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If all components of a unit quaternion (also known as versor) are multiplied by -1, so it still remains a versor, does the resulting versor is considered equivalent to the original versor?

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  • $\begingroup$ Surely it depends on what you mean by "equivalent". If you're using them only for rotations via x -> q^(-1)xq, then multiplying q by -1 doesn't change what it does. If you're doing something else with them, a factor of -1 may matter after all. $\endgroup$ – Gareth McCaughan May 18 '16 at 15:29
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If we think about rotation quaternion which are also unit quaternion then multiplying it with $-1$ will result it in additional $2\pi$ rotation and in consequence this will not affect original rotation hence it is equivalent to original unit quaternion. $$q = \cos{\frac{\theta}{2}}+\sin{\frac{\theta}{2}}\frac{\vec{u}}{\|\vec{u}||}$$$$-q= \cos{(\frac{\theta}{2}+\pi)}+\sin{(\frac{\theta}{2}+\pi)}\frac{\vec{u}}{\|\vec{u}||}$$ hence if $q$ is rotates through $\theta$ around $\vec{u}$ then $-q$ rotates $\theta +2\pi$.

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Equivalent for what purpose?

It is true that if $q$ is a unit quaternion, then $q$ and $-q$ represent the same rotation of 3-dimensional space.

But, a continuous loop of rotations around a fixed vector of $\mathbb{R}^3$ may start at $q$ and end at $-q$. Choose the fixed vector to be $v=(1,0,0)$; then this quaternion represents rotation by $\theta$ around $v$: $$q_{\theta} = \cos\left(\frac{\theta}{2}\right) + i \sin \left(\frac{\theta}{2}\right).$$ Note that $q_{2\pi} = -q_0$, so one loop around from $\theta=0$ and $\theta=2\pi$ takes us to the negative of where we started. We cannot deform this continuous loop to both start and end at $q_0$. So in this sense, $q_0$ and $-q_0$ are not equivalent.

This is similar to the question, "In the complex plane, are the angles $\theta=0$ and $\theta=2\pi$ equivalent angles?" Sure, they are both the same ray in the complex plane, but if you make a counterclockwise loop around the origin of the complex plane, and you start at $\theta=0$, then you must end at $\theta=2\pi$ (not 0).

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