# Does an asymmetric top return to its initial position &orientation, while rotating freely in space? (contradiction in Landau&Lifshitz mechanics book)

The Poinsot's ellipsoid tells us that the angular momentum ($$M$$) (in the rotating frame) should be lying on the intersection curve between the conservation of momentum sphere radius $$|M|$$: $$M_1^2 + M_2^2 + M_3^2=|M|^2$$ and the conservation of energy ($$E$$) ellipsoid:

$$1 = M_1^2/(2E I_1) + M_2^2/(2E I_2) + M_3^2/(2E I_3)\, .$$

This intersection curve is closed so the motion of $$M$$ relative to the body should be periodical, and as written in the Landau & Lifshitz mechanics book page 117:

First of all, we may note that, since the paths are closed, the motion of the vector M relative to the top must be periodic; during one period the vector M describes some conical surface and return to its original position.

and because $$M$$ is fixed (unchanged) in the world frame, there the body moves, but it will return to its original starting orientation (and place as the center of mass stationery) after some time. as the relative motion $$M$$ and the body is periodical.

But then on page 120 it was written:

This incommensurability has the result that the top does not at any time return exactly to its original position.

Additionally on page 119, regarding the angular velocity function of time, it was written:

After a time T the vector W returns to its original position relative to the axes of the top. The top itself, however, does not return to its original position relative to the fixed system of co-ordinates

So is there a mistake in this book or where do I wrong? assume no initial translational motion.

• If the body has any translational momentum, this is trivial (it doesn't return to the same place because the center of mass moves). Are you talking about a nonrotating center-of-mass frame when you say "the world frame"? Jul 22 '19 at 15:58
• Assuming no initial translational motion. world frame is the fixed system of co-ordinates with the center of mass at its center like you wrote. Jul 22 '19 at 16:22
• The key is the “incommensurability” comment. If, when the axis is back to the same orientation after time T, the top has not made an integer number of rotations, it doesn’t line up. And if it’s not a rational number, it never will line up. Jul 30 '19 at 0:59

There is no mistake in the book. To really understand the motion of a freely rotating top (Euler's top), you have to look at the full system of differential equations that describe the motion of the top.

Let us denote by capital letters $$\vec{X}$$ the coordinates with respect to the rotating frame (body-fixed frame) and the by lower-case letters $$\vec{x}$$ the coordinates with respect to the inertial world-fixed frame (the world frame is an inertial frame of reference attached to the center of the top, so the center of the top is stationary). Knowing the position(orientation) of the top at any moment of time $$t$$ in the world space means that you know the $$3 \times 3$$ rotational matrix (also called orthogonal matrix) $$U = U(t)$$ that transforms the coordinates $$\vec{X}$$ of any point on the top to its coordinates in $$\vec{x}$$ in the world frame: $$\vec{x}(t) = U(t)\, \vec{X}$$ Notice, in the body fixed frame the coordinates $$\vec{X}$$ of a point do not change with time, but in the world frame the coordinates $$\vec{x}$$ of the same point change with time, because the body is reorienting constantly. The velocity of our arbitrarily chosen point is $$\frac{d\vec{x}}{dt} = \frac{dU}{dt}\, \vec{X}$$ which combined with $$\,\vec{X} = U^T\, \vec{x}\,$$, where $$U^T$$ is the matrix $$U$$ transposed, which in the case of a rotational matrix is the inverse matrix, i.e. $$U^T = U^{-1}$$, gives the expression $$\frac{d\vec{x}}{dt} =\frac{dU}{dt}\, \left(U^T \,\vec{x}\right) = \left(\frac{dU}{dt}\,U^T \right)\vec{x}$$ As it turns out, in the case of rotational matrices, there always exists a unique vector $$\vec{\omega} = \vec{\omega}(t)$$ such that $$\frac{d\vec{x}}{dt} = \left(\frac{dU}{dt}\,U^T \right)\vec{x} = \vec{\omega} \times \vec{x}$$ This $$\vec{\omega}$$ is the angular velocity with respect to the inertial world frame. If we apply the inverse rotation matrix $$U^T$$ from the left to the latter equation we obtain $$\vec{V} = U^T\,\frac{d\vec{x}}{dt} = U^T \Big(\vec{\omega} \times \vec{x}\Big) = \big( U^T\vec{\omega} \big)\times \big( U^T\vec{x}\big) = \vec{\Omega} \times \vec{X}$$ where $$\vec{V}$$ are the coordinates of the velocity vector $$\frac{d\vec{x}}{dt}$$ in the body-fixed frame and $$\vec{\Omega} = U^T\, \vec{\omega}$$ are the coordinates of the angular velocity in the body-fixed frame. Another way of writing the latter chain of equations is $$\vec{V} = U^T\,\frac{d\vec{x}}{dt} = U^T\, \left(\frac{dU}{dt}\, \vec{X}\right) = \left(U^T\,\frac{dU}{dt}\right)\vec{X} = \vec{\Omega} \times \vec{X}$$ i.e. $$U^T\,\frac{dU}{dt} = \Big(\vec{\Omega}\times \cdot\Big)$$ where by $$\Big(\vec{\Omega}\times \cdot\Big)$$ I denote the skew-symmetric matrix that acts as follows: $$\Big(\vec{\Omega}\times \cdot\Big) \vec{X} = \vec{\Omega} \times \vec{X}$$ Let us denote by $$J$$ the $$3 \times 3$$ inertia matrix calculated in the body-fixed frame. In the body-fixed frame, $$J$$ is a constant matrix, i.e. it doesn't change with time because the body is fixed in that frame and doesn't change its orientation. In the inertial world frame however, the inertia tensor constantly changes with time because the body (and thus its mass distribution) changes orientation. So in the inertial world frame, the inertia matrix is $$j = j(t) = U(t)\, J\, U(t)^T$$ The angular momentum in world frame coordinates is $$\vec{m} = j\,\vec{\omega}$$ and by the conservation of angular momentum law: $$\frac{d \vec{m}}{dt} = \vec{0}$$

