# I Laplacian of retarded potential

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1. Jun 15, 2017

### DavideGenoa

Dear friends,
I have found a derivation of the fact that, under the assumptions made in physics on $\rho$ (to which we can give the physical interpretation of charge density) the function defined by
$$V(\mathbf{x},t):=\frac{1}{4\pi\varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\mathbf{y},t-c^{-1}\|\mathbf{x}-\mathbf{y}\|)}{\|\mathbf{x}-\mathbf{y}\|} d^3y,$$
where I think the integral to be a Lebesgue integral or limit of a Riemann integral (since the singularity at $\mathbf{y}=\mathbf{x}$ does not allow, as far as I know, to define a Riemann integral), which corresponds to the retarted electric potential of electrodynamics if the constant $\varepsilon_0$ is interpretated as vacuum permittivity, is such that
$$\nabla_x^2V(\mathbf{x},t)=-\frac{1}{\varepsilon_0} \rho(\mathbf{x},t)+ \frac{1}{c^2} \frac{\partial^2 V(\mathbf{x},t) }{\partial t^2} .$$
But I am not able to understand the mathematical meaning of the steps I find in that derivation. Could anybody help me in understanding it, with a rigourous justification of its steps by known theorems, or giving me an alternative mathematically rigourous proof?
To simplify the notation I will omit the arguments of the functions $V$ and $\rho$. The derivation that I have found is as follows (blue text):

$$\nabla_x V= \frac{1}{4\pi\varepsilon_0} \int \frac{1}{\|\mathbf{x}-\mathbf{y}\|} \nabla\rho+\rho\nabla\left(\frac{1}{\|\mathbf{x}-\mathbf{y}\|}\right) d^3y$$

I notice that differentiation under the integral sign is used, but I have no idea why it is legitimate. Given the physical nature of the problem, I think that $\rho(-,t):\mathbf{y}\mapsto \rho(\mathbf{y},t)$ can be assumed to be compacly supported as a function of $\mathbf{y}$ for all $t$ and such that $\rho\in C^2(\mathbb{R}^4)$, but the discontinuity of the integrand at $\mathbf{x}$ does not allow me to use a theorem I know*.

$\nabla_x\rho=-c^{-1}\dot\rho\nabla_x \|\mathbf{x}-\mathbf{y}\|$ (where the dot above denotes the partial derivative with respect to the second variable), $\nabla_x \|\mathbf{x}-\mathbf{y}\|=\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|}$ and $\nabla_x\|\mathbf{x}-\mathbf{y}\|^{-1}=-\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}$, so $$\nabla_x V= \frac{1}{4\pi\varepsilon_0} \int -\frac{\dot\rho}{c}\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^2} -\rho\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3} d^3y.$$Taking the divergence $$\nabla_x^2 V= \frac{1}{4\pi\varepsilon_0} \int -\frac{1}{c}\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^2}\cdot\nabla_x\dot\rho -\frac{\dot\rho}{c}\nabla_x\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^2} \right)$$$$-\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}\cdot\nabla_x \rho - \rho \nabla_x\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3} \right)\,d^3y$$

Where differentiation under the integral sign is assumed to be legitimate, but it is not clear to me why it is. The derivation follows:

But $$\nabla_x \dot\rho=-\frac{\ddot\rho}{c} \nabla_x \|\mathbf{x}-\mathbf{y}\|=-\frac{\ddot\rho}{c} \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|},$$ $$\nabla_x\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^2} \right)=\frac{1}{\|\mathbf{x}-\mathbf{y}\|^2}$$and $$\nabla\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3} \right)=4\pi\delta^3(\mathbf{y}-\mathbf{x}).$$

Where we see that here, unlike the other derivatives, the "derivatives" are not meant to be the ordinary derivatives (which would give $\nabla\cdot( (\mathbf{x}-\mathbf{y}) / \|\mathbf{x}-\mathbf{y}\|^3 )=0$ everywhere except for $\mathbf{x}=\mathbf{y}$, where it does not exist, but one point would not change the integral), if it were a Lebesgue integral. The derivation follows:

So $$\nabla_x^2 V=\frac{1}{4\pi\varepsilon_0} \int\frac{\ddot\rho}{c^2\|\mathbf{x}-\mathbf{y}\|}-4\pi\delta^3(\mathbf{y}-\mathbf{x})\,d^3y=\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2}-\frac{\rho}{\varepsilon_0}.$$

It is not the first time I find unexplained derivations under the integral sign, which suddenly becomes something different from a Lebesgue integral at some step, and introductions of $\delta$ without any explanation of why such introductions are legimitate. I know, as proved here, that, provided that we give the special sense used in functional analysis to the derivatives of the Laplacian, $\nabla^2\left( \frac{1}{\| \mathbf{x}-\mathbf{y}\|} \right)= -4\pi\delta^3(\mathbf{y}-\mathbf{x})$, **but** that is in the sense that $$\forall\varphi\in C_c^2(\mathbb{R}^3)\quad \int_{\mathbb{R}^3}\frac{\nabla_y^2\varphi(\mathbf{y})}{\| \mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}=-4\pi \varphi(\mathbf{x}),$$ while, if we read $\nabla_x^2$ as a Laplacian where the derivatives are what they are in calculus, $$\int_{\mathbb{R}^3} \varphi(\mathbf{y}) \nabla_x^2 \left( \frac{1}{\| \mathbf{x}-\mathbf{y}\|} \right) d\mu_{\mathbf{y}}=\int_{\mathbb{R}^3} \varphi(\mathbf{y})\cdot 0 \,d\mu_{\mathbf{y}}=0.$$
I also notice that the singularity at $\mathbf{y}=\mathbf{x}$ does not prevent to consider $-\frac{\dot\rho}{c}\nabla_x\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^2} \right) -\frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3}\cdot\nabla_x \rho=0$, while $\nabla\cdot\left( \frac{\mathbf{x}-\mathbf{y}}{\|\mathbf{x}-\mathbf{y}\|^3} \right)$ is not considered constantly $0$. (Honestly speaking, from the answers I get when talking about these issues with other students - I study by myself and I am not enrolled at a university yet -, I am beginning to think that many see that sort of differentiations under the integral sign and introductions of $\delta$ as tricks to pretend to do mathematics.)

