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Math Div expression help

  1. May 12, 2008 #1
    Is the following true?

    [tex]
    \newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }

    \nabla \cdot \vec{r}f(r) = \left ( \pd{f}{r}{} \pd{r}{x}{} \cdot x + f \right) + \left ( \pd{f}{r}{} \pd{r}{y}{} \cdot y + f \right) + \left( \pd{f}{r}{} \pd{r}{z}{} \cdot z + f \right)[/tex]

    where

    [tex]\vec{r} = (x, y, z)[/tex]
     
    Last edited: May 12, 2008
  2. jcsd
  3. May 12, 2008 #2
    no, it would just be the partial of r times f in each case
     
  4. May 12, 2008 #3
    Like this?

    [tex]
    \newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }

    \nabla \cdot \vec{r}f(r) = \pd{(rf)}{x}{} + f\pd{(rf)}{y}{} + f\pd{(rf}{z}{}
    [/tex]

    or just 3f?
     
    Last edited: May 12, 2008
  5. May 12, 2008 #4
    My initial expression seems right to me now...
     
  6. May 12, 2008 #5

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    Is f a scalar field or a vector field?
    Is it del.(r*f) or (del.r)(f)
     
  7. May 12, 2008 #6
    I wrote it out just how they gave it to me but I assumed del.(rf) based on the question they ask after it.

    f is scalar field, f(r), r = ||r||
     
  8. May 12, 2008 #7

    Vid

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    There's a type of product rule for a scalar field times a vector field.

    del.(sF) = (del.F)*s + grad(s).F
    where . is the vector dot product.
     
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