Find a general equation of geodesics on cylinder's surface.
What's the name of these curves?
just a guess, cycloids? or if they are unnamed we can call them "tehnos"!
They are not cycloids mathwonk.Funny thing :the tehnicians of various fields are familar with them.The things shaped that way have interesting properties and important applications.
A general param would be...
[tex]\gamma(t) = (a\cos(ct),a\sin(ct),bt)[/tex]
up to a rigid motion
Yes, because a cylinder is flat!
For a given value of "flat"
[tex]\kappa_1 \kappa_2 = 0[/tex] to be exact.
As a minor point of interest, if one considers that the gauusian curvature of the cylinder is zero, and thus that we can form a cylinder from a flat plane, then straight lines on the plane, become helixs on the cylinder.
But is it true that all geodesics are isomorphic under an isometry of this kind. If the gaussian curvature between two surface is equal, can we identify geodesics on one surface with those on the other?
Well, the simply-connected cover of both surfaces is the plane and each geodesic of a given surface is the image of a line in the plane under the covering map. So, I think the answer is, yes (sort of): given a geodesic on one surface, we look at its inverse image in R^2. This will be a family of lines (maybe infinite, maybe not). Then take the geodesics in the second surface corresponding to those lines.
So, we can associate a family of geodesics of one surface for every geodesic of the other.
i think i was thinking of helices. i gues cycloids are those curves you get when you mark a point on a penny and roll it right?
it is a pretty easy problem since all you have to do is unroll the cylinder to a rectangle.
But, what about surfaces with non zero curvature?
Yeah, I guess you could do the same procedure for the non-zero curvature as long as its constant.
For constant negative Gaussian curv., the simply-connected cover would be the hyperbolic plane. For positive, it would be the 2-sphere.
of course, i'm assuming the two surfaces have the same curvature *and* that there isn't anything too aberrant about either's topology, i.e. both are connected and complete (when unioned with its boundary) etc. etc.
That's why I classified the problem under "easy".
Helices ,as quasar987 said, is the correct answer.
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