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

In summary, the derivation of Maxwell's equations in the presence of matter involves modifying the classical equations to account for the effects of electric charges and currents within materials. This includes introducing polarization and magnetization to describe how electric fields and magnetic fields interact with matter. The resulting equations incorporate the dielectric constant and magnetic permeability of materials, allowing for a comprehensive understanding of electromagnetic phenomena in various media. The derivation emphasizes the importance of boundary conditions and the role of material properties in influencing electromagnetic behavior.
  • #1
LeoJakob
24
2
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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  • #2
Apply the divergence theorem to a surface outside the material where the polarization exists.
 
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