How is the Laplacian Applied to Retarded Potential?

[CMBR]

Homework Statement

See attachment.

Homework Equations

The Attempt at a Solution

I'm not understanding how the laplacian is creating those 3 terms in 5.4.5.

I just understand the basics that laplacian on f(x,y) = d²f/dx² + d²f/dy². Can someone elaborate?

Thanks in advance.

EDIT:
Just realized this is an identity of vector calculus (second derivative of 2 scalars)...

 

[vanhees71]

The Laplace operator in three dimensions is
$$\Delta=\vec{\nabla} \cdot \vec{\nabla}=\partial_x^2+\partial_y^2+\partial_z^2.$$
Now you have a product of functions, and you can simply use the product rule of differentiation to evaluate it.

This is, however not very clever, because then you have to handle the dependence of $\rho$ on $\vec{r}$ in the formula for the retarded potential. It's not undoable but inconvenient.

It's always wise to rewrite electromagnetic equations in relativistically covariant form. For the retarded potential this simply means to introduce a $\delta$ distribution,
$$\phi(t,\vec{x})=\frac{1}{4 \pi} \int_{\mathbb{R}^4} \mathrm{d}t' \mathrm{d}^3 \vec{x}' \delta(t'-t+R/c) \frac{\rho(t',\vec{x}')}{R}.$$
Now the Laplacian is much easier to evaluate. Also note that
$$\Delta \frac{1}{R}=-4 \pi \delta(R).$$