Saturday, February 9, 2013

An Interpretation of Curvature: Osculating Plane and Circle

To be very clear, this post follows Spivak Vol 2 pretty closely.
Suppose that we have some curve in space, and we have chosen a point $P$ on it. We are concerned with the following three questions:
  1. There is no guarantee that the curve lies in a plane, but if there were a plane through $P$ which came closest to containing the curve, how would you find it?
  2. Can we find a circle which serves as a good approximation to the curve near $P$?
  3. Is there a simple and reasonable interpretation of the curvature of the curve at $P$?
Of course, the answers are all "yes." The plane is called the osculating plane, the circle is the osculating circle, and the curvature gets an interpretation from the size of the osculating circle. The goal of this post is to show that these objects exist and have real geometric significance.


To make progress, we will employ a simple argument by using a powerful tool repeatedly.

The Mean Value Theorem (MVT)

Suppose that $f$ is a real-valued function which is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists a value $c \in (a,b)$ such that
\[ f'(c) = \dfrac{f(b) - f(a)}{b-a} .\]

One view of the MVT is that it tells us about the shape of planar curves written as graphs. Can we generalize it? Yes, some...

The Extended Mean Value Theorem (or EMVT)

Suppose that $f$ and $g$ are real-valued functions which are continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists a value $c \in (a,b)$ such that
\[ f'(c)( g(b) - g(a) ) = g'(x) ( f(b) - f(a) ) .\]

The extended mean value theorem means that for a curve in the plane described by $t \mapsto (f(t), g(t))$ the tangent at $t = c$ is parallel to the chord between $t=a$ and $t=b$. This is very useful in dealing with planar curves. It would be really nice if we could generalize it further...

No More Mean Value Theorems

There is not an effective form of MVT for curves in $\mathbb{R}^3$. One is tempted, but alas, the following simple example shows that there is no hope for a simple analogous statement for this case:
Consider a circular helix. For definiteness, try this simplest one.
\[ \gamma(t) = ( \cos(t), \sin(t), t) . \]
For the inputs $t=0$ and $t=2\pi$, we see that the chord is a multiple of $(0,0,1)$, but at no point does the helix have a vertical tangent.

This example teaches us that we will have to stick to real-valued functions to make our arguments with the Mean Value Theorem.

And Now... Curves in Space

When we last left our heros, we were searching for some way to understand the local geometry of a curve near a given point. At this point, we are searching for answers to the questions above. (Not quite the meaning and purpose of life but still...)


For what we do next to work, we make the following assumption: $\gamma$ is a $C^2$ curve with $\gamma''(s) \neq 0$. This makes it possible to define an osculating plane unambiguously:

Definition: If $\gamma$ is a space curve defined in some neighborhood of a point $s$ with $\gamma'(s)\neq 0$ and $\gamma''(s) \neq 0$, then the osculating plane of $\gamma$ at $s$ is the plane spanned by $\gamma'(s)$ and $\gamma''(s)$.

It should be noted that the osculating plane is really a plane through the origin by this definition. We are making a standard abuse by pretending that this plane is "centered" at the point $\gamma(s)$. If this really bothers you, redefine things by translating by $\gamma(s)$.

We want to show that the osculating plane (circle) is the plane (resp. circle) through 3 "consecutive" points on $\gamma$ in a neighborhood of $\gamma(s)$. 

Theorem: Let $\gamma$ be a unit speed curve defined on some neighborhood of the parameter value $s$. Assume that $\gamma''(s) \neq 0$. For $s_1, s_2, s_3$ sufficiently close to $s$, the points $\gamma(s_1)$, $\gamma(s_2)$ and $\gamma(s_3)$ do not lie on a single line in $\mathbb{R}^3$. Thus these three points uniquely define a plane and a circle through them. As $s_1, s_2, s_3$ tend to $s$, these planes and circles limit onto the osculating plane and a circle lying in that plane. The radius of that circle is the reciprocal of the curvature of $\gamma$ at $s$.

Proof:
We will fix $s$ for this whole discussion and work near it. Let $P = \gamma(s)$. So consider $s_1, s_2, s_3$ to be three values of the parameter near $s$. Without loss of generality, we take these values to be ordered $s_1 < s_2 < s_3$.

It should be remembered that since $\gamma$ is unit speed, we know that $\gamma'(s)$ and $\gamma''(s)$ are orthogonal, and hence they span a plane. By definition this is the osculating plane of $\gamma$ at $s$.

First, we wish to show that for the $s_i$'s sufficiently close to $s$, those three points cannot all lie on a single line.

Suppose that for a choice of $s_1, s_2$ and $s_3$ we have that $\gamma(s_1)$, $\gamma(s_2)$ and $\gamma(s_3)$ all lie on a single line, $\ell$. For any vector $V$ which is perpendicular to $\ell$, we shall consider the function
\[ f(t) = \langle \gamma(t) - \gamma(s_1), V \rangle . \]
This function is $C^2$ and takes the value $0$ at the points $s_1$, $s_2$, and $s_3$. So by the Mean Value Theorem, there exist parameter values $\xi_1$ and $x_2$ which satisfy $s_1 < \xi_1 < s_2 < \xi_2 < s_3$, and
\[ 0 = f'(\xi_1) = \langle \gamma'(\xi_1), V \rangle  = f'(\xi_2) = \langle \gamma'(\xi_2), V \rangle . \]
Applying the Mean Value Theorem to $f'$, we now see that there is a point $\eta$ with $\xi_1 < \eta < \xi_2$ and
\[ 0 = f''(\eta) = \langle \gamma''(\eta), V \rangle .\]
The set of vectors perpendicular to $\ell$ is an entire plane. If we could choose the $s_i$'s arbitrarily close to $s$, the continuity of $\gamma'$ and $\gamma''$ and the two equations above show us that there are a whole plane of vectors which are nearly orthogonal to both $\gamma'(s)$ and $\gamma''(s)$, hence nearly orthogonal to the osculating plane. But there is not enough room for this to happen in $\mathbb{R}^3$: the set of vectors nearly orthogonal to a given plane should be a cone centered around the true orthogonal complement, which is a line. We conclude that for $s_i$'s sufficiently close to $s$, we cannot have $\gamma(s_1)$, $\gamma(s_2)$, and $\gamma(s_3)$ collinear.


