Rigid-body configuration and inertia — Examples

Finite-atomic integration formula: let $f$ be $\nu$-integrable. If $s={\sum }_j{a}_j{\chi }_{E_j}$ is a nonnegative simple function, then $\int sd\nu ={\sum }_j{a}_j\nu (E_j)={\sum }_j{a}_j{m}_0{\sum }_k{\delta }_{x_k}(E_j)=m_0{\sum }_{k}s(x_k)$ by [L1]; taking the supremum over simple $0\le s\le f$ and using the minorant $s_0:={\sum }_{k}f(x_k){\chi }_{\{x_k\}}\le f$ together with the bound $\int sd\nu =m_0{\sum }_{k}s(x_k)\le m_0{\sum }_{k}f(x_k)$ valid for every simple $0\le s\le f$ gives $\int fd\nu =m_0{\sum }_{k}f(x_k)$ for nonnegative $f$; the real case follows from $f=f^{+}-f^{-}$ and linearity [L2], and the complex case by real/imaginary parts.

Trigonometric values and sums: put $\theta =2\pi \mathrm{/}3$ and $c=\cos \theta$. The period in [L4] gives $\cos 3\theta =1$. Repeated addition and the Pythagorean identity give $\cos 3\theta =4c^3-3c$, hence $(c-1)(2c+1{)}^2=0$. Since $0<\theta <\pi$ and cosine strictly decreases on $[0,\pi ]$, $c<\cos 0=1$, so $c=-1\mathrm{/}2$. Positivity of sine and its squared value $1-c^2=3\mathrm{/}4$ give $\sin \theta =\sqrt{3}\mathrm{/}2$. The addition formulas now give $(\cos 2\theta ,\sin 2\theta )=(-1\mathrm{/}2,-\sqrt{3}\mathrm{/}2)$. Together with $(\cos 0,\sin 0)=(1,0)$ these three explicit pairs yield all five finite sums in [L4] by addition.

The measure is a mass distribution: by [L1] it is finite and bounded with support $\{x_0,x_1,x_2\}$, and $\nu ({\mathbb{R}}^3)=3m_0\in (0,\mathrm{\infty })$. Its first moment is $\int \rho d\nu =m_0{\sum }_k{x}_k=m_{0}a({\sum }_k\cos (2\pi k\mathrm{/}3),{\sum }_k\sin (2\pi k\mathrm{/}3),0)=0$ by step 1.1 and [L4]: the reference coordinates are centred.

Inertia matrix: by step 1.1 and [L3], $I=\int (\parallel \rho {\parallel }_2^2{I}_3-\rho {\rho }^{\mathsf{T}})d\nu (\rho )=m_0{\sum }_k(\parallel x_k{\parallel }_2^2{I}_3-x_k{x}_k^{\mathsf{T}})=m_0(3a^2{I}_3-{\sum }_k{x}_k{x}_k^{\mathsf{T}}),$ because $\parallel x_k{\parallel }_2=a$ for every $k$. The matrix ${\sum }_k{x}_k{x}_k^{\mathsf{T}}$ has entries $a^2{\sum }_k{\cos }^2(2\pi k\mathrm{/}3)$, $a^2{\sum }_k\cos (2\pi k\mathrm{/}3)\sin (2\pi k\mathrm{/}3)$ and $a^2{\sum }_k{\sin }^2(2\pi k\mathrm{/}3)$ in the plane and zero in the third row and column, so by [L4] it equals $\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(\frac{3a^2}{2},\frac{3a^2}{2},0)$. Hence $I=m_0\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(3a^2-\frac{3a^2}{2},3a^2-\frac{3a^2}{2},3a^2)$, which is the displayed matrix of clause 2.

