# Covariant form of Green's function for wave equation

In J. D. Jackson's "Classical Electrodynamics", page 614 in the 3rd edition, he states that you can write the Green's functions for the wave equation in covariant form using the fact that \begin{align*}\delta(x-x')^2 &= \delta\left[\left(x_{0} - {x_{0}'}\right)^2-\left\lvert\vec x -\vec x'\right\rvert^2\right] \\ &= \delta\left[\left(x_{0}-x_{0}'-R\right)\left(x_{0}-x_{0}'+R\right)\right] \\ &=\frac{1}{2R}\left[\delta\left(x_{0}-x_{0}'-R\right)+\delta(x_{0}-x_{0}'+R)\right] \end{align*} I'm not seeing that last step. I have the feeling he's using something like $$\delta(f(x)) = \sum_{r\in f^{-1}(0)}\frac{\delta(x-r)}{f'(r)}$$ But since there are multiple variables here I'm not sure how to use that in this case.

For reference, this is for the wave equation $$\square f = (\frac{1}{c^{2}}\partial_{t}^{2}-\nabla^{2})f = \delta^{(4)}(x-x')$$ and the two Green's functions are:

• $$D_{r}(x-x') = \frac{\theta(x_{0}-x_{0}')}{4\pi|\vec x - \vec x'|}\delta(x_{0}-x_{0}'-|\vec x - \vec x'|)$$ or in easier notation, $$\frac{\theta(t-t')}{4\pi R}\delta\left[t' - \left(t-\frac{R}{c}\right)\right]$$ (retarded Green's function)

• $$D_{a}(x-x') = \frac{\theta(x_{0}'-x_{0})}{4\pi|\vec x - \vec x'|}\delta(x_{0}-x_{0}'+|\vec x - \vec x'|)\quad$$ (advanced Green's function)

And in covariant form they end up and $$D_{r} = \frac{\theta(x_{0}-x_{0}')}{2\pi}\delta[(x-x')^2]$$ and $$D_{a} = \frac{\theta(x_{0}'-x_{0})}{2\pi}\delta[(x-x')^2]$$.

• Usually in Green's function one distinguishes field variable $x$ and source variable $x'$. Physically it is a bit like Green's function specify how the solution (function of $x$) reacts to a point source at $x'$. Therefore, regard $x_0$ as $x$ in your $\delta(f(x))$ formula, and identify the zeros ($x_0'+R$ and $x_0'-R$) of the quadratic polynomial, you will get the third line by using your Dirac delta formula. – chichi Aug 18 '19 at 2:54

The formula for Dirac delta function has an absolute value in it, so it differs from yours (Dirac delta function - Composition with a function), that's why you are not getting the third step. Just take $$x_0-x_0'$$ as a variable and use the correct formula.