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A paradox about Maupertuis' principle (symplectic viewpoint)

  1. May 27, 2010 #1

    mma

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    Maupertuis' principle states in symplectic formulation that the integral of the tautological 1-form is extremal among its integrals on other phase space curves on the given level set of the Hamiltonian, connecting the same starting end ending fiber. Specifically, for closed phase trajectories this is equivalent of the extremallity of the symplectic area of a surface expanded on this closed curve compared to the simplectic area of the surfaces expanded on other closed phase space curves on the same level set and passing across two different fibers from the fibers intersected by the phase trajectory.

    For example, in the case of a 2-dimensional harmonic oscillator, the Hamiltonian is:

    [tex]H=\frac{1}{2}(x^2+y^2+p_x^2+p_y^2)[/tex].​

    The level sets are spheres with radius R = 2H.

    Now take a level set belonging to R = 1, and the following two points on this level set:

    a = (-1, 0, 0, 0) and b = (1, 0, 0, 0).​

    The reduced action integral (i.e., the integral of the tautological 1-form) on curves (cos t, sin t, 0, 0) and ( cos t, 0, sin t, 0) both will be extremal because the first circle lies in an isotropic plane while the second one in a symplectic plane, hence the action integral (i.e. the symplectic area of the disk bounded by these curves) are 0, and π respectivelly. The firs one is a minimum, while the second one is a maximum. And here comes the paradox. Arnold proves that the Mapertuis' principle yields unique solution (Arnold: Mathematical Methods of Classial Mechanics, p.244).

    Could anybody tell me, what is the resolution of this paradox?
     
  2. jcsd
  3. May 27, 2010 #2

    mma

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    Sorry, forget it. 0 isn't extremal. The symplectic area of the unit circle is between -π and π, and not between 0 and π.
     
  4. Jun 1, 2010 #3

    mma

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    Of course I mean unit disk and not unit circle. By the way, it's okay that among the disks bounded by great circles of the unit sphere, the value of -π and π are extremal, but is there a simple way to prove that they are also extremal among the symplectic areas of all surfaces bounded by an arbitrary closed curve on the unit sphere? (of course there is a not too simple way to prove: the Maupertuis principle itself. But it would be nice to prove this independently, because so we would get a new proof of the Mapertuis' principle for the special case of closed phase trajectories of a 2-dimensional harmonic oscillator)
     
    Last edited: Jun 1, 2010
  5. Jun 10, 2010 #4

    mma

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    I have doubts that it is true at all. Symplectic area is similar to the magnetic flux of Maxwell's electrodynamics. Let's consider this: take a solenoid (connected its wire-ends to each other to form a closed circuit) and blow up it to a sphere shape. Then the wire of the solenoid will form a closed curve on the sphere and the flux evidently increases with the number of the loops of the solenoid and there is no upper limit. What is the difference if i take the integral of the symplectic form instead of magnetic flux?
     
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