- #1

etotheipi

- Homework Statement
- Prove the Weiss Zone Law

- Relevant Equations
- N/A

For some context, a Bravais lattice in 3D is described by 3 basis vectors ##\vec{a}_1##, ##\vec{a}_2##, ##\vec{a}_3## s.t. a discrete point in the lattice can be generated with ##\vec{R} = \sum R_i \vec{a}_i##, with ##R_1##, ##R_2##, ##R_3## being integers. A direction is specified by a triplet ##[u_1 \, u_2 \, u_3]## (i.e. a vector along this direction is ##\vec{u} = \sum u_i \vec{a}_i##) and a plane is specified by a triplet ##(h_1 \, h_2 \, h_3)## s.t. the intersections of the plane with our coordinate system are at ##x_i = \phi a_i / h_i##, where ##a_i = |\vec{a}_i|## and ##\phi## is just some multiplicative constant (specifically, the lowest common denominator of all the ##a_i/x_i##) by which all of the ##a_i/x_i## are multiplied, so that the tuple ##(h_1 \, h_2 \, h_3)## contains integers).

The question is to show that if a direction ##[u_1 \, u_2 \, u_3]## lies in the plane ##(h_1, h_2, h_3)##, then ##u_1 h_1 + u_2 h_2 + u_3 h_3 = 0##. I thought my way was a bit clunky. The plane can be defined by two vectors i.e. ##\vec{\Pi}(\alpha, \beta) = \vec{\Pi}_0 + \alpha \vec{v}_1 + \beta \vec{v}_2## which are$$\vec{v}_1 = \frac{\vec{a}_3}{h_3} - \frac{\vec{a}_1}{h_1}, \quad \vec{v}_2 = \frac{\vec{a}_3}{h_3} - \frac{\vec{a}_2}{h_2}$$A normal to the plane is$$\vec{n} = \vec{v}_1 \times \vec{v}_2 = \frac{\vec{a}_1 \times \vec{a}_2}{h_1 h_2} - \frac{\vec{a}_1 \times \vec{a}_3}{h_1 h_3} - \frac{\vec{a}_3 \times \vec{a}_2}{h_3 h_2}$$If ##\vec{u} = u_1 \vec{a}_1 + u_2 \vec{a}_2 + u_3 \vec{a}_3## lies within ##(h_1, h_2, h_3)## then ##\vec{u} \cdot \vec{n} = 0##, i.e.$$\frac{u_3}{h_1 h_2} (\vec{a}_1 \times \vec{a}_2) \cdot \vec{a}_3 - \frac{u_2}{h_1 h_3} (\vec{a}_1 \times \vec{a}_3) \cdot \vec{a}_2 - \frac{u_1}{h_3 h_2} (\vec{a}_3 \times \vec{a}_2) \cdot \vec{a}_1 = 0 $$ $$\begin{align*}\left[ \frac{u_3}{h_1 h_2} + \frac{u_2}{h_1 h_3} + \frac{u_1}{h_2 h_3} \right] (\vec{a}_1 \times \vec{a}_2) \cdot \vec{a}_3 = 0 &\implies\frac{u_3}{h_1 h_2} + \frac{u_2}{h_1 h_3} + \frac{u_1}{h_2 h_3} = 0\\ \\

&\implies u_3 h_3 + u_2 h_2 + u_1 h_1 = 0

\end{align*}$$It works, but it seems long winded. Does anyone have a better way?

The question is to show that if a direction ##[u_1 \, u_2 \, u_3]## lies in the plane ##(h_1, h_2, h_3)##, then ##u_1 h_1 + u_2 h_2 + u_3 h_3 = 0##. I thought my way was a bit clunky. The plane can be defined by two vectors i.e. ##\vec{\Pi}(\alpha, \beta) = \vec{\Pi}_0 + \alpha \vec{v}_1 + \beta \vec{v}_2## which are$$\vec{v}_1 = \frac{\vec{a}_3}{h_3} - \frac{\vec{a}_1}{h_1}, \quad \vec{v}_2 = \frac{\vec{a}_3}{h_3} - \frac{\vec{a}_2}{h_2}$$A normal to the plane is$$\vec{n} = \vec{v}_1 \times \vec{v}_2 = \frac{\vec{a}_1 \times \vec{a}_2}{h_1 h_2} - \frac{\vec{a}_1 \times \vec{a}_3}{h_1 h_3} - \frac{\vec{a}_3 \times \vec{a}_2}{h_3 h_2}$$If ##\vec{u} = u_1 \vec{a}_1 + u_2 \vec{a}_2 + u_3 \vec{a}_3## lies within ##(h_1, h_2, h_3)## then ##\vec{u} \cdot \vec{n} = 0##, i.e.$$\frac{u_3}{h_1 h_2} (\vec{a}_1 \times \vec{a}_2) \cdot \vec{a}_3 - \frac{u_2}{h_1 h_3} (\vec{a}_1 \times \vec{a}_3) \cdot \vec{a}_2 - \frac{u_1}{h_3 h_2} (\vec{a}_3 \times \vec{a}_2) \cdot \vec{a}_1 = 0 $$ $$\begin{align*}\left[ \frac{u_3}{h_1 h_2} + \frac{u_2}{h_1 h_3} + \frac{u_1}{h_2 h_3} \right] (\vec{a}_1 \times \vec{a}_2) \cdot \vec{a}_3 = 0 &\implies\frac{u_3}{h_1 h_2} + \frac{u_2}{h_1 h_3} + \frac{u_1}{h_2 h_3} = 0\\ \\

&\implies u_3 h_3 + u_2 h_2 + u_1 h_1 = 0

\end{align*}$$It works, but it seems long winded. Does anyone have a better way?

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