# Divergence questions from Griffith's Electrodynamics

1. Jun 22, 2012

### Curtis15

Hi all

I basically have two questions that are very closely related to each other about divergence, specifically the divergence of a vector function 1/r2$\widehat{r}$

First, I will be referencing pages 17, 18, and 45 from the 3rd edition of Intro to Electrodynamics.

The first question comes from page 45 when Griffith states on the vector function above "if ever there was a function that ought to have a large positive divergence, this it it."

From my intuition based on pages 17 and 18 and other online material I have read, divergence measures how much a vector spreads out. A positive divergence represents a source where "more" is coming out, a source. A negative divergence, conversely, is where "more" is coming, a sink.

So based on the vector function, as r increases, the magnitude value of the vector decreases. It basically looks like the opposite of the first diagram on page 17 except instead of increasing the size of the vector as you radiate from the origin, you would decrease the size of the vector. That means that at any point besides the origin, if you draw a small circle, there are larger vectors going into the circle than going out, so this would be a net influx, or sink, meaning negative divergence.

So if this equation were to have a large divergence, I think it would be a large negative divergence. Is this correct or is Griffith correct?

Now the second part is just an extension of the first question.

I did the problem in the book and have seen for myself that the divergence of the above vector function is in fact 0 when you calculate it using the formula. However, from the diagram I visualize in my head, I do not see why. As above, I see that there is more going in than there is going out. How can divergence be 0?

The only thing I can think off is that the diagram I am looking at, since it is only 2 dimensional, doesn't represent the actual 3-d function and instead, when you do have the 3 dimensions accounted for, there is a spherical symmetry that cancels out similar to how Newton's shell theorem is only valid for a perfect sphere. Can anyone elaborate on this please?

Thank you all very much.

2. Jun 23, 2012

### Natey213

Griffiths is right.

If you look at the wiki (or anywhere else) for a qualitative explanation of divergence, you will find,

"More technically, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. "
-wiki

So, since r^-2 rhat, vanishes at infinity, if you draw an infenesimal sphere around the origin, there is no flux, because at infinity the value of all of the vectors is (the limit) zero.

3. Jun 23, 2012

### Hassan2

I don't understand the above statement. The divergence of the above vector field is zero everywhere except for at the origin. It is infinite at the origin. The net flux exiting the boundary of any volume which does not contain the origin, is zero.

For an intuitive understanding, imagine a cube with one of its faces toward the origin. The flux entering the cube through the near face is more than the flux leaving it through the far face. But the are four more faces through them flux leaves the cube. If you consider all the faces, the net flux becomes zero.

If you work in 2 dimensions, then you must get the same conclusion for vector field $F=\frac{1}{r}\hat{r}$

4. Jun 23, 2012

### arildno

1. The function, and hence its divergence, is only defined on sets that excludes the origin.
2. On whichever such set, the divergence is, indeed, zero.

5. Jun 23, 2012

### vanhees71

Better don't understand the statement, because it's wrong ;-).

I don't know Griffith's book very well, but he should explain this point very carefully, and indeed it might help to think of vector fields as velocity field of a fluid of constant density.

Now you consider the Coulomb field:

$$\vec{E}(\vec{r})=\frac{1}{r^2} \hat{r}=\frac{1}{r^3} \vec{r}.$$

From the latter form of the expression you see that this vector field is smooth everywhere except at the origin $\vec{r}=0$, where it has a singularity.

Now, when you calculate the divergence via the partial derivatives, e.g., in Cartesian coordinates:

$$\vec{\nabla} \cdot \vec{E}=\partial_j E^j,$$

you can do so only at points where $\vec{E}$ is a well-defined smooth function! This is the case for all $\vec{r} \neq 0$.

Now remember, how the divergence is defined (provided the book is good!): It's defined as the limit

$$\vec{\nabla} \cdot \vec{E}(\vec{r})=\lim_{\Delta V \rightarrow 0} \frac{1}{\Delta V} \int_{\partial \Delta V} \mathrm{d} \vec{A}' \cdot \vec{E}(\vec{r}').$$

Here $\Delta V$ is a small volume around the point $\vec{r}$, where you like to get the divergence, and $\partial \Delta V$ is the surface with normal vectors pointing outwards of the volume. The limit has to be read as "contracting" the volume to the one point $\vec{r}$.

If $\vec{E}$ is smooth in an open neighbourhood of $\vec{r}$, indeed the divergence according to the general definition is the same as taking the derivatives.

Now think again about your Coulomb field and depict it in your mind as a fluid flow. Now take a volume $V_1$, not containing the origin. There you clearly see that some fluid is flowing into the volume and some out. Gauß's integral theorem tells you that in this case there is as much fluid flowing into the volume as is flowing out, because everywhere inside this volume you can use the above formula to get the divergence, and it's 0.

