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I am not a physicist, but I am trying to maintain the level of my intuition and basic knowledge in physics. Recently I tried to look at thermal radiation and black-body radiation.

After I looked through Wikipedia articles, my intuitive understanding was that thermal radiation and black-body radiation were the same thing in the sense that if there is no incoming radiation to reflect or absorb (lights are off), then any hot object will only emit its thermal radiation like if it was a black body. I thought that black-body radiation is associated with black bodies simply because in black bodies it is easier to observe and to study, because there is no need to separate reflected radiation from thermal radiation in the total outgoing radiation.

Is my understanding wrong? It contradicts the claims I've found in the Internet here and here that the intensity of thermal radiation depends on the object's surface colour. Unless the authors of these claims actually meant to say instead that the intensity at which radiation is absorbed depends on the surface colour, and if more radiation is absorbed, more will eventually be emitted, my intuition fails to see how this is possible.

Will cast iron kitchen stove emit less heat (at the same temperature) if it is painted white (or, rather, covered with thin reflective coating)? This is rather counter-intuitive for me.

If the intensity of thermal radiation really depends not only on the object's temperature but also on its surface colour, I will appreciate any references to where I could read more about the role of surface colour in thermal radiation, and how it works.

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  • $\begingroup$ Thermal radiation $\neq$ blackbody radiation. $\endgroup$ – Rob Jeffries Aug 26 '18 at 14:47
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Emissivity is equal to absorptivity at each wavelength. Intuitive argument (not a real proof): otherwise an object in an oven would heat up or cool down to a temperature different from that of the oven walls, and one could make a perpetuum mobile. Reference: Gustav Kirchhoff.

Metals are reflective in infrared and visible, and are not emitting much thermal radiation.

White paint though (and snow and textiles and most stuff) is quite black for wavelengths of around 10 micrometers and most of these things are approximately black body emitters at room temperature. And as Leslie's cube shows: a thin layer of isenglas varnish on a metal surface is enough to change the emissivity.

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  • $\begingroup$ Possibly instead of "white paint" I should have said "reflective coating," this is what I meant. $\endgroup$ – Alexey Aug 26 '18 at 11:48
  • $\begingroup$ Kirchhoff's law seems to be about thermodynamic equilibrium... $\endgroup$ – Alexey Aug 26 '18 at 11:52
  • $\begingroup$ @Alexey A material with a reflective coating (for example emergency blankets) will emit much less. But a mirror with a glass surface (or the traditional isenglas varnish of Leslie's cube) will emit almost like a black body. $\endgroup$ – Pieter Aug 26 '18 at 11:54
  • $\begingroup$ I meant truly reflective coating (at least in the relevant wavelengths). $\endgroup$ – Alexey Aug 26 '18 at 11:56
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    $\begingroup$ Reference to Leslie's cube experiment is also helpful, maybe it should be included in the answer... $\endgroup$ – Alexey Aug 26 '18 at 13:13
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Thermal radiation is not the same thing as blackbody radiation. Thermal radiation simply means you can characterise the emission from a system with a single temperature that also describes the distribution of particle speeds, occupation of energy levels etc.

It is a necessary, but insufficient, condition for blackbody radiation that the radiation is "thermal". Blackbody radiators must also be "thick" to their own radiation field, such that the radiation field is in complete thermodynamic equilibrium with the emitting matter. This can only be approximately true for any observed object and means that all light of all wavelengths should be absorbed by the object.

Blackbody emission is the most efficiently that radiative energy can be lost by a thermal emitter. Blackbody radiation has a colour, determined by Wien's displacement law, and hotter blackbodies have a radiation flux at their surfaces which increases as $T^4$.

If you paint an object with paint that only absorbs certain wavelengths, then by definition it is no longer a blackbody and will not radiate as efficiently as a blackbody at the same temperature.

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Emissivity equals absorptivity because the same electronic transitions are involved. When such waves reach the interface they encounter the same (Fresnel) reflectivity at the material surface whether they come from inside or outside.

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  • $\begingroup$ What about the case of inelastic scattering (say, Raman scattering)? The incident wave has a different frequency than the reflected one. $\endgroup$ – thermomagnetic condensed boson Aug 26 '18 at 12:51
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    $\begingroup$ @user54826 I suggest that you submit this as a new question after checking the existing ones to avoid a duplicate. $\endgroup$ – my2cts Aug 26 '18 at 13:02
  • $\begingroup$ Yes, the same transition matrix elements, but there may also be other routes for decay, and the argument is more complicated. See for example hyperphysics.phy-astr.gsu.edu/hbase/optmod/eincoef.html and en.wikipedia.org/wiki/Einstein_coefficients for transitions in isolated atoms and molecules. $\endgroup$ – Pieter Aug 26 '18 at 20:58

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