📂Geometry

Properties of Vector Fields Parallel to a Curve

Properties

Let $\mathbf{X}(t)$ and $\mathbf{Y}(t)$ be vectors parallel to a regular curve $\alpha (t)$ on the surface $M$. Then the angle between $\mathbf{X}$ and $\mathbf{X}(t), \mathbf{Y}(t)$, and the magnitude of $\left\| \mathbf{X}(t) \right\|$ are constants.

Description

In other words, both the angle and magnitude are conserved.

Proof

Let $f(t) = \left\langle \mathbf{X}(t), \mathbf{Y}(t) \right\rangle$. Differentiating $f$, by the differentiation of inner products, we get:

$$\dfrac{d f}{d t} = \left\langle \dfrac{d \mathbf{X}}{d t}, \mathbf{Y} \right\rangle + \left\langle \mathbf{X}, \dfrac{d \mathbf{Y}}{d t} \right\rangle = 0 + 0 = 0$$

Here, $\mathbf{X}(t), \mathbf{Y}(t)$ is the tangent vector of $M$, and $\dfrac{d \mathbf{X}}{d t}(t), \dfrac{d \mathbf{Y}}{d t}(t)$, by definition, is orthogonal to the tangent vector, so the inner product is $0$. Hence $f(t)$ is constant. If we set $\mathbf{Y}=\mathbf{X}$, we get that $\left\| \mathbf{X}(t) \right\|$ is also constant.

Now, if we denote the angle between $\mathbf{X}(t)$ and $\mathbf{Y}(t)$ as $\theta$, we obtain:

$$\dfrac{f(t)}{\left\| \mathbf{X} (t) \right\| \left\| \mathbf{Y}(t) \right\|} = \cos \theta$$

Since $f(t), \left\| \mathbf{X} (t) \right\|, \left\| \mathbf{Y}(t) \right\|$ are both constants, $\cos \theta$ is also constant. Therefore, the angle between them is constant.

■