Monday Math: Curvature
Nov. 12th, 2012 11:17 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowWell, I bombed out on discussing my research on polygons. Let's see if I can do better with my research on prisms.
The specific topic I'm investigating is closed geodesics on rectangular prisms. Before I can talk about that, I need to discuss geodesics in general; and before I can do that, I need to talk about the notion of curvature.
Suppose you have a curve, and you want to discuss how fast, and in what direction, it's curving. What does it mean, to say that it's curving? A moment's thought suggests that curvature is change in direction - the more rapid the change, the greater the curvature. If we're talking about change, we need to bring calculus - the mathematics of change - into play; that, in turn, requires that we represent the curve by a function. This is straightforward. We imagine a particle moving along the curve, and define a vector-valued function f(t) which gives the location of the particle at time t: f(t)=(x(t),y(t),z(t)). The direction in which the particle is moving, at any instant, is indicated by f'(t) - the velocity of the particle, and hence the rate of change of the direction is given by the acceleration, f''(t). Now, if you remember your physics, acceleration means change in speed or direction or both. We want to isolate the change in direction, so we should make sure that the speed does not change. In other words, we have to choose f(t) so that the speed - the length of f'(t) - does not change. For simplicity's sake, we'll insist that the speed be unity - one (unit of length) per (unit of time). We can make this happen for any reasonably smooth curve; I won't go into the technicalities.
So, we're thinking of a particle moving along the curve at constant, unit, speed - we've parametrized the curve as a unit-speed curve. Then f''(t), the acceleration, indicates how fast, and in what direction, the curve is bending. The curvature is the length of the acceleration vector f''(t). Let's do a couple of examples.
First, consider a straight line. Our particle is moving from its starting point in a certain direction; if we let v be the unit-length vector in that direction and p be the starting point, then we can set f(t) = p + t v. Note that v is a constant vector, with length 1. Therefore f'(t) = v, and f''(t) = 0 - in other words, a line has curvature 0. Of course it does; but that's how we check that we're on the right track, by making sure our mathematical construction agrees with our rough, intuitive notions.
Next, consider a circle. Let's put it in the xy-plane, and center it at the origin; let the radius be r. We'll measure angles in radians. That way, if the particle travels through k radians, we'll know how far it's travelled - kr units. So, if we want the motion to be unit speed, the particle has to move 1/r radians per time-unit. Our parametrization then is f(t) = (r cos(t/r), r sin(t/r), 0). The velocity vector is f'(t) = (-sin(t/r), cos(t/r), 0), which has unit length (as desired). What about the acceleration? It's f''(t) = (-1/r cos(t/r), -1/ r sin(t/r), 0) - a vector of length 1/r, pointing straight in towards the center. That makes sense. If we think of a circular orbit under the influence of gravity, the force, and hence the acceleration, points towards the center of the attracting mass. Furthermore, motion along a small circle is more curved than motion along a large one; a large enough circle is indistinguishable by eye from a straight line.
Similar computations can be carried out with any smooth curve you might consider. But what does this have to do with geodesics? What are geodesics? Here's a loose definition, which I'll tighten up later: a geodesic is a curve, on a surface (or in a three-dimensional object, or wherever), which is as straight as possible. On the plane, or in ordinary three-space, that simply means straight - no curvature at all, or, more precisely, curvature 0 everywhere. We already saw that that means a straight line. But if we insist that the curve lie on, say, the surface of a sphere, that's simply going to be impossible - no straight line can lie on a sphere! What to do? I'll talk about that next time.
The specific topic I'm investigating is closed geodesics on rectangular prisms. Before I can talk about that, I need to discuss geodesics in general; and before I can do that, I need to talk about the notion of curvature.
Suppose you have a curve, and you want to discuss how fast, and in what direction, it's curving. What does it mean, to say that it's curving? A moment's thought suggests that curvature is change in direction - the more rapid the change, the greater the curvature. If we're talking about change, we need to bring calculus - the mathematics of change - into play; that, in turn, requires that we represent the curve by a function. This is straightforward. We imagine a particle moving along the curve, and define a vector-valued function f(t) which gives the location of the particle at time t: f(t)=(x(t),y(t),z(t)). The direction in which the particle is moving, at any instant, is indicated by f'(t) - the velocity of the particle, and hence the rate of change of the direction is given by the acceleration, f''(t). Now, if you remember your physics, acceleration means change in speed or direction or both. We want to isolate the change in direction, so we should make sure that the speed does not change. In other words, we have to choose f(t) so that the speed - the length of f'(t) - does not change. For simplicity's sake, we'll insist that the speed be unity - one (unit of length) per (unit of time). We can make this happen for any reasonably smooth curve; I won't go into the technicalities.
So, we're thinking of a particle moving along the curve at constant, unit, speed - we've parametrized the curve as a unit-speed curve. Then f''(t), the acceleration, indicates how fast, and in what direction, the curve is bending. The curvature is the length of the acceleration vector f''(t). Let's do a couple of examples.
First, consider a straight line. Our particle is moving from its starting point in a certain direction; if we let v be the unit-length vector in that direction and p be the starting point, then we can set f(t) = p + t v. Note that v is a constant vector, with length 1. Therefore f'(t) = v, and f''(t) = 0 - in other words, a line has curvature 0. Of course it does; but that's how we check that we're on the right track, by making sure our mathematical construction agrees with our rough, intuitive notions.
Next, consider a circle. Let's put it in the xy-plane, and center it at the origin; let the radius be r. We'll measure angles in radians. That way, if the particle travels through k radians, we'll know how far it's travelled - kr units. So, if we want the motion to be unit speed, the particle has to move 1/r radians per time-unit. Our parametrization then is f(t) = (r cos(t/r), r sin(t/r), 0). The velocity vector is f'(t) = (-sin(t/r), cos(t/r), 0), which has unit length (as desired). What about the acceleration? It's f''(t) = (-1/r cos(t/r), -1/ r sin(t/r), 0) - a vector of length 1/r, pointing straight in towards the center. That makes sense. If we think of a circular orbit under the influence of gravity, the force, and hence the acceleration, points towards the center of the attracting mass. Furthermore, motion along a small circle is more curved than motion along a large one; a large enough circle is indistinguishable by eye from a straight line.
Similar computations can be carried out with any smooth curve you might consider. But what does this have to do with geodesics? What are geodesics? Here's a loose definition, which I'll tighten up later: a geodesic is a curve, on a surface (or in a three-dimensional object, or wherever), which is as straight as possible. On the plane, or in ordinary three-space, that simply means straight - no curvature at all, or, more precisely, curvature 0 everywhere. We already saw that that means a straight line. But if we insist that the curve lie on, say, the surface of a sphere, that's simply going to be impossible - no straight line can lie on a sphere! What to do? I'll talk about that next time.
