# Explanation of the identities $\rho=\rho(\delta)$ ($\delta$ function) and $\rho=q\,\delta(\bar{r})$

Poisson's equation is $$\boldsymbol{\nabla}^{2}\varphi=-4\pi k_{e}\,\rho, \tag{*}$$ that in the case of a point charge $$q$$, already with spherical symmetry, has as solution

$$$$\varphi(r)=k_{e} \frac{q}{r}, \tag{**}$$$$

Replacing the (**) in the (*) we get:

$$$$k_{e}\, q \boldsymbol{\nabla}_{r}^{2}\left(\frac{1}{r}\right)=-4\pi \, k_{e}\, \rho$$$$

1. Why is the charge density $$\rho$$ also considered as a delta-function ($$\delta$$) over the whole classical space-time $$\mathbb{R}^4$$?
2. Why exist this identity

$$$$\rho=q\,\delta(\bar{r})\quad ?$$$$

• There is no time in this. Hence \rho is constant, and the delta function is a delta function in the regular 3D space. For the demonstration, simply use Gauss theorem! Alternatively one can make a (3D) Fourier transform and get \tilte{\phi}\propto 1/k^2, which is not convergent (expected if the real space version is a distribution). The 1/k^2 is easily "regularized" as a Lorentzian function 1/(k^2+a^2). Th inverse Fourier transform is then \propto \exp(-|x|/a) and taking a \to 0 gives the \delta distribution. – Jhor Mar 19 '19 at 21:50
• Keep in mind that these topics are the result of my research. I would like to say that I am a high school teacher and I do not deal with these topics. All the detailed answers are very welcome to me, as is your comment. – Sebastiano Mar 19 '19 at 21:53

As you noticed the Laplacian is ill-defined at the origin: such equations are usually solved by lifting them to distribution theory and using Green's functions.

Let $$-\nabla^2 \phi(\mathbf{x}) = -4\pi \rho(\mathbf{x})\tag{1}$$

one can show$$^1$$ that the solution $$\phi(\mathbf{x})$$ can always be written as$$^2$$ $$\phi(\mathbf{x}) = \frac{1}{4\pi} \int_V d^3 x' \rho(\mathbf{x})G(\mathbf{x}, \mathbf{x}') + \frac{1}{4\pi}\int_{\partial V}d\sigma\Big[G(\mathbf{x}, \mathbf{x}')\frac{\partial \phi}{\partial n} - \phi(\mathbf{x}') \frac{\partial}{\partial n}G(\mathbf{x}, \mathbf{x}')\Big] \tag{2}$$ where the Green's function $$G(\mathbf{x}, \mathbf{x}')$$ solves the associated Green's equation $$-\nabla^2 G(\mathbf{x}, \mathbf{x}') = -4\pi \delta(\mathbf{x}-\mathbf{x}')\tag{3}$$ Using appropriate boundary conditions (in the simple cases one can demand the functions to vanish at the boundaries and the first order derivative to be proportional to the surface element$$^3$$) equation $$(2)$$ can be solved by plugging the solution of $$(3)$$, which you already have recognised to be the potential of a single point charge.

Details on all of this can be found in the standard textbook of J. D. Jackson on Classical Electrodynamics.

$$^1$$ In order to show why this holds multiply $$(1)$$ by $$G(\mathbf{x}, \mathbf{x}')$$ and $$(3)$$ by $$\phi(\mathbf{x})$$ and integrate against $$\delta(\mathbf{x}-\mathbf{x}')$$.

$$^2$$ There might be some $$2\pi$$ being forgotten left and right, somewhere.

$$^3$$ I might be wrong about this.