Principal data: the displayed matrix is diagonal, so $I{e}_1=\frac{3m_0{a}^2}{2}{e}_1$, $I{e}_2=\frac{3m_0{a}^2}{2}{e}_2$ and $I{e}_3=3m_0{a}^2{e}_3$: $e_3$ is a principal axis with moment $3m_0{a}^2$, and every $g$-orthonormal pair in the plane $\{\rho \in {\mathbb{R}}^3:{\rho }_3=0\}$ is a pair of principal axes with the common moment $\frac{3m_0{a}^2}{2}$. The integral identities are checked directly: $\int \parallel e_3\times \rho {\parallel }_2^{2}d\nu =m_0{\sum }_k((x_k{)}_1^2+(x_k{)}_2^2)=m_0\cdot 3a^2=3m_0{a}^2$ and $\int \parallel e_1\times \rho {\parallel }_2^{2}d\nu =m_0{\sum }_k((x_k{)}_2^2+(x_k{)}_3^2)=m_0\cdot \frac{3a^2}{2}$ by [L4] and [L6], in agreement with the diagonal entries by clause 2.

Nondegeneracy and invertibility: no line through the origin contains two of the points $x_k$, because their arguments modulo $\pi$ are $0,\frac{\pi }{3},\frac{2\pi }{3}$ (using [L4]), which are pairwise distinct; hence for every line $L$ through the origin at least two of the three points lie outside $L$ and $\nu ({\mathbb{R}}^3\setminus L)\ge 2m_0>0$. The model is therefore nondegenerate, and by clause (d) of [L3] all principal moments are positive (as they visibly are) and $I$ is invertible.

Rotational term: by clause (b) of [L3] and clause 2, for a motion with body angular velocity $\omega ={\sum }_j{\omega }_j{e}_j$ the quadratic form is ${\omega }^{\mathsf{T}}I\omega ={\sum }_j{I}_j{\omega }_j^2=\frac{3m_0{a}^2}{2}({\omega }_1^2+{\omega }_2^2)+3m_0{a}^2{\omega }_3^2$; multiplying by $\frac{1}{2}$ and citing clause 3 of Total momentum, kinetic-energy splitting and centre-of-mass angular momentum of a rigid-material motion for the reading of the term as the rotational kinetic energy gives clause 4. In particular a rotation about the symmetry axis $e_3$ with rate $\varpi$ has rotational term $\frac{3}{2}{m}_0{a}^2{\varpi }^2$, and a rotation about an in-plane principal axis has term $\frac{3}{4}{m}_0{a}^2{\varpi }^2$.

Symmetry check: $S=R_z(2\pi \mathrm{/}3)$ is the rotation of the reference plane that carries $x_k$ to $x_{k+1}$ for $k=0,1,2$ (indices modulo $3$): its matrix acts on $a(\cos \theta ,\sin \theta ,0)$ as $a(\cos (\theta +\frac{2\pi }{3}),\sin (\theta +\frac{2\pi }{3}),0)$ by the addition formulas and [L5]. Since $S$ is orthogonal and preserves norms, $SI{S}^{\mathsf{T}}=m_0(3a^2{I}_3-{\sum }_k(S{x}_k)(S{x}_k{)}^{\mathsf{T}})=m_0(3a^2{I}_3-{\sum }_k{x}_{k+1}{x}_{k+1}^{\mathsf{T}})=I$ by step 2.2, the preservation of norms and the relabelling of the finite sum. Thus $SI=IS$, and the invariance of the inertia data under the threefold symmetry is exhibited; on the plane $\{\rho \in {\mathbb{R}}^3:{\rho }_3=0\}$, where $S$ acts as the rotation by $\frac{2\pi }{3}$, the matrix $I$ is scalar, which is exactly the equality of the two transverse principal moments in step 3.1.

Interpretation: the model is a rigid-material model with three equal point masses whose reference shape is symmetric under the threefold rotation; its inertia matrix, principal axes, principal moments and rotational term have been computed, and the symmetry invariance of the inertia matrix checked. The numbers are proportions of the model parameters $m_0$ and $a$; no measured moment of inertia and no concrete physical body is asserted, and the statements are conditional on the model of Adoption of the rigid-material model branch: reference mass distribution, rigid placements, orientation and scope.

