The axis that will not hold

Take a hardback book, tape it shut, and throw it spinning into the air. Spun about the long axis down its spine it comes back turning about the spine. Spun about the axis through its covers it comes back turning about that. Spun about the third axis — the short one across the pages — it does something else entirely: it turns half a rotation about the spin axis, flips completely over, turns another half rotation, and flips back, over and over, for as long as it is in the air. Nothing touches it during any of this.

The whole of it follows from three numbers and their order.

Three numbers, and nothing else about the body #

A rigid body’s response to a rotation is carried entirely by its inertia tensor, and every body has three mutually perpendicular directions in which that tensor is diagonal. Those are the principal axes; the three numbers on the diagonal are the principal moments $I_1 \le I_2 \le I_3$. For a uniform rectangular block of sides $a \times b \times c$ and mass $m$ they are

$$I_a = \frac{m}{12}(b^2 + c^2), \qquad I_b = \frac{m}{12}(a^2 + c^2), \qquad I_c = \frac{m}{12}(a^2 + b^2),$$

so the longest side, which appears in two of the three expressions, makes the two moments it appears in the largest, and the axis along it carries the smallest moment. A book is longest along its spine and thinnest through its covers, so the spine is $I_1$, the axis through the covers is $I_3$, and the short axis across the pages — the one nobody thinks about — is $I_2$.

The moment of inertia decides outcomes in the plainest case there is: three bodies of the same mass and the same radius, released together down the same slope, arrive in an order fixed entirely by how far each one’s mass sits from its axis. Nothing about the material enters, and nothing about the slope changes the order. Three numbers describe a body’s resistance to being spun, and this essay is about what happens when they are all different.

In the body’s own frame, with no torque acting, the angular velocity obeys Euler’s equations:

$$I_1\dot\omega_1 = (I_2 - I_3)\,\omega_2\omega_3, \qquad I_2\dot\omega_2 = (I_3 - I_1)\,\omega_3\omega_1, \qquad I_3\dot\omega_3 = (I_1 - I_2)\,\omega_1\omega_2.$$

They are three lines of algebra with no forces in them, and they are what Newton’s second law becomes when it is written in axes that turn with the body rather than in axes that stay still. Every statement below is a consequence of these three and of nothing else — no air, no hand, no imperfection in the body.

What happens to a small departure #

Suppose the body is spinning at $\Omega$ about one principal axis with tiny components on the other two. Put $\omega_2 = \Omega$ and treat $\omega_1, \omega_3$ as small. The first and third equations become linear in the small quantities, and differentiating one and substituting the other gives

$$\ddot\omega_1 = \frac{(I_2 - I_3)(I_1 - I_2)}{I_1 I_3}\,\Omega^2\,\omega_1.$$

Everything is in the sign of that coefficient. With $I_1 < I_2 < I_3$ both brackets are negative, their product is positive, and the solution is a growing exponential. Do the same about $I_1$ or about $I_3$ and one bracket flips sign, the coefficient is negative, and the solution is a sine. The three exponents are

$$\lambda_1 = \Omega\sqrt{\frac{(I_3 - I_1)(I_2 - I_1)}{I_2 I_3}}, \qquad \sigma = \Omega\sqrt{\frac{(I_3 - I_2)(I_2 - I_1)}{I_1 I_3}}, \qquad \lambda_3 = \Omega\sqrt{\frac{(I_3 - I_2)(I_3 - I_1)}{I_1 I_2}},$$

the first and last being frequencies of oscillation and the middle one a rate of growth.

Three features of that figure are worth dwelling on. The growth rate goes to zero at both ends, so a body with any two moments equal has no unstable axis at all — which is why a thrown discus, a spinning coin and a rifle bullet never do this. It is largest at the arithmetic midpoint of the outer two moments, which is not where any particular body’s $I_2$ happens to sit — an observation of the same kind as a torque depending on where the force is applied and not on how large it is. And it scales with $\Omega$: spinning the book twice as fast does not make it more stable, it makes the flip arrive twice as soon.

