Let the Green's function for the gauge field be given (after gauge fixing) as $$G_{\mu \nu}(x,y) = \delta_{\mu \nu}G(x-y) \tag{1}$$ where $$G(x-y)= \int \frac{d^dk}{(2\pi)^d} \frac{e^{ik \cdot (x-y)}}{k^2+m^2} \tag{1'}$$ and a source $$ J_{\mu}(x) = \delta_{0}^{\mu}q\delta(\vec{x})\tag{2}. $$ Then, the gauge field is given by $$ A_{\mu}(x)=\int d^dy \, G_{\mu \nu}J^{\nu} \tag{3}. $$

How do I derive $$A^{0}(x) = \frac{q}{4\pi}\frac{1}{|\vec{x}|} \tag{4}$$ for $d=4$? Obviously I have to plug into (3) equations (1) and (2) in 4 dimensions. By doing so though I have 4-dimensional quantities. If I set $m=0$, then I do get the Fourier transform of (1') and I do have a quantity like $1/x$ but how do I get only the spatial component? And how do I continue from there on?

My attempt: \begin{align} A_{\mu}(x)= \int d^dy G_{\mu \nu}(x,y)J^{\nu}(y) \end{align} and now set $\mu=0$ \begin{align} A_{0}(x)&= \int d^4y G(x-y)q \delta(\vec{y}) \\ &= q\int d^4y \int \frac{d^4k}{(2\pi)^4}\frac{e^{ik\cdot (x-y)}}{k^2} \delta(\vec{y}) \\ &= \int d^4y \int \frac{d^3\vec{k}}{(2\pi)^3}\frac{e^{i\vec{k}\cdot (\vec{x}-\vec{y})}}{|\vec{k}|^2} \int dk^0 \frac{e^{ik^0\cdot(x^0-y^0)}}{(k^0)^2}\delta(\vec{y}) \end{align} Now the $d^3\vec{k}$ integral should give me (after the $dy$ integration too) $\frac{1}{|\vec{x}|}$ while, having the requirement that $x^0=y^0$ the time-component integral simplifies.

But should I make the assumption $x^0=y^0$? And how to proceed?


closed as off-topic by JamalS, ACuriousMind, Brandon Enright, Kyle Kanos, Pranav Hosangadi Dec 15 '14 at 21:13

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  • 3
    $\begingroup$ Have actually tried just working it out? $\endgroup$ – Prahar Dec 15 '14 at 17:08
  • $\begingroup$ Yes, but I do not know what to do with the time parts. I do get the spatial part ok but then, I have the integral of the time components and even at "equal times" I do not see what to do with it. $\endgroup$ – Marion Dec 15 '14 at 20:06
  • $\begingroup$ I have added my first steps. $\endgroup$ – Marion Dec 15 '14 at 20:21
  • $\begingroup$ $k^2 \neq (k^0)^2 |\vec{k}|^2$, but $k^2 = (k^0)^2 - |\vec{k}|^2$. $\endgroup$ – Prahar Dec 15 '14 at 21:10
  • 1
    $\begingroup$ You do not need to set x0 = y0. The integral can be computed easily. Use contour integrals for the time part. $\endgroup$ – Prahar Dec 15 '14 at 21:41