Principal normal and curvature of a helix

In summary, the curvature of a 3-d unit-speed curve is given by ##\kappa=||\ddot{{\mathbf r}}||## and the principal unit normal vector is given by ##{\mathbf n}=\frac{\ddot{{\mathbf r}}}{||\ddot{{\mathbf r}}||}##However, the value of ##\kappa## is different in different coordinate systems.
  • #1
kkz23691
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Hi,

##x(s)=\cos\frac{s}{\sqrt{2}}##
##y(s)=\sin\frac{s}{\sqrt{2}}##
##z(s)=\frac{s}{\sqrt{2}}##,
it is a unit-speed helix. Its curvature is ##\kappa=||\ddot{r}||=\frac{1}{2}##. Principal unit normal is ##{\mathbf n}=(\cos\frac{s}{\sqrt{2}},\sin\frac{s}{\sqrt{2}},0)##. So far so good...

But the helix in cylindrical coordinates is
##r(s)=1##
##\theta(s)=\frac{s}{\sqrt{2}}##
##z(s)=\frac{s}{\sqrt{2}}##
It is still unit-speed, ##||\dot{r}||=1##, but ##||\ddot{r}||=0##. What's wrong? How does one calculate the curvature and the principal unit normal in cylindrical coordinates?...
 
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  • #2
Hi, the helix is a geodesic on the cylinder so the curvature is 0 , when you look the curve in a submanifold (cylinder) different from the
natural space (euclidean) this can be appen ...

see for example the concept of geodesic curvature: http://en.wikipedia.org/wiki/Geodesic_curvature,

hi,
Ssnow
 
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  • #3
Hi Ssnow, very interesting point. Let me make sure I understand you correctly.

The curvature of a 3-d unit-speed curve is given by
##\kappa=||\ddot{{\mathbf r}}||##
and the principal unit normal vector is given by
##{\mathbf n}=\frac{\ddot{{\mathbf r}}}{||\ddot{{\mathbf r}}||}##
Are these expressions true in any coordinate system or only in Cartesian coordinates?

Is then correct that the value of the curvature ##\kappa## depends on the coordinate system?

Many thanks!
 
  • #4
yes, formulas are for a curve in the natural euclidean space (in your case tridimensional), on the cylinder the helix is as a line in the plane so the curvature it is obvious 0.

In fact yours first coordinates were the '' cylindrical coordinates '' in ##\mathbb{R}^3##, and the second were '' coordinates on a cylinder '' so as in ##\mathbb{R}^{2}##.

Hi,
Ssnow
 
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  • #5
Ssnow said:
on the cylinder the helix is as a line in the plane so the curvature it is obvious 0.
I have managed to understand this. Then ##\kappa## has different values in Cartesian, spherical and cylindrical coordinates, but the formula is always ##\kappa=||\ddot{{\mathbf r}}||##, no matter what coordinates we are working with.

Ssnow said:
In fact yours first coordinates were the '' cylindrical coordinates '' in ##\mathbb{R}^3##, and the second were '' coordinates on a cylinder '' so as in ##\mathbb{R}^{2}##.
My apologies, I didn't quite get this...
 
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  • #6
Yes, it makes sense now. Since ##\ddot{\mathbf r}=0##, then the curve is a geodesic.
The conclusion then is:

(1) The expressions of differential geometry are always true for curves irrespective of the type of coordinate system used

(2) The values of mathematical entities are different in different spaces, e.g. in flat (Euclidian) space described with Cartesian coordinates, or in curved spaces such as those described with cylindrical coordinates and spherical coordinates.

Thanks for you input, Ssnow!
 
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  • #7
Uh-oh. It turns out, I didn't properly calculate ##||\ddot{\mathbf r}||##; it is still 1/2 in cylindrical coordinates. One needs to pay attention that
##\ddot{\mathbf r} = (\ddot{r}-r\dot{\theta}^2,2\dot{r}\dot{\theta}+r\ddot{\theta},\ddot{z} )##

This means that changing the coordinates doesn't change the manifold, but changes the way the same manifold is charted. Then conclusion (1) above is still true; but (2) is not. ##||\ddot{\mathbf r}||## is independent of coordinate system.
 
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Related to Principal normal and curvature of a helix

1. What is a helix and how is it different from a spiral?

A helix is a three-dimensional curve that follows a constant slope and circular path. It is different from a spiral in that a spiral gradually increases or decreases its distance from a central point, while a helix maintains a constant distance from its central axis.

2. What is the principal normal of a helix?

The principal normal of a helix is the line that is perpendicular to the tangent line at any point on the helix. It represents the direction in which the helix is curving at that point.

3. How is the curvature of a helix calculated?

The curvature of a helix is calculated using the formula k = 2π/λ, where k is the curvature and λ is the distance between two successive points on the helix that lie on the same horizontal plane.

4. Why is the principal normal of a helix important in physics?

The principal normal of a helix is important in physics because it helps determine the direction of forces acting on an object moving along a helical path. It is also useful in understanding the motion of particles in helical structures such as DNA.

5. Can the principal normal of a helix ever be parallel to the helix axis?

No, the principal normal of a helix can never be parallel to the helix axis because it is always perpendicular to the tangent line. However, it can be parallel to the helix axis at the point where the helix intersects with its own axis.

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