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I've been struggling with this one for about $3$ weeks:

What is the integral of a $\mathbb{R}^3$ curve with respect to its curvature?

I though about approaching it with the Ferret-S formulas, and doing a change of variables or lookings at the curvature as a $1$-form, but I'm stumped.

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Could you please put this question into a little more context? Your idea worries me. We integrate with respect to a parameter that parametrises the curve. What happens when $\kappa' = 0$? –  Fly by Night Oct 28 '12 at 22:24

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Are you trying to generalize what you would do in $\mathbb{R}^2$, where the integral of the curvature is an angle that represents the total change of direction along the curve?

If this is what you are asking, it's a very interesting question.

In $\mathbb{R}^3$ it can be done, but it is much, much more complicated. The reason why it is complicated is torsion. Without torsion, the curve is planar, and you can do exactly like in the plane.

To generalize to $\mathbb{R}^3$, let's look at the problem from a "holonomic" point of view.

What you actually do in $\mathbb{R}^2$ is trying to compose infinitesimal rotations, one after each other, along the curve. In "complex notation", if $k$ is the curvature, the rotation you are (probably) looking for is: $$ R=\exp(i\,k\,\Delta t_n)\exp(i\,k\,\Delta t_{n-1})\dots\exp(i\,k\,\Delta t_1). $$

where $i$ denotes the matrix: $$ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. $$

For example, if the curvature is constant, you simply obtain: $$ R=\exp(i\,k\,t)= \exp \begin{pmatrix} 0 & -kt \\ kt & 0 \end{pmatrix} = \begin{pmatrix} \cos(kt) & -\sin(kt) \\ \sin(kt) & \cos(kt) \end{pmatrix}. $$

If the curvature is not constant, you have: $$ R=\exp \bigg(i\int k\,dt \bigg). $$

The integral of the curvature is the total angle of rotation, and the total rotation transformation is written above.

You are absolutely right to consider $k\,dt$ a 1-form. You can think of it as an angle-valued 1-form. The problem is that in more dimension this picture does not generalize in this simple way! The reason is mostly because rotations in the plane belong to the group $SO(2)$, which is Abelian. There you can sum different rotations and interchange them. In three dimensions, the group $SO(3)$ is not abelian anymore, and you cannot interchange rotations around different axes (try rotating a book). So the composition of exponentials above does not become the exponential of an integral.

Even if the picture gets more complicated, we can solve your problem. Let's do it!

Angles are not numbers you can sum (except in 2 dimensions), because they do not form a vector space. Only infinitesimally (that is, in a suitable tangent space) they can. Think, in physics, of the "angular velocity" vector. That vector space is called Lie algebra. Every rotation group has its Lie algebra. $SO(2)$ has the algebra $so(2)$, and $SO(3)$ has $so(3)$.

The Lie algebra of rotations is made by anti-symmetric matrices.

Let's see why. We want them to be "infinitesimal transformations that preserve lengths" (that's what rotations do). If $v$ is a vector and $M=1+\epsilon$, where 1 is the identity and $\epsilon$ is a "small deviation", we want: $$v\cdot v = (Mv)\cdot (Mv)$$ $$v\cdot v = ((1+\epsilon)v)\cdot ((1+\epsilon)v)$$ $$v^t v = v^t(1+\epsilon)^t (1+\epsilon)v$$ $$1 = (1+\epsilon)^t (1+\epsilon)$$ $$1 = 1+\epsilon^t + \epsilon + O(\epsilon^2)$$ $$ \epsilon^t = - \epsilon + O(\epsilon^2)$$

So the $\epsilon$, which work at first order, have to be anti-symmetric.

In two dimensions, the only anti-symmetric matrix is our matrix $i$, and its scalar multiples. Therefore the Lie algebra of rotations in a plane is only 1-dimensional. We can see the form $i\,k\,dt$, or: $$ \begin{pmatrix} 0 & -k\,dt \\ k\,dt & 0 \end{pmatrix}, $$

as a Lie algebra-valued 1-form. To "integrate" such a 1-form you also have to exponentiate it. Intuitively, the infinitesimal rotation is $1+i\,k\,dt\approx \exp(i\,k\,\Delta t)$.

Composing many of these, you get exactly the expression above, and you can find the "integral" rotation $R$.

Now, instead of the matrix $i$, in 3 dimensions you have the matrix: $$ \begin{pmatrix} 0 & -k & 0 \\ k & 0 & -\tau \\ 0 & \tau & 0 \end{pmatrix}, $$

where $k$ is the curvature, and $\tau$ is the torsion. This is an anti-symmetric matrix, it belongs to $so(3)$. As you can see, $so(3)$ is 3-dimensional, so this (varying $k$ and $\tau$) spans a 2-dimensional subspace. We can write this matrix as $k\,i + \tau\,j$, where: $$ i= \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},\quad j= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 1 & 0 & 0 \end{pmatrix}. $$

If you look carefully, $i$ is a "small" rotation around the $z$ axis. $j$ around the $x$ axis! (Try to calculate the exponentials of $i\,k\,t$ and $j\,\tau\,t$ separately, and see). The problem is that $i$ and $j$ can not be interchanged! (Remember the trick with the book?)

For matrix exponentials, it is in general not true that: $$ \exp(A)\exp(B) = \exp(A+B), $$

this is true only if $A$ and $B$ commute. So, the composition of "small" rotations that you see in the first equation does not become the exponential of a global big integral. In rigor, to integrate the Lie-algebra valued 1-form $\omega=(k\,i+\tau\,j)\,dt$ we have to use another operation, called ordered exponential. This is obtained as: $$ \lim_{n\to\inf} \exp(\omega \Delta t_n/n) \exp(\omega \Delta t_{n-1}/n)\dots\exp(\omega \Delta t_1/n). $$

where $\omega$ is any Lie algebra-valued 1-form. It becomes the exponential of the integral if and only if the various pieces commute.

Notationwise, we write: $$ \operatorname{Pexp}\bigg( \int\omega\,dt \bigg), $$

but it is not true that we have first evaluated the integral of $\omega$, and then "exponentiated" in some way. It is just notation!

Our integral rotation in $\mathbb{R}^3$ will then be: $$ R = \operatorname{Pexp}\bigg( \int(k\,i+\tau\,j)\,dt \bigg). $$

This is not a "global angle" of rotation, because there is no such thing in three dimensions. But it is the global rotation matrix, and in a way, it generalizes exactly what you do in two dimensions.

The ordered exponential, as an operation, can be approximated using either a Dyson series, or a recursion. To see what it looks like, look at the second link.

I hope that this is what you were looking for, and that it is clear enough.

For more information: http://en.wikipedia.org/wiki/Lie_algebra http://en.wikipedia.org/wiki/Rotation_group_SO(3) http://en.wikipedia.org/wiki/Matrix_exponential http://en.wikipedia.org/wiki/Exponential_map#Lie_theory http://en.wikipedia.org/wiki/Ordered_exponential

...or ask me in a comment!

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