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Laplacian on retarded potential

  1. Aug 25, 2014 #1
    1. The problem statement, all variables and given/known data

    See attachment.

    2. Relevant equations



    3. 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 realised this is an identity of vector calculus (second derivative of 2 scalars)...
     

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  2. jcsd
  3. Aug 26, 2014 #2

    vanhees71

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    The Laplace operator in three dimensions is
    [tex]\Delta=\vec{\nabla} \cdot \vec{\nabla}=\partial_x^2+\partial_y^2+\partial_z^2.[/tex]
    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 [itex]\rho[/itex] on [itex]\vec{r}[/itex] 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 [itex]\delta[/itex] distribution,
    [tex]\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}.[/tex]
    Now the Laplacian is much easier to evaluate. Also note that
    [tex]\Delta \frac{1}{R}=-4 \pi \delta(R).[/tex]
     
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