Miller indices for hexagonal crystal systems

To find the draw the direction for a given Miller index say,  we first convert this miller index consisting of 4 indices into one containing 3 indices. To do so, we have a set of formulae prescribed in almost every book. Sadly I haven't been able to come across a single book the gives the derivation of those formulae!

I thought that I could use vector-component method to get the results but that gives totally weird formulae not even close to the ones I see in my textbooks. Here is an example, just to be clearer. (and have a look at the attached image)

So, can anyone suggest me a textbook, a link or anything that can help me understand the derivation? I'm not finding the enthusiasm for rote-memorising the formulae if i don't know where they come from... Before answering, please see our policy on resource recommendation questions. Please write substantial answers that detail the style, content, and prerequisites of the book, paper or other resource. Explain the nature of the resource so that readers can decide which one is best suited for them rather than relying on the opinions of others. Answers containing only a reference to a book or paper will be removed!

From the diagram it is obvious that the Miller-Bravais index has redundant information, as the indices point along three directions that are 120° apart. That means that you can do a simple geometrical derivation using this diagram as your guide: The third equation follows immediately from the vector addition of $\vec u$ and $\vec v$ - you can see that $\vec t$ points in the opposite direction.

It is equally obvious that, in the way I drew the picture, $\vec u = \vec{v'} + \frac12 \vec{u'}$ and $\vec v = \frac12 \vec{u'} - \vec v'$

Simple manipulation of these equations gets you to the expressions you quote.

• according to the formula i attached, it says u = (2u' - v')/3 though, that's where i'm stuck – Sakazuki Akainu Aug 9 '17 at 14:01
• i can't get that forumla, i in fact reached the same point you did – Sakazuki Akainu Aug 9 '17 at 14:01

In a hexagonal crystal system, just like in any other three dimensional system, every vector can be represented in a basis consisting of 3 linearly independent vectors. Thus, 3 such vectors would be sufficient to describe any direction we want in a crystal.

The image below shows a hexagonal unit cell with 4 axes (represented by vectors): $$\vec{a_1}$$, $$\vec{a_2}$$, $$\vec{a_3}$$ and $$\vec{z}$$, used to index directions in a crystal. Vectors $$\vec{a_1}$$, $$\vec{a_2}$$ and $$\vec{a_3}$$ are $$\mathit{not}$$ linearly independent. Conventionally, we choose $$\vec{a_3}$$ to be the "extra" vector, and $$\vec{a_1}$$, $$\vec{a_2}$$, and $$\vec{z}$$ the "main" vectors, common to both 4 and 3 index systems.

The vector $$\vec{a_3}$$ is defined as $$\vec{a_3} = -\left(\vec{a_1} + \vec{a_2}\right)$$, which gives us some intuition for requiring $$t = -(u+v)$$ to be obeyed, since u,v and t are components along $$\vec{a_1}, \vec{a_2}$$ and $$\vec{a_3}$$, respectively. We could have used a different relation between $$u$$, $$v$$ and $$t$$, but this one is the most straightforward (we need an extra equation for the vector components to be unique since $$\vec{a_1}$$, $$\vec{a_2}$$ and $$\vec{a_3}$$ are not linearly independent). The difference between the 3 index (denoted by [u'v'w']) and 4 index representations (denoted by [uvtw]) is that when using 3 indices, we ignore the vector $$\vec{a_3}$$, and only use $$\vec{a_1}$$, $$\vec{a_2}$$ and $$\vec{z}$$ (the $$\mathit{same}$$ $$\vec{a_1}$$, $$\vec{a_2}$$ and $$\vec{z}$$ in $$\mathit{both}$$ representations - the only difference is having or not having $$\vec{a_3}$$). Notice that $$\vec{a_1}$$ and $$\vec{a_2}$$ are $$\mathbf{not}$$ orthogonal. Instead, they make a $$120^\circ$$ angle, so they aren't the usual $$\hat{x}$$ and $$\hat{y}$$ vectors, but rather ones that respect the symmetry of the crsystal.

Now, any vector $$\vec{v}$$ can be written in both representations (again, note that the only difference between the two is $$\vec{a_3}$$ being present in 4-index notation and missing in 3-index notation):

$$\vec{v} = u'\vec{a_1} + v'\vec{a_2} + w'\vec{z} \tag{1} \label{1}$$ and $$\vec{v} = u\vec{a_1} + v\vec{a_2} + t\vec{a_3} + w\vec{z} \tag{2} \label{2}$$

As $$\vec{a_3} = -\left(\vec{a_1} + \vec{a_2}\right)$$, inserting this into equation $$\eqref{2}$$ we get:

$$\vec{v} = (u-t)\vec{a_1} + (v-t)\vec{a_2} + w\vec{z}$$

By substituting $$t = -(u+v)$$ we further get

$$\vec{v} = (2u+v)\vec{a_1} + (2v+u)\vec{a_2} + w\vec{z} \tag{3} \label{3}$$

Equating the components along the same vectors from $$\eqref{3}$$ and $$\eqref{1}$$ we get

\begin{aligned} u' &= 2u+v\\ v' &= 2v+u\\ w' &= w \end{aligned}

Assuming $$u'$$, $$v'$$ and $$w'$$ are known, we can solve the system for $$u$$, $$v$$ and $$w$$ and obtain

\begin{aligned} u &= \frac{1}{3} \left(2u'-v'\right)\\ v &= \frac{1}{3} \left(2v'-u'\right)\\ t &= -\left(u+v\right)\\ w &= w' \end{aligned} as desired.