There are certain curves that can slide rigidly along themselves - lines and circles for example. Some other curves of this type are cylindrical spirals and Archimedes spirals. The concept can be generalized to more than one dimension - the surface of a sphere for example or a Euclidean space of any dimension.

I was wondering if anyone has ever investigated such curves and come up with a way of enumerating or classifying them. This may be a trivial problem but I would not know how to begin solving it.

I haven't heard of the idea before, but here are some ideas that might help you get started in classifying them. Do you know about scalar curvature? I think that for any curve which can slide rigidly along itself the scalar curvature will have to be constant. Here is my reasoning, let me know if any part of it doesn't make sense to you:

Firstly, if you can slide one point on the curve anywhere on the curve, you must be able to slide it everywhere on the curve. Do you see why this is? Basically, if you can slide a point from A to B, then by the time A has reached point B, point B has reached point C (say). But you can repeat the slide that got A to B because the curve hasn't changed its outward appearance, and so you can slide A to C. Repeat this until you've got everywhere on the curve.

The next point is that the condition that you can slide the curve is just the same as being able to find a rigid transformation (rotation and translation) taking one point on the curve to any other point. Here's why: if I can slide the curve then I can certainly take any point to any other point, and the transformation at the end has to be rigid. Alternatively, if I can take any point to any other point on the curve by a rigid transformation, I can slide one point to any other point by repeatedly taking my starting point to all of the points between the starting point and the end point.

So, here's a way of generating a condition that any curve will have to satisfy: if you can find some property of a point on the curve that doesn't change if you apply a rigid transformation to the curve, then that property must be the same everywhere along the curve. A good example of such a property is the curvature of the curve at that point. (Have a look at http://mathworld.wolfram.com/Curvature.html for more on curvature.)

I'm not particularly knowledgeable about curvature so I'm not sure about this, but I think that in 2D the only constant curvature curves are lines and circles.

In 3D the problem is more difficult. For example, the spirals you mention, you could have spirals spiralling around circles, spirals spiralling around hyperbolae, and so on.

The problem can be generalised to n dimensions in more than one way. For surfaces (for example) you might only be looking for surfaces that can be slid around in one way, or you might be looking for surfaces which can be slid in every way. For surfaces in 3 dimensions, I suspect you'll find that spheres, planes and hyperboloids will be the only surfaces that will work sliding in every way. They each have constant curvature. If you only need to be able to slide in some way, but not every way, then surfaces like a torus (the surface of a doughnut with a hole, or a ring) also work, and so would any 2d curve extruded along a perpendicular axis infinitely (like a infinite circular or square cylinder). Either way, it is a more difficult problem in higher dimensions, definitely.

I hope that helps, and maybe someone will come along and tell us something about curvature that will help.

Dan Goodman, thank you for your answer. What you say makes sense and is in agreement with what I felt intuitively but was not able to put into words. The feeling that I have is that each point on the curves that I described is locally indistinguishable from any other point. The translation of this intuition may very well be completely described in terms of curvature. I will check your reference.

For higher dimensional surfaces I was only interested in curves that could be slid along themselves for any possible movement. It is interesting to ask, however, what the signifcance is that a curve like a torus can be rigidly slid along itself in some directions but not in others.

A correction to what I said. An Archimedes spiral can not be slid along itself. In addition to a sphere a cyllinder has the property among two dimensional surfaces.

I'm sorry to interfere, but what exactly do you mean by 'sliding'

Yatir

Sliding is not actually well-defined in those examples. In the case of a line, it's a translation parallel to the line. In the case of a circle, it's a rotation about its centre. For a helix it's a combination of both...

More precisely, a continuous motion of the whole space that is always a rigid transformation and preserves the curve. In other words for each $t$ in $[0,1]$ (for example), you have a map $A_t(x)$ from the plane to the plane that is a rigid transformation (i.e. a combination of a rotation and translation). You also have to have that the map $B(x,t)=A_t(x)$ is continuous in $x$ and $t$. If $\Gamma$ is the set of points in the curve then $A_t(\Gamma )=\Gamma$ for all $t$.

I'm pretty sure only a circle or line will work in 2D. Here's my argument why: consider a point A on the curve and move it to some other point B near it. The rigid transformation moving A to B we can call T say. Now, if we do the same slide again we get the same transformation T, so we can slide the whole space using the transformation T² . In other words, we do the same transformation that took A to B to take B to a third point C, and so on. If A, B and C lie on a straight line then all the translations of A are also going to be on this straight line. If the angle between AB and BC is not 0 then all of the translates of A will be on a circle. This suggests (but doesn't prove) that it has to be a straight line or a circle. I haven't got any time right now, but I think this shouldn't be too difficult to work into a proof. I'll get back to you. You might like to have a go at doing it yourself (try thinking about dividing the curve between A and B into two parts with midpoint D say, so that the transformation sliding A to D slides D to B, now subdivide this and so on), you might even find an easier way. Another thing you could try is using the same argument to extend it to curves in higher dimensions.

Another way of going about the proof would be to write down the 2nd order differential equation for constant curvature, and demonstrate that the equation of a circle satisfies it. Given that we have two constants in the solution (i.e. the x,y coordinate of the centre) this must capture all possible solutions.

Andre

Andre, isn't there a problem with that approach, that the differential equation you get is not linear?

Yes, sorry, that was a sweeping generalisation too far. In this case, though, we are fine, since this (nonlinear) DE is directly integrable. The first substitution is just z=dy/dx leaving a first order differential equation which can be integrated with one constant of integration. Getting rid of the z gives the second constant.

Andre