The Fundamental Theorem of Space Curves

# The Fundamental Theorem of Space Curves

We are now going to look at an extremely important theorem known as The Fundamental Theorem of Space Curves which says that two curves $C_1$ and $C_2$ with the same non-vanishing curvature function $\kappa (s)$ and the same torsion function $\tau (s)$ will be congruent - that is each curve can be rigidly shifted so that all points on the curve coincide.

 Theorem (The Fundamental Theorem of Space Curves): Let $\vec{r_1}(s)$ and $\vec{r_2}(s)$ be two vector-valued functions that represent the space curves $C_1$ and $C_2$ respectively, and suppose that these curves have the same non-vanishing curvature $\kappa (s)$ and the same torsion $\tau (s)$. Then $C_1$ and $C_2$ are congruent such that each can be rigidly shifted/rotated so that every point on $C_1$ coincides with every point on $C_2$.
• Proof: Let $C_1$ and $C_2$ be defined by the vector-valued functions $\vec{r_1}(t)$ and $\vec{r_2}(t)$ respectively, and without loss of generality, rigidly shift/rotate $C_1$ so that its initial point at $s = 0$ coincides with the initial point of $C_2$ at $s = 0$ and so that the Frenet frames of both points also coincide. Let $\hat{T_1}(s), \hat{T_2}(s)$ be the unit tangent vectors that correspond to $s$, $\hat{N_1}(s), \hat{N_2}(s)$ be the unit normal vectors that correspond to $s$, and let $\hat{B_1}(s) , \hat{B_2}(s)$ be the unit binormal vectors that correspond to $s$; respectively to the curves $C_1$ and $C_2$, and define $f(s)$ as follows:
(1)
\begin{align} \: f(s) = \hat{T_1}(s) \cdot \hat{T_2}(s) + \hat{N_1}(s) \cdot \hat{N_2}(s) + \hat{B_1}(s) \cdot \hat{B_2}(s) \end{align}
• If we differentiate both sides of this equation, we get that:
(2)
\begin{align} \quad \quad f'(s) &= \hat{T_1}(s) \cdot \left [ \frac{d}{ds} \hat{T_2}(s)\right ] + \left [ \frac{d}{ds} \hat{T_1}(s) \right ] \cdot \hat{T_2} (s) + \hat{N_1}(s) \\ \quad & \cdot \left [ \frac{d}{ds} \hat{N_2}(s)\right ] + \left [ \frac{d}{ds} \hat{N_1}(s) \right ] \cdot \hat{N_2} (s) + \hat{B_1}(s) \cdot \left [ \frac{d}{ds} \hat{B_2}(s)\right ] + \left [ \frac{d}{ds} \hat{B_1}(s) \right ] \cdot \hat{B_2} (s) \\ \quad \quad f'(s) &= \hat{T_1}(s) \cdot \left [ \kappa (s) \hat{N_2}(s) \right ] + \left [ \kappa(s) \hat{N_1}(s) \right ] \cdot \hat{T_2} (s) + \hat{N_1}(s) \cdot \left [ -\kappa (s) \hat{T_2}(s) + \tau (s) \hat{B_2}(s) \right ] \\ & + \left [ -\kappa (s) \hat{T_1}(s) + \tau (s) \hat{B_1} (s) \right ] \cdot \hat{N_2} (s) + \hat{B_1}(s) \cdot \left [ -\tau (s) \hat{N_2}(s)\right ] + \left [ -\tau (s) \hat{N_1}(s) \right ] \cdot \hat{B_2} (s) \\ \quad \quad f'(s) &= \kappa (s) \left [ \hat{T_1}(s) \cdot \hat{N_2}(s) + \hat{T_2}(s) \cdot \hat{N_1}(s) \right ] - \kappa(s) \left [ \hat{T_2}(s) \cdot \hat{N_1}(s) + \hat{T_1}(s) \cdot \hat{N_2}(s) \right ] \\ &+ \tau (s) \left [ \hat{N_1}(s) \cdot \hat{B_2}(s) + \hat{N_2}(s) \cdot \hat{B_1}(s) \right ] - \tau(s) \left [ \hat{N_2}(s) \cdot \hat{B_1}(s) + \hat{N_1}(s) \cdot \hat{B_2}(s) \right ] \\ \quad f'(s) &= 0 \end{align}
• Since $f'(s) = 0$, this implies that $f(s)$ is a constant. For $s = 0$ we have that $f(s) = 3$. Since each of these dot products cannot exceed $1$, we deduce that dot product must be equal to $1$ for all $s$.
(3)
\begin{align} \quad \quad \hat{T_1}(s) \cdot \hat{T_2}(s) + \hat{N_1}(s) \cdot \hat{N_2}(s) + \hat{B_1}(s) \cdot \hat{B_2}(s) = \| \hat{T_1}(s) \|^2 + \| \hat{N_1}(s) \|^2 + \| \hat{B_1}(s) \|^2 = 1 + 1 + 1 = 3 \end{align}
• In particular we have that for all $s$, $\hat{T_1}(s) = \hat{T_2}(s)$. If we integrate both sides of this equation, we get that:
(4)
\begin{align} \hat{T_1}(s) = \hat{T_2}(s) \\ \frac{d \vec{r_1}(s)}{ds} = \frac{d \vec{r_2}}{ds} \\ \int \frac{d \vec{r_1}(s)}{ds} \: ds = \int \frac{d \vec{r_2}}{ds} \\ \vec{r_1}(s) = \vec{r_2}(s) + C \end{align}
• But notice that for $s = 0$ we have that $\vec{r_1}(0) + \vec{r_2}(0) = 0$, which implies that $C = 0$.
• Since $\vec{r_1}(s)$ and $\vec{r_2}(s)$ start at the same initial point and are parameterized in terms of arc-length which traverses the curves at unit speed, we get that it must be that $\vec{r_1}(s) = \vec{r_2}(s)$ for all defined $s$, and so the curves $C_1$ and $C_2$ are congruent. $\blacksquare$

