Why does $\nabla\delta(t-\frac{r}{c}) = \frac{-\vec{e_r}}{c}\delta(t-\frac{r}{c})$?

Question

I am trying to show that when you multiply $$G(t,\textbf{r}) = -\frac{1}{4\pi r}\delta(t-\frac{r}{c}) $$ by the d'Alembertian you get $$\square G = \delta(t)\delta(\textbf{r})$$ and immensely struggling, so any help would be much appreciated!

Answer 1

I have been trying to answer your question since you posted it, this is my current state now. Not yet an answer! Maybe this helps you move forward while i struggle.

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$$\nabla f(x-x')=\left(\partial_xf,\partial_yf,\partial_zf\right)=f'(x-x')$$

$$\square G(x-x')=\delta(x-x') $$

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$$\phi=\delta(x-x_0,y-y_0)$$

This is the distribution sending $f$ to $f(x_0,y_0)$

$$\phi(f)=f(x_0,y_0)$$

rule: https://math.stackexchange.com/a/1413826/738388

$$\langle f,\delta'_a\rangle=-\langle f',\delta_a\rangle=-f'(a)$$

$$(\delta')(f)=-\delta(f')=-f'(0)$$

$$\frac{d\phi}{dx}(f)=\frac{d}{dx}\delta(x-x_0,y-y_0)(f)=-\delta(x-x_0,y-y_0)(f')=-\frac{d}{dx}f(x_0,y_0))$$

https://math.stackexchange.com/a/445092/738388

$$\langle\nabla f,\cdot\rangle=df$$

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$$\nabla\phi=\frac{d\phi}{dx}dx+\frac{d\phi}{dy}dy$$

where $dx$ and $dy$ are differentials

$$\vec{\nabla}_x \delta^{(3)}(\vec{x}-\vec{x}')=-\vec{\nabla}_{x'} \delta^{(3)}(\vec{x}-\vec{x}').$$

where $\delta^{(3)}$ is the generalized 3 dimensional Dirac-delta. Source https://www.physicsforums.com/threads/gradient-of-dirac-delta.880397/

Answer 2

You seem to be missing a derivative. Since by the chain rule$$\partial_if(r)=\frac{1}{2r}\partial_i(r^2)f^\prime(r)=\frac{x_i}{r}f^\prime\implies\nabla f=\vec{e}_rf^\prime$$for radially symmetric $f$ (in this case $f$ is a distributional measure rather than a function), the final result follows from another application of the chain rule, which gives the $-\tfrac1c$ factor.