I have a question about identifying regular curves in $ R^3 $ using Frenet-Serret apparatus. Considering the congruence of curves every unit-speed curves are congruent when they have same curvature and torsion.
Now I have a simple question. I wonder this is possible approach to prove the following
$ \alpha $ is a curve with $ \kappa >0 \ $and $ \tau \ $ both constant. Show that $ \alpha $ is a circular helix
(This is the problem in the O'neil's Differential Geometry textbook)
I think regular curves are always can be reparametrized to be a unit-speed curve and after the reparametrization, reparametrized $ \kappa \ $and $ \tau $ still have the same property with $\alpha $ , so I can construct the circular helix with same curvature and torsion. Thus by the congruence argument, $ \alpha $ is also a circular helix.
I doubt my argument at some points.
- Is it possible to define a helix with any constant $ \kappa >0 $ and $ \tau $ ?
- Can I argue that original $ \alpha $ is circular helix when its unit-speed reparametrization is circular helix(I think this is possible since their image is eqaul...)
And finally I wonder there is another way to prove this problem. I think I can use the cross-sectional curve of the cylinder (Since the fact $ \alpha $ is a cylindrical helix is trivial.), but I don't know how can I use this....
Please show me the right way to solve problems like this.