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M Quack
Mar7-12, 03:19 AM
P: 675
Torque-free precession

I see what you mean.

Try the following Hamiltonian:

[itex]\hat{H} = \frac{1}{2} \sum\limits_{ij} \hat{L}_i I^{-1}_{ij} \hat{L}_j[/itex]

where [itex]I^{-1}_{ij}[/itex] is the invserse of the inertia tensor. In the normal system [itex] I^{-1}_{ij} = \delta_{ij} \frac{1}{I_i}[/itex]

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.