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For $0<I_1<I_2<I_3$, the transverse linear stability rates about steady rotation with signed component $\Omega \ne 0$ areAxes one and three have transverse eigenvalues $\pm i{\nu }_1$ and $\pm i{\nu }_3$, producing bounded oscillations. Axis two has transverse eigenvalues $\pm {\lambda }_2$, giving growing and decaying exponential functions. A neutral longitudinal perturbation corresponds to a small change of $\Omega$. Generic intermediate-axis disturbances contain a growing component; a precisely chosen decaying direction is exceptional. For the normalized intermediate-axis model with $|\Omega |=1$, the values are ${\nu }_1=1/\sqrt{3}$, ${\nu }_3=1$ and ${\lambda }_2=1/\sqrt{3}$.

Proof outline: the linearization of the Euler equations for a torque-free rigid body about the third principal axis gives transverse perturbations obeyingand henceCyclic permutation gives the other rates. For nonlinear Lyapunov stability of the extreme rotations, use the torque-free rigid-body invariants: the conserved quadratic forms $L^2-2I_{1}E$ and $2I_{3}E-L^2$ have strictly positive transverse coefficients about axes one and three respectively. They bound transverse angular velocity, while $E$ bounds its longitudinal change near nonzero rotation. Imaginary transverse eigenvalues alone would not establish this nonlinear conclusion.^[ref]

The practical lesson is to choose a stable principal axis when using rotation to maintain an instrument's pointing direction. Both extreme principal moments of inertia are stable in the ideal torque-free model; if internal damping lowers rotational kinetic energy while approximately preserving angular momentum, the minimum energy $E=L^2/(2I_3)$ instead favors the largest principal moment of inertia. A control system must account for torque, changing mass distributions and flexibility before applying this ideal result.

Choose an inertia unit $I_{\ast }$ and an angular speed unit ${\Omega }_{\ast }>0$, so physical principal moments of inertia are $I_{\ast }(1,2,3)$, physical angular velocity is ${\Omega }_{\ast }\mathbf{\omega }$, and normalized time is $t={\Omega }_{\ast }{t}_{\mathrm{physical}}$. The Euler equations for a torque-free rigid body becomeThis is a physically admissible ideal planar mass distribution: $I_3=I_1+I_2$ is the perpendicular axis theorem. A book of finite thickness need not have these exact ratios.

For an original comparison, start with angular velocity $(1,\epsilon ,0)$, $(\epsilon ,1,0)$ or $(\epsilon ,0,1)$, respectively. With $\epsilon =0.01$, the transverse Euclidean norm stays small about axes one and three, but becomes order one about axis two. For the intermediate-axis start, its linearization givesThe hyperbolic functions contain a growing exponential function with rate ${\lambda }_2=1/\sqrt{3}$. Reducing $\epsilon$ tenfold delays reaching the same small transverse amplitude by approximately $\sqrt{3}\log 10\simeq 3.99$ normalized time units, once the growing term dominates. That delays the instability. Once the transverse angular velocity is comparable to the main component, the linearization ceases to describe the motion.

The components in Euler equations for a torque-free rigid body are evaluated in the body-fixed principal-axis frame, with $L_i=I_i{\omega }_i$. In an inertial space frame the angular momentum vector is constant; its components in the rotating body frame need not be.

Assume $\Omega \ne 0$ and distinct positive principal moments of inertia. Write $\omega =(\Omega +{\delta }_1,{\delta }_2,{\delta }_3)$. To first order,soThe coefficient is positive when $I_1$ lies between the other moments, giving an exponentially growing mode and instability. When $I_1$ is an extreme moment, it is negative and the transverse perturbations oscillate.

To establish actual nonlinear Lyapunov stability in the latter case, use the conserved kinetic energy $E=\frac{1}{2}\sum I_j{\omega }_j^2$ and squared angular momentum $L^2=\sum I_j^2{\omega }_j^2$. Their combination isFor an extreme $I_1$ its two coefficients have the same sign. Conservation controls both transverse components by their initially small values, while conservation of $E$ controls ${\omega }_1^2$ near ${\Omega }^2$. Continuity preserves its initial sign for sufficiently small perturbations. Thus nonzero steady rotation is stable exactly about the largest or smallest principal moment, the intermediate axis theorem. It is not asymptotically stable, since energy is conserved. If $\Omega =0$, energy makes the zero angular-velocity state stable for every ordering; the nonzero-rotation assumption is essential to the stated equivalence.