Now, the ultimate goal of the rigid-body dynamics is to obtain the rotation matrix $$U = U(t)$$ as a function of time. This matrix, via $$\vec{x} = U\,\vec{X}\,$$, gives you the position $$\vec{x}$$ in the world frame of every point $$\vec{X}$$ from the body at any moment of time $$t$$. The equation $$\frac{d \vec{m}}{dt} = \vec{0}$$ is the right equation to start from, but just knowing the angular momentum in the world frame doesn't immediately provide us with a way to find the rotation matrix $$U(t)$$. So the strategy is to transform the angular momentum equation into an equation for the angular velocity $$\vec{\Omega}$$ in the body fixed frame, solve it, find $$\vec{\Omega} = \vec{\Omega(t)}$$, and then solve the equation $$U^T\, \frac{dU}{dt} = \big(\vec{\Omega} \times \cdot\big)$$. Why in the body-fixed frame, you may ask. Because in the world frame the inertia matrix is complicated time-dependent matrix, while in the body-fixed frame it is a constant matrix, which can be even diagonal if the body frame is chosen appropriately.

The coordinates of the angular momentum in the body-fixed frame are \begin{align} \vec{M} &= U^T\, \vec{m} = U^T\, (j\, \vec{\omega}) \\ &= U^T \Big( U\, J\, U^T \, \vec{\omega}\Big) = U^T \Big( U\, J\, U^T \, U\, \vec{\Omega}\Big) = \big(U^T\, U\big)J\big(U^T\, U\big) \vec{\Omega}\\ &= J\, \vec{\Omega} \end{align} Thus \begin{align} \vec{0} &= \frac{d \vec{m}}{dt} = \frac{d}{dt}\big(U\, \vec{M}\big) \\ &=\frac{d}{dt} \big(U\, J\, \vec{\Omega}\big) = \frac{dU}{dt}\, J\,\vec{\Omega} + U\, \frac{d}{dt} \,\big( J\vec{\Omega}\big) \end{align}
If you multiply the latter equation by the orthogonal matrix $$U^T$$ from the right, you get \begin{align} \vec{0} &= U^{T}\,\vec{0} = U^T\left( \frac{dU}{dt}\, J\,\vec{\Omega} + U\, \frac{d}{dt} \,\big( J\vec{\Omega}\big)\, \right) \\ &= \left( U^T\,\frac{dU}{dt} \right)\, J\,\vec{\Omega} + \big(U^T\,U\big)\, \frac{d}{dt} \,\big( J\vec{\Omega}\big) \\ &= \left( U^T\,\frac{dU}{dt} \right)\, J\,\vec{\Omega} + \frac{d}{dt} \,\big( J\vec{\Omega}\big) \end{align}
The latter equation is \begin{align} \vec{0} &= \left( U^T\,\frac{dU}{dt} \right)\, J\,\vec{\Omega}\, +\, \frac{d}{dt} \,\big( J\vec{\Omega}\big) \\ & = \frac{d}{dt} \,\big( J\vec{\Omega}\big) \, + \, \left( U^T\,\frac{dU}{dt} \right)\, J\,\vec{\Omega} \end{align} and if we recall that $$\left( U^T\,\frac{dU}{dt} \right)\, J\,\vec{\Omega} = \vec{\Omega} \times \big( J\vec{\Omega}\big)$$ The conservation of angular moemntum equation becomes teh following equation for the angular velocity in the body-fixed frame: $$\frac{d}{dt} \big( J\vec{\Omega}\big) \, = \, - \,\, \vec{\Omega} \times \big( J\vec{\Omega}\big)$$ Thus, the full system of equation, that determines the rotation $$U = U(t)$$ that we really care about is \begin{align} J \, &\frac{d\vec{\Omega}}{dt} = \big( J\vec{\Omega}\big) \times \vec{\Omega}\\ &\frac{d}{dt} U = U\, \big(\vec{\Omega} \times \cdot \big) \end{align} Using the conservation laws, which say that the magnitude of $$\vec{M} = J\,\vec{\Omega}$$ is conserved (because the vector $$\vec{m}$$ itself is constant in the world frame, so its representation $$\vec{M}$$ in the bopdy-fixed frame should have a constant length, but not a constant direction) and that the energy is conserved, we arrive at the fact that the solutions of the first set of equations $$J \, \frac{d\vec{\Omega}}{dt} = \big( J\vec{\Omega}\big) \times \vec{\Omega}$$ always lie on the pair of ellipsoids \begin{align} &(J\,\vec{\omega} \, \cdot \, J\,\vec{\omega}) = c_1 \,\,\, \text{(conservation of momentum)}\\ &(J\,\vec{\omega} \, \cdot \, \vec{\omega}) = c_2 \,\,\, \text{(conservation of energy)}\\ \end{align} So any solution $$\vec{\Omega} = \vec{\Omega}(t)$$ lies on a closed curve obtained from the intersection of the two ellipsoids and is therefore periodic, i.e. there is a time period $$T>0$$ such that $$\vec{\Omega}(t+T) = \vec{\Omega}(t)$$. But only because $$\vec{\Omega}(t)$$ is periodic doens't mean that the solution to the second set of equations $$\frac{d}{dt} U = U\, \Big(\vec{\Omega}(t) \times \cdot \Big)$$ will be periodic. Think of the simpler scalar equation $$\frac{du}{dt} = w(t) u$$ where $$w(t) = 1 + \cos(t)$$. The function $$w(t)$$ is clearly $$2\pi-$$periodic, but the full solution is $$u(t) = u_0\, e^{t + \sin(t)}$$ and it is clearly not periodic. What happens in the case of the rigid body is that
the angular velocity can be decomposed as follows $$\vec{\Omega}(t) = |\vec{\Omega}(t)| \, \left(\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}\right)$$ where the unit vector $$\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}$$ determines the instantaneous axis of rotation of the body at time $$t$$ while the magnitute $$|\vec{\Omega}(t)|$$ is the angular speed, i.e. it represents the instantaneous angle of rotation of the solid body around the instantaneous axis $$\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}$$. Now, after time $$T$$, the axis of rotation $$\frac{\vec{\Omega}(t + T)}{|\vec{\Omega}(t + T)|}$$ comes back exactly to $$\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}$$, but the angle increments of rotation of the body around this axis, which are represented by the angular speed $$|\vec{\Omega}(t)|$$ from time $$t$$ to time $$t+T$$, generally do not exactly sum up to one full rotation around the axes $$\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}$$. In other words $$U(t + T) \neq U(t)$$ and this discrepancy between the period of $$\frac{\vec{\Omega}(t)}{|\vec{\Omega}(t)|}$$ and the accumulation of rotational increments $$|\vec{\Omega}(t)|$$ is what the authors probably mean by "incommensurability". The actual evolution of the top is in most cases quasi-periodic, i.e. at time $$t+T$$ the top's orientation in space returns very close to where it was at time $$t$$ but not exactly.