The form of the above argument is important. We design a function that measures something geometric and apply the Mean Value Theorem to deduce the existence of points where certain conditions hold. Then we take a limit and use smoothness of the curve to deduce that those conditions hold at $s$. We will repeat this structure for each of the osculating plane and osculating circle.


Next, we take up the osculating plane. We have seen that for $s_1, s_2, s_3$ sufficiently close to $s$, the points $\gamma(s_1), \gamma(s_2), \gamma(s_3)$ are not collinear. This means that they define a unique plane. We let $A(s_1, s_2, s_3)$ be a unit vector perpendicular to this plane, and consider the function

\[ f(t) = \langle \gamma(t) - \gamma(s_1), A(s_1, s_2, s_3) \rangle . \]
This function is $C^2$ and takes the value $0$ at the points $s_1$, $s_2$, and $s_3$. So by the Mean Value Theorem, there exist parameter values $\xi_1$ and $x_2$ which satisfy $s_1 < \xi_1 < s_2 < \xi_2 < s_3$, and
\[ 0 = f'(\xi_1) = \langle \gamma'(\xi_1), A(s_1, s_2, s_3) \rangle  = f'(\xi_2) = \langle \gamma'(\xi_2), A(s_1, s_2, s_3) \rangle . \]
Applying the Mean Value Theorem to $f'$, we now see that there is a point $\eta$ with $\xi_1 < \eta < \xi_2$ and
\[ 0 = f''(\eta) = \langle \gamma''(\eta), A(s_1, s_2, s_3) \rangle .\]

By the continuity of $\gamma'$ and $\gamma''$, we deduce that as $s_1, s_2$ and $s_3$ tend toward $s$, the vectors $A(s_1, s_2, s_3)$ must have a limit, and this limit is orthogonal to both $\gamma'(s)$ and $\gamma''(s)$. Therefore, the planes in question have a limit, and this limit is the osculating plane.


Now, we take up the osculating circle. The three points $\gamma(s_1), \gamma(s_2), \gamma(s_3)$ are not collinear and lie in a single plane, and also they lie on the circumference of a circle lying in that plane. Let $C(s_1,s_2,s_3)$ be the center of that circle. We consider the function
\[ f(t) = \langle \gamma(t) - C(s_1,s_2,s_3) , \gamma(t) - C(s_1,s_2,s_3)\rangle . \]
This function is $C^2$ and takes identical values at the points $s_1, s_2, s_3$ (namely, the square of the radius of the circle). So, by the Mean Value Theorem, we find points $\xi_1, \xi_2$ such that
\[ 0 = f'(\xi_i) = 2 \langle \gamma'(\xi_i), \gamma(\xi_i) - C(s_1,s_2,s_3) \rangle. \]
Again, we use the Mean Value Theorem on $f'(t)/2$ to find a point $\eta$ such that
\[ 0 = \langle \gamma''(\eta) , \gamma(\eta) - C(s_1,s_2,s_3) \rangle + \langle \gamma'(\eta), \gamma'(\eta) \rangle .\]
Following the general form of the argument, we use the continuity of the functions involved to deduce that the circles have a limit which is another circle lying in the osculating plane, and whose center $C$ must satisfy these equations:
\[  \langle \gamma'(s)  , \gamma(s) - C \rangle = 0   \tag{1} \]
\[  \langle \gamma''(s) , \gamma(s) - C \rangle = -1  \tag{2} \]
This circle lying in the osculating plane with center $C$ and passing through $\gamma(s)$ is called the osculating circle of $\gamma$ at $s$.

It can be helpful to remember that Equations (1) and (2) are linear equations satisfied by the unknown point $C$. If we add in the equation expressing that $C$ lies in the osculating plane:
\[ \langle \gamma'(s) \times \gamma''(s) , \gamma(s) - C \rangle  = 0 \]
we get a system of three equations in three unknowns (the coefficients of $C$). So determining $C$ is a simple matter of linear algebra technique.

There is a last piece of information to be had from all of this hard work. It is clear from the set-up that $\gamma(s) - C$ is a multiple of $\gamma''(s)$. Let us use this fact.
If $\gamma(s) - C = \alpha \gamma''(s)$, then Equation (2) helps us see that
\[ -1 = \langle \gamma''(s), \gamma(s) - C \rangle = \langle \gamma''(s), \alpha \gamma''(s) \rangle \]
so $\alpha = -1 / ||\gamma''(s) ||^2$. This tells us that the radius of the osculating circle is
\[ \left( ||\gamma(s) - C || \right)^{1/2} = \left( \alpha^2 || \gamma''(s) ||  \right)^{1/2} = \dfrac{1}{\kappa} \]
since $||\gamma''(s)|| = ||T'(s)|| = \kappa$.

This means that the radius of the osculating circle is the reciprocal of the curvature of $\gamma$ at $s$. But this is more conveniently taken as a geometric interpretation of curvature:

The curvature of $\gamma$ at $s$ is the reciprocal of the radius of the osculating circle to $\gamma$ at $s$.

And that makes all of this effort worthwhile.

No comments:

Post a Comment