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We know we can solve one of the Maxwell's equation using Green's function. More specifically, we can solve

$$\nabla ^2 \phi(\textbf{r}) = \rho(\textbf{r})$$

using $$\phi(\textbf{r}) = \int d\textbf{r}' G(\textbf{r},\textbf{r}') \rho(\textbf{r}) \qquad\text{where}\qquad \nabla ^2 G(\textbf{r},\textbf{r}') = \delta (\textbf{r}-\textbf{r}'),$$ and $$G(\textbf{r},\textbf{r}')=\frac{1}{|\textbf{r}-\textbf{r}'|}$$ follows.

However, I cannot understand how we obtain the delta function $\delta (\textbf{r}-\textbf{r}')$ by having the Laplacian act on $G(\textbf{r},\textbf{r}')$. Any thoughts on how can I obtain that?

J.-E. Pin
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sined
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    You would need to be careful and deal with distributions. The idea is do the integral on all of space with a little ball cut out at the origin, then use integration by parts and show that the limit as the radius of the ball goes to zero is $4\pi \rho(0)$ (with the function $\frac{1}{|r|}$ that is ) – Ninad Munshi Sep 06 '20 at 17:51
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    @LL3.14 This Answer, which I posted in June 2015, might be of interest also. – Mark Viola Sep 12 '20 at 15:38
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    Here's a calculation I had made earlier: https://math.stackexchange.com/a/2415552/168433 – md2perpe Sep 13 '20 at 21:52

2 Answers2

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Define a vector field $$ \mathbf{F} = \nabla \frac{1}{|\mathbf{r}|} = -\frac{\mathbf{r}}{|\mathbf{r}|^3} . $$

If we take the divergence of this, we see that it vanishes: $$ \nabla \cdot \mathbf{F} = - \frac{(\nabla\cdot\mathbf{r})|\mathbf{r}|^3 - \mathbf{r}\cdot3|\mathbf{r}|^2\mathbf{r}/|\mathbf{r}|}{|\mathbf{r}|^6} = -\frac{3|\mathbf{r}|-3|\mathbf{r}|^3}{|\mathbf{r}|^6} = 0. $$ But this calculation is only defined for $\mathbf{r} \neq \mathbf{0}.$ To cover origin we will use the divergence theorem: $$ \iiint_\Omega \nabla\cdot\mathbf{F} \, dV = \iint_{\partial\Omega} \mathbf{F}\cdot\mathbf{n}\,dS, $$ where $\Omega$ is some region with a smooth enough boundary $\partial\Omega.$ If $\Omega$ does not contain origin, both sides of the equality vanish. Now take $\Omega=B_r(\mathbf{0}),$ i.e. a ball with radius $r$ and center in origin. Then $\partial\Omega$ is the sphere $S_r(\mathbf{0})$ with radius $r$ and center in origin, and the right hand side becomes $$ \iint_{\partial\Omega} \mathbf{F}\cdot\mathbf{n}\,dS = \iint_{S_r(\mathbf{0})} \left(-\frac{\mathbf{r}}{|\mathbf{r}|^3}\right)\cdot\frac{\mathbf{r}}{|\mathbf{r}|} \,|\mathbf{r}|^2 d\omega = -\iint_{S_r(\mathbf{0})} d\omega = -4\pi . $$ (Here $\omega$ is the solid angle measure.)

Thus, $$ \iiint_\Omega \nabla\cdot\mathbf{F} \, dV = \begin{cases} 0, & \text{ if } \mathbf{0} \not\in \Omega \\ -4\pi, & \text{ if } \mathbf{0} \in \Omega \\ \end{cases} $$ Therefore, $\nabla\cdot\mathbf{F}(\mathbf{r}) = -4\pi\,\delta(\mathbf{r}).$

