# The Laue Equations

1. Apr 14, 2012

### Roo2

I can't really follow the template here since there's no specific question that I'm trying to answer; I'm going to give a presentation about diffraction next week and there's one particular line in my notes that doesn't make sense. I've attached the offending line to this post.

The fundamental presumption is that the scattering from a unit cell at coordinates (u,v,w) is equal to the scattering from a unit cell at the origin modulated by a complex exponential. This already confuses me; I understand that there is a scattering phase dependence on position (r*S) but I don't get why it has to be exponential and can't be modeled by a real space wave function. However, I'm willing to live with this as a postulate.

What really bothers me is the statement that the individual sums of the scattering factors go to zero unless a*S, b*S, and c*S are integers. Each wave component can be written, according to Euler's formula, as cos(2∏ua*S) + isin(2∏ua*S). I understand that in the case of perfectly constructive interference, where ua*S is indeed an integer, the real cosine term becomes 1 and the imaginary sine term becomes 0. However, if the waves are not perfectly antiphase but not completely in phase, you can get any number of values of both the real and imaginary term which are clearly non-zero. For example, if ua*S = n/2, then the real cosine term goes to zero and the imaginary sine term goes to 1. I have no idea what this physically corresponds to, but it's clearly non-zero.

I asked my professor about this and he said that while individually the equation does not evaluate to zero, the sum of all such equations for each unit cell dimension does. To me this seems like mathemagics. Could somebody please explain to me why the statement holds? It obviously does because the observed diffraction pattern (and therefore scattering angle) is discreet; however, I don't understand why. I would also like to know why scattering is modeled as a complex exponential rather than a real space wave; however, my other question is more important.

Thank you very much for your help.

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