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

AI Thread Summary
The discussion focuses on deriving the integral involving macroscopic polarization, specifically the calculation of the term involving the divergence of polarization and its relationship to Maxwell's equations. It highlights that the expression $$\overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot\left(\vec{P}\left(\overrightarrow{r^{\prime}}\right) \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|}\right) = 0$$ is valid due to the properties of divergence in the context of electrostatics. By applying the divergence theorem, the integral can be transformed, leading to a relationship between the polarization and the potential. This transformation is crucial for solving the equation $$\Delta \Phi(\vec{r})$$ in the presence of matter. Understanding these relationships is essential for applying Maxwell's equations effectively in materials with polarization.
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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