermal capacity (an idealized blackbody must absorb all radiation incident upon it, and must emit all radiation it absorbs). That's why cavity theory for idealized blackbodies is predicated upon all of the energy being in the radiation field in the cavity space, and none in the walls.

Graybody objects emit if their temperature is > 0 K above whatever is in their view factor.

Here's a handy graphic I created:

https://i.imgur.com/QErszYW.gif

Why? Because temperature is a measure of energy density, equal to the fourth root of radiation energy density divided by Stefan's Constant, 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)

And when we plug that into the S-B equation, we get:

q = ε σ (T_h^4 - T_c^4)

∴ q = (ε_h * (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)

You will note that σ / a = 5.6703744191844294539709967318892308758401229702913e-8 W m-2 K-4 / 7.5657332500339284719430800357226e-16 J m-3 K-4 = 74948114.502437694376419756266673 W m-2 / J m-3.

Well, what do you know... that's the conversion factor for radiant exitance (W m-2) and energy density (J m-3)!

It's almost as if the radiant exitance of graybody objects is determined by the energy density gradient, right?

One can see from the immediately-above equation that the Stefan-Boltzmann (S-B) equation for graybody objects is all about subtracting the energy density of the cooler object from the energy density of the warmer object to arrive at the energy density gradient, which determines radiant exitance of the warmer object.

Energy does not and cannot spontaneously flow up an energy density gradient. If it could (ie: if all objects > 0 K emitted at all times regardless of energy density gradient), that would violate 2LoT in the Clausius Statement sense. I expand upon this at the link above.

Which means for two graybody objects at thermodynamic equilibrium, no energy flows, no absorption nor emission takes place. The system reaches a state of quiescence (which is the definition of thermodynamic equilibrium). The photons remaining in the intervening space set up a standing wave, with the wavemode nodes at the object surfaces by dint of the boundary constraints. Nodes being a zero-crossing point (and anti-nodes being the positive and negative peaks), no energy can be transferred into or out of the objects. Photon chemical potential is zero, they can do no work, photon Free Energy is zero, they can do no work... there is no impetus for the photons to be absorbed. Should one object change temperature, the standing wave becomes a traveling wave with the energy flux equating to the energy density differential times the group velocity. That coincides with bog-standard cavity theory.

Do remember that a warmer object will have higher energy density at all wavelengths than a cooler object:

https://web.archive.org/web/20240422125305if_/https://i.stack.imgur.com/qPJ94.png

... so there is no physical way possible by which energy can spontaneously flow from cooler (lower energy density) to warmer (higher energy density).

In short, that text blurb clings to the long-debunked Prevost Principle from 1791, which postulates that an object's radiant exitance is only determined by that object's absolute temperature. A quick glance at the S-B equation for graybody objects, however, disproves the Prevost Principle... there's a second factor which affects radiant exitance:

[1] Idealized Blackbody Object form (assumes emission to 0 K and ε = 1 by definition):

q_bb = ε σ (T_h^4 - T_c^4)

= 1 σ (T_h^4 - 0 K)

= σ T^4

[2] Graybody Object form (assumes emission to > 0 K and ε < 1):

q_gb = ε σ (T_h^4 - T_c^4)

T_c^4 is the second term which affects radiant exitance of the warmer object. And:

q_gb = ε σ (0) = 0

q = (ε_h * (σ / a) * Δe)

q = (ε_h * (σ / a) * 0) = 0

... anything multiplied by zero is zero. Radiant exitance of graybody objects falls to zero as temperature differential (and hence energy density gradient) falls to zero.

76.30.103.137 (talk) 02:45, 27 October 2024 (UTC)[reply]