Action variables in a non inertial system

[l4teLearner]

Homework Statement

Find action angle variables for the following system:

A point particle of mass m is constrained to move on an ellipsoid of equation

$$\frac{\xi^2}{a^2} + \frac{\eta^2+\zeta^2}{b^2}=1$$

where $a > b > 0$. The ellipsoid rotates in space around the y-axis with angular velocity $\omega$. At the instant $t = 0$ the principal axes $\xi$, $\eta$ and $\zeta$ coincide with the axes $x$, $y$ and $z$.

Relevant Equations

After setting $$ξ = a cos θ, η = b sin θ sin ϕ, ζ = b sin θ cos ϕ$$, the kinetic energy of the point is $$T = T_2+ T_1+ T_0$$, where $$T_2= \frac{m}{2}[(a^2sin^2θ + b^2cos^2θ)\dot{θ}^2+ b^2sin^2θ\dot{ϕ}^2]$$

$$T_1= abmω[cos ϕ\dot{θ} − sinθcosθsinϕ\dot{ϕ}]$$

$$T_0= \frac{m}{2}[a^2ω^2cos^2θ + b^2ω^2sin^2θ cos^2ϕ]$$

I guess the first steps in the resolution of the problem are
- to calculate the expression for the conjugate momenta to $θ$ and $ϕ$
- to calculate the Hamiltonian of the problem
- to write the Hamilton-Jacobi equation and see if it is separable

I struggle in finding conserved quantities from which I could calculate the action variables. One constant of motion should be the overall Hamiltonian, as it does not depend on time? What is the other recognizable physical quantity that is conserved in this system (I need two integrals to find two action variables)? The presence of the constant rotation $\omega$ baffles me a bit, because the system θ
becomes non inertial.

thanks

 

[l4teLearner]

I have evaluated the conjugate momenta as $p_\theta=abm\omega cos\phi+m(a^2 sin^2\theta+b^2cos^2\theta)\dot{\theta}$ and $p_\phi abm\sin\theta cos\theta sin\phi+mb^2sin^2\theta \dot{\phi}$

hence the Hamiltonian, after some algebra, is

$$H=\frac{(p_\theta - abm\omega cos\phi)^2}{2m(a^2sin^2\theta+b^2cos^2\theta)}
+
\frac{(p_\phi+abm\omega sin\theta cos\theta \sin\phi)^2}{2mb^2sin^2\theta}
-
\frac{m}{2}\omega^2 a^2 cos^2\theta
-
\frac{m}{2}\omega^2 b^2 sin^2\theta cos^2\phi$$

from here on I cannot see how can I separate variables or find a conserved quantity.
Maybe I need to apply some canonical transformation?

 

[l4teLearner]

I am assuming that the $y$ component of the angular momentum is conserved. This should be my second integral.

I evaluate it as

$$L_y=mab[-\frac{p_\phi sin\theta cos\theta sin\phi + abm\omega sin^2\theta cos^2 \theta sin^2\phi}{mb^2sin^2\theta}+\frac{p_\theta cos\phi-abm\omega cos^2\phi}{m(a^2sin^2\theta+b^2cos^2\theta)}]+m\omega(a^2cos^2\theta+b^2sin^2\theta cos^2\phi)$$

and after I bit of algebra I can rewrite my Hamiltonian as

$$H = \frac{p_\theta^2}{2m(a^2sin^2\theta+b^2sin^2\theta)} - \frac{p_\theta a b m \omega cos\phi}{2m(a^2sin^2\theta+b^2 sin^2\theta)}
+ \frac{p_\phi^2}{2mbsin^2\theta}+
\frac{p_\phi a b m \omega sin\theta cos\theta sin\phi}{2mb^2sin^2\theta} - \frac{\omega L_y}{2}$$

I guess it would have been nicer if the terms non quadratic in the $p$s would disappear completely, but they don't seem to do. In any case I still fail to see how one could evaluate action variables from $L_y$ and $H$ if that is the spirit of the exercise.
Any idea is welcome, thanks

 

[TSny]

Your calculations for the kinetic energy and the Hamiltonian look good to me. The Hamiltonian will be a constant of the motion. For the ellipse to maintain constant $\omega$ about the y-axis, a time-dependent external torque will be required. This torque will generally have nonzero x, y, and z components. So, this torque will cause $E$, $L_x$, $L_y$, and $L_z$ to be time dependent. For this system, the Hamiltonian does not equal $E$ (unless $\omega = 0$). So there is no problem with $H$ being constant but $E$ not being constant.

I don't see any constants of the motion other than $H$. But I don't know how to prove that there aren't any others.

The only examples of finding action-angle variables I've seen are calculated for systems where the Hamiltonian is separable. But, as you noted, $H$ is not separable in $(\theta, p_{\theta})$ and $(\phi, p_{\phi})$. So, I don't know how to obtain the action-angle variables. Hopefully, someone else can help.

--------------------------------------

I used Mathematica to integrate the equations of motion numerically. The system has rich dynamics.

The picture below shows the particle's trajectory for the case where the ellipsoid is not rotating. In this case, $H = E$. The particle moves at constant speed along a geodesic of the ellipsoid. Even when $\omega = 0$, the trajectories can be complicated. The green dot is the starting point and the initial velocity was arbitrarily chosen.

The picture below is for the same initial conditions but for an arbitrarily chosen $\omega \neq 0$.

