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So, in three-dimensions we famously have the result that the Laplacian acting on $1/r$ is a distribution: $$\vec{\nabla}^2\frac{1}{4\pi r}=-\delta^3(\vec{r})$$ where $\delta^3(\vec{r})$ is the Dirac-delta function.

My question: how should one think of the mixed derivative $\partial_{i}\partial_{j}\frac{1}{4\pi r}=?$.

Naively, taking derivatives, one gets $$\partial_{i}\partial_{j}\frac{1}{4\pi r}=\frac{1}{4\pi}\left(\frac{3 r_i r_j}{r^5}-\frac{\delta_{ij}}{r^3}\right)$$ but, tracing over indices does not reproduce the $\delta$-function piece, of course. So instead, it naively seems that we should have something like $$\partial_{i}\partial_{j}\frac{1}{4\pi r}\stackrel{?}{=}\frac{1}{4\pi}\left(\frac{3 r_i r_j}{r^5}-\frac{\delta_{ij}}{r^3}\right)-\left(\frac{\delta_{ij}}{3}+c(\delta_{ij}/3-r_ir_j/r^2)\right)\delta^3(\vec{r})$$ which reproduces the original relation for any value of $c$ upon contracting indices. So, is something like the above correct? If so, is there a unique way of fixing $c$?

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  • $\begingroup$ Crossposted to physics.stackexchange.com/q/527692/2451 $\endgroup$
    – Qmechanic
    Jan 28, 2020 at 17:29
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    $\begingroup$ You can get displayed equations by using double instead of single dollar signs. This is especially relevant when mixing fractions and exponents. $\endgroup$
    – joriki
    Jan 28, 2020 at 21:56
  • $\begingroup$ You need to use derivatives in the sense of distributions in your computations. $\endgroup$ Jan 29, 2020 at 14:47
  • $\begingroup$ They certainly are well-defined distributions, by the definition of differentiation of distributions. If $r^{-n}$ in $\mathbb R^3$ is defined as here, then $$\frac {\partial^2} {\partial x_i \partial x_j} r^{-1} = 3 x_i x_j r^{-5} - \left( r^{-3} + \frac {4 \pi} 3 \delta(\boldsymbol x) \right) \delta_{i j},$$ where $\delta_{i j}$ is the Kronecker delta. $\endgroup$
    – Maxim
    Feb 11, 2020 at 17:06

2 Answers 2

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OP's differentiation formulas can of course be understood pointwise on $\mathbb{R}^3\backslash\{0\}$ where the functions are smooth. The interesting non-trivial question is whether they can be promoted to distributions on the full space $\mathbb{R}^3$? Well, let's see.

We regularize $1/r$ as a smooth function $$ u_{\varepsilon}(r)~:=~\frac{1}{(r^2+\varepsilon)^{1/2}} ~\rightarrow~ {\rm P.V.}\frac{1}{r} \quad\text{for}\quad\varepsilon\to 0^+ \tag{A}$$ in $C^{\infty}(\mathbb{R}^3)$, in the sense of generalized functions. Then the derivatives are well-defined: $$ \frac{\partial u_{\varepsilon}(r)}{\partial x_i}~=~-\frac{x_i}{(r^2+\varepsilon)^{3/2}},\tag{B} $$ $$ \frac{\partial^2 u_{\varepsilon}(r)}{\partial x_i\partial x_j}~=~3\frac{x_ix_j}{(r^2+\varepsilon)^{5/2}}-\frac{\delta_{ij}}{(r^2+\varepsilon)^{3/2}}~\rightarrow~ {\rm P.V.}\left(\frac{3x_ix_j}{r^5} -\frac{\delta_{ij} }{r^3}\right) \quad\text{for}\quad\varepsilon\to 0^+, \tag{C} $$ $$\nabla^2u_{\varepsilon}(r) ~=~-\frac{3\varepsilon}{(r^2+\varepsilon)^{5/2}}~\rightarrow~ -4\pi\delta^3({\bf r}) \quad\text{for}\quad\varepsilon\to 0^+. \tag{D}$$

In order to make sense of eq. (C) [which OP is inquiring about] we apparently need the principal value distributions $${\rm P.V.} \frac{1}{r^p}[f]~:=~\lim_{\varepsilon\to 0^+}\int_{\mathbb{R}^3} \mathrm{d}^3{\bf r}\frac{f({\bf r})}{(r^2+\varepsilon)^{p/2}}, \qquad p~\leq~3,\tag{E}$$ $${\rm P.V.} \frac{x_ix_j}{r^p}[f]~:=~\lim_{\varepsilon\to 0^+} \int_{\mathbb{R}^3} \mathrm{d}^3{\bf r}\frac{x_ix_jf({\bf r})}{(r^2+\varepsilon)^{p/2}}, \qquad p~\leq~5.\tag{F}$$ On one hand, eqs. (E) & (F) do not make sense for smooth test functions $f\in C^{\infty}_c(\mathbb{R}^3)$ with compact support but they do make sense if the test functions $f$ are restricted to vanish $f({\bf 0})=0$ at the origin ${\bf r}={\bf 0}$, because then the singularity is removable. On the other hand, applying this restriction $f({\bf 0})=0$, we are not able to detect Dirac delta contributions in eq. (C), which seems to be OP's main motivation to start with.

This issue does not affect eq. (D), which is a well-known representation for the 3D Dirac delta distribution.

