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Divergence in spherical coordinates

  1. May 13, 2010 #1
    I am stuck on this problem.

    Use these equations:

    [tex]\textbf{v}(\textbf{r}) = f(r)\textbf{r}[/tex]

    [tex]\frac{\partial r}{\partial x} = \frac{x}{r}[/tex]

    And the chain rule for differentiation, show that:

    [tex](\nabla\cdot\textbf{v}) = 2f(r) + r\frac{df}{dr}[/tex]
    (cylindrical coordinates)

    Any help greatly appreciated, I will post my progres so far in a following post.

    Cheers
     
  2. jcsd
  3. May 13, 2010 #2
    So far I have tried to crack this by subbing in

    [tex]\textbf{v}(\textbf{r}) = f(r)\textbf{r}[/tex]

    in the bottom equation I get:

    [tex](\nabla\cdot f(r)\textbf{r}) = 2f(r) + r\frac{df}{dr}[/tex]

    But I think that if we write out the left hand side of the equation and separate out the differentials we get this:

    [tex](\nabla\cdot f(r)\textbf{r}) = r_{x}\frac{\partial f(r)}{\partial x} + f(r)\frac{\partial r_{x}}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y} + f(r)\frac{\partial r_{y}}{\partial y}[/tex]

    Which I think reduces further to:

    [tex](\nabla\cdot f(r)\textbf{r}) = 2f(r) + r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y} = 2f(r) + r\frac{df}{dr}[/tex]

    Which IF that is right (and it's a big IF) basically means that I'm left to prove that:

    [tex]r\frac{df}{dr} = r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y}[/tex]

    That's where I'm stuck. I have a feeling that the total derivative is needed next and I am thinking along those lines, I still haven't used the chain rule yet so I think that will come in to play in the total derivative context. It could well be that I've strayed off and am heading nowhere with this, which is why I need a bit of guidance.

    Thanks
     
  4. May 13, 2010 #3

    LCKurtz

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    I assume you know, or can prove, that for scalar f and vector V in rectangular coordinates,

    [tex]\nabla\cdot f\vec V = \nabla f \cdot \vec V + f(\nabla \cdot\vec V)[/tex]

    Also, I assume the standard notation

    [tex]\vec r = \langle x, y \rangle,\ r = |\vec r|[/tex]

    Then by the first identity

    [tex]\nabla\cdot f(r)\vec r = \nabla f(r) \cdot \vec r + f(r)(\nabla \cdot \vec r)[/tex]

    The only place you need the chain rule is to calculate the first term on the right.

    [Edit] Corrected typos.
     
    Last edited: May 13, 2010
  5. May 14, 2010 #4
    Thank you for your reply.

    If I understand you correctly then I have more or less followed your reasoning thus far.

    I am stuck on the bit you mention in your last line, I am struggling to calculate the first term on the right.

    The way I see it is that

    [tex]\nabla f(r) \cdot \vec r = r_{x}\frac{\partial f(r)}{\partial x} + r_{y}\frac{\partial f(r)}{\partial y}[/tex]

    Is that correct?

    That's my last line in my earlier post. How do I invoke the chain rule here to get this into the form I am after?

    Thanks
     
  6. May 14, 2010 #5

    LCKurtz

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    When you differentiate f(r) with respect to, for example, x, you get f'(r)rx. And do the dot product last.
     
  7. May 16, 2010 #6
    Okay, I think I've got it now.

    Use the chain rule:

    [tex]\frac{\partial f(r)}{\partial x} = \frac{df}{dr}\times \frac{\partial r}{\partial x} [/tex]

    And the result:

    [tex]\frac{\partial r}{\partial x} = \frac{x}{r}[/tex]

    To get:

    [tex]r\frac{df}{dr} = r_{x}\frac{df}{dr}\frac{x}{r} + r_{y}\frac{df}{dr}\frac{y}{r}[/tex]

    Now rx=x. Use this result as well as the fact that in cylindrical coordinates r2 = x2+y2,and the term on the right reduces to the term on the left. Is this OK?

    Cheers
     
    Last edited: May 16, 2010
  8. May 16, 2010 #7

    LCKurtz

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    Yes, I think you have it figured out now.
     
  9. May 17, 2010 #8
    Ahhh good. For some reason that one had me scratching my head for a while, but it seems quite simple now.

    Cheers
     
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