Proof of Ampère's law from the Biot-Savart law for tridimensional current distributions

Question

Let us assume the validity of the Biot-Savart law for a tridimensional distribution of current:$$\mathbf{B}(\mathbf{x})=\frac{\mu_0}{4\pi}\int_V \mathbf{J}(\mathbf{y})\times\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}\: \mathrm d^3 y$$

where $\mathbf{J}$ is the density of the current distributed over the region $V$. How could we then prove Ampère's law, in the form $$\oint_{\gamma}\mathbf{B}\cdot d\mathbf{x}=\mu_0 I_{\text{linked}}\quad\text{or}\quad\nabla\times\mathbf{B}=\mu_0\mathbf{J}?$$ I would prefer a proof using multi-variable calculus only, not making use of Dirac's $\delta$, which is a mathematical tool, quite delicate to use, which would require much explanation for me to understand the equalities where it appears (I am a beginner, as you can read in my profile, but I just want to understand why physical laws are derived in a mathematically correct way and I am not interested in shortcuts unless knowing their mathematical proof) and whose utilisation sometimes found in textbooks causes enormous problems to me when the text does not say, for example, why it commutes integral and differentiation signs, like $\int$ and $\nabla\times$. Anyhow, I would be very grateful to anybody producing any proof, even using the $\delta$, but clearly explaining what theorems and mathematical results justify delicate steps such as those where integral (explicitly defined as Riemann, Lebesgue, symbolic signs for distributions or other) and differential operators commute. I heartily thank any answerer. Suggestions for proofs having such requisites that can be found on line or on printed books (better if not using the $\delta$) are very welcome! I heartily thank any answerer.

P.S.: A derivation of Ampère's law from the Biot-Savart one is here, but it encompasses the case of a linear distribution of current, which is different, as I thought to be blatantly obvious before seeing "duplicate" close votes.

---

Trials: Although I have been looking for a rigorous and understandable (by me) proof for more than a month, I have not been able to find, or to produce, one. The most common derivation of Ampère's law from the Biot-Savart one essentially uses the calculations found in Wikipedia's outline of proof, which are un-understandable to me: I know some properties of Dirac's $\delta$ like the fact that, if $(J_1,J_2,J_3)=\mathbf{J}\in C^\infty(\mathbb{R}^3)$ is compactly supported, then $$-\frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{\nabla_l^2 J_i(\mathbf{l})}{\|\mathbf{l}-\mathbf{r}\|}\; d\mu_{\mathbf{l}}=J_i(\mathbf{r})=:\int\delta(\mathbf{l}-\mathbf{r})\,J_i(\mathbf{l})\,\mathrm d^3l$$ where the integral on the left is of the Lebesgue type, while the integral sign on the right is just a symbolic notation for a linear functional, but I have got tremendous problems in understanding the steps found in that outline of proof, for the reasons I explain here in details for potential answerers itending to modify it, by adding the explanations of the mathematical facts used to justify the commutations between $\iiint$ and $\nabla$ and of the exact meaning of the integral signs, to produce a more detailed and complete proof. Just to make a brief resumee of the problems I found in Wikipedia's outline of proof, as I have been requested by a commenter, I do not understand what those integral signs mean (Lebesgue integrals, symbolic signs for functionals or what else), neither do I understand what the derivative components of $\nabla\times\mathbf{B}$ are: since theorems such as Stokes' are usually applied when integrating $\nabla\times\mathbf{B}$, I would believe that they are the ordinary derivatives of elementary multivariate calculus, but then the $\delta$, which is a tool of the theory of distributions, pops up in the outline of proof, and in the theory of distributions there exist derivatives of distributions which are a very different thing, but 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, the outline of proof starts with $\nabla_r\times \mathbf{B}$ with $r$ , while the integral appears with $\mathrm{d}^3\mathrm{l}$...

