Navigating 3D Structures - Crystallographic Directions

Now that we have points within a unit cell, we need to think about crystallographic directions. Why is this important?

Let's say you want to pull on a perfect crystal and see how it responds. Well, depending on how the crystal is oriented, the assembly of atoms may be different, and it might be easier to pull or break the crystal in one direction vs another.

You may want to apply an electric field or thermal gradient to a material so that you could transport charge or heat. Depending on the orientation of the crystal, we might expect the electrons or thermal vibrations to be different in different directions. And of course, if you want to process any of these crystals perfectly, you need to understand how the crystallize and grow, which different in different directions.

Perhaps you want to add additional atoms to the crystal. How do this atoms get in? They'll diffuse, and they'll have preferential diffusion along certain directions in the crystal.

The list of directional-dependent properties is endless, and it's critical in most areas of materials engineering that use highly perfect crystals, especially the semiconductor industry, the solar industry, sensors, optics, and high-temperature structural materials.

The way we find crystallographic directions in a unit cell is the exact same way you'd define any vector in 3D space, we just use a bit different notation. The algorithm for finding a crystallographic direction is:

  1. Find the two points that you will connect via the vector.
  2. Take the difference between the absolute positions of these points and normalize them to their respective $a$, $b$, and $c$ lattice parameters.
  3. Adjust to smallest integer values: we are interested in the general direction, the magnitude of the vector is not important here.
  4. Define the vector components as $u$, $v$, and $w$ and write them in square brackets with no seperating commas: i.e., $[uvw]$.
  5. Use an overbar to represent negative indices.

Mathematically, these steps are to define the absolute positions of two points:

\begin{align} (x_1, y_1, z_1) &= (q_1a, r_1b, s_1c)\ (x_2, y_2, z_2) &= (q_2a, r_2b, s_2c) \end{align}

Define $u$, $v$ and $w$ as:

$$\begin{align} u = n\left(\frac{x_2-x_1}{a} \right) &= n\left(\frac{(q_2-q_1)a}{a}\right) = n(q_2-q_1)\\\ v = n\left(\frac{y_2-y_1}{b} \right) &= n\left(\frac{(r_2-r_1)b}{b}\right) = n(r_2-r_1)\\\ w = n\left(\frac{z_2-z_1}{c} \right) &= n\left(\frac{(s_2-s_1)c}{c}\right) = n(s_2-s_1)\\\ \end{align} \tag{5.7.1}$$

Where $n$ is an integer that we will define later. We will choose it to ensure that $u$, $v$, and $w$ are all integers. Nicely, $u$, $v$, or $w$ can be defined cleanly in terms of the fractional positions of the points.

Then, we define the direction as $[uvw]$. If any of $u$, $v$, or $w$ are negative, we take away the negative sign and write a bar over the top.

A cubic unit cell with two positions indicated.

Figure 5.7.1 A cubic unit cell with two positions indicated.

That description is a bit abstract, so let's do an example. We'll take the two points in Figure 5.7.1 and find the direction from head ($(qrs)_1 = 000$) to tail ($(qrs)_2 = 10\frac{1}{2}$)

\begin{align} (x_1, y_1, z_1) &= (0a, 0b, 0c)\\ (x_2, y_2, z_2) &= (1a, 0b, \frac{1}{2}c) \end{align}

Let's run through the algorithm from Eq. 5.7.1:

\begin{align} u &= n(1-0) = 1n\\ v &= n(0-0) = 0n\\ w &= n(\frac{1}{2}-0) = \frac{1}{2}n\\ \end{align}

The value of $n$ that will get us the smallest set of integer values for $u$, $v$, and $w$ is $n=2$. So our crystallographic direction is:

$$[uvw] = [201]$$

Let's have you give it a try for the two points in Figure 5.7.2 by completing exercise Exercise 5.7.1.

A crystallographic direction from $(qrs)\_1 = 0\frac{1}{2}0$ to $(qrs)\_2 = -111$.

Figure 5.7.2 A crystallographic direction from $(qrs)_1 = 0\frac{1}{2}0$ to $(qrs)_2 = -111$.

Exercise 5.7.1: Deriving a Crystallographic Direction
Not Currently Assigned

  1. Find the $[uvw]$ direction between the two points in Figure 5.7.2.

    If you want an overbar, you can type in something like $[\bar{u}vw]$ for your values.

    This exercise should only take 2-3 minutes.