# Verifying Griffiths' Result on Coulomb Law

• quasar987
In summary: If so, then the delta function is just a convenient notation for saying that the charge density is infinite at r=0.In summary, Griffiths establishes that the divergence of a point charge's field is equal to the charge divided by the delta function. However, at r=0, the divergence is undefined.
quasar987
Homework Helper
Gold Member
The following result has been established by Griffiths (pp.50)

$$\nabla \cdot \left(\frac{\hat{r}}{r^2}\right) = 4\pi \delta^3(\vec{r})$$

Applying this result to Coulomb law, I get that

$$\nabla \cdot \vec{E} = \frac{q}{4 \pi \epsilon_0}4\pi \delta^3(\vec{r}) = \frac{q}{\epsilon_0}\delta^3(\vec{r})=\left\{ \begin{array}{rcl} q/\epsilon_0 & \mbox{for} & \vec{r} = \vec{0} \\ 0 & \mbox{elsewhere} \end{array}$$

Is this true? I'm wondering because Griffiths never makes mention of this result and I've never seen it anywhere else.

Note: Read $\vec{r}$ has the vector going from the point charge to some point P.

quasar987 said:
The following result has been established by Griffiths (pp.50)

$$\nabla \cdot \left(\frac{\hat{r}}{r^2}\right) = 4\pi \delta^3(\vec{r})$$

Applying this result to Coulomb law, I get that

$$\nabla \cdot \vec{E} = \frac{q}{4 \pi \epsilon_0}4\pi \delta^3(\vec{r}) = \frac{q}{\epsilon_0}\delta^3(\vec{r})$$

So far so good, it seems to me...

quasar987 said:
$$=\left\{ \begin{array}{rcl} q/\epsilon_0 & \mbox{for} & \vec{r} = \vec{0} \\ 0 & \mbox{elsewhere} \end{array}$$

Huh? Where do you get this result from?

Edit: check out pp. 69-70.

Last edited:
Your result ought to be in the book somewhere, considering it's the first Maxwell equation (also known as Gauss' law). It's usually expressed with respect to a charge density $$\rho$$, as opposed to a monopole charge q.

Huh? Yeah sure, about the "zero elsewhere" part, but where do you get this result from?

The zero elsewhere comes from the definition of the delta function.

rachmaninoff said:
The zero elsewhere comes from the definition of the delta function.

I know! I was saying Yeah SURE about the zero elsewhere part, meaning yeah sure, it is indeed correct. But the rest of his result is not. In any case I edited my post when I came across pp. 69-70.

cepheid said:
Edit: check out pp. 69-70.

This is the divergence of the field of a continuous distribution of charge. Because this field is obtained by evaluating an integral, we can make use of the property of the delta function exposed in equation (1.98) [pp.50]

However in calulating the divergence of a point charge's field, the delta function remains.

Exactly..that is the point I was getting at. You should not have removed the dirac delta function, as there is a singularity there. The divergence is undefined, but has been sort of "defined" in terms of the delta function.

$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$

and note that the charge density is undefined for a point charge (again, a singularity). So...since the two conclusions must be reconciled, it seems they've redefined the density of a point charge as:

$$q\delta^3(\vec{r})$$

But at r = 0, isn't q*delta indeterminant? The charge density and, therefore, the divergence of the field become arbitrarily large. The divergence can't be equal to a finite charge q/epsilon at r=0. Can it?

But at r = 0, isn't q*delta indeterminant? The charge density and, therefore, the divergence of the field become arbitrarily large. The divergence can't be equal to a finite charge q/epsilon at r=0. Can it?

Why not? The charge density is infinite at r=0, not the charge.

If I try to follow the reasoning of Griffiths exposed at pp.45-46, I understand that

$$\nabla \cdot \left(\frac{\hat{r}}{r^2}\right) = \frac{0}{r^2} = \left\{ \begin{array}{rcl} undefined & \mbox{for} & \vec{r} = \vec{0} \\ 0 & \mbox{elsewhere}\end{array}$$

On the other hand, the surface integral over a sphere centered at the origin is

$$\int_S \left(\frac{\hat{r}}{r^2}\right)\cdot d\vec{a} = 4\pi$$

Now, it is obvious that the divergence theorem does not apply, for the vector field is not even defined at 0.

