# Principal normal and curvature of a helix

1. May 17, 2015

### kkz23691

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?...

2. May 18, 2015

### Ssnow

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

3. May 18, 2015

### kkz23691

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. May 18, 2015

### Ssnow

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

5. May 18, 2015

### kkz23691

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.

My apologies, I didn't quite get this...

6. May 19, 2015

### kkz23691

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!

Last edited: May 19, 2015
7. May 28, 2015

### kkz23691

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.