Put $I=I_1=I_2$. The third of the Euler equations for a torque-free rigid body gives ${\dot{\omega }}_3=0$. The first two become ${\dot{\omega }}_1=-\Omega {\omega }_2$, ${\dot{\omega }}_2=\Omega {\omega }_1$, whereTherefore ${\omega }_1+i{\omega }_2=C{e}^{i\Omega t}$. In body coordinates, $L_1+i{L}_2=IC{e}^{i\Omega t}$ and $L_3=I_3{\omega }_3$ is constant. The angular momentum precesses about the body symmetry axis, with periodFor positive ${\omega }_3$ this is the printed expression. In the inertial frame the torque-free angular momentum is constant; the precession here is its changing representation in the rotating body frame. If $e=0$ or ${\omega }_3=0$, this body-frame precession rate is zero.

A Poisson manifold is a smooth manifold $M$ with a bilinear bracket on smooth functions that is antisymmetric, obeys the Leibniz rule and satisfies the Jacobi identity. Equivalently, a Poisson tensor $\pi$ gives $\{f,g\}=\pi (df,dg)$. Its rank need not be constant or maximal. Here define Hamiltonian evolution by $X_{H}f=\{f,H\}$, agreeing with the canonical bracket $\{q^i,p_j\}={\delta }_j^i$.

A smooth Lie group action sends each $x$ along its orbit $Gx$, with stabilizer $G_x$. If the action is free and proper, the orbit space is a smooth quotient manifold and the projection is a submersion. Without these hypotheses the quotient can have singular orbit-type strata, or even fail to be Hausdorff. For an action preserving the Poisson bracket, invariant functions form a Poisson subalgebra. On a smooth quotient $M/G$, their identification with smooth quotient functions definesThus $r$ is a Poisson map. The quotient is generally Poisson, not necessarily symplectic: different reduced momentum sectors can be different leaves. An arbitrary smooth group action alone does not guarantee this bracket descent.

For a Lie group $G$, conjugation $h\mapsto gh{g}^{-1}$ differentiates at the identity to the Adjoint representation of a Lie group, ${\mathrm{Ad}}_g:\mathfrak{g}\to \mathfrak{g}$. Its infinitesimal version is ${\mathrm{ad}}_{\xi }\eta =[\xi ,\eta ]$. The coadjoint representation on the dual space is defined byThe inverse in the first formula makes this a left action. These definitions distinguish the representation on infinitesimal generators from its dual action on momenta.

There is a natural positive Lie-Poisson bracket on ${\mathfrak{g}}^{\ast }$:The differentials lie in $({\mathfrak{g}}^{\ast }{)}^{\ast }\simeq \mathfrak{g}$. In a basis with $[e_a,e_b]=c_{ab}{}^c{e}_c$, linear coordinates satisfy $\{{\ell }_a,{\ell }_b{\}}_{+}=c_{ab}{}^c{\ell }_c$. The bracket extends by the chain rule and Leibniz rule to smooth functions. Its coordinate Jacobi condition reduces precisely to the Jacobi identity for $c_{ab}{}^c$, proving that it is Poisson. Reversing its overall sign also gives a Poisson bracket and is useful for body-frame conventions below.

On a general Poisson manifold, the vectors $X_f$ span a distribution. The Jacobi identity closes this family under commutators, and its orbits under Hamiltonian flows form the symplectic foliation. On each leaf the induced two-form satisfiesIt is well-defined after quotienting out differential covectors giving zero Hamiltonian vector field, is nondegenerate on the leaf, and is closed by the Jacobi identity. Leaf dimensions can vary, so ordinary constant-rank foliation terminology must be qualified. A Hamiltonian trajectory remains in one leaf; a Casimir function of a Poisson manifold is constant along all such trajectories, but its level set can contain more than one leaf.

