# Commutations and delta in deriving Ampère's law

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1. Feb 19, 2016

### DavideGenoa

Hi, friends! I have been able to understand, thanks to Hawkeye18, whom I thank again, that, if $\mathbf{J}$ is measurable according to the usual $\mathbb{R}^3$ Lebesgue measure $\mu_{\mathbf{l}}$ and bounded, a reasonable hypothesis if we consider it the density of current, if $V\subset\mathbb{R}^3$ is bounded and $\mu_{\mathbf{l}}$-measurable and if we define, for any $\mathbf{r}\in\mathbb{R}^3$,$$\mathbf{B}(\mathbf{r}):=\frac{\mu_0}{4\pi}\int_V\mathbf{J}(\mathbf{l}) \times\frac{\mathbf{r}-\mathbf{l}}{\|\mathbf{r}-\mathbf{l}\|^3}d\mu_{\mathbf{l}}\quad\text{ and }\quad\mathbf{A}(\mathbf{r}):=\frac{\mu_0}{4\pi}\int_V \frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}d\mu_{\mathbf{l}}$$then$$\nabla_r\times\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4\pi}\int_V\nabla_r\times\left[ \frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}\right]d\mu_{\mathbf{l}}=\mathbf{B}(\mathbf{r}).$$

Wikipedia's derivation of Ampère's law from the Biot-Savart law, which is the only one that I have been able to find, calculates $$\nabla\times\mathbf{B}=\frac{\mu_0}{4\pi}\nabla\iiint_Vd^3l\mathbf{J}(\mathbf{l})\cdot\nabla\left(\frac{1}{|\mathbf{r}-\mathbf{l}|}\right)-\frac{\mu_0}{4\pi}\iiint_Vd^3l\mathbf{J}(\mathbf{l})\nabla^2\left(\frac{1}{|\mathbf{r}-\mathbf{l}|}\right) .$$I suppose that this equality descends from the identity$$\nabla_r\times\left[\nabla_r\times\left[\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}\right]\right]=\nabla_r\left[ \mathbf{J}(\mathbf{l})\cdot\nabla_r\left[\frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right] \right]-\nabla_r^2\left[\frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right]\mathbf{J}(\mathbf{l})$$valid for any $\mathbf{r}\in\mathring{A}$, $\mathbf{r}\ne\mathbf{l}$, if $\mathbf{J}\in C^2(\mathring{A})$, with some commutation between integral and differential operators which I am not able to justify. How can such commutations be mathematically proved, provided that they make sense (see below)?

Moreover, it is worth of note that $$-\frac{\mu_0}{4\pi}\iiint_Vd^3l\mathbf{J}(\mathbf{l})\nabla^2\left(\frac{1}{|\mathbf{r}-\mathbf{l}|}\right)$$ is equated with $\mu_0\mathbf{J}(\mathbf{r})$ as if it were (following equality for ex. if $\mathbf{J}$ is a Schwarz function with compact support contained in $V$)$$\mu_0\int\delta(\mathbf{l}-\mathbf{r}) \mathbf{J}(\mathbf{l})d^3l:=\mu_0\mathbf{J}(\mathbf{r})=\frac{\mu_0}{4\pi}\int_{V}\frac{\nabla_l^2\mathbf{J}(l)}{\|\mathbf{r}-\mathbf{l}\|}d\mu_{\mathbf{l}}$$where the integral on the left is just a symbolic notation for a linear operator and the integral on the right, as the measure $\mu_{\mathbf{l}}$ indicates, is a Lebesgue integral, while we can notice that $$\forall\mathbf{r}\in\mathbb{R}^3\quad-\frac{\mu_0}{4\pi}\int_V\nabla_r^2\left[\frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right]\mathbf{J}(\mathbf{l})d\mu_{\mathbf{l}}=\mathbf{0}$$(and the same with $\nabla_l^2$ in place of $\nabla_r^2$): a quite different result.
How can the steps of Wikipedia's outline of proof be justified mathematically? I have even been told, by a friend of mine who is a graduate mathematics student, that such a derivation is not correct, but I have not been able to find another one...

I $\infty$-ly thank you for any answer!

P.S.: To illustrate something that I know that I do not exclude to be useful to prove the desired result, I think that, if $\phi\in C^2(\mathring{A})$, $\phi:\mathring{A}\to\mathbb{R}$, $\bar{V}\subset \mathring{A}$, $\mathbf{x}\in\mathring{V}$ and there is a $\delta$ such that, for all $\varepsilon\le \delta$, $\varepsilon>0$, the region $V\setminus B_\epsilon(\mathbf{x})$ satisfies the assumptions of Gauss' divergence theorem, then$$\phi(\mathbf{x})=\frac{1}{4\pi}\int_{\partial V}\left( \frac{\nabla'\phi(\mathbf{x}')}{\|\mathbf{x}-\mathbf{x}'\|}-\phi(\mathbf{x}')\frac{\mathbf{x}'-\mathbf{x}}{\|\mathbf{x}-\mathbf{x}'\|}\right)\cdot\mathbf{N}_e 'd\sigma' -\frac{1}{4\pi}\lim_{\varepsilon\to 0^{+}}\int_{V\setminus B_{\varepsilon}(\mathbf{x})}\frac{\nabla'^2\phi(\mathbf{x}')}{\|\mathbf{x}-\mathbf{x}'\|}dV'$$and$$\nabla^2\phi(\mathbf{x})=-\frac{1}{4\pi}\nabla^2\int_{V}\frac{\nabla'^2\phi(\mathbf{x}')}{\|\mathbf{x}-\mathbf{x}'\|}d\mu_{\mathbf{x}'}$$.

Last edited by a moderator: Feb 20, 2016
2. Feb 21, 2016

### Xiuh

In my opinion the way to go is to use distribution theory. In the book Applied Functional Analysis - Applications to Mathematical Physics by Eberhad Zeidler there is a discussion of the electrostatics case (page 183). Maybe you can try to use similar arguments to prove the magnetostatics case.

3. Feb 29, 2016

### DavideGenoa

@Xiuh I thank you so much for your bibliographic suggestion! I hope that my last trial to understand how Ampère's law is a consequence of the Biot-Savart law is not so incorrect. Nevertheless, I still cannot give a meaning to each of the steps appearing in Wikipedia's outline proof, which I must frankly admit to find very misleading for the inexperienced student and not only for him/her. In fact, while asking about what the integrals are (Lebesgue or what else) and why the commutations are mathematically licit in some sites and forums, I have received answers by people seeing commutations between integrals and derivatives as something so natural to prevent them understanding even my question and I am not only talking about undergraduate students, but even professionals...