However, on this topic, Griffiths says,

[...] if the divergence theorem is right (and it is), we should get

$$\int_V \nabla \cdot \left(\frac{\hat{r}}{r^2}\right) dV = 4\pi$$

for any sphere centered at the origin, no matter how small. Thus, the divergence has the bizarre property that it vanishes everywhere except at one point and yet its integral over any volume containing that point is $4 \pi$.
So it seems to me that he denied that the divergence theorem does not apply, and to FORCE it to apply, he implicitely defined the divergence as a delta function. (I thrust him that he has verified that the partial derivatives are continuous at the origin.).

He then goes on to say that the same apply for the density function of a point particle:

$$\rho(\vec{r}) = m \delta^3(\vec{r})$$

Or, for charge density,

$$\rho(\vec{r}) = q \delta^3(\vec{r})$$

But I don't understand why, cepheid, when you say

and note that the charge density is undefined for a point charge (again, a singularity). So...since the two conclusions must be reconciled, it seems they've redefined the density of a point charge as:
$$q\delta^3(\vec{r})$$
You don't take the next step and write

$$q\delta^3(\vec{r})=\left\{ \begin{array}{rcl} q & \mbox{for} & \vec{r} = \vec{0} \\ 0 & \mbox{elsewhere}\end{array}$$

Isn't this just what the notation $\delta^3$ means?

Last edited:
Isn't this just what the notation $\delta^3$ means?

I think that that's the delta function which is defined as equaling infinity at x=0 and 0 everywhere else. It's integral is 1. That's why it's useful for denoting pointlike things.

Berislav said:
I think that that's the delta function which is defined as equaling infinity at x=0 and 0 everywhere else.
Ugh! That hurts.

Ok, now everything cepheid said makes perfect sense.

Hey jdavel---you're right. The divergence is not equal to q/epsilon_0 at r = 0. quasar made a mistake initially. That's what we were just clearing up.

Hey quasar87: I think I was a bit abrupt before. Sure, so you forgot about the def'n of the dirac delta function but your "discovery" was not a total loss or anything. The fact that the divergence of the electric field of a point charge is zero everywhere except at the location of the charge is an expression of the fact that the electric field is "locally divergenceless", and in the absence of sources, $\nabla \cdot \vec{E} = 0$. That's why the flux of the electric field of a point charge through a spherical surface never changes, even as you consider surfaces at ever larger radii from the source. The point source is the only place from which the field is "diverging", and it's field strength goes as 1/r^2, whereas the surface area increases as r^2, so the total flux remains constant. In order for that not to be the case, ie for the flux to intensify or die out as you consider surfaces ever farther away, intermediate "sources" or "sinks" between your source at the origin and your surface would be required. In that case, the field in these intermediate locations would no longer be "locally divergenceless" due to the presence of those sources. That is the insight I got out of this stuff, but only with a little help and explanation from someone on this forum. I hope this helps...

Griffiths is simply wrong.
The divergence theorem has never, ever been proven on a region which includes a point where the vector function is undefined.
It never will be, either.

Ii is because the well-defined surface integral equals $$4\pi$$ on any surface enclosing the point source that we are JUSTIFIED in sliding over to the Dirac-delta formalism, so that the divergence theorem will SEEM to be fulfilled in this case as well.

It's just a convenient formalism, that's all there is to it.
No mathematical mystery, and no physical mystery, either.
Griffiths word "bizarre" shows that he is just plain dumb (or at least, a bad textbook author).

Last edited:
quasar987 said:
$$q\delta^3(\vec{r})=\left\{ \begin{array}{rcl} q & \mbox{for} & \vec{r} = \vec{0} \\ 0 & \mbox{elsewhere}\end{array}$$

Isn't this just what the notation $\delta^3$ means?

No, it's not! The delta function, in 3 dimensions, is a "distribution" (also called "generalized function") which is 0 every where except at the origin and, at the the origin, such that It's integral over every set containing the origin is 1. What IS true is that the integral of [itex]q\delta^3(\ver{r}) is q, not the value AT 0 itself.

