# Commutation integral/derivative in deriving Ampère's law

1. Feb 14, 2016

### DavideGenoa

Hi, friends! I have been struggling to understand the only derivation of Ampère's law from the Biot-Savart law for a tridimensional distribution of current that I have been able to find, i.e. Wikipedia's outline of proof, for more than a month with no result. I have also been looking for a proof not using Dirac's $\delta$, which complicates things with respect to the use of Riemann or Lebesgue integrals because I am less familiar with its use and the reasons it "pops out" in some proofs, but I have not succeeded until now.

Therefore I would like to ask here about the steps that I do not understand of that outline of proof, which I suppose by default to be correct, also because it also appears in J.D. Jackson's Classical Electrodynamics, which I know to be considered an outstanding text, and I am posting the question in this section because it focusses on the mathematical methods used to produce that derivation of Ampère's law.

My first doubt starts at the very first commutation between the integral and curl signs. I know that $$\mathbf{J}(\mathbf{l})\times\frac{\mathbf{r}-\mathbf{l}}{|\mathbf{r}-\mathbf{l}|^3}=\nabla_r\times\left(\frac{\mathbf{J}(\mathbf{l})}{|\mathbf{r}-\mathbf{l}|}\right)$$ at least provided that we read the components of $\nabla_r$ as ordinary derivatives in the usual 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?

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 $$\lim_{\varepsilon\to 0}\iiint_{V\setminus B(\mathbf{r},\varepsilon)}\mathbf{J}(\mathbf{l})\times\frac{\mathbf{r}-\mathbf{l}}{|\mathbf{r}-\mathbf{l}|^3}\,dl_1dl_2dl_3$$but, then, if the commutations between integral and differential operators used in 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 cannot be the case.

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

2. Feb 16, 2016

### blue_leaf77

The integral is over the volume occupied by the current, while the curl acts on the observation point. The coordinate within the integrated volume and the coordinate of the observation point are independent of each other, therefore the curl and integral commute.

3. Feb 16, 2016

### PeroK

The equation over which you have your first doubt is an identity that only applies for the special case of the function $1/r$. It won't be true in general. Simply take the curl to check.

These special identities involving derivatives of $1/r$ crop up frequently.

4. Feb 16, 2016

### PeroK

The basic result on which this depends is. If

$f(x) = \int h(x,y)dy$

Then

$f'(x) = \int \frac{\partial h}{\partial x} dy$

I'm not sure of the precise conditions on $h$ for this to hold. It must be well-behaved, some sort of uniform continuity of the partial derivative, perhaps. Look it up somewhere online: Leibnitz rule.

Last edited: Feb 16, 2016
5. Feb 16, 2016

### DavideGenoa

@everybody: I heartily thank you for your replies! There is a fundamental point that I am missing and that I think should clear before taking the issue on: what are those integral? Some sort of limit of a Riemann integral, Lebesgue integrals, distributions or what else? (and, if they are distributions, what do they map to what?)
As to the derivatives in the "components" of $\nabla$, are they the ordinary partial derivatives of multivariable calculus? Or is some other special notation I am not aware of? (I don't think they are derivative of distributions because that kind of derivative is calculated with respect to the "variables of integration" present in the integral symbolic notation).

Forgive me, I don't know any mathematical result using this fact to allow such a commutation. Could you state it and, if it is possible to you, post a link to a proof (or write a proof here if you have got the time)?

I know a formulation of Leibniz rule in the form: if $V\subset\mathbb{R}^n$ is compact and $f:V\times [a,b]\to \mathbb{R}$, $(\boldsymbol{x},t)\mapsto f(\boldsymbol{x},t)$ is such that $\frac{\partial f}{\partial t}\in C(V\times[a,b])$ (and therefore,by the Heine-Cantor theorem, uniformly continuous on $V\times[a,b]$, which is compact), then $$\forall t\in[a,b]\quad\frac{d}{dt}\int_Vf(\boldsymbol{x},t)d^nx=\int_V\frac{\partial f(\boldsymbol{x},t)}{\partial t}d^nx$$where the integrals are (proper) Riemann integrals.

