A problem with non evident first integral

[wrobel]

The following problem we considered with the students. Perhaps it would also be interesting for PF

A homogeneous rod can rotate freely in a plane about its (fixed) center of mass O . The corresponding moment of inertia is equal to J. Two identical particles of mass m can slide along the rod freely. Each particle is connected with the point O by a spring of stiffness k. The relaxed length of both springs is equal to zero.
This system has 3 degrees of freedom: the angle of rod's rotation $\varphi$ and the distances $x,y$ from the particles to the origin O. Two integrals are obvious: the energy and the angular momentum. But there is a third less banal integral: see the attachment.
This observation allows to reduce the system to one degree of freedom and then study it by means of the effective potential.

 

[vanhees71]

If I see it right, the coordinates $\psi$ and $\varphi$ are cyclic. Then $L$ is not explicitly time dependent, so that $H$ (energy) is conserved either. So you've three first integrals and an integrable system, right?

 

[wrobel]

exactly

 

[strangerep]

OK, this looks like an interesting exercise to dust off what's left of my brain...

I'll start by tidying it up a bit. The Lagrangian is
$$L ~:=~ \frac12 J \dot\phi^2 ~+~ \frac12 m \left(\dot r^2 + r^2 \dot\psi^2 + r^2 \dot\phi^2\right) ~-~ \frac12 k r^2 ~.$$ I compute the canonical momenta to be $$p_r =
m \dot r ~,~~~~ p_\phi = (J + m r^2) \dot\phi ~,~~~~ p_\psi = m r^2 \dot\psi ~.$$ and I compute the Hamiltonian as $$H ~=~
\frac{p_r^2}{2m} ~+~ \frac{p_\phi^2}{2(J+mr^2)}
~+~ \frac{p_\psi^2}{2mr^2}
~+~ \frac12 k r^2 ~.$$ From the Hamiltonian form of the EoM one quickly finds that $p_\phi$ and $p_\psi$ are constants.

I compute the Lagrangian (E-L) equations of motion to be $$
\ddot r ~=~ r(\dot\phi^2 + \dot\psi^2) + \frac{kr}{m} ~~=:~ w_r ~,$$$$
\ddot\phi ~=~ -\, \frac{2mr \dot r \dot\phi}{J+mr^2} ~~=:~ w_\phi ~,$$$$
\ddot\psi ~=~ -\, \frac{2\dot r \dot\psi}{r} ~~=:~ w_\psi ~.$$To find nontrivial first integrals (a.k.a constants of the motion) systematically we need the canonical on-shell total time derivative operator: $$A ~:=~
\frac{\partial}{\partial t} ~+~ \dot q^a \frac{\partial}{\partial q^a}
~+~ w^a \frac{\partial}{\partial \dot q^a} ~,$$where the $q_a$ are $r,\phi,\psi$, etc. First integrals are functions $f(t,r,\phi,\psi,\dot r, \dot\phi, \dot\psi)$ of the phase space variables satisfying $$A \, f ~=~ 0 ~.$$[I'll continue tomorrow. Have I made any mistakes so far?]

 

[vanhees71]

That looks good, but why are you looking for more "first integrals". You already have 3 independent ones, i.e., $p_{\phi}$, $p_{\psi}$, and $H$. From this you can find an effective 1D equation of motion for $r$.

 

[strangerep]

That looks good, but why are you looking for more "first integrals". You already have 3 independent ones, i.e., $p_{\phi}$, $p_{\psi}$, and $H$. From this you can find an effective 1D equation of motion for $r$.

I thought that was the point of the exercise -- to find all the (functionally independent) 1st integrals, not just the obvious ones. I presumed this is like in the Kepler problem where one has the additional 1st integral known as the Laplace-Runge-Lenz vector, which is not all obvious from the Kepler Lagrangian.

Moreover, I'd intended to (try and) find all the dynamical symmetries of the problem, even though that was not explicitly asked for in the problem statement. Again, this is like in the Kepler problem where one finds a dilation-like symmetry, responsible for Kepler's 3rd law, yet does not correspond to any conserved quantity.
@wrobel: if I've misunderstood your intent in this problem, please let me know.

 

[vanhees71]

That's of course an interesting exercise too!

 

[wrobel]

if I've misunderstood your intent in this problem, please let me know.

