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Obtaining vector identities

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data
    The vectors F and G are arbitrary functions of position. Starting w/ the relations F x (∇ x G) and G x (∇ x F), obtain the identity

    ∇(F . G) = (F . ∇)G + (G . ∇)F + F x (∇ x G) + G x (∇ x F)


    2. Relevant equations



    3. The attempt at a solution

    I started off with the relation F x (∇ x G) and used the BAC-CAB rule:

    F x (∇ x G) = ∇(F . G) - (G . ∇)F

    so ∇(F . G = (G . ∇)F + F x (∇ x G) which seems to contradict the identity I am supposed to get. What am I doing wrong?
     
  2. jcsd
  3. Mar 3, 2009 #2
    you can't apply the BAC-CAB rule here. That was derived for vectors by assuming that A X B = - B X A

    However with the del operator del X A is a vector, but A X del is an operator.....so there's a world of difference
     
  4. Mar 4, 2009 #3

    lurflurf

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    Homework Helper

    The trouble is commuting an opperator adds a commutator term
    In single variable calculus
    D(uv)=uDv+vDu not uDv
    we can use partial opperators to avoid this
    let a opperant in {} be fixed
    D(uv)=D({u}v)+D(u{v})={u}Dv+{v}Du=uDv+vDu

    ∇(F . G )=∇({F} . G )+∇(F . {G} )
    ∇({F} . G )=Fx(∇xG)+(F.∇)G
    ∇(F . {G} )=Gx(∇xF)+(G.∇)F
    ∇(F . G )=∇({F} . G )+∇(F . {G} )=Fx(∇xG)+(F.∇)G+Gx(∇xF)+(G.∇)F
     
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