md2perpe
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  • Thanks for your answer. Could you please comment whether or not we can establish a relation for $\frac{d^2}{dx^2} F_x = \frac{1}{3} \delta(\textbf{r})$ or something like that? – sined Sep 06 '20 at 23:05
  • Quick follow-up question, what happened to the minus sign? – sined Sep 07 '20 at 00:45
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    @denis. The minus sign was stolen. Now I have got it back, thanks to the math police. ;-) – md2perpe Sep 07 '20 at 13:36
  • thanks! @md2perpe Do you know whether we could affirm anything related to $\frac{d^2}{dx^2}F_x = \frac{1}{3} \delta(\textbf{r})$ ? – sined Sep 07 '20 at 13:43
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    @denis. That formula hardly is valid. It should have some derivative of $\delta.$ – md2perpe Sep 07 '20 at 14:02
  • I'm sorry @md2perpe, I meant $\frac{d}{dx} F_x = \frac{1}{3} \delta(\textbf{r})$. – sined Sep 07 '20 at 15:18
  • @md2perpe When I was a professor (1000 years ago) and teaching electromagnetic field theory, I used your approach. It's fast and does yield the correct result. But I know that you know that the Divergence Theorem is really inapplicable for $\vec F=\frac{\hat r}{r^2}$ and $\Omega =B_r(0)$. So, I've posted a solution with two approaches. The first uses a regularization of the Dirac Delta and the second uses classical analysis. As always, I enjoy our conversations and would like to hear from you. – Mark Viola Sep 12 '20 at 15:17
  • @MarkViola. The divergence theorem is inapplicable but if $\Omega\subset\mathbb{R}^3$ such that $\operatorname{sing supp}\vec{F} \cap \Omega = \emptyset$ then $\iint \vec{F}\cdot\vec{n},dS$ is defined and one can even define $\vec{F}\chi_\Omega$ as a distribution. Take $\rho\in C_c^\infty(\mathbb{R}^n)$ with $\rho\equiv 1$ on a neighborhood of $\Omega$. Then $\langle (\nabla\cdot\vec{F})\chi_\Omega, \rho \rangle = \iint \vec{F}\cdot\vec{n},dS.$ Abusing integral notation we can write $\langle (\nabla\cdot\vec{F})\chi_\Omega, \rho \rangle$ as $\iiint_\Omega \nabla\cdot\vec{F} , dV.$ – md2perpe Sep 12 '20 at 20:50
  • Computation: $$ 0 = -\langle \vec{F}\chi_\Omega \stackrel{\bullet}{,} \nabla\rho \rangle = \langle \nabla\cdot(\vec{F}\chi_\Omega), \rho \rangle = \langle (\nabla\cdot\vec{F})\chi_\Omega, \rho \rangle + \langle \vec{F}\cdot\nabla\chi_\Omega, \rho \rangle \ = \langle (\nabla\cdot\vec{F})\chi_\Omega, \rho \rangle - \iint \vec{F}\cdot\vec{n},dS . $$ – md2perpe Sep 12 '20 at 20:51
  • @md2perpe Even easier. Define the distribution $D_{\vec F}$ as $\langle \vec F, \phi\rangle=\int_{\mathbb{R}}^3 \vec F\phi ,dV$, since $\vec F=\frac{\hat r}{r^2}$ is integrable locally. Then using linearity and the definition of the distributional derivative, we have $$\langle \nabla \cdot \vec F, \phi\rangle=-\sum_{i}\langle F_i,\partial_i \phi\rangle =-\int_V \vec F\cdot \nabla \phi,dV$$Now, we follow the development in $(4)$ of my posted solution and we are done. ;-) – Mark Viola Sep 12 '20 at 22:05
  • @MarkViola. Well, that's the definition of $\nabla\cdot\vec{F}$. What I wanted to show was a distributional form of the divergence theorem. – md2perpe Sep 12 '20 at 22:36
  • I think I just showed that. Once we get to $-\int_V \nabla \phi \cdot \vec F,dV$ we are into purely classical analysis since $\vec F$ is locally integrable. – Mark Viola Sep 12 '20 at 22:40
  • @MarkViola. Yes, but with your approach you don't define $\iiint_\Omega \nabla\cdot\vec{F} , dV$ or $\iint_{\partial\Omega} \vec{F}\cdot\vec{n} , dS.$ (I see now that I missed index ${}_{\partial\Omega}$ on the double integral in my computation. – md2perpe Sep 12 '20 at 22:44
  • I define the distribution $\nabla \cdot \vec F$ which operates on test functions. Since the distribution itself has compact support ${0}$, then it's easy to define how it operates on $1$. – Mark Viola Sep 12 '20 at 23:04
  • @MarkViola. I this case $\nabla\cdot\vec{F}$ has compact support, but what I do is more general. I don't require compact support of $\nabla\cdot\vec{F}$. – md2perpe Sep 13 '20 at 09:07
  • @MarkViola. You might think that I define $\nabla\cdot\vec{F}$ in some complicated way, but I don't. I define it in the same way as you do. What I do is to show a distributional version of the divergence theorem, and this justifies the calculations I made in my post. – md2perpe Sep 13 '20 at 09:13
  • @md2perpe I do understand. Basically, the distribution transfers to $\nabla \chi_\Omega$, which is a Dirac Delta itself. The approach is fine. – Mark Viola Sep 13 '20 at 13:56
  • @MarkViola. $\nabla\chi_\Omega$ is a vector distribution supported on the boundary of $\Omega$. It's basically $-\iint \cdots \vec{n},dS.$ – md2perpe Sep 13 '20 at 14:18
  • Yes, I know what the object is: $\langle \nabla \chi_\Omega,\phi\rangle =\oint_S \phi(r,\theta,\phi) ,\hat r,r^2 ,\sin(\theta),d\theta,d\phi$So, if $\phi=1$ we have $\langle \vec F\cdot \nabla\chi_\Omega,\phi\rangle=4\pi$ – Mark Viola Sep 13 '20 at 15:20
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REGULARIZING THE DIRAC DELTA:

As I showed in This Answer, we can show that $\nabla \cdot \left(\frac{\hat r}{r^2}\right)=4\pi \delta (\vec r)$ by using a regularization of the Dirac Delta. To begin, let $\vec \psi$ be the function given by

$$\vec \psi(\vec r;a)=\frac{\vec r}{(r^2+a^2)^{3/2}} \tag 1$$

where we note that $\psi(\vec r;0)=\frac{\hat r}{r^2}$ for $\vec r\ne0$.

