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minor typo (the question has been bumped anyway)

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edited Nov 19, 2019 at 18:39

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Perhaps I'm too late here, but there's a simple treatment which does not refer to Euler's equations.

Let $A$ denote the inertia tensor of the body, that is, if $\omega_b$ is its angular velocity in the frame of the body, then $M_b:=A\omega_b$ is its angular momentum and $E:=(A\omega_b;\omega_b)$ is its kinetic energy. The position of the body at each moment $t$ is described by an orthogonal matrix $C_t$; if $\omega$ and $M$ denote its angular velosityvelocity and momentum in the laboratory frame, then $$ M=C^{-1}_tAC_t\omega, $$ and the kinetic energy is given by $$ E=(M;w)=(C_tM;A^{-1}C_t M). $$ However we know that both $M$ and $E$ are conserved. Hence, if we draw level lines of the quadratic form $v\mapsto (v;A^{-1}v)$ on the sphere $|v|=|M|$, then $C_t$ maps the fixed vector $M$ to some other vector on the same level line. Near maximum an minimum, these level lines are small closed loops, but the intermediate eigenvalue of $A$ corresponds to a saddle point; the nearby level line goes all the way around the sphere to the opposite saddle point. This proves that the former rotations are stable and explains why the latter is not. If you just know that there are some first-order dynamics for $\omega$ in the body frame and that the eigenvalues of $A$ are its only fixed points, then you get a complete qualitative picture.

Perhaps I'm too late here, but there's a simple treatment which does not refer to Euler's equations.

Let $A$ denote the inertia tensor of the body, that is, if $\omega_b$ is its angular velocity in the frame of the body, then $M_b:=A\omega_b$ is its angular momentum and $E:=(A\omega_b;\omega_b)$ is its kinetic energy. The position of the body at each moment $t$ is described by an orthogonal matrix $C_t$; if $\omega$ and $M$ denote its angular velosity and momentum in the laboratory frame, then $$ M=C^{-1}_tAC_t\omega, $$ and the kinetic energy is given by $$ E=(M;w)=(C_tM;A^{-1}C_t M). $$ However we know that both $M$ and $E$ are conserved. Hence, if we draw level lines of the quadratic form $v\mapsto (v;A^{-1}v)$ on the sphere $|v|=|M|$, then $C_t$ maps the fixed vector $M$ to some other vector on the same level line. Near maximum an minimum, these level lines are small closed loops, but the intermediate eigenvalue of $A$ corresponds to a saddle point; the nearby level line goes all the way around the sphere to the opposite saddle point. This proves that the former rotations are stable and explains why the latter is not. If you just know that there are some first-order dynamics for $\omega$ in the body frame and that the eigenvalues of $A$ are its only fixed points, then you get a complete qualitative picture.

Perhaps I'm too late here, but there's a simple treatment which does not refer to Euler's equations.

Let $A$ denote the inertia tensor of the body, that is, if $\omega_b$ is its angular velocity in the frame of the body, then $M_b:=A\omega_b$ is its angular momentum and $E:=(A\omega_b;\omega_b)$ is its kinetic energy. The position of the body at each moment $t$ is described by an orthogonal matrix $C_t$; if $\omega$ and $M$ denote its angular velocity and momentum in the laboratory frame, then $$ M=C^{-1}_tAC_t\omega, $$ and the kinetic energy is given by $$ E=(M;w)=(C_tM;A^{-1}C_t M). $$ However we know that both $M$ and $E$ are conserved. Hence, if we draw level lines of the quadratic form $v\mapsto (v;A^{-1}v)$ on the sphere $|v|=|M|$, then $C_t$ maps the fixed vector $M$ to some other vector on the same level line. Near maximum an minimum, these level lines are small closed loops, but the intermediate eigenvalue of $A$ corresponds to a saddle point; the nearby level line goes all the way around the sphere to the opposite saddle point. This proves that the former rotations are stable and explains why the latter is not. If you just know that there are some first-order dynamics for $\omega$ in the body frame and that the eigenvalues of $A$ are its only fixed points, then you get a complete qualitative picture.

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created Aug 22, 2014 at 22:25

Kostya_I

Perhaps I'm too late here, but there's a simple treatment which does not refer to Euler's equations.

Let $A$ denote the inertia tensor of the body, that is, if $\omega_b$ is its angular velocity in the frame of the body, then $M_b:=A\omega_b$ is its angular momentum and $E:=(A\omega_b;\omega_b)$ is its kinetic energy. The position of the body at each moment $t$ is described by an orthogonal matrix $C_t$; if $\omega$ and $M$ denote its angular velosity and momentum in the laboratory frame, then $$ M=C^{-1}_tAC_t\omega, $$ and the kinetic energy is given by $$ E=(M;w)=(C_tM;A^{-1}C_t M). $$ However we know that both $M$ and $E$ are conserved. Hence, if we draw level lines of the quadratic form $v\mapsto (v;A^{-1}v)$ on the sphere $|v|=|M|$, then $C_t$ maps the fixed vector $M$ to some other vector on the same level line. Near maximum an minimum, these level lines are small closed loops, but the intermediate eigenvalue of $A$ corresponds to a saddle point; the nearby level line goes all the way around the sphere to the opposite saddle point. This proves that the former rotations are stable and explains why the latter is not. If you just know that there are some first-order dynamics for $\omega$ in the body frame and that the eigenvalues of $A$ are its only fixed points, then you get a complete qualitative picture.