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Help on physical meaning

  1. May 15, 2008 #1
    Hello all,

    Could anyone please define the physical meaning of [Laplacian(f) = 0; f is a potential function of a vector field] ..

    I don't know whether it's easy or not, but I'm a noob in vector analysis, so I thought I'd better ask :)

  2. jcsd
  3. May 15, 2008 #2
    What function has 0 as its Laplace transform? Well, f=0 will work, and the inverse transforms are unique, so f=0 is the only solution. Or am I totally wrong?
  4. May 15, 2008 #3


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  5. May 15, 2008 #4
    Oops, my bad. Ignore everything I said.
  6. May 15, 2008 #5
    The laplacian of a function at a point, [tex]\Delta f(p)[/tex] measures how much f(p), deviates from the average of f on a small circle surrounding p. This is similar to how the second derivative measures whether a function of a single variable is concave up or concave down, except extended to functions of many variables. In a sense, it measures how much the function is "curving up" or "curving down" around a point.

    If the laplacian is zero, that means that f(p) is equal to the local average of f. Imagine f as a 2D surface, and (px,py,f(p)) is a point on the surface - If the surface is curving up in one direction around p, it must be curving down in another direction. Thus functions where laplacian f is zero everywhere are ones where every point looks like a saddle point.

    As a result, when the laplacian is zero, f can have no local maxima or minima - if f had a local maxima at q, then f(q) > average of f around q, which would make the laplacian nonzero.
  7. May 22, 2008 #6
    Thank you all ..

    maze, I don't quite get it, but I understand a little bit of what you're getting at. That's satisfactory for now .. finals start in two days so I will investigate in this later.

    Would appreciate it though if you, or anyone else could provide a graphical example or something to clarify it more.
  8. May 24, 2008 #7
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