Answer 1

I think I have a proof for you (though you may find it unsatisfying). For the most part, I'm following the proof on the wikipedia page you link to. I do avoid the dirac delta function however. Starting from the Biot-Savart law:

$$\mathbf{B}(\mathbf{r})=\frac{\mu_0}{4\pi}\int_V \mathbf{J}(\mathbf{l})\times\frac{\mathbf{r}-\mathbf{l}}{\|\mathbf{r}-\mathbf{l}\|^3}\: \rm d^3l$$

One of your criticisms of the proof was that the type of integral is left unspecified. The functions we need to integrate are Riemann integrable (and thus Lebesgue integrable as well), so for the purposes of choosing particular proofs, I'm going to proceed based on the Riemann definition (though this is an arbitrary choice on my part).

A problem we encounter already is that the Biot-Savart law is an improper integral. We solve this problem by saying the integral is the Cauchy principle value, which is defined in terms of a limit (this is relevant).

I make the substitution:

$$ \frac{\mathbf{r}-\mathbf{l}}{\|\mathbf{r}-\mathbf{l}\|^3}= - \nabla_\mathbf{r} \left( \frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right)$$

$$\mathbf{B}(\mathbf{r})= - \frac{\mu_0}{4\pi}\int_V \mathbf{J}(\mathbf{l})\times \nabla_\mathbf{r} \left( \frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right)\: \rm d^3l$$

Using a standard vector calculus identity:

$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int_V \nabla_\mathbf{r} \times \left( \frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}\right)\: \rm d^3l $$

Next, we're going to take the curl operator outside the integral. Since both the curl and integral are defined in terms of limits, this is equivalent to exchanging the order of limits, which is generally acceptable given certain convergence criteria (e.g. uniform convergence. There are a few different convergence theorems that may be appropriate).

$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \nabla_\mathbf{r} \times \int_V \left( \frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}\right)\: \rm d^3l $$

Applying curl to both sides of the equation, and the vector calculus identity for the curl of a curl:

$$\nabla_\mathbf{r} \times \mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \nabla_\mathbf{r} \int_V \nabla_\mathbf{r} \cdot \left( \frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}\right)\: \rm d^3l - \frac{\mu_0}{4\pi} \int_V \nabla_\mathbf{r}^2 \left( \frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}\right)\: \rm d^3l $$

I'll assume you're familiar with the argument for why the first integral term is 0. This leaves:

$$\nabla_\mathbf{r} \times \mathbf{B}(\mathbf{r}) = - \frac{\mu_0}{4\pi} \int_V \mathbf{J}(\mathbf{l}) \nabla_\mathbf{r}^2 \left( \frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right)\: \rm d^3l $$

In evaluating the Laplace operator, I'm using the typical expression for spherical coordinates, but shifted by $\mathbf{l}$.

$$\nabla_\mathbf{r}^2 \left( \frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right) = \frac{1}{\|\mathbf{r}-\mathbf{l}\|^2} \frac{\partial}{\partial r} \left(\|\mathbf{r}-\mathbf{l}\|^2 \frac{\partial}{\partial r} \frac{1}{\|\mathbf{r}-\mathbf{l}\|} \right)$$

You'll notice this evaluates to 0 everywhere, except at $\mathbf{r}=\mathbf{l}$, where the expression is not defined. Since the expression inside the integral evaluates to 0 everywhere except at $\mathbf{r}=\mathbf{l}$, we can make this substitution in the current term, and move it outside the integral.

$$\nabla_\mathbf{r} \times \mathbf{B}(\mathbf{r}) = - \frac{\mu_0}{4\pi} \mathbf{J}(\mathbf{r}) \int_V \nabla_\mathbf{r}^2 \left( \frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right)\: \rm d^3l $$

To evaluate the final integral, we apply the divergence theorem.

