In his book introduction to electrodynamics, Griffiths uses derives the identity $$\nabla \cdot \frac{\mathbf{\hat{r}}}{r^2} = 4\pi\delta^3(\mathbf{r})$$ Using the formula for divergence in polar coordinates. He then states that "more generally" $$\nabla \cdot \frac{\mathbf{\hat{r_s}}}{r_s^2} = 4\pi\delta^3(\mathbf{r_s})$$ Where $\mathbf{r_s}$ is the seperation vector, $\mathbf{r}-\mathbf{r'}$ and $\mathbf{r'}$ is a constant. I'm trying to find a way to derive this identity using polar coordinates, but am struggling to find a way to represent $\frac{\mathbf{r-r'}}{{|r-r'|^3}}$ in terms of the basis vectors in polar coordinates $(\mathbf{\hat{r}}, \hat{\boldsymbol{\phi}},\hat{\boldsymbol{\theta}})$ as $v_r\mathbf{\hat{r}}+v_\theta\boldsymbol{\hat{\theta}}+v_\phi\boldsymbol{\hat{\phi}}$ for some scalar functions $v_r, v_{\theta}, v_{\phi} $ to use in the formula for divergence in polar coordinates.


1 Answer 1


By chain rule $$ \frac{\partial}{\partial x}f(x-x_0)=\frac{\partial}{\partial x}f(x)\vert_{x=x-x_0} $$ You can do the same with the divergence operator $\nabla$.

Therefore, you only need to prove $$ \nabla \cdot \frac{\hat{\mathbf{r}}}{r^{2}}=4 \pi \delta^{3}(\mathbf{r}) $$ You could use whatever coordinate to prove this, given that $r\neq0$, since at $r=0$, $\frac{\hat{\mathbf{r}}}{r^{2}}$ blows up. Normal differentiation can't help you determine what happens at the origin, so you have to employ some other tool to do so.

We could imagine a small ball of radius $R$ around the origin, and construct the integral $$ \int_R d^3r \nabla \cdot \frac{\hat{\mathbf{r}}}{r^{2}} $$ This will give you a finite result no matter how small $R$ is, and it happens only around the origin. You can argue that the only type of "function" that has such property is a delta function centered at the origin.

By the way, if you really wish to expand something like $1/|\mathbf{r}-\mathbf{r'}|$ with polar coordinates, you will get a series of Lengendre polynomials

$$ \frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} = \frac{1}{r}\sum_{\ell=0}^{\infty}\left(\frac{r^{\prime}}{r}\right)^{\ell}P_{\ell}(\cos\theta) $$

  • $\begingroup$ Is there an intuitive reason that the $\frac{\partial}{\partial x}f(x-x_0)=\frac{\partial}{\partial x}f(x)|_{x-x_0}$ justifies this in polar coordinates? I see how it's trivial in cartesian coordinates, but in polar coordinates it seems like there'd be more nuance since the basis vectors change when you change the position. $\endgroup$
    – QED
    Dec 28, 2021 at 22:03

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