# 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:

- If we differentiate both sides of this equation, we get that:

- 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 $s$.

- 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:

- 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)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)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. |