Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Definition of Curl. Can anyone derive the gradient operator?

  1. Jul 25, 2010 #1
    "Definition" of Curl. Can anyone derive the gradient operator?

    Can anyone prove why this equality is true?
    http://en.wikipedia.org/wiki/Curl_(mathematics)#Definition

    Wikipedia says it is defined, however that's BS since the gradient operator was already defined so this needs to be proven. Either you take this for a definition and prove that the little "inverted triangle" is a derivative operator, or you prove the equality and don't call it a definition.

    I can't tell how to go about proving that differentiating a vector field with a weird determinant is EQUAL to the loop integral of F*dr divided by the area enclosed (as the are goes to zero).

    Its probably not hard, the cross product comes out of the "moment" of the field about a point, however I don't quite see how the derivative comes in.
     
  2. jcsd
  3. Jul 25, 2010 #2

    Char. Limit

    User Avatar
    Gold Member

    Re: "Definition" of Curl. Can anyone derive the gradient operator?

    Well, this is a slight abuse of notation, but here's another way that the operator is defined.

    [tex]\nabla = \frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k}[/tex]

    Obviously, this can be generalized to higher dimensions, and as you can see, this only makes sense when cross-producted with a vector...

    at least, I think that's how it goes.
     
  4. Jul 25, 2010 #3

    lurflurf

    User Avatar
    Homework Helper

    Re: "Definition" of Curl. Can anyone derive the gradient operator?

    Why is definition in quotes? When was the gradient operator was already defined? What does gradient operator have to do with curl? That definition of curl is senseable and standard, though other definitions are possible. Just because we write
    grad(something)=∇(something)
    and
    curl(something)=∇×(something)
    does not mean that
    curl(something)=∇×(something)=∇(×something)=grad(×something)
    or
    curl(something)=∇×(something)=∇(×)(something)=grad(×)(something)
    are valid as ×something and grad(×) are not meaningful.
    ∇× should be thought of as a symbol for curl, not a gradient of anything.
     
    Last edited: Jul 25, 2010
  5. Jul 25, 2010 #4
    Re: "Definition" of Curl. Can anyone derive the gradient operator?

    Okay then prove that grad X Field is equal to the limiting value of the loop integral F*dr divided by area enclosed by the path, as the area approaches zero.

    When we compute curl we differentiate the vector field using a determinant. How do we know that it gives the same answer as doing a tiny loop integral around the area of interest and dividing by the are enclosed by the loop?

    EDIT: Okay I finally found my book and its proven using stokes' theorem.
     
    Last edited: Jul 25, 2010
  6. Jul 26, 2010 #5
    Re: "Definition" of Curl. Can anyone derive the gradient operator?

    I think you're confused in the meaning of the "upside down" triangle.

    The upside down triangle is the del operator. http://en.wikipedia.org/wiki/Del

    It is invoked in the definition of grad (and div and curl). I think you are thinking that the upside down triangle is grad?
     
  7. Jul 26, 2010 #6

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Re: "Definition" of Curl. Can anyone derive the gradient operator?

    It is certainly important to be able to connect together different definitions, by means of proof!

    What you however should focus on, with the wiki-definition, is that they define something they call "curl" of a fluid as the net circulation contained in a unit area, as that area shrinks to zero.

    Thus, the curl is a measure of the local rotation rate of the fluid.

    That this can also be calculated directly by a swift differentiation operation, rather than by a tricky limiting operation upon an integral, is indeed, one of the many wonders of Stokes' theorem! :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook