Show that the following curve, $\alpha (s)$ is a circle centered at the origin

Question

Let $\alpha (s)$ for $s \in \mathbb{R}$ be a curve in $\mathbb{R}^3$, parametrized by arc length which does not pass through the origin.

Suppose for every $s\in \mathbb{R}$, the straight line through $\alpha(s)$ parallel to $\bf{n}$$(s)$ always passes through the origin $(0,0,0)$,

Show $\alpha (s)$ is a circle centered at the origin.

I dug around and found out that if the curvature is constant and torsion is $0$, then the curve is a circle. But now I have the following questions.

It makes sense for a circle to have constant curvature part, but I don't quite understand how torsion works aside from it's definition.

Answer 1

The torsion $\tau(s)=<-n(s),b'(s)>$ involves the derivative of $b(s)=t(s)\times n(s)$ which is vertical to $t(s),n(s)$ so it gives you the information on how "fast" the curve $\alpha(s)$ will "leave" from the plane that $t(s),n(s)$ are creating . Hence if $\tau(s)=0 \forall s$ then $b(s)$ will be constant so the curve will never leave from the plane that $n(s),t(s)$ are creating.

Note that $t(s)=\alpha'(s), n(s)=\frac{t'(s)}{k(s)}$.

Now in the exercise you have the line which passes from $\alpha(s)$ and its parallel to $n(s)$ passes through the origin and this is $\forall s$.

So let $s$ , and here is the equation of the line that passes through $\alpha(s)$ and parallel to $n(s)$ , $\epsilon(x)=\alpha(s) +xn(s)$, now we now that there is number $x(s)$ such that $\alpha(s)+x(s)n(s)=0(*)$ because the line passes through the origin.

Hence for every $s$ we have the equation $\alpha(s)+x(s)n(s)=0$ so you get $\alpha'(s)+x'(s)n(s)+x(s)n'(s)=0.$ And by applying the Frenet-Serret equations you get $(1-k(s))t(s)+x'(s)n(s)+\tau(s)b(s)=0$.

Because $\{t(s),n(s),b(s)\}$ is a basis for $\Bbb R^3$ $\forall s$ you get $k(s)=1$,$\tau(s)=0$ and $x(s)=c$.

Apply $x(s)=c$ in $(*)$ and you get $|\alpha(s)|=c$ and because $\tau(s)=0$ you have that $\alpha(s)$ belongs to a circle centered at the origin and not in a sphere.

Answer 2

I assume $\mathbf n(s)$ is the standard normal vector to $\alpha(s)$, and as such it is part of the Frenet-Serret apparatus of $\alpha(s)$.

For any point $\alpha(s)$, the line $\mathbf l(s, \rho)$ through $\alpha(s)$ parallel to $\mathbf n(s)$ is given by the equation

$\mathbf l(s, \rho) = \alpha(s) + \rho \mathbf n(s); \tag 1$

here $\rho$ is a free parameter, the running parameter along $\mathbf l(s, \rho)$. We are given that for every curve point $\alpha(s)$, there is some $\rho$ such that

$\mathbf l(s, \rho) = \alpha(s) + \rho \mathbf n(s) = 0; \tag 2$

$\rho$ will of course depend on $s$, so we write

$\alpha(s) + \rho(s) \mathbf n(s) = 0. \tag 3$

We may differentiate (3) with respect to $s$ and obtain

$\alpha'(s) + \rho'(s) \mathbf n(s) + \rho(s) \mathbf n'(s) = 0; \tag 4$

we recall the Frenet-Serret equations, noting that since $s$ is the arc-length along $\alpha$ we have the unit tangent vector

$\mathbf t(s) = \alpha'(s), \tag 5$

and

$\mathbf t'(s) = \kappa(s) \mathbf n(s), \tag 6$

$\mathbf n'(s) = - \kappa(s) \mathbf t(s) + \tau(s) \mathbf b(s), \tag 7$

$\mathbf b'(s) = -\tau(s) \mathbf n(s), \tag 8$

where

$\mathbf b(s) = \mathbf t(s) \times \mathbf n(s) \tag 9$

is the binormal vector to $\alpha(s)$, and $\tau(s)$ is the torsion. Substituting from (5) and (7) into (4) we find

$\mathbf t(s) + \rho'(s) \mathbf n(s) + \rho(s)(-\kappa(s) \mathbf t(s) + \tau(s) \mathbf b(s) = 0, \tag{10}$

which after a little algebraic re-arrangement becomes

$(1 - \rho(s)(\kappa(s)) \mathbf t(s) + \rho'(s) \mathbf n(s) + \rho(s) \tau(s) \mathbf b(s) = 0. \tag{11}$

Now since $0$ does not lie on $\alpha(s)$, it follows from (2) that $\rho(s) \ne 0$; and by the orthogonality, hence linear independence of $\mathbf t(s)$, $\mathbf n(s)$, and $\mathbf b(s)$ we conclude from (11) that

