Representing $\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^3}$ in polar coordinates

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.

• Would Mathematics be a better home for this question? Commented Dec 28, 2021 at 19:53
• Commented Dec 28, 2021 at 22:15
• Hi QED. So you mean polar coordinates with origin ${\bf r}=0$, not polar coordinates with origin ${\bf r}={\bf r}'$? Commented Dec 29, 2021 at 9:01
• Yes, I was trying to avoid the 'translation property' of the divergence, because my understanding of it was somewhat hand-wavy
– QED
Commented Dec 29, 2021 at 16:22

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

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

• 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.
– QED
Commented Dec 28, 2021 at 22:03