Finite-atomic integration formula: fix $m_0>0$, $a>0$ and $\nu =m_0{\delta }_{a{e}_3}+m_0{\delta }_{-a{e}_3}$. For a nonnegative simple function $s$, the defining simple integral gives $\int sd\nu =m_{0}s(a{e}_3)+m_{0}s(-a{e}_3)$ by [L1] and [L2]. If $f\ge 0$ is measurable and finite at the two atoms, $s_0=f(a{e}_3){\chi }_{\{a{e}_3\}}+f(-a{e}_3){\chi }_{\{-a{e}_3\}}\le f$ is a nonnegative simple minorant. Its integral gives the lower bound $m_{0}f(a{e}_3)+m_{0}f(-a{e}_3)$, while every simple $s\le f$ gives the corresponding upper bound. Taking the supremum proves $\int fd\nu =m_{0}f(a{e}_3)+m_{0}f(-a{e}_3)$. For a nonnegative function infinite at an atom, simple minorants with arbitrary finite value there have unbounded integrals, giving the same identity with value $+\mathrm{\infty }$. For a real measurable function finite at the atoms, applying the finite formula to its modulus establishes integrability, and positive/negative parts and linearity give the signed formula; the complex case follows by real/imaginary parts.

The model. Take $m_0>0$, $a>0$ and $\nu :=m_0{\delta }_{a{e}_3}+m_0{\delta }_{-a{e}_3},$ a finite weighted sum of Dirac measures at the distinct points $a{e}_3$ and $-a{e}_3$ with positive weights. By [L1] $\nu$ is a finite bounded measure; its total mass is $m=2m_0\in (0,\mathrm{\infty })$ and its first moment is $\int \rho d\nu =m_{0}a{e}_3+m_0(-a{e}_3)=0$ by the finite-atomic formula of step 1.1, so the reference coordinates are centred and $\nu$ is a mass distribution of the model branch.

Inertia matrix: by step 1.1 and [L3], $I=m_0(\parallel a{e}_3{\parallel }_2^2{I}_3-(a{e}_3)(a{e}_3{)}^{\mathsf{T}})+m_0(\parallel -a{e}_3{\parallel }_2^2{I}_3-(-a{e}_3)(-a{e}_3{)}^{\mathsf{T}})=2m_0{a}^2(I_3-e_3{e}_3^{\mathsf{T}}),$ which is $\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(2m_0{a}^2,2m_0{a}^2,0)$ in the standard basis by [L5]. In particular $I{e}_3=0$ with $e_3\mathrm{\ne }0$, so $I$ is not invertible.

Principal data and the vanishing criterion: the displayed matrix is diagonal, so the principal moments are $2m_0{a}^2$, $2m_0{a}^2$ and $0$, and $e_3$ is a principal axis with vanishing moment. This is consistent with clause (d) of [L3]: $\nu$ is supported on the line $\mathbb{R}{e}_3$ of the principal axis $e_3$, so the principal moment about $e_3$ vanishes; the model is degenerate, and the nondegeneracy hypothesis of the positivity statement fails.

A rotation about the line is invisible to the rotational term. Let the model move with constant-rate orientation path $R(t)=R_z(\varpi t)$ about the symmetry axis, so that the configuration path is $(c,R)$ with $c$ arbitrary of class $C^2$; by Body and spatial angular velocity of a rigid-material motion, and the material velocity field its body angular velocity is $\omega =\varpi e_3$ (the axis $e_3$ is fixed by $R_z$, and $R^{\mathsf{T}}{R}^{\mathrm{\prime }}=[\varpi e_3{]}_{\times }$ by the derivative clause of the coordinate-rotation matrices). Every material point lies on the rotation axis: $\omega \times (\pm a{e}_3)=\pm \varpi (e_3\times a{e}_3)=0$ by [L5], so the material velocity equals $c^{\mathrm{\prime }}$ on both mass-carrying labels and the rotational term is $\frac{1}{2}{\omega }^{\mathsf{T}}I\omega =\frac{1}{2}{\varpi }^2{e}_3^{\mathsf{T}}I{e}_3=0$ by [L4] and step 3.1. The angular momentum about the centre of mass is $L_c=RI\omega =0$ by [L4], so the motion's spin about the line produces neither rotational kinetic energy nor angular momentum.

