swansont Posted April 1 Share Posted April 1 4 hours ago, martillo said: I have given a wrong answer. The same material can present different velocities in transmitting "light" or "heat". Glass for instance is an excellent light transmitter, it is transparent to light while a very bad heat transmitter, it is an insulator to heat. The difference is in the frequency of the photons involved. "Light" frequency is in the optical range while "heat" is mainly in the infrared range. Glass is an excellent transmitter in the optical range while a very bad transmitter in the infrared range. But for your idea to be true, this must hold for all materials. There are materials that are transparent in the IR. 4 hours ago, martillo said: I must answer now why this happens with photons? It is because the glass' atoms cannot absorb the relative high energetic photons and they pass through the lattice or are reflected while they can well absorb and emit quite all the less energetic photons of heat. Now, how is that? Well, the absorption of photons by atoms depend on the available levels of energy in the atoms according to their particular characteristic energy spectrum with their characteristic levels of energies. I must also explain how the velocity of light in the glass is anyway less than in vacuum and this is because the light's photons pass through the lattice but have another interaction with atoms rather than absorption and emission. The interaction would be elastic collisions in which the photons are slowed down. The photons have an EM field which interacts with the EM field of the atoms with an EM force between them. The force act braking a bit the photons at first while accelerating them again after but a delay of time is involved in this process. But the time delay has to be the same for all photons of the same energy, because the photons are identical, regardless of how they are generated Link to comment Share on other sites More sharing options...
martillo Posted April 1 Author Share Posted April 1 (edited) 1 hour ago, swansont said: But for your idea to be true, this must hold for all materials. There are materials that are transparent in the IR. Each material has its own spectrum of absorption and emission depending on the spectrum of its atoms. Also the molecular structure and the lattice plays a role, f.i. the number of atoms the photons encounter in their way through depends on the density of the material and so could be more or less time delays and so different velocities and different intensities (quantity of photons) passing through the material. 1 hour ago, swansont said: But the time delay has to be the same for all photons of the same energy, because the photons are identical, regardless of how they are generated And it is that way. The time delay depend on the frequency and this the basis for the functioning of prismas presenting different angles of refraction of the incident light for different colors (frequency). This is the Newton's experiment with light. Edited April 1 by martillo Link to comment Share on other sites More sharing options...
swansont Posted April 1 Share Posted April 1 8 minutes ago, martillo said: Each material has its own spectrum of absorption and emission depending on the spectrum of its atoms. Also the molecular structure and the lattice plays a role, f.i. the number of atoms the photons encounter in their way through depends on the density of the material and so could be more or less time delays and so different velocities and different intensities (quantity of photons) passing through the material. And it is that way. The time delay depend on the frequency and this the basis for the functioning of prismas presenting different angles of refraction of the incident light for different colors (frequency). The dispersion effect in a prism is quite small. So basically you are predicting that any material that is transparent in the IR will have almost instantaneous heat flow through it. A material with an IR index of refraction of 1.5, IR light will travel at 2/3 c so heat will conduct at 2/3 c. Is that correct? Link to comment Share on other sites More sharing options...
martillo Posted April 1 Author Share Posted April 1 (edited) 24 minutes ago, swansont said: The dispersion effect in a prism is quite small. So basically you are predicting that any material that is transparent in the IR will have almost instantaneous heat flow through it. A material with an IR index of refraction of 1.5, IR light will travel at 2/3 c so heat will conduct at 2/3 c. Is that correct? According to the Snell refraction formula it is so, we have: v1/v2 = n2/n1 I don't know if in some case the formula is not verified, may be other effects could be present. F.i. note that refraction happens at certain angles of incidence only. Edited April 1 by martillo Link to comment Share on other sites More sharing options...
swansont Posted April 1 Share Posted April 1 56 minutes ago, martillo said: According to the Snell refraction formula it is so, we have: v1/v2 = n2/n1 I don't know if in some case the formula is not verified, may be other effects could be present. F.i. note that refraction happens at certain angles of incidence only. So you concur. Is it actually the case that e.g. polyethylene, which is transparent in the IR, with an index of about 1.5, conducts heat at this speed? Or various glasses that works in that range, such as “Zinc Selenide (ZnSe), Zinc Sulfide (ZnS), Calcium Fluoride (CaF2) and Magnesium Fluoride (MgF2). All of these operate from the visible spectrum up to 8-10μ” https://www.emf-corp.com/optical-materials/optical-material-infrared-optics/ According the chart, the two zinc options operate out to 14 microns. Link to comment Share on other sites More sharing options...
