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

In summary: I'm just curious about the notation now.In summary, the confusion about the notation of Dirac Delta arises from the different ways it is defined and represented in different sources. The tutorial presents a notation that uses d^3\textbf{r'} to represent the 3-dimensional coordinate system, while other sources use dV or d^3\textbf{r} for the same purpose. However, in a Cartesian coordinate system, these notations are equivalent and can be used interchangeably. The notation d^3\textbf{r'} is simply a convenient way to denote the volume element in the 3-dimensional coordinate system.
  • #1
genxium
141
2

Homework Statement



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 referring 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 :)

Homework 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]

The Attempt at a Solution



Mentioned above.
 
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  • #2
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
genxium said:

Homework Statement



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 referring 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 :)

Homework 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]

The Attempt at a Solution



Mentioned above.

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
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 :)
 

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

1. How is (d^3)r in Green's Function different from regular volume element?

The (d^3)r in Green's Function is a mathematical expression used in the calculation of electromagnetic fields, while the regular volume element is a physical quantity used to measure volume. They are different in terms of their purpose and application.

2. Why is (d^3)r used in Green's Function instead of a regular volume element?

(d^3)r is used in Green's Function because it represents a differential volume element, which is needed in the mathematical derivation of the Green's Function. It allows for a more precise calculation of the electric and magnetic fields in a given region.

3. How does (d^3)r in Green's Function relate to the distance from the source point?

The (d^3)r in Green's Function is a function of the distance from the source point, as it represents a differential volume element at a specific point in space. As the distance from the source point changes, so does the value of (d^3)r.

4. Is (d^3)r in Green's Function dependent on the shape of the source?

No, (d^3)r in Green's Function is not dependent on the shape of the source. It is a function of the distance from the source point and is used to calculate the fields at any point in space, regardless of the source's shape.

5. How is (d^3)r in Green's Function related to the overall volume of the source?

(d^3)r in Green's Function is not directly related to the overall volume of the source. It is a mathematical expression used in the calculation of fields and is not based on the physical size or shape of the source itself.

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