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In Hamiltonian mechanics, angular momentum is a certain momentum map and a component of the angular momentum is the generator function of the action of a one-parameter subgroup of the rotation group $SO(3)$.

In Newtonian physics, angular momentum is something that causes the precession of a bicycle wheel that is rotating about a horizontal axis suspended at one end.

There is no apparent relationship between the two. Is there any way to relate these two things without using sums and indices?

Motivation

I'd like to see Hamiltonian mechanics as a complete, standalone theory, without the crutches of the Newtonian model or Lagrangian formalism. So for example I'd like to see that angular momentum (defined in Hamiltonian theory) has the properties that we expect from it, namely it changes the behavior of a rigid body against external effects. Shortly: how can describe Hamiltonian mechanics phenomena like the bicycle wheel precession?

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2 Answers 2

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The angular momentum vector of Newtonian mechanics and the element of $\mathfrak{so}(3)$ in Hamiltonian mechanics represent the same quantity. $\mathfrak{so}(3)$ is canonically isomorphic to $\mathbb{R}^3$ where the bracket is the cross product. This isomorphism is given by the hodge star, which identifies the vector $\mathbf{L}$ with the antisymmetric operator $$\mathbf{v} \mapsto \mathbf{L}\times\mathbf{v}$$ The angular momentum vector is then the composition of the inverse of this isomorphism with the momentum map in the symplectic formalism.

Edit: It occurred to me that you might be a mathematician exploring classical mechanics. Within this formalism, particles follow trajectories given by integral curves of a hamiltonian vector field $X_H$ on $T^*\mathbb{R}^3$ where $H$ is typically (but not always) identified with the energy of the system. Precession like the kind involving a bicycle wheel happens when $\{L_{ij}, H\} = \omega(X_{L_{ij}}, X_H) \neq 0$, so $L(\gamma(t))$ is not constant along the trajectory.

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  • $\begingroup$ Fine, but where is the precession? $\endgroup$
    – mma
    Dec 3, 2021 at 7:31
  • $\begingroup$ @mma This is less a question about Hamiltonian mechanics and more a question about how torque works. In the case of a bicycle wheel fixed at a point, there is a torque due to gravity, and this torque causes the angular momentum to change uniformly. I don't fully understand why you're doing Hamiltonian mechanics before understanding Newtonian, but Kleppner & Kolenkow is a particularly nice textbook if you want to see the details of this. $\endgroup$
    – Omaredabed
    Dec 3, 2021 at 18:13
  • $\begingroup$ I hope that the extension of the question clarifies my motivation. $\endgroup$
    – mma
    Dec 4, 2021 at 5:26
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There is a straightforward description of spinning tops (or other spinning objects) via Hamiltonian dynamics:

Our generalized coordinates are the orientations of the three principal axes of the spinning object relative to an arbitrary axis. For a top under the influence of gravity, it's convenient to choose the axis along which the force of gravity points (i.e. the "$z$-axis"), and so we have generalized coordinates $z_1,z_2,z_3$ and generalized angular momenta $l_1,l_2,l_3$ along the principal axes with the Hamiltonian $$ H = \sum_i\frac{l^2_i}{2I_i} + mg \sum_i c_i z_i,$$ where the $I_i$ are the moments of inertia and the $c_i$ are determined by where the center of mass sits, i.e. how the orientation of the top affects its total gravitational potential energy. The $l_i$ are the angular momenta, i.e. the generators of rotation as determined by the momentum map.

If the top is symmetric $I = I_1 = I_2$ and the center of mass sits at height $h$ along the symmetry axis ($n_3$), we have a so-called Lagrange top with Hamiltonian $$ H = \frac{l_1^2 + l^2_2}{2I} + \frac{l_3^2}{2I_3} + mghz_3.$$ If you solve the equations of motion for this system (be mindful that the $z_i,l_i$ are not canonical coordinates, but have Poisson brackets $\{l_i,l_j\} = \epsilon_{ijk}l_k$ and $\{l_i,z_j\} = \epsilon_{ijk}z_k$), you will get that initial conditions that do not align the symmetry axis of the top with the $z$-axis show precession - the angular momenta along the z-axis and along the symmetry axis are conserved, but the symmetry axis itself will precess around the z-axis.

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  • $\begingroup$ Are you sure these equations are correct? I have doubts because if $(l_1,l_2,l_3)$ is replaced by $(-l_1,-l_2,-l_3)$ then nothing changes here, but we know that the direction of precession should change. $\endgroup$
    – mma
    Dec 5, 2021 at 6:58
  • $\begingroup$ @mma Yes (for one, the Wiki article agrees, and secondly these are just the correct expressions for energy - obviously rotational energy doesn't care in which direction something rotates). Your concern is unfounded: Symmetries of the Hamiltonian in general do not descend to symmetries of the solutions of the equations of motion. Consider the free particle with Hamiltonian $H\propto p^2$. The Hamiltonian is invariant under $p\mapsto -p$, but that doesn't mean the particle doesn't reverse direction when you reverse the momentum! $\endgroup$
    – ACuriousMind
    Dec 5, 2021 at 10:31
  • $\begingroup$ That Wikipedia article cites the Classical Mechanics of Goldstein, Poole, and Safko but I don't find your Hamiltonian in this book. Can you provide a valid source for it? $\endgroup$
    – mma
    Dec 7, 2021 at 6:54
  • $\begingroup$ @mma The Hamiltonian is just the expression for the energy of the top as it always is in unconstrained system and I'm not sure what your problem with it is, but if you want a source that actually writes this down, see "The heavy top: a geometric treatment" by Lewis, Ratio, Simo and Marsden, in particlular eq. (1.27). I just chose a different variable for the spatial orientation (the principal axis orientation instead of the vector from the top fixed point to the c.o.m.). You can also get it from 4.4. of Abraham/Marsden's book if you can stomach the notational differences. $\endgroup$
    – ACuriousMind
    Dec 7, 2021 at 8:51

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