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Phonon contribution to specific heat of solids

  1. Dec 7, 2014 #1

    CAF123

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    The low temperature heat capacity was experimentally shown to go as T^3 for low temeperatures (and as T for even lower temperatures for metals). This result came out of Debye's model after he modified Einstein's model, relaxing some of his assumptions. I have a few questions on Debye's model that I wish to better understand.

    (Ref. Ashcroft and Mermin P.458)

    He assumes that the dispersion relation for all branches can be approximated by a linear disperson ##\omega = ck##. Now, this would make sense to me at low k values, however this approximation is used to obtain the specific heat capacity for any ##T##.

    There is also the Debye temperature introduced and a Debye cut-off wave vector introduced so that we obtain the correct number of modes. I didn't really understand why we needed this. I see that we approximate the 1st Brillioun zone by a sphere with a radius such that it contains the same number of allowed modes (with radius equal to magnitude of Debye cut off wave vector). This sphere extends outside the 1st B.Z.

    Thanks.
     
  2. jcsd
  3. Dec 8, 2014 #2

    DrDu

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    The Debye model gives you both the correct low and high temperature behaviour, because at low temperatures, only the modes with low k will be excited while at high T, each mode will contribute kT to heat capacity independent of it's frequency.
    At intermediate temperatures, there will be some differences. It is a nice model, as it can be solved analytically.
     
  4. Dec 8, 2014 #3

    CAF123

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    Hi DrDu,
    I see, I believe this answers my first question. It gives the low T behaviour correctly since then we are in the acoustic branch region and for small enough k, the phonon dispersion satisfies w=ck? Is that right? And then for high T, we get the result obtained through the purely classical result since this was independent of the frequency of the oscillator, so the approximation of w=ck gives us the correct low and high T behaviour. I think I understand that now, thanks very much.

    Do you have any comments with regard to my second question? What does the Debye wave vector correspond to and what is the physics behind the Debye sphere (spherical cow approximation to 1st B.Z) spilling out of the 1st B.Z?

    Thanks!
     
  5. Dec 8, 2014 #4

    DrDu

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    It's quite simple: Once you assume the linear dispersion relation for all k vectors, you have to make sure that you still get 3N degrees of freedom. So you need a cut-off.
     
  6. Dec 8, 2014 #5

    CAF123

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    Ok so since all phonon wavevectors outside the 1st Brillouin zone can be mapped into the 1st Brillouin zone, this zone contains all distinct wavevectors. The 1st Brillouin zone is approximated as a sphere of radius ##k_D## such that all distinct allowed k vectors are equal to or below ##k_D##. So the cut off is a way to make sure we don't overcount the number of modes? Is that right? (Each mode has with it an associated wave vector, why is that the there are N allowed wavevectors for N ions in the crystal?) In 3D, each wavevector per mode can be decomposed, thereby giving the 3N degrees of freedom you mentioned. Thanks.
     
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