martillo Posted April 1 Author Share Posted April 1 (edited) I will answer now that question that surged some time ago: On 3/30/2024 at 9:05 PM, swansont said: This wouldn’t seem to explain the T vs T^4 difference we observe for conduction vs radiation. The answer is that the question is wrongly formulated. Stefan-Boltzmann law states that the energy density of the radiation of a system at a given temperature T is proportional to T4 while the heat energy transfer by conductivity is linearly proportional to T. The point is that they are different things and cannot be compared. They have different units. The question must be: If the radiance depends on T4 then why the energy transfer depends linearly on T only? I don't have to develop any formula for that, Boltzmann already solved the problem long time ago with his concept of Entropy. Boltzmann found the "fundamental hypothesis in statistical mechanics": This formula relates the micro states of the system (micro regions in the system) with the macro state of the system. It works under certain conditions like that all the micro states are equal and equally probable in the system. All the formulas of Thermodynamics can be derived from this formula. The Second Law of Thermodynamics is: and the Fundamental Thermodynamics Relation is: The other equations of Thermodynamics are derived form these equations and are all linear with T. The energy transfer we are thinking on is governed by these laws giving the linearity between dQ and T. Edited April 1 by martillo Link to comment Share on other sites More sharing options...
swansont Posted April 1 Share Posted April 1 16 minutes ago, martillo said: I will answer now that question that surged some time ago: The answer is that the question is wrongly formulated. Stefan-Boltzmann law states that the energy density of the radiation of a system at a given temperature T is proportional to T4 while the heat energy transfer by conductivity is linearly proportional to T. Nope S-B gives radiated power, which is a rate of energy transfer, i.e. energy per unit time. Newton’s law gives the heat transfer rate, which is also energy per unit time 16 minutes ago, martillo said: The point is that they are different things and cannot be compared. They have different units. They are different, since they apply to different mechanisms, but they have the same units (when appropriately stated; you might have to do a little algebraic rearrangement, depending on how the formula is written) Link to comment Share on other sites More sharing options...
martillo Posted April 1 Author Share Posted April 1 (edited) 18 minutes ago, swansont said: S-B gives radiated power, which is a rate of energy transfer, i.e. energy per unit time. Nope. From Wikipedia: The Stefan–Boltzmann law, also known as Stefan's law, describes the intensity of the thermal radiation emitted by matter in terms of that matter's temperature. It is named for Josef Stefan, who empirically derived the relationship, and Ludwig Boltzmann who derived the law theoretically. For an ideal absorber/emitter or black body, the Stefan–Boltzmann law states that the total energy radiated per unit surface area per unit time (also known as the radiant exitance) is directly proportional to the fourth power of the black body's temperature, T: The constant of proportionality, �, is called the Stefan–Boltzmann constant. It has a value σ = 5.670374419...×10−8 W⋅m−2⋅K−4. In the general case, the Stefan–Boltzmann law for radiant exitance takes the form: where � is the emissivity of the matter doing the emitting. The emissivity is generally between zero and one, although some exotic materials may have an emissivity greater than one. An emissivity of one corresponds to a black body. and it is specifically explained: The radiant exitance (previously called radiant emittance), �, has dimensions of energy flux (energy per unit time per unit area), and the SI units of measure are joules per second per square metre (J⋅s−1⋅m−2), Edited April 1 by martillo Link to comment Share on other sites More sharing options...
swansont Posted April 1 Share Posted April 1 30 minutes ago, martillo said: The radiant exitance (previously called radiant emittance), �, has dimensions of energy flux (energy per unit time per unit area), and the SI units of measure are joules per second per square metre (J⋅s−1⋅m−2), energy per unit time per unit area. So you multiply both sides by the area, and you have an equation for power: energy per unit time, just as I said. I said you might have to do a little algebraic manipulation, but I didn’t think that would be a barrier. Link to comment Share on other sites More sharing options...
martillo Posted April 1 Author Share Posted April 1 (edited) 23 minutes ago, swansont said: energy per unit time per unit area. So you multiply both sides by the area, and you have an equation for power: energy per unit time, just as I said. Not so easy, you must have a specific area to consider what depends on each case in particular to solve. S-B law is general and applies to any case. More important than that, do you really pretend to solve the energy rates with Newton Second Law of force? I don't think you could solve the thing that way. You know, Boltzmann had to develop the concept of entropy and find his laws of statistical mechanics to solve the thing. You can't bypass over all that subject that way. Edited April 1 by martillo Link to comment Share on other sites More sharing options...
