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How will I represent the scalar function?

  1. Jan 16, 2012 #1
    1. The problem statement, all variables and given/known data

    show that [itex]\nabla \times (f F)= f \nabla \times F+ (\nabla f) \times F[/itex]


    3. The attempt at a solution

    How will I represent the scalar function? Do I write [itex]f=\psi(x,y,z)[/itex] or

    [itex]f=A_x+A_y+A_z[/itex]

    I chose [itex]F=a_x \vec i +a_y \vec j +a_z \vec k[/itex]

    Using [itex]f=\psi(x,y,z)[/itex] I work out the LHS of question as

    [itex](\psi a_z)_y-(\psi a_y)_z-(\psi a_z)_x+(\psi a_x)_z+(\psi a_y)_x-(\psi a_x)_y[/itex].............How to go further?

    Thanks
     
  2. jcsd
  3. Jan 16, 2012 #2

    SammyS

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    Re: Identities

    Just saying that f is a function of x, y, and z would have been sufficient, but [itex]f=\psi(x,y,z)[/itex] is perfectly fine.

    Use the product rule (of differentiation) for each of [itex](\psi a_z)_y\,,\ (\psi a_y)_z\,,\ (\psi a_z)_x\,,\ (\psi a_x)_z\,,\ (\psi a_y)_x\,,\ (\psi a_x)_y\,.[/itex]

    Using subscripts for partial derivatives may not be such a good idea here. There may be some confusion as to the meaning of such quantities as [itex](a_x)_y[/itex] for example.
     
  4. Jan 16, 2012 #3
    Re: Identities

    Ok but why would we use the product rule though, looking at the first term for example

    if psi is a function of x,y and z but since we are partially differentiating wrt y doesn't it just become [itex]a_z \psi_y[/itex]. Ie, if the function was something like [itex]\psi=x^2y^2z^2[/itex] then [itex]\psi_y= 2yx^2z^2[/itex] etc
     
  5. Jan 16, 2012 #4

    SammyS

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    Re: Identities

    az is also a function of x, y, and z.

    [itex]\displaystyle\frac{\partial}{\partial y}\left(\psi a_y\right)=\left(\frac{\partial\psi}{\partial y}\right)a_y+\psi\left(\frac{\partial a_y}{\partial y}\right)[/itex]
     
  6. Jan 28, 2012 #5
    Re: Identities


    I got this, thank you.
     
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