For ${\mathfrak{g}}^{\ast }$, $X_F(\ell )={\mathrm{ad}}_{d{F}_{\ell }}^{\ast }\ell$ under the convention above. Hence the connected coadjoint orbits are exactly its symplectic leaves for connected $G$; for disconnected $G$, take connected orbit components. The orbit two-form is the Kirillov–Kostant–Souriau symplectic form:If a generator is in the stabilizer, its pairing with every bracket vanishes, proving independence of the choice of tangent generator. Conversely, vanishing pairing with all tangent generators puts it in that stabilizer, proving nondegeneracy. The cyclic derivative relation giving $d\omega =0$ is the Lie algebra Jacobi identity. This explains the symplectic structure of the orbit rather than just naming it.

Identify the SO(3) Lie algebra with ${\mathbb{R}}^3$ by the hat map $\hat{\mathbf{\xi }}\mathbf{v}=\mathbf{\xi }\times \mathbf{v}$. Its Lie bracket becomes the vector cross product, and the invariant Euclidean pairing identifies the dual with angular-momentum vectors $\mathbf{M}$. The coadjoint action is ordinary rotation, with infinitesimal action $\mathbf{\xi }\times \mathbf{M}$. Therefore its nonzero orbits are spheres $|\mathbf{M}|=r$; the origin is a zero-dimensional leaf. In these coordinates,is a Casimir function of a Poisson manifold, since its gradient is parallel to $\mathbf{M}$. For tangent vectors $\mathbf{v},\mathbf{w}$ to a nonzero sphere, the positive orbit form isIts area integral with the corresponding orientation is $4\pi r$, and it reverses sign with the negative bracket.

For a torque-free rigid body, choose body angular momentum $\mathbf{M}$ and principal moments $I_1,I_2,I_3>0$. Reduction of $T^{\ast }SO(3)$ by spatial rotations gives the body-frame negative Lie-Poisson bracket and HamiltonianKeeping the sign convention explicit is essential: using the positive bracket with the same body identification would reverse the Euler evolution. The negative bracket giveswith cyclic expressions for the other two components. These are the Euler equations for a torque-free rigid body. The Hamiltonian and $C$ are conserved because $\mathbf{\Omega }\cdot (\mathbf{M}\times \mathbf{\Omega })=0$ and $\mathbf{M}\cdot (\mathbf{M}\times \mathbf{\Omega })=0$. Trajectories therefore lie on intersections of energy ellipsoids with momentum spheres, the latter being the symplectic leaves.

The body's orientation is reconstructed from $\dot{R}=R\hat{\mathbf{\Omega }}$. Its spatial momentum $\mathbf{J}=R\mathbf{M}$ is constant, since $\dot{\mathbf{J}}=R(\mathbf{\Omega }\times \mathbf{M}+\mathbf{M}\times \mathbf{\Omega })=0$. Thus the body momentum need not be constant even though the spatial conserved momentum is. Poisson reduction retains the dynamics and its momentum leaves while eliminating orientation variables; the unreduced motion is recovered by reconstruction.

Set ${\omega }_1=u$, ${\omega }_2=\Omega +v$, ${\omega }_3=w$ and retain terms linear in the perturbations in Euler equations for a torque-free rigid body. Using $I_3=I_1+I_2$ givesConsequently $\ddot{u}={\beta }^2{\Omega }^{2}u$, where $\beta =\sqrt{(a^2-b^2)/(a^2+b^2)}>0$. For $\Omega >0$ the growing solution isThere is also a decaying mode with the opposite exponent. For negative $\Omega$, select the exponent $+\beta |\Omega |$ as the growing one. This is the intermediate axis theorem: steady rotation about the intermediate principal inertia axis is unstable whenever $\Omega \ne 0$.

Set $I=I_1=I_2$. The third of the Euler equations for a torque-free rigid body gives $I_3{\dot{\omega }}_3=0$, so ${\omega }_3$ is constant. Define $\Omega =(I_3-I){\omega }_3/I$. The first two Euler equations for a torque-free rigid body becomeso ${\omega }_1+i{\omega }_2=C{e}^{i\Omega t}$. Since the body-frame angular momentum has components $(I{\omega }_1,I{\omega }_2,I_3{\omega }_3)$, its transverse components rotate at $\Omega$ while its axial component is constant. Substitution of the principal moments of inertia givesThis is precession relative to the body axes; the angular momentum is constant in an inertial frame. If $h^2=3a^2$, the rate is zero even when ${\omega }_3\ne 0$. If the transverse components vanish, the precession cone degenerates to its axis.

The quadratic first integrals of the torque-free Euler equations for a torque-free rigid body areDifferentiation and substitution of all three Euler equations make each derivative vanish: the coefficients respectively sum to $(I_2-I_3)+(I_3-I_1)+(I_1-I_2)=0$ and $I_1(I_2-I_3)+I_2(I_3-I_1)+I_3(I_1-I_2)=0$. At either endpoint these constants are $2E=I_2{\Omega }^2$, $L^2=I_2^2{\Omega }^2$, proving $L^2=2E{I}_2$.

Eliminating the other squared components gives, with $s={\omega }_2/\Omega$,Hence $|s|\le 1$ and the second Euler equation gives $ds/d\tau =\pm (1-s^2)$, where $\tau =\Omega t\sqrt{(I_2-I_1)(I_3-I_2)/(I_1{I}_3)}$ increases with $t$. The specified orbit goes from $s=1$ to $s=-1$, so its sign must be negative. The sign in the original PDF's reduced equation is therefore inconsistent with its endpoint conditions. The corrected heteroclinic satisfiesIntegrating by separation gives this family for $|s|<1$; the integration constant is a time translation. The printed positive-sign equation instead has the increasing solutions $s=\tanh (\tau -{\tau }_0)$ and cannot meet the stated endpoints. For completeness its other real solutions are $s=\pm 1$ and the branches $s=\coth (\tau -{\tau }_0)$ on intervals avoiding their pole. These latter branches are excluded by the conserved-energy constraint $|s|\le 1$. The scalar heteroclinic is unique up to translation. For the full angular-velocity vector there are two sign-related branches: simultaneously reversing ${\omega }_1$ and ${\omega }_3$ preserves all three Euler equations and both endpoints. Full-vector uniqueness therefore also requires choosing one of those branches.

In principal axes, $2E=\sum I_i{\omega }_i^2$ and $L^2=\sum I_i^2{\omega }_i^2$. Substitute the Euler equations for a torque-free rigid body:For the squared angular momentum,Thus both $E$ and $L^2$ are conserved, even though the body-frame components of $L$ generally change.

Here $(I_1,I_2,I_3)=(20,17,5)M{a}^2/12$, so $I_1>I_2>I_3$ and $e_2$ is the intermediate axis. Eliminating the other components from $L^2=2I_{2}E$ and the conservation of energy relation gives, with $J=2E-I_2{\omega }_2^2$,Thus the second Euler equations for a torque-free rigid body reduces toThe sign is minus the sign of ${\omega }_1{\omega }_3$; it is constant along a nontrivial branch. Let $\Omega =\sqrt{24E/(17M{a}^2)}$. Since ${\omega }_2(0)=0$, integration yields ${\omega }_2(t)=\pm \Omega \tanh (3\Omega t/5)$, with the same sign as in the scalar equation. The remaining components areTheir initial squares are fixed by the preceding algebraic formulas with $J=2E$; their initial signs determine the branch. Hence $\boxed{\omega (t)⟶\pm \Omega e_2}$ as $t\to \infty$. This trajectory is an intermediate-axis separatrix: approaching intermediate-axis rotation on this special conserved-energy level does not make that rotation stable to generic perturbations. If $E=0$, all components vanish identically.

A Poisson bivector is a smooth antisymmetric contravariant two-tensorwhose bracket on smooth functions,satisfies the Jacobi identity. Bilinearity and antisymmetry are immediate, and the product rule makes the bracket a derivation in each argument. Together these properties define a Poisson manifold. Unlike a symplectic form, the Poisson bivector need not be nondegenerate.

