Misorientation Distributions and Symmetry

We now bring together all the tools developed so far — spherical geometry, rotation representations, and the Rodrigues space — to investigate misorientation distributions for objects with rotational symmetry, as in real crystals.

Point groups $1, 2, 3, 4, 6$

We begin with the simplest case: a single $n$-fold rotation axis. Specifically: “if many objects each having a single $n$-fold rotation axis are distributed in completely random orientations, what is the probability distribution of the misorientation angle $\theta$ between two such objects?” This is equivalent to fixing one object at the Rodrigues origin and finding the misorientation distribution with respect to the others. The orientation of the rotation axis in Rodrigues space (e.g., along the X-axis) can be chosen freely without loss of generality. As discussed on the previous page, rotational symmetry folds Rodrigues space so that points far from the origin become equivalent to nearby ones, and the region of interest must be restricted accordingly.

Point group $1$

For completeness, we repeat the result for the trivial case of point group $1$ (identity only). The sphere of radius $\rho=\|\mathbf{r}\|=\tan(\theta/2)$ centred at the Rodrigues origin has surface area $4\pi\rho^2$, and the $SO(3)$ density on it is $P(\rho)=\pi^{-2}(1+\rho^2)^{-2}$, giving
$$F(\rho)=\frac{4\rho^2}{\pi(1+\rho^2)^2}, \qquad F(\theta)=F(\rho)\frac{d\rho}{d\theta}=\frac{1-\cos\theta}{\pi}.$$
The graph¹ is shown on the right.

Misorientation distribution for point group $1$

Point groups $2, 3, 4, 6$

An $n$-fold rotation folds Rodrigues space into the region $|x|\le\tan(\pi/2n)$, as stated at the end of the previous page. The $SO(3)$ density $P(\rho)$ inside is multiplied by $n$ (the multiplicity, or group order) but its shape is unchanged. For $\rho\le\tan(\pi/2n)$ (equivalently $0\le\theta\le\pi/n$), the entire sphere lies within the region, so the result is simply $n$ times the point-group-1 formula.

Case $\rho\le\tan(\pi/2n)$²
(shown for $n=2$)

Case $\rho>\tan(\pi/2n)$
(shown for $n=2$)

When $\rho>\tan(\pi/2n)$, the sphere extends outside the region: the area of each protruding cap (shown in blue) must be subtracted. As shown on the Spherical Geometry page, each cap has area $2\pi\rho\{\rho-\tan(\pi/2n)\}$, and there are two such caps, so the area of the sphere inside the region (red) is
$$4\pi\rho^2 – 2\times2\pi\rho\left\{\rho-\tan\frac{\pi}{2n}\right\} = 4\pi\rho\tan\frac{\pi}{2n}.$$
The misorientation distribution function is then
$$F(\rho)=4\pi\rho\tan\frac{\pi}{2n}\,P_n(\rho)=\frac{4n\rho\tan(\pi/2n)}{\pi(1+\rho^2)^2},$$
$$F(\theta)=F(\rho)\frac{d\rho}{d\theta}=F(\rho)\frac{1+\rho^2}{2}=\frac{2n\rho}{\pi(1+\rho^2)}\tan\frac{\pi}{2n}=\frac{n}{\pi}\sin\theta\tan\frac{\pi}{2n},$$
where the folded density $P_n(\rho)=nP(\rho)$, and the final simplification uses standard trigonometric identities³. To summarise, for an object with a single $n$-fold rotation axis in a random orientation distribution:
$$F(\theta)=\frac{n}{\pi}(1-\cos\theta) \quad (\theta\le\pi/n),$$
$$F(\theta)=\frac{n}{\pi}\sin\theta\tan\frac{\pi}{2n} \quad (\pi/n<\theta<\pi).$$

Misorientation distribution graphs for point groups $2, 3, 4, 6$

The interval boundaries for each point group are summarised below.

