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Differentiable Scalar Field

  1. Mar 1, 2012 #1
    How to prove that [itex]\nabla[/itex] x ([itex]\phi[/itex][itex]\nabla[/itex][itex]\phi[/itex]) = 0?
    ([itex]\phi[/itex] is a differentiable scalar field)

    I'm a bit confused by this "differentiable scalar field" thing...
     
  2. jcsd
  3. Mar 1, 2012 #2

    HallsofIvy

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    Are you saying that you do not know what a "differentiable scalar field" means? A "scalar" is simply a number, rather than a vector. This is just saying that [itex]\Phi(x, y, z)[/itex] is a (differentiable) function that returns a number for each point (x,y,z)- exactly the kind of function you are used to working with!

    And, of course, [itex]\nabla \Phi[/itex] is the vector function
    [tex]\frac{\partial\Phi}{\partial x}\vec{i}+ \frac{\partial\Phi}{\partial y}\vec{j}+ \frac{\partial\Phi}{\partial z}\vec{k}[/tex]

    so that [itex]\Phi\nabla\Phi[/itex] is that vector multiplied by the number [itex]\Phi[/itex]:
    [tex]\Phi\frac{\partial\Phi}{\partial x}\vec{i}+ \Phi\frac{\partial\Phi}{\partial y}\vec{j}+ \Phi\frac{\partial\Phi}{\partial z}\vec{k}[/tex]
    so is a "vector valued" function- it returns that vector at each point (x, y, z).

    Finally,
    [tex]\nabla\times(\Phi\nabla\Phi)[/tex]
    is the "curl" of that vector valued function.
     
    Last edited: Mar 1, 2012
  4. Mar 1, 2012 #3

    I like Serena

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    Homework Helper

    Hi!

    The way to proof such an expression is to write it out into x, y, and z components and simplify it (as HallsofIvy is suggesting).

    As an alternative you can use the curl identities that you can find for instance on wiki:
    http://en.wikipedia.org/wiki/Curl_(mathematics)#Identities
     
  5. Mar 2, 2012 #4
    Thank you! Just figured out how to solve this!
     
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