As a part of the "rabbit hole" I am descending in order to understand the meaning of the integrals of a not-so-rarely found derivation of Ampère's law, I am trying to understand how to see the validity of the following result (found here):

Let $F$ be a $C^2$ scalar field defined on a region of $\mathbb{R}^3$ satisfying opportune assumptions [which I do not know in the details, but I think that $F\in C^2(\mathring{A})$, with $\bar{V}\subset\mathring{A}$ satisfying the conditions of the divergence theorem, and $r\in\mathring{V}$ is what is needed]; then $$\frac{1}{4\pi}\int_{V}\nabla^2 F(r')\frac{1}{|r-r'|}dV(r') =\lim_{\epsilon\downarrow 0}\frac{1}{4\pi}\int_{V\setminus B_{\epsilon}(r)}\nabla^2F(r')\frac{1}{|r-r'|}dV(r')$$$$ = \lim_{\epsilon\downarrow 0}\frac{1}{4\pi}\int_{\partial(V\setminus B_{\epsilon}(r))}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}dS(r') = -F(r)+\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}(r')dS(r').$$

I applied Green's second identity, by taking the identity $\forall r'\ne r\quad\nabla^2\left(\frac{1}{|r-r'|}\right)=0$ into account, to get (the differentiations are intended with respect to the components of $r'$; $\hat{N}_e$ is the external normal to the surface of $V\setminus B_{\epsilon}(r)$) $$\frac{1}{4\pi}\int_{V\setminus B_{\epsilon}(r)}\nabla^2F(r')\frac{1}{|r-r'|}-F(r')\nabla^2\left(\frac{1}{|r-r'|}\right)dV(r')$$$$=\frac{1}{4\pi}\int_{\partial(V\setminus B_{\epsilon}(r))}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}+ F(r')\frac{r-r'}{|r-r'^3|}\cdot \hat{N}_e\,dS(r')$$$$=\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}dS(r')+\frac{1}{4\pi}\int_{\partial V}F(r')\frac{r-r'}{|r-r'|^3}\cdot \hat{N}_e\,dS(r')$$$$+\frac{1}{4\pi}\int_{\partial B_{\epsilon}(r)}\frac{1}{|r-r'|}\frac{\partial F}{\partial n}dS(r')-\frac{1}{4\pi}\int_{\partial B_{\epsilon}(r)}\frac{F(r')}{\epsilon^2}dS(r')$$In this last member I see that, as $\epsilon\to 0$, the last addend approaches $-F(r)$, the last to one approaches $0$, but I do not understand why $\frac{1}{4\pi}\int_{\partial V}F(r')\frac{r-r'}{|r-r'^3|}\cdot \hat{N}_e\,dS(r')$ disappears. I thank both TrialAndError who told me about this identity and any other people willing to answer me.

Edit (warning): The user who first told me about this result corrected himself: its correct form is

$$\lim_{\epsilon\downarrow 0}\frac{1}{4\pi}\int_{V\setminus B_{\epsilon}(r)}\frac{\nabla^2F(r')}{|r-r'|}dV(r')$$$$=-F(r)+\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F}{\partial n'}(r')-F(r')\frac{\partial}{\partial n'}\frac{1}{|r-r'|}dS(r') $$

  • 2
    $\begingroup$ The last term will go to $4\pi F$. Then, take $V$ as all of space. Under suitable conditions on $F$, the surface integrals at infinity will vanish. $\endgroup$ – Mark Viola Jan 30 '16 at 21:11
  • $\begingroup$ @Dr.MV Thank you very much! Mmh.. well, I think that it would be enough for $F$ to be bounded, if I am understanding... Nevertheless, it seems quite unusual to me that no limit but $\lim_{\epsilon\downarrow 0}$ appears in the statement... $\endgroup$ – Self-teaching worker Jan 30 '16 at 21:24
  • 1
    $\begingroup$ Yes, sometimes the authors take that ommission lightly. Bounded is sufficient. In fact, if the scalar is a physical quantity, we assume that it is zero outside a finite volume $V'$ that is inside $V$. $\endgroup$ – Mark Viola Jan 30 '16 at 21:29
  • $\begingroup$ @Dr.MV Thank you so much! $\endgroup$ – Self-teaching worker Jan 31 '16 at 9:21
  • 1
    $\begingroup$ You're welcome! My pleasure. - Mark $\endgroup$ – Mark Viola Jan 31 '16 at 17:14

I think I have understood what the author intended and that the problem is solved. This is what I think the statement to mean:

If $F\in C^2(\mathring{A})$, $F:\mathring{A}\to\mathbb{R}$, $\bar{V}\subset \mathring{A}$, $r\in\mathring{V}$ and there exists a $\delta$ such that, for all $\epsilon\le \delta$, $\epsilon>0$, the region $V\setminus B_\epsilon(r)$ satisfies the condition of the divergence theorem, then $$F(r)=\frac{1}{4\pi}\int_{\partial V}\frac{1}{|r-r'|}\frac{\partial F}{\partial n'}(r')-F(r')\frac{\partial}{\partial n'}\frac{1}{|r-r'|}dS(r') $$$$-\lim_{\epsilon\downarrow 0}\frac{1}{4\pi}\int_{V\setminus B_{\epsilon}(r)}\frac{\nabla^2F(r')}{|r-r'|}dV(r')$$

Nevertheless I do not accept this answer of mine, reserving that to future answers confirming or refuting its correctness.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.