Now the trajectory over the ellipsoid can be very complicated due to the centrifugal and coriolis forces in the rotating frame of the ellipsoid.

 

[l4teLearner]

Thanks for your answer.
But I am super confused now.
What is $E$ for you? If by $E$ we mean the total mechanical energy measured in the rotating frame, I was under the impression that it is indeed conserved, as the fictitious forces measured (if $\omega$ is constant) are conservative.
As for the conservation of the component of angular momentum along the axis of rotation as measured in the non inertial frame, this should be conserved as well, as long $\omega$ is constant.
So we should have two integrals.
I am basing my statements also on what is written here:

https://physics.stackexchange.com/a/262146/242754

 

[l4teLearner]

Nevermind, that reference is for _linear momentum_, and not _angular momentum_. :) Apparently, the coriolis force that you mentioned causes angular momentum in the non inertial reference frame to be not conserved.

 

[TSny]

Thanks for your answer.
But I am super confused now.
What is $E$ for you? If by $E$ we mean the total mechanical energy measured in the rotating frame, I was under the impression that it is indeed conserved, as the fictitious forces measured (if $\omega$ is constant) are conservative.

Yes, if you define $E$ this way, it is conserved. In this case, $E$ equals the sum of the kinetic energy of the particle (in the ellipsoid frame) and the potential energy associated with the centrifugal force in this frame. Let me call this $E^{\rm rot}$, which denotes energy relative to the rotating frame.

I took $E$ to mean the particle's total energy in the inertial frame. Let me use the notation $E^{\rm in}$ for this. $E^{\rm in}$ is just the particle's KE in the inertial frame. There is no potential energy in this frame. $E^{\rm in}$ is not conserved.

The Hamiltonian in the inertial frame is $H^{\rm in} = p_{\theta} \dot \theta + p_{\phi} \dot \phi - L^{\rm in}$, where $L^{\rm in}$ is the Lagrangian in the inertial frame. $H^{\rm in}$ is conserved. $H^{\rm in} \neq E^{\rm in}$. But, you can show that $H^{\rm in} = E^{\rm rot} = H^{\rm rot}$. Here, $H^{\rm rot}$ is the Hamiltonian in the rotating frame constructed from the Lagrangian in the rotating frame.

Whew! I hope I haven't made things worse.

 

[TSny]

Using the Lagrangian for the rotating frame (with the centrifugal potential energy), the associated Hamiltonian has a much simpler form than the Hamiltonian in the inertial frame. The momenta conjugate to $\theta$ and $\phi$ in the rotating frame are different than for the inertial frame. It should be easier to find the action-angle variables associated with the Hamiltonian in the rotating frame than for the Hamiltonian in the inertial frame.

 

[l4teLearner]

Thanks again for your patience.
Before I dive into the calculations again, I would have two more (requests for) clarification to make.

1. I think the statement of the problem was a bit misleading in this sense: $\xi, \zeta, \xi$ and the parametrization with $\phi,\theta$ are coordinates/manifold defined in the (rotating) body frame relative to the ellipsoid, yet the formula given in the statement of the problem $T_0+T_1+T_2$ for kinetic energy is $E^{in}$. It has indeed two components depending on $\omega$. Correct?

2. When calculating $H^{rot}$, still keeping the same parametrization, $KE^{rot}$ should be just
$T_2$ as the other two terms vanish as we don't consider $\omega$ in the kinetic part. On the other hand, for the potential part of the Lagrangian, I should consider a generalized potential as in https://physics.stackexchange.com/a/498249/242754
to accoun for the Coriolis force. Correct?

thanks again!

 

[TSny]

Thanks again for your patience.
Before I dive into the calculations again, I would have two more (requests for) clarification to make.

1. I think the statement of the problem was a bit misleading in this sense: $\xi, \zeta, \xi$ and the parametrization with $\phi,\theta$ are coordinates/manifold defined in the (rotating) body frame relative to the ellipsoid, yet the formula given in the statement of the problem $T_0+T_1+T_2$ for kinetic energy is $E^{in}$. It has indeed two components depending on $\omega$. Correct?

Yes.

2. When calculating $H^{rot}$, still keeping the same parametrization, $KE^{rot}$ should be just
$T_2$ as the other two terms vanish as we don't consider $\omega$ in the kinetic part.

Yes

On the other hand, for the potential part of the Lagrangian, I should consider a generalized potential as in https://physics.stackexchange.com/a/498249/242754
to accoun for the Coriolis force. Correct?

Yes, that's correct. In post #8 of this thread, I forgot the need for the Coriolis term in $H^{\rm rot}$. Now I'm not sure there is any advantage in working with $H^{\rm rot}$. With the Coriolis term included, it might not be any simpler than working with $H^{\rm in}$. However, you could try it and see.

 

[TSny]

Regarding action variables, I'm only familiar with examples where the variables ($\theta$ and $\phi$ in our case) are periodic: oscillations or orbital motion. The general trajectory on the rotating ellipsoid seems to me to be too complicated for the use of action variables. But, I could be wrong.

We can get oscillations or orbits for the rotating ellipsoid if the trajectory lies in the x-z plane.

On the left, the particle oscillates on the blue arc in the x-z plane. On the right, the particle orbits around the ellipse in the x-z plane. Since $\phi$ is fixed at 0 here, $\theta$ is the only variable. So we would have only one action variable $J_{\theta}$.