Calculating Divergence of a Function: A Berliner's Struggle

[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

 

[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
$$v= \frac{\vec{r}}{r^2}$$
while the second is
$$\nabla \cdot v= \frac{\partial}{\partial x}\left(\frac{x}{r^3}\right)+\frac{\partial}{\partial y}\left(\frac{y}{r^3}\right)+\frac{\partial}{\partial y}\left(\frac{y}{r^3}\right)$$.
(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 $\frac{\vec{r}}{r}$ is itself a unit vector in the direction of the vector r, $\frac{\vec{r}}{r^2}$ is a vector with length $\frac{1}{r^2}$ 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
$$r= \sqrt{x^2+ y^2+ z^2}= \left( x^2+ y^2+ z^2)^\frac{1}{2}$$
so that
$$r^3= \left(\sqrt{x^2+ y^2+ z^2}\right)^3= \left(x^2+ y^2+ z^2\right)^\frac{3}{2}$$

of course, the fact that the r³ is in the denominator means we have
$$\frac{x}{r^3}= \frac{x}{\left(x^2+ y^2+ z^2\right)^\frac{3}{2}}= x\left(x^2+ y^2+ z^2\right)^\frac{-3}{2}$$
since the negative power corresponds to the term in the denominator (for example, $2^{-1}= \frac{1}{2}$).

 

[Architeuthis Dux]

Hello Daniel,

I can understand your struggle with this problem. The approach the author is taking is using the definition of divergence and applying it to the given function. The -3/2 power comes from the general formula for calculating divergence, which is:

see pic 3

This formula involves taking the partial derivatives of the components of the vector function with respect to each coordinate (x, y, z) and then summing them together. In this case, the function is a scalar function (since it has no direction) and the components are x, y, and z themselves. Therefore, when we take the partial derivatives, we get 1 for each component, resulting in the sum of 3. The -3/2 power comes from the fact that we are taking the inverse of the distance (x^2+y^2+z^2) to the power of 3/2, which is equivalent to taking the square root of the distance to the power of -3/2. This is a common occurrence when dealing with vector functions in electrodynamics.

I hope this explanation helps clarify the approach and formula used by the author. Keep up the good work in your studies and don't hesitate to reach out if you have any further questions.

Best regards,