Is there are any command to expand quaternion multiplication as below STEP by STEP.
Below is equation from my text.
I'm self learner and for some reason I can't wrap my head around quaternion multiplication.
I really need to see step by step workout.
I remember seeing step by step workout in wolframalpha but that's for calculus integration.
Just wondering if the same can be done in mathematica.
$$
\begin{equation}
\begin{split}
w =& qvq^* \\
=& (q_0+\vec{q})(0+\vec{v})(q_0-\vec{q}) \\
=&(2q_0^2-1)\vec{v}+2(\vec{q}\cdot\vec{v})\vec{q} + 2q_0(\vec{q}\times\vec{v})
\end{split}
\end{equation}
$$
$\begingroup$
$\endgroup$
3
-
$\begingroup$ The last equation is incorrect. This becomes obvious when you consider the special case $q_0+\vec{q}=0$ and $\vec{v}\ne \vec{0}$: then $w$ must be $0$, but the last expression equals $-\vec{v}$. $\endgroup$– whuberCommented Oct 26, 2012 at 13:17
-
$\begingroup$ the book mention about using triple vector product $a\times(b\times c)=(a\cdot c)b-(a\cdot b)c$ to derive the final equation. $\endgroup$– kyproniteCommented Oct 26, 2012 at 15:14
-
$\begingroup$ That's fine--but the answer is still obviously wrong, as my counterexample shows. $\endgroup$– whuberCommented Oct 26, 2012 at 16:29
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
You can create quaternions in Mathematica. Here is a limited implementation, sufficient for the kinds of calculations shown in the question. As in the question, it represents quaternions as ordered pairs of a real part and a (vector) quaternionic part and defines multiplication in terms of dot and cross products:
ClearAll[quaternion];
quaternion[q0_, q_] + quaternion[t0_, t_] ^= quaternion[q0 + t0, q + t];
quaternion[q0_, q_] ** quaternion[t0_, t_] ^= quaternion[q0 t0 - q.t, q0 t + q t0 + Cross[q, t]];
Conjugate[quaternion[q0_, q_]] ^= quaternion[q0, -q];
(* Some helpful procedures to let reals interact with quaternions: *)
quaternion[q0_] := quaternion[q0, {0, 0, 0}];
Re[quaternion[q0_, _]] ^= q0;
Im[quaternion[_, q_]] ^= q;
The question asks for the purely quaternionic part of $q v q^*$ when--apparently--$v$ is a pure quaternion (with zero real part).
Im[quaternion[q0, q] ** quaternion[0, v] ** Conjugate[quaternion[q0, q]]]
$q0\ (q0\ v+q\times v)+(q0\ v+q\times v)\times (-q)+q\ q\cdot v$
Now the working out can be traced:
Clear[q0, q, v];
traceView2[quaternion[q0, q] ** quaternion[0, v] ** Conjugate[quaternion[q0, q]]]
(This is an excerpt from deeper into the trace.)
