Principal axes and principal moments of the inertia data of a rigid-material model, axis moments and extremal values

By [L2] the matrix $I$ is symmetric, so the map $T:\omega \mapsto I\omega$ satisfies $⟨T\omega ,z⟩=(I\omega {)}^{\mathsf{T}}z={\omega }^{\mathsf{T}}Iz=⟨\omega ,Tz⟩$ for all $\omega ,z$: it is self-adjoint on the Euclidean space ${\mathbb{R}}^3$. Applying the spectral theorem [L4], or equivalently clause 1 of [L1], gives an orthonormal basis $(e_1,e_2,e_3)$ and real $I_j$ with $I{e}_j=I_j{e}_j$. This is clause (a).

Spatial axis moments: put $a:=a(t,p)=R(t{)}^{\mathsf{T}}(p-c(t))$ and $m:=R(t{)}^{\mathsf{T}}n$, so that $\parallel m{\parallel }_2=\parallel n{\parallel }_2=1$. By [L3] and bilinearity of the matrix product, $n^{\mathsf{T}}I(t,p)n=m^{\mathsf{T}}{I}_{\nu }(a)m={\sum }_{j,k}{m}_j{m}_k\int (\parallel \rho -a{\parallel }_2^2{\delta }_{jk}-({\rho }_j-a_j)({\rho }_k-a_k))d\nu (\rho )\mathrm{.}$ The integrand is the shifted quadratic polynomial of [L3], integrable by [L6]; pulling the finite sum inside the integral by linearity [L6] and evaluating the sum gives ${\sum }_{j,k}{m}_j{m}_k(\parallel \rho -a{\parallel }_2^2{\delta }_{jk}-({\rho }_j-a_j)({\rho }_k-a_k))=\parallel m{\parallel }_2^2\parallel \rho -a{\parallel }_2^2-⟨m,\rho -a{⟩}^2=\parallel m\times (\rho -a){\parallel }_2^2$ for each $\rho$, by the cross-norm identity of [L5] and $\parallel m{\parallel }_2=1$. Hence $n^{\mathsf{T}}I(t,p)n=\int \parallel m\times (\rho -a){\parallel }_2^{2}d\nu (\rho )$.

For $\omega ={\sum }_j{\omega }_j{e}_j$, linearity of $T$ and $I{e}_j=I_j{e}_j$ give $I\omega ={\sum }_j{I}_j{\omega }_j{e}_j$; taking the inner product with $\omega$ and using $⟨e_j,e_k⟩={\delta }_{jk}$ gives ${\omega }^{\mathsf{T}}I\omega ={\sum }_j{I}_j{\omega }_j^2$, and [L1] gives the further equality ${\sum }_j{I}_j{\omega }_j^2=\int \parallel \omega \times \rho {\parallel }_2^{2}d\nu (\rho )$. Setting $\omega =e_j$ gives $I_j=e_j^{\mathsf{T}}I{e}_j=\int \parallel e_j\times \rho {\parallel }_2^{2}d\nu (\rho )$. This is clause (b).

Changing description: for every $\rho$, $R(t)(m\times (\rho -a))=(Rm)\times (R\rho -Ra)$ by the equivariance clause of [L5], and $Rm=n$ while $R\rho -Ra=R\rho -(p-c)=x(t,\rho )-p$ by the defining relation $Ra=p-c$; since $R(t)$ preserves the norm [L5], $\parallel m\times (\rho -a){\parallel }_2=\parallel n\times (x(t,\rho )-p){\parallel }_2$. Substituting into step 1.2 gives the second identity of clause (c).

Invariance on the line: let $p^{\mathrm{\prime }}=p+sn$ with $s\in \mathbb{R}$, so that $a(t,p^{\mathrm{\prime }})=a+sm$ and $\rho -a(t,p^{\mathrm{\prime }})=(\rho -a)-sm$; then $m\times (\rho -a^{\mathrm{\prime }})=m\times (\rho -a)-s(m\times m)=m\times (\rho -a)$ by [L5], so the integral of step 1.2 is unchanged. Replacing $n$ by $-n$ replaces $m$ by $-m$, and $(-m)\times (\rho -a)=-m\times (\rho -a)$ has the same norm; the quadratic expression $n^{\mathsf{T}}I(t,p)n$ is unchanged under $n\mapsto -n$ by bilinearity. Hence ${\mathrm{\Theta }}_{\mathrm{\ell }}(t)$ depends only on the spatial line $\mathrm{\ell }$ and not on the choice of $p$ on it or on the sign of $n$. This completes clause (c).

