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

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1. Nov 17, 2014

### genxium

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 $\delta^3(x, y, z)=\delta(x)\delta(y)\delta(z)$ in http://mathworld.wolfram.com/DeltaFunction.html or $\delta(\textbf{x})=\delta(x_1)\delta(x_2)\delta(x_3)$ 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 $v(\textbf{r})=\int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})d^3\textbf{r'}$ and the $d^3\textbf{r'}$ is confusing me. I'm pretty sure that the tutorial is refering to a 3-dimensional coordinate system and I suppose that $\textbf{r'}=x' \cdot \textbf{i}+y' \cdot \textbf{j}+z' \cdot \textbf{k}$ is indicating the position vector. Thus how does $d^3\textbf{r'}$ work here?

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

$v(\textbf{r})=\int \int \int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})dx'dy'dz'$

or

$v(\textbf{r})=\int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})dV'$.

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

Any help is appreciated :)

2. Relevant equations

$v(\textbf{r})=\int \int \int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})dx'dy'dz'$
$v(\textbf{r})=\int \delta(\textbf{r}-\textbf{r'})v(\textbf{r'})dV'$

3. The attempt at a solution

Mentioned above.

2. Nov 17, 2014

### Orodruin

Staff Emeritus
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.

3. Nov 17, 2014

### Ray Vickson

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.

4. Nov 19, 2014

### genxium

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 $d^3\textbf{r} \stackrel{\Delta}{=} dV$ 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:

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

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

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

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