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  • $\begingroup$ Is principal value really needed for $1/r$ in three dimensions, and if so, how is it defined? $\endgroup$
    – md2perpe
    Feb 9, 2020 at 20:58
  • $\begingroup$ I'm sure this treatment is more correct than mine, but at the moment it doesn't seem to say as much. Is it complete? $\endgroup$
    – Ali
    Feb 11, 2020 at 0:49
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Feb 11, 2020 at 15:02
  • $\begingroup$ So what is your final answer to what is $\partial_i \partial_j \frac{1}{r}$? $\endgroup$
    – md2perpe
    Feb 12, 2020 at 8:21
  • $\begingroup$ As an alternative rigorous approach, see Example $2$ of THIS ANSWER. ;-) $\endgroup$
    – Mark Viola
    Jun 11, 2021 at 16:38
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Probably you should make an approach something like this. Let \begin{equation*} \Phi(x)=\frac{1}{4\pi r}\quad\text{with}\quad r(x)=\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)^{\frac{1}{2}}. \end{equation*} We hope to describe $\partial_{i} \partial_{j} \Phi(x)$ as some sort of distribution. A distribution must be integrated against a smooth function, so let's do that: \begin{equation*} \int_{\mathbb{R}^{3}}\partial_{i}\partial_{j}\Phi(x)f(x)dx =\int_{\mathbb{R}^{3}\setminus B_{\epsilon}} \partial_{i}\partial_{j}\Phi(x)f(x)dx + \int_{B_{\epsilon}} \partial_{i}\partial_{j}\Phi(x)f(x)dx. \end{equation*} We are isolating the singularity inside a small ball. I guess you are happy with the first term there (it can be evaluated using the expression you derived) so let's focus on the second. Integrating by parts: \begin{equation*} \int_{B_{\epsilon}} \partial_{i}\partial_{j}\Phi(x)f(x)dx =-\int_{B_{\epsilon}}\partial_{i}\Phi(x) \partial_{j}f(x)dx +\frac{1}{\epsilon}\int_{\partial B_{\epsilon}}\partial_{i}\Phi(x) f(x)x_{j}dS(x) \end{equation*}

The first term integrate by parts again \begin{equation*} -\int_{B_{\epsilon}}\partial_{i}\Phi(x) \partial_{j}f(x)dx= \int_{B_{\epsilon}}\Phi(x) \partial_{i}\partial_{j}f(x)dx -\frac{1}{\epsilon}\int_{\partial B_{\epsilon}}\Phi(x) \partial_{j}f(x)x_{i}dS(x). \end{equation*} But we can ignore all of that as $\epsilon\to 0$ because \begin{equation*} \left\lvert \int_{B_{\epsilon}}\Phi(x) \partial_{i}\partial_{j}f(x)dx\right\rvert \leq \left\lVert \partial_{i}\partial_{j}f \right\rVert_{L^{\infty}} \int_{B_{\epsilon}}\lvert\Phi(x)\rvert dx\leq C \epsilon^{2} \end{equation*} and \begin{equation*} \left\lvert \frac{1}{\epsilon}\int_{\partial B_{\epsilon}}\Phi(x) \partial_{j}f(x) x_{i} dS(x)\right\rvert \leq \left\lVert \partial_{j}f \right\rVert_{L^{\infty}} \int_{\partial B_{\epsilon}}\lvert\Phi(x)\rvert dS(x)\leq C \epsilon \end{equation*} We are left with \begin{equation*} \frac{1}{\epsilon}\int_{\partial B_{\epsilon}}\partial_{i}\Phi(x) f(x)x_{j} dS(x)= -\frac{1}{4\pi\epsilon^{4}}\int_{\partial B_{\epsilon}}x_{i}x_{j}f(x)dS(x). \end{equation*} I had to look it up but it seems that \begin{equation*} \int_{\partial B_{\epsilon}}x_{i}x_{j}dS(x)=\frac{4\pi}{3}\epsilon^{4}\delta_{ij}. \end{equation*} Therefore as $\epsilon\to 0$ \begin{equation*} \int_{B_{\epsilon}} \partial_{i}\partial_{j}\Phi(x)f(x)dx\to -\frac{1}{4\pi\epsilon^{4}}\int_{\partial B_{\epsilon}}x_{i}x_{j}f(x)dS(x)\to -\frac{1}{3}\delta_{ij}f(0). \end{equation*} So in conclusion one might write \begin{equation*} \partial_{i}\partial_{j}\Phi(x)= \begin{cases} \frac{1}{4\pi}\left(\frac{3 x_{i}x_{j}}{r^{5}}-\frac{\delta_{ij}}{r^{3}}\right)&\text{for }x\neq 0\\ -\frac{1}{3}\delta_{ij}\delta(x)&\text{for } x = 0. \end{cases} \end{equation*}

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  • $\begingroup$ You should actually start with $\int_{\mathbb{R}^3} \Phi(x) \, \partial_i \partial_j f(x) \, dx$ since $\partial_i \partial_j \Phi$ is defined by $$\int \partial_i \partial_j \Phi(x) \, f(x) \, dx := \int \Phi(x) \, \partial_i \partial_j f(x) \, dx.$$ $\endgroup$
    – md2perpe
    Feb 1, 2020 at 21:36
  • $\begingroup$ I don't think that adds anything. This is a fairly informal argument. $\endgroup$
    – Ali
    Feb 3, 2020 at 9:31
  • $\begingroup$ The development and result are not quite correct. See Example $2$ of THIS ANSWER for a rigorous development and precise result. ;-) $\endgroup$
    – Mark Viola
    Jun 11, 2021 at 16:37

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