Description of the angular velocity in the inertial frame. I should add here a description of the angular velocity in the inertial frame. The conservation of energy yields the equation $$\big(j\,\vec{\omega}(t) \cdot \vec{\omega}(t)\big) = c_2$$ but since the angular momentum $$j\,\vec{\omega}(t) = \vec{m}_0$$ in the inertial world frame is constant for all $$t$$, we obtain the equation of a plane $$\big(\vec{m}_0 \cdot \vec{\omega}(t)\big) = c_2$$ Therefore, in the inertial world frame, the angular velocity $$\vec{\omega}(t)$$ traverses a curve lying on the fixed plane $$\big(\vec{m}_0 \cdot \vec{x}\big) = c_2$$, where the plane is perpendicular to the angular momentum $$\vec{m}_0$$. At the same time, in the body-fixed frame, if we look at the ellipsoid $$\big(J\,\vec{X} \cdot\vec{X} \big) = c_2$$, we know that the angular velocity at any time $$t$$ lies on it, i.e. $$\big(J\,\vec{\Omega} \cdot \vec{\Omega} \big) = c_2$$. The equation of the ellipsoid's tangent plane at the point with position vector $$\vec{\Omega}$$ is given by the equation $$\big(J\,\vec{\Omega} \cdot \vec{X} \big) = c_2$$. Let us apply the change of coordinates $$\vec{x} = U \, \vec{X}$$, which takes us from the body-fixed frame at fixed time $$t$$ to the inertial world frame. It's inverse transformation is $$\vec{X} = U^T \, \vec{x}$$ and the body-fixed ellipsoid $$\big(J\,\vec{X} \cdot\vec{X} \big) = c_2$$ with the tangent plane $$\big(J\,\vec{\Omega} \cdot\vec{X} \big) = c_2$$ is transformed into the world ellipsoid $$\Big((U\,J \,U^T)\,\vec{x} \cdot\vec{x} \Big) = c_2$$ with the corresponding tangent plane $$\big(\vec{m}_0 \cdot \vec{x}\big) = c_2$$ ( because $$\vec{m}_0 = U\,J\,\vec{\Omega}\,\,$$).

Consequently, if we let time $$t$$ run, the angular velocity $$\vec{\omega}(t)$$ in the inertial world-frame traverses a curve always lying on the fixed plane $$\big(\vec{m}_0 \cdot \vec{x}\big) = c_2$$ with that plane being perpendicular to the fixed angular momentum $$\vec{m}_0$$ and tangent to the moving ellipsoid $$\Big(\big(U(t)\,J \,U^T(t)\big)\,\vec{x} \, \cdot\,\vec{x} \Big) = c_2$$ at the tip of the vector $$\vec{\omega}(t)$$. Consequently, the trajectory of $$\vec{\omega}(t)$$ can be seen as the planar curve traversed by the point of contact $$\vec{\omega}(t)$$ of the ellipsoid $$\Big(\big(U(t)\,J \,U^T(t)\big)\,\vec{x} \,\cdot \,\vec{x} \Big) = c_2$$, rolling without slipping on the plane $$\big(\vec{m}_0 \cdot \vec{x}\big) = c_2$$ so that the center of the ellipsoid is always fixed at the origin.

When the rolling ellipsoid has different principal axes, the curve $$\vec{\omega}(t)$$ is in general not closed, and thus the angular velocity is not periodic, because $$\vec{\omega}(t) = U(t)\, \vec{\Omega}(t)$$ and as discussed earlier $$\vec{\Omega}(t)$$ is periodic, but $$U(t)$$ in general is not, hence neither is $$\vec{\omega}(t)$$. In some cases though, when the parameters and the constants are chosen correctly, the time-evolution of the angular velocity $$\vec{\omega}(t)$$ will be periodic and the rotation matrix $$U=U(t)$$ will also be periodic.