Conclusion. The model is a rigid-material model of positive total mass, and its inertia matrix $\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(2m_0{a}^2,2m_0{a}^2,0)$ has a zero principal moment, is not invertible, and fails to detect rotation about the line carrying the masses. The refuted statement — that every positive-mass model of the branch has positive definite inertia — is therefore false. The correct criterion is clause (d) of Principal axes and principal moments of the inertia data of a rigid-material model, axis moments and extremal values: the moment about a principal axis vanishes exactly when the mass is supported on that axis; the model branch, the kinetic splitting and the principal-axis theorem all remain valid with their stated degeneracy caveats.

Pole values: by the composition law and $R_y(0)=I_3$ of [L1], $\mathrm{\Phi }(\alpha ,0,\gamma )=R_z(\alpha )R_z(\gamma )=R_z(\alpha +\gamma )$, so the value depends only on $\alpha +\gamma$. For every real $s$, $\mathrm{\Phi }(\alpha +s,0,\gamma -s)=R_z(\alpha +s+\gamma -s)=R_z(\alpha +\gamma )=\mathrm{\Phi }(\alpha ,0,\gamma )$, and the points $(\alpha +s,0,\gamma -s)$ approach $(\alpha ,0,\gamma )$ as $s\to 0$ while remaining distinct for $s\mathrm{\ne }0$. Hence $\mathrm{\Phi }$ is not injective on any neighbourhood of the pole point; this is clause 1.

Partial derivatives at the pole: differentiating with respect to $\alpha$ with $\beta ,\gamma$ fixed and using the product rule and [L1], ${\mathrm{\partial }}_{\alpha }\mathrm{\Phi }(\alpha ,\beta ,\gamma )=[e_3{]}_{\times }{R}_z(\alpha )R_y(\beta )R_z(\gamma )$, and at $\beta =0$ this is $[e_3{]}_{\times }{R}_z(\alpha +\gamma )$ by clause 1. Differentiating with respect to $\gamma$ gives ${\mathrm{\partial }}_{\gamma }\mathrm{\Phi }(\alpha ,\beta ,\gamma )=R_z(\alpha )R_y(\beta )R_z(\gamma )[e_3{]}_{\times }$; at $\beta =0$ the middle factor is $I_3$ and the factors $R_z(\gamma )$ and $[e_3{]}_{\times }$ commute by the identity $R_z(a)[e_3{]}_{\times }=[e_3{]}_{\times }{R}_z(a)$ of [L1], so ${\mathrm{\partial }}_{\gamma }\mathrm{\Phi }(\alpha ,0,\gamma )=R_z(\alpha )[e_3{]}_{\times }{R}_z(\gamma )=[e_3{]}_{\times }{R}_z(\alpha )R_z(\gamma )=[e_3{]}_{\times }{R}_z(\alpha +\gamma )$, the same value as ${\mathrm{\partial }}_{\alpha }\mathrm{\Phi }$.

Rank and absence of a local inverse: the Jacobian matrix $J\mathrm{\Phi }(\alpha ,0,\gamma )$ has columns ${\mathrm{\partial }}_{\alpha }\mathrm{\Phi },{\mathrm{\partial }}_{\beta }\mathrm{\Phi },{\mathrm{\partial }}_{\gamma }\mathrm{\Phi }$ [L3], and the first and third columns coincide by step 1.2. Thus the linear map $D\mathrm{\Phi }:{\mathbb{R}}^3\to M_3(\mathbb{R})\cong {\mathbb{R}}^9$ annihilates the nonzero vector $(1,0,-1)$ and has rank at most $2$, so it is not injective. Moreover, step 1.1 gives distinct points with equal images in every neighbourhood of the pole point, ruling out any local inverse, including a differentiable inverse onto a neighbourhood in $SO(3)$. This proves clause 2.

