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Vector Analysis

  1. May 31, 2006 #1
    Hi everybody,

    I'm preparing myself for the introduction to electrodynamics course and thus I
    go through vector analysis, but I hardly understand a problem given in my book
    (Griffiths, Introduction to Electrodynamics 3rd edition, page 18, Problem

    I have to calculate the divergence of the following function:

    see pic 1

    Ok my problem is, I can't figure it out how the author calculates the result of it. He has the following approach...see pic 2 (between del and v should be a dot, cause it is meant to be the dot product, not the cross product!!)

    OK...so far, so clear...I understand this step but now, he goes ahead by taking x^2+y^2+z^2 to the power of -3/2. Why?? Where does this fraction of -3/2 comes from? Is there any rule I didn't know? :confused:

    Anyway, thanks a lot in advance. Have a nice day...sincere greetings from Berlin, Germany

    Daniel Jackson

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  2. jcsd
  3. May 31, 2006 #2


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    I'm a bit confused. Are these supposed to be three stages in the solution of the problem? That's what they look like but the first is
    [tex]v= \frac{\vec{r}}{r^2}[/tex]
    while the second is
    [tex]\nabla \cdot v= \frac{\partial}{\partial x}\left(\frac{x}{r^3}\right)+\frac{\partial}{\partial y}\left(\frac{y}{r^3}\right)+\frac{\partial}{\partial y}\left(\frac{y}{r^3}\right)[/tex].
    (Click on the formulas to see the code used.)

    Was that supposed to be r3 in the first picture, as in the second? (And do you see the reason for the cube? Since [itex]\frac{\vec{r}}{r}[/itex] is itself a unit vector in the direction of the vector r, [itex]\frac{\vec{r}}{r^2}[/itex] is a vector with length [itex]\frac{1}{r^2}[/itex] in that same direction. While this is an inverse square law, we need the cube in order to account for the length of r itself.)

    Assuming that is the case then
    [tex]r= \sqrt{x^2+ y^2+ z^2}= \left( x^2+ y^2+ z^2)^\frac{1}{2}[/tex]
    so that
    [tex]r^3= \left(\sqrt{x^2+ y^2+ z^2}\right)^3= \left(x^2+ y^2+ z^2\right)^\frac{3}{2}[/tex]

    of course, the fact that the r3 is in the denominator means we have
    [tex]\frac{x}{r^3}= \frac{x}{\left(x^2+ y^2+ z^2\right)^\frac{3}{2}}= x\left(x^2+ y^2+ z^2\right)^\frac{-3}{2}[/tex]
    since the negative power corresponds to the term in the denominator (for example, [itex]2^{-1}= \frac{1}{2}[/itex]).
    Last edited: May 31, 2006
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