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A question about path independence and curl of a vector field

  1. Nov 28, 2004 #1
    If the curl of a vector field is zero, then we can that the vector field is path independent. But there are cases where this is not true, I was wondering how?
    Whats the explanation for this? Thanks in advance for any help.

    - harsh
     
  2. jcsd
  3. Nov 29, 2004 #2

    James R

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    There may be continuity requirements for the field to make the statement absolutely true (from memory it follows from Stokes' Theorem). However, in any physically well-behaved system, that will usually be true.
     
  4. Nov 29, 2004 #3

    arildno

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    If you've got a singularity in your velocity field, the quantity "curl" (i.e, local angular velocity) might be zero everywhere in the domain, yet the circulation around the singularity might still be non-zero.
    As an examle, consider the velocity field associated with a 2-D point vortex:
    [tex]\vec{v}(r,\theta)=\frac{\Gamma}{2\pi{r}}\vec{i}_{\theta}[/tex]
    The curl is then:
    [tex]\nabla\times\vec{v}=\vec{i}_{r}\times\frac{\partial}{\partial{r}}(\frac{\Gamma}{2\pi{r}}\vec{i}_{\theta})+\vec{i}_{\theta}\times\frac{1}{r}\frac{\partial}{\partial{\theta}}(\frac{\Gamma}{2\pi{r}}\vec{i}_{\theta})=-\frac{\Gamma}{2\pi{r}^{2}}(\vec{i}_{r}\times\vec{i}_{\theta}+\vec{i}_{\theta}\times\vec{i}_{r})=\vec{0}[/tex]

    However, the circulation on a (circular) path of radius R containing the origin is:
    [tex]C=\int_{0}^{2\pi}\vec{v}\cdot\vec{i}_{\theta}Rd\theta=\Gamma[/tex]

    The interpretation is as follows:
    Place a small paddlewheel with a vertical pole in it some distance from the vortex (the origin). The paddle-wheel will of course circle about the origin, BUT IT WILL NOT ROTATE ABOUT ITS OWN AXIS, due to zero curl..
     
    Last edited: Nov 29, 2004
  5. Nov 29, 2004 #4
    I believe this is usually stated by saying that the path along which you integrate must all be within simply connected region in which the vector function is continuous with continuous first derivatives.
    A region is simply connected if any loop you can form in it can be 'capped' by a plane which does not leave the region.
     
  6. Nov 29, 2004 #5

    mathwonk

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    the simplest example (and the simplest way to describe it) is dtheta, the angle form about the orign.

    here is an answer to the same question from another thread. (if they post multiple questions, i hope i may be forgiven for posting multiple answers.)

    there is a special class of integrals, i.e. differential forms, that DO give the same integral over any path joining the same two points. these are called "exact" differentials, and are precisely those of form df for some function in that region, i.e. a gradient.

    another related concept is of a "closed" differential, one such that its curl is zero. these are in fact the same as the exact differentials in any "simply connected" region.


    thus to measure how far a region is from being simply connected, one can ask how many closed differentials in it fail to be exact.

    for example, if we remove n points from the plane, there will still be exactly an n dimensional vector space of closed forms in that region, after we consider all exact forms to be zero.

    \this measuring device is a big tool in topology called derham cohomology.

    work done by gravity for example is exact so does not depend on the path taken by the object.
     
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