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Vector field uniquely determined by rot/div

  1. Jun 19, 2007 #1
    In physics one often uses the following: If the rotation of a vector field A vanishes, one can write A as the gradient of some scalar field, i.e. [tex]rot(A)=0 \Rightarrow A=\bigtriangledown \Phi[/tex].

    Is this true without further restrictions? If yes: Why?
    Thanks in advance...Cliowa
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  3. Jun 19, 2007 #2


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    well lets see, to write it as a gradient means we need to integrate it to find the function of which ti si the rgadient. for this we need path integartion to be independent of choice of path, in order to get a welld efined integral as a function only of endpoints.

    now it seems to me that rgeens theorem should give you that a vector fielkd with zero rotation, has path integral independent of path. is this too brief?
  4. Jun 19, 2007 #3


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    There is a slight caveat, that you need to be able to continuously transform every closed path in the domain to a point. For example, if a function was defined everywhere where z =/= 0 in R3, then the z-axis isn't in the domain so a path of the form x2+y2=1 can't be shrunk to a point (as it would need to pass through the z-axis... try drawing it out). In cases like this, you can have vector fields with a rotation of 0 that aren't conservative
  5. Jun 20, 2007 #4
    analogous thing for divergence?

    Thanks mathwonk and Office_Shredder, I guess I understand now. It basically comes down to whether the vector field has a simply connected domain.

    Now, if the divergence is zero: can I write my vectorfield as the rotation of something? I wouldn't now how to prove that, since I can't integrate. Gauss theorem tells me that the integral over any surface enclosing a volume must vanish. So what?
  6. Jun 23, 2007 #5
    If the divergence of a vector field is zero, the vector field is equal to the curl of some vector field. Mathematically,


    [tex]\nabla \cdot {\bf B} = 0[/tex]

    then there exists a vector field [itex]{\bf A}[/itex] such that

    [tex]{\bf B} = \nabla \times {\bf A}[/tex]

    This is because [itex]\nabla \cdot (\nabla \times {\bf v}) = {\bf 0}[/itex] for all twice differentiable vector fields. This is an identity: the divergence of a curl is zero. So if the divergence of something is zero, that something equals the curl of something else. From Gauss's Theorem then, you get that the volume integral of the divergence of B is zero which implies that the surface integral of B is independent of the bounding surface (of the volume) chosen to integrate.

    Similarly, the curl of a gradient is identically zero. So if the curl of something is zero, that something equals the gradient of some scalar field.

    I hope I have not made it more confusing.
    Last edited: Jun 23, 2007
  7. Jun 23, 2007 #6

    matt grime

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    You have made it slightly confusing to me, and I know what's going on - what you wrote seems to imply that you can always write B as del.A, and that is not true.

    Since del(curl(A)) is zero for all A, and since vector fields form a vector space, what is true is that the space of things of the form curl(A) is in the kernel of the map del.( ).

    Thus we can define a 'defect group' to be the space


    what we can show is that if we have something simply connected like R^3, then this defect group is trivial. It is in anycase a (de rham) cohomology group, and such ideas are fundamental in maths these days - we measure the obstruction to some construction. This is the basis of homological algebra: (co)homology groups, toda classes, chern classes, thom classes and so on.
  8. Jun 24, 2007 #7
    True, I agree.

    Could you be more precise? I.e.: How could we show this? Btw, you do mean setminus by / (no quotient space involved), don't you?

    Well, I haven't studied that up to now. Is there a way around this for the special case of R^3?

    Thanks alot...Cliowa
  9. Jun 24, 2007 #8

    matt grime

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    I mean the quotien space. If Im(curl)=ker(del), then the quotient is the trivial zero vector space.

    If you don't want to study de rham cohomology, then the R^3 or R^2 case is Stokes or Greens theorem, one of those integral over some object is the integral over the boundary.

    If you are interested in the de Rham cohomology, then you jusy need to work out the cohomology for R^3, which is straightforward by the standard results on cohomology theory. See e.g. Bott and Tu's book on the subject. Or forget about it entirely since it is probably unnecessary at this stage.
  10. Jun 25, 2007 #9
    Hello, I am sorry I didn't fully appreciate this. I know vector calculus from my analysis and physics classes, but I don't know anything about cohomology. Can you please suggest some text where I could learn more about this? I only know that if the curl of a vector field is zero, it can be written as the gradient of a scalar field and if the divergence of a vector field is zero, it can be written as the curl of some vector.
  11. Jun 25, 2007 #10

    matt grime

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    You don't need to know about cohomology - you can just do it by integrating things - in the case of R^3.

    I did suggest a text book earlier - Bott and Tu, one of the Springer Graduate Texts in Mathematics. Perhaps you should read the stuff in the differential geometry/tensor analysis subforum here, since that is what this is.

    There are many fancy ways of writing all this out, but all it comes down to is that some map of vector spaces is surjective, and you actually show this by doing the integration on partial derivatives if you want to do it directly. Other proofs will be indirect and don't illuminate why it's true at all.
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