Linear chain with m th nearest neighbor interactions

In summary, the dispersion relation must be generalized to ω=2(∑_(m>0)▒K_m ((〖sin〗^2 (1/2 mka)))/M )^(1/2) |k|. 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|.
  • #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.


Homework Equations

: Included in part 1



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.
 
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  • #2
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##.
 
  • #3
Thanks, I think that did it.
 
  • #4
When p = 3, How to show w~kln(k)?
 

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