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: Inverting the gradient operator

  1. Sep 4, 2009 #1
    1. The problem statement, all variables and given/known data

    I have a derivation for an equation here:

    Basically, I need to invert the gradient operator, so I have:

    [tex]\nabla B = k_z[/tex]

    k is known and I want to solve for B numerically. How do I get rid of the gradient operator? Do I integrate? If so, in 3D or 1D? With respect to what? Time or space?

    If [tex]k_z[/tex] is a vector that lies completely in the z-axis, does that change how I un-operate the gradient operator on B?

    2. Relevant equations

    [tex]\nabla = (\frac{d}{dx}, \! \frac{d}{dy}, \! \frac{d}{dz})[/tex]

    3. The attempt at a solution

    See the link for the derivation I've done:
  2. jcsd
  3. Sep 4, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    Pick a point and define B(x)=0. Define the value of B(y) at any other point y to be the path integral of k from x to y along ANY path connecting x and y. You do have to check that this definition is independent of the choice of path. How would you check that?
  4. Sep 4, 2009 #3


    User Avatar
    Science Advisor

    So you are saying that
    [tex]\nabla B= \frac{\partial B}{\partial x}\vec{i}+ \frac{\partial B}{\partial y}\vec{j}+ \frac{\partial B}{\partial z}\vec{k}= k_z \vec{k}[/tex]

    Is your "kz" constant or a function of x, y, and z?

    In any case, you have
    [tex]\frac{\partial B}{\partial x}= 0[/tex]
    which says that B does NOT depend on x,

    [tex]\frac{\partial B}{\partial y}= 0[/tex]
    which says that B does NOT depend on y, and

    [tex]\frac{\partial B}{\partial z}= k_z[/tex]
    It should be clear that kz cannot depend on either x or y (because then B would depend on them) so you must have
    [tex]\frac{dB}{dz}= k_z[/tex]
    B is just the anti-derivative of kz.
  5. Sep 4, 2009 #4

    Yes, that is exactly what I'm saying. The magnetic field, B, varies only along z, so it is constant in x and y. Your formalism for it makes a lot of sense to me now.

    Thank you for your help!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook