Is $g(\eta)$, as defined in the question, a Dirac delta function?

This is an interesting question, and the answer to it is definitely YES. Let us deal with the problem using expansions in spherical harmonics.

First note that the expansion of the Laplacian's Green's function in terms of Legendre polynomials is known

$$\frac{1}{|\mathbf{r}-\mathbf{r}'|}=\begin{Bmatrix}\frac{1}{r}\sum_{n=0}^{\infty}P_n(\hat{r}\cdot\hat{r}')\Big(\frac{R}{r}\Big)^{n}~~~~,r> R\\ \frac{1}{R}\sum_{n=0}^{\infty}P_n(\hat{r}\cdot\hat{r}')\Big(\frac{r}{R}\Big)^{n}~~~~, r<R\end{Bmatrix}$$

Then we compute the derivative of this function with respect to the radial coordinate $r$ which will help us compute the electric field:

$$f(r,\eta)=\frac{\partial}{\partial r}\frac{1}{|\mathbf{r}-\mathbf{r}'|}=\begin{Bmatrix}-\frac{1}{R^2}\sum_{n=0}^{\infty}(n+1)P_n(\hat{r}\cdot\hat{r}')\Big(\frac{R}{r}\Big)^{n+2}&,~r> R\\ \frac{1}{R^2}\sum_{n=0}^{\infty}nP_n(\hat{r}\cdot\hat{r}')\Big(\frac{r}{R}\Big)^{n-1}&,~r<R\end{Bmatrix}$$

Finally we deal with the quantity claimed to be a delta-function:

$$g(\eta)=\frac{\partial}{\partial r}\frac{1}{|\mathbf{r}-\mathbf{r}'|}\Bigg|_{r\to R^+}-\frac{\partial}{\partial r}\frac{1}{|\mathbf{r}-\mathbf{r}'|}\Bigg|_{r\to R^-}=-\frac{1}{R^2}\sum_{n=0}^{\infty}(2n+1)P_n(\hat{r}\cdot\hat{r}')$$

Using the addition theorem for spherical harmonics, which states that

$$P_{n}(\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}')=\frac{4\pi}{2n+1}\sum_{m=-n}^{n}Y_{nm}(\hat{\mathbf{r}})(Y_{nm})^*(\hat{\mathbf{r}}')$$

and the expansion of the delta function in spherical harmonics:

$$\delta(\theta-\theta')\delta(\phi-\phi')=\sin\theta'\sum_{lm}Y_{lm}(\theta, \phi)Y^*{}_{lm}(\theta', \phi')$$

we have proven that

$$\frac{\partial}{\partial r}\frac{1}{|\mathbf{r}-\mathbf{r}'|}\Bigg|_{r\to R^+}-\frac{\partial}{\partial r}\frac{1}{|\mathbf{r}-\mathbf{r}'|}\Bigg|_{r\to R^-}=-\frac{4\pi}{R^2}\frac{\delta(\theta-\theta')\delta(\phi-\phi')}{\sin\theta'}$$

and thus

$$\begin{align}E_r(r\to R^{+},\theta, \phi)-E_r(r\to R^{-},\theta, \phi)&=-\frac{1}{4\pi\epsilon_0}\int R^2\sin\theta'd\theta'd\phi'\sigma(\theta', \phi')\Big(-\frac{4\pi}{R^2}\frac{\delta(\theta-\theta')\delta(\phi-\phi')}{\sin\theta'}\Big)\\&=\frac{\sigma(\theta,\phi)}{\epsilon_0}\end{align}$$

and our faith to mathematical consistency of physical theories has been successfully restored.

EDIT: Upon @mrc ntn's prompt,and for completeness, I want to close this by showing that $g(\eta)\propto\delta(\eta-1)$.

Without repeating any of the calculations above because they are identical (substituting $\eta=\hat{r}\cdot\hat{r}'$, it is true that

$$g(\eta)=-\frac{1}{R^2}\sum_{n=0}^{\infty}(2n+1)P_n(\eta)$$

But $P_n(1)=1$ and due to the completeness relation of the Legendre polynomials given here we conclude that

$$g(\eta)=-\frac{1}{R^2}\sum_{n=0}^{\infty}(2n+1)P_n(\eta)P_n(1)=-\frac{2}{R^2}\delta(\eta-1)$$

Tags:

Physics

Multivariable Calculus

Dirac Delta

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