Crystal structures are typically determined by x-ray diffraction. X-rays are directed at a crystal, and the waves scatter from every atom in the crystal. Because of the periodic arrangement of the atoms, the scattered waves add constructively in certain scattering directions. The diffraction pattern has sharp peaks in certain directions. By analyzing the positions and the intensities of the peaks, it is possible to determine how the atoms are arranged in a crystal.

To keep things simple, let's first consider two scatterers, one at $\vec{r}_1$ and one at $\vec{r}_2$.

In the two-dimensional simulation below, a collimated beam of plane waves travels from left to right. Red is a positive wave amplitude, blue is a negative wave amplitude, and black is no wave amplitude. Two point scatterers are placed in the beam. These scatterers emit circular waves with the wavelength of the incident beam. The scatterer emits a wave peak when a wave peak in the incident beam arrives.

If both of the scattering strengths $F_1$ and $F_2$ are set to zero, there are no scattered waves. If the scattering strength of only one scatterer is set to zero, you can see the circular waves being emitted coherently with the incident beam.

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F1 =  [Å3/2] F2 =  [Å3/2] x1 =  [Å] x2 =  [Å] y1 =  [Å] y2 =  [Å] λ =  [Å]

Notice that the scattered waves add constructively in some directions and destructively in others. For waves in three dimensions, the amplitude of the scattered waves at position $\vec{r}$ is simply the sum of the waves radiated by the two scatterers,

$$F(\vec{r})=F_1\frac{\cos\left(\vec{k}_1'\cdot(\vec{r} - \vec{r}_1)-\omega t + \vec{k}\cdot\vec{r}_1\right)}{|\vec{r} - \vec{r}_1|}+F_2\frac{\cos\left(\vec{k}_2'\cdot(\vec{r} - \vec{r}_2)-\omega t +\vec{k}\cdot\vec{r}_2\right)}{|\vec{r} - \vec{r}_2|}.$$

Here $F_1$ and $F_2$ are the scattering amplitudes, $|\vec{r}-\vec{r}_1|$ is the distance from scatterer 1 to $\vec{r}$ and $|\vec{r}-\vec{r}_2|$ is the distance from scatterer 2 to $\vec{r}$. The wave vector $\vec{k}$ is the wave vector of the incident waves, the phase factors $\vec{k}\cdot\vec{r}_1$ and $\vec{k}\cdot\vec{r}_2$ ensure that the peak of the outgoing spherical wave leaves the scatterer when the peak of the plane wave arrives, $\vec{k}_1'$ and $\vec{k}_2'$ are the wave vectors of the scattered waves. Wave vector $\vec{k}_1'$ is in the direction $\vec{r}-\vec{r}_1$, and wave vector $\vec{k}_2'$ is in the direction $\vec{r}-\vec{r}_2$. The angular frequency $\omega$ is related to the period $T$ of the oscillations, $\omega=\frac{2\pi}{T}$.

If the detector is much further from the scatterers than $\vec{r}_1 - \vec{r}_2$, then this is the far field limit and $|\vec{r}-\vec{r}_1|\approx |\vec{r}-\vec{r}_2|=R$ and $\vec{k}_1'\approx \vec{k}_2'=\vec{k}'$.

$$F(\vec{r})=\frac{F_1}{R}\cos\left(\vec{k}'\cdot\vec{r}-\omega t + (\vec{k}-\vec{k}')\cdot\vec{r}_1\right)+\frac{F_2}{R}\cos\left(\vec{k}'\cdot\vec{r} -\omega t +(\vec{k}-\vec{k}')\cdot\vec{r}_2\right).$$

X-rays have a very high frequency, and the detector at position $\vec{r}$ typically cannot distinguish the rapid oscillations in the amplitude of the wave. Usually, the detector can only measure the average intensity of the waves at position $\vec{r}$. To calculate the intensity of the scattered waves, it is convenient to use a complex representation of harmonic motion. This is discussed in the next subsection.