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I often see claims that the negative of a quaternion represents the same rotation, just that the axis and angle have both been reversed. However, if I look at the axis-angle representation of the quaternion $q=[\cos(\theta/2),\vec{n}\sin(\theta/2)]$ and then reverse the rotation angle $\theta=-\theta$ and rotation axis $\vec{n}=-\vec{n}$ and plug it in I get $$q(-\theta,-\vec{n})=[\cos(-\theta/2),-\vec{n}\sin(-\theta/2)]=[\cos(\theta/2),\vec{n}\sin(\theta/2)]$$ but this is not the same as $-q=[-\cos(\theta/2),-\vec{n}\sin(\theta/2)]$. I must be missing something obvious but cannot figure it out. To me, negating the axis and angle gives $q$ not $-q$, so where am I going wrong?

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Your calculations are correct.

However, if you replace $\theta$ by $\theta + 2\pi$ (which is equivalent to $\theta$), then you will get $-q$.

In your example, if you replace $n$ by $-n$, and $\theta$ by $-\theta + 2\pi$ (which is equivalent to $-\theta$), then you will also get $-q$. That's where the claims are coming from.

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  • $\begingroup$ Ah ok I was going to ask where the claims came before your edit, but this makes sense now. Thank you! $\endgroup$ – compmatsci Nov 16 '16 at 5:46
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In addition to what Ted said that statement may mean the following. The rotation $R$ related to quaternion $q$ sends a vector $\vec{v}$ to the vector $$R(\vec{v})=q\,\vec{v}\,\overline{q}.$$ Because $\overline{-q}=-\overline{q}$, the rotation $R'$ related to $-q$ sends the vector $\vec{v}$ to $$ \begin{aligned} R'(\vec{v})&=(-q)\,\vec{v}\,\overline{-q}\\ &=(-1)^2q\,\vec{v}\,\overline{q}\\ &=q\,\vec{v}\,\overline{q}\\ &=R(\vec{v}). \end{aligned} $$ The multiplication of quaternions is bilinear, so we can move those $-1$ factors to the front when they cancel each other.

Anyway, the calculation shows that $R$ and $R'$ are the same rotation.

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