I believe on moral grounds that the following three definitions are equivalent, and determine "the" uniform distribution on rotations in three dimensions.

  1. The Haar measure on $SO(3)$.

  2. The uniform distribution on the unit sphere in $\mathbb R^4$, modulo antipodal points, each point being interpreted as a quaternion determining a 3D rotation.

  3. Consider the distribution that arises from the following sampling scheme. Pick independently and uniformly a unit vector $\hat w$ in $\mathbb R^3$ and a real number $\theta$ in $[0, 2\pi)$. Define the corresponding rotation as (i) an arbitrary rotation that maps the $\hat z$ axis to $\hat w$, followed by (ii) a rotation by $\theta$ about $\hat w$. (The effect of the choice in (i) is smoothed away by the subsequent rotation in (ii).)

I would like to see an (ideally, elegant and succinct) proof or disproof of the equivalence of these three distributions. I can probably work through it myself by turning the calculus crank, but that would not be elegant.

  • $\begingroup$ Some rationale for scheme #3: The image of the $z$ axis under a random rotation should be uniformly distributed over the unit sphere, so we make that happen. This leaves one degree of freedom in the orientation of the $xy$ plane, so we pick that uniformly as well. $\endgroup$
    – user856
    Aug 18, 2012 at 19:44

1 Answer 1


For the equivalence of 1 and 2, we can show that (left or right) translation of the unit quaternions by a unit quaternion is an isometry of $S^3$:

Given vectors $x,y\in S^3$, translating them both by $a\in S^3$ in the sense of quaternion multiplication leads to the dot product

$$ \begin{align} (ax)\cdot(ay) &= \pmatrix{a_0x_0-\vec a\cdot\vec x\\a_0\vec x+x_0\vec a-\vec a\times\vec x} \cdot \pmatrix{a_0y_0-\vec a\cdot\vec y\\a_0\vec y+y_0\vec a-\vec a\times\vec y} \\ &=a_0^2x_0y_0+\vec a^2x_0y_0+a_0^2\vec x\cdot\vec y+(\vec a\cdot\vec x)(\vec a\cdot\vec y)+ (\vec a\times\vec x)(\vec a\times\vec y) \\ &=(a_0^2+\vec a^2)x_0y_0+(a_0^2+\vec a^2)\vec x\cdot\vec y \\ &=x_0y_0+\vec x\cdot\vec y \\ &=x\cdot y\;, \end{align} $$

so the translation is an isometry of $S^3$ and leaves the distribution 2 invariant. (I'm taking the more technical aspects of the Haar measure involving inner and outer regularity for granted in this context.)

If we split the rotations up as in 3, then for the distribution to be invariant under translation by arbitrary rotations, in particular the distribution of $\theta$ must be invariant under arbitrary rotations about $\hat w$, so $\theta$ must be uniformly distributed. Also the distribution of $\hat w$ clearly has to be rotationally symmetric about $\hat z$. Thus, it remains only to show that the uniform distribution of $\hat w\cdot\hat z$ in distribution 3 is correct.

I don't have an elegant argument without calculus for this, but here's a calculation that tries to do without too much brute force. For a rotation by $\alpha$ about $\hat u$, the rotated $z$ axis is

$$ \cos\alpha\,\hat z+\sin\alpha\,\hat u\times\hat z+(1-\cos\alpha)\hat u(\hat u\cdot\hat z)\;, $$

and forming the dot product with $\hat z$ yields a relationship between the three cosines,


where $a=\cos\alpha$, $b=\hat u\cdot\hat z$ and $c$ is the cosine between the original and rotated $z$ axes.

To calculate the cumulative distribution function for $c$, note that $c\ge a$ and that $b$ is uniformly distributed according to distribution 2, so we can solve for the upper limit of $b$,

$$ b=\sqrt{\frac{c-a}{1-a}}\;, $$

and find the density for $a$ according to distribution 2,

$$ \sin^2\frac\alpha2\mathrm d\alpha\propto\frac{1-\cos\alpha}{\sin\alpha}\mathrm d(\cos\alpha)=\frac{1-a}{\sqrt{1-a^2}}\mathrm da=\sqrt{\frac{1-a}{1+a}}\mathrm da\;, $$

to find the cumulative distribution function for $c$ proportional to

$$ \int_{-1}^c\sqrt{\frac{c-a}{1-a}}\sqrt{\frac{1-a}{1+a}}\mathrm da=\int_{-1}^c\sqrt{\frac{c-a}{1+a}}\mathrm da\;. $$

Now we can differentiate with respect to $c$ to get the density for $c$. The term from varying the integration limit vanishes since the integrand vanishes at $a=c$, so the result is proportional to

$$ \int_{-1}^c\frac1{\sqrt{(c-a)(1+a)}}\mathrm da=\left[2\arctan\sqrt{\frac{1+a}{c-a}}\right]_{-1}^c=\pi\;, $$

a constant, which establishes the equivalence between distributions 2 and 3.

  • $\begingroup$ Thanks for the nice answer! I see that the only tricky part is proving that a random rotation of a fixed unit vector is uniformly distributed over the sphere, which is useful to know. $\endgroup$
    – user856
    Aug 22, 2012 at 17:09

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