# Dirac-Delta Function

1. Sep 12, 2004

### brianparks

In introducing the dirac-delta function, my electrodynamics discusses the following function:

$$v = 1 / r^2$$r

The text states that "at every location, v is directed radially outward; if there ever was a function that ought to have a large positive divergence, this is it. And yet, when you actually calculate the divergence (using Eq. 1.71), you get precisely zero."

At least on the positive portion of the x-y-z axis, it seems to me that v would have a negative divergence, meaning that as you move away from the origin, the magnitude of the vectors (which all point radially outward from the origin), would get smaller. That is, for every point other than the origin, the amount going in would exceed the amount coming out (the definition of a sink). For example:

Origin---------->-------->------>---->-->

Does this not indicate a negative divergence?

This question has caused me to wonder if my understanding of the divergence function is flawed.

Any help is greatly appreciated,

--Brian

Last edited: Sep 12, 2004
2. Sep 12, 2004

### Tide

The divergence is zero for r > 0 because the field never diverges from the radial direction.

3. Sep 12, 2004

### brianparks

But doesn't the field weaken as you move radially outward? If so, wouldn't that mean that the divergence is negative?

Maybe my understanding of divergence is wrong.

Consider this function:

$$v = 1 / x^2$$x

From 0 to infinity, it would look like this:

---------->--------->-------->------->------>----->---->--->-->

and so on...

Wouldn't it have a negative divergence?

Thanks again for any help,
--Brian

4. Sep 12, 2004

### pervect

Staff Emeritus

See http://hyperphysics.phy-astr.gsu.edu/hbase/diverg.html#c3

for the appropriate formulas for divergence in spherical coordiantes.

5. Sep 13, 2004

### ehild

Let us try.

$$\Vec{v}(x,y,z)=\frac{1}{x^2+y^2+z^2}(x\Vec{i}+y\Vec{j}+z\Vec{k})$$

$$\frac{\partial v_x}{\partial x} =-\frac{2x^2}{(x^2+y^2+z^2)^2}+\frac{1}{x^2+y^2+z^2}= \frac{-2x^2}{r^4} + \frac{1}{r^2}$$

In the same way,

$$\frac{\partial v_y}{\partial y} =-\frac{2y^2}{r^4}+\frac{1}{r^2}$$

and

$$\frac{\partial v_z}{\partial z} =-\frac{2z^2}{r^4}+\frac{1}{r^2}$$

The divergence is

$$\mbox{Div}(\Vec{v})=\frac{\partial v_x}{\partial x} +\frac{\partial v_y}{\partial y} +\frac{\partial v_z}{\partial z}=$$

$$=\frac{-2x^2-2y^2-2z^2}{r^4}+\frac{3}{r^2}=\frac{-2r^2}{r^4}+\frac{3}{r^2}=\frac{1}{r^2}$$

It is the function
$$\Vec{v}(\Vec{r}) = \frac{\Vec{r}}{r^3}$$
whose divergence is zero everywhere except at the origin.

ehild