Angular velocity along a curve through the pole: let ${\mathrm{\Phi }}_t:=\mathrm{\Phi }(\alpha (t),\beta (t),\gamma (t))$; by the product rule and [L1] at $t=0$, where $\beta (0)=0$ and $R_y(0)=I_3$, ${\mathrm{\Phi }}_0^{\mathrm{\prime }}=({\alpha }^{\mathrm{\prime }}(0)+{\gamma }^{\mathrm{\prime }}(0))[e_3{]}_{\times }{R}_z(\alpha (0)+\gamma (0))+{\beta }^{\mathrm{\prime }}(0)R_z(\alpha (0))[e_2{]}_{\times }{R}_z(\gamma (0)),$ using the two computations of step 1.2 for the $\alpha$- and $\gamma$-directions and the middle derivative ${\mathrm{\partial }}_{\beta }\mathrm{\Phi }=R_z(\alpha )[e_2{]}_{\times }{R}_z(\gamma )$ at the pole. Multiplying on the left by ${\mathrm{\Phi }}_0^{\mathsf{T}}=R_z(-\gamma (0))R_z(-\alpha (0))={\mathrm{\Phi }}_0^{-1}$ (the matrix is orthogonal [L1]) and using the conjugation law $R[\omega {]}_{\times }{R}^{\mathsf{T}}=[R\omega {]}_{\times }$ of [L2] together with $R_z(-\gamma (0))e_2$ for the second term gives ${\mathrm{\Phi }}_0^{\mathsf{T}}{\mathrm{\Phi }}_0^{\mathrm{\prime }}=[({\alpha }^{\mathrm{\prime }}(0)+{\gamma }^{\mathrm{\prime }}(0))e_3+{\beta }^{\mathrm{\prime }}(0)R_z(-\gamma (0))e_2{]}_{\times },$ so by the injectivity clause of [L2] and the characterisation of the body angular velocity [L4], $\omega (0)=({\alpha }^{\mathrm{\prime }}(0)+{\gamma }^{\mathrm{\prime }}(0))e_3+{\beta }^{\mathrm{\prime }}(0)R_z(-\gamma (0))e_2$. The rates ${\alpha }^{\mathrm{\prime }}$ and ${\gamma }^{\mathrm{\prime }}$ enter only through the sum, so $({\alpha }^{\mathrm{\prime }}+h,{\beta }^{\mathrm{\prime }},{\gamma }^{\mathrm{\prime }}-h)$ and $({\alpha }^{\mathrm{\prime }},{\beta }^{\mathrm{\prime }},{\gamma }^{\mathrm{\prime }})$ give the same angular velocity: the rate map is not injective at the pole. This is clause 3.

The regular physical path: $R(t)=R_y(\kappa t)$ has $R(0)=I_3$ and is the composition of the smooth coordinate-rotation curve with the smooth map $t\mapsto \kappa t$, hence of class $C^{\mathrm{\infty }}$ [L1]; by the derivative clause of [L1], $R^{\mathrm{\prime }}(t)=\kappa R_y(\kappa t)[e_2{]}_{\times }$, so $R(t{)}^{\mathsf{T}}{R}^{\mathrm{\prime }}(t)=[\kappa e_2{]}_{\times }$ for every $t$ and the body angular velocity is the constant vector $\kappa e_2$ by [L2] and [L4]; in particular the orientation path passes through the pole orientation $R(0)=I_3=\mathrm{\Phi }(0,0,0)$ with a well-defined, constant angular velocity. The failure of clause 2 is thus a failure of the parameter triple to be a chart at the pole, not a singularity of the orientation path or of its angular velocity. This proves clause 4.

Summary: the parametrization $\mathrm{\Phi }$ is smooth as a map, but at every pole point it is non-injective (clause 1), its derivative has rank at most $2$ and it admits no differentiable local inverse (clause 2), and the angular velocity does not uniquely determine the Euler-angle rates at a fixed pole triple (clause 3), while a physical orientation path through the same pole orientation is regular with a perfectly well-defined angular velocity (clause 4).