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About vector fields

  1. Apr 4, 2014 #1
    Every conservative vector field is irrotational? Every irrotational vector field is conservative?
    Every solenoidal vector field is incompressible? Every incompressible vector field is solenoidal?
     
  2. jcsd
  3. Apr 4, 2014 #2

    micromass

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    Please tell us what you think and give your reasoning.
     
  4. Apr 4, 2014 #3
    I think that every exact form is closed, but not all closed form is exact.

    However, I see irrotational vector field and conservative vector field be treated of same way and incompressible vector field and selenoidal vector field be treated like if they are the same thing too.

    Edit: those are the definitions that I use...
    image.png
     
    Last edited: Apr 4, 2014
  5. Apr 4, 2014 #4

    Ben Niehoff

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    To answer your question, you need to think about de Rham cohomology classes on ##\mathbb{R}^3##.
     
  6. Apr 4, 2014 #5

    Matterwave

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    Before you get off into Ben's post (which is the correct answer of course), you should first relate your original statements with the questions "are all closed forms exact? and "are all exact forms closed?".

    Your four statements in your original post correspond to four the statements:

    Are the following two statements always true?:

    $$\nabla\times\nabla f=0$$
    $$\nabla\cdot(\nabla\times\vec{f})=0$$

    And the statements, if:
    $$\nabla\times\vec{f}=0$$

    Then is it true that:
    $$\vec{f}=\nabla a$$
    For some function a?

    And if:
    $$\nabla\cdot\vec{f}=0$$

    Then is it true that:
    $$\vec{f}=\nabla\times\vec{a}$$

    These 4 statements can be simplified into two statements that I mentioned above (respectively). Why? Because of how the gradient, curl, and divergence can be related to the exterior derivative. Recall the relations:

    $$\nabla f=(df)^\sharp$$
    $$\nabla\times\vec{f}=[\star(d\vec{f}^\flat)]^\sharp$$
    $$\nabla\cdot\vec{f}=\star d(\star\vec{f}^\flat)$$

    Then our first two statements are, are the following always true?:
    $$\nabla\times\nabla f = [\star(ddf)]^\sharp=0$$
    $$\nabla\cdot(\nabla\times\vec{f})=\star d(\star\star d\vec{f}^\flat)=0$$

    We can immediately see that these two statements are really the statement "are all exact forms closed?", which is a true statement by construction. The exterior derivative is constructed in such a way that this is true.

    The other two statements are a little bit more subtle to prove, they are, in the language of exterior derivatives, the statement "are all closed forms exact?" (You can work this out quite easily yourself given the information provided). It is here that one evokes the de Rham cohomology classes. The Poincare Lemma tells us that all closed forms are locally exact. That is in the neighborhood of each point, one can find a function f such that da=0 implies a=df. In order to make this statement global, one has to care about the global topology of the underlying manifold (which is where de Rham cohomology comes into play). The general result is that globally all closed forms are exact if the manifold is contractable (able to be continuously shrunken to a point) which is true on R^3, but might not be true on say a toroid.
     
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