A better way of proving the Weiss Zone Law?

[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?

 

[Office_Shredder]

I would just use the definition of the plane and the definition of being a direction in the plane directly. If u is a direction in the plane then it has to be the difference of two points in the plane. You know exactly one property those points need to satisfy. Can you do anything with it?

 

[Charles Link]

I think the vector $\vec{v}=h_1 \vec{a}_1 + h_2 \vec{a}_2 + h_3 \vec{a}_3$ is normal to the plane(s) regardless of whether the coordinates are orthogonal or not. If you write the equation for the plane passing through the origin that is parallel to the set of planes, the zone law follows immediately.
Edit: Upon further consideration, I think the first sentence may be in error, but the methodology in post 4 should still work.

 

[Charles Link]

To add to post 3, note one parallel plane will have the form $h_ 1 u_1+h_2 u_2 +h_3 u_3=1$, where I'm using the $u's$ as coordinates. Note when $u_1= 1/h_ 1$, then $u_2=0$, and $u_3=0$ for the axis intercept. Similarly for $u_2$ and $u_3$ intercepts. I think you only need to look at the plane that is parallel to this one that passes through the origin for the result you need.

 

[etotheipi]

I would just use the definition of the plane and the definition of being a direction in the plane directly. If u is a direction in the plane then it has to be the difference of two points in the plane. You know exactly one property those points need to satisfy. Can you do anything with it?

Thanks, that works! I interpreted that to mean, for some $\alpha, \beta$,$$u_1 \vec{a}_1 + u_2 \vec{a}_2 + u_3 \vec{a}_3 = \alpha \vec{v}_1 + \beta \vec{v}_2 = \alpha \left( \frac{\vec{a}_3}{h_3} - \frac{\vec{a}_1}{h_1} \right) + \beta \left( \frac{\vec{a}_3}{h_3} - \frac{\vec{a}_2}{h_2} \right)$$if the set $\{ \vec{a}_1, \vec{a}_2, \vec{a}_3 \}$ is linearly independent then we can deduce$$u_1 = -\frac{\alpha}{h_1}, \quad u_2 = -\frac{\beta}{h_2}, \quad u_3 = \frac{1}{h_3}(\alpha + \beta)$$from which it immediately follows that
$$u_1 h_1 + u_2 h_2 + u_3 h_3 = -\alpha - \beta + (\alpha + \beta) = 0$$

I think the vector $\vec{v}=h_1 \vec{a}_1 + h_2 \vec{a}_2 + h_3 \vec{a}_3$ is normal to the plane(s) regardless of whether the coordinates are orthogonal or not.

I think this only holds in a cubic lattice, for the direct lattice vectors. On the other hand for the reciprocal lattice vectors $\{\vec{b}_1, \vec{b}_2, \vec{b}_3 \}$, it is always true that the plane $(h_1 \, h_2 \, h_3)$ is orthogonal to $h_1 \vec{b}_1 + h_2 \vec{b}_2 + h_3 \vec{b}_3$. There is some justification here: https://en.wikipedia.org/wiki/Miller_index

 

[Charles Link]

Yes, I think my statement of the vector always being normal is incorrect. (I will mark it with an "Edit" in post 3). In any case, I think my post 4 still applies, where parallel planes can be used to come up with the Weiss zone law formula. The direction vector's coefficients become the coordinates when the plane passes through the origin.

 

[etotheipi]

In case anyone was interested, I realized earlier that this law is actually pretty trivial if we use the reciprocal basis. The plane with Miller indices $(h_1 h_2 h_3)$ is orthogonal to $h_1 \vec{a}_1^* + h_2\vec{a}_2^* + h_3 \vec{a}_3^*$, where $\vec{a}_i^* = \frac{\vec{a}_j \times \vec{a}_k}{[\vec{a}_i, \vec{a}_j, \vec{a}_k]}$ are the reciprocal basis. The zone law follows from constraining$$(u_1\vec{a}_1 + u_2\vec{a}_2 + u_3 \vec{a}_3) \cdot (h_1\vec{a}_1^* + h_2\vec{a}_2^* + h_3\vec{a}_3^*) = 0$$and we can simply use that $\vec{a}_i \cdot \vec{a}_j^* = \delta_{ij}$ to end up with$$u_1 h_1 + u_2 h_2 + u_3 h_3 = 0$$