$$ \int_V \nabla_\mathbf{r}^2 \left( \frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right)\: \rm d^3l = \int_{\partial V} \mathbf{\hat n} \cdot \nabla_\mathbf{r} \left( \frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right) \: \rm d^2l = -\int_{\partial V} \mathbf{\hat n} \cdot \left(\frac{\mathbf{r}-\mathbf{l}}{\|\mathbf{r}-\mathbf{l}\|^3}\right) \: \rm d^2l $$

Because the value inside the volume integral is 0 everywhere except at $\mathbf{r}=\mathbf{l}$, we can choose the volume to be spherically symmetric about $\mathbf{l}$, and finally evaluate the integral.

$$\int_{\partial V} \mathbf{\hat n} \cdot \left(\frac{\mathbf{r}-\mathbf{l}}{\|\mathbf{r}-\mathbf{l}\|^3}\right) \: \rm d^2l = 4\pi$$

This leaves us with Ampere's law.

$$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} $$

Edit: As pointed out, if we were to use our definition of integral, the answer would just come out to 0. Remember that exchanging limits is allowed given certain convergence criteria? If the integrals or integrands don't converge, then exchanging the limits is an error based on the formulations I've chosen. In this case, the integrand doesn't converge.

We have shown however, that excluding the divergent point, the integral is 0. We can divide our original integral into two domains, $V\setminus B(\mathbf{r},\epsilon)$, and $B(\mathbf{r},\epsilon)$, which is a ball of radius $\epsilon$ centered at $\mathbf{r}$. We've already shown that the integral over the first domain is 0. Based on this:

$$\nabla_r \times \mathbf{B}(\mathbf{r})= -\frac{\mu_0}{4\pi} \nabla_\mathbf{r} \times \int_{B(\mathbf{r},\epsilon)} \nabla_\mathbf{r} \times \left( \frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}\right)\: \rm d^3l$$

The current density is constant in the limit of $\epsilon \rightarrow 0$, so we can take it outside the integral. Additionally, the integral/integrand here do converge, considering the Cauchy principle value, so we can move all the derivatives outside the integral. The steps are largely the same as before, except all derivative operators are outside the integral. This leads to the following equation:

$$\nabla_\mathbf{r} \times \mathbf{B}(\mathbf{r}) = - \frac{\mu_0}{4\pi} \mathbf{J}(\mathbf{r}) \nabla_\mathbf{r}^2 \int_{B(\mathbf{r},\epsilon)} \left( \frac{1}{\|\mathbf{r}-\mathbf{l}\|}\right)\: \rm d^3l $$

The only difference compared with our initial proof is that the Laplacian operator is outside the integral. The integral here can be evaluated, followed by the Laplacian operator. This should produce $-4\pi$, finishing the proof.

One subtle difficulty in evaluating this integral, even though we've written the domain as $B(\mathbf{r},\epsilon)$, this is an abuse of notation. The domain isn't really a function of $\mathbf{r}$, and so should remain constant for the purposes of evaluating $\nabla_\mathbf{r}^2$. Otherwise, you get the odd result that the integral isn't a function of $\mathbf{r}$, and the whole expression becomes 0.

Answer 2

I hope I have been able to find a proof using Lebesgue integrals, where the differentiations under the integral signs are justified, if I am not wrong, by proved mathematical results.

Let $\mathbf{J}:\mathbb{R}^3\to \mathbb{R}^3$ be of class $C^2 (\mathbb{R}^3)$ and let its support be contained in the bounded compact subset $V\subset\mathbb{R}^3$, $\mu$-measurable according to the usual Lebesgue measure $\mu$ defined in $\mathbb{R}^3$.

I will use 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.

Let us define $$\mathbf{A}(\mathbf{x}):=\frac{\mu_0}{4\pi}\int_{V}\frac{\mathbf{J}(\mathbf{y})}{\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{y})}{\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}$$where we notice that the point $\mathbf{y}=\mathbf{x}$ does not prevent, for any $\mathbf{x}\in\mathbb{R}^3$, the summability of $\mathbf{y}\mapsto \|\mathbf{x}-\mathbf{y}\|^{-1}$ on any bounded measurable subset of $\mathbb{R}^3$, as we can verify by using spherical coordinates around $\mathbf{x}$ and by taking the properties of the Lebesgue integral and its relationship with the Riemann integral into account.

