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I'm trying to understand how the integral form is derived from the differential form of Gauss' law.
I have several issues:

1) The law states that $ \nabla\cdot E=\frac{1}{\epsilon 0}\rho$, but when I calculate it directly I get that $ \nabla\cdot E=0$ (at least for $ r\neq0$).

2) Now $ \iiint\limits_\nu \nabla\cdot E d\tau $ should be zero no matter what the value of the divergence is at 0, since the divergence is zero everywhere but 0 (in contrast to the law which states it is non-zero).

3)

a. The proof itself goes on to use the divergence theorem to state that for any volume $\nu$, $ \iiint\limits_\nu \nabla\cdot E d\tau = \iint\limits_{\partial\nu} E d a $, however the divergence theorem requires E to be continuously differentiable everywhere in $\nu$ (it is not differentiable at 0, let alone continuously differentiable there).

b. The function cannot be corrected in any way at 0 since the derivative goes to infinity around 0.

c. The point 0 cannot be removed from the integrated volume because the divergence theorem requires that the volume of integration be compact.

d. In light of the former I don't see how the divergence theorem can be used here.

TomM
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    https://en.wikipedia.org/wiki/Dirac_delta_function – Giuseppe Negro Sep 15 '15 at 16:09
  • You wrote "The law states that $\nabla \cdot \vec E=\frac{\rho}{\epsilon_0}$, but when I calculate it directly I get that $\nabla \cdot \vec E=0$ (at least for $r\ne0$)." What are you calculating that gives $0$? Gauss's Law is as stated, $\nabla \cdot \vec E=\frac{\rho}{\epsilon_0}$ for all $\vec r$. – Mark Viola Sep 15 '15 at 16:16
  • @Dr.MV Assuming the formula $ \nabla \cdot v = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2v_r) + \frac{1}{r\sin\theta}\frac{\partial v_\theta}{\partial \theta}(\sin\theta v_\theta) + \frac{1}{r\sin\theta}\frac{\partial v_\phi}{\partial \phi} $, my calculation was: (latter two terms vanish) $ \frac{1}{r^2}\frac{\partial}{\partial r}(r^2v_r) = \frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{1}{r^2}) = 0$. – TomM Sep 15 '15 at 16:30
  • @Dr.MV Sorry, I am speaking about a point charge throughout. – TomM Sep 15 '15 at 16:51
  • Yes. I understand. So, one needs to define a point charge. What does that mean to you mathematically? – Mark Viola Sep 15 '15 at 16:52
  • What do you mean? My textbook defines the electrical field due to a point charge at a point $(r,\phi,\theta)$ to be $\frac{q}{r^2}\vec{r}$ – TomM Sep 15 '15 at 16:54
  • OK. So, you know the electric field from a point charge is given by Coulomb's Law. And now, you take the divergence of that field and find it is zero for $\vec r\ne 0$. Great. From that field, can you calculate the flux integral over a surface that contains the origin? – Mark Viola Sep 15 '15 at 16:57
  • @Dr.MV Yes. $\frac{1}{\epsilon 0}q$ – TomM Sep 15 '15 at 17:02
  • @TomM OK. Fantastic! Well done. I just posted an answer. Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Sep 15 '15 at 17:27
  • Crossposted from http://physics.stackexchange.com/q/207134/2451 – Qmechanic Sep 16 '15 at 09:56

1 Answers1

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We have that the electric field from a "point charge" $q$ located at the origin is

$$\vec E=\frac{q}{4\pi \epsilon_0}\frac{\vec r}{|\vec r|^3} \tag 1$$

Clearly from $(1)$, we have $\nabla \cdot \vec E=0$ for $\vec r\ne 0$. We note that the divergence is undefined (using classical analysis) at the origin.

Now, let $V$ be any region, whose "outer boundary" contains the origin, that excludes the "small" spherical region $0<\epsilon <r$ (i.e., $V$ has a hole to exclude the origin). Denote the "outer boundary" of $V$ to be $S$ and the boundary of the "small" spherical region to be $S_{\delta}$. Clearly, we have from the Divergence Theorem

$$\begin{align} \int_V \nabla\cdot \vec E\,dV&=\oint_{S} \vec E\cdot \hat n\,dS-\oint_{S_{\delta}} \vec E\cdot \hat n\,dS\\\\ &=0 \tag 2 \end{align}$$

since $\nabla \cdot \vec E=0$ throughout $V$. Equation $(2)$ implies that

$$\oint_{S} \vec E\cdot \hat n\,dS=\oint_{S_{\delta}} \vec E\cdot \hat n\,dS \tag 3$$

for any $S$ surrounding the origin. We can evaluate the integral on the right-hand side of $(3)$ using $(1)$. Proceeding we have

$$\begin{align} \oint_{S_{\epsilon}} \vec E\cdot \hat n\,dS &=\int_0^{2\pi}\int_0^\pi \left(\frac{q}{4\pi \epsilon_0} \frac{\hat r}{\delta^2}\right)\,\cdot \hat r \,\delta^2\,d\theta\,d\phi\\\\ &=\frac{q}{\epsilon_0} \tag 4 \end{align}$$

which yields the integral-form of Gauss's Law for a point charge.

