Rotation of macroscopic magnetization = average (Magnetization current density)

[LeoJakob]

Homework Statement

rotation of macroscopic magnetization = averege (Magnetization current density )

Relevant Equations

$(\vec{\nabla} \cdot \vec{j}_{\text{mag}}^{(i)}=0$

,$\vec{M}(\vec{r})=\frac{1}{v} \sum_{i=1}^{N} \vec{m}_{i}$

,$\vec{m}_{i}=\frac{1}{2} \int\left(\vec{r}^{\prime}-\vec{R}_{i}\right) \times \vec{j}_{mag}^{(i)}\left(\vec{r}^{\prime}\right) d^{3} \vec{r}^{\prime}$

, $\vec{\nabla} \times(\vec{A} \times \vec{B})=(\vec{B} \cdot \vec{\nabla}) \vec{A}-\vec{B}(\vec{\nabla} \cdot \vec{A})+\vec{A}(\vec{\nabla} \cdot \vec{B})-(\vec{A} \cdot \vec{\nabla}) \vec{B}$

In the headline to the question the statement should have been:
rotation of macroscopic magnetization = averege (Magnetization current density )

The Magnetization current densities $\vec{j}_{\text{mag}}^{(i)}$ of individual particles $i$ are stationary $(\vec{\nabla} \cdot \vec{j}_{\text{mag}}^{(i)}=0$) and generate the magnetic moments
$$
\vec{m}_{i}=\frac{1}{2} \int\left(\vec{r}^{\prime}-\vec{R}_{i}\right) \times \vec{j}_{mag}^{(i)}\left(\vec{r}^{\prime}\right) d^{3} \vec{r}^{\prime}
$$
at the locations$\vec{R}_{i}$. Analogous to the macroscopic electric polarization, introduce the macroscopic magnetization as the average magnetic dipole moment per mesoscopic volume $v$:
$$
\vec{M}(\vec{r})=\frac{1}{v} \sum_{i=1}^{N} \vec{m}_{i}
$$

Proof:
$$
\vec{\nabla} \times \vec{M}=\overline{\vec{j}_{mag}}
$$

Attempt
$$\begin{array}{l}\vec{\nabla} \times \vec{M}=\vec{\nabla} \times\left(\frac{1}{V} \sum \limits_{i=1}^{N} \vec{m}_{i}\right) \\ =\frac{1}{V} \vec{\nabla} \times\left(\sum \limits_{i=1}^{n} \frac{1}{2} \int\left(\vec{r}^{\prime}-\vec{R}_{i}\right) \times \vec{j}_{\text {mag }}^{(i)}\left(\vec{r}^{\prime}\right) \partial^{3} \vec{r}^{\prime}\right) \\ =\frac{1}{2 V} \sum \limits_{i=1}^{n} \underbrace{\vec{\nabla} \times\left(\left(\overrightarrow{r^{\prime}}-\overrightarrow{R_{i}}\right) \times \overrightarrow{j_{\text {mag }}^{(i)}}\left(\overrightarrow{r^{\prime}}\right)\right)}_{=(i)} d^{3} \vec{r}^{\prime} \\
= ?\end{array}$$

$$
\begin{array}{l}(i)=\left(\vec{j}_{\text {mag }}^{(i)}\left(\vec{r}^{\prime}\right) \cdot \vec{\nabla}\right)\left(\vec{r}^{\prime}-\vec{R}_{i}\right)- \vec{j}_{\text {mag }}^{(i)}\left(\vec{r}^{\prime}\right) \left(\vec{\nabla} \cdot\left(\vec{r}^{\prime}-\vec{R}_{i}\right)\right) \\ +\left(\vec{r}^{\prime}-\vec{R}_{i}\right)(\underbrace{\vec{\nabla} \cdot \vec{j}_{\text {mag }}^{(i)}\left(\vec{r}^{\prime}\right)}_{=0})-\left[\left(\vec{r}^{\prime}-\overrightarrow{R_{i}}\right) \cdot \vec{\nabla}\right] \vec{j}_{\text {mag }}^{(i)}\left(\vec{r}^{\prime}\right)
\\
\\
=\left(\vec{j}_{\text {mag }}^{(i)}\left(\vec{r}^{\prime}\right) \cdot \vec{\nabla}\right)\left(\vec{r}^{\prime}-\vec{R}_{i}\right)- \vec{j}_{\text {mag }}^{(i)}\left(\vec{r}^{\prime}\right) \left(\vec{\nabla} \cdot\left(\vec{r}^{\prime}-\vec{R}_{i}\right)\right) \\
-\left[\left(\vec{r}^{\prime}-\overrightarrow{R_{i}}\right) \cdot \vec{\nabla}\right] \vec{j}_{\text {mag }}^{(i)}\left(\vec{r}^{\prime}\right) \end{array} $$

I don’t have more ideas right now, can someone help me?

 

[TSny]

The Magnetization current densities $\vec{j}_{\text{mag}}^{(i)}$ of individual particles $i$ are stationary $(\vec{\nabla} \cdot \vec{j}_{\text{mag}}^{(i)}=0$) and generate the magnetic moments
$$
\vec{m}_{i}=\frac{1}{2} \int\left(\vec{r}^{\prime}-\vec{R}_{i}\right) \times \vec{j}_{mag}^{(i)}\left(\vec{r}^{\prime}\right) d^{3} \vec{r}^{\prime}
$$
at the locations$\vec{R}_{i}$. Analogous to the macroscopic electric polarization, introduce the macroscopic magnetization as the average magnetic dipole moment per mesoscopic volume $v$:
$$
\vec{M}(\vec{r})=\frac{1}{v} \sum_{i=1}^{N} \vec{m}_{i}
$$

As indicated, $\vec M (\vec r)$ is a function of the position vector $\vec r$ for a point in the medium. So, the right-hand side of the above equation for $\vec M(\vec r)$ must be a function of $\vec r$. Note, however, that $\vec m_i$ for the magnetization of the $i^{th}$ particle located at $\vec R_i$ does not depend on the vector $\vec r$.

However, the expression $\frac 1 v \sum_{i=1}^N \vec m_i$ does depend on $\vec r$ because the “mesoscopic” volume $v$ is assumed to be centered on $\vec r$. The sum is over all particles located in $v$. So, the sum does depend on $\vec r$ even though $m_i$ for a particular particle does not depend on $\vec r$.

When forming the curl (“rotation”) of $\vec M(\vec r)$, the derivative operators are derivatives with respect to the components of $\vec r$. These derivative operators do not act on $\vec r’$ appearing in the expression for $\vec m_i$.

The change in $\vec M(\vec r)$ for a small change $\vec {dr}$ in $\vec r$ is due to shifting the location of the volume $v$ by $\vec {dr}$ and performing the sum $\sum_{i= 1}^N \vec m_i$ over the particles in this shifted volume.

So, it’s pretty tricky. The text says “Analogous to the macroscopic electric polarization, introduce the macroscopic magnetizarion…”. Perhaps the textbook has already done a similar calculation for electric polarization $\vec P(\vec r)$ in deriving $\vec{\nabla} \cdot \vec P(\vec r) = - \rho_{\text {pol}}(\vec r)$, where $\rho_{\text {pol}}$ is the polarization charge density. If so, it might be helpful to study this derivation before tackling the magnetization problem.