# Question in Helmholtz Theorem

1. Aug 9, 2013

### yungman

In page 555, Appendix B of Intro to electrodynamics by D Griffiths:
$$\nabla\cdot\vec F=D,\;\nabla\times\vec F=\vec C\;\Rightarrow\;\nabla\cdot \vec C=0$$
Griffiths let $\vec F=-\nabla U+\nabla\times \vec W$
$$\nabla\cdot \vec F=-\nabla^2U=-\frac{1}{4\pi}\int D\nabla^2\left(\frac{1}{\vec{\vartheta}}\right)d\tau'=\int D(\vec r')\delta^3(\vec r-\vec r')d\tau'=D(\vec r) \;\hbox { where }\;\;\vec{\vartheta}=\vec r-\vec r'$$

My questions are:
(1)Why the theorem use $D(\vec r),\; and \;\vec C(\vec r)$ where $\vec r$ is a position vector pointing at the OBSERVATION point $P$?
In electrostatic, $\nabla\cdot\vec E=\frac{\rho(\vec r')}{\epsilon}$ where $\rho(\vec r')$ represented the charge density at the SOURCE point pointed by position vector $\vec r'$. $\nabla\cdot\vec F$ is the divergence of the field at the OBSERVATION point $P(\vec r)$ where position vector $\vec r$ pointing at the point $P$. Why in Helmholtz Theorem using $D(\vec r)$?

(2) Why $D(\vec r)$ and $\vec C(\vec r)$ goes to zero when $|\vec r| \rightarrow \;\infty$? Which, pretty much back to question (1)......why use $D(\vec r)$?

Thanks

Last edited: Aug 9, 2013