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Rayleigh-Jeans law derivation

  1. Feb 25, 2015 #1
    My quantum course starts with the Rayleigh-Jeans law derivation. Whilst I understand most of this, I'm not sure why the average energy of each mode is equal to what they say it is. For instance here is one explanation from hyperphysics:
    Assigning energy to the electromagnetic standing waves in a cavity draws on the principle of equipartition of energy. Each standing wave mode will have average energy kT where k is Boltzmann's constant and T the temperature in Kelvins.

    Why must each node have the same average energy? I understand why the number of nodes varies with wavelength and volume, just not why each has the same constant average energy. For instance my textbook says use the average energy from a Boltzmann distribution, which is kT, however why can you do so? I don't see the connection. I get what a Boltzmann distribution is, however why does each particle emit each node to an equal energy, as I seem to think using kT implies.
     
    Last edited: Feb 25, 2015
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  3. Feb 25, 2015 #2

    DrClaude

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    Staff: Mentor

    Do not confuse "mode" and "node." A mode is simply one allowed frequency in the cavity, i.e., a frequency where the corresponding wave can satisfy the proper boundary conditions.

    The equipartition theorem states that each quadratic degree of freedom has an average energy of kT/2. This theorem was formulated in the middle of the 19th century, and we can demonstrate it now with the tools of statistical physics. It was natural for Rayleigh and Jeans to use it to try to describe the blackbody spectrum.

    Note that a harmonic oscillator has two quadratic degrees of freedom. Hence the average energy of a cavity mode is kT.
     
  4. Feb 25, 2015 #3
    Sorry I really should have said mode, not node. Strange as I started using mode correctly. Anyway I've looked at the equipartition theorem in the past. As far as I'm aware every degree of freedom absorbs on average kT/2 worth of energy. However what makes each possible mode obtain such an energy.

    I don't understand what is natural about extending the amount of energy in a degree of freedom to that of energy in each mode. What causes this relation?

    Am I right in thinking that they considered each possible mode a method to which energy can be extracted from the thermal system. Hence each mode contributes a degree of freedom? If so why did they think that was true? Surely not every mode has to extract energy from the system?
     
  5. Feb 25, 2015 #4

    DrClaude

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    It is related to the idea that every microstate of a system (particular distribution of a fixed amount of energy among the different degrees of freedom) is equally probable. This is a postulate of statistical physics, and so far agrees with all experimental results, so is taken as valid.

    If I'm not mistaken, it originally comes from observations in gases, that lead physicist to postulate that the thermal energy of molecules goes not only into translational motion, but also into rotation and vibration, and in an equal (hence "equipartition") way. Again, we have so far no reason to doubt that this is correct, especially since quantum mechanics allowed to solve some difficulties at low temperature. Using statistical physics, you can also show that the equipartition theorem actually derives from the postulate of equiprobability of microstates.

    The idea is that each mode is a degree of freedom (or actually two). Hence, by virtue of equiprobability/equipartition, when energy is put into the cavity, each mode will on average get an equal bit of that energy. Now you turn the problem around: considering that the cavity is at a certain temperature, how much energy must have been put into it? Pre-Planck answer: it has an infinite number of modes, therefore it contains an infinite amount of energy! The problem was solved by Planck, who assumed that modes get excited by discrete amounts, depending on their frequency, hence some modes will be "frozen out" at any temperature, and the energy contained in the cavity is actually finite.
     
  6. Feb 25, 2015 #5
    Thanks, that's a perfect answer. No textbook went further than the fact each mode absorbed energy independently. Never mentioned that each could act as a degree of freedom. (Well two for each polarised state).
     
  7. Feb 26, 2015 #6

    DrClaude

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    To be clear: it is two degrees of freedom per mode, not counting any degeneracy due to polarization. In the harmonic oscillator, you have two quadratic terms, one for kinetic energy and one for potential energy, so ##\bar{E} = 2 \times k T/2 = kT##. When considering polarization, you either add a degeneracy factor of 2 to each mode when you count them, or consider the different polarizations to be independent modes; the result is of course the same.
     
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