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Black Body Radiation

  1. Mar 10, 2010 #1

    ibc

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    Hey

    In several books, the derivation of black body radiation is done by considering the energy density inside a cavity surrounded by walls in a certain temperature.
    The derivation is described as one of the first steps in quantum mechanics, where all considerations are purely classical but the limitation of possible energy values offered by plank.
    I understand the derivation, but can't understand why does this derivation represents the radiation emitted.
    It is mentioned that the energy density inside the cavity is proportional to the radiation emitted, yet I don't see the justification for that.
    Furthermore, this proportionality disturbs me for the following reasons:
    1. The only reason for the result being as we want it to be (after considering plank's revision to the possible energies) is because we are looking at standing waves, which are purely a consequence of the geometry of the cavity. So I don't see what does this have to do with the energy outside.
    2. for such a proportionality to occure, we need a proportionality constant, which should be with unites of [time]^-1 (since it relates energy density to radiation density), yet there is no phyisical quantity in the problem with units of time.

    Thanks
     
  2. jcsd
  3. Mar 12, 2010 #2

    ibc

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    Perhaps I'll try to rephrase it compactly:

    1) Why is the problem of a radiating black body equivalent to that of an inwards radiating cavity?
    2) How is the energy density inside this cavity proportional to the radiation intensity (units of energy density / time)?

    Thanks again
    ibc
     
  4. Mar 12, 2010 #3
    1) Since the radiation from a surface is independent of the geometry, we can choose any geometry we like to solve for. The rectangular cavity is the easiest geometry, but the solution we find must apply to any surface.

    2. Energy per unit volume divided by the speed of light equals power per square meter.
     
  5. Mar 12, 2010 #4

    SpectraCat

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    A black-body is a theoretical construct, which has a continuous spectral distribution that only depends on its temperature. Planck modeled this by thinking about a cavity as you have discussed ... what you seem to be missing is that in this model, the radiation from this blackbody is emitted from a pinhole in the cavity, not from the outer surface of the cavity itself.

    The http://en.wikipedia.org/wiki/Planck%27s_law" [Broken] on Planck's law gives an easy to follow derivation of this, including proper treatment of the units. If you still have questions after reading that, then I'll do my best to address them.
     
    Last edited by a moderator: May 4, 2017
  6. Mar 12, 2010 #5

    Matterwave

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    The perfect cavity is just there so that there aren't any effects of reflection, and to ensure that the cavity is in local thermodynamic equilibrium.
     
  7. Mar 13, 2010 #6
    As for the energy density being proportional to the radiation emitted...that seems like a reasonable assumption to me. If I take one thing with some energy A and another thing with the same energy A and bring them close together I should still have a total energy of 2A (unless there is some position dependent force or binding energy involved like with two charged spheres). What should I expect? 7A, e^A, sqrt A? When you pack in twice the radiation with energy A I would expect the new energy density to be 2A...three times the radiation would give me 3A so it seems like a good assertion.
     
  8. Mar 13, 2010 #7

    ibc

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    The main reason for which we eventually get our results, in this derivation, is the standing waves, which is a direct result of the geometry of the problem. I wouldn't expect any standing waves if I had a body radiating outwards into space.

    Also, what does it mean that the radiation is emitted from a pinhole? how do I get any information about the intensity of the radiation from this?


    Yes, the total energy would be 2A, but what does it have to do with the radiation intensity?
    Two bodies with energy A, can be radiating at each other with intesity of 7*sqrt(e^A), and still remain both with energy A.
     
    Last edited by a moderator: May 4, 2017
  9. Mar 13, 2010 #8

    Matterwave

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    Perhaps it is best to define what you mean by intensity. The intensity astronomers use is the total energy radiated per second per meter squared per steradian per Hertz. In this way, it is directly proportional to the energy.

    Others may simply call intensity the total energy per square meter (like is usual for sound waves). Or simply the total energy radiated.

    In any case, the intensity is always directly proportional to energy. It is never energy squared, or some weird function of energy.
     
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