Body-fixed axis moments: applying step 2.1 with $\omega =n$ gives $n^{\mathsf{T}}In=\int \parallel n\times \rho {\parallel }_2^{2}d\nu (\rho )$ for every $n\in {\mathbb{R}}^3$, in particular for unit vectors. This is the first identity of clause (c).

Nonnegativity and vanishing criterion: by step 2.1, $I_j=\int f_{j}d\nu$ with $f_j(\rho )=\parallel e_j\times \rho {\parallel }_2^2\ge 0$; by A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere and Monotonicity and nonnegative homogeneity of the nonnegative integral the integral vanishes exactly when $f_j=0$ $\nu$-almost everywhere, and by Measure-null sets and almost-everywhere statements relative to a measure this holds exactly when $\nu (\{\rho :f_j(\rho )\mathrm{\ne }0\})=0$. Since $e_j\mathrm{\ne }0$, the identity $e_j\times \rho =0$ is equivalent to $\rho \in \mathbb{R}{e}_j$ (The cross product in ${\mathbb{R}}^3$, The cross product is bilinear, alternating, and orthogonal to both factors); hence $I_j=0$ exactly when $\nu ({\mathbb{R}}^3\setminus \mathbb{R}{e}_j)=0$, that is, exactly when $\nu$ is supported on the line of the principal axis $e_j$.

Nondegeneracy: assume $\nu ({\mathbb{R}}^3\setminus \mathbb{R}\omega )>0$ for every $\omega \mathrm{\ne }0$. For each $j$, the line $\mathbb{R}{e}_j$ is a line through the origin and $\mathbb{R}{e}_j=\mathbb{R}\omega$ for $\omega =e_j\mathrm{\ne }0$, so $\nu ({\mathbb{R}}^3\setminus \mathbb{R}{e}_j)>0$ and step 3.2 gives $I_j\mathrm{\ne }0$, hence $I_j>0$ by nonnegativity. If $I\omega =0$ for some $\omega$, then ${\omega }^{\mathsf{T}}I\omega ={\omega }^{\mathsf{T}}0=0$, so $\int \parallel \omega \times \rho {\parallel }_2^{2}d\nu =0$ by [L2] and the argument of step 3.2 gives $\nu ({\mathbb{R}}^3\setminus \mathbb{R}\omega )=0$ for that $\omega$; if $\omega \mathrm{\ne }0$ this contradicts the nondegeneracy hypothesis, so the nullspace of $I$ is $\{0\}$ and $I$ is invertible (Invertible matrix theorem: invertibility, full pivot rank, RREF $I$, trivial nullspace and unique solvability are equivalent).

Degenerate support: suppose $\nu$ is supported on a line through the origin, with direction vector $u\mathrm{\ne }0$, that is $\nu ({\mathbb{R}}^3\setminus \mathbb{R}u)=0$. Then $u\times \rho =0$ for $\nu$-almost every $\rho$, so $u^{\mathsf{T}}Iu=\int \parallel u\times \rho {\parallel }_2^{2}d\nu =0$ by [L2]. For every $k$ and every real $s$, applying [L2] to $u+s{e}_k$ and expanding the quadratic form by bilinearity and symmetry of $I$ gives $u^{\mathsf{T}}Iu+2s{e}_k^{\mathsf{T}}Iu+s^2{e}_k^{\mathsf{T}}I{e}_k=\int \parallel (u+s{e}_k)\times \rho {\parallel }_2^{2}d\nu =s^2\int \parallel e_k\times \rho {\parallel }_2^{2}d\nu ,$ because $u\times \rho =0$ $\nu$-almost everywhere; the terms not involving $s^2$ therefore vanish for all $s$, so $e_k^{\mathsf{T}}Iu=0$ for every $k$ and hence $Iu=0$ with $u\mathrm{\ne }0$. Thus $I$ is not invertible. If the support of a centred $\nu$ is contained in a line, that line passes through the origin, since $0=\int \rho d\nu \mathrm{/}m$ is the mean of a measure supported there; hence the stated case covers every centred mass distribution supported on a line. This proves clause (d).

Extremal values: for a unit vector $n={\sum }_j{c}_j{e}_j$, orthonormality gives ${\sum }_j{c}_j^2=\parallel n{\parallel }_2^2=1$ by [L7], and step 2.1 gives $n^{\mathsf{T}}In={\sum }_j{I}_j{c}_j^2$; since the coefficients $c_j^2$ are nonnegative and sum to $1$ while each $I_j\ge 0$ by step 3.2, the value lies between ${\min }_j{I}_j$ and ${\max }_j{I}_j$; setting $n$ equal to a unit eigenvector whose principal moment is minimal (respectively maximal) attains the minimum (respectively maximum). This is clause (e).