Scattering amplitude of diffracted beam by a crystal

[Andreasdreas]

Homework Statement

The scattering amplidtude, F, of a, by a crystal, diffracted beam is defined to be:

(1) F=\sum{_\vec{G}}\int n_{\vec{G}}e^{i(\vec{G}-\delta\vec{k})\cdot \vec{r}}\mathrm{d}V

The integral is over all the volumeelements, dV, of the crystal.
n_{\vec{G}} is the local electron concentration of the crystal in dV, \vec{G} is a reciprocal lattice vector and the sum is over the set of all the reciprocal lattice vectors. \vec{r} is the position vector of dV
further -\delta\vec{k}=\vec{k}-\vec{k'} where \vec{k} is the wave vector of the incomming beam and \vec{k'} is the wave vector of the outgoing scattered beam.

The beam could be
fotons, electrons neutrons etc. It is descibed as a plane wave.

It is easy to see that F=n_{\vec{G}}V when \delta \vec{k}=\vec{G}

But how can it be shown that F is negligibly small when \delta \vec{k} differs sginificantly from any \vec{G}?

Homework Equations

I thought some expression for the summation could be used.

Maybe

(2) \sum{_m=0}^{M-1}x^m=\frac{1-x^{M}}{1-x}

The Attempt at a Solution

I can't find a away to make (1) fit into (2). And if i just look at (1) i really am lost.

Homework Equations

The Attempt at a Solution

 

[member 553083]

I thought of using the fact that the sum only goes over all the reciprocal lattice vectors and not \delta \vec{k}. This means that for \delta \vec{k} to be significantly different from any \vec{G}, then F should be small for the term in the summation that involves \delta \vec{k}.But how can one prove this mathematically?