I'm trying to verify and plot some "strange attractors" based on quaternions. I found a paper (The Quaternion Phi Spiral Interference Patterns) and was trying to verify and understand the thought process, but I'm having trouble understanding how quaternions can be presented in terms of sine and cosine, can someone explain this a little better.
I did try and experiment with plotting quaternions just to see the different effects it has on objects .
I've also made some plots based on his theories but I'm still unsure about quaternions being presented in terms of sine and cosine, is this his process correct? link to the animated images and plots I generated based on his theory
Here's the quote in the paper:
" Although the algebraic representation of the quaternion is widely used in modern engineering, the explicit geometrical representation (akin to the complex numbers) is often overlooked. For a normal complex number $z = w + ix = \mu \cos \theta + i \mu \sin \theta$ i.e. \begin{align} w &= \mu \cos θ\\ x &= \mu \sin θ \end{align} A quaternion can also be presented in terms of sine and cosine functions but in this case the imaginary part is a vector in $x$, $y$ and $z$ and the non-commutative relations between I, j and k give rise to: \begin{align} w &= \mu \cos θ\\ x &= \mu \sin θ \cos φ\\ y &= \mu \sin θ \sin φ \cos ψ\\ z &= \mu \sin θ \sin φ \sin ψ \end{align} Therefore $$ q = \mu (\cos θ + \sin θ (i \cos φ +\sin φ (j \cos ψ + k \sin ψ)))" $$
An example image of what is generated.
The trick is to understand the general square root in the quaternions is $bi+cj+dk$ with$$b,\,c,\,d\in\Bbb R,\,b^2+c^2+d^2=1.$$Since $l:=j\cos\psi+k\sin\psi$ is a square root of $-1$ anticommuting with $i$, and$$AB+BA=0\implies(A+B)^2=A^2+B^2,$$the definition $z:=i\cos\phi+l\sin\phi$ obtains$$z^2=-\cos^2\phi-\sin^2\phi=-1,$$so $\mu\cos\theta+z\mu\sin\theta=\mu\exp z\theta$ is isomorphic to the modulus-$|\mu|$ complex number $\mu\exp i\theta$. This argument is a nice motivation for the $2$-angle form of spherical polar coordinates using $\phi,\,\psi$, since we associate elements of $\Bbb R^3$ with the case $\cos\theta=0$.