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Constant angular velocity: the lift $q (t) = \cos (ω_{0} t / 2) + φ (\hat{n}) \sin (ω_{0} t / 2)$ is the rotation about $\hat{n}$ through $ω_{0} t$ , and a full turn lifts to the antipode

Example

For a curve on a closed interval $[0, T]$ , this item uses " $C^{1}$ on $[0, T]$ " to mean continuous on $[0, T]$ and $C^{1}$ on $(0, T)$ ; no endpoint derivatives are required.

Fix $ω_{0} \in R$ and a unit vector $\hat{n} \in R^{3}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ), put $ω : = ω_{0} \hat{n}$ , and consider

$q : R ⟶ S^{3}, q (t) : = \cos (ω_{0} t / 2) + φ (\hat{n}) \sin (ω_{0} t / 2),$

with $φ$ and $S^{3}$ as in The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts. Then:

$q$ is of class $C^{1}$ with $q (0) = 1$ , and $q$ solves the lift equation $q^{'} (t) = \frac{1}{2} q (t) φ (ω)$ on $R$ .
The curve $R : = ρ \circ q : R \to S O (3)$ is of class $C^{1}$ with $R^{'} (t) = R (t) Ω_{ω}$ for all $t$ ; that is, $ω$ is the constant body angular velocity of $R$ . Explicitly, $R (t)$ is the rotation about $\hat{n}$ through the angle $ω_{0} t$ :

$R (t) φ (v) = φ (\cos (ω_{0} t) v + (1 - \cos (ω_{0} t)) ⟨ \hat{n}, v ⟩_{2} \hat{n} + \sin (ω_{0} t) (\hat{n} \times v)) (v \in R^{3}) .$

If $ω_{0} \neq 0$ , set $T : = 2 π / ∣ ω_{0} ∣ > 0$ . Then $R (0) = R (T) = I_{3}$ while $q (T) = - 1 \neq q (0)$ , and the closed loop $t \mapsto R (t)$ on $[0, T]$ has no closed lift through $1$ with the closed-interval regularity just defined.

Facts & Assumptions

Given: The number $ω_{0}$ , the unit vector $\hat{n}$ , the curve $q (t) = \cos (ω_{0} t / 2) + φ (\hat{n}) \sin (ω_{0} t / 2)$ , and the objects $φ, S^{3}, ρ$ of The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts.

[L1]

$(\sin x)^{'} = \cos x$ , $(\cos x)^{'} = - \sin x$ , and $\sin^{2} x + \cos^{2} x = 1$ ; the addition formulas hold for all real arguments, and the chain rule for differentiating a composite with an affine inner function is available (The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine, The addition formulas for sine and cosine, The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , $C^{k}$ Euclidean maps and diffeomorphisms).

[L2]

For $q = s + φ (u)$ and $q^{'} = t + φ (w)$ one has $q q^{'} = (s t - ⟨ u, w ⟩_{2}) + φ (s w + t u + u \times w)$ ; in particular $φ (\hat{n})^{2} = - 1$ for $∥ \hat{n} ∥_{2} = 1$ , and $N (q) = s^{2} + ∥ u ∥_{2}^{2}$ (The product rule for quaternion and matrix curves, multiplicativity of the quaternion norm, and the conjugation formula $q φ (v) \overset{ˉ}{q} = φ ((a^{2} - ∥ u ∥^{2}) v + 2 ⟨ u, v ⟩ u + 2 a (u \times v))$ clause 1, The quaternions $H$ : real quadruples with componentwise addition and an explicit multiplication formula matching the table on $1, i, j, k$ , The cross product in $R^{3}$ ).

[L3]

If $N (q) = 1$ and $q = \cos (θ / 2) + φ (\hat{n}) \sin (θ / 2)$ with $∥ \hat{n} ∥_{2} = 1$ , then $ρ (q) (φ (v)) = φ (\cos θ v + (1 - \cos θ) ⟨ \hat{n}, v ⟩_{2} \hat{n} + \sin θ (\hat{n} \times v))$ (The product rule for quaternion and matrix curves, multiplicativity of the quaternion norm, and the conjugation formula $q φ (v) \overset{ˉ}{q} = φ ((a^{2} - ∥ u ∥^{2}) v + 2 ⟨ u, v ⟩ u + 2 a (u \times v))$ clause 4).

[L4]

For continuous $ω$ the lift equation $q^{'} = \frac{1}{2} q φ (ω (t))$ with $q (t_{0}) \in S^{3}$ has a unique solution on the whole interval, and for any of its solutions the associated path $ρ \circ q$ is $C^{1}$ with $(ρ \circ q)^{'} = (ρ \circ q) Ω_{ω}$ , so that $ω$ 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 body angular velocity of a $C^{1}$ rotation path: $Ω = R^{T} R^{'}$ is skew, $Ω z = ω \times z$ determines a unique continuous $ω$ , and $R^{'} = R Ω$ ).

[L5]

$ρ (- p) = ρ (p)$ for $p \in S^{3}$ , and $ρ$ maps into $S O (3)$ ; in particular $ρ (- 1)$ is the identity of $S O (3)$ (The quaternion rotation map $ρ$ : conjugation on the imaginary quaternions, the unit sphere $S^{3}$ , and quaternion lifts).

[L6]

For a $C^{1}$ rotation path on a nondegenerate open interval and a prescribed initial unit quaternion over its initial value, the $C^{1}$ quaternion lift is unique, and every $C^{1}$ lift of that path equals one of the two lifts $\pm q$ (Existence, uniqueness and sign ambiguity of the quaternion lift of a $C^{1}$ rotation path clauses 1-2).