martillo Posted April 1 Author Share Posted April 1 (edited) 5 hours ago, swansont said: So you concur. Is it actually the case that e.g. polyethylene, which is transparent in the IR, with an index of about 1.5, conducts heat at this speed? Or various glasses that works in that range, such as “Zinc Selenide (ZnSe), Zinc Sulfide (ZnS), Calcium Fluoride (CaF2) and Magnesium Fluoride (MgF2). All of these operate from the visible spectrum up to 8-10μ” https://www.emf-corp.com/optical-materials/optical-material-infrared-optics/ According the chart, the two zinc options operate out to 14 microns. As I said for the case of the prism, it is possible for the velocity of heat to be different at different frequencies. The velocity would be dependent on the frequency of the photons. This is not a problem in my approach since it only implies that the time delay in the absorption with posterior emission of the photons vary with the frequency of the photons. Edited April 1 by martillo Link to comment Share on other sites More sharing options...
martillo Posted April 1 Author Share Posted April 1 (edited) 14 hours ago, Mordred said: You don't particularly have to worry about it you can simply use the field treatments without specifying the mediator by using operator states. All mediator bosons are massless all you need from there is sufficient energy/momentum for mediation. However you don't need to learn the relations via the quantum world most statistical mechanics textbooks barely touch on the quantum regime. However at the quantum level then even at short range you will need a mediator. Let's put it this way you can mathematically describe everything you need using nothing more than classical mechanics. When you get to the quantum regime the number density of your mediator can then be determined by whatever the energy level of field or state your measuring by simply applying the Bose-Einstein statistics for bosons.So it's something you can worry about later provided your not breaking any laws such as the speed limit. I appreciate the good lectures you provided on the subatomic particles. You know, the world of the subatomic particles under the basic level of the electrons, protons, neutrons, neutrinos and photons is something weird for me. hope I would not need to enter too much in the area but who knows if I would need to make an incursion at some time. I think one of the main reasons to introduce the mediators in the model of the elementary particles is the finite velocity c of propagation of the EM fields. A far as I know the experimental evidence for it was Hertz experiment about standing waves which allowed an experimental evidence for the existence of the EM waves. I want to ask you if there are other experiments demonstrating the finite velocity c of propagation of the EM fields or if it is the only one. I'm managing the possibility for them to actually be instantaneous. I have a new interpretation for the Hertz experiment in which what the antennas emit and absorb would be photons. This way what Hertz has detected would be intensity of photons and not waves. The same patterns of waves can be obtained this way, particularly the pattern of standing waves. Then, photons would do travel at the finite velocity c but the EM fields would be instantaneous. Edited April 1 by martillo Link to comment Share on other sites More sharing options...
Mordred Posted April 2 Share Posted April 2 (edited) No worries I'm always glad to help anyone trying to better understand physics. As far as experiments for what equates to c the speed limit and there have been a huge number of tests and the error margin resulting from all those tests the constancy of c. This falls under Lorentz invariance so subsequently any tests for Lorentz invariance involves the constancy of c. This lists some of the more modern tests https://arxiv.org/pdf/2111.02029.pdf however its nowhere near a complete listing of test variations. the error margin is incredibly low something on the order of 1 part in 10^(18} for error margin but it would take some digging to find the current error margin Edited April 2 by Mordred Link to comment Share on other sites More sharing options...
martillo Posted April 2 Author Share Posted April 2 (edited) 3 hours ago, Mordred said: No worries I'm always glad to help anyone trying to better understand physics. As far as experiments for what equates to c the speed limit and there have been a huge number of tests and the error margin resulting from all those tests the constancy of c. This falls under Lorentz invariance so subsequently any tests for Lorentz invariance involves the constancy of c. This lists some of the more modern tests https://arxiv.org/pdf/2111.02029.pdf however its nowhere near a complete listing of test variations. the error margin is incredibly low something on the order of 1 part in 10^(18} for error margin but it would take some digging to find the current error margin But I'm not talking about the Lorentz invariance and the constancy of the speed of light. I'm thinking if separately the Electric Field alone and the Magnetic Field alone propagate at velocity c or not. Which experiments can prove that? Edited April 2 by martillo Link to comment Share on other sites More sharing options...
