Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Deriving the effective potential due to screening

  1. Feb 4, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm supposed to show that the Lindhard dielectric functions gives a contribution to the effective potential of a metals as

    [tex]U_{eff}( \vec{r} )[/tex] [tex]\propto[/tex] [tex]\frac{cos( 2 k_{F}r)}{r^{3}}[/tex]

    in the limit of [tex]r\rightarrow\infty[/tex] for d = 3 (3 dimensions)
    2. Relevant equations

    Lindhard dielectric function:
    [tex]\epsilon(\vec{k},0) = 1 + \frac{\kappa^{2}_{TF}}{2k^{2}} ( 1 + \frac{1}{4k_{F}}\frac{4k^{2}_{F}-k^{2}}{2k}\ln \frac{2k_{F}+k}{2k_{F}-k}) = \epsilon(\vec{k}) [/tex]

    [tex]U_{eff}[/tex]([tex]\vec{k}[/tex]) = [tex]\frac{U(\vec{k})}{\epsilon(\vec{k},0)}[/tex]

    [tex]U(\vec{k}) = \frac{4 \pi e^{2}}{k^{2}}[/tex]

    [tex]U_{eff}(\vec{r})[/tex] is the inverse (spatial) Fourier transform of [tex]U_{eff}(\vec{k})[/tex]

    [tex]k_{F}[/tex] is the Fermi wavevector

    [tex]\kappa^{2}_{TF}[/tex] is the Thomas-Fermi wavevector (constant)

    3. The attempt at a solution
    I've tried to Taylor expand [tex]\frac{1}{\epsilon(\vec{k})}[/tex] around [tex]2k_{F}[/tex] but the first derivative contains the logarithm which is divergent. I tried this because one of my classmates recommended it.

    I tried to perform the Fourier transform by
    [tex] U_{eff}(\vec{r}) \propto \int d\vec{k} e^{i \vec{k}\bullet\vec{r}}\frac{U(\vec{k})}{\epsilon(\vec{k},0)} \propto \int k^{2} dk d(cos(\theta)) d\phi e^{i k r cos(\theta)}\frac{U(\vec{k})}{\epsilon(\vec{k},0)} \propto \int k^{2} dk \frac{U(\vec{k})}{\epsilon(\vec{k},0)} \frac{e^{ikr} - e^{-ikr}}{ikr} \propto \frac{1}{r}\int dk \frac{1}{k \epsilon(\vec{k})} sin(kr)[/tex]

    This is kinda where I'm stuck.
    How do I proceed from here?
    Or am I supposed to have done something else?
     
  2. jcsd
  3. Feb 6, 2010 #2
    The logarithm diverges, but it is multiplied by 0 (for the first term in the Taylor expansion).
     
  4. Feb 10, 2010 #3
    I know that.

    The problem is how to simplify/rewrite that expression (the last one) into something solveable...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook