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Linear chain with m th nearest neighbor interactions

  1. Oct 13, 2013 #1
    1. Ashcroft and Mermin 22.1
    Reexamine the theory of the linear chain without making the assumption that only nearest neighbors interact, using the harmonic potential energy of the form:
    U^harm=∑_n▒∑_(m>0)▒1/2 K_m [u(na)-u([n+m]a) ]^(1/2)
    Show that the dispersion relation must be generalized to
    ω=2(∑_(m>0)▒K_m ((〖sin〗^2 (1/2 mka)))/M )^(1/2)
    Show that, provided the sum converges, the long wavelength limit of the dispersion relation must be generalized to:
    ω=a(∑_(m>0)▒〖m^2 K_m/M)^(1/2) |k| 〗
    Show that if Km = 1/mp (1<p<3), so that the sum does not converge, then in the long wavelength limit
    ω∝ k^((p-1)/2)

    Hint: it is no longer permissible to use the small-k expansion of the sine in equation a, but one can replace the sum by an integral in the limit of small k.

    2. Relevant equations: Included in part 1

    3. The attempt at a solution
    I have no problem getting parts a and b, but part c is eluding me. I first replaced the summation with an integral and got:

    ω=2(∫_0^k▒〖m^(-p) (〖sin〗^2 (1/2 mka))/M〗 〖dm)〗^(1/2)
    I then expanded the sin^2
    ω=2(∫_0^k▒〖m^(-p) (1-cos(1/2 mka))/M〗 〖dm)〗^(1/2)
    I attempted an integration by parts, but quickly realized that I would end up in a never ending cycle. I then read a tip online (not a solution, but a hint) that one should try expanding cos x. But I keep ending up with

    ω∝∫_0^k▒〖m^(-p) (1-(1-m^2 k^2+m^4 k^4-m^6 k^6+m^8 k^8-…)〗 〖dm)〗^(1/2)An
    Which, when I integrate, doesn’t seem to give me the right answer. I feel I’m missing something very basic, but I’ve been out of school for several years, and I’ve been banging my head against a wall on this for days now. Any help is appreciated.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 13, 2013 #2


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    Homework Helper
    Gold Member

    Try a substitution where you let ##u## = the argument of the sine function. Convert the integral over ##m## to a numerical factor times an integral with respect to ##u##. All of the ##k## dependence will be in the numerical factor. So you will not need to worry about doing the integral over ##u## if you just want to find how ##\omega## depends on ##k##.
  4. Oct 14, 2013 #3
    Thanks, I think that did it.
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