- #1
LeoJakob
- 21
- 1
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:
$$
0=\int \overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot\left(\vec{P}\left(\overrightarrow{r^{\prime}}\right) \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|}\right) d^{3} \overrightarrow{r^{\prime}} =\int \vec{P}\left(\overrightarrow{r^{\prime}}\right) \cdot \vec{\nabla}_{\overrightarrow{r^{\prime}}} \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} d^{3} \overrightarrow{r^{\prime}}+\int \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} \overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot \vec{P}\left(\overrightarrow{r^{\prime}}\right) d^{3} \overrightarrow{r^{\prime}}\\
\Rightarrow \int \vec{P}\left(\overrightarrow{r^{\prime}}\right) \cdot \vec{\nabla}_{\overrightarrow{r^{\prime}}} \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} d^{3} \overrightarrow{r^{\prime}}=\int \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} \overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot \vec{P}\left(\overrightarrow{r^{\prime}}\right) d^{3} \overrightarrow{r^{\prime}}
$$
$$\Delta \Phi(\vec{r})=\Delta \left( \frac{1}{4 \pi \varepsilon_{0}} \int(\frac{\varrho\left(\overrightarrow{r^{\prime}}\right)}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|}+\vec{P}\left(\overrightarrow{r^{\prime}}\right) \cdot \vec{\nabla}_{\overrightarrow{r^{\prime}}} \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|}) d^{3} \overrightarrow{r^{\prime}}\right)$$
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:
$$
0=\int \overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot\left(\vec{P}\left(\overrightarrow{r^{\prime}}\right) \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|}\right) d^{3} \overrightarrow{r^{\prime}} =\int \vec{P}\left(\overrightarrow{r^{\prime}}\right) \cdot \vec{\nabla}_{\overrightarrow{r^{\prime}}} \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} d^{3} \overrightarrow{r^{\prime}}+\int \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} \overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot \vec{P}\left(\overrightarrow{r^{\prime}}\right) d^{3} \overrightarrow{r^{\prime}}\\
\Rightarrow \int \vec{P}\left(\overrightarrow{r^{\prime}}\right) \cdot \vec{\nabla}_{\overrightarrow{r^{\prime}}} \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} d^{3} \overrightarrow{r^{\prime}}=\int \frac{1}{\left|\vec{r}-\overrightarrow{r^{\prime}}\right|} \overrightarrow{\nabla_{\overrightarrow{r^{\prime}}}} \cdot \vec{P}\left(\overrightarrow{r^{\prime}}\right) d^{3} \overrightarrow{r^{\prime}}
$$