Paths, Normal Vector, Osculating Plane

Normal Vector, Osculating Plane

Recall our direction vector, d(t), which is velocity divided by speed. Thus d(t) = p′/|p′|. This is a unit vector in n space that indicates the direction of travel, without regard to speed. Picture this direction vector moving along the curve as the path progresses. We can see, geometrically, that the change in direction is always perpendicular to the direction. Let's show this algebraically.

Let the normal vector n(t) be the derivative of d(t), recording the instantaneous change in direction at each point in time.

If the path is straight, d(t) is constant, and n(t) = 0. I guess the 0 vector is (technically) perpendicular to the path. But n(t) is always perpendicular, even if the direction is changing, and here's a simple proof.

Since the direction vector is a unit vector, write d(t).d(t) = 1. Differentiate to get 2×d(t).n(t) = 0, hence d(t) and n(t) are orthogonal.

The osculating plane at p(t) is the plane determined by d(t) and n(t). Osculating? Yes, osculate is a fancy word for kiss. The osculating plane just kisses the path. The path may break out of this plane later, but for now, the path is turning within the osculating plane.

Assume p′′ is continuous, and p′ and p′′ are nonzero at t. In other words, the particle is moving, and turning, at time t. Choose three points near t and watch what happens to the plane determined by these three points as we approach t.

Draw a chord between the first two points in the path. This defines a direction, which is arbitrarily close to d(t), since the path is continuously differentiable. If you want to be technical, the connecting segment, divided by time, approaches p′, the instantaneous velocity at t, but velocity and direction are parallel. One is a scaled version of the other. We're only interested in the plane determined by these line segments, so their lengths are not important. So I say again, the first line segment approaches d(t) as its endpoints approach t.

Since p′ is continuous, the speed is nearly constant close to t, hence direction vectors and velocity vectors are proportional. Subtract the two velocity vectors determined by the three points, the two connecting line segments, and we have a scaled version of the difference between direction vectors near t. Remember that d(t) is continuously differentiable. The direction is changing, turning, as indicated by n(t). And the direction is changing smoothly; it isn't a sharp turn. The difference between direction vectors near t is related to the change in direction near t, which may not be the same as the change in direction at t. However, p′′ is continuous, so it is almost the same. Thus the difference between direction vectors approaches n(t).

If a plane is determined by three points near t, it is defined by two vectors; one approaches d(t) and the other approaches n(t). The planes near t approach the osculating plane. All this can be made rigorous, with lots of deltas and epsilons flying about, but then this proof would expand into six pages. We can skip that, as long as you understand the concept.

Write the velocity as (p′/s)×s, where s is speed. This seems pointless, until you differentiate using the product rule. The result is n×s + d×s′. These are two perpendicular components of acceleration, and they both lie in the osculating plane. The first changes the particle's direction, and is proportional to speed. Thus it takes more force to change an object's direction when it is moving fast. The second component runs parallel to the direction of travel, and represents the change in speed. If the particle moves at constant speed, the acceleration, if any, is perpendicular to the path, and changes the particle's direction. This is illustrated by a space ship in a circular orbit. Gravity constantly changes the ship's direction as it moves around the planet, yet the speed of the ship is unchanged.