Derivatives of conjugate momenta wrt generalised coordinates in Lagrangian mechanics

Question

I am struggling understanding the validity of the solution that my professor gave to the following exercise.

A heavy symmetric top rotating about a fixed point has Lagrangian: $$ \mathcal L=\frac{I_1}{2}\left(\dot \theta ^2 + \dot \phi ^2 \sin^2 \theta\right) + \frac{I_3} 2 \left(\dot \psi+ \dot \phi \cos \theta \right)^2 - mgl \cos \theta $$ where $I_1=I_2, I_3$ are its principal moments of inertia and $(\phi, \theta, \psi)$ are the Euler angles. Write down the two conserved generalised momenta $p_ \phi$ and $p_\psi$ and show that $\theta$ obeys the equation $$ I_1 \ddot \theta = - \frac{\partial V_{eff}}{\partial \theta}$$ where $$V_{eff}(\theta)= \frac{\left(p_ \phi - p_ \psi \cos \theta\right)^2}{2I_1 \sin ^2 \theta} + mgl \cos \theta$$

The solution my professor offers, says the following:

$$p_ \phi= \frac{\partial \mathcal L}{\partial \dot \phi}= I_1 \dot \phi \sin ^2 \theta + I_3 (\dot \psi + \dot \phi \cos \theta)\cos \theta$$ $$p_ \psi= \frac{\partial \mathcal L}{\partial \dot \psi} = I_3 (\dot \psi + \dot \phi \cos \theta)$$ From which we get the relation: $$\dot \phi = \frac{p_ \phi - p_ \psi \cos \theta}{I_1 \sin^2 \theta}$$ Then he proceeds to use the E-L equation in theta: $$\frac {d}{dt}\frac{\partial \mathcal L }{\partial \dot \theta}= \frac{\partial \mathcal L}{\partial \theta}$$ And after some algebra and derivatives: $$I_1 \ddot \theta = \frac{(p_ \phi - p_ \psi \cos \theta)^2}{I_1 \sin^3 \theta} \cos \theta - \frac{p_ \psi(p_ \phi - p_ \psi \cos \theta)}{I_1 \sin^2 \theta} \sin \theta - \frac{\partial }{\partial \theta} (mgl \cos \theta)$$ Then, he asserts that the RHS is indeed $$- \frac{\partial }{\partial \theta} \left \{ \frac{\left(p_ \phi - p_ \psi \cos \theta\right)^2}{2I_1 \sin ^2 \theta} + mgl \cos \theta \right \}$$

However, I believe that this last step is controversial. This equality can be easily proved under the assumption $\frac{\partial p_ \phi}{\partial \theta}=\frac{\partial p_ \psi}{\partial \theta} = 0 $. This would be the case in Hamiltonian mechanics where we consider momenta and positions to be independent, but in the Lagrangian formalism the independent variables are positions and velocities. One could calculate directly the derivatives of the momenta wrt theta which are non-zero. Is my professor just being sloppy with the notation of the partial derivatives (*), or is there some more fundamental physics which I am unaware of?

(*) I.e. he means in the E-L equation: $$ \frac d {dt} \left (\frac{\partial \mathcal L}{\partial \dot \theta}\right) _{q_i, \dot \psi, \dot \phi} = \left(\frac{\partial \mathcal L}{\partial \theta}\right) _{\psi, \phi, \dot q_i}$$ But then $$I_1 \ddot \theta = - \left (\frac{\partial V_{eff}}{\partial \theta} \right) _{\phi, \psi, p_i}$$

Answer 1

It seems that the solution offered is using the Routhian. If you follow this link you will find exactly your professor's analysis under the Heavy Symmetrical Top section.

Answer 2

mike stone's answer is exactly right: OP's professor is implicitly using the Routhian $$ R(\theta,\dot{\theta},p_{\psi},p_{\phi})~=~\dot{\psi}p_{\psi}+\dot{\phi}p_{\phi}-L(\theta,\dot{\theta},\dot{\psi},\dot{\phi}),$$ which is a partial Legendre transform of the Lagrangian $L(\theta,\dot{\theta},\dot{\psi},\dot{\phi})$ wrt. the 2 cyclic variables $\psi$ and $\phi$ only; thereby turning a 3D problem into a 1D problem in $\theta$ (since the corresponding momenta $p_{\psi}$ and $p_{\phi}$ are COMs.)

More generally, the Routhian mechanics harvests the best features of both the Lagrangian and Hamiltonian mechanics in a hybrid formulation, cf. e.g. my Phys.SE answer here.

Answer 3

with EL

$$I_1\,\ddot\theta=f(\dot\phi~,\dot\psi~,\theta)$$

additional two first Integrals

$$\frac{\partial\mathcal L}{\partial\phi}=0\quad\Rightarrow\frac{\partial\mathcal L}{\partial\dot\phi}=c_\phi\tag 1$$ $$\frac{\partial\mathcal L}{\partial\psi}=0\quad\Rightarrow\frac{\partial\mathcal L}{\partial\dot\psi}=c_\psi\tag 2$$

where $~c_i~$ are constants

solve equations $~(1)~,(2)~$ for $~\dot\phi~,\dot\psi~$ and substitute to the equation of motion you obtain:

\begin{align*} I_1\,\ddot\theta=f(\theta)\quad\Rightarrow V_{\rm eff}=-\int f(\theta)\,d\theta\quad,f(\theta)=-\frac {\partial V_{\rm eff}}{\partial \theta} \end{align*}

