Thread: Torque-free precession View Single Post

## 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.