Nevertheless, our $\iiint_V d^3x$ cannot be a Riemann integral (at least a proper one) because the integrand is not bounded if $\mathbf{r}\in V$.
Moreover, we should identify $f:(\boldsymbol{x},t)\mapsto f(\boldsymbol{x},t)$ with the various component functions of the integrand whose derivatives are taken, for ex. $f:(\mathbf{l},r_2)\mapsto \frac{J_3(\mathbf{l})}{|\mathbf{r}-\mathbf{l}|}$ (where I write numbered components: $(r_1,r_2,r_3)=\mathbf{r}$), but $\frac{\partial f(\mathbf{l},r_2)}{\partial r_2}=\frac{J_3(\mathbf{l})(l_2-r_2)}{|\mathbf{r}-\mathbf{l}|^3}$ and we have no continuity where $\mathbf{r}=\mathbf{l}$...

I also know that a corollary of Lebesgue's dominated convergence theorem guaranteeing that:
- if $f:V\times [a,b]\to \mathbb{R}$, $(\boldsymbol{x},t)\mapsto f(\boldsymbol{x},t)$ with $V$ $\mu_{\boldsymbol{x}}$-measurable is such that $\forall t\in[a,b]\quad f(-,t)\in L_1(V,\mu_{\boldsymbol{x}})$, i.e. the function $\boldsymbol{x}\mapsto f(\boldsymbol{x},t)$ is Lebesgue summable on $V$,
- and if there is a neighbourhood $B(t_0,\delta)$ of $t_0$ such that, for almost all $\boldsymbol{x}\in V$ and for all $t\in B(t_0,\delta)$, $\left|\frac{\partial f(\boldsymbol{x},t)}{\partial t}\right|\le\varphi(\boldsymbol{x})$, where $\varphi\in L_1(V,\mu_{\boldsymbol{x}})$ is a Lebesgue summable non negative function, then
$$\frac{d}{dt}\int_V f(\boldsymbol{x},t) d\mu_{\boldsymbol{x}}\bigg|_{t=t_0}=\int_V\frac{\partial f(\boldsymbol{x},t_o)}{\partial t}d\mu_{\boldsymbol{x}}$$where the integrals are itended as Lebesgue integrals with respect to measure $\mu_{\boldsymbol{x}}$. In the case of $V\subset\mathbb{R}^n$, $\mu_{\boldsymbol{x}}$ is often chosen as the usual Lebesgue measure on $\mathbb{R}^n$.

I have been searching for a long time for a summable $\varphi$ such that $\left|\frac{\partial f(\mathbf{l},r_i)}{\partial r_i}\right|\le\varphi(\mathbf{l})$, but with no result.
Let us also notice that, as I said in the original post, if all the commutations between differential operators and integrals were licit and the integrals were all of the Lebesgue type, then the last integral I wrote in the original post would be $\mathbf{0}$...

I $\infty$-ly thank you all!

6. Feb 16, 2016

### Hawkeye18

First, you need to specify the assumptions. For example, the formula is true if $V$ is a bounded domain and $J$ is a bounded measurable function (there is no need to assume the continuity of $J$. The standard Leibniz rule, and even its advanced version like here does not apply in your case, since the integrand has singularity at $l=r$.

Let me show one possible proof. Denote $$\Phi(r)=\iiint_V \frac{\phi(l)}{|r-l|}d^3l, \qquad \Phi_k(r) = \iiint_V \phi(l)\frac{l_k-r_k}{|l-r|^3} d^3l,$$ where $\phi$ is one of the components of $J$ and $l_k$, $r_k$, $k=1, 2, 3$ are the coordinates of $l$ and $r$. Note that $\Phi_k$ is a formal derivative of $\Phi$. We want to show that $$\frac{\partial \Phi}{\partial r_k}=\Phi_k.$$ Let us do it for $k=1$. Fix $r_2$ and $r_3$, and let $r_1\in [a,b]$. Consider the integral $$\int_a^b \iiint_V \phi(l)\frac{l_1-r_1}{|l-r|^3} d^3l \,\,dr_1 = \int_a^b \Phi_1(r) dr_1.$$ The function n the integral is integrable, so the Fubini theorem applies and we can change the order of integration; integrating first with respect to $r_1$ and then with respect to $d^3l$ we get that $$\int_a^b \Phi_1(r) dr_1 =\Phi(r^b )-\Phi(r^a),$$ where $r^a=(r_1, r_2, a)$ and $r^b=(r_1, r_2, b)$ (recall that $r_1$ and $r_2$ are fixed).

