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2nd law of Thermodynamics and black body brightness function

  1. Oct 3, 2008 #1
    Hi,

    I feel extremely stupid for having to have this explained to me, but I am really confused by it. I recently encounter a proof of the fact that the brightness (energy/(time x area x solid angle x frequency)) of a black body can only depend on its temperature, not the properties of the enclosure itself. Here is the argument:

    "Two black body cavities at the same temperature are brought near each other. They each have a tiny hole so that they can exchange radiation with each other. A filter which only allows radiation of frequency [tex]\nu[/tex] to pass between them is placed between the holes. If the brightness function of the two is not the same, energy will spontaneously flow from one cavity to another, which violates the second law of thermodynamics. Thus, the brightness functions of the two black bodies must be the same at a given temperature."

    The part I don't understand is why the spontaneous flow of energy between two systems in thermal equilibrium violates the second law of thermodynamics. If I casually think to myself, "the second law gives the time arrow of energy, and energy only flows from hot to cold, then heat flow between bodies at equal temperatures must be a violation," then I get what is meant. But, if I write,

    [tex]\frac{\delta Q_{total}}{T}\geq 0[/tex] and [tex]\delta Q_{body1}[/tex]=-[tex]\delta Q_{body2}[/tex], then the second law reads [tex]\frac{\delta Q_{body1}+\delta Q_{body2}}{T}\geq 0[/tex] so [tex]\delta Q_{body1} \times 0 \geq 0[/tex]

    I don't see the contradiction, or how this puts a constraint on [tex]\delta Q_{body1}[/tex]

    pastro
     
  2. jcsd
  3. Oct 3, 2008 #2

    Andrew Mason

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    If the brightness function is not the same then one body will have more energy at frequency [itex]\nu[/itex] than the other. Therefore, one will lose more energy to the other body than it gains from that body. This means that one body will get hotter and one body will get colder. This violates the second law (Clausius statement).

    AM
     
  4. Oct 5, 2008 #3

    Andy Resnick

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    As AM points out, the argument works because the prohibition is on *net* flow. Spontaneous non-zero fluctuations in energy flow are allowable, but in thermodynamics must average to zero at equilibrium.
     
  5. Oct 29, 2008 #4

    atyy

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    The confusing thing to me this seems to imply that a greybody cannot be in thermal equilibrium with a blackbody.

    I looked up an article by Duley (Blackbody radiation in small cavities, Am J Phys 1972) which says that a perfect blackbody must be a large cavity, and that equilibrium cannot be established between a large and a small cavity at the same "temperature". The cavities' energy densities are different for large wavelengths. If the cavities are connected by a small hole, only small wavelengths can pass through the hole, and in this particular case they can be in equilibrium. If this is correct, then in addition to the second law, isn't the zeroth law about the existence of "temperature" also needed for the argument?

    I found some notes by Brown (http://www.strw.leidenuniv.nl/~brown/college_sterren/BlackbodyThermodynamics.pdf) which make more sense to me by filling in gaps in the argument quoted in the OP.
     
    Last edited: Oct 29, 2008
  6. Oct 29, 2008 #5

    atyy

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    Garcia-Garcia, Finite size corrections to the blackbody radiation laws, http://arxiv.org/abs/0709.1287

    Looking at Fig. 1 in the above paper, it seems the argument is not going to work unless we cannot filter with infinite precision?
     
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