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?

  • 1
    $\begingroup$ $c\to\infty$ doesn't make any sense to me as $c=2.9979\times10^{10}$ cm/s. $\endgroup$
    – Kyle Kanos
    Apr 18, 2014 at 13:27
  • 3
    $\begingroup$ Well, it's a way you could get the non-relativistic limit. The more formal procedure would be take limits like $v/c << 1$. But, $c\rightarrow \infty$ is something equivalent. Anyway, it's just like when we make $\hbar \rightarrow 0$ even though $\hbar$ is a constant. $\endgroup$
    – Erich
    Apr 18, 2014 at 14:19
  • $\begingroup$ Hmm, I suppose $c\to\infty$ could be equivalent to $\hbar\to0$ (though I am not much of a fan of doing that either). $\endgroup$
    – Kyle Kanos
    Apr 18, 2014 at 14:23

1 Answer 1


The wave equation is a partial differential equation for a field in space and time, the Laplace equation is an ordinary differential equation for functions of space only. The natures of these equations are really different. However, if you have a Green's function of the wave equation that vanishes at infinite time, and which derivative also vanishes you may integrate from $t=0$ to $t=\infty$ and get a relation that could be what you want. Note that this would not work in dimensions 1 or 2 because the Green's functions of the wave equation do not satisfy these prerequisites.

  • $\begingroup$ Well, I think I'm getting the point... I was making an analogy with the passage from $-\nabla^2 + m^2$ to $-\nabla^2$. But in these two cases we have equations of the same nature, as is evident by the form of the Dirac Delta function in the Green diferential equation... I was thinking about the $\partial^2_t$ as controlled by a parameter $1/c^2$ but didn't looked to the Dirac Delta's. I'll analyse it a little bit more. $\endgroup$
    – Erich
    Apr 18, 2014 at 14:32

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