I just wanted to draw your attention that this system has two trivial first integrals: energy and momentum; but it has another integral that does not follow from general theorems of dynamics. It happens sometimes but as for me every such a case is a little miracle

 

[strangerep]

I just wanted to draw your attention that this system has two trivial first integrals: energy and momentum; but it has another integral that does not follow from general theorems of dynamics.

Which theorem(s)? Noether? Other? Do you include the LRL vector of the Kepler problem in this category?

Probably I must finish working it out before I'll see your meaning properly.

 

[strangerep]

I'll post a continuation of my detailed computations tomorrow (it's too late here now). But the result I got for the unexpected 1st integral is $$\frac{(J+mr^2)^{J+mr^2}}{(mr^2)^{mr^2}}$$which looks too utterly weird to let me think it could possibly be correct.

Has anyone else besides @wrobel tried to solve this yet?

 

[vanhees71]

Which theorem(s)? Noether? Other? Do you include the LRL vector of the Kepler problem in this category?

Probably I must finish working it out before I'll see your meaning properly.

Noether's theorem works in "both directions": If you have a one-parameter symmetry group there's a conservation law, and the conserved quantity is the generator of the symmetry and if there's a conserved quantity it defines a one-parameter symmetry group being its generator.

The Runge-Lenz vector in the (non-relativistic) Kepler problem is an example. Depending on the three cases $E<0$ (bound, elliptic orbits) the symmetry group it generates together with angular momentum (which generates the rotation group of course) is and SO(4), for $E=0$ (unbound parabolic orbits) it creates a Galilei group, and for $E>0$ (unbound hyperbolic orbits) it generates the Lorentz group $\mathrm{SO}(1,3)^{\uparrow}$. The latter two have of course nothing to do with the spacetime symmetries of Newtonian or special relativistic physics but are the "dynamical symmetries" of the non-relativistic Kepler problem.

 

[vanhees71]

I'll post a continuation of my detailed computations tomorrow (it's too late here now). But the result I got for the unexpected 1st integral is $$\frac{(J+mr^2)^{J+mr^2}}{(mr^2)^{mr^2}}$$which looks too utterly weird to let me think it could possibly be correct.

Has anyone else besides @wrobel tried to solve this yet?

I've not tried it yet...

 

[wrobel]

unexpected 1st integral is (J+mr2)J+mr2(mr2)mr2which looks too utterly weird to let me think it could possibly be correct.

So you claim that r is a first integral. Noway!

 

[strangerep]

So you claim that r is a first integral.

No, I don't "claim" that. I merely wrote down what I had at the end of yesterday, realizing I must surely have cocked it up somewhere.

Noway!

Indeed. A few minutes after I arose this morning and looked at it again, I saw my mistake...

I'll keep going today...

 

[strangerep]

I get the following 1st integrals: $$ p_\phi ~=~ (J + m r^2) \dot\phi ~,$$$$ p_\psi ~=~ m r^2 \dot\psi ~,$$$$E ~=~ \frac12 m\dot r^2 ~+~ \frac{p_\phi^2}{2(J+mr^2)} ~+~ \frac{p_\psi^2}{2m r^2} ~+~ \frac{k r^2}{2} ~.$$But I can't find any more.

Is there really another one?

 

[vanhees71]

Good question. It may be, but you don't need it, because with the 3 standard first integrals, you've already found you have reduced the EoM. to an effective 1D equation of motion in an effective potential, and that can be solved up to a quadrature.

 

[strangerep]

Good question. It may be, but you don't need it, because with the 3 standard first integrals, you've already found you have reduced the EoM. to an effective 1D equation of motion in an effective potential, and that can be solved up to a quadrature.

But knowing all the dynamical symmetries of a problem might allow features of the solution to be revealed more easily than solving the EoM. E.g., in the Kepler problem, one gets a very inconvenient transcendental equation for the orbit. But besides the symmetries associated with conservation of energy, angular momentum, and the LRL vector, one also finds a (non-conserved) dilation-like symmetry that is responsible for Kepler's 3rd law. One finds this without exhaustively solving the EoM.

I don't know if that sort of thing is applicable to the current double spring-loaded beads on a rotating rod problem. I suspect not, but I'd like to know for sure.

 

[vanhees71]

For the orbit you get a simple equation of a shifted harmonic oscillator, leading to the ellipse, parabola, or hyperbola, dependening on the initial conditions. The additional symmetry explains why you have closed orbits, which is a very special case, since usually for central-potential problems not all bound orbits are closed. Only the symmetric harmonic oscillator and the $1/r$ potential have the feature that all bound orbits are closed.