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Thermal phonon conductivity and umklapp process

  1. Apr 12, 2007 #1

    neu

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    This is as I currently understand it:

    Phonon thermal conductivity is dependent on the phonon mean free path. To define phonon thermal conductivity a mechanism whereby phonons can be brought into thermal equalibruim is required.

    This is what I have a problem with:

    Phonon interactions with the crystal boundaries, imperfections, and if an anharmonic crystal, phonon-phonon scattering limit the mean free path but are not sufficient to bring phonons into thermal equalibrium.

    this is because these interactions are elastic and do not bring about an energy change in individual phonons, and so cannot bring about equalibrium.

    ie K1 + K2=K3

    so Umklapp process are proposed which do cause thermal resistivity:

    K1 + K2 = K3 + G

    is this because K3-K2-K1=G=delta K ?

    This means that all collisions outside 1st brillouin zone can be translated back into it.?? so what?

    and here im lost..
     
  2. jcsd
  3. Apr 13, 2007 #2

    neu

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    Ok, what if I re-state my question to simply:

    Why are Umklapp processes required?
     
  4. Apr 26, 2007 #3
    I'd like an answer to this too. My impression of Umklapp processes was that when two phonons' momenta add, it's sometimes outside of the first Brillouin zone, so you get a third phonon at some k that's less than what you'd expect if you didn't have to worry about Brillouin zones, but to balance things out, you need to add G. But I don't know for sure, I'm currently taking the class and I think I missed the one where we talked about Umklapp processes.
     
  5. Aug 8, 2007 #4
    I'll join the choirs of the masses asking that question.

    I failed an exam for not knowing the answer :cry:. I'd rather not see that repeat itself.
     
  6. Aug 8, 2007 #5

    olgranpappy

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    Homework Helper

    Because otherwise the total momentum will be the same before and after the collision. If this is the case and, say, there is a current flowing then there will always be a current flowing. Thus, there can be no relaxation towards equilibrium in which there can be no current flowing.
     
  7. Aug 8, 2007 #6

    olgranpappy

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    Consider
    [tex]
    \vec J = \sum_{i=1}^{N}\vec p_i
    [/tex]
    after a collision (involving, say, particles one two three and four)
    [tex]
    \vec J' = \left(\sum_{i=5}^N\vec p_i\right) + \vec p'_1+\vec p'_2+\vec p'_3+\vec p'_4
    [/tex]
    but by conservation of momentum (non-umklapp process)
    [tex]
    p_1+p_2+p_3+p_4=p'_1+p'_2+p'_3+p'_4
    [/tex]
    so after the collision the current is still
    [tex]
    \vec J' = \sum_{i=1}^N \vec p_i = \vec J
    [/tex]
     
  8. Aug 9, 2007 #7
    What could I be overlooking when I fail to see how non-conservation of momentum facilitates thermal resistance?
     
  9. Aug 9, 2007 #8

    olgranpappy

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    Homework Helper

    Well, if there is only always conservation of momentum, then the distribution of momenta can not change. Thus there is no way for a distribution which is not an equilibrium distribution to ever because an thermal distribution.

    For example, pretend that somehow I was able to start off with a system of electrons whose momentum distribution looked just like a fermi sphere (of radius p_F) except it was not centered at zero. Say it was centered at [tex]\vec p=(p_0,0,0)[/tex]. Then there would be a net current density in the negative x direction (since the electrons are negatively charges) given by an integral over a sphere centered at [tex](p_0,0,0)[/tex]
    [tex]
    \int \frac{d^3p}{2^3\pi^3}{\frac{e}{m}\vec p=
    \frac{e}{m}\hat x p_0\frac{N}{V}
    [/tex]

    Collisions among particles which conserve momentum can never change the above value since they can only change some two momentum that were occupied, say p_1 and p_2, to different momentum p'_1 and p'_2 consistent with p'_1+p'_2=p_1+p_2. But the individual momentum only appear as a sum (integral above) of all the momentum.

    So, it is clear that if all collisions conserve momentum then there will be no relaxation towards equilibrium and no resistence (the current never changes).

    What is not so clear is how exactly the Umklapp processes bring about this resistence, and for the explaination I had better just reference Abrikosov "Theory of Metals" Section 4.4.

    Cheers.
     
  10. Aug 10, 2007 #9
    Thank you. Your answer was helpfull to me, though I'll likely need to take a peak in the book you refer to to have all the pieces fall into place.
     
  11. Aug 10, 2007 #10
    May I suggest a short answer:

    Heat transferred by a flux of phonons is the sum of (velocity) x (energy) for each phonon. Non-umklaap processes conserve this quantity, thus cannot diminish a heat flow initially set in motion.
     
  12. Jun 22, 2010 #11
    hi
    i know some thing about your question
    why umklapp process is necesseray?
    Here necessery is nothing to do with u process
    because u process is the way in which heat transfer takes place through the soilds,where anharmonicity presents because no solid is ideal ,that means no atom in the soli bahaves as independent harmonic oscillator
     
  13. Jun 22, 2010 #12
    hi
    i know some thing about your question
    why umklapp process is necesseray?
    Here necessery is nothing to do with u process
    because u process is the way in which heat transfer takes place through the soilds,where anharmonicity presents because no solid is ideal ,that means no atom in the soli bahaves as independent harmonic oscillator
     
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