Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Defining a generalized coordinate system

  1. May 17, 2016 #1
    (Note that the title of this thread might be incorrect - I'm just drawing on the vocabulary people use when discussing Lagrangian Mechanics...)

    Hi, I'm trying to set up a coordinate system to represent points in space where one of the coordinates is the distance along a parametric curve, one is the shortest distance from a point to the curve, and one is an angle defined relative to some direction.

    I have some curve in space defined by:

    [itex] x = x(t) [/itex]
    [itex] y = y(t) [/itex]
    [itex] z = z(t) [/itex]

    and I want to define a coordinate system [itex] (s, \rho, \omega) [/itex] relative to this curve.

    [itex] \rho [/itex] is defined by the shortest distance from a point [itex] p' = (x',y',z') [/itex] (I will use primes to denote points not on the curve) in space to the curve (assume this is unique).

    [itex] s [/itex] is defined by the distance along the curve, starting from some initial point [itex] \left (x(t_0), y(t_0), z(t_0) \right) [/itex] to the point [itex] \left (x(t), y(t), z(t) \right) [/itex] such that

    [tex] d \left (x', y', z', x(t), y(t), z(t) \right ) = \rho [/tex]

    where [itex] d [/itex] is the euclidean distance.

    In other words, say I have some point in space [itex] p' = (x',y',z') [/itex], then [itex] \rho [/itex] is the length of the smallest line segment between [itex] p' [/itex] and some point on the curve [itex] p(t) = (x(t), y(t), z(t)) [/itex]. I want to represent the vector from [itex] p(t) [/itex] to [itex] p' [/itex] by an orthogonal coordinate system that is attached to the curve.

    Let [itex] \hat{s}(t) [/itex] be the unit tangent vector to the curve at the point [itex] p(t) [/itex]. This defines a plane where the point [itex] p' [/itex] lies on a circle of radius [itex] \rho [/itex] in the plane. The only other thing I need to uniquely define the point [itex] p' [/itex] in this plane is some reference direction to measure the angle at which the point [itex] p' [/itex] lies on the circle. Call this direction [itex] \hat{\omega}(t) [/itex].

    Here's the kicker though - I want [itex] \hat{\omega}(t) [/itex] to be defined in such a way that when the curve is a straight line, and in the z-direction, that the coordinate system becomes plain old cylindiral coordinates, and [itex] \hat{\omega}(t) = \hat{x}[/itex] . However, when the curve is not a straight line, then [itex] \hat{\omega} [/itex] should rotate with the curve in such a way that it is always orthogonal to [itex] \hat{s}(t) [/itex], and locally if [itex] \hat{s}(t) \simeq \hat{z} [/itex] then [itex] \hat{\omega}(t) \simeq \hat{x} [/itex].

    My question is how to define [itex] \hat{\omega}(t) [/itex] in such a way that satisfies these constraints.

    Does this make sense? I hope at least the idea of what I want to do is clear. I further hope this problem is not ill-posed. Any help would be appreciated and please let me know if something is unclear.
  2. jcsd
  3. May 17, 2016 #2


    User Avatar
    Gold Member

  4. May 17, 2016 #3
    Hi, yes I remembered that as I was writing this, but I'm still not sure how to enforce that the Frenet-Serret frame becomes normal cylindrical coordinates in the limit that the curvature goes to zero (assuming that any straight line will be in the z-direction).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted