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Principal-axis uniform rotations of the torque-free rigid-material model: existence, a fixed spatial direction, the invariants, and the converse classification

Physical scope

Conditional statements about the torque-free branch of a nondegenerate rigid-material model of Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope: for every principal axis $e_{j}$ and every real rate $Ω$ there is a motion whose orientation path is a uniform rotation with constant body angular velocity $Ω e_{j}$ ; for $Ω \neq 0$ the spatial direction $R (t) e_{j}$ of the body-fixed line through $c (t)$ parallel to the reference axis $e_{j}$ is fixed in the inertial frame, so the spatial angular velocity $ω_{s} (t) = Ω R (t) e_{j}$ is constant in that direction while its spatial location may translate with the arbitrary centre-of-mass curve $c$ ; the invariants take the values $C = Ω^{2} I_{j}^{2}$ , $E = \frac{1}{2} Ω^{2} I_{j}$ ; and conversely, a torque-free motion with constant body angular velocity must spin about a principal axis. The statements carry the countable-choice assumption inherited for finite mass moments and inertial integrals and for the double-cover surjectivity used to choose an initial unit quaternion, assert no measurement, no empirical law and no claim that any apparatus has been observed.

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), carried through Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope clause (i) and Inertia matrix, spatial inertia tensor, inertia tensor about a point and moment of inertia about an axis of a rigid-material model for finite mass moments, through Total momentum, kinetic-energy splitting and centre-of-mass angular momentum of a rigid-material motion and Adoption of the external-torque branch of the rigid-material model: declared torque data and the balance of angular momentum about the centre of mass clause (i) for the finite inertial integrals, and through the double-cover supplier of The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts (its second remark) for the surjectivity used to choose an initial unit quaternion. Adopt a nondegenerate model of the rigid-material branch Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope: a reference mass distribution $ν$ with total mass $m$ , centred reference coordinates and no line through the origin carrying the whole mass (Inertia matrix, spatial inertia tensor, inertia tensor about a point and moment of inertia about an axis of a rigid-material model clause (a)). Fix an orthonormal principal basis $(e_{1}, e_{2}, e_{3})$ with $I e_{j} = I_{j} e_{j}$ and $I_{j} > 0$ (Principal axes and principal moments of the inertia data of a rigid-material model, axis moments and extremal values clauses (a) and (d)), use the model in the torque-free branch of Adoption of the external-torque branch of the rigid-material model: declared torque data and the balance of angular momentum about the centre of mass clause (iv), and fix an index $j \in {1, 2, 3}$ and a rate $Ω \in R$ . Use the model's nondegenerate open time window $I_{0}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length and Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope clause (ii)), and fix a time $t_{0} \in I_{0}$ and an orientation $R_{0} \in S O (3)$ . Then:

