📂Geometry

The Definition of Strings

Definition ¹

1. Let there be a given curve $\alpha : (c,d) \to \mathbb{R}^{3}$. When $c < a < b < d$, for all $t \in [a,b]$ that satisfies $\alpha (t) = \gamma (t)$, $\gamma : [a,b] \to \mathbb{R}^{3}$ is called a Chord or Curve Segment.
2. The length of chord $\gamma : [a,b] \to \mathbb{R}^{3}$, denoted $l_{[a,b]}(\gamma)$, is defined as follows. $$l_{[a,b]}(\gamma) := \int_{a}^{b} \left| {{ d \gamma } \over { d t }} \right| dt$$
3. The Length of Arc is defined as follows for $s = h(t)$ according to $\alpha$. $$h(t) := \int_{t_{0}}^{t} \left| {{ d \alpha } \over { dt }} \right| dt$$

Explanation

In the definition, the curve segment $\gamma$ is naturally defined as a part of $\alpha$, and its definition of length readily follows from the concept of the Jacobian. The idea of the length of an arc is considered as a convenient way to avoid tediously redefining and discussing the length of a chord for the interval $\left[ t, t_{0} \right]$ we are interested in.

Reparameterization of the Length of an Arc

Especially, the function $h$ is a reparameterization itself and is called the reparameterization of the length of an arc if it satisfies the following for $g: (c,d) \to (a,b)$ and $\displaystyle h(t) = \int_{t_{0}}^{t} \left| {{ d \alpha } \over { du }} \right| du$. $$g(s) = h^{-1} (s)$$ This is nothing but a well-known variable transformation. Let’s understand why this is used. $$\begin{align*} \beta =& \alpha \circ g \\ =& \alpha \circ h^{-1} \end{align*}$$ If it is given that $\alpha = \beta \circ h$, and since $h$ is an increasing function, then $\left| h^{\prime} \right| = h^{\prime}$, thus $$\begin{align*} s =& \int_{t_{0}}^{t} \left| \alpha^{\prime} (u) \right| du \\ =& \int_{t_{0}}^{t} \left| \left( \beta \circ h \right)^{\prime} \right| du \\ =& \int_{t_{0}}^{t} \left| \left( \beta^{\prime} \circ h \right) \right| h^{\prime}(u) du \\ =& \int_{0}^{s} \left| \beta^{\prime} (v) \right| dv \end{align*}$$ Differentiating both sides with respect to $s$, according to the Fundamental Theorem of Calculus, $$1 = \left| \beta^{\prime} (s) \right|$$ This means the Speed is constant at $1$, making it easier to handle.

Furthermore, from the following theorem, one can understand that the reparameterization of a curve does not alter the length of the curve. Since $g$ moves by $(c,d)$, $\alpha$ moves by $(a,b)$, which is obvious. By analogy, the path from the starting point to the destination remains the same, only the mode of transportation changes, therefore the distance remains unchanged while only the speed changes.

Theorem

For reparameterization, the length of the chord is invariant.

Proof

For the chord $\alpha$ and reparameterization $g : [c,d] \to [a,b]$, suppose $\beta := \alpha \circ g$, then the length of $\beta$ is as follows. $$\begin{align*} \int_{c}^{d} \left| {{ d \beta } \over { d t }} \right| dr =& \int_{c}^{d} \left| \left( {{ d \alpha } \over { d t }} \right) \left( {{ d g } \over { d r }} \right) \right| dr \\ =& \int_{c}^{d} \left| {{ d \alpha } \over { d t }} \right| \left| {{ d g } \over { d r }} \right| dr \end{align*}$$

---

Case 1. If $g$ is an increasing function

Since $g(c) = a, g(d) = b$ and $\left| {{ d g } \over { d r }} \right| = {{ d g } \over { d r }}$, $$\begin{align*} \int_{c}^{d} \left| {{ d \alpha } \over { d t }} \right| \left| {{ d g } \over { d r }} \right| dr =& \int_{c}^{d} \left| {{ d \alpha } \over { d t }} \right| {{ d g } \over { d r }} dr \\ =& \int_{c}^{d} \left| {{ d \alpha } \over { d t }} \right| d g \\ =& \int_{a}^{b} \left| {{ d \alpha } \over { d t }} \right| d t \end{align*}$$

---

Case 2. If $g$ is a decreasing function

Since $g(c) = b, g(d) = a$ and $\left| {{ d g } \over { d r }} \right| = - {{ d g } \over { d r }}$, $$\begin{align*} \int_{c}^{d} \left| {{ d \alpha } \over { d t }} \right| \left| {{ d g } \over { d r }} \right| dr =& \int_{c}^{d} \left| {{ d \alpha } \over { d t }} \right| \left( - {{ d g } \over { d r }} \right) dr \\ =& - \int_{c}^{d} \left| {{ d \alpha } \over { d t }} \right| d g \\ =& - \int_{b}^{a} \left| {{ d \alpha } \over { d t }} \right| d t \\ =& \int_{a}^{b} \left| {{ d \alpha } \over { d t }} \right| d t \end{align*}$$

---

Therefore, whether $g$ is an increasing or decreasing function, the length of the chord remains constant.

■