Vector Analysis

Daniel Jackson

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
1.16):

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?

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

Daniel Jackson

Attached Thumbnails

     

HallsofIvy

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]).