- #1
latentcorpse
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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?
\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?
