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The body angular velocity of a C1 rotation path: Ω=RTR′ is skew, Ωz=ω×z determines a unique continuous ω, and R′=RΩ

Statement

Let I⊆R be a nondegenerate open interval, let SO(3) be as in The orthogonal group O(3) and the rotation group SO(3), and let R:I→SO(3) be of class C1, that is, all nine entries of R have continuous first derivatives; write R′(t) for the entrywise derivative and put

Ω(t):=R(t)TR′(t)∈M3(R)

for the product of the transpose of R(t) with R′(t) (Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes, The vector space Mm×n(F):=F m×n of m by n matrices over a field, with entrywise operations). Then:

  1. Ω(t) is skew-symmetric for every t∈I, that is Ω(t)T=−Ω(t).
  2. There is exactly one map ω:I→R3 with

R(t)TR′(t) z=ω(t)×z(z∈R3, t∈I),

namely ω(t)=(−Ω23(t), Ω13(t), −Ω12(t)), and this ω is continuous. It is called the body angular velocity of R. 3. For every t∈I, writing Ωω(t) for the matrix z↦ω(t)×z, one has R′(t)=R(t) Ωω(t).

Facts & Assumptions

Given: A C1 curve R:I→SO(3) and Ω=RTR′, with cross product and Euclidean inner product as in The cross product in R3 and The cross product is bilinear, alternating, and orthogonal to both factors.

[L1]

SO(3) consists of the 3×3 real matrices with RTR=I3 and det⁡R=1; for R∈SO(3) one has RT=R−1 and RRT=I3 (The orthogonal group O(3) and the rotation group SO(3), SO(3) is a subgroup of the general linear group, with R−1=RT).

[L4]

The cross product is u×v=(u2v3−u3v2, u3v1−u1v3, u1v2−u2v1), so it is bilinear and u×u=0 (The cross product in R3, The cross product is bilinear, alternating, and orthogonal to both factors).

Proof

technique · direct; differentiate the orthogonality identity, then the finite skew-matrix/cross-product correspondence
1.1L1L2L3algebra

By [L3], differentiating the identity RTR=I3 of [L1] gives 0=(RTR)′=(R′)TR+RTR′. Transposing the resulting identity (R′)TR=−RTR′ with [L2] gives ΩT=(RTR′)T=R′T(RT)T=R′TR=−Ω. This proves clause 1.

1.2L4algebra

Let Ω∈M3(R) be skew-symmetric and put ω:=(−Ω23,Ω13,−Ω12). For z∈R3, the first coordinate of Ωz is Ω11z1+Ω12z2+Ω13z3=Ω13z3−Ω12z2=(ω×z)1, and the second and third coordinates are the same computation with the remaining entries; hence Ωz=ω×z for all z. Conversely, if ω′×z=0 for all z, then testing z=e1,e2,e3 gives ω′=0 by the coordinate formula of [L4]; in particular the ω above is the only vector representing Ω.

2.1step 1.1step 1.2L1L2algebra

By [step 1.1] the matrix Ω(t) is skew for every t, so by [step 1.2] the vector ω(t):=(−Ω23(t),Ω13(t),−Ω12(t)) satisfies Ω(t)z=ω(t)×z for all z, and it is the only such vector; the entries of Ω=RTR′ are finite sums of products of the continuous entries of R and R′, so t↦ω(t) is continuous. This proves clause 2.

3.1step 2.1L1L2algebra∎

Multiplying the identity Ω=RTR′ on the left by R(t) and using the associativity of matrix multiplication and RRT=I3 from [L1] gives RΩ=R(RTR′)=(RRT)R′=R′. By [step 2.1] the matrix Ω is exactly Ωω, so R′=RΩω; this proves clause 3.

Remarks

  • The convention here is the body-fixed one: Ω=RTR′ and R′=RΩ. The spatial angular velocity of the same path is the vector attached to R′RT, and the two differ by the rotation R; consumers must state which convention they use. For C2 paths this ω is the same vector that appears in the rotating-frame acceleration identity of Angular velocity of a moving frame and the rotating-frame acceleration identity, whose clause 2 records it under a twice-differentiability hypothesis; the present lemma covers the C1 case and proves only what is used here.

  • The lemma is a statement about curves of matrices. No physical object, frame or time scale is introduced; an interpretation of ω as the angular velocity of a rigid body belongs to the physical items that adopt a rigid-body model.

Depends on

Dependency tree · 2 levels

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