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Delta Function Identity in Modern Electrodynamics, Zangwill

  1. Jan 15, 2016 #1
    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.
     
  2. jcsd
  3. Jan 15, 2016 #2

    Demystifier

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    Hint: Write ##r## in terms of ##r_k## and use (1.121).
     
  4. Jan 17, 2016 #3
    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?
     
  5. Jan 18, 2016 #4

    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##.
     
  6. Jan 18, 2016 #5

    stevendaryl

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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*}
     
  7. Jan 18, 2016 #6

    Demystifier

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    Stevendaryl, your guess is correct.
     
  8. Jan 18, 2016 #7
    Yeah, that was just a typo. The ##\frac{1}{r}## definitely needs to be there. Sorry about that.
     
  9. Jan 18, 2016 #8
    Thanks for all of your help, I was finally able to figure it out because of your useful directions!
     
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