When we think of the hard topics in physics, quantum mechanics and general relativity spring to mind. Although those topics are incredibly complex and non-intuitive, I personally feel that the motion of an asymmetric top trumps both QM and DR in complexity, making it one of the hardest concepts to grasp in physics. In this article, we will explore how to analyse the motion of the asymmetric top, of course with some constraints to make our analysis tractable.
We consider the free rotation of an asymmetric top, for which we let all three moments of inertia be different. However, for simplicity, we impose the following condition on the moments of inertia:
Anyone who has dabbled in classical mechanics will be familiar with Euler Equations, a set of equations that allow us to analyse rotational systems. In the case of free rotation, in other words, when net moments about all the axes are zero, Euler Equations reduce to the following form:
Now our task will be to in some way find closed form solutions (or at least some simpler representation) of the angular velocities of the free rotating top. If we find the angular velocities of the top, we will have a fully determined system as we can find any other quantity using the angular velocities. Note that if the motion of the top included translation, then we will also need to know the linear velocities to fully determine the system.
Approaching the solution
We know from elementary physics that there are two main conservation theorems: total energy and momentum. In the case of rotation, momentum is replaced by angular momentum. Therefore, we can already produce two integrals of the equations of motion:
Where E and M are the total energy and angular momentum respectively. We can actually write the first equation in terms of the components of angular momentum instead to make our analysis easier.
We can already draw some conclusions about the relationship between the various angular velocities and moments of inertia. The energy conservation equation produces an ellipsoid with semi-axes sqrt(2EI1), sqrt(2EI2), sqrt(2EI3), and the angular momentum conservation equation produces a sphere with radius M. So when the angular momentum vector M moves around the component space, it moves along the intersection lines between the ellipsoid and sphere. The following figure illustrates that:
If we want to be rigorous, we can prove that intersections between the ellipsoid and sphere exist using the following inequality:
We can prove this inequality by considering the initial condition we imposed on the moments of inertia and the surface equations of the ellipsoid and sphere.
Let us examine the intersection curves in a bit more detail. When M² is slightly bigger than 2EI_1, the intersection curve is a small ellipse near axis x_1, and as M² tends to 2EI_1, the curve gets smaller until it shrinks to the axis x_1 itself. When M² gets larger, the curve also expands, until M² equals 2EI_2, where the curves become plane ellipses that intersect at the pole x_2. Now as M² increases past 2EI_2, the curves become closed ellipses near the x_3 poles. Conversely, when M² is slightly less than 2EI_3, the intersection curve is a small ellipse near the axis x_3, and as M² tends to 2EI_3, the curve shrinks to a point on x_3.
We can also note a few things by looking at the nature of the curves. Firstly, since the all of the intersection curves are closed, there must be some sort of a period to those precession/rotation. Secondly, if we look at the size of the curve near the axes, we get an interesting result. Near axes x_1 and x_3, in other words, near the smallest and largest moment of inertia, the intersection curves are small and entirely in the neighborhood of the poles. This can be interpreted to be stable precession about the axis of largest and smallest moment of inertia. However, near the axis x_2, the intermediate moment of inertia, the intersection curves are non local and large. This means that deviations in rotation around the intermediate axis is unstable. This is consistent with the famous Tennis Racket Theorem (read more about it here), where it can be proved that the perturbation of motion about the intermediate axis is unstable. It is quite a remarkable way to prove the tennis racket theorem, purely graphically, with minimal mathematics.
Analysing Angular Velocity
Now that we have an idea of how the angular momentum and energy of the asymmetric top are interrelated, we can proceed and try to understand how the angular velocity evolves over the rotation of the top. We can first try and represent the angular velocities in terms of one another, and in terms of the constants of equation of motion, aka the energy and momentum. We get the following set of equations:
Now, we can substitute these two equations into the Euler equation component for Omega_2:
This expression hints to us that the integral for the angular velocity would be some form of an elliptic integral. Now we can add another condition to make our life easier:
We now suggest the following change of variables to make the solution more tractable:
Note that if the inequality is reversed, we can just interchange the signs of the moments of inertia in the substitutions. It is also useful to define a positive parameter k²<1, as
Finally, we get a familiar integral
Note that the origin of time is taken to be when Omega_2 is zero. This integral is non analytic, but inverting the integral gives us the Jacobian elliptic functions s=sn(tau). Now we can finally write our angular velocities as ‘closed’ form solution:
Obviously these are periodic functions and we know that the period is
Where K is the complete Jacobian integral of the first kind
Note that at time T, the angular velocity vector returns back to its original position, but that does not mean that the asymmetric top itself returns to its original position.
The solutions for the angular velocities might be elegant, but they do little for us mortals to understand the actual motion (maybe Landau was smart enough to visualise it). We can attempt to try and understand it by converting the angular velocities into equations involving Euler angles instead. However, the mathematics to do so are quite long and tricky, so I have omitted it.
A Simpler Case
Since the asymmetric top doesn’t really allow us to intuitively understand how the Euler equations work and how to interpret the results, we turn to a simpler problem. The following problem was actually set in Landau’s Mechanics textbook.
Reduce to quadratures the problem of the motion of a heavy symmetrical top whole lowest point is fixed.
We can represent this using the following diagram
We know that the Lagrangian for this system is
Since phi and psi are both cyclic coordinates, in other words, their derivative in the Lagrangian is zero, we can already write down two integrals of the equations of motion
We also know that the total conserved energy is
Using our two integrals of equations of motion, we obtain
Now, substituting those into our energy conservation equation, we obtain
Note that the energy equation now resembles the sum of kinetic (as dictated by the parallel axis theorem) and potential energy (now called the effective potential). We know from standard analysis that this can now be represented as
Evaluating this integral will give us the necessary solutions that we seek for the various angles. Note that this is also an elliptic integral. We know that E’ must be more than or equal to the effective potential. Also, the effective potential tends to infinity when theta is equal to either 0 or pi, and has a minimum between those points. So, the equation E’=U_eff must have two roots. We denote those two roots theta_1 and theta_2.
When theta changes from theta1 to theta2, the derivative of phi might change sign depending on whether or not the difference M_z-M3cos(theta) changes sign. The different scenarios can result in the following type of motion:
When the derivative of phi does not change direction, we get the scenario in 49a, and this type of motion is known as nutation. Note that the curve shows the path of the axis of the top while the center of the sphere shows the fixed point of the top. If the derivative of phi does change direction, we get 49b, where the top moves in the opposite direction for a brief amount of time for phi. Lastly, if the theta1 or theta2 is equal to the difference M_z-M3cos(theta), then both derivatives of phi and psi vanish, resulting in the motion from 49c.
Hopefully this article has given you some insight on how the mind boggling field of rigid body dynamics works, in particular, how asymmetric tops rotate in free space. Notice that sometimes, looking at graphical representations of quantities can give us a lot of information instead of delving directly into the mathematics and getting stuck inside. Also, it sometimes helps to tackle simpler problems to understand how we can visualise the solutions, albeit for non-realistic scenarios.