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Integrability (meaning)

  1. Jan 18, 2013 #1
    The title is self-explanatory. What is it meant in the physics and maths community by the words integrability and integrable system?
     
  2. jcsd
  3. Jan 18, 2013 #2
    Have you seen an explanation like this:

    http://en.wikipedia.org/wiki/Integration_(mathematics [Broken])

    where they first discuss integrating a smooth function.....

    or is this what really interests you:

    http://en.wikipedia.org/wiki/Integrable_system

    like maybe one of the systems listed at the end of the article??
     
    Last edited by a moderator: May 6, 2017
  4. Jan 18, 2013 #3
    So, is it correct to state in general that: "an integrable system is a system which thanks to certain properties its dynamics is exactly solvable" ?
     
  5. Jan 18, 2013 #4
    You should wait for someone who is more up to date on math and current terminology than I....but I'll give you my 2 cents:

    first, you posted this under Quantum Physics,so if you are looking for a specific answer, check here in the Wikipedia article:

    Quantum integrable systems

    that seems different from you latest post.

    second, You may have to define what 'solvable' means to you because the section in Wikipedia says this:

    General dynamical systems

    and under the 'Chaos' link

     
  6. Jan 18, 2013 #5

    Bill_K

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    No, integrability means: can a given relationship between derivatives be integrated to yield a relationship between functions. For example, given the system

    ∂f/∂x = F(x,y)
    ∂f/∂y = G(x,y)

    does f(x,y) exist? Answer, only if an integrability condition is satisfied: ∂2f/∂x∂y = ∂2f/∂y∂x,

    that is, ∂F/∂y = ∂G/∂x.
     
    Last edited: Jan 18, 2013
  7. Jan 19, 2013 #6
    it is more suited with classical section,integrability of system is classified according to it's holonomicity.In classical dynamics a system which is non holonomic has at least one non-integrable eqn.they look like
    Ʃaidqi +atdt=0
    this eqn should not be a total differential(or can be converted).there are many examples of it.One simple and particular is rolling of a sphere on a rough surface.Point of contact satisfy a non integrable relation.
     
  8. Jan 19, 2013 #7

    vanhees71

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    2016 Award

    That's only half of the truth! Your integrability conditions are sufficient only for simply connected regions in the [itex](x,y)[/itex] plane, where [itex]F[/itex] and [itex]G[/itex] are free of singularities and smoothly differentiable.

    A simple but eluminating example is the potential curl
    [tex]\vec{F}(\vec{x})=\frac{-y \vec{e}_x+x \vec{e}_y}{r^2}.[/tex]
    It's everywhere curl free, except in the origin, i.e.,
    [tex]\partial_x F_y-\partial_y F_x=0,[/tex]
    but it does not have a unique potential in every region in the plane that contains the origin, where the singularity sits.

    Indeed, integrating the vector field along any circle around the origin gives [itex]2 \pi[/itex].

    To make the potential unique, one has to cut the plane by a ray starting from the origin. A standard choice is the negative [itex]x[/itex]-axis. I.e., you take out the points [itex](x,0)[/itex] with [itex]x \leq 0[/itex].

    It's most easy to find the corresponding potential by introducing polar coordinates. Here, we use
    [tex](x,y)=r (\cos \varphi,\sin \varphi)[/tex]
    with [tex]\varphi \in (-\pi,\pi),[/tex]
    which automatically excludes the negative x axis. The function [itex]\vec{F}[/itex] then reads
    [tex]\vec{F}=\frac{\vec{e}_{\varphi}}{r}.[/tex]
    The potential thus can be a function of only [itex]\varphi[/itex], and the gradient reads
    [tex]\vec{F} \stackrel{!}{=}-\vec{\nabla} V(\varphi)=-\frac{1}{r} V'(\varphi).[/tex]
    This gives, up to a constant
    [tex]V(\varphi)=-\varphi.[/tex]
    The potential is indeed unique everywhere except along the negative [itex]x[/itex] axis, along which it has a jump
    [tex]V(\varphi \rightarrow \pi-0^+)=-\pi, \quad V(\varphi \rightarrow -\pi + 0^+)=+\pi.[/tex]
    In Cartesian Coordinates this potential is given by
    [tex]V(\vec{x})=-\mathrm{sign} y \arccos \left (\frac{x}{\sqrt{x^2+y^2}} \right ).[/tex]
     
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