2

I am reading through this notes on quaternions. I am trying to understand how they work as rotations as long as their norm is always 1.

First, in page 4 above the figure, the quaternion rotation operator is defined as follows

$$q = \cos(\theta) + u\sin(\theta)$$

But later, in equation 4 it is defined with half the angle: $$q = \cos(\frac{\theta}{2}) + u\sin(\frac{\theta}{2})$$ Is this a mistake or are they telling the same?

ElPotac
  • 31

2 Answers2

2

It doesn't matter if your quaternion has norm $1$ or not. It always induces a rotation on the subset of pure quaternions.

This is because if $q$ is nonzero, $qxq^{-1}=\frac{|q|}{|q|}qxq^{-1}=\frac{q}{|q|}x\left(\frac{q}{|q|}\right)^{-1}$ and of course $\frac{q}{|q|}$ is a unit quaternion.

So saying "it's a rotation as long as it's a unit quaternion" does not really carry any weight.

But choosing a unit-length representative reduces the number of quaternions being considered all the way down to two (instead of uncountably many.) In fact one can go a step further and insist that the real part is positive to get a single representative for a nontrivial rotation.

Strictly speaking, there is no mistake, but there is an awkwardness being caused by using a $\theta$ in both places. The question is "how is $\theta$ related to the angle of rotation caused by the unit quaternion?"

One could write $q=\cos(\theta)+u\sin(\theta)$ with $\theta\in [0,\pi/2]$ and rightly say "it rotates by an angle of $2\theta$".

OR one could have written $q=\cos(\theta/2)+u\sin(\theta/2)$ with $\theta\in [0,2\pi]$ to begin with and said "this rotates by an angle of $\theta$.

rschwieb
  • 153,510
  • Aah, I see. Thanks for the answer. Although i don't get why $\theta$'s domain is bounded to $[0, \pi]$ and not $[-\pi, \pi]$ or $[0, 2\pi]$ so to represent all quadrants. – ElPotac Jan 12 '21 at 19:53
  • @ElPotac I don't understand why you think it doesn't cover all quadrants (angles, I guess you mean.) In the first scheme, as $\theta$ ranges over $[0, \pi]$, you have the angle rotated $2\theta$ ranging over $[0,2\pi]$ which does cover all quadrants. (In the second scheme I described, it does range over all quadrants by ranging over $[0,2\pi]$.) – rschwieb Jan 12 '21 at 21:11
0

In addition to @rschwieb's explanation, there is always the possibility that two different actions of quaternions (on themselves...) are in play. One is simply by multiplication, the other by conjugation. The latter "doubles" a natural angular parameter...

In summary, one has to figure this out from context... and allow for typos.

paul garrett
  • 52,465