David Wych


Introduction to Crystallography: Part 3

“The Ewald Sphere”

If you need enticement to stick with this – this is probably the most beautiful piece of theory I’ve ever come across ✨it’s really something special.

Last time, we found that when an X-ray beam scatters through a volume of electron density, the intensity we measure in the far field is proportional to the squared magnitude of the form factor:

\[I(\mathbf{q}) \propto \| F(\mathbf{q}) \|^{2}\]

Let’s review a little…

Introduction to Crystallography: Part 2

“The Phase Problem”, or, “You didn’t think Mother Nature was gonna let us off that easy, did you?”

Last time we talked about how an incident X-ray beam is scattered by the electrons around atoms, and why the cumulative effect of the scattering is such that the electrons perform the Fourier Transform!

Let’s discuss what this means in practice.

We started with an X-ray wave traveling in a direction, defined by a wavevector, \(\mathbf{k}\).

Electrons scatter that wave, so it radiates out spherically, at all angles defined by \(\mathbf{k}'\) (a vector at an angle \(2\theta\) to the original wavevector, \(\mathbf{k}\))

spherical wave w vecs

We choose to express the scattering through the “scattering vector” \(\mathbf{q}\), the vector that points from the tip of \(\mathbf{k}\) to the tip of \(\mathbf{k}'\).

Sweeping through all angles, the collection of these \(\mathbf{k}\)s forms a sphere.

Introduction to Crystallography: Part 1

“How electrons perform the Fourier Transform”

Warning: this is going to require a good bit of physics and math.

If that’s not your thing, fair enough, feel free to bail.

But I’m going to try to make this as palatable as I can.

So give it a go, if you’re feeling up to it.

Let’s start with the math of how crystallography really works.

X-rays (wavelength λ=10⁻¹¹-10⁻⁸ m; energy 0.02-20 keV) scatter, elastically, off of electrons bound to atoms, producing spherical waves of the same wavelength.

spherical wave

We can describe the direction of traveling wave using a “wavevector” \(\mathbf{k}\), with the magnitude of the vector equal to the “wavenumber”, \(\frac{2\pi}{\lambda}\), and the direction pointing in the direction of propagation. For the incident X-ray plane wave, there is a single, definite wavevector (\(\mathbf{k}\)). For the scattered wave, there are many, equivalent wavevectors (\(\mathbf{k}'\)), one for each scattering angle.

spherical wave w vecs

The difference in direction between the incident (\(\mathbf{k}\)) and scattered (\(\mathbf{k'}\)) wavevectors is given by the “scattering vector”, \(\mathbf{q} = \mathbf{k'} - \mathbf{k}\).

Its magnitude is given by \(\lvert \mathbf{q} \rvert = \frac{4 \pi \sin( \theta )}{ \lambda }\), where \(2 \theta\) is the scattering angle.

If there are multiple electrons scattering the incident wave, separated by a vector \(\mathbf{r}\), the scattered waves will differ by a phase shift given by the dot product of \(\mathbf{q}\) and \(\mathbf{r}\): \(\Delta \phi = \mathbf{q} \cdot \mathbf{r}\).

Hey There

Hey, welcome to my site. work in progress