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Annoying vectors

  1. Apr 16, 2009 #1
    for a constant magnetic field [itex]\vec{B}[/itex] everywhere, [itex]\vec{A}=\frac{1}{2} \vec{B} \times \vec{r}[/itex]

    because (im not going to use vector notation to save time)

    [itex]\nabla \times (B \times r)= (\nabla \cdot r)B + (r \cdot \nabla)B - (\nabla \cdot B)r - (B \cdot \nabla)r[/itex]

    the first term gives 3B
    the third term vanishes
    the fourht term gives -B
    so to get the answer i need the second term to vanish but i can't get it to go away - how do i do this?
     
  2. jcsd
  3. Apr 16, 2009 #2

    dx

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    The second term vanishes because the gradient of a constant vector field is zero. None of the components are changing, so their space derivatives and hence the gradient is zero.
     
    Last edited: Apr 16, 2009
  4. Apr 16, 2009 #3
    how can you take the gradient of a vector - you need index notation yes?

    consider [itex](m \cdot \nabla)r[/itex]
    is this just [itex]m_j \partial_j r_i=m_i[/itex]
     
  5. Apr 16, 2009 #4

    dx

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    The gradient of a vector (a,b,c) just (grad a, grad b, grad c).
     
  6. Apr 16, 2009 #5
    but [itex]\nabla \phi=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y},\frac{\partial \phi}{\partial z})[/itex] so doesn't that give us a nine component vector - is that just a tensor?

    e.g. [itex]\partial_i r_j =\delta_{ij}[/itex]?
     
  7. Apr 16, 2009 #6

    dx

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    Sorry, I misread the equation. The second term is [tex] (r\cdot\nabla) B [/tex], which is

    [tex] \sum_{i = 1}^3 (r_x \partial_x + r_y \partial_y + r_z \partial_z) B_i \mathbf{e}_i[/tex]
     
  8. Apr 16, 2009 #7
    how is that 0 though?

    why can't this be done using eisntein summation convention i.e.
    [itex]r_j \partial_j B_i=...[/itex] i can't get it to go any further?
     
  9. Apr 16, 2009 #8

    dx

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    The derivatives of [tex] B_i [/tex] are zero, so what you just wrote must be zero.
     
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