Can someone explain the 3D Dirac Delta Function in Griffiths' Section 1.5.3?

In summary, Griffiths' section 1.5.3 discusses the divergence of the vector function r/r^2 = 4*Pi*δ^3(r). The divergence theorem is used to show that the integral of the divergence of this function is equal to 4π for a spherical surface enclosing the origin. The divergence is then calculated explicitly using the formula for divergence in spherical coordinates and is found to be zero everywhere except at the origin, where it is infinite. This leads to the conclusion that the vector function must be equal to some unknown normal function f(\vec{r}) times a delta function, which is solved for by setting the integral equal to the value calculated using the divergence theorem. While there is no physical interpretation for
  • #1
cordyceps
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Griffiths' section 1.5.3 states that the divergence of the vector function r/r^2 = 4*Pi*δ^3(r). Can someone show me how this is derived and what it means physically? Thanks in advance.
 
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  • #2
cordyceps said:
Griffiths' section 1.5.3 states that the divergence of the vector function r/r^2 = 4*Pi*δ^3(r). Can someone show me how this is derived and what it means physically? Thanks in advance.

First, use the divergence theorem to show that [tex]\int_{\mathcal{V}}( \vec{\nabla}\cdot\frac{\hat{r}}{r^2})d\tau=4\pi[/tex] for any surface enclosing the origin (use a spherical surface centered at the origin for simplicity).

Then, calculate the divergence explicitly using the formula on the inside of the front cover for divergence in spherical coords. You should find that it is zero everywhere except at the origin where it blows up (because the 1/r terms correspond to dividing by zero).

Finally, put the two together by using eq 1.98 with [tex]f(\vec{r})\delta^3(\vec{r})\equiv \vec{\nabla}\cdot\frac{\hat{r}}{r^2}[/tex]

As for a physical interpretation, there really isn't one (although there are physical consequences as you'll see in chapter 2 and beyond); but there is a geometric interpretation...if you sketch the vector function [tex]\frac{\hat{r}}{r^2}[/tex] (figure 1.44 in Griffiths) you'll see why there must be an infinite divergence at the origin. This is all discussed in section 1.5.1
 
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  • #3
Sorry, I'm kinda slow- why does LaTeX Code: f(\\vec{r})\\delta^3(\\vec{r})\\equiv \\vec{\\nabla}\\cdot\\frac{\\hat{r}}{r^2}?
 
  • #4
Can't use latex. Why does f(R)δ^3(R) = the divergence of R/r^2?
 
  • #5
cordyceps said:
Can't use latex. Why does f(R)δ^3(R) = the divergence of R/r^2?

The fact that the divergence of [tex]\frac{\hat{r}}{r^2}[/tex] is zero everywhere, except at the origin where it is infinite, but when integrated gives a constant, means that it must be some unknown normal function [itex]f(\vec{r})[/itex] times a delta function so, you call it [itex]f(\vec{r})\delta^3(\vec{r})[/itex] and solve for [itex]f[/itex] by setting the integral (eq. 1.98) equal to the value you calculated using the diverergence theorem ([itex]4\pi[/itex])
 
  • #6
Ok. I think I got it. Thanks a lot!
 

1. What is the 3D Dirac Delta Function?

The 3D Dirac Delta Function is a mathematical function that is used to represent a point source of a substance or quantity in three-dimensional space. It is a generalization of the Dirac Delta Function, which is a mathematical concept used in the field of calculus.

2. How is the 3D Dirac Delta Function different from the Dirac Delta Function?

The 3D Dirac Delta Function takes into account the three-dimensional nature of space, whereas the Dirac Delta Function only considers one dimension. This means that the 3D version can represent point sources in a three-dimensional space, while the 1D version can only represent point sources on a line.

3. What is the purpose of using the 3D Dirac Delta Function in scientific research?

The 3D Dirac Delta Function is commonly used in scientific research, particularly in fields such as physics and engineering, to model and analyze point sources in three-dimensional space. It is also used in mathematical calculations and equations to simplify and solve problems involving point sources.

4. Can the 3D Dirac Delta Function be graphed?

Technically, the 3D Dirac Delta Function cannot be graphed since it represents a point source in three-dimensional space. However, it can be visualized as a spike or spike-like shape at the origin of a three-dimensional coordinate system.

5. Are there any real-world applications of the 3D Dirac Delta Function?

Yes, the 3D Dirac Delta Function has various real-world applications, such as in fluid mechanics to model point sources of mass or energy, in electromagnetic theory to represent point charges, and in signal processing to analyze point sources in three-dimensional space. It is also used in image processing and computer vision to detect and analyze point features in images.

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