Apply the Jacobi identity to the coordinate functions. Since $\{x^i,x^j\}={\omega }^{ij}$,Changing ${\omega }^{im}$ to $-{\omega }^{mi}$ and rearranging the three summands gives the printed coordinate Jacobi condition for a Poisson bivector:This is also sufficient: expanding the Jacobiator of three arbitrary smooth functions, all terms involving second derivatives cancel in pairs by antisymmetry. The remaining coefficient of ${\partial }_{i}F{\partial }_{j}G{\partial }_{k}H$ is the coordinate-function Jacobiator displayed above.

For a Lie algebra, the natural global space carrying the proposed linear bracket is its dual space ${\mathfrak{g}}^{\ast }$. If $v^i$ is the chosen basis, define its linear coordinate function by $x^i(\ell )=\ell (v^i)$. The Lie-Poisson bracket iswhere $d{F}_{\ell },d{G}_{\ell }$ are elements of $\mathfrak{g}=({\mathfrak{g}}^{\ast }{)}^{\ast }$. On a general manifold the same coordinate expression gives a local construction; a global one requires compatible transition rules. The use of ${\mathfrak{g}}^{\ast }$ supplies that compatibility intrinsically.

Here ${\omega }^{ij}=c_r^{ij}{x}^r$ and ${\partial }_m{\omega }^{jk}=c_m^{jk}$. The left side of the coordinate Jacobi condition for a Poisson bivector is thereforeThe final coefficient is the negative of the coefficient of $v^r$ in the Lie algebra identity $[v^i,[v^j,v^k]]+[v^j,[v^k,v^i]]+[v^k,[v^i,v^j]]=0$. Thus the Lie-Poisson bracket satisfies the Jacobi identity.

It remains to find the Lie algebra structure constants for the printed rotation fields. Distinguish their original spatial coordinates $(x,y,z)$ from the coordinates $(x^1,x^2,x^3)$ on the dual space. Use the conventional Lie bracket of vector fields. Their component vectors are $(0,-z,y)$, $(z,0,-x)$ and $(-y,x,0)$, respectively. For example,Similarly,The negative sign is essential: these are the fundamental fields of a left rotation action with the stated Lie bracket of vector fields, and consequently have the infinitesimal left-action sign convention discussed above. The first field here has component $y{\partial }_z$, as printed in the PDF.

The Lie-Poisson bracket on the dual space consequently hasFor the evolution convention $\dot{F}=\{F,H\}$, the Hamiltonian function has derivatives $(2A{x}^1,2B{x}^2,2C{x}^3)$. Substitution gives the quadratic rotational Lie-Poisson dynamicsThese are Euler-type Hamilton's equations on a noncanonical Poisson manifold. When $A=1/(2I_1)$, $B=1/(2I_2)$ and $C=1/(2I_3)$ with positive principal inertias, they are the Euler equations for a torque-free rigid body in body angular-momentum coordinates. As a check, $H$ is conserved by antisymmetry of the Poisson bracket, and $C_0=(x^1{)}^2+(x^2{)}^2+(x^3{)}^2$ is a Casimir function of a Poisson manifold. Direct differentiation of $C_0$ in the three equations cancels the terms $4[(C-B)+(A-C)+(B-A)]x^1{x}^2{x}^3$. The Hamiltonian flow therefore lies on both an energy level and a sphere, a symplectic leaf of this signed rotational bracket.

The Poisson bracket is $\{f,g\}=\mathbf{r}\cdot (\nabla f\times \nabla g)$. Since $\nabla {\rho }^2=2\mathbf{r}$, the scalar triple product vanishes, andEquivalently, check the identity on $x,y,z$ and extend to every polynomial using the Leibniz rule. Thus ${\rho }^2$ is a Casimir function of a Poisson manifold of this rotational Lie-Poisson structure on R3.

