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Div B = 0 implies B = curl A

  1. Mar 6, 2010 #1
    Is this a theorem, that for a vector field B satisfying

    div B = 0 everywhere

    then there is a vector field A such that B = curl A? If so, is it hard to prove? Of course, the converse is obviously true.
     
  2. jcsd
  3. Mar 7, 2010 #2

    quasar987

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    The answer is that it depends on the domain of definition of the vector field. For vector fields defined on all of R³, it is true that there exists a vector field A such that B = curl A. If B is merely defined on R³\{0} (and cannot be extended smoothly to all of R³), there may not exist such a vector field.

    You could also have asked, given that curlA=0, does there exist a function f such that A=grad(f) ?

    The general setting for answering questions such as these concerning the existence of a "primitive" in some sense or another is the subject of the de Rham theory of differential forms. In this context, to ask whether every smooth vector field of vanishing divergence defined on some open subset U of R³ is equal to the curl of some other vector fields is the same as asking whether the second de Rham cohomology group of U is the trivial group: [itex]H^2_{\mbox{de Rham}}(U)=0[/itex]. Now, it is known that the de Rham cohomology groups are homotopy invariants: two homotopy equivalent spaces have isomorphic de Rham cohomology groups. In particular, R³ is homotopy equivalent to a point, which have vanishing de Rham cohomology, and so [itex]H^2_{\mbox{de Rham}}(\mathbb{R}^3)=0[/itex], which translated into the fact that for vector fields defined on all of R³ of vanishing divergence, it is true that there exists a vector field A such that B = curl A. But for R³\{0}, there are known counter-examples. One of them translates in the following way in the terms that interests you.

    Let B:R³\{0}-->R³ be the vector field

    [tex]B(x,y,z)=\frac{x\hat{x}+y\hat{y}+z\hat{z}}{(x^2+y^2+z^2)^{3/2}}=\frac{r}{|r|^3}[/tex]

    Suppose that B=curl(A) for some vector field A:R³\{0}-->R³. Then, by Stoke's theorem, we would have

    [tex]\int_{S^2}B\cdot\hat{r}dA=\int_{S^2}\mbox{curl}(A)\cdot\hat{r}dA=\int_{\partial S^2}A\cdot dl = 0[/tex]

    (because the sphere has no boundary: [itex]\partial S^2=\emptyset[/itex]). But, on the other hand, a direct calculation using spherical coordinates gives

    [tex]\int_{S^2}B\cdot\hat{r}dA=\int_0^{2\pi}\int_0^{\pi}\sin(\phi)d\phi d\theta = 4\pi[/tex]

    This is a contradiction that shows that B is not the curl of any vector field A.
     
  4. Mar 8, 2010 #3
    Thank you, quasar987. I didn't know this result had such a deep connection with the topology of the domain of B. deRham cohomology sounds a little over my head for right now, but it is cool to learn that it comes into play here. And you chose a great counterexample. The divergence of r/|r|3 is a commonly used representation of the delta function for 3-space in electrostatics and other places.

    Just to make sure I understand right, it would be possible to find a vector field A of which B=r/|r|3 is the curl if we restrict B's domain to some open subset U of R3 that does not contain the origin, since a closed surface in any such subset can be shrinked to a point without going outside U. Is this right?
     
  5. Mar 8, 2010 #4

    quasar987

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    It is a bit ambiguous to say "an open set that does not contain the origin", but your statement

    it would be possible to find a vector field A of which B=r/|r|3 is the curl if we restrict B's domain to some open subset U of R3 that does not contain the origin

    is correct if by "an open set that does not contain the origin" you mean an open set which is contained in a contractible set not containing the origin.

    For instance, if you restrict the domain to the open "spherical shell" S define by 1<r<2 around the origin, then B still isn't the curl of any A. But if you restrict to the open "spherical shell" S around (3,0,0), then B is the curl of some A, because S is contained in the ball of radius 2 centered at (3,0,0) which is contractible.
     
    Last edited: Mar 8, 2010
  6. Mar 9, 2010 #5
    Ah yes. Thanks for that clarification. I had the right mental picture, but what I wrote was wrong.

    In the meantime I searched the scriptures* and found that in Book II, chapter 12, verse 16, a way is given for constructing a vector field from its curl. This provides a limited proof, namely that a solenoidal vector field defined on some open interval in R3 is the curl of a some vector field in that region. The Author mentions the much stronger result you gave about contractible open sets, but said the proof is "difficult".

    Thanks for the help!



    * Tom Apostol's Calculus
     
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