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How is (d^3)r in Green's Function equivalent to volume element?

  1. Nov 17, 2014 #1
    1. The problem statement, all variables and given/known data

    This is part of the online tutorial I'm reading: http://farside.ph.utexas.edu/teaching/em/lectures/node49.html

    I'm so confused about the notation of Dirac Delta. It's said that 3-dimensional delta function is denoted as [itex]\delta^3(x, y, z)=\delta(x)\delta(y)\delta(z)[/itex] in http://mathworld.wolfram.com/DeltaFunction.html or [itex]\delta(\textbf{x})=\delta(x_1)\delta(x_2)\delta(x_3)[/itex] in http://en.wikipedia.org/wiki/Dirac_delta_function#Properties_in_n_dimensions , which was taken by me as granted before.

    However in the tutorial, it's said that [itex]v(\textbf{r})=\int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})d^3\textbf{r'}[/itex] and the [itex]d^3\textbf{r'}[/itex] is confusing me. I'm pretty sure that the tutorial is refering to a 3-dimensional coordinate system and I suppose that [itex]\textbf{r'}=x' \cdot \textbf{i}+y' \cdot \textbf{j}+z' \cdot \textbf{k}[/itex] is indicating the position vector. Thus how does [itex]d^3\textbf{r'}[/itex] work here?

    In my understanding, for Cartesian Coordinate, the traditional delta property is

    [itex]v(\textbf{r})=\int \int \int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})dx'dy'dz'[/itex]

    or

    [itex]v(\textbf{r})=\int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})dV'[/itex].

    It's not obvious to me that [itex]v(\textbf{r})=\int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})d^3\textbf{r'}[/itex] is equivalent to either of them.

    Any help is appreciated :)

    2. Relevant equations

    [itex]v(\textbf{r})=\int \int \int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})dx'dy'dz'[/itex]
    [itex]v(\textbf{r})=\int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})dV'[/itex]

    3. The attempt at a solution

    Mentioned above.
     
  2. jcsd
  3. Nov 17, 2014 #2

    Orodruin

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    This is the definition of dV and d^3r. Not writing out the triple ingegral is merely convention. Also, naturally, the volumr dV of a box of side lengths dx, dy, and dz is dx dy dz.
     
  4. Nov 17, 2014 #3

    Ray Vickson

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    Your confusion has nothing to do with Dirac; it is just notational confusion in integration. There is nothing to be confused about: in a cartesian coordinate system, ##dx \, dy \, dz = dV = d^3 \vec{r}##, by definition. And, as Orodruin has said, ##\int = \int \int \int## is a common contraction.
     
  5. Nov 19, 2014 #4
    Thank you for the replies.

    @Ray Vickson, yes it's fine for me to take it as a definition :)

    In fact I'm expecting to find consistency between notations, like that [itex]d^3\textbf{r} \stackrel{\Delta}{=} dV[/itex] is introduced by "reasoning" instead of "convenience". However if it's indeed just a convenient notation I'll still be happy to go on reading and learning the tutorial.

    Here's what I've tried but failed to find the "consistency" in Cartesian coordinate:

    [tex]d\textbf{r} = \frac{\partial \textbf{r}}{\partial x} \cdot dx + \frac{\partial \textbf{r}}{\partial y} \cdot dy + \frac{\partial \textbf{r}}{\partial z} \cdot dz = dx \cdot \textbf{i} + dy \cdot \textbf{j} + dz \cdot \textbf{k}[/tex]

    [tex]d^2 \textbf{r} = d(d\textbf{r}) = \frac{\partial (d\textbf{r})}{\partial x} \cdot dx + \frac{\partial (d\textbf{r})}{\partial y} \cdot dy + \frac{\partial (d\textbf{r})}{\partial z} \cdot dz[/tex]

    I'm stuck here because there's no explicit term [itex]x[/itex] or [itex]y[/itex] or [itex]z[/itex] in [itex]d\textbf{r}[/itex].

    Anyway this is no longer a big problem for me as it was 1 week ago :)
     
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