Property of a generalised helix

[CAF123]

Homework Statement

A generalized helix is a space curve whose unit tangent makes $T$ makes a constant angle $\theta$ with the a fixed unit vector $A$ in Euclidean space, I.e $T \cdot A = \cos \theta = \text{const}$. Prove that if the torsion $\tau \neq 0$ everywhere then the space curve is a generalized helix iff $\kappa/\tau = \text{const}$

2. Homework Equations

Frenet-Serret Formulae

The Attempt at a Solution

$(\Rightarrow)$WLOG, assume curve is unit speed parametrised. Then $$\frac{d}{ds} (T \cdot A) = T' \cdot A = \kappa N \cdot A \Rightarrow N \cdot A = 0$$ Then $$ \frac{d}{ds} (N \cdot A) = (- \kappa T + \tau B) \cdot A = 0$$ using the Frenet serret formulae. This is equal to $-\kappa \cos \theta + \tau B \cdot A = 0$. I would nearly be there if I can show that $B \cdot A = \text{const}$. I managed to see that $A$ lives in the plane spanned by T and B. So $A = aT + bB$ for some a and b. Then $A \cdot A = 1 = a \cos \theta + b B \cdot A \Rightarrow B \cdot A = (1 - a \cos \theta)/b$ which certainly is constant provided a and b are constant. But since T and B revolve around the curve, I think a and b are parameter dependent. So I don't think my argument works. Thanks for any help here. Any hints on how to start the reverse implication would be great too, thanks.

 

[CAF123]

Has anyone any pointers? thanks.

 

[voko]

I would nearly be there if I can show that $B \cdot A = \text{const}$.

Consider ${d \over ds} \left( B \cdot A \right)$.

 

[CAF123]

Hi voko! Thanks, I got it from that. Do you have any hints on the reverse implication please?

 

[voko]

Since you have $N \cdot A = 0$, $A$ must be in the T-B plane, as you concluded earlier. Your coefficients $a$ and $b$ are simply $\cos \theta$ and $\sin \theta$. What condition do you get for $\kappa, \tau$ and $\theta$ from that? Can you use it to complete the proof?