Could anybody help me with a rigourous proof of $\nabla_x^2V=-\frac{1}{\varepsilon_0} \rho+ \frac{1}{c^2} \frac{\partial^2 V }{\partial t^2}$ including the mention of the theorems (I think that this and this may help, but I am not able to produce a complete proof) used to justify the not-so-trivial steps?
I $\infty$-ly thank any answerer!

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

Last edited: Jun 15, 2017
2. Jun 15, 2017

### WWGD

I suggest you summarize your results and shorten your post then expand on them as you get replies; it seems like a heavy read.

3. Jun 23, 2017

### DavideGenoa

Thank you, @WWGD! I have begun to fragmentate the question by asking here why (and whether) we can differentiate under the integral sign the first time it is done....

4. Jun 23, 2017

### jasonRF

You clearly know a lot more math than I do. However, you may want to consider a different angle. Since delta distributions seem to naturally pop up with this problem, it may be easier to simply use distribution theory from the outset. At least from my view (as a non-mathematician) it seems simpler, but it could just be that I am too naive.

Anyway, the basic idea is this. To solve
$$\nabla^2 V - c^{-2} \partial_{tt} V = - \rho / \epsilon_0,$$
we find the Green's "function" $g$ by solving (in the sense of distributions),
$$\nabla^2 g - c^{-2} \partial_{tt} g = - \delta(\mathbf{x})\delta(t)/\epsilon_0.$$
Using Fourier analysis we find,
$$g(\mathbf{x},t) = \frac{\delta(t - |\mathbf{x}|/c )}{4\pi \epsilon_0 |\mathbf{x}|}.$$
The solution to the original problem is then the convolution
$$V = g \ast \rho.$$
Note that this convolution is both spatial and temporal (4 dimensional). If $\rho$ is nice enough to be a test function, then this expression reduces to the integral at the beginning of your post. But the convolution is also valid when $\rho$ is a class of distributions including delta functions, although in this case it should not be represented by an integral if you want to be rigorous. Of course, when physicists and engineers use the retarded potential integral we feel free to allow $\rho$ to be a delta function (see the typical derivation of the Lienard-Wichert potentials: https://en.wikipedia.org/wiki/Liénard–Wiechert_potential).

Your original problem is then equivalent to demonstrating that the convolution does indeed satisfy the PDE. To show this simply apply the wave operator,
$$\begin{eqnarray*} \left( \nabla^2 - c^{-2} \partial_{tt} \right) V & = & \left( \nabla^2 - c^{-2} \partial_{tt} \right) \left( g \ast \rho\right) \\ & = & \left( \left( \nabla^2 - c^{-2} \partial_{tt} \right) g\right) \ast \rho\\ & = & - \delta(\mathbf{x})\delta(t)/\epsilon_0 \ast \rho \\ & = & -\rho / \epsilon_0. \end{eqnarray*}$$
All of the equalities above are to be understood in the sense of distributions. Convolution and differentiation of distributions always commute, so there is nothing special needed to justify going from line 1 to line 2; note that this step is equivalent to swapping the order of differentiation and integration when considering functions instead of distributions. I skipped the algebra to show the 3rd line follows from the 2nd line.

This framework is usually (often unknowingly) used by engineers and scientists all the time, which is why the sloppy math of swapping limit processes often (but not always!) works out fine if the final result is interpreted as a generalized function.

Jason

5. Jun 24, 2017

### DavideGenoa

Not at all. I have only studied Kolmogorov-Fomin's Элементы теории функций и функционального анализа, roughly corresponding to the English version Introductory Real Analysis, by myself, since I have not enrolled at university, at least not yet, but I study mathematics and physics, while working, just to be less ignorant. Therefore I know very little about convolutions and Fourier analysis, in one dimension only, and about distributions (symbolically represented by one dimensional integrals, $\int_{-\infty}^{\infty}...dx$, again), and nothing at all about Green's functions (and I would be very grateful for any suggestion of texts containing the proofs necessary to understand these issues).

Cannot we prove the desired result just by using the theory of Lebesgue integration and ordinary derivatives? After all, if $f:\mathbb{R}\to\mathbb{R}$ is a function Lebesgue-summable on any finite interval $[a,b]\subset\mathbb{R}$, then for any test function $\varphi$ the distribution $T_f$ defined by $$T_f(\varphi)= \int_{-\infty}^{\infty}f(x)\varphi(x) dx$$is written as an integral, but to calculate the value of $T_f(\varphi)$ we really calculate the ordinary Lebesgue integral $\int_{-\infty}^{\infty}f(x)\varphi(x)$ (which is the same as $\int_{\mathbb{R}}f\varphi d\mu$, a notation that I prefer), and, when solving a problem from a Physics 2 textbook that proposes to calculate a certain $V$, a student does calculate a Lebesgue integral (as the limit of a Riemann integral).
$\int_{-\infty}^{\infty}dx$ again!