# Torsion and curvature of generalized helix

A regular curve $$\textbf{\gamma}$$ in $$\mathbb{R}^3$$ with curvature $$> 0$$ is called a generalized helix if its tangent vector makes a fixed angle $$\theta$$ with a fixed unit vector $$\textbf{a}$$. Show that the torsion $$\tau$$ and curvature $$\kappa$$ of $$\textbf{\gamma}$$ are related by $$\tau = ±\kappa \cot \theta$$. Show conversely that, if the torsion and curvature of a regular curve are related by $$\tau = \lambda \kappa$$ where $$\lambda$$ is a constant, then the curve is a generalized helix. Note that we have the Frenet equations $$\textbf{t}'=\kappa \textbf{n}$$ and $$\textbf{b}'= -\tau \textbf{n}$$.

I have done the first half of the proof. In the second half, I claim that $$\textbf{a}=\textbf{t} \cos \theta ± \textbf{b}\sin \theta$$ where $$\textbf{t}$$ and $$\textbf{b}$$ are the unit tangent and binormal vectors, respectively, satisfies the conditions for a general helix. I'm trying to show that the derivative of $$\textbf{a}$$ is $$0$$ if we assume that $$\tau = \lambda \kappa$$, proving that $$\textbf{a}$$ is constant. I tried showing that $$\mathbf{a' \cdot a'}=0$$, which would prove that $$\mathbf{a'}=0$$, but I was unable to make it work. Any suggestions or different ideas on how to approach the converse would be appreciated.

My attempt: $$\textbf{a}'=\textbf{t}' \cos \theta ± \textbf{b}' \sin \theta =\kappa \textbf{n} \cos \theta ± \tau \textbf{n} \sin \theta = \kappa \textbf{n} \cos \theta ± \lambda \kappa \textbf{n} \sin \theta$$

• Don't over-complicate! Just show $\mathbf a’=0$ directly. Commented Mar 2, 2020 at 21:31
• Any idea how? I took the derivative but couldn't get anywhere.
– A.B
Commented Mar 3, 2020 at 2:36
• Obviously, you need to use the Frenet equations. Please edit your question to include this. Commented Mar 3, 2020 at 2:37
– A.B
Commented Mar 3, 2020 at 2:43
• No, I meant your computation of $\mathbf a'$. Don't forget to use the hypothesis! Commented Mar 3, 2020 at 2:44

Since $$\mathbf a$$ and $$T$$ are unit vectors we may write

$$\mathbf a \cdot T = \Vert a \Vert \Vert T \Vert \cos \theta = \cos \theta, \tag 1$$

where $$\theta$$ angle 'twixt $$\mathbf a$$ and $$T$$; we may differentiate this equation with respect to the arc-length $$s$$ along our curve $$\gamma(s)$$, yielding

$$\dot {\mathbf a} \cdot T + \mathbf a \cdot \dot T = 0, \tag 2$$

and since

$$\dot {\mathbf a} = 0 \tag{2.5}$$

and we have the first Frenet-Serret relation

$$\dot T = \kappa N, \tag{2.6}$$

(2) becomes

$$\mathbf a \cdot \dot { \kappa N} = 0, \tag 3$$

and with

$$\kappa > 0 \tag{3.5}$$

we find that

$$\mathbf a \cdot N = 0, \tag 4$$

which we may differentiate yet again with respect to $$s$$:

$$\dot{\mathbf a} \cdot N + \mathbf a \cdot {\dot N} = 0, \tag 5$$

and yet again via (2.5) we may write

$$\mathbf a \cdot {\dot N} = 0; \tag{5.5}$$

we now deploy the second Frenet-Serret equation

$$\dot N = -\kappa T + \tau B \tag 6$$

to obtain

$$\mathbf a \cdot {(-\kappa T + \tau B)} = 0, \tag 7$$

or

$$-\kappa \mathbf a \cdot T + \tau \mathbf a \cdot B = 0, \tag 8$$

whence, using (1),

$$-\kappa \cos \theta + \tau \mathbf a \cdot B = 0; \tag 9$$

we expand $$\mathbf a$$ in terms of $$T$$, $$N$$, $$B$$ using (1) and (4) as follows:

$$\mathbf a = (\mathbf a \cdot T)T + (\mathbf a \cdot N)N + (\mathbf a \cdot B)B = (\cos \theta) T +(\mathbf a \cdot B)B; \tag{10}$$

since

$$\Vert \mathbf a \Vert = \Vert T \Vert = \Vert B \Vert = 1, \tag{11}$$

and

$$T \cdot B = \mathbf a \cdot N = 0, \tag{12}$$

we infer from (10) that

$$1 = \Vert \mathbf a \Vert^2 = \cos^2 \theta \Vert T \Vert^2 + (\mathbf a \cdot B)^2 \Vert B \Vert^2$$ $$= \cos^2 \theta + (\mathbf a \cdot B)^2, \tag{13}$$

which implies that

$$\mathbf a \cdot B = \pm \sin \theta; \tag{14}$$

substituting this into (9) yields

$$-\kappa \cos \theta \pm \tau \sin \theta = 0, \tag{15}$$

and then

$$\kappa \cos \theta = \pm \tau \sin \theta, \tag{16}$$

whence

$$\tau = \pm \kappa \cot \theta, \tag{17}$$

as per request.

