Differential geometry notes 4

by Emin Orhan

Formulae of Frenet: These describe the derivatives of the moving trihedron of a curve in terms of the moving trihedron itself:

\begin{bmatrix}    \mathbf{\dot{t}} \\    \mathbf{\dot{p}} \\    \mathbf{\dot{b}}    \end{bmatrix} = \begin{bmatrix}    0 & \kappa & 0 \\    -\kappa & 0 & \tau \\    0 & -\tau & 0    \end{bmatrix} \begin{bmatrix}    \mathbf{t} \\    \mathbf{p} \\    \mathbf{b}    \end{bmatrix}

Note that the coefficients inside the skew-symmetric matrix need not be constant, hence the system described by this equation is not necessarily linear.

Definition 16.1: A vector \mathbf{d} is called a rotation vector of a rotation if it has the following properties:

  1. \mathbf{d} has the direction of the axis of rotation.
  2. The sense of \mathbf{d} is such that the rotation has the clockwise sense if one looks from the initial point of \mathbf{d} to its terminal point.
  3. The magnitude |\mathbf{d}| of \mathbf{d} equals the angular velocity \omega of the rotation, that is, the velocity of points at distance 1 from the axis of rotation.

Theorem 16.2: The rotation vector of the trihedron of a curve C: \mathbf{x}(s) of class r \geq 3 with non-vanishing curvature, when a point moves along C with constant velocity 1 is given by the expression:

\mathbf{d} = \tau \mathbf{t} + \kappa \mathbf{b}

This vector is called the vector of Darboux.

The curves traced out by the terminal points of \mathbf{t}(s), \mathbf{b}(s) and \mathbf{p}(s) on the unit sphere as the trihedron of a curve moves are called the tangent, binormal and principal normal indicatrix respectively. The linear elements of these curves ds_T, ds_P and ds_B are given by:

ds_T^2 = \kappa^2 ds^2

ds_P^2 = (\tau^2 + \kappa^2) ds^2

ds_B^2 = \tau^2 ds^2

Hence, we can write ds_P^2 = ds_T^2 + ds_B^2 (Equation of Lancret).

Shape of a curve in the neighborhood of any of its points: We can Taylor expand the representation of the curve around s = 0 (note that the zero point can be chosen arbitrarily):

\mathbf{x}(s) = \mathbf{x}(0) + \sum_{v = 1}^3 \frac{s^v}{v!} \frac{d^v \mathbf{x}(0)}{ds^v} + o(s^3)

We have \mathbf{\dot{x}} = \mathbf{t}, \mathbf{\ddot{x}}=\mathbf{\dot{t}} = \kappa \mathbf{p}, and \mathbf{\dddot{x}} = \mathbf{\ddot{t}}= \dot{\kappa}\mathbf{p} + \kappa \mathbf{\dot{p}} = \dot{\kappa} \mathbf{p} - \kappa^2 \mathbf{t} + \kappa \tau \mathbf{b}. We can choose the coordinate system such that \mathbf{t}(0)=(1,0,0), \mathbf{p}(0)=(0,1,0), \mathbf{b}(0)=(0,0,1).

Plugging these expressions in the Taylor expansion, we get:

x_1(s) = s - \frac{\kappa_0^2}{3!}s^3 + o(s^3)

x_2(s) = \frac{\kappa_0}{2}s^2 + \frac{\dot{\kappa_0}}{3!}s^3 + o(s^3)

x_3(s) = \frac{\kappa_0 \tau_0}{3!}s^3 + o(s^3)

where \kappa_0 and \tau_0 are the values of the curvature and torsion at s=0. If we keep only the leading terms, we get:

\mathbf{x}(s) \approx (s, \frac{\kappa_0}{2} s^2, \frac{\kappa_0 \tau_0}{6}s^3)              (\kappa_0 >0, \tau_0 \neq 0)

 

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