Point group	Interval 1	Interval 2
$2$	$\rho\le 1$ or $\theta\le\pi/2=90^\circ$	$1<\rho$ or $\pi/2=90^\circ<\theta\le\pi=180^\circ$
$3$	$\rho\le\sqrt{3}/3$ or $\theta\le\pi/3=60^\circ$	$\sqrt{3}/3<\rho$ or $\pi/3=60^\circ<\theta\le\pi=180^\circ$
$4$	$\rho\le\sqrt{2}-1$ or $\theta\le\pi/4=45^\circ$	$\sqrt{2}-1<\rho$ or $\pi/4=45^\circ<\theta\le\pi=180^\circ$
$6$	$\rho\le 2-\sqrt{3}$ or $\theta\le\pi/6=30^\circ$	$2-\sqrt{3}<\rho$ or $\pi/6=30^\circ<\theta\le\pi=180^\circ$

In interval 1, $F(\theta)$ rises rapidly. Once the sphere protrudes beyond the boundary (interval 2), the increase slows, reaching a maximum at $\theta=\pi/2$, after which $F$ decreases and reaches zero at $\theta=\pi$.

Misorientation distribution for point group $2$

Misorientation distribution for point group $3$

Misorientation distribution for point group $4$

Misorientation distribution for point group $6$

Point groups $222, 32, 422, 622$

Next we consider point groups with one principal $n$-fold axis and $n$ perpendicular 2-fold axes. Multiple intervals in $\rho$ (or $\theta$) are needed. We label the sphere area and distribution function for each interval as $A_1,A_2,\ldots$ and $F_1,F_2,\ldots$ respectively.

Point group $222$

Point group $222$ has three mutually orthogonal 2-fold axes. Aligned with the X, Y, Z axes, the folded region is the cube $|x|\le1\;\&\;|y|\le1\;\&\;|z|\le1$. The multiplicity is 4, so the density function is uniformly 4 times larger.

Interval 1

For $\rho\le1$ (i.e., $0\le\theta\le\pi/2$) the entire sphere lies inside the cube⁴:
$$A_1(\rho)=4\pi\rho^2,$$
$$F_1(\theta)=F_1(\rho)\frac{d\rho}{d\theta}=\frac{4}{\pi}(1-\cos\theta)\quad(0<\theta\le\pi/2).$$

Interval 1: $\rho\le1$
(shown for $\rho=0.8$)

Interval 2

For $1<\rho\le\sqrt{2}$ (i.e., $\pi/2<\theta\le2\arctan\sqrt{2}$), the sphere protrudes through 6 faces of the cube. Subtracting 6 caps: $$A_2(\rho)=A_1(\rho)-6\times2\pi\rho(\rho-1)=12\pi\rho-2A_1(\rho),$$ $$F_2(\theta)=\frac{12\sin\theta}{\pi}-2F_1(\theta)\quad(\pi/2<\theta\le2\arctan\sqrt{2}).$$

Interval 2: $1<\rho\le\sqrt{2}$
(shown for $\rho=1.2$)

Interval 3

For $\sqrt{2}<\rho\le\sqrt{3}$ (i.e., $2\arctan\sqrt{2}<\theta\le2\pi/3$), overlapping caps must be accounted for. From the Spherical Geometry page, the overlap area of two orthogonal caps of equal half-angle $\Theta$ is
$$S=2\cos^{-1}(\cot^2\Theta)-4\cos\Theta\cos^{-1}(\cot\Theta).$$
There are 12 such overlaps (one per edge of the cube). Thus
$$A_3(\rho)=A_2(\rho)+12\left[2\rho^2\cos^{-1}(\cot^2\Theta)-4\rho\cos^{-1}(\cot\Theta)\right],$$
$$F_3(\theta)=F_2(\theta)+\frac{24}{\pi^2}\left[\cos^{-1}\!\left(-\frac{1+\cos\theta}{2\cos\theta}\right)(1-\cos\theta)-2\cos^{-1}\!\left(\sqrt{-\frac{1+\cos\theta}{2\cos\theta}}\right)\sin\theta\right]$$
$$\quad(2\arctan\sqrt{2}<\theta\le2\pi/3).$$

Interval 3: $\sqrt{2}<\rho\le\sqrt{3}$
(shown for $\rho=1.6$)

Point groups $32, 422, 622$

These point groups all share the same derivation structure. Each has a principal $n$-fold axis and $n$ perpendicular 2-fold axes. The folded region is a regular $2n$-gonal prism: distance from centre to top/bottom faces is $\tan(\pi/2n)$, and to side faces is 1. The group order is $2n$. Interval diagrams for each point group are shown below.

