Differential geometry notes 3

Let $\mathbf{x}(t)$ be a parametric representation of a curve $C$ . Two points of $C$ , say $P$ and $P_1$ , determine a straight line $S$ . If $P_1$ tends to $P$ , then $S$ tends to the tangent to $C$ at $P$ .

Let us ask similarly for the limit position of a plane $E$ passing through three points $P$ , $P_1$ , $P_2$ of $C$ when both $P_1$ and $P_2$ tend to $P$ .

Let $\mathbf{x}(t)$ , $\mathbf{x}(t+h_1)$ and $\mathbf{x}(t+h_2)$ be the parametric representations of $P$ , $P_1$ , $P_2$ respectively. The chords $PP_1$ and $PP_2$ are given by the vectors $\mathbf{a}_i = \mathbf{x}(t+h_i) - \mathbf{x}(t)$ . These two vectors span $E$ . $E$ is thus also spanned by $\mathbf{v}^{(i)} = \mathbf{a}_i / h_i$ and thus by the vectors:

$\mathbf{v}^{(i)}$ and $\mathbf{w}= \frac{2(\mathbf{v}^{(2)}-\mathbf{v}^{(1)})}{h_2-h_1}$

Using the Taylor expansion:

$\mathbf{v}^{(i)} = \mathbf{x^\prime}(t) + \frac{1}{2}h_1 \mathbf{x^{\prime \prime}}(t) + o(h_1)$

$\mathbf{w} = \mathbf{x^{\prime \prime}}(t) + o(1)$

Thus as $h_i \rightarrow 0$ , $\mathbf{v}^{(i)} \rightarrow \mathbf{x^\prime}(t)$ and $\mathbf{w} \rightarrow \mathbf{x^{\prime \prime}}(t)$ .

Osculating plane: The plane spanned by $\mathbf{x^\prime}(t)$ and $\mathbf{x^{\prime \prime}}(t)$ is called the osculating plane of the curve at $P$ . For any point $\mathbf{z}$ on the osculating plane: $|(\mathbf{z}-\mathbf{x})\mathbf{x^\prime} \mathbf{x^{\prime \prime}}|=0$ where:

$|\mathbf{a}\mathbf{b}\mathbf{c}|$ $= \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ $= \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$

where $\cdot$ denotes the dot product and $\times$ denotes the vector product.

Principal normal: The intersection of the osculating plane with the corresponding normal plane is called the principal normal.

Unit principal normal vector: The unit vector $\mathbf{p}(s) = \frac{\mathbf{\dot{t}}(s)}{|\mathbf{\dot{t}}(s)|}$ is called the unit principal normal vector. This vector lies both in the osculating and the normal planes.

Curvature: the length of the vector $\mathbf{\dot{t}}(s)$ is called the curvature at point $\mathbf{x}(s)$ :

$\kappa(s) = |\mathbf{\dot{t}}(s)| = \sqrt{\mathbf{\ddot{x}}(s)\cdot \mathbf{\ddot{x}}(s)}$ ( $\kappa >0$ )

or in terms of any arbitrary parametric representation:

$\kappa(s) = |\mathbf{\dot{x}}\times \mathbf{\ddot{x}}| = |\mathbf{x^\prime} \times \mathbf{x^{\prime \prime}}|\big( \frac{dt}{ds}\big)^3$ . This is equal to:

$\kappa(s) = \frac{\sqrt{ (\mathbf{x^\prime}\cdot \mathbf{x^\prime}) (\mathbf{x^{\prime \prime}} \mathbf{x^{\prime \prime}}) - (\mathbf{x^\prime} \cdot \mathbf{x^{\prime \prime}})^2 }}{(\mathbf{x^\prime}\cdot \mathbf{x^\prime})^{3/2}}$

Primes denote derivatives with respect to $t$ , dots denote derivatives with respect to the natural parameter $s$ .

Radius of curvature: The reciprocal of curvature is called the radius of curvature:

$\rho(s) = \frac{1}{\kappa(s)}$

Center of curvature: The point $M$ on the positive ray of the principal normal at distance $\rho(s)$ from $\mathbf{x}(s)$ . The position vector of $M$ is given by $\mathbf{z} = \mathbf{x} + \rho \mathbf{p} = \mathbf{x} + \rho^2 \mathbf{\dot{t}}$ .

Binormal: To every point of the curve, we associate three orthogonal unit vectors:

1) $\mathbf{t}(s) = \mathbf{\dot{x}}(s)$ (unit tangent vector)

2) $\mathbf{p}(s) = \frac{\mathbf{\dot{t}}(s)}{|\mathbf{\dot{t}}(s)|} = \rho(s) \mathbf{\ddot{x}}(s)$ (unit principal normal vector)

The third vector is the vector product of these two unit vectors:

3) $\mathbf{b}(s) = \mathbf{t}(s) \times \mathbf{p}(s)$

This is called the unit binormal vector (make sure you understand why this vector is a unit vector and it is orthogonal to the other two vectors).

Moving trihedron: The triple $\mathbf{t}, \mathbf{p}, \mathbf{b}$ is called the moving trihedron of the curve.

$\mathbf{t}$ and $\mathbf{p}$ span the osculating plane: $(\mathbf{z} - \mathbf{x})\cdot \mathbf{b} = 0$ .
$\mathbf{p}$ and $\mathbf{b}$ span the normal plane: $(\mathbf{z} - \mathbf{x})\cdot \mathbf{t} = 0$ .
$\mathbf{t}$ and $\mathbf{b}$ span the rectifying plane: $(\mathbf{z} - \mathbf{x})\cdot \mathbf{p} = 0$ .

Torsion: The vector $\mathbf{\dot{b}}$ is orthogonal to $\mathbf{t}$ and $\mathbf{b}$ (make sure you understand why), consequently it lies in the principal normal. Thus, we may set: $\mathbf{\dot{b}}(s)=-\tau(s) \mathbf{p}(s)$ or $\tau(s) = -\mathbf{p}(s) \cdot \mathbf{\dot{b}}(s)$

$\tau(s)$ is called the torsion of the curve at point $\mathbf{x}(s)$ . In terms of $\mathbf{x}(s)$ and its derivatives, torsion is given by:

$\tau = \rho^2 |\mathbf{\dot{x}}\mathbf{\ddot{x}}\mathbf{\dddot{x}}| = \frac{|\mathbf{\dot{x}}\mathbf{\ddot{x}}\mathbf{\dddot{x}}|} {\mathbf{\ddot{x}}\cdot \mathbf{\ddot{x}}}$ .

The torsion, roughly speaking, measures the magnitude and direction of deviation of a curve from the osculating plane in the neighborhood of a point of the curve.

Differential geometry notes 3

Published by Emin Orhan

Leave a comment Cancel reply

Share this:

Related

Published by Emin Orhan

Leave a comment Cancel reply