In the case of an inertial ellipsoid with two equal principal axes, however, we can see that the angular velocity $$\vec{\Omega}(t)$$ in the body-frame moves along a circle and so its length $$|\vec{\Omega}(t)| = c_3$$ is constant. But length is independent quantity from Cartesian frames. Therefore, in the world frame $$|\vec{\omega}(t)| = |\vec{\Omega}(t)| = c_3$$ so $$\vec{\omega}(t)$$ also has a constant length, which means that $$\vec{\omega}(t)$$ points to a point that always lies on the sphere $$|\vec{x}| = c_3$$. Therefore, as time runs, the trajectory of $$\vec{\omega}(t)$$ always lies on the plane $$\big(\vec{m}_0 \cdot \vec{x}\big) = c_2$$ and on the sphere $$|\vec{x}| = c_3$$, so the said trajectory must be a circle (because it is the intersection of a sphere and a plane). Therefore, both $$\vec{\Omega}(t)$$ and $$\vec{\omega}(t)$$ are periodic, but in general they will have different periods, i.e. there will be two different numbers $$T_0 > 0$$ and $$T_1 > 0$$ such that $$\vec{\Omega}(t + T_0) = \vec{\Omega}(t)$$ and $$\vec{\omega}(t + T_1) = \vec{\omega}(t)$$. In general, these periods will be "incommensurable" in the snece that there is no positive rational number $$\frac{m}{n}$$ such that $$T_1 = \frac{m}{n}\, T_0$$. Then, the rotation matrix $$U(t)$$ will not be periodic and the symmetric top will not be returning periodically to its original orientation in space. However, in the rare occasions when there is a positive rational number $$\frac{m}{n}$$ such that $$T_1 = \frac{m}{n}\, T_0$$, the two representations of the angular velocity will have a common period $$T = n\, T_1 = m\, T_0$$ and then the rotation matrix $$U(t)$$ will be periodic $$U(t+T) = U(t)$$ and the symmetric top will be returning periodically to its original orientation in space.

• @GuyAb I did not use $\vec{X}$ mistakenly. It is deliberate. $\vec{X}$ is a general point in body-fixed coordinates, so the ellipsoid is indeed $$\{\, \vec{X} \,\, : \,\, (J\vec{X} \cdot \vec{X}) = c_2\, \}$$ The angular velocity $\vec{\Omega}$ is just one point on that ellipsoid, i.e. $(J\vec{\Omega} \cdot \vec{\Omega}) = c_2$. Jul 24 '19 at 13:49
• @GuyAb Your points 1), 2) and 3) are correct. Point 4) is wrong. I did some more careful thinking and math and corrected the end of my post, about the ellipsoid rolling on a plane in the case of a symmetric top. Read it very carefully, it explains what happens. Basically, the change form one perspective to the other ($U(t)$ from body-fixed to world-inertial) depends on time $t$, so it constantly changes, which complicates things. And finally, my analysis in terms of angular velocities and not angular momenta is crucial for understanding all these phenomena. Jul 24 '19 at 13:59
• @GuyAb Think about the following. You are in a transparent elevator, which is moving uniformly upwards. You are walking in the elevator in a circle. However, a person outside the elevator will see you moving along a helix, which is an open curve. Analogous phenomenon occurs with the top. Jul 24 '19 at 14:02
• @GuyAb If you think about the top, its motion in the inertial world-frame decomposes into two motions: 1) the precession of its angular velocity around the fixed angular momentum; 2) the rotation of the top around the angular velocity. The total motion is the composition of these two motions. If the precession in 1) is not periodic, then the total motion is not periodic. even if it is periodic, as in the case of the symmetric top, the two periodic motions from 1) and 2) may have non-matching periods, so that the total motion is not periodic. Jul 24 '19 at 14:25
• @GuyAb No, I don't think I wrote that. In body fixed frame, it is always true $\vec{\Omega}(t+ T_0) = \vec{\Omega}(t)$, so not only the direction is periodic, but the whole angular velocity is periodic. For the inertial world-frame it is more often that $\vec{\omega}(t)$ is not periodic (even the direction is not periodic). That is why what I wrote is that the rotation matrix $U(t+T) \neq U(t)$ for any $T$. So $\vec{\Omega}(t)$ always returns to its original position relative to the top's axes, but $\vec{\omega}(t)$ does not return to its original position in world frame. Jul 24 '19 at 18:16