There is a way to obtain the Green's Function for the Laplacian as a limit of the Green's function of the D'Alembertian?
For the Laplacian ($-\nabla^2$) we have
$$ G_1(\vec X) = \frac{1}{4\pi X}$$
And for the D'Alembertian ($\Box$) using the retarded prescription we have
$$ G_2(T,\vec X) = \frac{1}{4\pi X} \delta(c T-|\vec X|) $$
I suppose that a way to go from $G_2$ to $G_1$ would be take the limit $c\rightarrow \infty$. That supposition is based on:
$$ \left. \Box \right|_{c\rightarrow \infty}= \left. \frac{1}{c^2} \partial_t^2\right|_{c\rightarrow \infty} - \nabla^2 = - \nabla^2$$
The problem is that I can't see a way to have something like
$$ \lim_{c\rightarrow \infty} \delta(c T-|\vec X|) = 1 $$
What's going on?
Anyway, it's just like when we make $\hbar \rightarrow 0$ even though $\hbar$ is a constant.
– Erich Apr 18 '14 at 14:19