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Spliting a Vector

  1. Mar 24, 2004 #1
    The question is about the decomposition of any divergenceless vector field, B, in Poloidal and Toroidal parts. It says in one paper that I've been reading
    "Since B is solenoidal, it can be split into toroidal and poloidal parts, BT and Bp: B=curl(Tr)+curlcurl(Pr)"

    I cannot find the way of proving that the scalar potencial P must really exist, I mean, I cannot prove that if we have a vector field divergenceless like B we have to have a decomposition like that.


    This were my steps
    .div (B)=0 therefore B=curl(A)

    .A is the vector potencial and can be decomposed in 2 parts, one parallel to r and other perpendicular to r,i.e., A=Tr+Qxr

    .Then B comes like B=curl(Tr)+curl(Qxr)
    .Now, if Q were irrotacional then Q=grad P and the thing was done(B=curl(Tr)+curlcurl(Pr))

    But how can I prove Q is really Q=gradP???


    I've sent an email to the author and she said that Q is not required to be Q=gradP but isntead it should have the more general form
    Q=gradP +Sr
    Ok, it works fine and I get the final result as I want.
    But again, can I really write Q like Q=gradP +Sr?? It doesnt seem obvious for me...

    If you could give me a hand on this I would apreciate a lot.
     
  2. jcsd
  3. Mar 24, 2004 #2

    matt grime

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    Do you want to know why this is true? I mean the serious reason behind the why? Learn about de Rham cohomology then, the de Rham complex is exact, and that tells you when things lie in the image of grad, say. You might want to explain what all the symbols mean as well: what is Sr, for instance?
     
  4. Mar 24, 2004 #3
    Sorry not making it clear since the beginning;

    T,S,P are scalar functions

    B,A,Q are vector functions

    r stands for radial vector




    Rham cohomology ??? I never heard about it... I dont think it would required such a complex thing, anyway...
     
  5. Mar 24, 2004 #4

    matt grime

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    de Rham cohomology explains these phenomena, there may be an elementary reason, but remember what you're doing requires you to be in R^3.

    Perhaps recall the case of integration - it is only defined up to addition of a constant - that is something that differentiation kills off.
     
  6. Mar 24, 2004 #5
    But so, what do you think about it?

    What's the reason to have that kind ov vector decomposition?!
     
  7. Mar 24, 2004 #6
    What is de Rham cohomology?..where we can get the information regarding the same.please let me know
     
  8. Mar 25, 2004 #7

    matt grime

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    Bott and Tu's book, the name escapes me, is supposed to be good.

    It is related to differential forms and there might be something in the differential geometry forum.
     
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