Lagrangian of the Euler equations [duplicate]

Question

Do the Euler equations (where $I_1,I_2,I_3$ are principal moments of inertia):

$$I_1\dot{\omega}_1+(I_3-I_2)\omega_2\omega_3=M_1$$ $$I_2\dot{\omega}_2+(I_1-I_3)\omega_3\omega_1=M_2$$ $$I_3\dot{\omega}_3+(I_2-I_1)\omega_1\omega_2=M_3$$

in their above general form have a Lagrangian? If not, does a specific case of $\omega_1=\omega_2=0$ (and so $M_1=M_2=0$) have a general Lagrangian? ($M_3$ is the torque coming from a central gravitational potential - a planet - keeping the body (a satellite) on an elliptic orbit.)

Answer 1

It's some time since I worked this, but I believe that the Eqs of motion for $\omega$ ${\bf p}$, ${\bf r}$ from the action functional
$$ S[\omega, {\bf p}, {\bf r}]= \int \left(\frac 12 I_1\omega_1^2+\frac 12 I_2\omega_2^2+\frac 12 I_3\omega_3^2+ {\bf p}\cdot (\dot {\bf r}+ \omega \times {\bf r})\right)dt, $$ after the elimination of ${\bf p}$ and $ \dot {\bf r}$ in favour of the angular velocity vector $\omega$ gives the Euler equations for $\omega$ in the case $M_1=M_2=M_3=0$. A check strongly recommended as I posted a totally stupid wrong ``answer'' this morning, and am doubting my comptetence.

OK a check: The equations of motion that come from varying the vectors $\omega$, ${\bf p}$, ${\bf r}$, are (in that order) $$ {\bf L} \equiv {\bf I}\omega= -({\bf r}\times {\bf p})\\ \dot {\bf r}= - (\omega\times {\bf r})\\ \dot {\bf p}= - (\omega \times {\bf p}) $$ Then $$ \dot {\bf L} = -(\dot {\bf r}\times {\bf p})-({\bf r}\times\dot {\bf p})\\ = -{\bf p}\times(\omega\times {\bf r})- {\bf r}\times({\bf p}\times \omega)\\ = - \omega\times ({\bf p}\times {\bf r})\\ = - \omega\times {\bf L}. $$ In the last-but-one step I have use the vector triple-product Jacobi identity $$ {\bf a}\times ({\bf b}\times {\bf c}) +{\bf b}\times ({\bf c}\times {\bf a})+{\bf c}\times ({\bf a}\times {\bf b})=0. $$ Writing out $$ \dot{\bf L}+\omega\times {\bf L}=0 $$ with ${\bf L}= (I\omega_1, I_2\omega_2,I_3\omega_3)$ yields the torque-free Euler equations.

The idea of using Lin constraints to derive Euler's equations isquite modern. I think I saw this done in Cendra and Marsden "Lin Constraints Clebsh potentials and Variational Principles" Physica 27D (1987) 63-89.

Answer 2

They do. In the case of the symmetric top with $I_1=I_2\ne I_3$ for instance, the kinetic energy is given by (if my algebra is right) \begin{align} T_{\hbox{rot}}&=\textstyle\frac{1}{2}\sum_k I_k\omega_k^2\, ,\\ &=\textstyle\frac{1}{2}I_1\left((\dot{\theta}\cos\psi+\dot{\psi}\sin\theta\sin\psi)^2+ (\dot{\phi}\cos\psi\sin\theta -\dot{\theta}\sin\psi)^2\right) +\textstyle\frac{1}{2}I_3(\dot{\psi}+\dot{\phi}\cos(\theta))^2\, ,\\ &=\textstyle\frac{1}{2}I_1\left(\dot{\theta}^2+\sin^2(\theta)\dot{\phi}^2\right) +\textstyle\frac{1}{2}I_3(\dot{\psi}+\dot{\phi}\cos(\theta))^2\, . \end{align} where the expressions \begin{align} \omega_1&= \dot{\theta}\cos\psi+\dot{\psi}\sin\theta\sin\psi \\ \omega_2&= \dot{\phi}\cos\psi\sin\theta -\dot{\theta}\sin\psi \\ \omega_3&= \dot{\psi}+\dot{\phi}\cos(\theta)\, , \end{align} transition from the the angular frequencies to the time-derivatives of Euler angles.