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Scalar Function on a Surface

  1. May 1, 2008 #1
    Hi guys and gals

    This is a conceptual question. Lets say I have a scalar function, [tex]f(x,y,z)[/tex] defined throughout [tex]\mathbb{R}^3[/tex]. Further I have some bounded surface, S embedded in [tex]\mathbb{R}^3[/tex].

    How would I find the function f, defined on the surface S?

    Would it be the inner product of f and S, [tex]<f|S>[/tex] or a functional composition like [tex]f \circ S[/tex]?
  2. jcsd
  3. May 1, 2008 #2


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    f is a scalar, so inner product of S and f makes no sense. I don't know what you have in mind by functional composition
  4. May 1, 2008 #3
    you mean you want parameterize f by s? as in restrict f to s? like for the purposes of a surface integral?
  5. May 2, 2008 #4
    From what I understand the inner product <f|g> is

    The mistake I made was to think that they are scalar functions aswell even though f and g are complex functions. Sorry about that. The closest thing I've come to inner products for functions was the orthonormality of the basis functions for Fourier series. :blushing:

    This is exactly what I had in mind. Sorry I should have been more explicit where I was going with it.

    I understand what we are doing if we have a vector field [tex]\textbf{F}[/tex] and want to find out how it permeates (eg. flux through a surface) a surface S but dotting it with the unit normal of the surface and integrating on the surface. This is actually what made me think of the inner product:

    Thanks for the replies :smile:
  6. May 2, 2008 #5


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    That is the "inner product" only if you are thinking of f and g as vectors in L2.

    You have a function, f(x,y,z), and are given a surface S. You don't say how you are "given" the surface but since it is two dimensional, it is always possible to parameterize it with two variables: on S, x= x(u,v), y= y(u,v), z= z(u,v). Replace x, y, and z in f with those: f(x(u,v),y(u,v),z(u,v).

    For example, suppose you have the parabolic surface z= x2+ 2y2 and some function f(x,y,z). Then you can take x and y themselves as parameters and, restricted to that surface, your function is f(x,y,x2+ 2y2).
  7. May 2, 2008 #6
    Thanks HallsofIvy that cleared things up :)
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