Then, if we show that the function $\Phi_1(r)$ is continuous, the fundamental theorem of Calculus will imply that $$\frac{\partial\Phi}{\partial r_1} = \Phi_1.$$ Continuity of can be easily proved by pretty standard reasoning. Let $|\phi(l)|\le M$ It is easy to see that $$\iiint_{|l-r|<\delta }\frac{|l_1-r_1|}{|l-r|^3} d^3 l \le \iiint_{|l-r|<\delta }\frac{1}{|l-r|^2} d^3l = \iiint_{|l|<\delta }\frac{1}{|l|^2} d^3l \to 0$$ as $\delta\to 0$. It is also not hard to see that for any ball $B_\delta$ of radius $\delta$ we have $$\iiint_{B_\delta }\frac{|l_1-r_1|}{|l-r|^3} d^3 l \le \iiint_{|l-r|<\delta }\frac{1}{|l-r|^2} d^3l = \iiint_{|l|<\delta }\frac{1}{|l|^2} d^3l \to 0$$ as $\delta\to 0$.

So, given $\varepsilon>0$ we can find $\delta_1>0$ such that $$M\iiint_{|l|<\delta }\frac{1}{|l|^2} d^3l <\varepsilon/3.$$ Then to estimate $\Phi_1(r)-\Phi_1(r')$ we split the integral into integral over the set $|l-r|<\delta_1$ and the integral over the rest. The above reasoning shows that the integral over the set $|l-r|<\delta_1$ is bounded by $2\varepsilon/3$. The integral over the rest can be made arbitrarily small by taking $r'$ close to $r$.

PS. Note that a similar trick would not work with your example with Laplacian: when you take the second derivative of the integrand, you will get a function that is not integrable, and you cannot apply Fubini.

7. Feb 16, 2016

### blue_leaf77

By writing $\mathbf{l} = x_l \hat{x} + y_l \hat{y} +z_l \hat{z}$ and $\mathbf{r} = x_r\hat{x} + y_r \hat{y} +z_r \hat{z}$, the expression
$$\frac{\mathbf{J}(\mathbf{l})}{|\mathbf{r}-\mathbf{l}|}$$
can be denoted as a function of 6 variables, $f(x_l,y_l,z_l,x_r,y_r,z_r)$. Meanwhile the index $r$ following the nabla indicates that the partial derivatives involved there are to be performed against $(x_r,y_r,z_r)$. This means
$$\nabla_r = \frac{\partial}{\partial x_r} \hat{x} +\frac{\partial}{\partial y_r} \hat{y} +\frac{\partial}{\partial z_r} \hat{z}$$
The the integral in the equation of the magnetic field is over $dx_l\hspace{1mm}dy_l\hspace{1mm}dz_l$. Therefore, the curl and the integral are not intertwined at all.

8. Feb 17, 2016

### PeroK

You are over-thinking this. The integral of $1/r$ converges in 3 dimensions, so that although the integral in question is improper, it's easy to show that it converges:

$\iiint_{B(0, R)} \frac{1}{r}dV = \int_{0}^{R} \int_{0}^{2\pi} \int_{0}^{\pi} \frac{1}{r} r^2 sin \theta \ d \theta d \phi dr = 2 \pi R^2$

You really shouldn't be worrying about these integrals converging and going back to first principles.

The commutation of $\nabla_r$ and $\int d^3l$ similarly will hold for each volume excluding the singularity, so you can simply take limits (knowing that both integrals converge).

For the divergence equality, you don't have to use the Dirac delta function:

$\nabla_r . (\frac{\vec{r} - \vec{l}}{|\vec{r} - \vec{l}|^3}) = 4 \pi \delta(\vec{l})$ (1)

Instead, you can use the Divergence Theorem to show that:

$\iiint_V \nabla .(\frac{\vec{r}}{r^3}) dV = \iint_S (\frac{\vec{r}}{r^3}).\vec{n} \ dS = 4\pi$

Where V is a sphere around 0. And:

$\iiint_V f(\vec{r}) \ \nabla .(\frac{\vec{r}}{r^3}) dV = 4\pi f(\vec{0})$

By taking V as a smaller and smaller sphere. And, as long as $f$ is well-behaved, it will be close to a constant around 0.

You can then use this to derive the Dirac delta equation (1) as a corollary.

If you are learning electromagnetism, I'm not sure it's worth doing a full rigorous proof to establish exactly the criteria on $f$ or $J$ for these equalities to hold. You have to go with the physicists flow a bit!

9. Feb 18, 2016

### DavideGenoa

I did know that, under usual assumptions on $\mathbf{J}$ and $V$, it converges as an improper Riemann integral, or as a Lebesgue integral, but it was the commutation that I did not understand. Reviewing proofs is always a good thing: thank you so much for showing me that!

Well, but commuting limits and derivative is not always mathematically legitimate...

Mmh... the divergence theorem does not hold true in general if the function $\mathbf{F}$ (notation used in the link) is not of class $C^1(\mathring{A})$, with $\bar{V}\subset\mathring{A}$, and, in fact, if we intend $\iiint_V$ to be the improper integral you allude to, we see that $$\iiint_Vf(\vec{r})\,\nabla\cdot\left(\frac{\vec{r}}{r^3}\right)dV:=\lim_{\delta\to 0}\iiint_{V\setminus B_{\delta} (\vec{r})}f(\vec{r})\,\nabla\cdot\left(\frac{\vec{r}}{r^3}\right)dV=\lim_{\delta\to 0}\iiint_{V\setminus B_{\delta}(\vec{r})}0\,dV=0$$

You may be right, but, unless I know that it is one of those cases, whose set I know to be non-empty, where physics renounces the rigour of mathematics, I try to mathematically justify what I read in books.
I am a self-learner attending no university course, at least not yet, and I have got some time to go deep into what fascinates me about physics and mathematics.
But you may be right: if I stumble in one of those cases where a booktext of physics does not use mathematics rigourously ($\equiv$correctly), I risk to lose a lot of time trying to prove something unprovable...

Wow, what a precise and beautiful proof, where I see you define the integral as a Lebesgue integral, which perfectly answers my question. I think it was a wise idea to study a little theory of measure and Lebesgue integration on Kolmogorov-Fomin's before starting my self-learning path in physics. Very enlightening. I heartily thank you!

I had never seen independence of variables of integration and differentiation as a sufficient conditions to commute integral and derivative signs. Have you got a link to a proof of that?
Since I do know some theorems, like the Leibniz rule that I quoted and some corollaries to Lebesgue's dominated convergence theorem, which allows what you say, whose proofs are not trivial, I fear that the universal legitimacy of commutating integral and derivative signs (provided that the variables are independent, as you say, a quite "weak" assumption) must have a proof based on very advanced mathematics...

Last edited: Feb 18, 2016
10. Feb 18, 2016

### PeroK

I'm from a pure maths background and I do like to fill in the "gaps" in the physics I'm learning, but not to the extent that you do. There must be a place for rigorously justifying everything but you're going to severely limit the amount of physics you can learn. My advice would be to meet the physicists half way. Look under the surface for additional justification but stop short of requiring a watertight mathematical proof for everything.

Your maths is better than mine but maybe that's getting you bogged down in unnecessary maths detail.

11. Feb 18, 2016

### DavideGenoa

@PeroK Thank you for all your suggestions!

@Hawkeye18 I also see that the continuity proof can be easily adapted to prove that both$$\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4\pi}\int_V\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}d\mu_{\mathbf{l}}\quad\text{ and }\quad\mathbf{B}(\mathbf{r})=\frac{\mu_0}{4\pi}\int_V\mathbf{J}(\mathbf{l})\times\frac{\mathbf{r}-\mathbf{l}}{\|\mathbf{r}-\mathbf{l}\|}d\mu_{\mathbf{l}}=\nabla_r\times \mathbf{A}(\mathbf{r})$$are continuous functions of $\mathbf{r}$ on all $\mathbb{R}^3.$ Very enlightening proof, I $\infty$-ly thank you again. I will go ahead to see whether I can understand how to prove Ampère's law from the Biot-Savart one...

12. Feb 19, 2016

### DavideGenoa

I thought I had understood that, but I did not. We have to prove that $\forall\varepsilon>0\quad\exists\eta:$ $$\forall r\in B_{\eta}(r_0)\quad\left|\iiint_V\phi(l)\frac{l_k-r_k}{|l-r|^3} d^3 l -\iiint_V\phi(l)\frac{l_k-r_{0k}}{|l-r_0|^3} d^3 l \right|<\varepsilon$$
I thought that we can decomposed the above difference into $$\Bigg|\iiint_{V\setminus B_{\delta}(r)}\phi(l)\frac{l_k-r_k}{|l-r|^3} d^3 l -\iiint_{V\setminus B_{\delta}(r_0)}\phi(l)\frac{l_k-r_{0k}}{|l-r_0|^3} d^3 l$$ $$+\iiint_{B_{\delta}(r)}\phi(l)\frac{l_k-r_k}{|l-r|^3} d^3 l -\iiint_{B_{\delta}(r_0)}\phi(l)\frac{l_k-r_{0k}}{|l-r_0|^3} d^3 l \Bigg|$$where I have chosen a value of $\delta$ such that $|\iiint_{B_{\delta}(r_0)}\phi(l)\frac{l_k-r_{0k}}{|l-r_0|^3} d^3 l|<\varepsilon/3$ and $|\iiint_{B_{\delta}(r)}\phi(l)\frac{l_k-r_k}{|l-r|^3} d^3 l|<\varepsilon/3$. It remains to me to understand why$$\left|\iiint_{V\setminus B_{\delta}(r)}\phi(l)\frac{l_k-r_k}{|l-r|^3} d^3 l -\iiint_{V\setminus B_{\delta}(r_0)}\phi(l)\frac{l_k-r_{0k}}{|l-r_0|^3} d^3 l\right|<\frac{\varepsilon}{3}$$I thought I had understood, but I was wrong because I did not distinguish $B_{\delta}(r_0)$ and $B_{\delta}(r_0)$ (and then used the uniform continuity of the function $(r,l)\mapsto\frac{l_k-r_k}{|l-r|^3}$ on the compact set $\overline{B_{\eta}(r_0)}\times(V\setminus B_{\delta})$), but I now realise that $B_{\delta}(r_0)\ne B_{\delta}(r)$... How can we prove the last inequality? I heartily thank you again!

13. Feb 19, 2016

### Hawkeye18

@DavideGenoa:
You should split both integrals the same way, i.e. consider a ball over your "fixed" point and its complement. For example, you want to prove continuity at $r$, so $r$ is fixed and $r_0$ approaches $r$ (in may original post I used $r'$ instead of $r_0$). Then we split both integrals into integrals over $V\setminus B_\delta(r)$ and over $B_\delta(r)$.

To estimate $\iiint_{B_\delta(r)} |l-r_0|^{-2} d^3l$ I used a little trick: I noticed that since $|l|^{-2}$ is decreasing radial function, then for a ball $B$ of radius $\delta$ the integral $\iiint_B |l|^{-2}d^3l$ is maximal when $B$ is centered at $0$. And from there I could conclude that $$\iiint_{B_\delta(r)} |l-r_0|^{-2} d^3l \le \iiint_{B_\delta(r)} |l-r|^{-2} d^3l\le \varepsilon/3M.$$ You can avoid using this trick by picking $\delta$ such that $$M\iiint_{B_{2\delta}(0)} |l|^{-2}d^3 <\varepsilon/3.$$ Then if $|r-r_0|<\delta$ then $B_\delta(r)\subset B_{2\delta}(r_0)$ and you can estimate $$\iiint_{B_\delta(r)} |l-r_0|^{-2} d^3l \le \iiint_{B_{2\delta}(r_0)} |l-r_0|^{-2} d^3l\le \varepsilon/3M.$$ So the proof I presented works for more general kernels, where you do not have the decreasing radial estimate.

After you pick $\delta$ you choose $\delta_1$ such that the difference of the integrals over $V\setminus B_\delta(r)$ is small whenever $|r_0-r|<\delta_1$. So you will get that for $|r_0-r|<\delta_1$ the difference of integrals over $V$ will be less than $\varepsilon$.

14. Feb 19, 2016

### DavideGenoa

@Hawkeye18 Since $\phi$ can also be $J_k$ when we consider $\Phi_k$, your proof also also shows that $\mathbf{A}\in C^1(\mathbb{R}^3)$, as well as $\mathbf{B}\in C(\mathbb{R}^3)$. I $\aleph_1$-ly thank you!!!!

