Deriving the effective potential due to screening

[SkeZa]

Homework Statement

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

$$U_{eff}( \vec{r} )$$ $$\propto$$ $$\frac{cos( 2 k_{F}r)}{r^{3}}$$

in the limit of $$r\rightarrow\infty$$ for d = 3 (3 dimensions)

Homework Equations

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

$$U_{eff}$$($$\vec{k}$$) = $$\frac{U(\vec{k})}{\epsilon(\vec{k},0)}$$

$$U(\vec{k}) = \frac{4 \pi e^{2}}{k^{2}}$$

$$U_{eff}(\vec{r})$$ is the inverse (spatial) Fourier transform of $$U_{eff}(\vec{k})$$

$$k_{F}$$ is the Fermi wavevector

$$\kappa^{2}_{TF}$$ is the Thomas-Fermi wavevector (constant)

The Attempt at a Solution

I've tried to Taylor expand $$\frac{1}{\epsilon(\vec{k})}$$ around $$2k_{F}$$ 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
$$U_{eff}(\vec{r}) \propto \int d\vec{k} e^{i \vec{k}\bullet\vec{r}}\frac{U(\vec{k})}{\epsilon(\vec{k},0)} \propto \int k^{2} dk d(cos(\theta)) d\phi e^{i k r cos(\theta)}\frac{U(\vec{k})}{\epsilon(\vec{k},0)} \propto \int k^{2} dk \frac{U(\vec{k})}{\epsilon(\vec{k},0)} \frac{e^{ikr} - e^{-ikr}}{ikr} \propto \frac{1}{r}\int dk \frac{1}{k \epsilon(\vec{k})} sin(kr)$$

This is kinda where I'm stuck.
How do I proceed from here?
Or am I supposed to have done something else?

 

[badphysicist]

The logarithm diverges, but it is multiplied by 0 (for the first term in the Taylor expansion).

 

[SkeZa]

I know that.

The problem is how to simplify/rewrite that expression (the last one) into something solveable...