[L7]

The quarter-turn identities give $\sin π = 0$ and $\cos π = - 1$ ; applying the shift formulas twice and using parity gives $\sin (\pm 2 π) = 0$ and $\cos (\pm 2 π) = 1$ (Quarter-turn values and shifts by pi/2 and pi, Parity and the Pythagorean identity for sine and cosine).

Verification

technique · direct; differentiate the explicit trigonometric curve and substitute into the axis-angle formula

1.1L1L2algebra

The curve $q$ has $N (q (t)) = \cos^{2} (ω_{0} t / 2) + ∥ \hat{n} ∥_{2}^{2} \sin^{2} (ω_{0} t / 2) = 1$ by [L2] and the unit condition; hence $q$ takes values in $S^{3}$ , and $q$ is $C^{1}$ on $R$ by [L1]. At $t = 0$ one has $q (0) = \cos 0 + φ (\hat{n}) \sin 0 = 1$ .

2.1step 1.1L1L2algebra

Differentiating $q$ with [L1] gives $q^{'} (t) = - \frac{ω_{0}}{2} \sin (ω_{0} t / 2) + φ (\hat{n}) \frac{ω_{0}}{2} \cos (ω_{0} t / 2)$ . On the other hand, by [L2] $\frac{1}{2} q (t) φ (ω) = \frac{1}{2} (\cos (ω_{0} t / 2) + φ (\hat{n}) \sin (ω_{0} t / 2)) φ (ω_{0} \hat{n}) = \frac{ω_{0}}{2} (\cos (ω_{0} t / 2) φ (\hat{n}) + \sin (ω_{0} t / 2) φ (\hat{n})^{2})$ , and $φ (\hat{n})^{2} = - ⟨ \hat{n}, \hat{n} ⟩_{2} + φ (\hat{n} \times \hat{n}) = - 1$ , so this equals $- \frac{ω_{0}}{2} \sin (ω_{0} t / 2) + \frac{ω_{0}}{2} \cos (ω_{0} t / 2) φ (\hat{n}) = q^{'} (t)$ . Hence $q$ solves $q^{'} = \frac{1}{2} q φ (ω)$ with $q (0) = 1$ , which proves clause 1.

3.1step 2.1L3L4

By [L4] the path $R = ρ \circ q$ is $C^{1}$ with $R^{'} = R Ω_{ω}$ , so $ω$ is its body angular velocity; and by [L3] applied with $θ = ω_{0} t$ one has $R (t) φ (v) = ρ (q (t)) (φ (v)) = φ (\cos (ω_{0} t) v + (1 - \cos (ω_{0} t)) ⟨ \hat{n}, v ⟩_{2} \hat{n} + \sin (ω_{0} t) (\hat{n} \times v))$ . This proves clause 2.

4.1step 1.1step 3.1L1L5L6L7algebra∎

If $ω_{0} \neq 0$ , put $σ : = ω_{0} / ∣ ω_{0} ∣ \in {- 1, 1}$ and $T : = 2 π / ∣ ω_{0} ∣ > 0$ , so $ω_{0} T = 2 π σ$ . By [L7], clause 2 gives $R (T) φ (v) = φ (v)$ for every $v$ , hence $R (T) = I_{3} = R (0)$ ; the same quarter-turn values give $q (T) = \cos (π σ) + φ (\hat{n}) \sin (π σ) = - 1 \neq q (0) = 1$ . Suppose $\tilde{q} : [0, T] \to S^{3}$ is a lift, continuous on $[0, T]$ and $C^{1}$ on $(0, T)$ , with $\tilde{q} (0) = \tilde{q} (T) = 1$ . On the nondegenerate open interval $(0, T)$ , [L6] applies to $R ∣_{(0, T)}$ ; every $C^{1}$ lift there is one of the two lifts $\pm q ∣_{(0, T)}$ , so either $\tilde{q} = q$ throughout $(0, T)$ or $\tilde{q} = - q$ throughout $(0, T)$ . In the first case continuity at $T$ gives $\tilde{q} (T) = q (T) = - 1$ , a contradiction; in the second case continuity at $0$ gives $\tilde{q} (0) = - q (0) = - 1$ , also a contradiction. Hence no closed lift with the stated regularity through $1$ exists. This proves clause 3.

Remarks

The example realizes the half-angle relation explicitly: at time $t$ the quaternion angle from $1$ is $ω_{0} t / 2$ , while the rotation angle is $ω_{0} t$ ; the factor of two is the same factor that appears in the lift equation and in the definition of the body angular velocity.
Clause 3 is the explicit closed-loop instance referred to by Scope of the quaternion lift: the global sign ambiguity, closed loops, and the absence of a physical spin claim: the loop returns to the identity rotation, but its lift through the identity quaternion ends at the antipode. The example does not claim that no other lift is closed for some other initial value, and it makes no statement about any physical rotation.

Constant angular velocity: the lift $q (t) = \cos (ω_{0} t / 2) + φ (\hat{n}) \sin (ω_{0} t / 2)$ is the rotation about $\hat{n}$ through $ω_{0} t$ , and a full turn lifts to the antipode

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

Constant angular velocity: the lift q(t)=cos⁡(ω0t/2)+φ(n^)sin⁡(ω0t/2)q(t)=cos(ω0​t/2)+φ(n^)sin(ω0​t/2) is the rotation about n^n^ through ω0tω0​t, and a full turn lifts to the antipode

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

Constant angular velocity: the lift $q (t) = \cos (ω_{0} t / 2) + φ (\hat{n}) \sin (ω_{0} t / 2)$ is the rotation about $\hat{n}$ through $ω_{0} t$ , and a full turn lifts to the antipode