Existence of the uniform rotation. There is a path $R : I_{0} \to S O (3)$ of class $C^{1}$ with $R (t_{0}) = R_{0}$ and $R (t)^{T} R^{'} (t) = [Ω e_{j}]_{\times} (t \in I_{0}),$ that is, with constant body angular velocity $ω \equiv Ω e_{j}$ (Body and spatial angular velocity of a rigid-material motion, and the material velocity field clause (a)).
It is a torque-free motion. For every such path $R$ and every curve $c : I_{0} \to R^{3}$ of class $C^{2}$ , the map $x (t, ρ) : = c (t) + R (t) ρ$ is a motion of the torque-free model of the branch: its angular momentum about the centre of mass is $L_{c} (t) = Ω I_{j} R (t) e_{j},$ it is constant on $I_{0}$ , and the adopted balance $L_{c}^{'} = 0$ holds. (Here $R$ is automatically of class $C^{2}$ , because $R^{'} = R [Ω e_{j}]_{\times}$ is the product of the $C^{1}$ curve $R$ with a constant matrix, so the configuration path $(c, R)$ has the regularity required of a motion.)
Geometry and invariants. In the motions of clause 2 one has $l = I ω = Ω I_{j} e_{j}$ constant, $C = ∥ L_{c} ∥_{2}^{2} = Ω^{2} I_{j}^{2}, E = \frac{1}{2} ω^{T} I ω = \frac{1}{2} Ω^{2} I_{j};$ if $Ω \neq 0$ , define the spatial line of the body-fixed axis by $ℓ_{j} (t) : = {c (t) + s R (t) e_{j} : s \in R}$ and its unit direction by $n_{j} (t) : = R (t) e_{j}$ . Then $n_{j} (t) = L_{c} / (Ω I_{j})$ for every $t$ , so this direction is fixed in the inertial frame and the spatial angular velocity $ω_{s} (t) : = R (t) ω (t) = Ω n_{j}$ of Body and spatial angular velocity of a rigid-material motion, and the material velocity field clause (b) is constant; the line $ℓ_{j} (t)$ need not be fixed because $c$ is arbitrary. If $Ω = 0$ , then $R$ is constant on $I_{0}$ .
Converse classification. Conversely, let a motion of a nondegenerate torque-free model have constant body angular velocity $ω \equiv ω_{0}$ on its time window. Then $I ω_{0} = λ ω_{0}$ for some $λ \in R$ : either $ω_{0} = 0$ and the orientation path is constant, or $ω_{0} \neq 0$ and $ω_{0}$ lies along a principal axis; in the latter case an orthonormal principal basis can be chosen with one vector $e_{j} = ω_{0} / ∥ ω_{0} ∥_{2}$ and $Ω = ∥ ω_{0} ∥_{2} \neq 0$ , and the conclusions of clauses 2 and 3 hold in that adapted basis. It can be made positively oriented by negating a different basis vector. No alignment with the initially fixed basis is asserted when eigenvalues repeat.

The statement is conditional on the adopted model and zero-torque balance; it asserts no force, no measurement and no empirical law, and the existence clause is a statement about the model's kinematic class, not about nature.

Facts & Assumptions

Given: A nondegenerate model of Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope with principal basis $(e_{j})$ and positive principal moments $I_{j}$ , the zero-torque datum $τ \equiv 0$ , the balance $L_{c}^{'} = 0$ , and the data $j$ , $Ω$ , $I_{0}$ , $t_{0}$ , $R_{0}$ of the statement.

[L1]

Model data: $L_{c} (t) = R (t) I ω (t)$ for every motion, $C = ∥ L_{c} ∥_{2}^{2} = ⟨ l, l ⟩$ and $E = \frac{1}{2} ⟨ l, I^{- 1} l ⟩$ are the torque-free invariants, and $l^{'} = l \times (I^{- 1} l)$ under the zero-torque balance (Total momentum, kinetic-energy splitting and centre-of-mass angular momentum of a rigid-material motion clause 4, Torque-free motion of the rigid-material model: conservation of the angular momentum about the centre of mass, the energy-momentum invariants, the principal-axis equations and global existence of the spin dynamics clauses 1-2).

[L2]

Quaternion lifts: for a continuous $ω : I_{0} \to R^{3}$ and $q_{0} \in S^{3}$ there is exactly one $C^{1}$ curve $q : I_{0} \to S^{3}$ with $q (t_{0}) = q_{0}$ and $q^{'} = \frac{1}{2} q φ (ω (t))$ ; the curve $R : = ρ \circ q$ is $C^{1}$ with values in $S O (3)$ and $R^{'} (t) = R (t) Ω_{ω (t)}$ , so $ω$ is its body angular velocity (The quaternion lift of a prescribed continuous angular velocity: $q^{'} = \frac{1}{2} q φ (ω (t))$ keeps $S^{3}$ invariant, its rotation path has body angular velocity $ω$ , and the initial value problem has a unique solution on the whole interval clauses 1-2, The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts). The covering map $ρ : S^{3} \to S O (3)$ is surjective, every $R \in S O (3)$ having exactly two preimages $q$ and $- q$ , under ${A C}_{ω}$ (The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts second remark).

