[Math] the radius of torsion, geometrically

curvesdifferential-geometry

In studying calculus of space curves, we calculate the quantities "curvature" ($\kappa$) and "torsion"($\tau$). Both have inverse-length as units, so their reciprocals $\frac{1}{\kappa}$ and $\frac{1}{\tau}$ have units of length, and are called "radius of curvature" and "radius of torsion".

I understand that radius of curvature is the radius of a curve's osculating circle at a point. That's is a pretty clear notion geometrically, but I struggle to obtain a corresponding notion for radius of torsion.

Can anyone share a geometric intuition behind this length, and what it tells us about a non-planar curve? According to http://mathworld.wolfram.com/OsculatingSphere.html, the osculating sphere does not have radius $\frac{1}{\tau}$, so it's not that. Calling it a "radius" seems to imply that it's a radius of something.

Thanks in advance for any insight.


Edit: If a helix is given by $\left<a\cos t,a\sin t,bt\right>$, ($a$ and $b$ positive), then the curvature is $\frac{a}{a^2+b^2}$ and the torsion is $\frac{b}{a^2+b^2}$. There's a lovely duality there, and if you define a dual helix by swapping $a$ and $b$, then the curvature of one is the torsion of the other, and vice versa. Thus, we could say that the radius of torsion is the radius of curvature for a "dual" helix, but I hesitate to define a whole new kind of duality just to awkwardly impart meaning to a phrase I saw in a book. I don't know whether people who know a lot about helices think this way.

I'm still hoping there's a more natural answer out there.

Best Answer

I had also came across the notion of "radius of torsion" in the context of a lecture on Frenet-Serret system of equations, and had great difficulties in trying to conceptualize it. The notion of "radius of torsion" is a little more abstract then the "radius of curvature", and cannot be given a direct geometric interpretation as in the case of the letter. One reason for this difference is that torsion cannot exist without curvature; if one tries to define a space curve with non-zero torsion but zero curvature, then the torsion degenerates into curvature.

However, great background and guidance is given to this subject simply by reading a little about the history of differential geometry of space curves. Much earlier than the publication of Frenet-Serret formulas, Alexis Clariault reffered to general space curves as "curves of double curvature". This simple renaming gives some clue about conceptualization of torsion and curvature of general space curves.

Interpretation:

Assume $\gamma$ is a space curve traced by a point particle. Let $TN$ be it's local osculating plane at some point $X$ (the plane spanned by the tangent vector $T$ and the normal vector $N$), and $BT$ also be a local plane perpendicular to it (it's the plane spanned by the tangent vector $T$ and the binormal vector $B$. Now, let us project the curve $\gamma$ into each of the planes $TN$ and $BT$. One can do this simply by illuminating on the particle from the (local) $B$ and $N$ directions, and then looking at the curves which the particle's shadow traces in each plane.

Obviously, the local curvature of the curve traced by the shadow in the $TN$ plane is $\kappa$. In the same way, the local curvature of of the curve traced in the $TB$ plane is the torsion $\tau$; this gives a straightforward interpretation of the "radius of torsion" - this is radius of the circle which has second order contact with the projection of $\gamma$ into the local $TB$ plane.

A partial illustation of this interpretation can be seen in this picture; it shows the osculating ($TN$) plane and the $BT$ plane.

enter image description here

Caution:

I have not proved yet this claim with full rigourosity; but but nevertheless i wrote it down to make some contribution to the resolution of this question. I think that proving this claim will be my final attempt before i give up the attempts to gain intuition about the "radius of torsion" (and i hope my claim is correct!).

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