Therefore, if we define $\mathbf{B}(\mathbf{x}):=\frac{\mu_0}{4\pi}\int_V \mathbf{J}(\mathbf{y})\times\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}d\mu_{\mathbf{y}}=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\nabla_x\times\left[\frac{\mathbf{J}(\mathbf{y})}{\|\mathbf{x}-\mathbf{y}\|}\right]d\mu_{\mathbf{y}}$, I would say, by applying the lemma above with $\varphi=\frac{\mu_0}{4\pi}J_i$ ($i=1,2,3$), that $$\nabla\times\mathbf{A}=\mathbf{B}$$and, by using a known identity for the curl of the curl, which can be applied since this argument, with $f(z)=\|z\|^{-1}$ and $g=\frac{\mu_0}{4\pi} J_i$ ($i=1,2,3$), shows that $$\mathbf{A}\in C^2(\mathbb{R}^3),$$ we see, by applying this argument again, that $$\nabla_x\times\mathbf{B}(\mathbf{x})=\frac{\mu_0}{4\pi} \nabla_x\left[\nabla_x\cdot\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{y})} {\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}\right]-\frac{\mu_0}{4\pi} \nabla_x^2\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{y})} {\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}$$$$=\frac{\mu_0}{4\pi} \nabla_x\int_{\mathbb{R}^3}\frac{\nabla_y\cdot\mathbf{J}(\mathbf{y})} {\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}-\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\frac{\nabla_y^2\mathbf{J}(\mathbf{y})} {\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}$$where the first addendum is null if $\forall \mathbf{x}\in\mathbb{R}^3\quad\nabla_x\cdot\mathbf{J}(\mathbf{x})=0$, which holds for a stationary current, and where the second addendum I think to be able to be proved, analogously to what is done here for a $C^{\infty}$ compactly supported function, to be $$-\frac{\mu_0}{4\pi} \int_{\mathbb{R}^3}\frac{\nabla_y^2\mathbf{J}(\mathbf{y})} {\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}=\mu_0\mathbf{J}(\mathbf{x})=:\mu_0\int\delta(\mathbf{y}-\mathbf{x})\mathbf{J}(\mathbf{y})d^3y$$which obviously is not $-\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\nabla_x^2[\|\mathbf{x}-\mathbf{y}\|^{-1}]\mathbf{J}(\mathbf{y})d\mu_{\mathbf{y}}\equiv\mathbf{0}$ in general; I am saying this in order to clarify the notation of a step used in Wikipedia's outline of proof, which is a very commonly found derivation of Ampère's law from the Biot-Savart law, which also appears in Jackson's Classical Electrodynamics.

Let us also notice, with reference to that outline of proof, that in the term $\nabla_x\cdot\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{y})} {\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}$ we can differentiate, by using the lemma quoted above, with respect to the components of $\mathbf{x}$ under the integral sign and therefore $$\int_{\mathbb{R}^3}\frac{\nabla_y\cdot\mathbf{J}(\mathbf{y})} {\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}=\nabla_x\cdot\int_{\mathbb{R}^3}\frac{\mathbf{J}(\mathbf{y})} {\|\mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}$$$$=\int_{\mathbb{R}^3}\mathbf{J}(\mathbf{y})\cdot\nabla_x\left[\frac{1} {\|\mathbf{x}-\mathbf{y}\|}\right]d\mu_{\mathbf{y}}$$so I suppose that one possible interpretation of the integrals of that outline of proof is to read them as Lebesgue integrals except for the integral where $\nabla^2\left(\frac{1}{|\mathbf{r}-\mathbf{l}|}\right)$ appears.