The Hamilton's equations areLinearization about $(1,0,0)$ gives $\dot{\alpha }=0$, $\dot{\beta }=(C-A)\gamma$, $\dot{\gamma }=(A-B)\beta$. The two nonzero exponents satisfy ${\lambda }^2=(C-A)(A-B)$. ConsequentlyWhen $A$ is outside this interval the transverse motion is oscillatory. At equality the linearization is nonhyperbolic; if just one transverse coupling vanishes, the other permits secular linear growth, rather than an exponential instability. This is the familiar intermediate-axis instability described by the Euler equations for a torque-free rigid body.

A principal body frame is an orthonormal basis ${\mathbf{e}}_a(t)$ fixed in the rotating body and aligned with its principal axes. If $R(t)$ is the rotation matrix whose columns are these vectors, differentiating $R^{T}R=I$ shows that $\dot{R}{R}^T$ is skew-symmetric. Every skew-symmetric linear map in three dimensions is cross product with a unique vector $\mathbf{\omega }$, so

For any vector $\mathbf{A}=A_a{\mathbf{e}}_a$,Applying this to the conserved angular momentumgives the Euler equations for a torque-free rigid body:

Multiplying these equations respectively by ${\omega }_i$ and summing proves conservation of kinetic energy; multiplying by $I_i{\omega }_i$ and summing proves conservation of squared angular momentum. Thus the torque-free rigid-body invariants areSolving these two linear equations for the first two squared components givesUsing $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2$ yields the quartic equation for one component of torque-free angular velocity:

Let $e_1,e_2,e_3$ be a principal body frame, let $I_1,I_2,I_3$ be the corresponding principal moments of inertia, and write the angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{e}_i$. The body-frame components of the angular momentum areNo external torque acts, so $\mathbf{L}$ is constant in the fixed space frame. Since a body-fixed basis vector satisfies ${\dot{e}}_i=\mathbf{\omega }\times e_i$,Taking body-frame components gives the Euler equations for a torque-free rigid body:

Multiply the three Euler equations for a torque-free rigid body by ${\omega }_1,{\omega }_2,{\omega }_3$ and add. The cubic terms cancel, giving conservation of the rotational kinetic energyInstead multiply the equations by $I_1{\omega }_1,I_2{\omega }_2,I_3{\omega }_3$ and add. The cubic terms again cancel, givingthe squared magnitude of the conserved angular momentum. For an asymmetric body these two quadratic forms are not proportional, so the torque-free rigid-body invariants are independent.

Let $R$ be the center of mass and $g$ the uniform gravitational acceleration. The gravitational torque about $R$ is

Choose body-fixed principal axes of inertia. The inertial derivative of the angular momentum isFor torque-free motion $N=0$ and $L_i=I_i{\omega }_i$, giving the Euler equations for a torque-free rigid body

For a symmetric top, let $I_1=I_2=I$ and let $I_3$ be the axial moment. Then ${\omega }_3$ is constant andThus ${\omega }_1^2+{\omega }_2^2$ is constant and the transverse angular-velocity vector rotates uniformly about the body symmetry axis. Its rate is $|\Omega |$; it precesses in the positive axial sense when $(I_3-I){\omega }_3>0$ and in the opposite sense when this product is negative.

A fixed space frame is an inertial orthonormal frame whose axes remain fixed in space. A principal body frame is attached to the rigid body and aligned with the principal axes of its inertia tensor, so that the tensor is diagonal with entries $I_1,I_2,I_3$.

If $e_i(t)$ are the body axes and $\omega$ is the angular velocity, then the derivative of a body-fixed basis vector isIn the principal frame,For torque-free motion, the angular momentum has zero space derivative. ThereforeTaking body-frame components gives the Euler equations for a torque-free rigid body:

For an axisymmetric body, put $I_1=I_2=I$ and let $e_3$ be its symmetry axis. Decompose $\omega ={\omega }_1{e}_1+{\omega }_2{e}_2+{\omega }_3{e}_3$. ThenThus $L$ is a linear combination of $\omega$ and $e_3$, proving that the angular momentum, angular velocity, and symmetry axis are always coplanar. This is the coplanarity in a torque-free axisymmetric rigid body.