Answer 4

I think you're right. While in Lagrange equations we use generalized coordinates, their time derivatives (and time if Lagrangian function explicitly depends on time) and we mean

$$ \frac d {dt} \left (\frac{\partial \mathcal L}{\partial \dot \theta}\right) _{q_i, \dot \psi, \dot \phi} = \left(\frac{\partial \mathcal L}{\partial \theta}\right) _{\psi, \phi, \dot q_i}$$

The expression of $V_{eff}$ is a function of $\theta$ only, given the value of the generalized momenta, that are integrals of motion, i.e. constants, that can be determined as an example knowing the initial conditions of the system. So, the right way to interpret that partial derivative is performing derivative w.r.t. $\theta$ taking constant of motions $p_\phi$, $p_\psi$ constant,

$$V_{eff}(\theta; p_\phi, p_\psi)$$

Use of effective potential in mechanics. The definition of an effective potential as a function of a set of constant of motions - written as a function of the generalized coords of the problem - and one free generalized coordinate is a way to evaluate the evolution of that free generalized coordinate in a problem with more than one coordinate. It's a procedure similar to what is usually done in the evolution of 2-body problem. In general, this problem has two free coordinates: as an example you can take polar coordinates (this motion is a planar motion) $r$, $\theta$; angular momentum is a constant of motion and you can write it as a function of $r$ and $\dot\theta$; thus, if you know angular momentum, you can get $\dot\theta$ as a function of it and derive a dynamical equation for $r$ only with an effective potential

$$V_{eff}(r;\ell) = -\frac{\ell^2}{mr^3} + \frac{c}{r^2} \ , $$

with $c=GMm$ for a gravitational field.

I'll leave you a link here about central motions, in Italian language , but you should understand it if your nickname tells us where you come from - otherwise rely on the translator of your browser

Answer 5

I did manage to solve the question thanks to Qmechanic's answer and Mike Stone's answer and I thought I'd type out the work for clarity, please correct any error I make.

The idea is to introduce the Routhian, which is basically a Lagrangian that didn't believe enough to completely transform into a Hamiltonian. So we define it in the following way: $$ \mathcal R(\theta, \dot \theta, \psi , p_ \psi, \phi, p_ \phi) = p_ \psi \dot \psi + p_\phi \dot \phi-\mathcal L $$

The utility of introducing this quantity is that it satisfies the E-L equation in the variables that were not hamiltonianised and it satisfies the Hamiltonian equations for the rest. In our case: $$ \begin{align} \frac d{dt} \frac{\partial \mathcal R}{\partial \dot \theta}\bigg|_{\theta, \psi , p_ \psi, \phi, p_ \phi}&=\frac{\partial \mathcal R}{\partial \theta}\bigg|_{\dot \theta, \psi , p_ \psi, \phi, p_ \phi} \\ \dot \psi &= \frac{\partial \mathcal R}{\partial p_\psi}\bigg|_{\theta, \dot \theta, \psi , \phi, p_ \phi} \\ \dot p_\psi &= - \frac{\partial \mathcal R}{\partial \psi}\bigg|_{\theta, \dot \theta, p_ \psi , \phi, p_ \phi}\\ \dot \phi &= \frac{\partial \mathcal R}{\partial p_\phi}\bigg|_{\theta, \dot \theta, \phi , \psi, p_ \psi} \\ \dot p_\phi &= - \frac{\partial \mathcal R}{\partial \phi}\bigg|_{\theta, \dot \theta, p_ \phi , \psi, p_ \psi} \end{align} $$

For the Lagrangian given, after a bit of algebra, one gets the hamiltonian-resembling Routhian: $$ \begin{align} \mathcal R = \mathcal R(\theta, \dot \theta, \psi , p_ \psi, \phi, p_ \phi) &= \frac {I_1} 2 \dot \phi ^2 \sin ^2 \theta + \frac{I_3} 2 \left( \dot \psi + \dot \phi \cos \theta\right)^2 - \frac{I_1} 2 \dot \theta ^2 + m g l \cos \theta \\ &= \frac{\left( p_ \phi - p_ \psi \cos \theta \right)^2}{2 I_1 \sin ^2 \theta} + \frac{p_ \psi ^2}{2 I_3} - \frac{I_1} 2 \dot \theta ^ 2 + mgl \cos \theta \end{align} $$

Finally, making use of the E-L equation in theta, we now have the correct form. The LHS is: $$ \frac d{dt} \frac{\partial \mathcal R}{\partial \dot \theta}\bigg|_{\theta, \psi , p_ \psi, \phi, p_ \phi} = - I_1 \ddot \theta $$ And the RHS is: $$ \begin{align} \frac{\partial \mathcal R}{\partial \theta}\bigg|_{\dot \theta, \psi , p_ \psi, \phi, p_ \phi} & = \frac{\partial }{\partial \theta} \left[\frac{\left( p_ \phi - p_ \psi \cos \theta \right)^2}{2 I_1 \sin ^2 \theta} + \frac{p_ \psi ^2}{2 I_3} - \frac{I_1} 2 \dot \theta ^ 2 + mgl \cos \theta\right] \\ & = \frac{\partial }{\partial \theta} \left[\frac{\left( p_ \phi - p_ \psi \cos \theta \right)^2}{2 I_1 \sin ^2 \theta} + mgl \cos \theta\right] \\ & = \frac{\partial V_{eff}}{\partial \theta} \end{align} $$