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I Trouble understanding the idea of a cavity radiator being a Black Body

  1. Dec 28, 2017 #1

    kal

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    I have been trying to understand the role of a cavity as a black body radiator in the derivation of planks black body radiation law but it has left me with 5 main questions:

    1. If an object is a perfect absorber it must also be a perfect emitter, meaning that (allowing for a cavity not being a perfect representation) most of the light that enters the cavity should come back out through the hole. But given the size of the hole compared to the cavity wont most of the radiation escape through the walls and barely any through the hole.

    2. if the cavity is being kept at a constant temperature by an external heat source wont this heat permeate the walls and add to the radiation inside the cavity so that it isn't all energy from the light entering the cavity?

    3. Apparently the electric field of the wave must be 0 at the walls of the cavity but I don't understand the reasoning.

    4. how does the boundary condition that the electric field must be 0 at the walls lead to the formation of standing waves?

    5. When deriving planks law for blackbody radiation it calculates the density of modes inside the cavity as a function of frequency, assuming that all modes at that frequency are filled. Why is it the case that all the modes must be filled and is that still the case for cavities at a low temperature?

    Any help would be much appreciated :)

    I understand why the a ray of light that enters the cavity has little chance of getting out and therefore must be absorbed, making the hole and cavity a near perfect absorber. I also understand from a thermodynamics point of view that if in thermal equilibrium it must be a perfect emitter.

    My first question is what is the physical reason for why the cavity and hole is a perfect emitter? (e.g for being a perfect absorber it was that the radiation cant escape)

    My second is what does being a perfect emitter even mean because a perfect absorber absorbs all radiation that hits it but all objects will eventually radiate away all their heat.

    Thanks in advance :)
     
    Last edited by a moderator: Dec 30, 2017
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  3. Dec 28, 2017 #2

    Charles Link

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    In the derivation of the Planck function, (5) is incorrect. The average occupation number is given by the Bose factor ## \bar{n}_s=\frac{1}{e^{E_s/kT}-1} ##. In addition, I don't think (3) is necessary in the derivation. The derivation I like best is the one in F.Reif, Statistical and Thermal Physics. To count the cavity modes, you simply assume periodic boundary conditions. ## \\ ## See also post 206 of this rather lengthy thread: https://www.physicsforums.com/threads/emission-spectra-of-different-materials.913274/page-11 Also see the "link" in post 208 of this same thread for counting the modes: To repeat that "link" here: https://www.physicsforums.com/threads/boltzmann-vs-maxwell-distribution.918232/ See in particular, posts 2 and 4 to count the modes.
     
    Last edited: Dec 28, 2017
  4. Dec 28, 2017 #3

    kal

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    My undersanding was that ##\frac{8\pi \nu^2}{c^3}## was the possible number of states of a certain frequency and that ##h\nu## multiplied by the bose-einstien factor was the quantum equivalent to KT the average energy since that is what it approximates too. Is this wrong?
     
  5. Dec 28, 2017 #4

    Charles Link

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    I believe the result that "this equates to kT " is a high temperature result of the Bose factor for ## kT>>E_s ##. The Bose factor is correct regardless of the energy of the mode.
     
  6. Dec 28, 2017 #5

    kal

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    aha okay thanks a bunch, those links were also very helpful :)
     
  7. Dec 28, 2017 #6

    kal

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    One more thing, ## \frac{1}{e^{E_s/kT}-1} ## asymptotes with the x axis, so if this is the average occupancy of a state wont all states be occupied since the function never equals 0
     
  8. Dec 28, 2017 #7

    Charles Link

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    For low temperatures and high energies, the denominator gets quite large, making ## \bar{n}_s ## very small. ## \bar{n}_s ## is the average occupancy number. That number may be .000001 or less for a mode where ## E_s>> kT ##. It never goes precisely to zero, but can be very near zero.
     
  9. Dec 29, 2017 #8

    Charles Link

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    I think the reason has to do with the fact that when the radiation is confined to an enclosure, the occupancy of the modes of the electromagnetic waves is always at their thermal equilibrium value of ## \bar{n}_s=\frac{1}{e^{E_s/kT}-1} ##. e.g. A gas in free space at that temperature that occupies the same volume as the enclosure will in general have a very low emissivity, and the thermal motion of the particles, coupled to the E-M waves (E-M modes), simply isn't able to supply the energy to keep the EM states occupied, unless the gas is confined to an enclosure. As soon as an E-M mode gets some energy, in the form of photons, the photons travel at the speed of light away from the gas, and those photons are gone. ## \\ ## With some solids, and also with ionized gases, the emissivity is generally higher, and in some cases emissivities can be nearly 1.0. In the case of ionized gases and solids, there can be considerably more coupling between the thermal motion of the particles and the modes of the electromagnetic waves. (The occupancy number ## n_s ## value can be maintained closer to the equilibrium value with more coupling). ## \\ ## The enclosure with a small aperture is the best way though of getting an emissivity that is very nearly 1.0. ## \\ ## Additional item is if you make the aperture too large, you will no longer have emissivity equal to 1.0. This can also be seen from Kirchhoff's law, where some of the energy incident on the aperture will enter the enclosure and reflect back out, if the aperture is not very small. ## \\ ## Perhaps someone else has a better explanation, but this is how I understand it. ## \\ ## And to answer your second question, you do need to supply heat to your ideal blackbody source that is radiating energy away in order to keep it at the same temperature, so it can continue to radiate at that temperature.
     
    Last edited: Dec 29, 2017
  10. Dec 30, 2017 #9

    sophiecentaur

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    That is not the right conclusion. Any incident radiation will be absorbed because it will not be reflected but the inside will still have a temperature and radiate in its own right. The rate of emission will be equal to the rate of absorption when the inside and (effective) outside temperatures are equal. 'Obviously' there can be no overall build up of energy in the cavity beyond the equilibrium condition.
     
  11. Dec 30, 2017 #10

    vanhees71

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    The most clear derivation of the Planck law in the 21st century is to use thermal QED. Of course, as always in statistics of gases you have to use a finite quantization volume and then take the infinite-volume limit in the correct physical way. You find this derivation in my unfinished manuscript on photons:

    https://th.physik.uni-frankfurt.de/~hees/pf-faq/photon.pdf
     
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