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I have been struggling to understand the only derivation of Ampère's law from the Biot-Savart law for a tridimensional distribution of current (which, needless to say, is not the case of a linear distribution of current discussed here) that I have been able to find, i.e. Wikipedia's outline of proof, for more than a month with no result.

I was not excluding that Wikipedia's outline of proof is one of those cases, whose set I have been told to be non-empy, where physics, at least at some level, renounces the rigour of mathematics, until I was told that the proof is indeed rigourous. Although I have recently asked for an alternative proof of the same entailment, the fact that it is not to be excluded that Wikipedia's outline of proof is rigourous has made me decide, as it has been suggested to me in these comments, to ask for an explanation of the single steps that I do not understand of that outline of proof.

My first doubt starts at the first commutation between the integral and curl signs. I know that $$\mathbf{J}(\mathbf{l})\times\frac{\mathbf{r}-\mathbf{l}}{|\mathbf{r}-\mathbf{l}|^3}=\nabla_r\times\left(\frac{\mathbf{J}(\mathbf{l})}{|\mathbf{r}-\mathbf{l}|}\right)$$provided that we read the components of $\nabla_r$ as ordinary derivatives (see below) in the sense of elementary multivariable calculus. The outline of proof says that $$\mathbf{B}(\mathbf{r}):=\iiint_V\,d^3l\,\mathbf{J}(\mathbf{l})\times\frac{\mathbf{r}-\mathbf{l}}{|\mathbf{r}-\mathbf{l}|^3}=\nabla_r\times\iiint_V\,d^3l\,\frac{\mathbf{J}(\mathbf{l})}{|\mathbf{r}-\mathbf{l}|}.$$What mathematical result justifies that commutation between integral and curl? I heartily thank you for any answer.

In order to take the problem on, it is obviously needed to understand what the integral and the derivatives in $\nabla$ are, but I am not sure about what they are. Since theorems such as Stokes' are usually applied when integrating $\nabla\times\mathbf{B}$, I would tempted to believe that the components of $\nabla_r$ are ordinary derivatives of elementary multivariable/vector calculus, but Dirac's $\delta$, which is a tool of the theory of distributions, "pops out" at a certain point in the outline of proof, and in the theory of distributions there exist derivatives of distributions which are a very different thing, although they are taken, as far as I know, with respect to the variables written as "variables of integration" in the distribution integral notation, while, here, we start with $\nabla_r\times \mathbf{B}$ with $r$, while the integral is $\iiint_V d^3l$ with $l$. As to the integral sign, I would tend to interpretate it as the Lebesgue integral $$\int_V\mathbf{J}(\mathbf{l})\times\frac{\mathbf{r}-\mathbf{l}}{|\mathbf{r}-\mathbf{l}|^3}\,d\mu_{\mathbf{l}}$$ or as the limit of a Riemann integral (where $B(\mathbf{r},\varepsilon)$ is the ball centred in $\mathbf{r}$ with radius $\varepsilon$) $$\lim_{\varepsilon\to 0}\iiint_{V\setminus B(\mathbf{r},\varepsilon)}\mathbf{J}(\mathbf{l})\times\frac{\mathbf{r}-\mathbf{l}}{|\mathbf{r}-\mathbf{l}|^3}\,dl_1dl_2dl_3$$but, then, if the commutations between integral and differential operators used Wikipedia's outline of proof were licit for this interpretation, we would then conclude that $\iiint_V\,d^3l\, \mathbf{J}(\mathbf{l})\nabla_l^2\left(\frac{1}{|\mathbf{r}-\mathbf{l}|}\right)$ $=\int_V \mathbf{J}(\mathbf{l})\nabla_l^2\left(\frac{1}{|\mathbf{r}-\mathbf{l}|}\right)\,d\mu_{\mathbf{l}}$ $=\int_{V\setminus{\{\mathbf{r}}\}} \mathbf{J}(\mathbf{l})\nabla_l^2\left(\frac{1}{|\mathbf{r}-\mathbf{l}|}\right)\,d\mu_{\mathbf{l}}=\mathbf{0}$ even where $\mathbf{J}(\mathbf{r})\ne\mathbf{0}$, which must not be the case.

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    $\begingroup$ A question so specifically asking about a mathematical detail should be on Math.SE. Besides that, exchanging derivatives with integrals really comes down to exchanging the order of two limits. This is an incredibly well studied/described topic. $\endgroup$
    – DanielSank
    Feb 13, 2016 at 19:33
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    $\begingroup$ surf from a google search on Leibniz Rule or Leibniz Integral Rule. Indeed , it's well documented. $\endgroup$
    – igael
    Feb 13, 2016 at 19:56
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    $\begingroup$ If you're worried about the compactness of the domain just make a sequence of ever-larger domains, construct the limit, and commute that limit around just like the others. Or, better yet, think of improper integrals as a limit with a partition of unity. $\endgroup$
    – DanielSank
    Feb 13, 2016 at 20:06
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    $\begingroup$ did you account of the approximations of magnetostatics ? not sure that it is a question for math.se $\endgroup$
    – igael
    Feb 13, 2016 at 20:16
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    $\begingroup$ Ampere law is not accurate with wires of complex shapes, short wires, misshapen solenoids or toroids. It is not surprising to get inconsistencies when equaling the 2 calculus. Chapters 9.10.1 & 9.10.2 of direct pdf course may help to retreive the equality in the special cases. source $\endgroup$
    – igael
    Feb 13, 2016 at 21:25

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I hope I have found a proof. In fact, if the components of $\mathbf{J}:V\subset \mathbb{R}^3 \to\mathbb{R}^3$, are bounded and $\mu$-measurable, with $\mu$ as the $3$-dimensional Lebesgue measure, and $V$ is bounded and measurable according to the same measure, the following lemma

Let $\varphi:V\subset\mathbb{R}^3\to\mathbb{R}$ be bounded and $\mu_{\mathbf{y}}$-measurable, with $\mu_{\mathbf{y}}$ as the usual $3$-dimensional Lebesgue measure, where $V$ is bounded and measurable (according to the same measure). Let us define, for all $\mathbf{x}\in\mathbb{R}^3$, $$\Phi(\mathbf{x}):=\int_V \frac{\varphi(\mathbf{y})}{\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}$$then $\Phi\in C^1(\mathbb{R}^3)$ and, for $k=1,2,3$, $$\forall\mathbf{x}\in\mathbb{R}^3\quad\quad\frac{\partial \Phi(\mathbf{x})}{\partial x_k}=\int_V\frac{\partial}{\partial x_k} \left[\frac{\varphi(\mathbf{y})}{\|\mathbf{x}-\mathbf{y}\|}\right]d\mu_{\mathbf{y}}=\int_V \varphi(\mathbf{y})\frac{y_k-x_k}{\|\mathbf{x}-\mathbf{y}\|^3}d\mu_{\mathbf{y}}$$

which is proved here, with $\varphi=J_i$, $i=1,2,3$, would allow the desired commutation between integral and curl signs.

For a complete proof of how Ampère's law is entailed by the Biot-Savart, I hope I have reached one here.

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