Formulas to represent band edges in Kikuchi pattern
[English / Japanese]

Herein, the origin is considered as the projection center, and the axes and the unit of the length is fixed so that the phosphor screen is included in the plane z=1. The following vectors are used in the following discussion:

- ${\mathbf z} = (0, 0, 1)^T$,
- a reciprocal lattice vector ${\mathbf a}^*$,
- ${\mathbf u} = {\mathbf a}^* / | {\mathbf a}^* |$.

The band center line in a Kikuchi pattern is the intersection of the phosphor screen ${\mathbf z} \cdot {\mathbf x} = 1$ and the diffracting plane ${\mathbf a}^* \cdot {\mathbf x} = 0$. The Bragg equation $2 d \sin \theta = n \lambda$ ($n=1, 2, \ldots$) provides the relation between the Bragg angle θ and the d-spacing $d = 1/|{\mathbf a}^* |$.

For any point ${\mathbf x}$ of the Kossel Cone as in the following figure, $({\mathbf x} \cdot {\mathbf u}) {\mathbf u}$ is the projection of ${\mathbf x}$ in the direction of ${\mathbf a}^*$, and ${\mathbf x} - ({\mathbf x} \cdot {\mathbf u}) {\mathbf u}$ is the projection in the plane orthogonal to ${\mathbf a}^*$. Hence, their lengths satisfy the right-hand equality:

| $$ \frac{ | ({\mathbf u} \cdot {\mathbf x}) {\mathbf u} |^2 }{ | {\mathbf x} - ({\mathbf u} \cdot {\mathbf x}) {\mathbf u} |^2 } = \tan^2 \theta. $$
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Therefore, the Kossel cone is a conical surface defined by: $$ ({\mathbf u} \cdot {\mathbf x})^2 = | {\mathbf x} |^2 \sin^2 \theta. $$ The band edges are the intersections of the Kossel cone and ${\mathbf z} \cdot {\mathbf x} = 1$. For simplicity, we rotate the coordinate axes of the phosphor screen, and assume ${\mathbf u} = (-\cos \sigma, 0, \sin \sigma)$. In this case, from ${\mathbf u} \cdot {\mathbf x} = 0$, the equation of the band center line is provided by: $$ x = \tan \sigma. $$ As for the equation of the band edges, we have: $$ (-x \cos \sigma + \sin \sigma)^2 = (x^2 + y^2 + 1) \sin^2 \theta. $$ As a result, the band edges are represented as the following hypebola: $$ (\cos^2 \sigma-\sin^2 \theta) \left( x - \frac{ \cos \sigma \sin \sigma }{\cos^2 \sigma-\sin^2 \theta} \right)^2 - (\sin^2 \theta) y^2 = \frac{ \cos^2 \theta \sin^2 \theta }{\cos^2 \sigma-\sin^2 \theta}. $$ The above is also equivalent to: $$ \left( \frac{\cos 2 \sigma + \cos 2 \theta}{ \sin 2 \theta } \right)^2 \left( x - \frac{ \sin 2 \sigma }{ \cos 2 \sigma + \cos 2 \theta } \right)^2 - \frac{ \cos 2 \sigma + \cos 2 \theta }{ 2 \cos^2 \theta } y^2 = 1. $$ In the case of EBSD images, $\cos 2 \sigma + \cos 2 \theta = \cos 2 \sigma + \cos 2 \theta > 0$ holds. The band width equals the distance between the two intersections $\sigma_{begin}, \sigma_{end}$ of the $x$-axis and the band edges: Furthermore, we have: \begin{eqnarray*} \tan \sigma_{begin} &=& \frac{ \sin 2\sigma - \sin 2 \theta }{ \cos 2 \sigma + \cos 2 \theta } = \frac{ (e^{i(\sigma + \theta)} + e^{-i(\sigma + \theta)})(e^{i(\sigma - \theta)} - e^{-i(\sigma - \theta)}) } { i (e^{i(\sigma + \theta)} + e^{-i(\sigma + \theta)})(e^{i(\sigma - \theta)} + e^{-i(\sigma - \theta)}) } = \tan (\sigma - \theta), \\ \tan \sigma_{end} &=& \frac{ \sin 2\sigma + \sin 2 \theta }{ \cos 2 \sigma + \cos 2 \theta } = \frac{ (e^{i(\sigma + \theta)} - e^{-i(\sigma + \theta)})(e^{i(\sigma - \theta)} + e^{-i(\sigma - \theta)}) } { i (e^{i(\sigma + \theta)} + e^{-i(\sigma + \theta)})(e^{i(\sigma - \theta)} + e^{-i(\sigma - \theta)}) } = \tan (\sigma + \theta). \end{eqnarray*} As a result, $\sigma_{begin} = \sigma - \theta$, $\sigma_{end} = \sigma + \theta$ are also obtained. [Back to top] [Go to the page of the EBSD ab-initio indexing software] [Go to my research page (list of software)]