Periodicity in 3D Crystals

In the previous section we worked with simple 2D patterns for their simplicity. However, most crystal structures we encounter are 3D. In this section, we'll show the different unit cells (we'll actually show all the Bravais lattices for completeness) that we have for 3D crystals. These are of course more complex due to the additional dimensionality compared to 2D. We'll also classify each pattern in terms of its crystal family, which is defined by the symmetry of the crystal.

At this point, we'll focus on crystal systems that have orthogonal basis vectors, meaning that all of the lattice vectors (there are now three as opposed to two) are orthogonal (at 90° angles) with respect to each other. We do this to simplify calculations - we want you to think about the patterns that atoms have in these systems, we don't want you spending time struggling with the trigonometric relationships that arise when calculating (e.g.) volumes of the unit cells.

First, let's look at a 3D unit cell in Figure 5.5.1. The figure shows a parallelepiped with three lattice vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$, the directions of which are represented by the coordinate system at the lower right. The edges of the parallelepiped have lengths $a$, $b$, and $c$, while the angles between the lattice vectors are $\alpha$, $\beta$ , and $\gamma$:

• $\alpha$ is the angle between $\mathbf{b}$ and $\mathbf{c}$
• $\beta$ is the angle between $\mathbf{a}$ and $\mathbf{c}$
• $\gamma$ is the angle between $\mathbf{a}$ and $\mathbf{b}$.

Remember, in order to describe a space-filling crystal, this unit cell needs to tile to fill space via the translation vector $\mathbf{T} = n\mathbf{a}+m\mathbf{b}+k\mathbf{c}$, where the vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are the lattice vectors and $n, m, k$ are integers. Figure 5.5.2 tessellation of the unit cell in Figure 5.5.1 to produce a space-filling volume of eight unit cells. We can perform this space-filling tessellation as far as we'd like in each of the lattice directions to create our crystal. At this point, the unit cells are just empty space, but don't worry, we'll start to add atoms soon.

In Figure 5.5.1, we've shown a general 3D unit cell and have not defined any particular equalities or values between the lattice parameters. We can, however, determine all possible lattices in 3D and the associated relationships between their lattice parameters (again, Bravais, 1850). Those lattices, on the constraints on the lattice parameters of each of the crystal types are shown in Figure 5.5.3.

I want to highlight a few things here before we get too far into the weeds.

1. These lattices describe the underlaying periodic structure for all crystals. Because of our interest in structure-property relationships, it's important to know this broad overarching concept in crystallography.
2. We're going to use crystals to understand diffusion, mechanical properties, defects, electronic properties, and other behaviors.
3. In this class, we'll only be working with lattices with orthogonal basis sets ($\alpha = \beta = \gamma = 90^{\circ}$) because the math and visualization are simpler.
4. (Optional Consideration) In the last page's optional reading, we noticed there were only 5 possible lattices in 2D. In 3D there are 14. Why there are 14 Bravais lattices requires more advanced mathematics. Indeed, this wasn't figured out until 1850 even with very smart individuals working on the problem. Mortiz Frenkenheim, Bravais' contemporary, thought there were 15 lattices! Bravais found that two of them are equivalent... and now history barely mentions Frenkenheim.

Let's briefly look at a few of these. The lattice that has the fewest constraints on its lattice parameters (and effectively the lowest symmetry) is the triclinic lattice. Here, $a \neq b \neq c$ and $\alpha \neq \beta \neq \gamma$. There's one lattice point in the unit cell, which makes the lattice a primitive (represented with a $P$) lattice. The source of the name triclinic is interesting. "Tri-" means three, of course. "Clinic" has a Proto-Indo-European root meaning "to lean" or "to recline". So, we have three leaning things: the three lattice vectors are not equal to $90^{\circ}$. Similarly, for monoclinic, we have one leaning thing and $\alpha = \gamma= 90^{\circ}$ while $\beta \neq 90^{\circ}$ (or $120^{\circ}$). "Clinic" also has a more colloquial meaning this days for a medical facility. The roots are the same, it's where sick people go to lean or recline! Neat.

In any case, these lattices can be classified into crystal systems of cubic, tetragonal, orthorhombic, hexagonal, trigonal, monoclinic, and triclinic. Each of these crystal systems have unique relationships and constraints on their lattice parameters (and different symmetries).

We'll only work with the three crystal systems with orthogonal basis vectors: cubic, tetragonal, and orthorhombic. Note that there's all sorts of patterns that may show up in these structures. For example:

• The primitive lattice $P$, which has spheres only at the corners of the unit cell, shared between the corners. Primitive lattices only have one sphere per unit cell.
• The body-centered lattice $I$, which has another sphere at the center of the unit cell.
• The face-centered lattice $F$, which has three additional spheres shared between the six unit cell's faces.
• The base-centered lattice $C$, which has one additional sphere shared between two faces.

Because only part of these spheres are located within the unit cell, we need to be careful in counting the number of spheres present in the cell itself! For any system with orthogonal lattice vectors (which is what we'll limit ourselves to), the number of spheres per unit cell is:

where $N_{\text{Tot}}$ is the total number of spheres (or atoms, or ions, etc.) in cell, $ N_{\text{i}}$ is the number of atoms that are fully inside the cell, $N_{\text{f}}$ are the number of atoms on he faces of the cell, and $N_{\text{c}}$ are the number of atoms on the corner of the cell.

In the example in Section 5.5.7, we have:

$$N_{\text{Tot}} = 0 + \frac{6}{2}+\frac{8}{8} = 4$$

Let's explore this a bit more with another lattice system. Try the problem below, taking about 5 minutes to complete it.