# Intermediate axis theorem - why can't we have exponential decay? [duplicate]

I was reading about the intermediate axis theorem and its mathematical proof. Typically one starts with the torque-free Euler's equations \begin{align} 0&=I_1\dot\omega_1 + (I_3-I_2)\omega_3\omega_2\\ 0&=I_2\dot\omega_2 + (I_1-I_3)\omega_1\omega_3\\ 0&=I_3\dot\omega_3 + (I_2-I_1)\omega_2\omega_1\\ \end{align} with $$I_3 > I_2 > I_1$$. Then we start with some initial angular velocity with components $$(\epsilon_1,\,\omega_2,\,\epsilon_3)$$ in the rotating body frame of reference, with $$\epsilon_1,\epsilon_3 \ll \omega_2.$$

We first find that $$\dot\omega_2 \approx 0$$ for all times such that $$\epsilon_1,\epsilon_3 \ll \omega_2.$$ Then, taking a time scale on which $$\omega_2$$ is approximately constant, we attain a differential equation of the form $$\ddot\epsilon_1=k_1^2\epsilon_1,$$ and similarly $$\ddot\epsilon_1=k_3^2\epsilon_1.$$ (Take $$k_1, k_3 > 0.$$)

At this point every text I have found states that the solutions to these differential equations are exponential growth and thus the rotation is unstable. While I agree that the exponentially growths $$\epsilon_1\sim e^{k_1t},\,\epsilon_3\sim e^{k_3t}$$ may be valid solutions to the above differential equations, so too are the exponential decays $$\epsilon_1\sim e^{-k_1t},\,\epsilon_3\sim e^{-k_3t}.$$ In particular, if $$\omega_2>0,$$ these exponential decays seem to also be valid solutions to Euler's equations above.

So my question is: why can't axis 1 and axis 3 both have exponentially decaying solutions instead of exponentially growing solutions when rotating near the intermediate axis? If exponentially decaying solutions were possible, this would mean that the intermediate axis can undergo stable rotations even with a small error initially, but this contradicts what I have read in my textbook and on the internet (and what I have seen with my own eyes).

• Reading which text? Which year/edition? Which page? Commented Apr 4, 2023 at 0:43
• Morin's introduction to classical mechanics, page VIII-54 Commented Apr 4, 2023 at 0:58
• Commented Apr 4, 2023 at 6:43
• Thanks for the links - none of them answered my specific question though. Commented Apr 5, 2023 at 1:02

When a rigid body is rotating freely, with no external torques, its angular velocity in body-fixed coordinates follows lines that are the intersection of an ellipsoid and a sphere. (surfaces of constant kinetic energy and constant angular momentum magnitude, respectively.) For initial rotation axes close to the minimum and maximum axes of inertia, the angular velocity circles the axis. But at the intermediate axis the lines cross. There is a saddle point there. Two of the lines point inwards to the centre point, the other two point outwards. The movement slows down the closer you get to the intermediate axis.

The two lines outgoing from one crossing point are the incoming lines approaching the other crossing point on the opposite side, and vice versa.

We have a number of different trajectories in the neighbourhood. If off to the side of the crossing lines, the rotation axis approaches the intermediate axis, slowing, and drifts off to the side, speeding up. If exactly on one of the inbound pair of lines, the rotation axis approaches the intermediate axis asymptotically, never quite reaching it. If on one of the outbound pair of lines, it moves away from the intermediate axis, speeding up. And if exactly at the crossing point, it stays there. However, any slight perturbation in the direction of the outgoing lines, and it moves away at an accelerating rate.

(You tend to get the rotation hopping from the neighbourhood of one crossing point to the other, pointing in the opposite direction. It quickly approaches the axis along one incoming line, slows nearly to a halt, appearing to wait there for a while, then accelerates away and closes in on the opposite axis, where it waits again.)

It is like balancing a ball on a saddle point. It is considered unstable because virtually any perturbation you can make will result in divergence. If it is pushed away exactly along the line joining the two peaks, it will get pushed back, but any perturbation from that precise ridge line will be unstable.

• Please register yourself, so your answers can be catalogued by stackexchange in one place; omnia verba tua. Commented Apr 4, 2023 at 1:35
• I see. So you're saying if both axis 1 and axis 3 are nonzero, then it is impossible for the rotation about the intermediate axis to be stable. Commented Apr 4, 2023 at 1:37

The general solution, which is a linear combination of the two cases you have provided, is still unstable. The presence of decaying terms in the solution doesn't mean the whole solution decays. On the other hand, if one term blows up then the whole sum blows up.

• Is it unlikely that we could have a purely decaying solution? Commented Apr 4, 2023 at 0:57
• @LanceLampert Nothing's ever exactly zero. The coefficient of the growing term might start small if you're careful, but then after a short time it'll overwhelm the other term. Commented Apr 4, 2023 at 1:20