15. Feb 20, 2016

### DavideGenoa

Nevertheless, can we exclude that $\mathbf{A}:\mathbf{r}\mapsto\frac{\mu_0}{4\pi}\int_V\frac{\mathbf{J}(\mathbf{l})}{\|\mathbf{r}-\mathbf{l}\|}d\mu_{\mathbf{l}}$ is of class $C^2(\mathbb{R}^3)$ (not only $C^1(\mathbb{R}^3)$)?
If that is not excluded, I think that it is so interesting that it might be worth a separate thread...

I am asking that because I suspect that, if it could be proved that $\mathbf{A}\in C^2(\mathbb{R}^3)$, since$$\forall\mathbf{F}\in C^2(\mathring{A})\quad\nabla\times(\nabla\times\mathbf{F})=\nabla(\nabla\cdot\mathbf{F})-\nabla^2\mathbf{F},$$that might lead to derive that $\nabla\times\mathbf{B}=\mu_0\mathbf{J}$ (which would constitute a proof of Ampère's law from the Biot-Savart law).

Last edited: Feb 20, 2016
16. Feb 21, 2016

### Hawkeye18

If $J$ is not continuous then trivially $A$ cannot be $C^2$. However even continuity of $J$ is not sufficient for $A$ being in $C^2$, a bit stronger condition is needed. For example, if $J$ is Holder continuous, then $A\in C^2$; it is a known fact, but the proofs I know are not exactly easy.

On the other hand, if you assume that $J\in C^2$, then it is very easy to show that $A\in C^2$, and that will be probably enough for most of the physical applications. And after that you can treat a more general case when $J$ is a discontinuous function, and even a distribution, because smooth compactly supported functions are dense in the space of distributions. Of course, the derivative in this general case should be understood in the sense of distributions, i.e. as a weak derivative.

17. Feb 23, 2016

### DavideGenoa

A usual assumption in physics... I have written a specific post.
I suppose you mean $\mathbf{J}\in C^2(\mathbb{R}^3)$, $\forall x\in\mathbb{R}^3\setminus V\quad$ $\mathbf{J}(x)=0$, but, if you have the time to answer there, feel free to change the assumtpions on $\mathbf{J}$ if you intended something different. I $\aleph_1$-ly thank you in any case!

18. Feb 24, 2016

### Hawkeye18

I assumed that $J\in C^2(\mathbb R^3)$ and compactly supported, and just ignored your $V$. (It is common with the Lebesgue integration theory to consider only integrals over the whole space: if one needs to consider the integral over some set $V$, then one just multiplies the integrand by the characteristic function of $V$. )

Under these assumptions we can write $$A(x) = \iiint_{\mathbb R^3} J(l) \frac{1}{|x-l|} d^3 l = \iiint_{\mathbb R^3} J(x-y) \frac{1}{|y|} d^3 y$$ (change of variables $y=x-l$; the $-$ sign disappears, because the change of variables change orientation).

To differentiate the rightmost integral you can use the standard advanced version of the Leibniz rule found here. Note that you do not need the estimate for all $x\in \mathbb R^3$; you only need it in a small neighborhood of the point $x_0$ where you want to compute the derivative, and such estimate is easy.

19. Feb 24, 2016

### DavideGenoa

I heartily thank you!
I see that my idea to translate the domain was not so bad... But in the linked page I used $y=l-x$ which, if I am not wrong, would imply $\forall x_0,y_0\in \mathbb{R}^3\quad$ $\frac{\partial J(x_0+y_0)}{\partial x_k}=\frac{\partial J(x_0+y_0)}{\partial y_k}$: I think that to be correct as well... am I right?
I have thought about a possible $\varphi(y)$ (for the variable change $y=l-x$): once chosen a neighbourhood $B(x_0,\delta)$ of $x_0$, we choose a bounded measurable set, like a parallelepiped $P$, such that $P\supset \bigcup_{x\in B(x_0,\delta)}V-x$, where $V-x:=\{y\in\mathbb{R}^3: y+x\in V\}$, and use $$\varphi(y):=\max_{l\in\mathbb{R}^3}\left\|\frac{\partial^2 J(l)}{\partial l_i \partial l_j}\right\|\frac{\chi_P(y)}{\|y\|}$$where $\chi_P$ is the characteristic function of $P$. Do you think that to be correct?