Mordred Posted April 2 Share Posted April 2 Lol the EM field is used as one of the more common Lorentz invariant and constancy of c tests. We've even shown in tests that the EM fields do produce photons via the Aharohm Bohm effect where the particle production in the math was often considered just considered a math sleight of hand Link to comment Share on other sites More sharing options...
martillo Posted April 2 Author Share Posted April 2 (edited) 4 hours ago, Mordred said: Lol the EM field is used as one of the more common Lorentz invariant and constancy of c tests. We've even shown in tests that the EM fields do produce photons via the Aharohm Bohm effect where the particle production in the math was often considered just considered a math sleight of hand Is that I'm used to think with the classical physics approach. I'm reviewing concepts. In Particle Physics the EM force is carried by virtual photons. I mean, the Lorentz force is carried by virtual photons. The concept of a delayed action at a distance of a force is replaced by the concept of a "virtual particle" force carrier called a "virtual photon". The same but differently. To find the first experimental evidences of this delayed action I must go back to the years of Lorentz and Maxwell. The concept appears first while associating "light" to an EM wave, not to a particle. The concept of "light" as composed by photons particles comes with Planck and Einstein. Edited April 2 by martillo Link to comment Share on other sites More sharing options...
martillo Posted April 2 Author Share Posted April 2 (edited) 1 hour ago, martillo said: In Particle Physics the EM force is carried by virtual photons. I mean, the Lorentz force is carried by virtual photons. The concept of a delayed action at a distance of a force is replaced by the concept of a "virtual particle" force carrier called a "virtual photon". The same but differently. To find the first experimental evidences of this delayed action I must go back to the years of Lorentz and Maxwell. The concept appears first while associating "light" to an EM wave, not to a particle. The concept of "light" as composed by photons particles comes with Planck and Einstein. I consider that once "light", "heat", any kind of EM radiation is recognized to be the radiation of a particle (flux of "photons") there is no need to consider a delayed action in the EM fields. I can think in the photons as an "electromagnetic particle" carrying an EM energy and an EM force. I can consider then instantaneous EM fields acting on travelling "electromagnetic particles". This means to abandon the "EM wave" theory of light and remain with an "EM particle" theory for light. This means to stay with the particles' model of protons, neutrons, electrons, neutrinos, photons but with instantaneous fields. The "force carriers" as "virtual photons" are not needed. This is the next step in my approach. With this approach the things get simplified a lot. Maxwell equations remain for the E and B fields, of course, but not the "EM waves" that simply would not exist. To be demonstrated, of course, what would fall in demonstrating that the E and B fields are instantaneous. Edited April 2 by martillo Link to comment Share on other sites More sharing options...
swansont Posted April 2 Share Posted April 2 22 hours ago, martillo said: As I said for the case of the prism, it is possible for the velocity of heat to be different at different frequencies. The velocity would be dependent on the frequency of the photons. This is not a problem in my approach since it only implies that the time delay in the absorption with posterior emission of the photons vary with the frequency of the photons. But in science and engineering we take the step of quantifying things. The spread in index of refraction(the dispersion) is small - perhaps a percent or two over the range of frequencies in question. Not the 10 orders of magnitude required for the heat propagation time to match the light propagation time. (which is essentially zero variation on this scale) On 4/1/2024 at 11:16 AM, martillo said: Not so easy, you must have a specific area to consider what depends on each case in particular to solve. S-B law is general and applies to any case. And any case will have an area. Or you can just do it on a “per unit area” basis. The point being that this is a simple mathematical manipulation that should be well within the capability of an electrical engineer. On 4/1/2024 at 11:16 AM, martillo said: More important than that, do you really pretend to solve the energy rates with Newton Second Law of force? I don't think you could solve the thing that way. You know, Boltzmann had to develop the concept of entropy and find his laws of statistical mechanics to solve the thing. You can't bypass over all that subject that way. No, I’m referring to Newton’s law of cooling, since we’re discussing that topic. The law that says Q depends on the temperature difference for conduction and convection https://en.m.wikipedia.org/wiki/Newton's_law_of_cooling Link to comment Share on other sites More sharing options...