Now put the volume around the origin. Then you see that there's only fluid coming out. From the consideration above, however, you see that the shape of the volume doesn't matter: As long as it contains the origin in its interior the same amount of fluid comes out of any such volume. So you can as well make you life easy and take a sphere with the center at the origin. It's trivial to calculate the flow going through:

$$\int_{\partial K_{R}(0)} \mathrm{d} \vec{A} \cdot \vec{E}=4 \pi.$$

But what about the divergence? Again you may take a sphere in the general definition with the integral, and you immediately realize that it must diverge! All spheres give the same surface integral value, $4 \pi$. Now you devide through the volume and make the radius $R \rightarrow 0$. Thus, the divergence gives infinity at the origin and 0 everywhere else.

Further you want Gauß's theorem also valid for this singular case. Thus you must set

$$\vec{\nabla} \cdot \vec{E}=4 \pi \delta(\vec{r}).$$

That's what you expect from electromagnetics! The given Coulomb field is the solution of the electrostatic Maxwell Equations for the case of a charge of $4 \pi$ at the origin (in Heaviside-Lorentz units) since the electrostatic Maxwell Equations read

$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\rho,$$

and for a point charge of $4 \pi$ you have

$$\rho(\vec{r})=4 \pi \delta(\vec{r}).$$

6. Jun 23, 2012

### arildno

Incorrect, vanhees71.
The volume integral including zero is, quite simply, NOT defined.

However, the SURFACE integral around an arbitrary volume containing zero IS defined, and has the peculiar property that for any such enclosing surface, the value is a constant.

That MOTIVATES the introduction of the delta "function", but by no means necessitates that.

The divergence is, wherever it is defined, equal to..0

7. Jun 23, 2012

### Curtis15

Alright guys I appreciate the replies, but there is something that I feel I need to elaborate on.

I am not disputing the fact the mathematics of the divergence, but instead the diagramming of the divergence. I have attached a picture I drew to illustrate my point.

Here there is a green circle at the origin of the electrostatic field. The charged particle acts as a source and so the divergence at r = 0 is positive. However, at the red circle larger vectors are coming into the circle than there are coming out. Based on the definition of divergence, this would seem to represent a negative divergence, not a large positive divergence as Griffith hypothesizes before introducing the delta function, nor a zero divergence as the math shows.

From Wikipedia - "In physical terms, the divergence of a three dimensional vector field is the extent to which the vector field flow behaves like a source or a sink at a given point. It is a local measure of its "outgoingness"—the extent to which there is more exiting an infinitesimal region of space than entering it. If the divergence is nonzero at some point then there must be a source or sink at that position."

There is more outgoingness at the origin, so that point has positive divergence. There is more ingoingness at the red circle, so that point has negative divergence. Can someone please elaborate on this?

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8. Jun 23, 2012

### vanhees71

Ok, you are right that my argument is not a real mathematical proof of my assertion, but it's nevertheless correct.

To make this mathematically strict, you have to proof that for any test function $f \in C_0^{\infty}(\mathbb{R}^3)$

$$-\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{r} \frac{\vec{r}}{r^3} \cdot \vec{\nabla} f(\vec{r})=4 \pi f(0).$$

This is not too difficult (see, e.g., Sommerfeld, Lectures on Theoretical Physics Vol. 2 (fluid dynamics)).

9. Jun 23, 2012

### arildno

We are not in any disagreement about both the utility, and indeed mathematical jusitifiability of the delta function(al).

However, I have a strong dislike for Griffiths' treatment of this, where he seeks to "prove" that it is somehow wrong to say the divergence is zero.

There are better treatments of this than Griffiths, more mathematical sound.

10. Jun 23, 2012

### vanhees71

As I said, I don't know Griffiths's book, and one might argue, whether it is better to use classical vector analysis and functions only. Then it's clear that the here considered Coulomb field has a singularity at the origin and of course none of its derivatives or any of the vector operators like curl or div make sense there. The function is simply not defined. Everywhere else both the divergence and the curl are zero at each point except in the origin, where it is not defined.

On the other hand, it's also good to introduce distributions (i.e., functionals on appropriate dense subsets of Hilbert or Banach function spaces). This is the modern mathematical language which makes a lot of calculations also in classical field theory much easier than the ones using the classical vector analysis.

E.g., then you have the full machinery of Green's functions at hand, which is a quite delicate issue in classical vector analysis but appears just as the distribution which is the solution for a field problem with $\delta$-distribution like sources.

It also gives the $\delta$ distribution and its derivatives a quite intuitive meaning in terms of point charges and multipole sources.

11. Jun 23, 2012

### Muphrid

You're basing your assertion on a diagram instead of a mathematical calculation. I assure you, if you integrate over any closed surface that doesn't contain the origin, you will find that the electric flux is zero, and you can take the limit of the volume becoming arbitarily small and still find that the flux is zero.

If you insist on having a conceptual explanation for this, consider that more of the area of the sphere sees outgoing instead of incoming flux (think about where the electric field becomes tangent to the sphere and how that cuts off a section that is smaller than a hemisphere for incoming flux). So while the outgoing field vectors tend to be smaller than the incoming ones, they make up for this with a larger area to exit the sphere and thus the fluxes balance.

The Coulomb field is exactly the free space Green's function for the vector derivative $\nabla$. It is constructed such that its divergence is zero everywhere except for the one point where it is undefined--such that the source can be described by a delta function/distribution.

12. Jun 23, 2012

### Curtis15

Alright Muphrid I think I get what your saying, thank you.