Delta Function Identity in Modern Electrodynamics, Zangwill

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I am currently reading Modern Electrodynamics by Andrew Zangwill and came across a section listing some delta function identities (Section 1.5.5 page 15 equation 1.122 for those interested), and there is one identity that really confused me. He states:
\begin{align*}
\frac{\partial}{\partial r_k}\frac{\partial}{\partial r_m}=\frac{3r_k r_m - r^2 \delta_{km}}{r^5}-\frac{4\pi}{3}\delta_{km}\delta(\mathbf{r})\\
\end{align*}
I am having trouble with figuring out how to show this identity is true. If anyone can help get me on the right track to see how to achieve this identity I would greatly appreciate it.
 

Demystifier

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Hint: Write ##r## in terms of ##r_k## and use (1.121).
 
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Hint: Write ##r## in terms of ##r_k## and use (1.121).
So I wrote ##r## in terms of ##r_{i}## , but I just needed some dummy index so really any can work. The problem I ran into was in finding the final term:
\begin{align*}
-\frac{4\pi}{3}\delta_{km}\delta(\mathbf{r})
\end{align*}
I know the derivative I performed only works with ##r \neq 0##, and the Laplacian of ##\frac{1}{r}## gives you a ##-4\pi\delta(\mathbf{r})##, but I am just struggling to see how this particular term comes up...any thoughts?
 

Demystifier

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Suppose that for ##r=0## there is an additional term ##c \delta_{km}\delta({\bf r})##, where ##c## is a constant that you need to determine. (You are allowed to assume that because, a priori, ##c## can even be zero, which would be the same as if that term was not present at all.) Now put ##k=m##, sum over ##k##, and use (1.141). This will give you the non-zero value of ##c##.
 

stevendaryl

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I am currently reading Modern Electrodynamics by Andrew Zangwill and came across a section listing some delta function identities (Section 1.5.5 page 15 equation 1.122 for those interested), and there is one identity that really confused me. He states:
\begin{align*}
\frac{\partial}{\partial r_k}\frac{\partial}{\partial r_m}=\frac{3r_k r_m - r^2 \delta_{km}}{r^5}-\frac{4\pi}{3}\delta_{km}\delta(\mathbf{r})\\
\end{align*}
I am having trouble with figuring out how to show this identity is true. If anyone can help get me on the right track to see how to achieve this identity I would greatly appreciate it.
Is there something missing in this equation? It seems to me that the left-hand side is an operator, while the right-hand side is a function. I'm guessing that it's supposed to be:

\begin{align*}
\frac{\partial}{\partial r_k}\frac{\partial}{\partial r_m} \frac{1}{r} =\frac{3r_k r_m - r^2 \delta_{km}}{r^5}-\frac{4\pi}{3}\delta_{km}\delta(\mathbf{r})\\
\end{align*}
 

Demystifier

Science Advisor
Insights Author
2018 Award
10,378
3,197
Stevendaryl, your guess is correct.
 
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Is there something missing in this equation? It seems to me that the left-hand side is an operator, while the right-hand side is a function. I'm guessing that it's supposed to be:

\begin{align*}
\frac{\partial}{\partial r_k}\frac{\partial}{\partial r_m} \frac{1}{r} =\frac{3r_k r_m - r^2 \delta_{km}}{r^5}-\frac{4\pi}{3}\delta_{km}\delta(\mathbf{r})\\
\end{align*}
Yeah, that was just a typo. The ##\frac{1}{r}## definitely needs to be there. Sorry about that.
 
10
0
Suppose that for ##r=0## there is an additional term ##c \delta_{km}\delta({\bf r})##, where ##c## is a constant that you need to determine. (You are allowed to assume that because, a priori, ##c## can even be zero, which would be the same as if that term was not present at all.) Now put ##k=m##, sum over ##k##, and use (1.141). This will give you the non-zero value of ##c##.
Thanks for all of your help, I was finally able to figure it out because of your useful directions!
 

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