Intervals for point group $32$

Interval 1
$\rho\le\sqrt{3}/3$
($\rho=0.5$)

Interval 2
$\sqrt{3}/3<\rho\le1$
($\rho=0.8$)

Interval 3
$1<\rho\le2\sqrt{3}/3$
($\rho=1.1$)

Interval 4
$2\sqrt{3}/3<\rho\le\sqrt{5/3}$
($\rho=1.2$)

Intervals for point group $422$

Interval 1
$\rho\le\sqrt{2}-1$
($\rho=0.4$)

Interval 2
$\sqrt{2}-1<\rho\le1$
($\rho=0.6$)

Interval 3
$1<\rho\le\sqrt{4-2\sqrt{2}}$
($\rho=1.05$)

Interval 4
$\sqrt{4-2\sqrt{2}}<\rho\le\sqrt{7-4\sqrt{2}}$
($\rho=1.1$)

Intervals for point group $622$

Interval 1
$\rho\le2-\sqrt{3}$
($\rho=0.25$)

Interval 2
$2-\sqrt{3}<\rho\le1$
($\rho=0.4$)

Interval 3
$1<\rho\le\sqrt{6}-\sqrt{2}$
($\rho=1.02$)

Interval 4
$\sqrt{6}-\sqrt{2}<\rho\le\sqrt{15-8\sqrt{3}}$
($\rho=1.05$)

Intervals 1, 2, 3

Interval 1: the entire sphere lies inside the prism:
$$A_1(\rho)=4\pi\rho^2, \qquad F_1(\theta)=\frac{2n}{\pi}(1-\cos\theta).$$
Interval 2: only the top and bottom faces cause protrusions (analogous to the $n$-fold single-axis case):
$$A_2(\rho)=4\pi\rho\tan\frac{\pi}{2n}, \qquad F_2(\theta)=\frac{2n}{\pi}\sin\theta\tan\frac{\pi}{2n}.$$
Interval 3: the $2n$ side faces also cause protrusions:
$$A_3(\rho)=A_2(\rho)-2n\times2\pi\rho(\rho-1)=-nA_1(\rho)+A_2(\rho)+4n\pi\rho,$$
$$F_3(\theta)=-nF_1(\theta)+F_2(\theta)+\frac{2n^2}{\pi}\sin\theta.$$

Interval 4

In interval 4, overlaps between side-face caps and top/bottom caps must be accounted for. With half-angles $\Theta_1$ (side face) and $\Theta_2$ (top/bottom face):
$$\cos\Theta_1=\frac{1}{\rho},\quad\sin\Theta_1=\frac{\sqrt{\rho^2-1}}{\rho},\quad\cos\Theta_2=\frac{\tan(\pi/2n)}{\rho},\quad\sin\Theta_2=\frac{\sqrt{\rho^2-\tan^2(\pi/2n)}}{\rho}.$$
Since side and top/bottom faces are perpendicular, the overlap area is
$$S_1(\rho)=2\rho^2S_{11}(\rho)-2\rho S_{12}(\rho),$$
where
$$S_{11}(\rho)=\cos^{-1}\left(\frac{\tan\frac{\pi}{2n}}{\sqrt{(\rho^2-1)(\rho^2-\tan^2\frac{\pi}{2n})}}\right),\quad S_{12}(\rho)=\cos^{-1}\left(\frac{\tan\frac{\pi}{2n}}{\sqrt{\rho^2-1}}\right)+\tan\frac{\pi}{2n}\cos^{-1}\left(\frac{1}{\sqrt{\rho^2-\tan^2\frac{\pi}{2n}}}\right).$$
For side-face cap overlaps (angle $\alpha=\pi/n$, equal half-angle $\Theta_1$):
$$S_2(\rho)=2\rho^2S_{21}(\rho)-2\rho S_{22}(\rho),$$
where
$$S_{21}(\rho)=\cos^{-1}\left(\frac{1-\rho^2\cos\frac{\pi}{n}}{\rho^2-1}\right),\quad S_{22}(\rho)=2\cos^{-1}\left(\frac{1-\cos\frac{\pi}{n}}{\sqrt{\rho^2-1}\sin\frac{\pi}{n}}\right).$$
There are $4n$ overlaps of type $S_1$ and $2n$ of type $S_2$, so
$$A_4(\rho)=A_3(\rho)+4n\left[\rho^2\{2S_{11}+S_{21}\}-\rho\{2S_{12}+S_{22}\}\right],$$
$$F_4(\theta)=F_3(\theta)+\frac{2n^2}{\pi^2}\left[(1-\cos\theta)\{2S_{11}(\theta)+S_{21}(\theta)\}-\sin\theta\{2S_{12}(\theta)+S_{22}(\theta)\}\right],$$
with $\rho(\theta)=\tan(\theta/2)$.

