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Operation that will undo a curl operation?

  1. Feb 16, 2004 #1
    Does anyone know if there is an operation that will undo a curl operation?
     
  2. jcsd
  3. Feb 16, 2004 #2

    matt grime

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    No, curl isn't bijective.
     
  4. Feb 16, 2004 #3
    That isn't good news. Hmm, if it isn't bijective, what is it? I'm going to figure this would have to do with the fact that the matrix operator for curl is a singular matrix?
     
  5. Feb 16, 2004 #4

    matt grime

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    Why ought it to be bijective? A simple explanation from a decent differential geometer would be handy right now, but I'm not one of those.

    It is not bijective as there are plenty of things with curl equal to zero (irrotational fields)

    As the matrix operator is not a linear map on R^3, and is just a matrix of partial derivatives, I'm not sure how you'd even begin to define singularity.
     
  6. Feb 16, 2004 #5
  7. Feb 16, 2004 #6

    matt grime

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  8. Feb 16, 2004 #7

    Hurkyl

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    The other problem is that the curl of a vector field is always divergence free, so unless your field is divergence free, you cannot find an anticurl.


    Nonuniqueness of anticurl isn't a problem; just like with ordinary antiderivatives you return a class of solutions. In this case, you'd have a base solution plus an arbitrary irrotational field. Sure, it's a lot messier than just adding an arbitrary constant, but it can be done.


    Anyways, there are almost certainly better ways to do it, but you can find an anti curl just by looking at the partial differential equations. The procedure is pretty much in the same spirit as when you're looking for the antigradient of an exact vector field.

    (NOTE: I'm assuming that everything involved is well-behaved)

    Suppose that you're given this vector field:

    [tex]F = <0, y e^z - z \cos (xz), 2x - e^z>[/tex]

    (exercise: show that this is divergence free)

    You want to find [itex]A = <f, g, h>[/itex] such that [itex]\nabla \times A = F[/itex].

    First, note that [itex]f[/itex] can be anything, because our answer is only unique up to adding an arbitrary irrotational field. I'm going to hunt for the most generic answer, but if you only need one answer, you could simplify life by setting [itex]f = 0[/itex].

    From the definition of curl, we have:

    [tex]
    \begin{equation*}\begin{split}
    h_y - g_z &= 0 \\
    f_z - h_x &= y e^z - z \cos (xz) \\
    g_x - f_y &= 2x - e^z
    \end{split}\end{equation*}
    [/tex]

    So now solve.

    [tex]
    \begin{equation*}\begin{split}
    g_x &= f_y + 2x - e^z \\
    g &= x^2 - x e^z + \int f_y \, dx \\
    h_x &= f_z - y e^z + z \cos (xz) \\
    h &= \sin (xz) - y e^z + \int f_z \, dx
    \end{split}\end{equation*}
    [/tex]

    Now, remember that the "constants" of integration here are arbitrary functions of both y and z. You now use the third equation, [itex]h_y - g_z &= 0[/itex] to fix what those "constants" may be.

    One particular solution is [itex]A = <0, x^2 - x e^z, \sin (xz) - x y e^z>[/itex], so the general solution can be found by adding an arbitrary irrotational field to this.


    (incidentally, the solution I used to generate the initial vector field was [itex]A = <y e^z, x^2, \sin (xz)>[/itex])
     
    Last edited: Feb 16, 2004
  9. Feb 16, 2004 #8

    matt grime

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    Strictly speaking it is just the non-injectivity that matters: undo is a vague term. For a strict inverse one would require bijectivity, however the word undo could be interpreted to mean F undoes G if

    FG(x) =x,

    without requiring F to be defined on the whole of the set in which G maps to, merely defined on the image of G.

    Anyway, curl(f) =0 for all constant fields (and a whole lot more besides) so even that relaxation won't do.

    edit, this was written as Hurkyl's post appeared. the phrase 'so it's only injectivity that counts' is not a contradiction of his post, merely my presonal opinion about interpreting the word 'undo', which means inverse in my mind.
     
    Last edited: Feb 16, 2004
  10. Feb 16, 2004 #9
    Hmm, very intriguing. In the case I am looking at, the vector field is indeed divergence free. And I think in this case the irrotational fields will probably wind up canceling out, or not having much of a contribution at any rate, but that does indeed help a bit.

    My next question is, suppose you had a vector equation of the form

    [tex]u_t = curlu[/tex]. If I were to find the general solution to the case where [tex]\nabla \times A = u[/tex], would then the property of [tex]u_t \times A = u[/tex] hold?

    Note: the vector [tex]u[/tex] is such that each component of [tex]u[/tex] is defined by the form [tex]u_1(x,y,z,t)[/tex] and so on for [tex]u_2[/tex] and [tex]u_3[/tex].
     
  11. Feb 16, 2004 #10

    Hurkyl

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    FYI, the (generally very large) field

    [tex]
    F = <e^{x+y+z}, e^{x+y+z}, e^{x+y+z}>
    [/tex]

    is irrotational, as is the field

    [tex]
    F = <10^{10^{10}}, 10^{10^{10}}, -10^{10^{10}} >
    [/tex]

    You can't assume if you get one solution that it will be close to the desired solution.
     
  12. Feb 16, 2004 #11

    matt grime

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    ignore the t thing

    you want to know what

    [tex](\nabla\wedge u) \wedge A[/tex] is when

    [tex]\nabla\wedge A = u[/tex]

    how are you with summation convention?

    [tex](\nabla\wedge u) \wedge A)_r = \epsilon_{riq}\epsilon_{ijk}\partial_ju_kA_q = (\delta_{qj}\delta_{rk}-\delta_{qk}\delta_{rj})\partial_ju_kA_q[/tex]

    which simplifes to

    [tex]\partial_qu_rA_q - \partial_ru_qA_q[/tex]

    so you want to know what

    [tex]u(\nabla.A) - \nabla(u.A)[/tex]

    is if it is also true that [tex] \nabla \wedge A = u [/tex]
     
    Last edited: Feb 16, 2004
  13. Feb 16, 2004 #12

    matt grime

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    Ok, it's coming back to me now.

    [tex]u.A = (\nabla\wedge A).A=0[/tex] cos its the product of symmetric and anti symmetric things,

    hence we only have to examine [tex]\nabla . A[/tex] I see no reason why this should be 1.
     
  14. Feb 16, 2004 #13

    Hurkyl

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    Have you tried any examples?
     
  15. Feb 16, 2004 #14
    Touche'. I'll try working on it some more now that classes are done for the day.
     
  16. Feb 17, 2004 #15
    One thing of interest might be Helmholtz's theorem:

    If the divergence and the rotation of a vector field, which varies asymptotically as at least r^(-2) and is sufficiently smooth, is given, then there exists a unique vector field (barring an additive constant) with that particular curl and divergence.

    You can actually write it down, but I'm I don't have time to type it up in TeX right now. Watch this space!
     
  17. Feb 17, 2004 #16

    Hurkyl

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    Last edited: Feb 17, 2004
  18. Feb 18, 2004 #17
    That's what I meant with the correct asymptotic behaviour. And a really important pesky thing it is, if you can write down a gazzilion forms for a constant B-field that doens't vanish at infinity.
     
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