You might be interested in writing a proof here yourself, but, if you aren't and if my choice of $\varphi$ is not wrong, I will try to write a proof.

I really deply thank you. I feel I am approaching a proof of Ampère's law from the Biot-Savart one, which I have been struggling to find for more than two months, during which I have stopped any other physics study. I study physics and mathematics by myself, without attending university classes, at least not yet, and I have not got examinations "chasing after me", so I spend all the spare from work time I wish trying to understand ($\equiv$learning a proof) at least the most important things that I find in the book I am following even when the book does not contain a proof, at least if I find a proof using tools that are accessible to me. In this particular case all the proofs I have found (essentially identical to this) use Dirac's $\delta$, whose properties I do not know very much, but which is a tool not totally inaccessible to me (I have followed Kolmogorov-Fomin's Элементы теории функций и функционального анализа, more or less coinciding with Introductory Real Analysis), but they use in a way that is so nonchalant and without explanations of what happens between the steps, how they are mathematically justified, that I cannot understand it...

Last edited: Feb 24, 2016
20. Feb 28, 2016

### Hawkeye18

@DavideGenoa:
Your idea about translating the domain was absolutely correct. I used the change of variables $y= x-l$ to keep everything agree with the convolution notation. By convolution of $f$ and $g$ I mean the function $f * g(x):\int f(x-l) g (l) dl$. The change of variables $y-x-l$ shows that $f*g=g*f$.

Yes, you are correct here, both with the derivative and with the dominating $\varphi$.

The outline of proof presented in Wikipedia is indeed a bit imprecise, but is can be easily made completely rigorous. The theory of distributions used there is not really necessary for the proof, but it can give one some additional intuition.

Let us consider tempered distributions. The convolution of a distribution $f$ and a test function $g$ is well defined. From the definitions of the derivative of the distribution and of the convolution it can be easily shown that for the derivative we have $(f*g)' = f'*g = f*g'$. So, if you want to be rigorous, you need to differentiate the test function, i.e. $J$ is your case.

Namely, you write $$A(r) =\frac{\mu_0}{4\pi} \iiint J(r-l) \frac1{|l|} d^3 l = \frac{\mu_0}{4\pi} \iiint J(l) \frac{1}{|r-l|} d^3 l.$$ Differentiating you get $$B(r) =\frac{\mu_0}{4\pi} \iiint\nabla_r\times J(r-l) \frac1{|l|} d^3 l = \frac{\mu_0}{4\pi} \iiint J(l)\times \frac{r-l}{|r-l|^3} d^3 l ,$$ and both identities can be rigorously justified.

Now taking the curl of the first identity and using the formula for $\nabla\times \nabla$ and the fact that $J$ is divergence free you get that $$\nabla\times B(r) = - \frac{\mu_0}{4\pi} \iiint\nabla_r^2 J(r-l) \frac1{|l|} d^3 l .$$ And then you just need to use the Green's second identity to get for a test function $J$ (for example for a compactly supported smooth $J$) we have $$- \frac{1}{4\pi} \iiint\nabla_r^2 J(r-l) \frac1{|l|} d^3 l = J(r) ,$$ so we get the desired result. Note that the above identity is the rigorous statement for the formula for distributions $$\nabla^2 (|r-l|^{-1}) = -4\pi\delta (r-l)$$ (with $l$ treated as a constant).

21. Feb 29, 2016

### DavideGenoa

Thank you so much again!
I must admit that I have not understood how the change of variable works to get identities like $\iiint J(r-l)|l|^{-1}d^3l=\iiint J(l)|r-l|^{-1}d^3l$, while I do understand that $\iiint J(l)|r-l|^{-1}d^3l=\iiint J(-l)|r+l|^{-1}d^3l=\iiint J(-r-l)|l|^{-1}d^3l$ because the Lebesgue integral on $\mathbb{R}^3$ under the assumed conditions on $J$ is invariant to the change of sign and of the variable of integration $l$.
Nevertheless, with $J(r+l)$ in place of $J(r-l)$, I have got this, which I hope to be correct...

Last edited: Feb 29, 2016