- #1

yucheng

- 232

- 57

In Vanderlinde page 171-172, the author derives the vector potential for the magnetic dipole (and free currents)

\begin{align}

\vec{A}(\vec{r}) &=\frac{\mu_{0}}{4 \pi} \int_{\tau} \frac{\vec{J}\left(\vec{r}^{\prime}\right) d^{3} r^{\prime}}{\left|\vec{r}-\vec{r}^{\prime}\right|}+\frac{\mu_{0}}{4 \pi} \int_{\tau} \frac{\vec{M}\left(\vec{r}^{\prime}\right) \times\left(\vec{r}-\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|^{3}} d^{3} r^{\prime} (1) \\

&= \frac{\mu_{0}}{4 \pi} \int_{\tau} \frac{\vec{J}\left(\vec{r}^{\prime}\right) d^{3} r^{\prime}}{\left|\vec{r}-\vec{r}^{\prime}\right|}+\frac{\mu_{0}}{4 \pi} \int_{\tau} \vec{M}\left(\vec{r}^{\prime}\right) \times \vec{\nabla}^{\prime}\left(\frac{1}{\left|\vec{r}-\vec{r}^{\prime}\right|}\right) d^{3} r^{\prime} \\

&= \frac{\mu_{0}}{4 \pi} \int \frac{\vec{J}\left(\vec{r}^{\prime}\right)+\vec{\nabla}^{\prime} \times \vec{M}\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|} d^{3} r^{\prime}\\

&= \frac{\mu_{0}}{4 \pi} \int_{\tau - S^{\prime}} \frac{\vec{J}\left(\vec{r}^{\prime}\right)+\vec{\nabla}^{\prime} \times \vec{M}\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|} d^{3} r^{\prime}+\frac{\mu_{0}}{4 \pi} \oint_{S^{\prime}} \frac{\vec{M}\left(\vec{r}^{\prime}\right) \times d \vec{S}^{\prime}}{\left|\vec{r}-\vec{r}^{\prime}\right|}

\end{align}

Using integration by parts, we see that the second term of (2) is

$$\int_{\tau} \vec{M}\left(\vec{r}^{\prime}\right) \times \vec{\nabla}^{\prime}\left(\frac{1}{\left|\vec{r}-\vec{r}^{\prime}\right|}\right) d^{3} r^{\prime} = \int_{\tau} \frac{\vec{\nabla}^{\prime} \times \vec{M}\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|} d^{3} r^{\prime}+\oint_{S^{\prime}} \frac{\vec{M}\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|} \times d \vec{S}^{\prime} \tag{5}$$

However, in (5) since the volume on the LHS is inclusive of the surface as well, isn't the volume on the RHS also inclusive of the surface?

In (3), the surface integral term drops out because we integrate over a larger volume such that the magnetization there is zero.

However, for (3) the author claims that the curl is undefined at the boundaries of the magnetized material, hence it is easier to integrate over the volume of the magnetized integral, (4): the volume integral excluding the surface ##\tau - S^{'}## where it is undefined, and the surface integral taking care of the surface ## S^{'}##. However, why can we exclude the surface from the volume integral (now we have an 'open' volume)?

Does this mean that for

$$

\vec{E}(\vec{r})=\frac{1}{4 \pi \varepsilon_{0}} \int_{\tau} \frac{\left[-\vec{\nabla}^{\prime} \cdot \vec{P}\left(\vec{r}^{\prime}\right)\right]\left(\vec{r}-\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|^{3}} d^{3} r^{\prime}+\frac{1}{4 \pi \varepsilon_{0}} \oint_{S} \frac{(\vec{P} \cdot \hat{n})\left(\vec{r}-\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|^{3}} d S^{\prime}

$$

we ought to exclude the bounding surface from the volume integral as well?

Thanks in advance!

\begin{align}

\vec{A}(\vec{r}) &=\frac{\mu_{0}}{4 \pi} \int_{\tau} \frac{\vec{J}\left(\vec{r}^{\prime}\right) d^{3} r^{\prime}}{\left|\vec{r}-\vec{r}^{\prime}\right|}+\frac{\mu_{0}}{4 \pi} \int_{\tau} \frac{\vec{M}\left(\vec{r}^{\prime}\right) \times\left(\vec{r}-\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|^{3}} d^{3} r^{\prime} (1) \\

&= \frac{\mu_{0}}{4 \pi} \int_{\tau} \frac{\vec{J}\left(\vec{r}^{\prime}\right) d^{3} r^{\prime}}{\left|\vec{r}-\vec{r}^{\prime}\right|}+\frac{\mu_{0}}{4 \pi} \int_{\tau} \vec{M}\left(\vec{r}^{\prime}\right) \times \vec{\nabla}^{\prime}\left(\frac{1}{\left|\vec{r}-\vec{r}^{\prime}\right|}\right) d^{3} r^{\prime} \\

&= \frac{\mu_{0}}{4 \pi} \int \frac{\vec{J}\left(\vec{r}^{\prime}\right)+\vec{\nabla}^{\prime} \times \vec{M}\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|} d^{3} r^{\prime}\\

&= \frac{\mu_{0}}{4 \pi} \int_{\tau - S^{\prime}} \frac{\vec{J}\left(\vec{r}^{\prime}\right)+\vec{\nabla}^{\prime} \times \vec{M}\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|} d^{3} r^{\prime}+\frac{\mu_{0}}{4 \pi} \oint_{S^{\prime}} \frac{\vec{M}\left(\vec{r}^{\prime}\right) \times d \vec{S}^{\prime}}{\left|\vec{r}-\vec{r}^{\prime}\right|}

\end{align}

Using integration by parts, we see that the second term of (2) is

$$\int_{\tau} \vec{M}\left(\vec{r}^{\prime}\right) \times \vec{\nabla}^{\prime}\left(\frac{1}{\left|\vec{r}-\vec{r}^{\prime}\right|}\right) d^{3} r^{\prime} = \int_{\tau} \frac{\vec{\nabla}^{\prime} \times \vec{M}\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|} d^{3} r^{\prime}+\oint_{S^{\prime}} \frac{\vec{M}\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|} \times d \vec{S}^{\prime} \tag{5}$$

However, in (5) since the volume on the LHS is inclusive of the surface as well, isn't the volume on the RHS also inclusive of the surface?

In (3), the surface integral term drops out because we integrate over a larger volume such that the magnetization there is zero.

However, for (3) the author claims that the curl is undefined at the boundaries of the magnetized material, hence it is easier to integrate over the volume of the magnetized integral, (4): the volume integral excluding the surface ##\tau - S^{'}## where it is undefined, and the surface integral taking care of the surface ## S^{'}##. However, why can we exclude the surface from the volume integral (now we have an 'open' volume)?

Does this mean that for

$$

\vec{E}(\vec{r})=\frac{1}{4 \pi \varepsilon_{0}} \int_{\tau} \frac{\left[-\vec{\nabla}^{\prime} \cdot \vec{P}\left(\vec{r}^{\prime}\right)\right]\left(\vec{r}-\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|^{3}} d^{3} r^{\prime}+\frac{1}{4 \pi \varepsilon_{0}} \oint_{S} \frac{(\vec{P} \cdot \hat{n})\left(\vec{r}-\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|^{3}} d S^{\prime}

$$

we ought to exclude the bounding surface from the volume integral as well?

Thanks in advance!

Last edited: