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

AI Thread Summary
Griffiths' section 1.5.3 discusses the divergence of the vector function r/r^2, which equals 4*Pi*δ^3(r). The divergence theorem is used to show that the integral of the divergence over a volume enclosing the origin equals 4π. Explicit calculations reveal that the divergence is zero everywhere except at the origin, where it becomes infinite. This behavior indicates that the divergence can be represented as a function f(r) multiplied by a delta function, leading to the conclusion that f(r)δ^3(r) corresponds to the divergence of r/r^2. The physical interpretation is limited, but the geometric representation highlights the infinite divergence at the origin.
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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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 \int_{\mathcal{V}}( \vec{\nabla}\cdot\frac{\hat{r}}{r^2})d\tau=4\pi 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 f(\vec{r})\delta^3(\vec{r})\equiv \vec{\nabla}\cdot\frac{\hat{r}}{r^2}

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 \frac{\hat{r}}{r^2} (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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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}?
 
Can't use latex. Why does f(R)δ^3(R) = the divergence of R/r^2?
 
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 \frac{\hat{r}}{r^2} 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 f(\vec{r}) times a delta function so, you call it f(\vec{r})\delta^3(\vec{r}) and solve for f by setting the integral (eq. 1.98) equal to the value you calculated using the diverergence theorem (4\pi)
 
Ok. I think I got it. Thanks a lot!
 
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