Using the Green's function for Maxwell

[gentsagree]

Homework Statement

I need to calculate, in D=4, the time component of the vector potential, A_{0}, given the equation (below) for A_{\mu} with the Green's function, also given below.
The answer is given to be
A^{0}=+\frac{q}{4\pi}\frac{1}{\mid\overline{x}\mid}

Homework Equations

A_{\mu}=\int d^{D}y G_{\mu\nu}(x,y)J^{\nu}
I think G_{\mu\nu} can be used in Feynman gauge as G_{\mu\nu}=\delta_{\mu\nu}G(x-y), where G(x-y)=\frac{\Gamma(\frac{1}{2}D-2)}{4\pi^{D/2}(x-y)^{(D-2)}} is the massless scalar Green's function.

I am given the current as J^{\mu}(x)=\delta^{\mu}_{0}q\delta(\overline{x})

The Attempt at a Solution

What I have done is the following:

A_{0}=\int d^{4}y \delta_{0\nu}G(x-y)\delta^{\nu}_{0}q\delta(\overline{y})=\delta_{0\nu}\delta^{\nu}_{0}G(\overline{x})q
=-\frac{\Gamma(\frac{1}{2}(2))}{4\pi^{2}\mid\overline{x}\mid^{2}}q
A^{0}=-A_{0}=+\frac{q}{4\pi^{2}}\frac{1}{\mid\overline{x}\mid^{2}}

where I treated \delta_{0\nu}\delta^{\nu}_{0}=\eta_{00}=-1

In words, I plug in the definition for the current, solve the integral for the delta function and substitute the definition for the Green's function, evaluated at modx instead of x-y, by means of the change of variable induced by the delta function.
So I get an extra power of pi and of mod(x). Where am I goin wrong?

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

 

[Orodruin]

You have used a 3-dimensional delta distribution to perform a 4D integral.