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The Nabla Operator

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data
    I need to find [itex]\nabla[/itex]f(r). I am given r = |R| where R is a vector, R =(x,y,z). I also have the function f(r) which is a differentiable function of r.

    2. Relevant equations

    So i know [itex]\nabla[/itex](g) = ([itex]\partial[/itex]g/[itex]\partial[/itex]x, [itex]\partial[/itex]g/[itex]\partial[/itex]y, [itex]\partial[/itex]g/[itex]\partial[/itex]z)

    3. The attempt at a solution

    So I've got;

    r=√(x^2 + y^2 + z^2)

    [itex]\nabla[/itex](r) = (2x,2y,2z)

    Do i need to apply [itex]\nabla[/itex] operator again to the above [itex]\nabla[/itex](r)?

    Does [itex]\nabla[/itex]([itex]\nabla[/itex](r)) equal [itex]\nabla[/itex]f(r)?
     
  2. jcsd
  3. Nov 19, 2011 #2

    HallsofIvy

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    No, of course not. For one thing, [itex]\nabla \nabla r[/itex] makes no sense. [itex]\nabla r[/itex] is a vector function and "[itex]\nabla f[/itex] has to be applied to a scalar function. Also, there is NO "f" in that. You would be saying that [itex]nabla f(r)[/itex] is the same for all functions! You need the chain rule:
    [tex]\nabla f(r)= \frac{df}{dr}\nabla r[/tex]
    df/dr will be a scalar function aand [itex]\nabla f[/itex] a vector function so this is scalar multiplication.

    Or, equivalently, replace r in the function with [itex](x^2+ y^2+ z^2)^{1/2}[/itex] and find the gradient.

    What is f?
     
  4. Nov 19, 2011 #3
    Thanks for your reply. So far ive got;

    [itex]\nabla[/itex](r) = (1/r) (x,y,z).

    And then for df/dr = [itex]\partial[/itex]f[itex]_{x}[/itex]/[itex]\partial[/itex]x + [itex]\partial[/itex]f[itex]_{y}[/itex]/[itex]\partial[/itex]y + [itex]\partial[/itex]f[itex]_{z}[/itex]/[itex]\partial[/itex]z

    Im unsure about what df/dr is.
     
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