martillo Posted April 2 Author Share Posted April 2 (edited) 1 hour ago, swansont said: But in science and engineering we take the step of quantifying things. The spread in index of refraction(the dispersion) is small - perhaps a percent or two over the range of frequencies in question. Not the 10 orders of magnitude required for the heat propagation time to match the light propagation time. (which is essentially zero variation on this scale) In the link you provided the range of wavelength is about 1 - 10 microns but there is no mention in a change in the velocity of propagation. The velocity would remain the same. Why do you consider it varies? We have λf approximately c. While both λ and f vary accordingly the velocity remains the same: approximately c. May be there could be a little variation, I don't know exactly but there's no "10 orders of magnitude required for the heat propagation time to match the light propagation time" as you say. Your calculation is wrong. 1 hour ago, swansont said: And any case will have an area. Or you can just do it on a “per unit area” basis. The point being that this is a simple mathematical manipulation that should be well within the capability of an electrical engineer. Well, Boltzmann solved it in "per unit area" basis and the Thermodynamics equations derived give a final linear variation of rate transfer in T. I stay with this. I already solved the thing this way. Now is that you want to solve the thing in a different manner, well, do it. You can't ask me to solve the things in your way. At the end we reach to the same result isn't it? 1 hour ago, swansont said: No, I’m referring to Newton’s law of cooling, since we’re discussing that topic. The law that says Q depends on the temperature difference for conduction and convection As I said, you do it in your way. I already did it in my way and there is agreement in a linear relation with T. The discussion in this topic has ended for me. Just to mention, Newton's law of cooling works for small variations in T only. Edited April 2 by martillo Link to comment Share on other sites More sharing options...
swansont Posted April 2 Share Posted April 2 1 hour ago, martillo said: In the link you provided the range of wavelength is about 1 - 10 microns but there is no mention in a change in the velocity of propagation. The velocity would remain the same. Why do you consider it varies? We have λf approximately c. While both λ and f vary accordingly the velocity remains the same: approximately c. May be there could be a little variation, I don't know exactly but there's no "10 orders of magnitude required for the heat propagation time to match the light propagation time" as you say. Your calculation is wrong. Well, then, you do the calculation. How long does it take for heat to propagate through a material? I’ve listed several that are transparent to IR near room temperature. Assume the index of refraction is 1.5 How long does it take light of the same wavelength to propagate through the same material? (I had picked 2 cm for the thickness to make the calculation easy) 1 hour ago, martillo said: Well, Boltzmann solved it in "per unit area" basis and the Thermodynamics equations derived give a final linear variation of rate transfer in T. I stay with this. I already solved the thing this way. Now is that you want to solve the thing in a different manner, well, do it. You can't ask me to solve the things in your way. At the end we reach to the same result isn't it? No, absolutely not. Your result is very different from mainstream physics. 1 hour ago, martillo said: As I said, you do it in your way. I already did it in my way and there is agreement in a linear relation with T. The discussion in this topic has ended for me. Just to mention, Newton's law of cooling works for small variations in T only. I messed it, then. How does the Stefan-Boltzmann law give you a linear relation with T? Link to comment Share on other sites More sharing options...
martillo Posted April 2 Author Share Posted April 2 (edited) 2 hours ago, swansont said: Well, then, you do the calculation. How long does it take for heat to propagate through a material? I’ve listed several that are transparent to IR near room temperature. Assume the index of refraction is 1.5 How long does it take light of the same wavelength to propagate through the same material? (I had picked 2 cm for the thickness to make the calculation easy) The velocity of propagation for an index of refraction is 1.5 is v = c/(1.5)c = 2c/3. The time is t = d/v where d is the thickness so for thickness = 2 cm = .02 m is t = .02/(2c/3) = .03/c = .03x10-8 sec. While the index of refraction remains the same this calculation remains the same so same time of propagation. λf = v is always verified so for each λ you just have f = v/λ just varying accordingly. For same index of refraction 1.5 it is same v = 2c/3 and so same delay of propagation t = .03x10-8 sec. 2 hours ago, swansont said: No, absolutely not. Your result is very different from mainstream physics. No, the result is the same. What is different is just that I sustain that there are photons always present in the system (there would be a "photons gas" as is called). 2 hours ago, swansont said: I messed it, then. How does the Stefan-Boltzmann law give you a linear relation with T? No it is not the S-F law that gives me a linear relation with T. I have already told you that "energy transfer" and "energy transfer rate" are different things with different units. The S-B law gives the "energy transfer rate" of the radiation through some area at some place proportional to T4 while the Thermodynamics laws give the total "energy transfer" between two bodies proportional to T. How many times I must repeat this? Edited April 2 by martillo Link to comment Share on other sites More sharing options...
