Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Commutations and delta in deriving Ampère's law

  1. Feb 19, 2016 #1
    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. jcsd
  3. Feb 21, 2016 #2
    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.
     
  4. Feb 29, 2016 #3
    @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...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Commutations and delta in deriving Ampère's law
  1. Delta notation (Replies: 1)

  2. Delta function (Replies: 1)

Loading...