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B Photon Box

  1. Aug 9, 2018 #1
    If I have a box evacuated of air with 5 of the sides mirors and one side a heat conductor. will the photon gas inside have photons that get absorbed by the heat conductor and re-emitted when the photons strike the heat conductor
     
  2. jcsd
  3. Aug 9, 2018 #2

    PeterDonis

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    Why do you think there will be a photon gas inside the box?

    Also, what is a "heat conductor"?
     
  4. Aug 10, 2018 #3
    If the temprature inside the box is the same as the temprature outside the box there has to be a photon gas. Look up photon gas on wikipedea. and a heat conductor is a material with a hight thermal conductivity coeficiant the oposite of a insulator like a blanket
     
  5. Aug 10, 2018 #4

    vanhees71

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    Of course if you have a box with walls at non-zero temperature, you'll have a "photon gas" inside. This must be so, because that's what comes out analyzing the equilibrium state of the system consisting of the box and the electromagnetic field. The result is the Planck radiation law for the energy spectrum of photons (which is the right distribution to consider in this case, because it can be defined in a Lorentz-covariant way).

    The equilibrium comes about by continuous absorption and emission of photons by/from the walls. The rate of emission and absorption is the same, i.e., on average the mean photon-number density (which is for a given reference frame equivalent to energy density of the em. field) stays constant. Since photon number is not conserved the only possible equilibrium state is the canonical ensemble, and that's why for an ideal cavity the energy spectrum of the photons is a universal function of temperature only. That's why Planck was so eager to find the solution to this most puzzling problem of physics in his time (he worked on the problem for more than 10 years, before he found the solution in 1900 in terms of discovering an entire new theory, which later lead to the development of quantum theory in 1925).

    A "heat conductor" is any material that admits the exchange of (thermal) energy with the environment. Of course, to get a good thermal equilibrium you have to make all the walls to be at the same temperature and thus the photons in the equilibrium state at this given temperature. If you heat up one wall more than the others, it's not an equilibrium situation anymore, and you have effectively an energy transport from the hotter to the cooler walls, i.e., the system tries to get into thermal equilibrium by exchanging thermal energy, and the flow is mostly from the hotter to the colder walls (2nd Law of thermodynamics). If the box is evacuated the main mechanism is indeed radiation, i.e., on avarage the photon-emission rate at the hotter wall is larger than the emission rate at the colder one. If the conducting wall is coupled to a heat bath through this energy transport through the radiation field also the non-conducting walls get heated up to the same temperature, until equilibrium is reached again.
     
  6. Aug 10, 2018 #5
    I beleve you answered my question. I will just like to make sure. If a photon strikes the conductor surly it must be absorbed, but then the conductor will heat up, so then surely it will be instantly re-radiated again. Conferm for me please
     
  7. Aug 10, 2018 #6

    vanhees71

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    If you have an ideal conductor the photons are reflected. For the more realistic case of finite electric conductivity you have dissipation and thus photons can get absorbed. The corresponding energy transers into heat of the electron gas (for usual metals). The electrons of course also emit photons again because due to their random thermal motion they are accelerated.

    I think a very good way to understand black-body radiation intuitively is to look at Einstein's derivation of the Planck spectrum of 1916/17. For a first explanation, see the Wikipedia article

    https://en.wikipedia.org/wiki/Einstein_coefficients

    I think this work of Einstein's marks the true discovery for the necessity to quantize the electromagnetic field and thus the necessity of photons in the modern sense due to the necessity of introducing spontaneous emission. Despite claims still around in the modern textbook literature, to understand the photoeffect as well as the Compton effect at leading order perturbation theory of QED you can as well only quantize the charged particles (electrons) and deal with classical electromagnetic fields. This is different with anything involving spontaneous emission, which cannot be described in this semiclassical approximation.
     
  8. Aug 10, 2018 #7
    But then if you have a red hot laser in an Ice cold frige shining onto a room temprature conductor, It will reflect. That does not make sense
     
  9. Aug 10, 2018 #8

    vanhees71

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    A good conductor indeed reflects light. You find the corresponding treatment in any good textbook on (classical!) electrodynamics or optics. It's formally not so different for quantum optics by the way!
     
    Last edited: Aug 10, 2018
  10. Aug 10, 2018 #9
    So hyperleticaly speeking if all the photons were paralel and you had a lens that focused some of them, will the focused rays absorb and re-radiate
     
  11. Aug 12, 2018 at 4:39 AM #10
    These days you get laser material that can absorb IR and emit a beam. My question is will the this laser behave the same as a conductor or will it convert some of the photon gas if it were in the box.Please can someone answer this question
     
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