A problem with non evident first integral

  • I
  • Thread starter wrobel
  • Start date
  • Tags
    Integral
In summary: I just missing something?I haven't tried it myself, but I've seen a few posts on forums where people have had similar problems. I'm not sure what the problem is, but it seems like it might be related to a conserved quantity that is not actually conserved.
  • #1
wrobel
Science Advisor
Insights Author
1,104
961
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.
 

Attachments

  • rod and springs.png
    rod and springs.png
    14.7 KB · Views: 98
Physics news on Phys.org
  • #2
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?
 
  • Like
Likes wrobel
  • #3
exactly
 
  • Like
Likes vanhees71
  • #4
OK, this looks like an interesting exercise to dust off what's left of my brain... :oldbiggrin:

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?]
 
Last edited:
  • Like
Likes Hamiltonian and vanhees71
  • #5
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##.
 
  • Like
Likes wrobel
  • #6
vanhees71 said:
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.
 
  • Like
Likes vanhees71
  • #7
That's of course an interesting exercise too!
 
  • Like
Likes sysprog
  • #8
strangerep said:
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
 
  • Like
Likes vanhees71
  • #9
wrobel said:
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.
 
  • Like
Likes vanhees71
  • #10
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?
 
  • #11
strangerep said:
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.
 
  • #12
strangerep said:
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...
 
  • #13
strangerep said:
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!
 
  • #14
wrobel said:
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.
wrobel said:
Noway!
Indeed. A few minutes after I arose this morning and looked at it again, I saw my mistake...:doh:

I'll keep going today... :headbang:
 
  • #15
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?
 
Last edited:
  • Like
Likes vanhees71
  • #16
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.
 
  • #17
vanhees71 said:
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.
 
  • Like
Likes vanhees71
  • #18
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.
 
  • Like
Likes wrobel

Related to A problem with non evident first integral

1. What is a first integral in a scientific context?

A first integral is a mathematical expression that remains constant throughout a system's motion. In physics, it is often referred to as a conserved quantity, meaning its value does not change over time. It is an important concept in understanding the behavior of physical systems.

2. How does a non evident first integral differ from a regular first integral?

A non evident first integral is a first integral that cannot be easily derived from the equations of motion of a system. It is not immediately obvious and requires additional mathematical techniques to be identified. In contrast, a regular first integral can be easily found by inspection of the equations of motion.

3. What are some examples of systems with a problem of non evident first integral?

One example is the motion of a pendulum with a damping force. The energy of the pendulum is not conserved, making it difficult to identify a first integral. Another example is the three-body problem in celestial mechanics, where the equations of motion cannot be solved analytically and a first integral is not apparent.

4. Why is a problem with non evident first integral important in scientific research?

Identifying a first integral in a system can greatly simplify its analysis and provide insights into its behavior. However, in systems with a non evident first integral, this is not possible, making it more challenging to understand and predict their behavior. This is a common problem in many scientific fields, including physics, chemistry, and biology.

5. How do scientists deal with a problem of non evident first integral?

Scientists use various mathematical techniques, such as perturbation theory and numerical simulations, to identify and approximate first integrals in systems with a non evident first integral. They also rely on empirical data and experimental observations to gain a better understanding of the system's behavior. Additionally, scientists may try to simplify the system or make certain assumptions to make the problem more tractable.

Similar threads

Replies
13
Views
2K
  • Mechanics
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
31
Views
1K
Replies
3
Views
755
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
927
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Back
Top