The gradient of the d'Alembertian Green's function

Question

So I have to prove that the d'alembertian of the associated green's function $G(t,t',\vec{r},\vec{r}')$ is equal to zero when given that $\vec{r}\neq\vec{r}'$ $$\left(\frac{1}{c^2}\partial^2 t-\Delta\right)\frac{\delta\left(t-t'-\frac{|\vec{r}-\vec{r}'|}{c}\right)}{R}=0$$ Well for the sake of effectiveness I substituted $t-t'=:\tau$ and $|\vec{r}-\vec{r}'|=:R$. Well then the problem looks like this: $$\left(\frac{1}{c^2}\partial^2 \tau-\Delta\right)\delta\left(\tau-\frac{R}{c}\right)=0$$ The time derivative is very easy but my real problem is, I am confused how to calculate the gradient of the delta-function, as it's dependent on the time and the space coordinates.

So how shall I compute $$\nabla\delta\left(\tau-\frac{R}{c}\right)=?$$ In my course book the result for this term seem to be $$\nabla\delta\left(\tau-\frac{R}{c}\right)= \dot{\delta}\left(\tau-\frac{R}{c}\right)\cdot\left(-\frac{1}{c}\hat{R}\right)$$ with $\hat{R}$ as the unit vector of R.

Answer 1

We need frequently to differentiate expressions of Dirac $\delta$-function when the argument of the latter is a function $f\left(z\right)$ of the variable $z$ with respect to which we want to differentiate. So \begin{equation} \dfrac{\partial\delta\bigl[f\left(z\right)\bigr]}{\partial z} \boldsymbol{=}\dfrac{\mathrm d\delta\bigl[f\left(z\right)\bigr]}{\mathrm d f\left(z\right)}\dfrac{\partial f\left(z\right)}{\partial z}\boldsymbol{=}\boldsymbol{-}\dfrac{\delta\bigl[f\left(z\right)\bigr]}{ f\left(z\right)}\dfrac{\partial f\left(x\right)}{\partial z} \nonumber \end{equation} that is \begin{equation} \boxed{\:\:\dfrac{\partial\delta\bigl[f\left(z\right)\bigr]}{\partial z} \boldsymbol{=}\boldsymbol{-}\dfrac{\delta\bigl[f\left(z\right)\bigr]}{ f\left(z\right)}\dfrac{\partial f\left(z\right)}{\partial z}\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{01}\label{01} \end{equation} In our case \begin{equation} f\left(x_1,x_2,x_3\right)\boldsymbol{=}c\tau\boldsymbol{-}r\boldsymbol{=}c\tau\boldsymbol{-}\sqrt{x^2_1\boldsymbol{+}x^2_2\boldsymbol{+}x^2_3} \tag{02}\label{02} \end{equation} so \begin{equation} \dfrac{\partial\delta\bigl(c\tau\boldsymbol{-}r\bigr)}{\partial x_\jmath} \boldsymbol{=}\dfrac{\delta\bigl(c\tau\boldsymbol{-}r\bigr)}{\bigl(c\tau\boldsymbol{-}r\bigr)}\dfrac{\partial r}{\partial x_\jmath}\boldsymbol{=}\dfrac{\delta\bigl(c\tau\boldsymbol{-}r\bigr)}{\bigl(c\tau\boldsymbol{-}r\bigr)}\dfrac{x_\jmath}{r} \tag{03}\label{03} \end{equation} hence \begin{equation} \boldsymbol{\nabla}\delta\bigl(c\tau\boldsymbol{-}r\bigr) \boldsymbol{=}\dfrac{\delta\bigl(c\tau\boldsymbol{-}r\bigr)}{\bigl(c\tau\boldsymbol{-}r\bigr)}\dfrac{\mathbf{r}}{r}\boldsymbol{=}\dfrac{\delta\bigl(c\tau\boldsymbol{-}r\bigr)}{\bigl(c\tau\boldsymbol{-}r\bigr)}\mathbf{n}_r \tag{04}\label{04} \end{equation} where $\mathbf{n}_r$ the unit vector in the direction of $\mathbf{r}$. Finally \begin{equation} \boxed{\:\:\boldsymbol{\nabla}\delta\Bigl(\tau\boldsymbol{-}\dfrac{r}{c}\Bigr) \boldsymbol{=}\dfrac{\delta\Bigl(\tau\boldsymbol{-}\dfrac{r}{c}\Bigr)}{c\Bigl(\tau\boldsymbol{-}\dfrac{r}{c}\Bigr)}\dfrac{\mathbf{r}}{r}\boldsymbol{=}\dfrac{\delta\Bigl(\tau\boldsymbol{-}\dfrac{r}{c}\Bigr)}{c\Bigl(\tau\boldsymbol{-}\dfrac{r}{c}\Bigr)}\mathbf{n}_r\:\:} \tag{05}\label{05} \end{equation}

Answer 2

So thanks to @Nikodem's Tutorial notes I figured out what my problem was: One uses the normal chain rule; so with the index notation we get:

$$\partial r_i\delta\left(\tau-\frac{R}{c}\right)= \frac{\partial \delta}{\partial (\tau-\frac{R}{c})}\frac{\partial(\tau-\frac{R}{c})}{\partial r_{i}}$$

So what we now have to prove is: $$\frac{\partial}{\partial(\tau-a)}=\frac{\partial \tau}{\partial(\tau-a)}\frac{\partial}{\partial\tau}=\frac{\partial[(\tau-a)+a]}{\partial(\tau-a)}\frac{\partial}{\partial\tau}=\frac{\partial}{\partial\tau}$$

So it follows $$\partial r_i\delta\left(\tau-\frac{R}{c}\right)= \frac{\partial \delta}{\partial \tau}\frac{\partial(\tau-\frac{R}{c})}{\partial r_{i}}=\dot{\delta}\left(\tau-\frac{R}{c}\right)\cdot\left(-\frac{1}{c}\hat{R}\right)$$