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I need to prove del cross f= 0

  1. Sep 11, 2016 #1
    1. The problem statement, all variables and given/known data

    F(x) has the form F(x)=f(r)x where r=|x|

    A.) prove that del cross f =0
    B.) Now suppose also Del •F = O. What is the most general form allowed for f(r)?

    2. Relevant equations


    3. The attempt at a solution
    I have done part b but what do I need for A

    F(X)= r(hat)
    Fr = 1, F(theta) = F(phi)= O.

    Got this out of book.

    Using equation from the book ( del cross F) r
    I get zero.

    But what is the proof behind it
    Have my answer for part a uploaded
     

    Attached Files:

  2. jcsd
  3. Sep 11, 2016 #2

    andrewkirk

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    The formulas for ##\nabla\times## in polar coordinates make calculations with radial vectors easy because all the hard work has been done in deriving the formulas. If you have not been given those formulas, and hence are required to derive them if you use them, there is no point in using them because the work involved in deriving them is greater than the work involved in just doing the whole calc in Cartesian coordinates.

    Given that, the simplest way is just to do the calc in Cartesian coordinates. Since the usual coordinate names in Cartesian are ##x,y,z## I suggest you instead call your point ##\vec r##. Then the coordinates of ##\vec F(\vec r)## are ##(f(r)x,f(r)y,f(r)z)##, writing ##r## for ##\|\vec r\|##.

    Then use the Cartesian formula for ##\nabla\times##:
    $$\nabla\times\vec u=(\partial_x,\partial_y,\partial_z)\times (u_x,u_y,u_z)=(\partial_y u_z-\partial_zu_y)\hat i
    +(\partial_z u_x-\partial_xu_z)\hat j
    +(\partial_x u_y-\partial_yu_x)\hat k$$
    where ##\partial_x## denotes ##\frac{\partial}{\partial x}## and ##u_x## is the ##x## coordinate of ##\vec u##.

    The calc is quite short.
     
    Last edited: Sep 11, 2016
  4. Sep 11, 2016 #3

    LCKurtz

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    Your notation is terrible. It appears you are mixing up F and f in A. Apparently what you are calling x is what would normally be denoted ##\vec r = \langle x,y,z\rangle##. And in your work you use x as above and also as one of its components, which is very confusing. Also, your posting images instead of typing your work makes it difficult to edit and violates forum policy.

    I wouldn't normally bother to answer such a post, but it's a slow day here so...
    I think your question is, given ##\vec r = \langle x,y,z\rangle## and ##r = |\vec r| =\sqrt{x^2+y^2+z^2}##, and ##f(r)## is a differentiable function, calculate ##\nabla \times f(r)\vec r##. Write that out carefully and you should get the zero vector. If you get stuck and don't get that, type your work here for additional help.

    Edit: I see Andrew posted while I was typing. Obviously, he and I agree about how to work it.
     
  5. Sep 12, 2016 #4
    I must first ask, why does r=|r-hat|?

    I understand that |r-hat| equals sqrt(rx^2+ry^2+rz^2)
    Rx^2 is just r with respect to X.
     
  6. Sep 12, 2016 #5
    Also f(r) is a scalar function.
    When doing del X (f(r) (r-hat)
    I know that you can't multiply Del with a scalar. You can only cross with a vector which is r-hat.

    So in order to do this it would be
    Del•(f(r)) X r-hat + f(r)•(Del X r-hat)
     
  7. Sep 12, 2016 #6
    Wait, you can multiply Del with a scalar function but you can't cross Del with a scalar function
     
  8. Sep 12, 2016 #7

    LCKurtz

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    Of ##\phi## is a scalar and ##\vec V## is a vector, a standard formula (which you can prove yourself) is$$
    \nabla \times \phi \vec V = \nabla \phi \times \vec V + \phi ( \nabla \times \vec V)$$
     
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