Misorientation distribution graphs for point groups $222, 32, 422, 622$

Point group	Interval 1	Interval 2	Interval 3	Interval 4
$222$	$\rho\le1$ $\theta\le90^\circ$	$1<\rho\le\sqrt{2}$ $90^\circ<\theta\le109.47^\circ$	$\sqrt{2}<\rho\le\sqrt{3}$ $109.47^\circ<\theta\le120^\circ$
$32$	$\rho\le\sqrt{3}/3$ $\theta\le60^\circ$	$\sqrt{3}/3<\rho\le1$ $60^\circ<\theta\le90^\circ$	$1<\rho\le2\sqrt{3}/3$ $90^\circ<\theta\le98.21^\circ$	$2\sqrt{3}/3<\rho\le\sqrt{5/3}$ $98.21^\circ<\theta\le104.48^\circ$
$422$	$\rho\le\sqrt{2}-1$ $\theta\le45^\circ$	$\sqrt{2}-1<\rho\le1$ $45^\circ<\theta\le90^\circ$	$1<\rho\le\sqrt{4-2\sqrt{2}}$ $90^\circ<\theta\le94.53^\circ$	$\sqrt{4-2\sqrt{2}}<\rho\le\sqrt{7-4\sqrt{2}}$ $94.53^\circ<\theta\le98.42^\circ$
$622$	$\rho\le2-\sqrt{3}$ $\theta\le30^\circ$	$2-\sqrt{3}<\rho\le1$ $30^\circ<\theta\le90^\circ$	$1<\rho\le\sqrt{6}-\sqrt{2}$ $90^\circ<\theta\le91.99^\circ$	$\sqrt{6}-\sqrt{2}<\rho\le\sqrt{15-8\sqrt{3}}$ $91.99^\circ<\theta\le93.84^\circ$

In interval 1, $F(\theta)$ rises rapidly. When the sphere protrudes through the top/bottom faces (interval 2 for $32,422,622$), the increase becomes gradual. Protrusion through the side faces (interval 2 for $222$ or interval 3 for $32,422,622$) causes a rapid decrease. The final onset of cap overlaps (interval 3 for $222$ or interval 4 for $32,422,622$) slows the decrease until $F=0$.

Misorientation distribution for point group $222$

Misorientation distribution for point group $32$

Misorientation distribution for point group $422$

Misorientation distribution for point group $622$

Point groups $23, 432$

Finally, we treat the cubic point groups $23$ and $432$.

Point group $23$

For point group $23$, four 3-fold rotation axes fold Rodrigues space into a regular octahedron. The distance from the centre to each face is $\sqrt{3}/3$. The 2-fold axes along the octahedron vertices do not affect the shape. The multiplicity is 12, so the density is uniformly 12 times larger.

Interval 1
$\rho\le\sqrt{3}/3$
($\rho=0.5$)

Interval 2
$\sqrt{3}/3<\rho\le\sqrt{2}/2$
($\rho=0.65$)

Interval 3
$\sqrt{2}/2<\rho\le1$
($\rho=0.8$)

Intervals 1 and 2

Interval 1 (entire sphere inside octahedron):
$$A_1(\rho)=4\pi\rho^2, \qquad F_1(\theta)=\frac{12}{\pi}(1-\cos\theta).$$
Interval 2 (8 faces cause protrusions):
$$A_2(\rho)=A_1(\rho)-8\times2\pi\rho\left(\rho-\tan\frac{\pi}{6}\right)=-3A_1(\rho)+\frac{16}{\sqrt{3}}\pi\rho,$$
$$F_2(\theta)=-3F_1(\theta)+\frac{48}{\sqrt{3}\pi}\sin\theta.$$