Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Angular velocity in Euler angles: ω3=ψ′+φ′cos⁡θ and ω12+ω22=θ′2+φ′2sin⁡2θ

Statement

Let J⊆R be an interval with at least two points, let φ,θ,ψ:J→R be differentiable with θ(J)⊆(0,π), and put R(t):=Θ(φ(t),θ(t),ψ(t))=Rz(φ(t)) Ry(θ(t)) Rz(ψ(t)), the Euler-angle parametrization of The Euler-angle parametrization and the heavy symmetric top system. Then R is a differentiable SO(3)-valued curve, and its body angular velocity ω:J→R3 in the sense of Rigid placements, body and spatial angular velocity, and material velocities (the unique vector with R(t)TR′(t)=[ω(t)]×) is, in the standard basis, ω=(−φ′sin⁡θcos⁡ψ+θ′sin⁡ψ)e1+(φ′sin⁡θsin⁡ψ+θ′cos⁡ψ)e2+(ψ′+φ′cos⁡θ)e3, all functions evaluated at t. Consequently ω3=ψ′+φ′cos⁡θ,ω12+ω22=θ′2+φ′2sin⁡2θ,∥ω∥22=θ′2+φ′2sin⁡2θ+(ψ′+φ′cos⁡θ)2. Moreover, if A,C>0 and I:R3→R3 is the linear map with Ie1=Ae1, Ie2=Ae2, Ie3=Ce3, then ωTIω=A(θ′2+φ′2sin⁡2θ)+C(ψ′+φ′cos⁡θ)2.

Facts & Assumptions

Given: Differentiable φ,θ,ψ:J→R with θ∈(0,π), the curve R=Rz(φ)Ry(θ)Rz(ψ), and the matrices Rz,Ry of The coordinate rotation matrices Rz(a) and Ry(a): values in SO(3), composition law, and derivatives.

[L1]

Rz(a),Ry(a)∈SO(3) for all a, Rz(a)Rz(b)=Rz(a+b), Ry(a)Ry(b)=Ry(a+b), and for differentiable a one has (Rz∘a)′=a′Rz(a)[e3]×=a′[e3]×Rz(a) and (Ry∘a)′=a′Ry(a)[e2]×=a′[e2]×Ry(a) The coordinate rotation matrices Rz(a) and Ry(a): values in SO(3), composition law, and derivatives.

[L2]

The product rule for matrix-valued curves, the skewness of RTR′ for an SO(3)-valued curve, and the equivalence RTR′=[ω]×⇔R′=R[ω]× hold; moreover [ω]×z=ω×z and RT[u]×R=[RTu]× Derivative rules for differentiable matrix-valued curves, and skewness of RTR′ for a curve in SO(3), The skew-matrix/cross-product correspondence [ω]×z=ω×z and its conjugation law R[ω]×RT=[Rω]×, Rigid placements, body and spatial angular velocity, and material velocities.

[L3]

For R∈SO(3), RTR=I3; the matrices e1e1T+e2e2T+e3e3T=I3, [e3]×e3=0 and [e2]×e2=0 The skew-matrix/cross-product correspondence [ω]×z=ω×z and its conjugation law R[ω]×RT=[Rω]×, Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes.

[L5]

Evaluation of Rz(a)e3=e3, Rz(a)e1=cos⁡a e1+sin⁡a e2, Rz(a)e2=−sin⁡a e1+cos⁡a e2 and Ry(a)e1=cos⁡a e1−sin⁡a e3, Ry(a)e3=sin⁡a e1+cos⁡a e3 is the column reading of the displayed matrices The coordinate rotation matrices Rz(a) and Ry(a): values in SO(3), composition law, and derivatives.

Proof

technique · direct
1.1givenL1L2

By [L1] each of the three factors is a differentiable SO(3)-valued curve; by the product rule [L2] and closure of SO(3) under multiplication, R is differentiable with values in SO(3). Hence RTR′ is skew and there is a unique ω with RTR′=[ω]× by [L2].

1.2givenL2L5algebra

By the conjugation identity ST[u]×S=[STu]× of [L2] with S=Ry(θ) and S=Rz(ψ), and the column readings of [L5], Ry(θ)T[e3]×Ry(θ)=[Ry(θ)Te3]×=[−sin⁡θ e1+cos⁡θ e3]×,Rz(ψ)T[e2]×Rz(ψ)=[sin⁡ψ e1+cos⁡ψ e2]×.

2.1step 1.1givenL1L2algebra

Differentiating the product and applying [L1], R˙=φ′ Rz(φ)[e3]×Ry(θ)Rz(ψ)+θ′ Rz(φ)Ry(θ)[e2]×Rz(ψ)+ψ′ Rz(φ)Ry(θ)Rz(ψ)[e3]×.

3.1step 2.1givenL2L3algebra

Multiplying on the left by RT=Rz(ψ)TRy(θ)TRz(φ)T and using Rz(φ)TRz(φ)=I3 [L3] term by term, RTR˙=φ′ Rz(ψ)TRy(θ)T[e3]×Ry(θ)Rz(ψ)+θ′ Rz(ψ)T[e2]×Rz(ψ)+ψ′ [e3]×.

4.1step 3.11.2givenL2L5algebra

Conjugating the first vector of step 1.2 by Rz(ψ)T and using [L5], Rz(ψ)T(−sin⁡θ e1+cos⁡θ e3)=−sin⁡θcos⁡ψ e1+sin⁡θsin⁡ψ e2+cos⁡θ e3; substituting into step 3.1 gives RTR˙=[ω]× with ω exactly as displayed in the statement.

5.1step 4.1givenL4algebra

Reading off the components gives ω3=ψ′+φ′cos⁡θ; expanding the squares of the first two components and using sin⁡2ψ+cos⁡2ψ=1 from [L4] cancels the mixed terms and gives ω12+ω22=φ′2sin⁡2θ+θ′2, whence ∥ω∥22=θ′2+φ′2sin⁡2θ+(ψ′+φ′cos⁡θ)2.

6.1step 4.15.1givenL3algebra∎

For the diagonal map I with Ie1=Ae1, Ie2=Ae2, Ie3=Ce3, the quadratic form is ωTIω=A(ω12+ω22)+Cω32, and substituting step 5.1 gives the displayed identity.

Remarks

Depends on

Dependency tree · 2 levels

Sources