If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)

Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie

Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Anyone handy with Fourier transforms?

Morbert · 2012-01-15 19:13:04

I'm wondering what the fourier transform of 1/|r1 - r2| is, where |r1 - r2| = square root of {(x2 - x1^2) + (y2-y1)^2 + (z2-z1)^2} Normally it is just 4 Pi / K^2, but there are two catches. The first is not much of an issue: 1/|r1 - r2| is only defined over a region of fine volume V = {0,1}{0,1}{0,1}. A box, in other…

15-01-2012 7:13pm

#1

Morbert

Registered Users, Registered Users 2 Posts: 3,457 ✭✭✭

Join Date: September 2004

Posts: 3397

I'm wondering what the fourier transform of 1/|r1 - r2| is, where

|r1 - r2| = square root of {(x2 - x1^2) + (y2-y1)^2 + (z2-z1)^2}

Normally it is just 4 Pi / K^2, but there are two catches.

The first is not much of an issue: 1/|r1 - r2| is only defined over a region of fine volume V = {0,1}{0,1}{0,1}. A box, in other words. That's no big deal. The transform becomes

$tex2png--10.cgi?\frac{1}{8}%20\int_{-1}^{1}%20\int_{-1}^{1}%20\int_{-1}^{1}%20\frac{e^{-i\mathbf{g}\cdot%20\mathbf{r}}}{|\mathbf{r}|}%20d\mathbf{r}$

Click

which can be numerically solved.

What I am stuck on is how the above form of the transform is changed when |r1-r2| is

square root of {(x2-x1)^2 + A (y2-y1)^2 + B (z2-z1)^2}

where A and B are positive real numbers.

0