Rate of change of norm of a quaternion

Question

I am new to quaternions and currently trying to learn it. Here is my doubt. Let $\bf{q}$ be a quaternion whose norm is defined as $\|\bf{q}\| = \sqrt{q\otimes q^*}$. As you see, this norm is a scalar quantity. We'ld expect that the rate of change of this scalar quantity is also scalar (or, am I wrong in assuming that). The rate of change of this norm is computed as: \begin{equation} \begin{split} \frac{d}{dt}\|\bf{q}\| &= \frac{d}{dt}\left(\bf{q}\otimes \bf{q}^*\right)^{1/2}\\ &= \frac{1}{2}\frac{1}{\|\bf{q}\|}\left(\dot{\bf{q}}\otimes \bf{q}^* + \bf{q}\otimes \dot{\bf{q}^*}\right) \end{split} \end{equation} We see that the denominator is a scalar quantity, but since $\bf{q}$ is vector, the numerator is also vector, resulting into the rate of change of a scalar quantity being vector. Am I doing something wrong?

Answer 1

Too long for a comment and correction to one of my comments.

The author of the paper you cited in the comments uses $\mathbf{p}\otimes\mathbf{q}$ to denote the quaternion product. He writes each quaternion as $$\mathbf{q}=\begin{bmatrix}q_w\\\mathbf{q}_v\end{bmatrix}\,,\;q_w\in\mathbb R\,\,,\;\mathbf{q}_v\in\mathbb{R}^3\,.$$ Then

$$\tag{13}\mathbf{p}\otimes\mathbf{q}=\begin{bmatrix}p_wq_w-\mathbf{p}_v^\top\mathbf{q}_v\\[2mm] p_w\mathbf{q}_v+q_w\mathbf{p}_v+\mathbf{p}_v\times\mathbf{q}_v\end{bmatrix}\,.$$

The complex conjugate of a quaternion $\mathbf{q}$ flips the sign of its imaginary part $\mathbf{q}_v\,.$ Since $(-\mathbf{p}_v)\times(-\mathbf{q}_v)=\mathbf{p}_v\times\mathbf{q}_v=-\mathbf{q}_v\times\mathbf{p}_v$ we get from (13)

$$\tag{1} (\mathbf{p}\otimes\mathbf{q})^*=\mathbf{q}^*\otimes\mathbf{p}^*\,. $$ Differentiation of $\mathbf{q}\otimes\mathbf{q}$ gives the sum of

\begin{align}\tag{2} \dot{\mathbf{q}}\otimes\mathbf{q}^*\quad\text{ and }\quad\mathbf{q}\otimes\dot{\mathbf{q}}^* \end{align} which are both quaternions having possible non zero imaginary parts. In other words: in your language they are vectors in general.

From (1) $$ (\dot{\mathbf{q}}\otimes\mathbf{q}^*)^*=\mathbf{q}\otimes \dot{\mathbf{q}}^*\,. $$ That is: the "vectors" in (2) are conjugate to each other. Therefore their sum is the sum of their scalar parts and hence a scalar as it must.