Maxwell's equations in the presence of matter -- Derivation

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SUMMARY

This discussion focuses on the derivation of Maxwell's equations in the presence of matter, specifically addressing the calculation of the integral involving macroscopic polarization, ##\vec{P}\left(\overrightarrow{r^{\prime}}\right)##. The key transformation discussed is the application of the divergence theorem, which leads to the conclusion that ##\overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot\left(\vec{P}\left(\overrightarrow{r^{\prime}}\right) \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|}\right) = 0##. This result is crucial for understanding the behavior of electric fields in polarized media and directly relates to the formulation of Maxwell's equations.

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LeoJakob
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I want to calculate ##\int \vec{P}\left(\overrightarrow{r^{\prime}}\right) \cdot \vec{\nabla}_{\overrightarrow{r^{\prime}}} \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} d^{3} \overrightarrow{r^{\prime}}## with macroscopic polarization ##\vec{P}\left(\overrightarrow{r^{\prime}}\right)## because I want to solve:

$$\Delta \Phi(\vec{r})=\Delta \left( \frac{1}{4 \pi \varepsilon_{0}} \int(\frac{\varrho\left(\overrightarrow{r^{\prime}}\right)}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|}+\vec{P}\left(\overrightarrow{r^{\prime}}\right) \cdot \vec{\nabla}_{\overrightarrow{r^{\prime}}} \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|}) d^{3} \overrightarrow{r^{\prime}}\right)$$

There is a note that one can use the fact that: $$\overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot\left(\vec{P}\left(\overrightarrow{r^{\prime}}\right) \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|}\right) = 0$$

Why is this true? Does this have anything to do with the maxwell equations?

If I use the hint, I can do the following transformations:
$$
0=\int \overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot\left(\vec{P}\left(\overrightarrow{r^{\prime}}\right) \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|}\right) d^{3} \overrightarrow{r^{\prime}} =\int \vec{P}\left(\overrightarrow{r^{\prime}}\right) \cdot \vec{\nabla}_{\overrightarrow{r^{\prime}}} \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} d^{3} \overrightarrow{r^{\prime}}+\int \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} \overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot \vec{P}\left(\overrightarrow{r^{\prime}}\right) d^{3} \overrightarrow{r^{\prime}}\\

\Rightarrow \int \vec{P}\left(\overrightarrow{r^{\prime}}\right) \cdot \vec{\nabla}_{\overrightarrow{r^{\prime}}} \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} d^{3} \overrightarrow{r^{\prime}}=\int \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} \overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot \vec{P}\left(\overrightarrow{r^{\prime}}\right) d^{3} \overrightarrow{r^{\prime}}
$$
 
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Apply the divergence theorem to a surface outside the material where the polarization exists.
 
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