# Homework Help: Scattering from a rough surface

1. Nov 29, 2012

### Irid

1. The problem statement, all variables and given/known data
I'm studying scattering from a rough surface, and my textbook defines h(x,y) as a small vertical deviation from a flat surface. Then they proceed calculations by assuming that the height difference between two points h(x,y)-h(x',y')=some f(x-x', y-y'), i.e. it depends only on the relative position of the two points. In other words, the surface is isotropic.

2. Question
I don't see what kind of surface would ever fulfill this condition, except some very special one, like a constant inclination. If there are any bumps or dips, obviously Δh will not be the same as me move around the surface using the same bar of length (x-x')... Could anybody explain this assumption?

2. Nov 29, 2012

### haruspex

I agree, it would imply that all points with rational offsets from a given point would form a plane. Are you sure that's what is being assumed, rather than some statistical relationship?

3. Nov 29, 2012

### Irid

I'm referring to the book 'Modern X-Ray Physics'. They first assume this isotropic surface to evaluate a 4-D integral, and the statistical correlation between different points comes in later (uncorrelated surface and Gaussian correlation are treated in detail). My Prof. hinted that the isotropic assumption is valid on a scale intermediate between the rapid oscilations at atomic level and the flat surface at macroscopic distances. Apparently this is an isotropic roughness assumption, but I can't find any clear info on it :(

4. Nov 29, 2012

### haruspex

It still only makes sense to me as a statistical statement, e.g. that the probability distribution of h(x,y)-h(x',y')=some f(x-x', y-y'). Can you quote the book in some detail?

5. Nov 29, 2012

### Irid

I think I got it figured out. The assumption in that crap book is plain wrong, but the final answer is correct, even though for a completely different reason... since the roughness is isotropic, the integral over x' over a sufficiently large distance will not depend on where x was chosen, which renders integral over x trivial. Thanks for inspiration anyway :)