2
$\begingroup$

I am new to the topic of quaternions, but I have an excellent understanding of complex numbers and linear algebra.

I read that unitary quaternions (or versors) can be written in the form: $q = \exp(a\textbf{r}) = \cos(a) + \textbf{r}\sin(a)~,$ where $r^2 = -1$ and has no real part. This is similar to the complex formalism that is: $c = A \exp(i\theta) = A[cos(\theta) + i\sin(\theta)]$.

Can any quaternion be written in the same manner, by extending the formulation to: $q = A\exp(a\textbf{r}) = A[\cos(a) + \textbf{r}\sin(a)]$ ? If so, does the multiplication still work the same?

I have been trying to perform a quaternion rotation described here by hand, but I'm puzzeled by how to handle commutativity...

$\endgroup$
4
  • 2
    $\begingroup$ Certainly any quaternion can be written $q=Aexp(ar)$ where $r^2=-1$. I am not sure I understand what does it mean "the multiplication still work the same". $\endgroup$
    – asv
    Dec 10, 2013 at 16:46
  • $\begingroup$ Look at: mathoverflow.net/questions/17264/… $\endgroup$ Dec 10, 2013 at 16:50
  • $\begingroup$ @semyonalesker If quaternions can be written that way, then multiplying $q_1=A_1\exp(a_1 r_1)$ and $q_2$ (defined the same) into $q_1\cdot q_2 = A_1 A_2 exp(a_1 r_1 + a_2 r_2)$. Same thing if I multiply three quaternions. Then, any rotation operation defined with $p' = qpq^{-1}$ would be equal to $p$ (because arguments inside the exponential would simply cancel themselves). $\endgroup$
    – PhilMacKay
    Dec 10, 2013 at 17:06
  • 2
    $\begingroup$ After a quick examination, maybe this should have been posted on math.stackoverflow. I didn't know these were different sites... $\endgroup$
    – PhilMacKay
    Dec 10, 2013 at 17:13

1 Answer 1

2
$\begingroup$

Multiplication ... Note $$ \exp(a_1\textbf{r}_1) \exp(a_2\textbf{r}_2)= \exp(a_1\textbf{r}_1+a_2\textbf{r}_2) $$ provided $\textbf{r}_1$ and $\textbf{r}_2$ commute, which is not always true.

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.