NOTE:

Analysis can be facilitated using Generalized Functions as follows. In THIS ANSWER, I discuss regularizing the Electric Field of a point charge to assign meaning to the Dirac Delta for use in the Divergence Theorem. This provides a rigorous way forward where Dirac Delta is interpreted in terms of the limit of the regularized function $\vec \psi$ given by

$$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} \tag 1$$

Taking the divergence of $(1)$ reveals that

$$\nabla \cdot \vec \psi(\vec r; a)=\frac{3a^2}{(r^2+a^2)^{5/2}}$$

Now, in the Aforementioned Answer, I showed that for any sufficiently smooth test function $\phi$, we have that

$$\lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)dV= \begin{cases} 0&, \text{V does not include the origin}\\\\ 4\pi \phi(0)&,\text{V includes the origin} \end{cases}$$

and it is in this sense that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)=4\pi \delta(\vec r)}$$

Heuristically, let the volume charge density of a point charge be given by $q\delta (\vec r)$, where $\delta(\vec r)$ is the Dirac Delta. Then, the point-form of Gauss's Law is

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

and we have from the divergence theorem for Generalized Functions

$$\int_V \nabla \cdot \vec E\,dV=\frac{q}{\epsilon_0}=\oint_S\vec E\cdot \hat n\,dS$$

where $\delta(\vec r)$ interpreted in terms of the limit of the regularized function. And we are done!

Other answers I've posted on the subject of the Dirac Delta are HERE, HERE, and HERE. This latter post provides a good primer on Distributions.

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  • What is the divergence theorem for generalized functions? – TomM Sep 15 '15 at 17:32
  • Additionally, I don't see how you arrived in the note to the divergence formula. The divergence for the original E exists everywhere but 0 as a function while delta is not a function. – TomM Sep 15 '15 at 17:35
  • @TomM Tom, I never said the Dirac Delta was a function. I stated, and provide a reference, that it is a Generalized Function. To proceed rigorously, one can use a Distribution Theoretic Approach. In that case, the "integrals" are actually linear functionals and neither Riemann nor LeBesgue Integrals. The equivalent way forward looks formally as in the note. How else can I help here? – Mark Viola Sep 15 '15 at 18:33
  • Here is a book on Generalized Functions That Discusses The Divergence Theorem in that Context. However, in practice, the physicist, engineer, and applied mathematician typically adopts the formal approach outlined in the note. – Mark Viola Sep 15 '15 at 18:40
  • The problem is that the gradient is a function and a proper one. You can use it anywhere and not only as a delta function. Therefore saying it is equal to something which is not a function implies that either the gradient is not a function or that delta is a function. – TomM Sep 15 '15 at 20:06
  • What does the theoretical mathematician do? – TomM Sep 15 '15 at 20:07
  • @tomm As mentioned, the way forward would be a Distribution Theoretic Approach, but even then formal operations known to be equivalent might still be used. – Mark Viola Sep 15 '15 at 20:09
  • I can't upvote (not enough reputation), but unfortunately I can't mark it as the right answer since I am not convinced that the divergence is indeed as Gauss' law says (I am looking for full mathematical rigor).

    I am a mathematician (graduate) and have a proclivity to rigor generally, and specifically in physics. So if you can refer me to a place that solves it in a rigorous way that would be the best answer for me.

     – TomM Sep 15 '15 at 20:13
  • Yes, the first part was good but I intended specifically the problem which we're talking about (I already knew the first part). I looked through the link but I can't see anything other than a couple of pages. The specific page you linked has an example for something which I can't understand because I can't get to the previous pages which define the symbols. Besides that, it seems that they contain an alternative formulation for the divergence theorem (perhaps for distributions?) it still remains to be seen how/why the divergence is a distribution. – TomM Sep 15 '15 at 20:26
  • @TomM In THIS ANSWER, I discuss regularizing the Electric Field of a point charge to assign meaning to define the Dirac Delta for use in the Divergence Theorem. This should provide you a rigorous way forward. – Mark Viola Sep 15 '15 at 23:30
  • Other answers I've posted on the subject of the Dirac Delta are HERE, HERE, and HERE. This latter post provides a good primer on Distributions. – Mark Viola Sep 15 '15 at 23:30
  • Ok, thanks for all your help. – TomM Sep 16 '15 at 17:48