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\begin{align} - \nabla^2 G(r) &= \delta(r) \\ G(r) &\rightarrow 0 \mbox{ as } r \rightarrow \infty \end{align} where $ r = \sqrt{x_1^2 + x_2^2 + x_3^2}$

In order to find the Green's function I take the Fourier transform
of the equation. This gives me

\begin{align} \mathrm{k}^2 \hat{G}(\mathrm{k}) = 1 \tag{1} \end{align}

where $\mathrm{k} = \sqrt{k_1^2 + k_2^2 + k_3^2} $

The idea is to divide the equation by $\mathrm{k}^2$ then take the back transform of the equation. This gives me the Green's function $G(r)$. However $\mathrm{k}$ can be equal to $0$. Implying equation $(1)$ to be false.

The question is why is the Green's function $G(r)$ found by this method correct?

1 Answers1

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$$\mathscr{F}\{\nabla^2 G\}(\vec k)=\iiint_{\mathbb{R}^3}\nabla^2G(\vec r)e^{i\vec k\cdot\vec r}\,dx\,dy\,dz=-k^2\mathscr{F}\{G(\vec r)\}=-1$$

Hence,

$$\mathscr{F}\{G\}(\vec k)=\frac{1}{ k^2}$$

for $k\ne 0$. Since the value of the integral is independent of the value of the integrand at a single point, the value of $\mathscr{F}\{G\}(\vec k=0)$ is irrelevant. Proceeding, we find

$$\begin{align} G(\vec r)&=\frac{1}{(2\pi)^3}\iiint_{\mathscr{R}^3}\frac{e^{-i\vec k\cdot\vec r}}{k^2}\,dk_x\,dk_y\,dk_z\\\\ &=\frac{1}{(2\pi )^3}\int_0^{2\pi}\int_0^\pi \int_0^\infty e^{-i kr\cos(\theta)}\,\sin(\theta)\,dk\,d\theta\,d\phi\\\\ &=\frac{1}{(2\pi)^2}\int_0^\pi \int_0^\infty e^{-i kr\cos(\theta)}\,\sin(\theta)\,dr\,d\theta\\\\ &=\frac{1}{(2\pi)^2}\int_0^\infty \frac{2\sin(kr)}{kr}\,dk\\\\ &=\frac{1}{4\pi r} \end{align}$$

Mark Viola
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  • Thank you. As I understand it you argue that the value F(G)(k=0) is irrelevant. Since you show the integral to be exactly $\frac{1}{4 \pi r}$ –  Jan 09 '17 at 19:06
  • You're welcome. My pleasure. Yes, indeed; the value $\mathscr{F}{G}(\vec k=0)$ is not relevant to the value of its (inverse) Fourier Transform. – Mark Viola Jan 09 '17 at 19:08