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 $D_{a} = \frac{\theta(x_{0}'-x_{0})}{2\pi}\delta[(x-x')^2]$.

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    $\begingroup$ 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. $\endgroup$ – 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.

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