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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 $\log \left( \begin{bmatrix} \cos \alpha & \mathbf n \sin \alpha \end{bmatrix} \right)$ is equal to $\begin{bmatrix} 0 & \alpha \mathbf n \end{bmatrix}$. Can anyone help me out?

Thanks.

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2 Answers 2

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.

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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.

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