# Dirac Delta source for vectorial equation

1. Feb 19, 2014

### EmilyRuck

Hello!
By manipulating Maxwell's equation, with the potential vector $\mathbf{A}$ and the Lorentz' gauge, one can obtain the following vector wave equation:

$∇^2 \mathbf{A}(\mathbf{r}) + k^2 \mathbf{A}(\mathbf{r}) = -\mu \mathbf{J}(\mathbf{r})$

The first step for the solution is to consider a point source $- \delta (\mathbf{r} - \mathbf{r}')$. But what about its meaning in a vector form?
A vector is a 1-dimensional object in the space; a point is a 0-dimensional object, according to linear algebra. How can we relate a 0-dimensional object to a vector context?
With such a source, the unknown function is a Green function and the previous equation becomes

$∇^2 G(\mathbf{r}, \mathbf{r}') + k^2 G(\mathbf{r}, \mathbf{r}') = - \delta (\mathbf{r} - \mathbf{r}')$

but now we don't have a vector function $\mathbf{A}(\mathbf{r})$ any more, but just a scalar function $G(\mathbf{r}, \mathbf{r}')$. Why?
This is like saying that, when the source is a point, the vector potential $\mathbf{A}(\mathbf{r})$ becomes scalar...?!
$$G(\vec{r},\vec{r}') \propto \frac{\exp(+\mathrm{i}k|\vec{r}-\vec{r}'|)}{|\vec{r}-\vec{r}'|}.$$