swansont Posted April 2 Share Posted April 2 42 minutes ago, martillo said: The velocity of propagation for an index of refraction is 1.5 is v = c/(1.5)c = 2c/3. The time is t = d/v where d is the thickness so for thickness = 2 cm = .02 m is t = .02/(2c/3) = .03/c = .03x10-8 sec. c is 3 x 10^8, so this would be 0.01 x 10^-8 sec 42 minutes ago, martillo said: While the index of refraction remains the same this calculation remains the same so same time of propagation. λf = v is always verified so for each λ you just have f = v/λ just varying accordingly. For same index of refraction 1.5 it is same v = 2c/3 and so same delay of propagation t = .03x10-8 sec. So that’s how you’re saying fast heat is conducted through the material? In a tenth of a nanosecond? Through 2 cm of glass? 42 minutes ago, martillo said: No it is not the S-F law that gives me a linear relation with T. I have already told you that "energy transfer" and "energy transfer rate" are different things with different units. The S-B law gives the "energy transfer rate" of the radiation through some area at some place proportional to T4 while the Thermodynamics laws give the total "energy transfer" between two bodies proportional to T. q in Newton’s law is an energy transfer rate, per unit area. They describe the same thing From the link I provided earlier “q is the heat flux transferred out of the body (SI unit: watt/m2)” (watts being joules/sec) which is exactly what the units are in the S-B equation, which you posted Link to comment Share on other sites More sharing options...
martillo Posted April 2 Author Share Posted April 2 3 minutes ago, swansont said: c is 3 x 10^8, so this would be 0.01 x 10^-8 sec No it is .03 x 10-8 sec. 5 minutes ago, swansont said: c is 3 x 10^8, so this would be 0.01 x 10^-8 sec So that’s how you’re saying fast heat is conducted through the material? In a tenth of a nanosecond? Through 2 cm of glass? No, I'm saying that in a material with refractive index of 1,5 heat takes .03 x 10-8 sec to pass through. The refractive index of glass for "light" is different from that for "heat". This strongly depends on frequency. 1 hour ago, swansont said: q in Newton’s law is an energy transfer rate, per unit area. They describe the same thing From the link I provided earlier “q is the heat flux transferred out of the body (SI unit: watt/m2)” (watts being joules/sec) which is exactly what the units are in the S-B equation, which you posted From Wikipedia in the link you provided: "Another situation that does not obey Newton's law is radiative heat transfer. Radiative cooling is better described by the Stefan–Boltzmann law in which the heat transfer rate varies as the difference in the 4th powers of the absolute temperatures of the object and of its environment." Link to comment Share on other sites More sharing options...
martillo Posted April 3 Author Share Posted April 3 1 hour ago, swansont said: So that’s how you’re saying fast heat is conducted through the material? In a tenth of a nanosecond? Through 2 cm of glass? I think the real problem we have here is in which kind of "heat" is considered. One case is the infrared range as described in the link of "infrared optics" you provided. It is considered yet an "optical case" even in your link. So I would not call it "heat" range and yet call it "light" range, not visible but yet a "light" range would be more appropriated. Other case is yes the "heat" in the range which we can feel with our fingers for instance which is not an optical case at all. In this case much lower frequencies are involved and the glasses quite not transmit any "heat" and that's why they are good thermal insulators. Link to comment Share on other sites More sharing options...
swansont Posted April 3 Share Posted April 3 12 hours ago, martillo said: No it is .03 x 10-8 sec. .03/3 x 10^8 ≠ .03 x 10^-8 12 hours ago, martillo said: No, I'm saying that in a material with refractive index of 1,5 heat takes .03 x 10-8 sec to pass through. The refractive index of glass for "light" is different from that for "heat". This strongly depends on frequency. The frequencies are the same. The wavelength are the same. 12 hours ago, martillo said: From Wikipedia in the link you provided: "Another situation that does not obey Newton's law is radiative heat transfer. Radiative cooling is better described by the Stefan–Boltzmann law in which the heat transfer rate varies as the difference in the 4th powers of the absolute temperatures of the object and of its environment." Yes. Do you understand this is what we have been telling you and you keep denying? The radiative heat transfer depends on T^4 11 hours ago, martillo said: I think the real problem we have here is in which kind of "heat" is considered. One case is the infrared range as described in the link of "infrared optics" you provided. It is considered yet an "optical case" even in your link. So I would not call it "heat" range and yet call it "light" range, not visible but yet a "light" range would be more appropriated. Other case is yes the "heat" in the range which we can feel with our fingers for instance which is not an optical case at all. In this case much lower frequencies are involved and the glasses quite not transmit any "heat" and that's why they are good thermal insulators. Thermal radiation near 300K is centered at about 10 microns. The wavelengths are the same. Link to comment Share on other sites More sharing options...
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