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This question is somewhat similar to this, but in higher dimension.

I have two non-collinear unit vectors in $\mathbb{R}^4$, $\bf a$ and $\bf b$. I know from Wikipedia that there is at least one pair of quaternions $Q_L$ and $Q_R$ that is able to rotate $\bf a$ to $\bf b$, i.e.: $$ {\bf b} = Q_L {\bf a} Q_R $$ (where multiplication is the Hamilton product)

Questions:

  1. What is an easy to understand and fast to compute way to find $Q_L$ and $Q_R$?
  2. What are the rotation quaternions that moves $\bf a$ towards $\bf b$, but by a given angle $\theta$ ?

Edit: the comments made me realize I need further constraints to get the rotation I need. So, $\bf a$ and $\bf b$ spans a subspace. The rotation I need is closed in this subspace. I think this restriction reduces the number of possible solutions from infinite to two, (as there are two pairs of quaternions that describes the same rotation).

lvella
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  • If you view $a,b$ as quaternions, then $b=(ba^{-1})a(1)=(1)a(a^{-1}b)$. There are infinitely many pairs that work (just like in 3D there are infinitely many rotations that take one unit vector to another). – anon May 19 '20 at 23:51
  • @runway44 Well, that was too fast. If I restrict the problem to rotations that doesn't spin $\bf a$ around its own axis, do I manage to get only two solutions? What is a better term? "Minimum energy rotation?" – lvella May 20 '20 at 00:02
  • There are no "axes," that's not how 4D rotations work. (Even in 3D I'm not sure what "spin a vector around its own axis" is supposed to mean.) Any 4D rotation is a pair of simultaneous 2D rotations in two orthogonal 2D planes. But a "minimum energy rotation" likely makes sense. – anon May 20 '20 at 00:22

1 Answers1

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Let $a,b$ be two unit quaternions, for simplicity $a,b\ne\pm1$ and $b\ne\pm a$.

There are many pairs $(p,q)$ of quaternions for which $b=paq$, e.g. $(ba^{-1},1)$ or $(1,a^{-1}b)$.

You can even parametrize them all by solving for $p$ in terms of $q$ or vice-versa.

There is a unique pair (up to $\pm$) for which $F(x)=pxq$ acts as a 2D rotation in the plane spanned by $a,b$ but fixes all points in the plane orthogonal to it, and that is $(\sqrt{ba^{-1}},\sqrt{a^{-1}b})$.

The square roots are defined by halving the convex angle of polar form, i.e. $\sqrt{e^{\theta\mathbf{u}}}=e^{(\theta/2)\mathbf{u}}$ if $\theta\in[0,\pi)$; in practice you can calculate this using half-angle formulas on real/imag parts:

$$ \sqrt{r+\mathbf{v}}=\sqrt{\frac{1+r}{2}}+\sqrt{\frac{1-r}{2}}\frac{\mathbf{v}}{\|\mathbf{v}\|}. $$

(I define quaternions as sums of scalars $r$ and 3D vectors $\mathbf{v}$; these are real/imaginary parts.)

Here's how to get the formula. First define the left/right multiplication maps $L_p(x)=px$ and $R_p(x)=xp$. If we can create a rotation $G(x)$ which sends $1$ to $a^{-1}b$ and is a nonrotation in the complent, then we can define $F=L_a\circ G\circ L_a^{-1}$. The effect of the $L_a^{-1}$ is to send the $ab$-plane to the $1(a^{-1}b)$-plane, and the former's complement to the latter's complement. After we rotate purely in the $1(a^{-1}b)$-plane with $G$ we go back to the $ab$-plane with $L_a$.

If $p=e^{\theta\mathbf{u}}$ then $L_p$ and $R_p$ both act as a rotation by $\theta$ in the $1\mathbf{u}$-plane and its complement, the only difference being opposite directions in the complement (this follows algebraically from the fact orthogonal vectors anticommute). Therefore if we were to compose $L_{\sqrt{p}}$ and $R_{\sqrt{p}}$ we get a rotation from $1$ to $p$ in the $1\mathbf{u}$-plane and a nonrotation in the complement.

Thus $F(x)=a\big(\sqrt{a^{-1}b}(a^{-1}x)\sqrt{a^{-1}b}\big)$. Let $q=a\sqrt{a^{-1}b}a^{-1}$. Then $q^2=ba^{-1}$ so $q=\pm\sqrt{ba^{-1}}$ (square roots of nonreal quaternions are unique up to $\pm$; you can prove this with polar form). Since $\sqrt{a^{-1}b}$ has positive real part, so does its conjugate $q$, so $q$ must be $+\sqrt{ba^{-1}}$.

Conclude $F(x)=\sqrt{ba^{-1}}x\sqrt{a^{-1}b}$.

A more symmetric version of the above is how I intuited the formula: $L_{\sqrt{ba^{-1}}}$ and $R_{\sqrt{ab^{-1}}}$ both rotate $a$ halfway to $b$, but rotate in opposite directions in the complement.

anon
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