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4. Moving Frames

In the previous section, we introduced the notion of a frame in the unit normal bundle of a space curve. We now consider a slightly more general notion. By a frame, or more precisely a right-handed rectangular frame with origin, we mean a point X and a triple of mutually orthogonal unit vectors E1, E2, E3 forming a right-handed system. The point X is called the origin of the frame. Note that Ei Ej = 1 if i = j and 0 if i j. Moreover, E1 x E2 = E3, E2 = E3 x E1, and E3 = E1 x E2. In the remainder of this section, we will always assume that small Latin letters run from 1 to 3.
Note that given two different frames, X E1 E2 E3 and X E1 E2 E3, there is exactly one affine motion of Euclidean space taking X to X and taking Ei to Ei.
When X(t) E1(t) E2(t) E3(t) is a family of frames depending on a parameter t, we say we have a moving frame along the curve.
Proposition 1:A family of frames X(t) E1(t) E2(t) E3(t) satisfies a system of differential equations:
X'(t) = pi(t) Ei(t)
Ei'(t) = qij(t) Ej(t),
where pi(t) = X'(t)Ei(t) and qij(t) = Ei'(t)Ej(t).
Since Ei(t)Ej(t) = 0 for i j, it follows that qij(t) + qji(t) = Ei'(t)Ej(t) + Ei(t)Ej'(t) = 0, i.e. the coefficients qij(t) are anti-symmetric in i and j. This can be expressed by saying that the matrix (( qij(t) )) is an anti-symmetric matrix, with 0 on the diagonal.
In a very real sense, the function pi(t) and qij(t) completely determine the family of moving frames. Specifically we have:
Proposition 2: If X(t) E1(t) E2(t) E3(t) and X(t) E1(t) E2(t) E3(t) are two families of moving frames such that pi(t) = pi(t) and qij(t) = qij(t) for all t, then there is a single affine motion that takes X(t) E1(t) E2(t) E3(t) to X(t) E1(t) E2(t) E3(t) for all t.
Proof: Recall that for and specific value t0, there is an affine motion taking X(t0) E1(t0) E2(t0) E3(t0) to X(t0) E1(t0) E2(t0) E3(t0). We will show that this same motion takes X(t) E1(t) E2(t) E3(t) to X(t) E1(t) E2(t) E3(t) for all t. Assume that the motion has been carried out so that the frames X(t0) E1(t0) E2(t0) E3(t0) and X(t0) E1(t0) E2(t0) E3(t0) coincide.
Now consider (Ei(t)Ei(t))' = Ei'(t)Ei(t) + Ei(t)Ei'(t)
= qij(t)Ej(t)Ei(t) + Ei(t)qij(t)Ej(t)
= qij(t)Ej(t)Ei(t) + qij(t)Ei(t)Ej(t)
= qij(t)Ej(t)Ei(t) + qji(t)Ej(t)Ei(t) = 0.
It follows that Ei(t)Ei(t) = Ei(t0)Ei(t0) = Ei(t0)Ei(t0) = 3 for all t. But since |Ei(t)Ei(t)| 1 for any pair of unit vectors, we must have Ei(t)Ei(t) = 1 for all t. Therefore Ei(t) = Ei(t) for all t.
Next consider (X(t) - X(t))' = pi(t) Ei(t) - pi(t) Ei(t) = pi(t) Ei(t) - pi(t) Ei(t) = 0. Since the origins of the two frames coincide at the value t0, we have X(t) - X(t) = X(t0) - X(t0) = 0 for all t.
This completes the proof that two families of frames satisfying the same set of differential equations differ at most by a single affine motion.

Exercises

  1. Prove that the equations Ei'(t) = qij(t) Ej(t) can be written Ei'(t) = D(t) x Ei(t), where D(t) = q23(t)E1(t) + q31(t) E2(t) + q12(t)E3(t). This vector is called the instantaneous axis of rotation.
  1. Under a rotation about the x3-axis, a point describes a circle X(t) = ( a cos(t), a sin(t), b ). Show that its velocity vector satisfies X'(t) = D x X(t) where D = ( 0, 0, 1). (Compare Exercise 1 above.)
  1. Prove that (VV)(WW) - (VW)2 = 0 if and only if the vectors V and W are linearly dependent.
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