("and, at the the origin, such that It's integral over every set containing the origin is 1". No, that doesn't quite make sense- that's why it is a 'generalized' function rather than a true function. The definition of generalized functions is much more complicated than regular functions.)

Berislav

"Why not? The charge density is infinite at r=0, not the charge."

But the divergence is proportional to the charge density, not the charge.

Yeah,so what?Remember that this typically a Green function problem for the Poisson equation.

$$\nabla^{2}_{\vec{r}}\varphi\left(\vec{r}\right)=-\frac{q}{\epsilon_{0}}\delta^{3}(\vec{r})$$

Can u solve such equation...?

Daniel.

arildno said:
Griffiths is simply wrong.
The divergence theorem has never, ever been proven on a region which includes a point where the vector function is undefined.
It never will be, either.

Ii is because the well-defined surface integral equals $$4\pi$$ on any surface enclosing the point source that we are JUSTIFIED in sliding over to the Dirac-delta formalism, so that the divergence theorem will SEEM to be fulfilled in this case as well.

It's just a convenient formalism, that's all there is to it.
No mathematical mystery, and no physical mystery, either.
Griffiths word "bizarre" shows that he is just plain dumb (or at least, a bad textbook author).

I assume this was in response to post # 9, and not what I had posted directly above you (which I am still waiting for some feedback on, because it would be nice to know if my assessment is valid).

[Rant mode]

HallsofIvy said:
("and, at the the origin, such that It's integral over every set containing the origin is 1". No, that doesn't quite make sense- that's why it is a 'generalized' function rather than a true function. The definition of generalized functions is much more complicated than regular functions.)

Aha! Halls makes a point here! After all, contrary to what I said sarcastically just now, the Dirac Delta Function is not some made up nonsense after all! It has been rigorously treated (which makes me wonder why mathematicians scorn physicists whenever they use it). To quote Boyce and Diprima:

"It is often convenient to introduce the delta function when working with impulse problems and to operate formally on it as though it were a function of the ordinary kind. [...] It is important to realize however that the ultimate justification of such procedures must rest on a careful analysis of the limiting operations involved. Such a rigorous mathematical theory has been developed, but we do not discuss it here."

Well then answer me this HallofIvy: If we were to apply this high fallutin, uber complicated mathematics of generalized functions to the scenario put forth by quasar987, would we then finally have a completely consistent and rigourous mathematical description of of electrostatics? Is that really what it takes? If so...sheesh! Okay, I'm done now.
[/Rant mode]

cepheid:
It was not in response to your post (which I thought good), but in response to the quoted passage from Griffiths.
The "mystery" is all Griffiths' making, not mine.
The electric field isn't defined at the origin, neither is its divergence.

So, why on Earth should we at the outset believe that the divergence theorem holds on a region which includes a point where the function isn't defined?

The theorem doesn't apply to the region in the first place.
But Griffiths make this big fuss about that the divergence theorem OUGHT to hold here.
What he should have done, IMO(possibly he does that as well), was:

1) To make the student understand WHY we get $$4\pi$$ from the SURFACE integral (which IS well-defined, in contrast to the volume integral) when the source is inside, and 0 when it is outside.
Since the surface integral has the immensely neat geometrical interpretation of calculating the (net) solid angle the surface region makes with respect to the source point, 1) should be treated at some length

2) Then, he could observe that a really nice feature about this, is that it holds for an region of ARBITRARY form.
It would then be natural to ponder:
Since we have seen that this works when dealing with surface integrals, wouldn't it be convenient if we had something similar for volume integrals?
That is, wouldn't it be cool to have a function $$\delta$$ which fulfilled for ARBITRARY V, say:
$$\int_{V}\delta(\vec{r}){dV}=1, \vec{0}\in{V},\int_{V}\delta(\vec{r}){dV}=0 otherwise$$
That is, the introduction of the delta function should be MOTIVATED, as something we would like to have.
3) Then, a few comments could suffice to say, alas, that doesn't seem possible to have.
4) Then, we could say that from a different mathematical perspective, it is actually possible to give meaning to such a wishful object all the same, and that it is possible to show, that as long as we proceed carefully, we can use the delta function AS IF it were an ordinary function.
It would also be nice if the author provides some references for interested students for a rigourous treatment.
5) We could then introduce the delta function as the divergence of the electric field from the point source, and point out that by doing this, the divergence theorem retains it form even in this case, which is cool.