Going the other way, given that

$$\tau = \lambda \kappa \tag{18}$$

for some constant

$$\lambda \in \Bbb R, \tag{19}$$

we may choose $$\theta$$ such that

$$\lambda = \cot \theta = \dfrac{\cos \theta}{\sin \theta}; \tag{20}$$

next, we set

$$\mathbf a = (\cos \theta) T + (\sin \theta) B, \tag{21}$$

and note this implies;

$$\Vert a \Vert = \sqrt{\cos^2 \theta \Vert T \Vert^2 + \sin^2 \theta \Vert B \Vert^2}$$ $$= \sqrt{\cos^2 \theta + \sin^2 \theta} = \sqrt 1 = 1; \tag{21.1}$$

and apply $$d/ds$$ to (21):

$$\dot {\mathbf a} = (\cos \theta) \dot T + (\sin \theta) \dot B; \tag{22}$$

we substitute (2.6) and the third Frenet-Serret equation

$$\dot B = -\tau N, \tag{23}$$

and obtain

$$\dot{\mathbf a} = (\cos \theta)\kappa N - (\sin \theta)\tau N = (\kappa \cos \theta - \tau \sin \theta) N; \tag{24}$$

in light of (18) and (20),

$$\tau = \lambda \kappa = \dfrac{\cos \theta}{\sin \theta} \kappa, \tag{25}$$

and thus

$$\tau \sin \theta = \kappa \cos \theta, \tag{26}$$

which in concert with (24) shows that

$$\dot{\mathbf a} = 0. \tag{27}$$

We have shown the existence of a constant vector $$\mathbf a$$ and a constant angle $$\theta$$ such that (1) binds; $$\gamma(s)$$ is a generalized helix.

Note Added in Edit, Monday 20 January 2020 6:31 PM PST: As we transit 'twixt (15) and (17), we have occasion to divide through by $$\sin \theta$$; thus we should address the question of just when

$$\sin \theta = 0. \tag{28}$$

Now (28) occurs precisely when

$$\theta = 0, \pi, \tag{29}$$

that is, when $$T$$ is aligned parallel or anti-parallel to $$\mathbf a$$. (We observe that

$$0 \le \theta \le \pi \tag{30}$$

since it is the angle between the vectors $$\mathbf a$$ and $$T$$.) But (29) implies

$$T = \pm \mathbf a, \tag{31}$$

which further implies that $$\gamma(s)$$ is a straight line; as such,

$$\kappa = 0, \tag{32}$$

which contradicts our hypothesis that the curvature of $$\gamma(s)$$ is positive. Therefore

$$\sin \theta \ne 0, \tag{33}$$

and the quotient

$$\cot \theta = \dfrac{\cos \theta}{\sin \theta} \tag{34}$$

may legitimately be formed. End of Note.

• What does 'twixt mean? Commented May 11, 2021 at 1:20
• @Ramanujan: between Commented May 11, 2021 at 6:29

OK, now that you added the relevant information. You never specified how you would choose $$\theta$$. I.e., you haven't yet related $$\lambda$$ and $$\theta$$. Everything you typed suggested that you should take $$\theta$$ so that $$\lambda = \cot\theta$$. If you do that, what is $$\cos\theta - \lambda\sin\theta$$? Now simplify. (Obviously, you need the correct sign choice.)

• What do you mean by correct sign choice? Shouldn't both work?
– A.B
Commented Mar 3, 2020 at 4:18
• No, there can only be one direction that works. Commented Mar 3, 2020 at 5:23
• So is it incorrect to claim $\textbf{a}=\textbf{t} \cos \theta ± \textbf{b}\sin \theta$? Should I have instead said that $\textbf{a}=\textbf{t} \cos \theta + \textbf{b}\sin \theta$?
– A.B
Commented Mar 3, 2020 at 5:40
• Figure out which sign makes it work and use that sign. Who told you to put $\pm$ in there? Commented Mar 3, 2020 at 5:42
• I derived that $\textbf{a}=\textbf{t} \cos \theta ± \textbf{b}\sin \theta$ when showing the first part of the proof. See this post. They show $C = ± \sin \theta$. Is that wrong? math.stackexchange.com/questions/1478391/…
– A.B
Commented Mar 3, 2020 at 5:47