# Delta Function Identity in Modern Electrodynamics, Zangwill

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1. Jan 15, 2016

### chi_rho

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. Jan 15, 2016

### Demystifier

Hint: Write $r$ in terms of $r_k$ and use (1.121).

3. Jan 17, 2016

### chi_rho

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?

4. Jan 18, 2016

### Demystifier

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 $c$.

5. Jan 18, 2016

### stevendaryl

Staff Emeritus
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*}

6. Jan 18, 2016

### Demystifier

7. Jan 18, 2016

### chi_rho

Yeah, that was just a typo. The $\frac{1}{r}$ definitely needs to be there. Sorry about that.

8. Jan 18, 2016

### chi_rho

Thanks for all of your help, I was finally able to figure it out because of your useful directions!