[L3]

Skew/cross correspondence: $Ω_{ω} z = [ω]_{\times} z = ω \times z$ , and for an $S O (3)$ -valued curve the body angular velocity is the unique $ω$ with $R^{T} R^{'} = [ω]_{\times}$ (The skew-matrix/cross-product correspondence $[ω]_{\times} z = ω \times z$ and its conjugation law $R [ω]_{\times} R^{T} = [R ω]_{\times}$ , Body and spatial angular velocity of a rigid-material motion, and the material velocity field clause (a), The orthogonal group $O (3)$ and the rotation group $S O (3)$ ).

[L4]

Matrix products: $(R l)^{'} = R^{'} l + R l^{'}$ for the $C^{1}$ curves $R : I_{0} \to S O (3)$ and $l : I_{0} \to R^{3}$ ; $R^{'} = R [ω]_{\times}$ ; and a continuous curve on an interval with vanishing derivative is constant (Derivative rules for differentiable matrix-valued curves, and skewness of $R^{T} R^{'}$ for a curve in $S O (3)$ , A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

[L5]

Algebra: $u \times v$ is bilinear and alternating with $u \times u = 0$ , and $⟨ u, u ⟩ = ∥ u ∥_{2}^{2}$ (The cross product is bilinear, alternating, and orthogonal to both factors, The cross product in $R^{3}$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L6]

Torque-free Euler equations: for the zero-torque branch the body angular velocity satisfies $I ω^{'} + ω \times I ω = 0$ ; in particular, if $ω$ is constant, then $ω_{0} \times I ω_{0} = 0$ (Euler's equations with torque for the rigid-material model, the rotational power identity, the torque-free reduction and frame covariance clause 1 and clause 4).

Proof

technique · direct; construct the rotation path by the quaternion lift of the constant body angular velocity, verify the zero-torque balance, read off the geometry, and prove the converse from the Euler equations

1.1givenL2L3

By [L2] the covering map $ρ$ is surjective; choose $q_{0} \in S^{3}$ with $ρ (q_{0}) = R_{0}$ , and let $ω : I_{0} \to R^{3}$ be the constant curve $ω (t) : = Ω e_{j}$ , which is continuous. Clause 2 of [L2] supplies a $C^{1}$ curve $q : I_{0} \to S^{3}$ with $q (t_{0}) = q_{0}$ and $q^{'} = \frac{1}{2} q φ (ω)$ , and clause 1 of [L2] shows that $R : = ρ \circ q : I_{0} \to S O (3)$ is $C^{1}$ with $R (t_{0}) = R_{0}$ and $R^{'} (t) = R (t) Ω_{ω (t)}$ ; by [L3] this is $R^{T} R^{'} = [Ω e_{j}]_{\times}$ , so the body angular velocity of $R$ is the constant map $Ω e_{j}$ by the uniqueness in [L3]. This is clause 1.

2.1step 1.1givenL1L4L5algebra

Let $R$ be a path as in clause 1 and let $c : I_{0} \to R^{3}$ be of class $C^{2}$ . Its body angular velocity is the constant $ω \equiv Ω e_{j}$ by step 1.1, so $l = I ω = Ω I_{j} e_{j}$ is constant, and by [L1] $L_{c} = R l$ ; with [L4], $(R l)^{'} = R^{'} l + R l^{'} = R [ω]_{\times} l + 0 = R (ω \times l) = R (Ω^{2} I_{j} e_{j} \times e_{j}) = 0$ by [L5]. Moreover $R^{'} = R [ω]_{\times}$ is of class $C^{1}$ because $R$ is and $[ω]_{\times}$ is constant, so $R$ is of class $C^{2}$ by definition of second differentiability and $(c, R)$ is a configuration path of class $C^{2}$ ; hence, on the model's time window $I_{0}$ , $x (t, ρ) = c (t) + R (t) ρ$ is a motion of the model by Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope clause (ii). Since $L_{c}$ is constant, the zero-torque balance $L_{c}^{'} = 0$ holds and the motion belongs to the torque-free branch; this is clause 2.