The flip, integrated rather than argued #

A linearisation says only what happens while the departure is small. What happens afterwards has to be got from the full equations, and they are three coupled quadratics with no elementary solution — so they are integrated.

The shape of that trace is the thing to notice. It is not a wobble that grows into a tumble. It is a plateau, then a sudden reversal, then another plateau of the same length, then another reversal — a square wave with rounded corners. Between flips the body is spinning almost exactly about its intermediate axis, doing nothing interesting; the transitions are quick and the intervals between them are long.

The intervals are long because the growth is exponential. Starting a fraction $\epsilon$ off the axis, the departure needs a time of order $\ln(1/\epsilon)/\sigma$ to become order one. Halving $\epsilon$ therefore does not halve anything; it adds $\ln 2/\sigma$ to the waiting time, a fixed interval, once.

That logarithm is the reason this cannot be arranged away. To postpone the first flip by a factor of ten in time the initial departure would have to be reduced by a factor of $e^{10\sigma\tau}$, which for this book and a ten-second throw is a number with fourteen zeros in it. The instability is not fussy about how carefully the body is released. It is fussy about nothing at all.

Where the flip comes from: two conserved quantities, and their intersection #

The integration says what happens. It does not say why, and the reason is geometric.

With no torque, two things are conserved. The angular momentum vector $\mathbf{L}$ is fixed in space, so in the body’s frame its length is fixed and the vector moves on a sphere. The rotational energy is fixed too, and in terms of the momentum components it is

$$2E = \frac{L_1^2}{I_1} + \frac{L_2^2}{I_2} + \frac{L_3^2}{I_3},$$

which is an ellipsoid. The motion is confined to both surfaces at once, so it runs along their intersection — a closed curve on the sphere, called the polhode.

Now sweep the energy at fixed $|\mathbf{L}|$. The lowest possible energy is $L^2/2I_3$, all the momentum on the greatest moment; the highest is $L^2/2I_1$. Just above the bottom, the ellipsoid barely pokes through the sphere near the $I_3$ axis and the intersection is a small loop around it. Just below the top, likewise around $I_1$. In between there is exactly one energy, $L^2/2I_2$, at which the two families of small loops have to hand over to one another — and at that energy the intersection is not a loop at all but four arcs meeting at the intermediate axis.

This is the same picture as a pendulum’s phase portrait, and not by analogy. There too the level sets of a conserved energy are closed curves at low energy and open ones at high, and there too a single level set separates the two and passes through an equilibrium — the same structure the small-angle account of a pendulum throws away by keeping only the bottom of the well.

A pendulum’s level sets make the same point in a system with one degree of freedom. The curve separating swinging from going over the top passes through the inverted position, and the inverted pendulum and the intermediate axis are the same object: a fixed point that is a saddle rather than a centre, with one direction leading in and one leading out. Everything the flip does is what a saddle does to a trajectory that starts near it.

The correspondence is exact in the part that matters. An equilibrium at the crossing of a separatrix is a saddle: displacement along one direction returns and displacement along the other runs away, so the equilibrium survives no disturbance with any component in the second direction. The other two axes sit at the centres of families of closed curves, and a displacement there simply moves to a neighbouring curve and stays on it.

Reduced to one dimension, it is a hill between two wells. Energy exactly at the top is the separatrix energy, and a body placed exactly there stays there — while anything at all nearby leaves, slowly at first and then not. The intermediate axis is stable in the sense that a pencil balanced on its point is stable, and unstable in the only sense that matters to anything real.

Reading the flip in the space frame #

Everything above is in the body’s frame, which is a frame that tumbles. In the laboratory the angular momentum vector points in a fixed direction and never moves; what moves is the body around it. The flip, seen from outside, is the body turning over so that the axis which had been aligned with $\mathbf{L}$ ends up anti-aligned with it — while $\mathbf{L}$ itself is untouched.

