I got confused when deriving electric displacement $\bf D$ following Wolfgang Nolting's Electrodynamics book. Here is his derivation:
My problem is about integral ① in red. Property of derivative of delta function tell us:
$\int{\delta'(x-x_0)f(x)dx = -f'(x_0)}$ and $\delta'(-x)=-\delta'(x)$.
Aware that here ① integral over r', thus should be:
$$\int{d^3r'P(r')·\nabla_{r'}\delta(r-r')} = \int{d^3r'P(r')·-\nabla_{r'}\delta(r'-r) } = -[-\nabla ·P(r)] = \nabla · P(r)$$
Here I got positive sign $\nabla · P(r)$. But why in the book he got negative sign? Where's my mistake?
Best Answer
As seen in the book you should start by convincing yourself that $\nabla_{\vec r\,'}=-\nabla_{\vec r}$. Then your integral is $$4\pi\int d^{3}\vec r\,'\vec P(\vec r\,')\cdot\nabla_{\vec r\,'}\delta(\vec r-\vec r\,')=-4\pi\int d^{3}\vec r\,'\vec P(\vec r\,')\cdot\nabla_{\vec r}\delta(\vec r-\vec r\,')$$ Now you can use the identity $$\nabla\cdot(f\vec A)=f(\nabla\cdot\vec A)+\vec A\cdot\nabla f$$ Note that in this case $\nabla=\nabla_{\vec r}$ and $\vec A=\vec P(\vec r\,')$ such that $\nabla_{\vec r}\cdot \vec P(\vec r\,')=0$. Thus, $$\nabla_{\vec r}\cdot[\delta(\vec r-\vec r\,')\vec P(\vec r\,')]=\vec P(\vec r\,')\cdot\nabla_{\vec r}\delta(\vec r-\vec r\,')$$ Consequently, the integral is modified to $$-4\pi\int d^{3}\vec r\,'\nabla_{\vec r}\cdot[\delta(\vec r-\vec r\,')\vec P(\vec r\,')]=-4\pi\nabla_{\vec r}\cdot\int d^{3}\vec r\,'\delta(\vec r-\vec r\,')\vec P(\vec r\,')$$ since the integration is on primed coordinates. The final integral in the above equation equates to $\vec P(\vec r)$ and you end up with $-4\pi\nabla_{\vec r}\cdot \vec P(\vec r)$. I hope this helps. Let me know if you have any questions or the solution seems incorrect.