- #1
ognik
- 643
- 2
I'm not 100% confident of my approach to the 2 exercises below:
Orbital angular momentum of i'th element is $\vec{L_i} = \vec{r_i} \times \vec{p_i} = m_i \vec{r_i} \times (\omega \times \vec{r_i}) $
a) Find the inertia matrix $I$ such that (omitting vector signs from here on) $L = I \omega, |L \rangle = I|\omega \rangle$
$=> L_i = - m_i[r_i \times r_i \times \omega]$ (cyclic anti-commute)
$ =-m_i[r_i(r_i \cdot \omega) - \omega (r_i \cdot r_i)] $ (BAC-CAB)
$ = m_i\omega r_i^2 $ ($r_i$ and $\omega$ orthogonal)
$ L = \sum L_i = \sum (m_i r_i^2)\omega =\sum I_i \omega = I \omega, i.e. |L \rangle = I| \omega \rangle$
b)Similarly derive I starting with Kinetic Energy $T_i = \frac{1}{2}m_i(\omega \times r_i)^2 $
$ = \frac{1}{2}m_i [|\omega|^2 |r_i|^2 - (\omega \cdot r_i)^2] $ (identity)
$ = \frac{1}{2} \omega (m_i r_i^2) \omega $ ($r_i$ and $\omega$ orthogonal)
$ \therefore T= \sum T_i = \frac{1}{2} \omega \sum(m_i r_i^2) \omega = \frac{1}{2} \omega I \omega = \frac{1}{2}\langle \omega |I| \omega \rangle$
Orbital angular momentum of i'th element is $\vec{L_i} = \vec{r_i} \times \vec{p_i} = m_i \vec{r_i} \times (\omega \times \vec{r_i}) $
a) Find the inertia matrix $I$ such that (omitting vector signs from here on) $L = I \omega, |L \rangle = I|\omega \rangle$
$=> L_i = - m_i[r_i \times r_i \times \omega]$ (cyclic anti-commute)
$ =-m_i[r_i(r_i \cdot \omega) - \omega (r_i \cdot r_i)] $ (BAC-CAB)
$ = m_i\omega r_i^2 $ ($r_i$ and $\omega$ orthogonal)
$ L = \sum L_i = \sum (m_i r_i^2)\omega =\sum I_i \omega = I \omega, i.e. |L \rangle = I| \omega \rangle$
b)Similarly derive I starting with Kinetic Energy $T_i = \frac{1}{2}m_i(\omega \times r_i)^2 $
$ = \frac{1}{2}m_i [|\omega|^2 |r_i|^2 - (\omega \cdot r_i)^2] $ (identity)
$ = \frac{1}{2} \omega (m_i r_i^2) \omega $ ($r_i$ and $\omega$ orthogonal)
$ \therefore T= \sum T_i = \frac{1}{2} \omega \sum(m_i r_i^2) \omega = \frac{1}{2} \omega I \omega = \frac{1}{2}\langle \omega |I| \omega \rangle$