The divergence of $(1)$ is

$$\nabla \cdot \vec \psi(\vec r; a)=\frac{3a^2}{(r^2+a^2)^{5/2}}$$

And as shown in the referenced answer, for any test function $\phi$ we have

$$\begin{align} \lim_{a \to 0}\int_V \nabla \cdot \vec \psi(\vec r; a)\phi(\vec r)\,dV=\begin{cases}4\pi \phi(0)&, \{0\}\in V\\\\ 0&,\{0\}\notin V \end{cases} \end{align}$$

and it is in this sense that

$$\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r;a)\sim 4\pi \delta(\vec r)$$

Enforcing the translation $\vec r\mapsto \vec r-\vec r'$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\lim_{a\to 0} \nabla \cdot \vec \psi(\vec r-\vec r';a)\sim 4\pi \delta(\vec r-\vec r')}$$

CLASSICAL ANALYSIS:

We need not use the Dirac Delta to prove that $\nabla^2\int_{V}\rho(\vec r')G(\vec r,\vec r')\,dV'=\rho(\vec r)$.

For $\rho(\vec r)\in C^\infty_C$ the gradient of $\phi(\vec r)$ can be written

$$\begin{align} \nabla \int_{V}\rho(\vec r')G(\vec r,\vec r')\,dV'&=\int_{V}\rho(\vec r')\nabla G(\vec r,\vec r')\,dV'\\\\ &=-\int_{V}\rho(\vec r')\nabla' G(\vec r,\vec r')\,dV'\\\\ &=-\oint_{\partial V}\rho(\vec r') G(\vec r,\vec r')\hat n'\,dS'+\int_{V}\nabla' \rho(\vec r')G(\vec r,\vec r')\,dV'\tag2 \end{align}$$

Taking the divergence of $(2)$ reveals

$$\begin{align} \nabla^2 \int_{V}\rho(\vec r')G(\vec r,\vec r')\,dV'&=\oint_{\partial V}\rho(\vec r') \frac{\partial G(\vec r,\vec r')}{\partial n'}\,dS'-\int_{V}\nabla' \rho(\vec r')\cdot \nabla 'G(\vec r,\vec r')\,dV'\tag3 \end{align}$$

We may write the integrand of the integral on the right-hand side of $(3)$ as

$$\nabla' \rho(\vec r')\cdot \nabla 'G(\vec r,\vec r')=\nabla' \cdot (\rho(\vec r')\nabla' G(\vec r,\vec r'))-\rho(\vec r')\nabla'^2 G(\vec r,\vec r')$$

but cannot apply the Divergence Theorem since $\nabla'G(\vec r,\vec r')$is not continuously differentiable in $V$. We can work around this issue by proceeding as follows.

We exclude the singularity at $\vec r'=\vec r$ from $V$ with a spherical volume $V_\varepsilon$ with center at $\vec r$ and with radius $\varepsilon$. Then, using $\nabla'^2 G(\vec r,\vec r')=0$ for $\vec r'\in V-V\varepsilon$, we can write

$$\begin{align} \int_{V}\nabla' \rho(\vec r')\cdot \nabla 'G(\vec r,\vec r')\,dV'&=\lim_{\varepsilon\to 0^+}\int_{V-V_\varepsilon}\nabla' \rho(\vec r')\cdot \nabla 'G(\vec r,\vec r')\,dV'\\\\ &=\lim_{\varepsilon\to 0}\int_{\partial V+\partial V_\epsilon}\rho(\vec r') \frac{\partial G(\vec r,\vec r')}{\partial n'}\,dS'\\\\ &=\int_{\partial V}\rho(\vec r') \frac{\partial G(\vec r,\vec r')}{\partial n'}\,dS'\\\\ &+\lim_{\varepsilon\to 0^+}\int_0^{2\pi}\int_0^\pi \rho(\vec r')\frac{\vec r-\vec r'}{\varepsilon^3}\cdot \frac{\vec r'-\vec r}{\varepsilon}\,\varepsilon^2\,\sin(\theta)\,d\theta\,d\phi\\\\ &=\int_{\partial V}\rho(\vec r') \frac{\partial G(\vec r,\vec r')}{\partial n'}\,dS'-4\pi \rho(\vec r)\tag4 \end{align}$$

Substituting $(4)$ into $(3)$, we find that

$$\bbox[5px,border:2px solid #C0A000]{\nabla^2 \int_{V}\rho(\vec r')G(\vec r,\vec r')\,dV'=4\pi \rho(\vec r)}\tag5$$

Inasmuch as $(5)$ is true for any test function $\rho(\vec r)$, we see that in the sense of distributions

$$\bbox[5px,border:2px solid #C0A000]{\nabla^2 G(\vec r,\vec r')=4\pi \delta(\vec r-\vec r')}$$

Mark Viola
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