## Example 1

Prove that if a curve $C_1$ has constant curvature $\kappa (s) = c \neq 0$ and zero torsion, $\tau (s) = 0$ then the curve is a circle.

Consider a circle parameterized as $\vec{r}(t) = \left ( \frac{1}{c} \cos t, \frac{1}{c} \sin t \right)$ for $0 ≤ t ≤ 2 \pi$. Let's first find the arc length parameterization of this curve. We first differentiate $\vec{r}(t)$ to get $\vec{r'}(t) = \left (-\frac{1}{c} \sin t, \frac{1}{c} \cos t \right )$. The norm $\| \vec{r'}(t) \| = \sqrt{\frac{1}{c^2} \sin ^2 t + \frac{1}{c^2} \cos ^2 t} = \frac{1}{c}$. Therefore we have that the arc length $s$ measured from $0$ is:

(5)
\begin{align} \quad s = \int_0^t \frac{1}{c} \: du = \frac{1}{c} \int_0^t 1 \: du = \frac{t}{c} \end{align}

Therefore we have that $s = \frac{t}{c}$ so $t = sc$. The arc length parameterization of our circle for $0 ≤ t ≤ \frac{2 \pi}{c}$ is therefore:

(6)
\begin{align} \quad \vec{r}(s) = \left ( \frac{1}{c} \cos (sc), \frac{1}{c} \sin (sc) \right ) \end{align}

Now this circle has radius $\frac{1}{c}$ so its curvature is $c$.

Now we can calculate the torsion of this curve with the formula $\hat{B'}(s) = - \tau (s) \hat{N}(s)$. If we do so, we'll see that $\tau (s) = 0$. So we have that this circle has curvature $C$ and torsion $0$.

By The Fundamental Theorem of Space Curves, any curve $C_2$ with the same curvature and torsion is congruent to this circle $C_1$, that is, $C_2$ is a circle.

 Remark 1: It can be shown that if $C_1$ is a curve for which $\kappa (s) = c \neq 0$ and $\tau (s) = d \neq 0$ then the curve is a circular helix. This can be proven in the same manner as above.