# Interesting question about Screening in a metal

1. Dec 6, 2009

### tjny699

Hello,

Chemistry grad student here about to start a physics-y project...so trying to learn about condensed matter physics.

I get how screening works when you add a single charge to a gas of electrons, but what happens in a metal when you have a whole lattice, ie, a background charge density? Specifically, say the background charge density is

$$n(x)=n_0+\delta n e^{-A*|x|}$$

where $$\delta n$$ is small.

I've tried applying the Thomas-Fermi screening function (dielectric) and using a linear response approximation but can't get a sensible result when I try to calculate the self-consistent potential, distribution of electron charge, and electric field for x>>1/A

Does anyone know any references that contain a discussion of screening for background distributions that aren't just point particles?

Any help or suggestions would be great!

Last edited: Dec 6, 2009
2. Dec 8, 2009

### weejee

I think we should know how the positive background move when a force is applied to it, to answer your question.

However, it seems that only the electron motion is important since electrons move a lot faster than positve ions. The lattice structure still plays an important role in screening, since it determines the band structure and thereby properties of the Fermi surface.

3. Dec 9, 2009

### tjny699

Hi weejee,

thanks a lot for your reply. i think you are correct: i assumed that the motion of the positive background can be neglected.

I think I've figured out how to get the potential and electron charge distribution using one of the Maxwell Equations, something called the "Thomas-Fermi screening dielectric" and by fourier transforming the potential and charge density.

However, Im a bit stuck on how to determine the electric field for x>>1/A. Perhaps it is easier to find the electric field assuming that there is no screening first? I have a biochemistry background so i'm not quite sure how to go about calculating the E-field.

Thanks again and, as always, any help would be very appreciated.