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Prove this vector identity

  1. May 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that:
    [itex]curl(r \times curlF)+(r.\nabla)curlF+2curlF=0[/itex], where r is a vector and F is a vector field.

    (Or letting [itex]G=curlF=\nabla \times F[/itex]
    i.e. [itex]\nabla \times (r \times G) + (r.\nabla)G+2G=0[/itex])

    3. The attempt at a solution
    I used an identity to change it to reduce (?) it to
    [itex](\nabla.G)r+(G.\nabla)r-(\nabla.r)G-(r.\nabla)G+(r.\nabla)G+2G[/itex]
    [itex](\nabla.G)r+(G.\nabla)r-(\nabla.r)G+2G[/itex]

    I'm not sure where to go from here to show that it's equal to zero. At the moment the only approach I know of is to compute all the components an hope they sum up to zero but surely there's another identity that can simplify this a bit further.
     
    Last edited: May 15, 2014
  2. jcsd
  3. May 14, 2014 #2

    Simon Bridge

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    Sub back G=curl.F

    div.curl.F=?
     
  4. May 14, 2014 #3
    0!
    Which gives
    [itex]((\nabla \times F).\nabla)r-(\nabla.r)(\nabla \times F)+2\nabla \times F[/itex]
    or
    [itex](G.\nabla)r-(\nabla.r)G+2G[/itex]

    One term less = a bunch of less components to deal with - I'll try expanding it out now and see where I get.
     
  5. May 15, 2014 #4
    Now I've got:

    [itex](G_1\frac{\partial r}{\partial x}+G_2\frac{\partial r}{\partial y}+G_3\frac{\partial r}{\partial z})-(G\frac{\partial r_1}{\partial x}+G\frac{\partial r_2}{\partial y}+G\frac{\partial r_3}{\partial z})+2G_1+2G_2+2G_3[/itex].

    When I expand out the vectors([itex]\frac{\partial r}{\partial x}[/itex] into [itex]\frac{\partial r_1}{\partial x}[/itex], etc. and [itex]G[/itex] into [itex]G_1, G_2, G_3[/itex]), the diagonal terms cancel, i.e. [itex]G_1\frac{\partial r_1}{\partial x}i, G_2\frac{\partial r_2}{\partial y}j, G_3\frac{\partial r_3}{\partial z}k[/itex] . What I'm left with doesn't look like it sums to zero, however.
     
    Last edited: May 15, 2014
  6. May 15, 2014 #5

    vela

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    Why did you change ##(\vec{r}\cdot\nabla)\vec{F}## into ##(\vec{r}\cdot\nabla)\vec{G}##?

     
  7. May 15, 2014 #6
    Sorry, my mistake. IT should be ##(\vec{r}\cdot\nabla)\nabla \times \vec{F}##
     
  8. May 16, 2014 #7

    vela

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    I think you're supposed to use the fact that ##\vec{r}## is not just any vector but that it's equal to ##\vec{r} = (x, y, z)##.
     
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