$1 - \rho(s) \kappa(s) = \rho'(s) = \rho(s) \tau(s) = 0; \tag{12}$

since $\rho(s) \ne 0$ it follows that $\tau = 0$; since $\rho'(s) = 0$ we have $\rho(s) = \rho$ is constant; since $1 - \rho \kappa(s) = 0$ we have that $\kappa = \rho^{-1}$ is constant as well; from (5) and (6) we conclude that

$\alpha''(s) = \mathbf t'(s) = \kappa \mathbf n(s), \tag{13}$

or

$\mathbf n(s) = \rho \alpha''(s); \tag{14}$

inserting (14) into (3) we find

$\alpha(s) + \rho^2 \alpha''(s) = 0, \tag{15}$

whence

$\alpha''(s) + \kappa^2 \alpha(s) = \alpha''(s) + \rho^{-2} \alpha(s) = 0. \tag{16}$

(16) is a linear, constant-coefficient, second-order, ordinary differential equation for the vector-point $\alpha(s)$; as such, it's solution through $\alpha(s_0)$ is easily seen to be of the general form

$\alpha(s) = \beta_1 \cos \kappa (s - s_0) + \beta_2 \sin \kappa (s - s_0); \tag{17}$

if we take $s = s_0$ we find that

$\beta_1 =\alpha(s_0); \tag{18}$

furthermore we may take the derivative of (17):

$\alpha'(s) = -\beta_1 \kappa \sin(s - s_0) + \beta_2 \kappa \cos (s - s_0), \tag{19}$

which evaluated at $s_0$ yields

$\alpha'(s_0) = \kappa \beta_2, \tag{20}$

or

$\beta_2 = \kappa^{-1} \alpha'(s_0) = \rho \alpha'(s_0); \tag{21}$

inserting (18) and (21) into (17) we obtain

$\alpha(s) = \alpha(s_0) \cos \kappa(s - s_0) + \rho \alpha'(s_0) \sin \kappa (s - s_0); \tag{22}$

we may recall (5) and (3) to finally write

$\alpha(s) = - \rho \mathbf n(s_0) \cos \kappa(s - s_0) + \rho \mathbf t(s_0) \sin \kappa (s - s_0), \tag{23}$

which shows that $\alpha(s)$ lies in the subspace of $\Bbb R^3$ spanned by $\mathbf t(s_0)$ and $\mathbf n(s_0)$; it is a thus a planar curve. Also, since

$\Vert \mathbf t(s) \Vert = \Vert \mathbf n(s) \Vert = 1, \; \mathbf t(s) \cdot \mathbf n(s) = 0, \tag{24}$

(23) yields

$\Vert \alpha(s) \Vert^2 = \alpha(s) \cdot \alpha(s) = \rho^2 \mathbf n(_0) \cdot \mathbf n(s_0) \cos^2 \kappa(s - s_0) + \rho^2 \mathbf t(s_0) \cdot \mathbf t(s_0) \sin^2 \kappa (s - s_0)$ $= \rho^2 \Vert \mathbf n(s_0) \Vert^2 \cos^2 \kappa (s - s_0) + \rho^2 \Vert \mathbf t(s_0) \Vert^2 \sin^2 \kappa (s - s_0)$ $= \rho^2 \cos^2 \kappa (s - s_0) + \rho^2 \sin^2 \kappa (s - s_0) = \rho^2, \tag{25}$

and thus

$\Vert \alpha(s) \Vert = \rho \tag{26}$

We have now demonstrated that $\alpha(s)$ is a plane curve whose distance from the origin is everywhere $\rho > 0$; thus, it must lie in a circle of radius $\rho$. The circumference of such a circle is $2 \pi \rho$; since $\alpha(s)$ is a unit speed curve, it will traverse such a circle in a time interval of precisely $2 \pi \rho$.

It is worthy of note that (26) also follows directly and more simply from (3):

$\Vert \alpha(s) \Vert^2 = \alpha(s) \cdot \alpha(s) = \rho^2 \mathbf n(s) \cdot \mathbf n(s) = \rho^2, \tag{27}$

since $\mathbf n(s)$ is a unit vector. It is in fact possible to show that $\alpha(s)$ is lies entirely in a plane by invoking the well-known result that the vanishing of torsion implies the planarity of the curve; this follows from (8), which then leads to $\mathbf b(s)$ being constant, and since according to (9) $\mathbf b(s) \cdot \mathbf t(s) = \mathbf b(s) \cdot \mathbf n(s) = 0$, the subspace spanned by $\mathbf t(s)$ and $\mathbf n(s)$ is invariant as well; see my answer to this question. Such an approach may provide a less lengthly demonstration, since it would allow to bypass the argument given in (13)-(23); I chose a different route here, mostly because I wanted to see how it worked out.

As far as "how torsion works", the best thing I can say in a small space is that it measures the way the $\mathbf t(s)$-$\mathbf n(s)$ plane turns out of itself as $s$ varies. Since $\mathbf b(s)$ uniquely determines this plane, $\mathbf b'(s) = -\tau(s) \mathbf n(s)$ quantifies said turning. The study of many examples helps develop intuition as to how this works in practice.