I've been told that one could build rotational Hamiltonian based on Lagrangian of general form: $\mathcal{L} = \mathcal{L} (\vec{\Omega})$. By introducing Euler angles one could rewrite Lagrangian in terms of Euler angles and their derivatives: $\mathcal{L} = \mathcal{L} (\vec{e}, \dot{\vec{e}})$. Angular velocity is expressed in terms of Euler angles in a following way: $$\vec{\Omega} = \begin{bmatrix} \sin(\theta) \sin(\psi) & \cos(\psi) & 0 \\ \sin(\theta) \cos(\psi) & - \sin(\psi) & 0 \\ \cos(\theta) & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} \dot{\phi} \\ \dot{\theta} \\ \dot{\psi} \end{bmatrix}.$$ (Vector $\begin{bmatrix} \phi \\ \theta \\ \psi \end{bmatrix}$ is denoted by $\vec{e}$. ) Big problem that I encountered is that angular velocity depends on general velocity vector ($\dot{\vec{e}}$) in a linear way (by linear operator $\mathbf{M}(\vec{e})$ matrix of which is presented above), so general momentum $\vec{p_e} = \frac{\partial \mathcal{L}}{\partial \dot{\vec{e}}}$ is not an explicit function of general velocity vector($\dot{\vec{e}}$). That is Donkin's theorem or Legendre transform cannot be applied because general velocity components cannot be explicitly written as functions of $\vec{e}$ and $\vec{p_e}$.
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$\begingroup$ I once took a class on Lie algebras where the focus was rigid body rotation in SO(3), and we developed discrete versions of Hamilton's equations using variational techniques directly for each problem; we then did computer simulations with these, obtaining a very high degree of stability. The purpose of the class was to study techniques suitable for orbital calculations. We used attitude rotation matrices as the working element; these are skew-symmetric matrices forming a Lie algebra in so(3). The class was very involved & cool. $\endgroup$– Peter DiehrCommented Mar 24, 2016 at 22:48
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$\begingroup$ Peter Diehr, and how discrete-time Hamiltonian be of any use here? $\endgroup$– artfinCommented Mar 25, 2016 at 5:00
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$\begingroup$ We got from Lagrangian to Hamiltonian; the method was different than yours, but I don't recall any theoretical impediments. But the Lagrangian was expressed ddifferently. $\endgroup$– Peter DiehrCommented Mar 25, 2016 at 11:00
1 Answer
The Lagrangian is in fact an equation of $\vec \Omega$, however, in general it will be a quadratic function of $\vec \Omega$, as the rotational kinetic energy would be given by $$\frac{1}{2}\vec \Omega ^T \mathbf{I}\ \vec \Omega$$ This would give you the desired generalized momenta as a function of the general velocity vectors, as the diagonal entries of the moment of inertia tensor must not be zero, guaranteeing square velocity terms will appear in the Lagrangian. The exact form of the Lagrangian will depend on the object which is spinning, and its moment of inertia.
David Tong's notes go over the case of a rotational Lagrangian in terms of Euler angles for a symmetric top quite lucidly. From the explicit Lagrangian it is easy to see that $\frac{\partial^2 \mathcal{L}}{\partial \dot{\vec{e}}^2}$ is non-zero, allowing the application of the Legendre Transform.
