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The intermediate axis theorem states that the rotation of an object around its first and third principal axes is stable, while rotation around its second principal axis (or intermediate axis) is not. What is the analogue of this theorem in higher dimensions, where rotations are no longer described by axes but instead by bivectors?

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    $\begingroup$ For anyone interesting in learning about the intermediate axis theorem I highly recommend this YouTube video on the subject. $\endgroup$ – John Rennie Oct 4 at 15:24
  • $\begingroup$ ...and this related Phys.SE post and links therein. $\endgroup$ – Qmechanic Oct 4 at 15:50
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    $\begingroup$ The higher-dimensional intermediate axis theorem is the obvious generalization of the 3D case. $\endgroup$ – Qmechanic Oct 4 at 16:08
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This answer doesn't show the whole derivation, but it indicates how to set it up and what the result looks like. (I haven't seen this in the literature before, so let the reader beware: nobody has double-checked my derivation.)

Treat the rigid body as a conglomerate of pieces. Let $m_n$ be the mass of the $n$th piece, and let $\mathbf{b}_n$ denote its displacement from the body's center of mass (in body-fixed coordinates). Define the square matrix $$ M = \sum_n m_n \mathbf{b}_n\mathbf{b}_n^T $$ where $T$ means transpose. This definition makes sense in any number $D$ of spatial dimensions. When $D=3$, it's different than what we usually call the moment-of-inertia tensor, but it's closely related.

The stability analysis uses a $D$-dimensional version of Euler's equation, which can be written $$ \{\dot W,M\}+[W^2,M]=0 $$ with $\{A,B\} = AB+BA$ and $[A,B]=AB-BA$ and $W=R^T\dot R$, where $R$ is the time-dependent $D\times D$ rotation matrix that relates the body-fixed coordinate system to an inertial coordinate system, and $\dot R$ is the time-derivative of $R$. This is the equation of motion for a freely rotating rigid body (no external torque), expressed in a body-fixed coordinate system. The quantity $W$ is the angular velocity bivector. The square matrix $W$ is antisymmetric, and the square matrix $M$ is symmetric.

Work in a basis where $M$ is diagonal, and assume that the $D$ eigenvalues of $M$ are all distinct so that stability can be analyzed using first-order perturbation theory. According to first-order perturbation theory (if I didn't make any mistakes), rotation in the $j$-$k$ plane is stable only if the quantities $$ \lambda_\ell := \frac{(M_{\ell\ell}-M_{jj})(M_{\ell\ell}-M_{kk})}{ (M_{\ell\ell}+M_{jj})(M_{\ell\ell}+M_{kk})} $$ are positive for all $\ell\neq j,k$. This is possible only if the two factors in the numerator are either both positive or both negative for all $\ell\neq j,k$, which in turn is possible only if $M_{jj}$ and $M_{kk}$ are either the two largest or the two smallest components of $M$ (in a basis where $M$ is diagonal).

In other words, assuming no degeneracies, there are only two planes in which rotational motion is stable: One is the $j$-$k$ plane for which $M_{jj}$ and $M_{kk}$ are the two largest components of $M$ (in a diagonal basis), and one is the $j$-$k$ plane for which $M_{jj}$ and $M_{kk}$ are the two smallest components of $M$. This is consistent with the familiar situation in $D=3$.

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