I would like to verify my understanding on why it is that we can represent a charge density $\rho(\mathbf{r})$ as delta functions.

If we start from $$ \nabla^2V=-4\pi\rho(\mathbf{r})\tag{1} $$ then we conclude that $$ V=\frac{1}{|\mathbf{r}-\mathbf{r}'|}+\Gamma(\mathbf{r}, \mathbf{r}')\tag{2} $$ where $\nabla^2\Gamma=0$. We then find that [plugging (2) in for (1)] $$ -4\pi\delta(\mathbf{r}-\mathbf{r}')=-4\pi\rho(\mathbf{r}) $$ or, equivalently, $$ \rho(\mathbf{r})=\delta(\mathbf{r}-\mathbf{r}'). $$ This result is intuitive to me. The charge density $\rho$ is the charge per unit volume, zero everywhere, but has some value at $\mathbf{r}$. The specific value can be obtained by recognizing that, if $\mathbf{r}$ is the only non-vanishing point, it must have a small volume (as to not interfere with other points). Hence, $$ \rho(\mathbf{r})=\lim_{\mathrm{volume}\to0}\frac{\mathrm{charge}}{\mathrm{volume}}\to\pm\infty $$

But how could this be so if $$ V=\int d^3r'\,\frac{\rho(\mathbf{r})}{|\mathbf{r}-\mathbf{r}'|}=\int d^3r'\,\frac{\delta(\mathbf{r}-\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\to\pm\infty $$

What assumptions have I made that are incorrect? (Is it that $\rho(\mathbf{r})$ is only a function of $\mathbf{r}$ and not $\mathbf{r}'$ so in fact $\rho(\mathbf{r})=\delta(\mathbf{r})\neq \delta(\mathbf{r}-\mathbf{r}')$?)

  • $\begingroup$ Not sure how you conclude (2) from (1)? $\endgroup$
    – NDewolf
    Commented Feb 27, 2021 at 15:33
  • $\begingroup$ @NDewolf the Laplacian of $1/|\mathbf{r}-\mathbf{r}'|$ is $-4\pi\delta(\mathbf{r}-\mathbf{r}')$. $\endgroup$
    – user143
    Commented Feb 27, 2021 at 15:34
  • $\begingroup$ So you're deriving (equation 4) that $\rho = \delta$ by already assuming it? $\endgroup$
    – NDewolf
    Commented Feb 27, 2021 at 15:39
  • $\begingroup$ @NDewolf From (2) and (1) we obtain the third equation and hence the fourth one. $\endgroup$
    – user143
    Commented Feb 27, 2021 at 15:40
  • $\begingroup$ You can assume (2) and use it with (1) to obtain $\rho(r) = \delta(r-r')$, but you cannot just conclude (2) from (1). $\endgroup$
    – NDewolf
    Commented Feb 27, 2021 at 15:44

1 Answer 1


The potential $V_0(\textbf{r},\textbf{r}') = \frac{1}{|\textbf{r}-\textbf{r}'|}$ is a solution to (1) only if $\rho_0(\textbf{r}.\textbf{r}') = \delta(\textbf{r} - \textbf{r}')$ for some fixed $\mathbf{r}'$. Also note that $\nabla^2$ acts with respect to the unprimed coordinates only.

Because the equation is linear, we can use this to construct a general solution to an arbitrary charge distribution $\rho(\textbf{r})$ (which now is a given function, and not a Dirac delta). First, we multiply the equation by $\rho(\textbf{r}')$. Here $\textbf{r}'$ is fixed, so $\rho(\textbf{r}')$ is just a number. This yields \begin{equation} \rho(\textbf{r}') \nabla^2 V_0(\textbf{r},\textbf{r}') = - \rho(\textbf{r}') 4\pi \delta(\textbf{r} - \textbf{r}'). \end{equation} Now integrate both sides of the equation with respect to the primed coordinates over all of space. This gives \begin{equation} \int d\textbf{r}' \rho(\textbf{r}') \nabla^2 V_0(\textbf{r},\textbf{r}') = -4\pi \int d\textbf{r}' \rho(\textbf{r}') \delta(\textbf{r}-\textbf{r}') = - 4\pi \rho(\textbf{r}). \end{equation} Now on the left-hand side of the equation, we note that the integral is with respect to the primed coordinates while the differential operator $\nabla^2$ acts with respect to the unprimed coordinates. These are independent, so we can move the Laplacian outside the integral. Then we obtain \begin{equation} \nabla^2 \int d\textbf{r}' \rho(\textbf{r}') V_0(\textbf{r},\textbf{r}') = -4\pi \rho (\textbf{r}) \end{equation} But this means that the function \begin{equation} V(\textbf{r}) = \int d\textbf{r}' \rho(\textbf{r}') V_0(\textbf{r},\textbf{r}') \end{equation} solves the equation $\nabla^2 V(\textbf{r}) = - 4\pi \rho(\textbf{r})$. This is the origin of the integral formula that you have at the end.

  • 1
    $\begingroup$ Please don't use MathJax for formatting of text (such as \textit) -- use the Markdown constructs (like *italics* or _italics_) instead. $\endgroup$ Commented Feb 27, 2021 at 16:07
  • $\begingroup$ I've edited in some examples $\endgroup$ Commented Feb 27, 2021 at 16:29

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