Reading it in the space frame removes the last of the strangeness. The reversals in the previous figures are components measured in the body’s frame, and the vector they are components of does not reverse at all — the angular momentum is constant throughout, as it must be with no torque acting. What flips is the body relative to the vector, not the vector. A straight path drawn in a rotating frame is a curve, and that is the whole of what is happening.

This is why the effect has been rediscovered so often by people with nothing to spin but themselves. It was written down by Louis Poinsot in 1834 as the geometry of the polhode, and it was noticed again in 1985 by Vladimir Dzhanibekov aboard the Salyut 7 station, where a wing nut spun off a threaded rod drifted down the module flipping regularly end over end. The nut was in free fall, in vacuum, and had nothing acting on it, and it flipped anyway. The recording circulated for years as something unexplained, which it had not been for a hundred and fifty years.

Half a turn, exactly #

There is a detail of the flip that the component traces do not show and that anybody who has thrown a racket has seen: it comes back with the other face toward them.

Watch the handle rather than the components. Between one flip and the next, the body does not merely turn over — it also rotates about the axis it is spinning around, and in the limit of a very small initial departure that rotation is exactly half a turn. Not approximately: the twist tends to $\pi$ as the departure tends to zero, and the result is a theorem rather than an observation.

Which is why the effect is called the tennis racket theorem. Toss a racket, spinning about the axis across the face, and it flips and lands with the opposite face upward — every time, at any spin rate, from any careful throw. The half-twist is as reliable as the flip, and it is more surprising, because nothing about the equations obviously singles out a half turn.

The reason it is exactly a half turn rather than some other angle is the separatrix. In the limiting case the motion follows the separatrix arcs themselves, and those arcs run from the intermediate axis, round the sphere, and back to the intermediate axis on the opposite side. Integrating the orientation along that path gives a rotation that is fixed by the geometry rather than by the moments, and it comes out at $\pi$.

Departures from the limit shift it. A larger initial deviation gives a twist that differs from a half turn by an amount growing with the deviation, and the drift accumulates from flip to flip — which is what makes a real thrown racket eventually come down in an orientation that was not predictable, even though each individual flip was.

The rocks that stopped tumbling #

The dissipation argument has a large-scale consequence that is measured rather than derived, and it explains a fact about the solar system.

Almost every asteroid observed rotates about its axis of greatest moment of inertia, cleanly, with a single well-defined period. That is the state of lowest energy at fixed angular momentum, and the argument above says that any body able to flex will end up there. Rock flexes: a tumbling body is worked by its own inertial stresses on every wobble, and a tiny fraction of the energy goes into heat each time.

The timescale is what makes the observation informative. It depends steeply on the body’s size and on its spin rate — roughly as the inverse square of the diameter and the inverse cube of the rotation rate — so a large, fast-rotating body damps in a few million years and a small, slow one takes longer than the solar system has existed.

So the bodies that are still tumbling are a selected population, and their existence says something. A small asteroid found in non-principal-axis rotation either has an unusually low internal dissipation, or is rotating unusually slowly, or was set tumbling recently — by a collision, or by a close pass that raised a tide on it. Each of those is a piece of history, recoverable from a light curve that repeats with two periods rather than one.

The same argument runs on cometary nuclei, which are less rigid and more likely to be tumbling, and on spacecraft debris, which is small enough that some of it never settles. In every case the reasoning is the one Explorer 1 demonstrated: a body left alone finds its greatest-moment axis, and how long it takes is a measurement of how lossy it is.

What a designer does about it #

The Explorer 1 lesson produced three standard answers, and they are worth listing because they are three quite different responses to one instability.

The simplest is to obey it: build the spacecraft so that its spin axis is the axis of greatest moment — short and fat rather than long and thin. That works and it constrains the shape severely, which for a vehicle that has to fit inside a rocket fairing is a real cost.

The second is to damp actively rather than passively. Fit a nutation damper — in the crudest form a tube of viscous fluid with a ball in it, mounted so that any wobble drives the ball back and forth and dissipates the wobble’s energy rather than the spin’s. That converts a slow drift toward tumbling into a fast decay back toward the intended axis, provided the intended axis is the greatest-moment one.

