I've been working on (self-studying) Geometric Algebra for Physicists which, sadly, has no solutions manual. This is not a problem in general, but I feel like one of my solutions for a question asked in the textbook is incomplete.
The question is:
A particle in three dimensions moves along a curve $x(t)$ such that $|v|$ is constant. Show that there exists a bivector $Ω$ such that $$ \dot v = Ω\cdot v $$ and give an explicit formula for $Ω$. Is this bivector unique?
My solution:
Since we're in three dimensions, we can construct a vector $\dot v$ with the following property: $$ v^2=v_0^2 \implies \dot v v+v \dot v = 0 \implies v\cdot \dot v = 0 $$ (This is obvious from elementary multivariable calculus)
As $\dot v$ must always be perpendicular to $v$, we can always come up with such a vector by forming a plane with it and some arbitrary vector and then take this resulting bivector's dual. That is, $$ \dot v = I(v\wedge b) $$ Where $I=e_1e_2e_3$ is the unit pseudoscalar. We can re-write this in the following form: $$ I(v\wedge b) = v\cdot (Ib)=-(Ib)\cdot v $$If we allow $b$ to absorb the constant, then we can claim: $$ \Omega = Ib(t) $$ Where $b(t)$ is any vector-valued function of $t$. Clearly, then, the bivector is not unique.
Is this all? It seems like the question implies there should be a more restrictive condition on $\Omega$, but I haven't been able to find any (and, intuitively, it doesn't seem like it could be made more restrictive).
Thank you.
For the sake of completeness, I'll just post up the full answer, which is relatively simple after ahmetselcuk's response.
Since $v^2 = v_0^2$, then we can describe $v$ purely by means of rotations. That is, for some unit vector $u$, we have:
$$ v = v_0RuR^\dagger $$
where all of the information of motion is purely on the rotor $R$.
Then, we must have $$\dot v = v_0\frac{d}{dt}(RuR^\dagger)=v_0\dot RuR^\dagger+v_0Ru\dot R^\dagger.$$
Since $u = \frac{1}{v_0}R^\dagger v R$, it follows that $$ \dot v = \dot RR^\dagger vRR^\dagger +RR^\dagger v R\dot R^\dagger =\dot RR^\dagger v+vR\dot R^\dagger. $$
We know that $\dot RR^\dagger = -R\dot R^\dagger$. Hence, $\dot RR^\dagger$ is a pure bivector. Therefore, the above reduces to:
$$ \dot v = (2\dot RR^\dagger)\cdot v = \Omega \cdot v. $$
Clearly, $R$ depends on the choice of $u$, thus $\Omega$ is similarly dependent, yet $v$ is not. Hence $\Omega$ is not unique.