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Solid State - Phonons at Brillouin Zone Boundary

  1. Apr 1, 2013 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=57364&stc=1&d=1364839743.jpg

    2. Relevant equations
    {3.9b}
    [tex]A[2\mu -m\omega ^2 ]=2\mu Bcos(\frac{ka}{2})[/tex]
    [tex]B[2\mu -M\omega ^2 ]=2\mu Acos(\frac{ka}{2})[/tex]

    3. The attempt at a solution

    All I can think of is setting [tex]k =\frac{\pi}{a}[/tex] so that

    [tex]B[2\mu -M\omega ^2 ]=A[2\mu -m\omega ^2 ][/tex]

    solve for omega squared and take the negative root as the accoustic phonon frequency which I imagine to be =0 to show that the phonons are stationary in the accoustic branch. What I'm not entirely sure of is how to initially manipulate the amplitude Eq's, I guess I somehow have to sub part of the equation representing the small mass particles into the equation representing the large mass particles (as the optic phonons are dependent of the small mass particles and visa versa) and manipulate, some guidance would be great.
     

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  3. Apr 5, 2013 #2

    TSny

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    Since you are only interested in the zone boundary, go ahead and substitute ##k = \pi/a## into the above equations. So:
    [tex]A[2\mu -m\omega ^2 ]=0[/tex]
    [tex]B[2\mu -M\omega ^2 ]=0[/tex]
    To get a non-trivial solution for ##A## and ##B##, you need to assume at least one of ##A## or ##B## is nonzero. What can you conclude if you let ##A \neq 0##?
     
  4. Apr 5, 2013 #3
    Well if A does not equal zero then once you equate the equations B cannot equal zero? :S
     
  5. Apr 5, 2013 #4

    TSny

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    Start with ##A[2\mu -m\omega ^2 ]=0## and the assumption that ##A## is not zero. What does this equation tell you about the frequency ##\omega##?
     
  6. Apr 5, 2013 #5
    I guess frequency squared is inversely proportional to the mass so the smaller the particle mass the larger the frequency? And visa versa, the optical branch frequency > accoustic branch so it would imply the optical branch should affect light particles? Not sure
     
  7. Apr 5, 2013 #6

    TSny

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    Note that if ##A \neq 0## then the equation ##A[2\mu -m\omega ^2 ]=0## can only be satisfied if ##\omega## has a particular value. With the value of ##\omega## now fixed, move on to the equation ##B[2\mu -M\omega ^2 ]=0## and see what that tells you about ##B##.
     
  8. Apr 5, 2013 #7
    Ok, to figure that out I'll need to understand the first bit, how come A being non-trivial can only be satisfied if omega is fixed? Not really understanding that bit :S
     
  9. Apr 5, 2013 #8

    TSny

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    Suppose you have two two numbers x and y and you know two things: x≠0 and xy=0.

    What can you conclude about y?

    Apply that to ##A[2\mu -m\omega ^2 ]=0##
     
  10. Apr 5, 2013 #9
    ok yeh, so [2*mu - m*w^2] must equal 0, but then m AND omega should be a specific value such that m*w^2 = 2*mu right, not just omega? Cause i'm assuming the mass is a variable
     
  11. Apr 5, 2013 #10
    Also I understand now, since omega is fixed and m does not equal M. . . [2*mu - M*w^2] does not equal 0 in the 2nd equation like it does in the first?
     
  12. Apr 5, 2013 #11

    TSny

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    No, the masses m and M are considered fixed masses for a given lattice. They are not variables.
     
  13. Apr 5, 2013 #12
    Yeh my bad :), so basically all this implies that B=0 and A does not
     
  14. Apr 5, 2013 #13
    So, B = [2*mu - m*w^2]

    ((B-2*mu)/m)^1/2 = w, am I on the right lines?
     
  15. Apr 5, 2013 #14

    TSny

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    No, I'm afraid that isn't right at all.

    Earlier, you made the correct deduction that when A ≠ 0 then 2*mu - m*w^2 = 0 and B = 0 . The equation 2*mu - m*w^2 = 0 can be used to find w for this mode.

    Repeat similar arguments for the other mode where B ≠ 0.
     
  16. Apr 5, 2013 #15
    Ok, well thanks for all the help, much appreciated :)
     
  17. Apr 6, 2013 #16
    Ok sorry, so I get w = (2*mu/m)^1/2, but I don't see how this is indicative of the small mass particles being stationary in the accoustic branch?
     
  18. Apr 6, 2013 #17

    TSny

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    What do the numbers A and B represent physically?
     
  19. Apr 6, 2013 #18
    Amplitudes, so B = 0 means the amplitude for the particles in the accoustic branch is zero?
     
  20. Apr 6, 2013 #19

    TSny

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    B is the amplitude of oscillation for which particles? Particles with mass m or particles with mass M?

    You have found that if A ≠ 0 then B = 0.

    You also have found that if A ≠ 0, then ω = (2*mu/m)^1/2. Does this frequency correspond to the acoustic branch or the optical branch?

    In order to answer this, you might want to work out the other case where you start with the assumption that B≠ 0.
     
  21. Apr 6, 2013 #20
    well I imagine that in the sake of answering the question, B=0 would suggest the amplitude of 'm' mass particles is zero considering those are the ones that are supposed to be stationary, but yes I'll attempt setting B to not equal 0
     
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