# How do we get $p_{\psi}=M_3$ and $p_{\phi}=M_z$ for a symmetrical top?

I am working on Landau and Lifshitz's Mechanics, 3rd edition. In Section 35 Problem 1 it studies a heavy symmetrical top. We know that the Lagrangian is given by $$L=\frac{1}{2}(I_1+\mu l^2)(\dot{\theta}^2+\dot{\phi}^2\sin^2\theta)+\frac{1}{2}I_3(\dot{\psi}+\dot{\phi}\cos\theta)^2-\mu gl\cos \theta.$$ Since $$\psi$$ and $$\phi$$ are cyclic coordinates, we have two integrals $$p_{\psi}=\frac{\partial L}{\partial \dot{\psi}}=I_3(\dot{\psi}+\dot{\phi}\cos\theta)\equiv M_3$$ and $$p_{\phi}=\frac{\partial L}{\partial \dot{\phi}}=((I_1+\mu l^2)\sin^2\theta+I_3cos^2\theta)\dot{\phi}+I_3\dot{\psi}\cos\theta\equiv M_z.$$

I understand that $$p_{\psi}$$ and $$p_{\phi}$$ are constants. On the other hand, I understand that by physics argument, the projection of the angular momentum to the $$z$$ axis and the $$x_3$$ axis are both constants.

My question is: how do we know that $$p_{\psi}=M_3$$ and $$p_{\phi}=M_z$$? Do we get it from some general result in mechanics?

Edit: by (33.2) of Mechanics $$M_3=I_3\Omega_3$$ and by (35.1) $$\Omega_3=\dot{\psi}+\dot{\phi}\cos\theta$$. Therefor $$M_3=I_3(\dot{\psi}+\dot{\phi}\cos\theta)$$ is clear.

As for $$M_z$$, we can use (33.2) and $$(35.1)$$ to get the expression of the angular momentum vector $$\mathbf{M}$$ as $$\mathbf{M}=(I_1+\mu l^2)(\dot \phi\sin\theta\sin\psi+\dot \theta\cos \psi)\mathbf{e_1}+(I_1+\mu l^2)(\dot \phi\sin\theta\cos\psi-\dot \theta\sin \psi)\mathbf{e_2}+I_3(\dot{\psi}+\dot{\phi}\cos\theta)\mathbf{e_3}$$

Then we can use Eulerian angles to project $$\mathbf{e_1}$$, $$\mathbf{e_2}$$, and $$\mathbf{e_3}$$ to the $$z$$-axis and obtain the projection of $$\mathbf{M}$$ to the $$z$$-axis. But the computation is complicated.

My edited question is: can we get $$M_z=\frac{\partial L}{\partial \dot\phi}$$ in a simple way?

• What is the definition of angle $\psi$?
– ytlu
Jul 28, 2021 at 6:56
• @ytlu It is one of the Eulerian angles, you can see the convention on this webpage galileoandeinstein.phys.virginia.edu/7010/…. Jul 28, 2021 at 14:18

The transformation matrix between body fixed system and inertial system is

$$S=S_z(\varphi)\,S_x(\theta)\,S_z(\psi)$$

form here you obtain the angular velocity vector (components in body system)

$$\vec\omega= \left[ \begin {array}{c} \sin \left( \theta \right) \sin \left( \psi \right) \dot\varphi +\cos \left( \psi \right) \dot\theta \\ \sin \left( \theta \right) \cos \left( \psi \right) \dot\varphi -\sin \left( \psi \right) \dot\theta \\ \dot\psi +\dot\varphi \,\cos \left( \theta \right) \end {array} \right]$$

the components of the angular momentum vector in body system are:

$$\vec L=\begin{bmatrix} I_1 & 0 & 0 \\ 0 & I_1 & 0 \\ 0 & 0 & I_3 \\ \end{bmatrix}\,\vec\omega$$

thus the components of angular momentum vector in inertial system are:

$$\begin{bmatrix} L_x \\ L_y \\ L_z \\ \end{bmatrix}=S\,\vec L$$

from here you obtain that

$$L_z=(I_1\,\sin^2(\theta)+I_3\,\cos^2(\theta))\,\dot\varphi+ I_3\,\cos(\theta)\,\dot\psi=\frac{\partial T}{\partial\dot\varphi}$$

where T is the kinetic energy

$$T=\frac 12 \vec\omega^T\,\begin{bmatrix} I_1 & 0 & 0 \\ 0 & I_1 & 0 \\ 0 & 0 & I_3 \\ \end{bmatrix}\,\vec\omega$$

• I understand that $p_{\psi}$ and $p_{\phi}$ are constants, but my question is why they coincide with the projection of the angular momentum to the $z$-axis and the $x_3$-axis. Jul 28, 2021 at 14:20
• @ZhaotingWei this is my new answer
– Eli
Jul 28, 2021 at 20:28