The third is to give up on being a single rigid body. A dual-spin vehicle has a spinning section for gyroscopic stiffness and a despun platform carrying the instruments, and the energy dissipation is deliberately concentrated on the despun part. That arrangement can be stable spinning about the least moment, which the single-body argument forbids, because the relevant energy accounting is now about which section is losing the energy rather than about the vehicle as a whole.

The third answer is the one worth noticing, because it does not evade the physics — it changes the system the physics applies to. The rule that a flexible body ends up on its greatest-moment axis is a statement about a body dissipating its own energy, and a vehicle built in two parts that rotate relative to each other is not one body in the sense the rule requires.

Where the model stops: rigid, and nothing is #

Every word above assumes the body is rigid. Drop that and the conclusion changes, in a way the linear analysis has no term for.

At fixed angular momentum, the rotational energy $L^2/2I$ is smallest when $I$ is largest. A body that can flex — a real body, with joints, fuel, or merely elasticity — will dissipate energy internally while conserving angular momentum, because internal forces cannot change $\mathbf{L}$ any more than they can change the momentum of a pair of colliding bodies. So it slides down in energy at fixed momentum, and the state it slides to is rotation about the axis of greatest moment.

Rotation about the least moment, which the rigid analysis calls perfectly stable, is therefore the highest-energy state available and is unstable to any dissipation whatever. This was discovered expensively. Explorer 1, launched in 1958, was a long thin cylinder set spinning about its long axis — the axis of least inertia, chosen because that is what a rocket produces. Its four flexible whip antennas dissipated a little energy on every wobble, and within a few hours the satellite was tumbling end over end about its transverse axis. Nothing had gone wrong with it. The design had asked a body to remain in its highest-energy state.

The two failures are opposites and are easy to confuse. The intermediate axis is unstable within rigid-body dynamics, on a timescale of seconds, and dissipation is irrelevant to it. The least axis is stable within rigid-body dynamics and unstable to dissipation, on a timescale set by how lossy the body is — hours for a satellite, geological ages for a planet. The second is not a mechanics question at all but a question about where energy goes when it stops being useful.

What the pictures cannot show #

The polhode figures draw the direction of the angular momentum in the body’s frame and say nothing about the body’s orientation in space. Recovering that requires a second integration — the polhode has to be rolled on the invariable plane — and two bodies with identical polhodes can be in quite different attitudes. The trace of $\omega_2(t)$ likewise shows a component reversing and does not show the object turning over; those are the same event described in two frames, and the figures show one of them.

None of the figures shows the transition being sharp. It looks sharp on a fourteen-turn axis and it is not: the reversal takes about one e-folding time, and the plateau takes a logarithm. Compressing the time axis to fit several flips is what makes the corners look square.

And nothing here shows energy. The reversals happen at constant energy and constant angular momentum, so a plot of either against time is a horizontal line, which is exactly why those two lines are quoted as numbers in the caption instead: the interesting thing about them is that they do not move.

Where the ladder goes next #

The ordering of three numbers has decided an outcome here, and the same structure appears wherever a conserved quantity confines motion to a surface and a second conserved quantity cuts it. It is the same reasoning that makes a pendulum’s separatrix separate swinging from circulating, that makes an orbit stable or not at a given angular momentum, and that decides when a coupled pair exchanges its energy completely rather than partially.

Two further rungs are visible from here. One is the gyroscope: a body with two equal moments, deliberately built so that no unstable axis exists, and the precession that a torque then produces. The other is the flexible body’s slow slide to its greatest-moment axis, which is a question about dissipation rather than about rotation and which connects to the arithmetic of irreversibility rather than to the mechanics here.

The habit worth carrying away is smaller than either. Three numbers were given, and the answer depended on their order and not on their values — the growth rate is zero at both ends of the range and positive between, so it is the fact of $I_2$ being in the middle that does the work. Quantities that decide an outcome by their ordering rather than their size are rare, and they are worth noticing when they appear.