1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook