Spatial curve with certain condition

Question

In a certain exercise I have been asked to find all spatial curves such that, given the curve $\gamma$, it is satisfied that $\gamma'' = \gamma' \times a$, $a$ is a certain constant vector. I am aware of the fact that this might not be nothing really determinant, but I tried to do something regarding tangent, normal and binormal vectors. I would be grateful if somebody could give me a hint.

Answer 1

The mapping $\ x\mapsto x\times a\ $ is linear with eigenvalues $\ 0\ ,i\|a\|\ $ and $\ {-}i\|a\|\ ,$ corresponding to eigenvectors $\ a, u-iv\ $ and $\ u+iv\ ,$ respectively, where $\ u\ $ and $\ v\stackrel{\text{def}}{=}$$\,{-}\|a\|^{-1}(a\times u)\ $ are orthogonal unit vectors orthogonal to $\ a\ .^\color{red}{\dagger}$ This means that the coordinates of $\ \gamma\ $ with respect to the basis $\ a, u, v\ $ will each satisfy a linear differential equation of second degree.

If you set $$ \gamma=g_1u+g_2v+g_3a\ , $$ where $\ g_1,g_2,g_3\ $ are each real-valued functions, then your differential equation becomes \begin{align} g_1''u+g_2''v+g_3''a&=\big(g_1'u+g_2'v+g_3'a\big)\times a\\ &=g_1'\|a\|v-g_2'\|a\|u\ . \end{align} Equating the coordinates of each side of this equation with respect to the basis $\ a, u, v\ $ gives \begin{align} g_1''&={-}\|a\|g_2'\ ,\\ g_2''&=\|a\|g_1'\ \text{, and}\\ g_3''&=0\ . \end{align} The solution of the third of these equations is $\ g_3(t)=g_3(0)+g_3'(0)t\ .$ Integrating the first gives $\ g_1'=-\|a\|g_2+g_1'(0)+\|a\|g_2(0)\ ,$ and then substituting this expression for $\ g_1'\ $ in the second gives $$ g_2''=-\|a\|^2g_2+\|a\|g_1'(0)+\|a\|^2g_2(0)\ , $$ a second-order linear differential equation whose general solution is $$ g_2(t)=A\sin(\|a\|t)+B\cos(\|a\|t)+\|a\|^{-1}g_1'(0)+g_2(0)\ . $$ Initial conditions for $\ g_2\ $ give $\ B={-}\frac{g_1'(0)}{\|a\|}\ $ and $\ A=\frac{g_2'(0)}{\|a\|}\ ,$ from which we get $$ g_2(t)=\|a\|^{-1}\big(g_2'(0)\sin(\|a\|t)-g_1'(0)\cos(\|a\|t)+g_ 1'(0)\big)+g_2(0)\ . $$ A similar calculation will give you a similar expression for $\ g_1(t)\ .$

It's evident from all of this that if $\ g_3'(0)=0\ ,$ then the path of $\ \gamma\ $ will be a circle lying in a plane perpendicular to $\ a\ .$ On the other hand, if $\ g_3'(0)\ne0\ ,$ the path will be a spiral of fixed radius and constant speed around an axis parallel to $\ a\ .$

$\left.\right.^\color{red}{\dagger}$ The vectors $\ u,v $ and $\ a\ $ could be taken as defining the directions of the $x,y$ and $z$ axes, respectively, of a right-handed Cartesian coordinate system.