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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.

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    $\begingroup$ For existence, is there any reason you can't take $\Omega = \overset{\cdot}{v}v^{-1}$ ? The answer seems to over-complicate things. For (non)uniqueness, you can add $\Omega+(B\wedge v)v$ for any bivector $B$. $\endgroup$ – mr_e_man May 24 '18 at 4:12
  • $\begingroup$ @mr_e_man That seems correct to me. Here non-uniqueness seems to just be a proportionality constant (since $B\wedge v$ is either zero or proportional to the unit pseudoscalar—this corresponds to the non-uniqueness condition I indicated in the question, but not in the solution, since the curve drawn is the same). Note that it's been a while since I last stared at this question, so it's completely likely that the two presented solutions are much more complicated than they should be. $\endgroup$ – Guillermo Angeris May 29 '18 at 16:38
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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.

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    $\begingroup$ You can write $v(t) = R(t) v(0) R^\dagger(t)$, and then $\Omega(t)$ is determined. Any reparameterization can change the function $\Omega(t)$, but at any point on the curve, the value of $\Omega$ there will be the same. $\endgroup$ – Muphrid Jan 2 '15 at 18:18
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The restriction comes from the fact that $v$ is on a curve. If you look at the Rotating frames section in the same chapter you will find the answer.

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  • $\begingroup$ Oh, wow. That was quite a bit more obvious than I thought. Thank you! $\endgroup$ – Guillermo Angeris Jan 1 '15 at 22:36

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