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Annoying vectors

  • #1
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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?
 

Answers and Replies

  • #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.
 
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  • #3
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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]
 
  • #4
dx
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The gradient of a vector (a,b,c) just (grad a, grad b, grad c).
 
  • #5
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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]?
 
  • #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]
 
  • #7
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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?
 
  • #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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