Thread: Torque-free precession View Single Post
 P: 669 Torque-free precession I see what you mean. Try the following Hamiltonian: $\hat{H} = \frac{1}{2} \sum\limits_{ij} \hat{L}_i I^{-1}_{ij} \hat{L}_j$ where $I^{-1}_{ij}$ is the invserse of the inertia tensor. In the normal system $I^{-1}_{ij} = \delta_{ij} \frac{1}{I_i}$ The angular momentum operator L is well defined, and the moment of inertia can be taken as constant. If I am not mistaken, then L does not commute with the Hamiltonian, so that you get precession.