Why Does Helmholtz Theorem Use Position Vector at Observation Point?

[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

 

[eljose79]

for the questions! I can understand your confusion with the use of $D(\vec r)$ in the Helmholtz Theorem. Let me try to explain it in a simpler way.

Firstly, we need to understand that in the Helmholtz Theorem, $\vec F$ is a vector field, which means that at every point in space, it has a magnitude and direction. In the case of electrostatics, $\vec E$ is a vector field, and at every point in space, it has a magnitude and direction. This is why we use $\rho(\vec r')$, which represents the charge density at the source point, to calculate the divergence of the electric field at the observation point $\vec r$.

Now, in the Helmholtz Theorem, we are dealing with a more general vector field, so we can't just use the charge density at the source point to calculate the divergence. Instead, we use the function $D(\vec r)$, which represents the source of the vector field $\vec F$ at every point in space. In other words, at every point in space, $D(\vec r)$ tells us how much of the vector field is coming from that point. This is why we use $D(\vec r)$ to calculate the divergence of $\vec F$ at the observation point $\vec r$.

As for your second question, the reason why $D(\vec r)$ and $\vec C(\vec r)$ go to zero as $|\vec r| \rightarrow \infty$ is because we are assuming that the vector field $\vec F$ is localized, meaning that it is only non-zero in a finite region of space. This is a reasonable assumption for most physical systems, as they tend to have a finite extent. Therefore, as we move away from this finite region, the effect of the vector field becomes smaller and smaller, and eventually, it goes to zero. This is why we say that $D(\vec r)$ and $\vec C(\vec r)$ go to zero as $|\vec r| \rightarrow \infty$.

I hope this helps to clarify your questions. Please let me know if you need any further explanation.