I was trying to take the following derivative:


from page $230$, section $6$ of the article Interpreting the Kustaanheimo–Stiefel transform in gravitational dynamics, with the help of the definitions from the artice : A Quaternion Gradient Operator and Its Applications, page $49$, equation $14$

with respect to the quaternion components, where $\boldsymbol{Q_1}=\dfrac{x\boldsymbol{i}+y\boldsymbol{j}+Z\boldsymbol{k}}{\sqrt{2Z}}$

with $Z=z+\sqrt{x^2+y^2+z^2}$

I then proceeded with the quaternion multiplication and got to:


and now I have to take the partial derivatives $\dfrac{\partial}{\partial q_a}\boldsymbol{Q_1}^2$, with $a$ from $1$ to $3$

where $q_1=\dfrac{x}{\sqrt{2Z}}$, $q_2=\dfrac{y}{\sqrt{2Z}}$, $q_3=\dfrac{Z}{\sqrt{2Z}}$

Unfortunately I can't imagine a procedure to go any further with my calculations, and I am stuck at this point.

Any help would be appreciated.


1 Answer 1


$ \newcommand\q\mathbf \newcommand\grade[1]{\langle#1\rangle} \newcommand\PD[2]{\frac{\partial#1}{\partial#2}} \newcommand\re{\mathrm{re}} $


I realized that my original post likely didn't address your actual question: how to compute $\PD{}{q_a}\q Q_1^2$, where I assume you mean $a=1,2,3$. With partial derivatives, we always have to keep in mind what is varying and what is being held constant. In this case, since $(x, y, z)$ is one set of variables and $(q_1, q_2, q_3)$ is another, I would assume in taking a derivative like $\PD{}{q_1}\q Q_1^2$ that $q_2, q_3$ are held constant and $x, y, z$ are allowed to vary as function of $q_1, q_2, q_3$. So then, since $\q Q_1 = q_1\q i + q_2\q j + q_3\q k$, we see $$ \PD{\q Q_1}{q_1} = \PD{}{q_1}(q_1\q i + q_2\q j + q_3\q k) = \q i $$ and so we get $$ \PD{}{q_1}\q Q_1^2 = \PD{\q Q_1}{q_1}\q Q_1 + \q Q_1\PD{\q Q_1}{q_1} = \q i\q Q_1 + \q Q_1\q i = \frac12\re[\q i\q Q_1], $$ the last equality following because both $\q i$ and $\q Q_1$ are pure imaginary. In the case that you want to compute something like $\PD{}x\q Q_1^2$, first we note that $$ q_1 = \frac x{2\sqrt z},\quad q_2 = \frac y{2\sqrt z},\quad q_3 = \frac12\sqrt z, $$ and then by the chain rule we get $$ \PD{}x\q Q_1^2 = \sum_{a=1}^3\PD{q_a}x\PD{}{q_a}\q Q_1^2 = \frac1{2\sqrt z}\PD{}{q_1}\q Q_1^2 = \frac1{4\sqrt z}\re[i\q Q_1]. $$

Addendum to my original post:

Having spent some time looking at your second link, I think I understand what is going on now. Firstly, the operator $\nabla_{\q Q_1}$ from your first link is not the gradient $\nabla$ from your second link. It can't be, since in your second link the gradient is a 4D vector of quaternions (i.e. an element of $\mathbb H^4$) and the identity $\nabla_{\q Q_1}\q Q_1^2 = 2\q Q_1$ wouldn't make any sense. What the operator $\nabla_{\q Q_1}$ appears to correspond to is the $\PD{}q$ from your second link: $$ \nabla_{\q Q_1} = \PD{}q = \frac14\left[q_0 - \q i\PD{}{q_1} - \q j\PD{}{q_2} - \q k\PD{}{q_3}\right], $$ where I've used $0,1,2,3$ instead of their $a,b,c,d$. This is exactly $\frac14$ times the even multivector derivative of $\mathrm{Cl}_3(\mathbb R)$ I refer to below. However, as I also pointed out below, in this case the identity $\nabla_{\q Q_1}\re[\q Q_1^*\q A] = A$ is incorrect, and should instead be $\nabla_{\q Q_1}\re[\q Q_1^*\q A] = A^*$. In fact, the only consistent possibilities are (for imaginary $\q Q_1$) $$ \nabla_{\q Q_1}\q Q_1^2 = 2\q Q_1,\quad \nabla_{\q Q_1}\re[\q Q_1^*\q A] = A^* $$ or $$ \nabla_{\q Q_1}\q Q_1^2 = 2\q Q_1^*,\quad \nabla_{\q Q_1}\re[\q Q_1^*\q A] = A $$ in light of the fact that $\q Q_1^2 = \re[\q Q_1\q Q_1]$. (And what these correspond to is taking $\re[\q A\q B]$ or $\re[\q A^*\q B]$ as our desired inner product.)