I do NOT mean that distributions should be rigourously treated in an introductory course in electrodynamics, but should we sweep under the carpet that the manner in which we proceed is lax, but that it CAN be done rigourously?

EDIT:
This, quote however, from Boyce, is excellent:
"It is often convenient to introduce the delta function when working with impulse problems and to operate formally on it as though it were a function of the ordinary kind. [...] It is important to realize however that the ultimate justification of such procedures must rest on a careful analysis of the limiting operations involved. Such a rigorous mathematical theory has been developed, but we do not discuss it here."
It can't be said better.

Last edited:
Hey arildno,

Thanks for clarifying. That actually makes a hell of a lot more sense now, and is probably a better way of proceeding than Griffiths, if only to inform the student that there is something out there to deal with this situation, from which we are only taking the result, not the details. Oh, and thanks for putting up with my smart alek remarks too.

btw (off topic), but one more thing I wanted to say: your little (er...not so little) discussion of the formulation of physical laws in arbitrary geometric systems as opposed to material systems was invaluable to me in understanding the "control volume" analysis section of a recent course in fluid mechanics, yet another subject not so well explained in class. So if you thought Griffiths was bad, you better not check out our fluid mech text, lol. But I used your thread from several months ago as a reference...I noticed you brought it back again. I'll have to check that out, because I still get confused with the example of the rocket equation.

cepheid:
Your smart alek remarks were quite appropriate since I didn't clarify what I responded to in the first place. In addition, I may have been too judgmental of Griffiths in general; I'm not familiar with that book, but the quoted reference reminded me of some truly horrendous treatments of these issues, and I simply assumed (possibly wrongly) that Griffiths' text would in all particulars be of the same low quality than these texts.
Possibly, the quote from him is an extremely unfortunate exception within an otherwise excellent book.

Secondly, concerning my verbose thread on geometric systems:
If there are some particular details you want to discuss, just PM me, or place it inGokul's thread in General Physics.
And yes, there are lots of bad texts on fluid mechanics..

## 1. What is Griffiths' result on Coulomb Law?

Griffiths' result on Coulomb Law states that the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

## 2. How can I verify Griffiths' result on Coulomb Law?

To verify Griffiths' result on Coulomb Law, you can perform experiments and measure the electrostatic force between two point charges at different distances and with different magnitudes of charge. You can also use mathematical calculations to confirm the relationship between the force, charges, and distance.

## 3. Why is it important to verify Griffiths' result on Coulomb Law?

Verifying Griffiths' result on Coulomb Law is important because it is a fundamental law in electromagnetism and is used to understand and predict the behavior of charged particles in various systems. Confirming its validity helps to ensure the accuracy and reliability of future experiments and applications.

## 4. What are the implications of disproving Griffiths' result on Coulomb Law?

If Griffiths' result on Coulomb Law were to be disproven, it would mean that our current understanding of electrostatics is incorrect. This could have significant implications for various fields, such as electronics, engineering, and physics, which rely on the principles of Coulomb Law.

## 5. Are there any known limitations to Griffiths' result on Coulomb Law?

Griffiths' result on Coulomb Law is considered a classical law and does not take into account quantum effects. It also assumes that the charges are stationary, which may not always be the case. Additionally, the law may not hold true for extremely small distances or very high charges, requiring the use of more advanced theories such as quantum electrodynamics.

• Classical Physics
Replies
8
Views
675
• Classical Physics
Replies
8
Views
646
• Classical Physics
Replies
8
Views
913
• Classical Physics
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
187
• Classical Physics
Replies
3
Views
1K
Replies
3
Views
631
• Engineering and Comp Sci Homework Help
Replies
0
Views
140
• Electromagnetism
Replies
1
Views
566