- #1
DavideGenoa
- 155
- 5
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$$\nabla^2\phi(\mathbf{x})=-\frac{1}{4\pi}\nabla^2\int_{V}\frac{\nabla'^2\phi(\mathbf{x}')}{\|\mathbf{x}-\mathbf{x}'\|}d\mu_{\mathbf{x}'}$$.
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$$\nabla^2\phi(\mathbf{x})=-\frac{1}{4\pi}\nabla^2\int_{V}\frac{\nabla'^2\phi(\mathbf{x}')}{\|\mathbf{x}-\mathbf{x}'\|}d\mu_{\mathbf{x}'}$$.
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