# Homework Help: Dirac delta function definition

1. Feb 9, 2009

### latentcorpse

By definition of the Dirac delta function, we have:

$\int f(x) \delta(x-a) dx=f(a)$

This is fair enough. But in ym notes there is a step that goes like the following:

$\mathbf{\nabla} \wedge \mathbf{B}(\mathbf{r})=-\frac{\mu_0}{4 \pi} \int_V dV' \nabla^2(\frac{1}{|\mathbf{r}-\mathbf{r'}|}) \mathbf{J}(\mathbf{r'}) = \mu_0 \mathbf{J}(\mathbf{r})$

where we have used that $\nabla^2(\frac{1}{|\mathbf{r}-\mathbf{r'}|}) =-4 \pi \delta(\mathbf{r}-\mathbf{r'})$

clearly the minus signs and the $4 \pi$'s cancel so it's now just

$\mathbf{\nabla} \wedge \mathbf{B}(\mathbf{r})=\mu_0 \int_V dV' \mathbf{J}(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'})$

i don't see how that goes from there to give $\mu_0 \mathbf{J}(\mathbf{r})$ as we are integrating with respect to $V'$ are we not? and so i would assume that the delta function would need to be of the form $\delta(\mathbf{r'}-\mathbf{r})$ in order to give the desired answer.

the onnly explanation i can come up with is that $\delta(\mathbf{r'}-\mathbf{r})=\delta(\mathbf{r}-\mathbf{r'})$ since the delta function is symmetric about the spike. however the spike would be at different positions in these two cases. i'm kind of lost-any advice?

2. Feb 9, 2009

### Tom Mattson

Staff Emeritus
Yes, that's true.

No, the delta function has a spike whenever it's argument is zero. In both cases this occurs at $r=r'$.