On a tangent, I think care needs to be taken with the $\PD{}q, \PD{}{q^i},\PD{}{q^j},\PD{}{q^k}$ from your second link. While it is indeed true that, for example, $\PD q{q^i} = \PD{q^i}q = 0$, they are not independent, for consider $$ \PD{}qiq^i = -\PD{}qi^2qi = \PD{}qqi = i. $$ As far as I can tell, the authors do not comment on this.

Original post:

I'm having a hard time figuring out how $\nabla_{\q Q_1}$ is being defined, but if it's reasonable enough then this should be on the right track.

We need only consider arbitrary pure imaginary $\q Q_1$. The trick is that $\q Q_1^2 = \re[\q Q_1^2]$ and $\re[\q A\q B] = \re[\q B\q A]$. Hence, using the product rule and dots to indicate what variables are being differentiated, $$ \nabla_{\q Q_1}\q Q_1^2 = \nabla_{\q Q_1}\re[\q Q_1^2] = \dot\nabla_{\q Q_1}\re[\dot{\q Q_1}\q Q_1] + \dot\nabla_{\q Q_1}\re[\q Q_1\dot{\q Q_1}] = 2\dot\nabla_{\q Q_1}\re[\dot{\q Q_1}\q Q_1]. $$ However, the second identity given in your first link, $\nabla_{\q Q_1}\re[\q Q_1^*\q A] = \q A$, would then suggest that $$ \nabla_{\q Q_1}\q Q_1^2 = 2\dot\nabla_{\q Q_1}\re[\dot{\q Q_1}\q Q_1] = 2\q Q_1^* = -2\q Q_1 $$ This appears to be an inconsistency. Again, on one hand I might not be understanding how $\nabla_{\q Q_1}$ is defined, but on the other hand I didn't assume very much. If we define $\nabla_{\q Q_1}$ for $\q Q_1 = q_0 + q_1\q i + q_2\q j + q_3\q k$ as $$ \nabla_{\q Q_1} = \PD{}{q_0} - \q i\PD{}{q_1} - \q j\PD{}{q_2} - \q k\PD{}{q_3}, $$ then I am certain that my derivation is correct up until application of the $\nabla_{\q Q_1}\re[\q Q_1^*\q A] = \q A$ identity. In fact, in this case I would say that $\nabla_{\q Q_1}\q Q_1^2 = 2\q Q_1$ is correct and $\nabla_{\q Q_1}\re[\q Q_1^*\q A] = \q A$ is wrong.

I should point out that $\re[\q Q_1^*\q A]$ is exactly the Euclidean inner product on $\q Q_1$ and $\q A$ considered as 4D vectors, and the two identities $\nabla_{\q Q_1}\q Q_1^2 = 2\q Q_1$ and $\nabla_{\q Q_1}\re[\q Q_1^*\q A] = \q A$ are identities of the vector derivative in geometric calculus, calculus done over real Clifford algebras (but the $\q Q_1^2$ identity would require using the Clifford product instead of the quaternion product). See Chapter 2 of Clifford Algebra to Geometric Calculus (1984) by David Hestenes and Garret Sobczyk, or Chapter 6 of Geometric Algebra for Physicists (2003) by Chris Doran and Anthony Lasenby; I recommend the latter as an introduction.

What does work with quaternion multiplication in this framework, though, is the case $\nabla_{\q Q_1}\re[\q Q_1^*\q A] = \q A^*$. This corresponds to treating quaternions as elements of the even subalgebra of $\mathrm{Cl}_3(\mathbb R)$ and using the even multivector derivative; this is arguably the more natural approach. With this approach, for pure imaginary $\q Q_1$ we do get the requisite $\nabla_{\q Q_1}\q Q_1^2 = 2\q Q_1$. In this case we can express the derivative of an arbitrary $\q Q_1$ as $$ \nabla_{\q Q_1} = \PD{}{q_0} - \q i\PD{}{q_1} - \q j\PD{}{q_2} - \q k\PD{}{q_3}. $$

  • $\begingroup$ I emailed the author and he said that on his paper, we should consider $Q_1^2=\dfrac{x^2+y^2+Z^2}{\sqrt{2Z}^2}$, as defined in equation $3$ of the first article, and $\nabla_\boldsymbol{Q_1}$ as the ordinary derivative, with respect to the quaternion coeficients. So indeed the derivative would be $\nabla_\boldsymbol{Q_1}Q_1^2=2\boldsymbol{Q_1}$, without the boldface on the $Q_1^2$ and $\nabla_\boldsymbol{Q_1}re[\boldsymbol{Q_1^*A}]=A$, with $A$ being purely imaginary. Anyway, thank you very much Nicholas Todoroff for your help. $\endgroup$ Jul 21, 2022 at 23:18
  • $\begingroup$ @NilsonFernandes I have to say, I don't understand how that works, but I'll trust you have a lot more context than me. Glad I could help at all! $\endgroup$ Jul 21, 2022 at 23:33

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