# Choosing principal axes for symmetric top

I am following Landau. Here $\mathbf{L}$ is angular momentum and $\mathbf{\Omega}$ is the angular velocity. The qualitative treatment for symmetric top in absence of gravity starts by choosing principal axes of body such that $\mathbf{L\cdot e_2}=0$, where {$\mathbf{e_1,e_2,e_3}$} are the principal axes directions and $\mathbf{e_3}$ is axis of symmetry

Now, since the body is going to precess about $\mathbf{L}$, the motion is such that $\mathbf{e_3, L, \Omega, e_1}$ all form a plane which rotates, as seen in inertial frame. However, that means, $\mathbf{e_2\cdot L}$ remains zero. I wanted to show this, before I proceeded.

$$\frac{d}{dt}(\mathbf{L\cdot e_2}) = \mathbf{L}\cdot(\Omega\times \mathbf{e_2}) = -I_1\Omega_1\Omega_3+I_3\Omega_3 \Omega_1 \neq 0 \quad\text{since } I_1 = I_2\neq I_3$$

Can you point out where I made a mistake? Since if $\Omega_2$ is actually zero for all time, this doesn't make sense. Incidentally, if I write similar equation for 3rd component it gives me $L_3= const.$, which is correct.

Now that I read Landau & Lifshitz again I truly got your mistake. In section 33 (page 106 of the last english edition) they consider the symmetric top in the inertial frame. At a given instant, the fixed axes in space are chosen to coincide with the principal axis of the body and they chose the principal axis $x_2$ to be perpendicular to $\vec L$. Then $$L_2=I_2\Omega_2=0\Rightarrow \Omega_2=0.$$ The symmetrical top has infinitely many sets of principal axis - any pair of axes mutually perpendicular with the symmetry axis (and with same origin) is a pair of principal axes - so it is possible to find principal axes that are not rigidly attached to the body. At every instant we can chose $x_2$ satisfying so that $\Omega_2=0$, such that $x_3$, $\vec L$ and $\vec \Omega$ are in the same plane. The angular velocity and $x_3$ are precessioning about $\vec L$. Notice that this is viewed in the inertial frame.
On the other hand, in section 36 (page 115) they consider again the same problem but this time in another frame of reference, namely the non-inertial frame rigidly attached to the body and given by the principal axis. The equations of motions read $$\dot\Omega_1+(I_3-I_2)\Omega_2\Omega_3/I_1=0,$$ $$\dot\Omega_2+(I_1-I_3)\Omega_3\Omega_1/I_2=0,$$ $$\dot\Omega_3+(I_2-I_1)\Omega_1\Omega_2/I_3=0\Rightarrow\dot\Omega_3=0.$$ The first two equations have the solution, $$\Omega_1=A\cos\omega t,\quad \Omega_2=A\sin\omega t,$$ where $$\omega=\frac{\Omega_3(I_3-I_1)}{I_1}\neq 0.$$ The precession of the angular velocity, as viewed here, is with respect to the axis fixed to the body which rotate by themselves with angular velocity $\Omega$. It is therefore hard to visualize this motion.
• I refer you to the same book. The solutions you have derived in fact, hold for, system attached to the body. An obvious way to see that is, $\Omega$ is going to precess about e_3 according to this solution. Now, when can that happen? Nov 16 '17 at 14:38
• I didn't assume $\Omega_2=0$. Only thing I meant was at t=0, it is zero. Nov 16 '17 at 14:52
• @physicophilic But if you did not assume $\Omega_2=0$ for any time, then there is no problem at all, since this means $\dot\Omega_2\neq 0$. Nov 16 '17 at 15:32
• It makes sense. Thank you. I realized that it also explains why $\Omega$ and (thus L) must precess in body frame, since this new "non-body frame" which always has $\Omega_2=0$, is rotating with respect to fixed axis of body (considered in derivation of Euler's equation) Nov 17 '17 at 4:26