the logarithm of quaternion

Question

i'm reading 3D math primer for graphics and game development By Fletcher Dunn and Ian Parberry, on page 170, the logarithm of quaternion is defined as

\begin{align} \log \mathbf q &= \log \left( \begin{bmatrix} \cos \alpha & \mathbf n \sin \alpha \end{bmatrix} \right) \\ &\equiv \begin{bmatrix} 0 & \alpha \mathbf n \end{bmatrix} \end{align}

i don't see how does $\log \left( \begin{bmatrix} \cos \alpha & \mathbf n \sin \alpha \end{bmatrix} \right)$ equals to $\begin{bmatrix} 0 & \alpha \mathbf n \end{bmatrix}$, can anyone help me out?

thanks.

Answer 1

I can't see the page in Google books, but what you apparently have there is the logarithm of a unit quaternion q, which has scalar part $\cos(\theta)$ and vector part $\sin(\theta)\vec{n}$ where $\vec{n}$ is a unit vector.

Since the logarithm of an arbitrary quaternion $q=(s\;\;v)$ is defined as

$\ln q=(\ln\,|q|\;\;\frac{v}{\|v\|}\arccos\frac{s}{|q|})$

where $|q|$ is the norm of the quaternion and $\|v\|$ is the norm of the vector part, applying that formula to a unit quaternion yields a scalar part of 0 (the logarithm of the norm of a unit quaternion is zero), and you should now be able to derive the formula for the vector part.

Answer 2

Just recall $\exp(\alpha i) = \cos \alpha + i \sin \alpha$ for complex numbers, the quaternion (remember a quaternion is just 3 complex numbers which all have the same real part) version is by direct analogue and take logarithm of both sides.