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:
\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:
\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 $s$.
\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:
\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:

\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:

\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.
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