3.1step 2.1givenL1L4L5algebra

In the motions of step 2.1, $l = Ω I_{j} e_{j}$ and by [L1] and [L5] $C = ⟨ l, l ⟩ = Ω^{2} I_{j}^{2}$ and $E = \frac{1}{2} ⟨ l, I^{- 1} l ⟩ = \frac{1}{2} Ω^{2} I_{j}$ . If $Ω \neq 0$ , then $Ω I_{j} \neq 0$ and the identity $L_{c} = Ω I_{j} R (t) e_{j}$ of step 2.1 gives $R (t) e_{j} = L_{c} / (Ω I_{j})$ with $L_{c}$ constant, so the material axis $R (t) e_{j}$ is fixed in the inertial frame and the motion is the uniform rotation about that direction; if $Ω = 0$ , then $R^{'} = R [ω]_{\times} = R [0] = 0$ , so $R$ is constant by [L4]. This is clause 3.

4.1step 2.1givenL4L5L6algebra∎

For the converse, let a torque-free motion have constant body angular velocity $ω \equiv ω_{0}$ . By [L6] its body angular velocity satisfies $I ω^{'} + ω \times I ω = 0$ ; with $ω^{'} = 0$ this reads $ω_{0} \times I ω_{0} = 0$ . If $ω_{0} \neq 0$ , in the positively oriented principal coordinates, the coordinate formula for the cross product in [L5] makes this the vanishing of all $2 \times 2$ minors $u_{i} v_{k} - u_{k} v_{i}$ for $u = ω_{0}$ and $v = I ω_{0}$ . Since $ω_{0} \neq 0$ , choose $k$ with $(ω_{0})_{k} \neq 0$ ; then $v_{i} = (v_{k} / (ω_{0})_{k}) (ω_{0})_{i}$ for every $i$ , so $I ω_{0} = λ ω_{0}$ for $λ = v_{k} / (ω_{0})_{k} \in R$ and $ω_{0}$ is an eigenvector of $I$ . Within its eigenspace choose the unit vector $e_{j} = ω_{0} / ∥ ω_{0} ∥_{2}$ . Starting with the original principal vectors in that eigenspace, subtract their projection onto $e_{j}$ , and normalize a nonzero remainder if the eigenspace has dimension greater than one; subtract the projections onto both selected vectors and normalize a nonzero remainder if its dimension is three. Such remainders exist because the original vectors span the eigenspace; linear combinations remain eigenvectors for the same eigenvalue, and the projection subtractions give orthonormality by [L5]. Retain the original orthonormal vectors in the other eigenspaces. This supplies an adapted orthonormal principal basis directly in dimension three. Negating one of the other two basis vectors if necessary makes it positively oriented. In this adapted basis $ω_{0} = Ω e_{j}$ , $Ω = ∥ ω_{0} ∥_{2} \neq 0$ , without a simple-spectrum hypothesis; if $ω_{0} = 0$ , then $R^{'} = R [0] = 0$ and $R$ is constant by [L4]. In the nonzero case the path satisfies the identity of clause 1 with rate $Ω$ , so clauses 2 and 3 apply to it. This is clause 4.

Remarks

Existence is a kinematic statement. Clause 1 constructs the orientation path; clause 2 then verifies the adopted zero-torque balance for it. Neither clause produces a force or a datum, and the curve $c$ is arbitrary of class $C^{2}$ : the torque balance constrains the orientation path alone, in accordance with the absence of a momentum balance in Adoption of the external-torque branch of the rigid-material model: declared torque data and the balance of angular momentum about the centre of mass clause (vi).
Constancy of $l$ and of $ω$ is equivalent. The body inertia matrix $I$ is fixed and invertible, so $l = I ω$ and $ω = I^{- 1} l$ show that either vector is constant exactly when the other is. Thus clause 4 also classifies motions with constant body angular momentum. Constancy of the magnitude alone is insufficient: the companion page exhibits a torque-free motion of a symmetric model with constant $∥ ω ∥_{2}$ and nonconstant $ω$ . The classification does not claim that every torque-free motion is a uniform rotation.

Principal-axis uniform rotations of the torque-free rigid-material model: existence, a fixed spatial direction, the invariants, and the converse classification

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources