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[personal profile] stoutfellow
Differential geometry is the study of geometry using the techniques of calculus. The simplest sort of geometric object, for these purposes, is a regular curve.

Geometrically, what we're interested in is the curve itself - the circle, ellipse, helix, or what-have-you. But in order to apply calculus, we need to parametrize it. The simplest way to think of this is to envision a particle moving (or if you must, a person running) along the curve. The motion has to be smooth, so that we can make sense of concepts like velocity, but that's not quite enough. There's one additional requirement, that the particle never comes to a stop, even momentarily. Why is this? If the particle is allowed to stop, the curve itself may be jagged - even though the particle is moving smoothly. Picture, for example, the impressionistic curve with which one draws a bird in flight, consisting of two quick arcs joined together. The "corner" where the two arcs meet is an undesirable feature, but a smoothly moving particle could negotiate that corner by slowing to a stop on the approach and then accelerating out along the other arc.
Technically, what we have is a vector-valued function α:(a,b)→R3, where we have a parameter t (think of it as time) varying over a range from a to b; α(t) is the position of the particle at time t. We'll require this function to be C3; that is, if we write α(t)=(α123), we want the three component functions each to be three times differentiable, and for those third derivatives to be continuous. The second condition is that α'(t) never be zero for any value of t. The function (t2,t3,0) fails to meet this second condition; when t is 0, the particle is momentarily motionless, and there is a corner

Any given curve can be parametrized in many different ways; that is, there are many ways to run along a track. As suggested in my previous post, what we really want to find are quantities that don't depend on the parametrization. Set that aside for the moment, though; we're going to, instead, pick a parametrization that simplifies things a great deal (rather like setting the right coordinate system for an ellipse). We'll assume for the time being that the particle is moving at a steady speed of 1. (1 what? Feet per second, miles per hour, furlongs per fortnight; you pick the units.) Then, at each point along the curve, we have a velocity vector, pointing in the direction that the curve is, at that instant, taking. That vector has length 1 (its length is the speed). Now, at each point, look at the acceleration vector - the rate of change of the velocity vector. A vector can change in two ways, by changing its length or changing its direction (or some combination of the two, of course). But this velocity vector isn't changing its length; the only way it can change is by changing direction. It's not hard to show that the acceleration vector then has to be perpendicular to the velocity vector, and its length indicates how rapidly the curve is changing direction. We call that length the curvature of the curve. (For example: if the curve is a straight line, the acceleration vector is 0, and so the curvature is 0. If the curve is a circle with radius r, the acceleration vector points straight in towards the center of the circle, and its length is 1/r. In general, the tighter the "turning radius" of the curve, the larger the curvature.)
We call the length-1 velocity vector T. Now, the length of T is a constant, equal to 1. That is, the dot product T·T=1. The product rule applies just as well to dot products (and to cross products) as it does to ordinary multiplication, so the derivative of T·T is 2T·T' - which must be 0. That is, T' is perpendicular to T.

Now suppose that, on our curve, the curvature is never 0 - so we're ruling out straight lines, and also certain other sorts of curve, like the letter S. (Right at the midpoint of that curve, where it switches from bending left to bending right, the curvature is 0.) That's not too much of a hardship; a general curve can usually be broken into segments where the curvature isn't 0. Then we can set up another vector, the principal normal vector, which points in the same direction as the acceleration vector but has length 1 again. At this point, we have three quantities at any given point: the velocity vector, indicating the direction the curve is headed; the curvature, indicating how rapidly the curve is bending; and the principal normal vector, indicating the direction the curve is bending in. Each of these is easy to think about intuitively; the important thing here is that we've made them precise.

There's one more feature of the curve that needs to be addressed; in addition to where it's going and how it's bending away from a straight line, we have to look at whether it's staying in one plane - like a circle - or twisting out - like a helix. I'll hold off on that until my next post.
The symbol for the curvature is κ; the principal normal N is 1/κT'.

Date: 2006-01-16 10:45 pm (UTC)
From: [identity profile] countrycousin.livejournal.com
That's not too much of a hardship; a general curve can usually be broken into segments where the curvature 0.

That's is not 0, I presume?

Date: 2006-01-17 12:02 am (UTC)
From: [identity profile] stoutfellow.livejournal.com
Yes. Oops. Corrected.

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