1.2.2 Effective Field and Brown's Equations

Thus, to summarize the previously derived results, we can write the expression for the first-order variation of the free energy functional (1.53):

$$\begin{split} \delta G=-\int_\Omega \left[ 2\nabla\cdot(A\na... ...{n}}\cdot\delta \textbf{{m}}\right] dS =0 \quad. \end{split}$$ (1.64)

Now we claim the fact that the variation $\delta\textbf{{m}}$ has to satisfy the constraint $\vert\textbf{{m}}+\delta\textbf{{m}}\vert=1$. For this reason, it can be easily observed that the most general variation is a rotation of the vector field $\textbf{{m}}$, that is

$$\delta \textbf{{m}}=\textbf{{m}}\times \vec{\delta \theta} \quad,$$ (1.65)

where the vector $\vec{\delta \theta}$ represents an elementary rotation of angle $\delta \theta$. By substituting this expression in Eq. (1.64) and remembering that $\textbf{v}\cdot(\textbf{w}\times\mathbf{u})=\mathbf{u}\cdot(\textbf{v}\times\textbf{w})=-\mathbf{u}\cdot(\textbf{w}\times\textbf{v})$, one obtains:

$$\begin{split} \delta G=\int_\Omega \textbf{{m}}\times \left[... ...{m}}\right]\cdot \vec{\delta \theta} dS =0\quad. \end{split}$$ (1.66)

Since the elementary rotation $\delta \theta$ is arbitrary, Eq. (1.66) can be identically zero if and only if:

$$\left\{ \begin{aligned}&\textbf{{m}}\times \left[2\nabla\cdot(... ...ight]_{\partial \Omega}=\mathbf{0} \end{aligned} \right. \quad.$$

In the second equation the fact that $\frac{\partial \textbf{{m}}}{\partial \textbf{n}} \times \textbf{{m}}=\mathbf{0}$ implies that $\frac{\partial \textbf{{m}}}{\partial \textbf{n}}=\mathbf{0}$, as the vectors $\textbf{{m}}$ and $\frac{\partial \textbf{{m}}}{\partial \textbf{n}}$ are always orthogonal; in fact, the only way their vector product can vanish is that $\frac{\partial \textbf{{m}}}{\partial \textbf{n}}$ is identically zero. We introduce now the effective field

$$\textbf{H}_{\text{eff}}=\frac{2}{\mu_0 M_s}\nabla\cdot(A\nabla\te... ...{an}}{\partial \textbf{{m}}} + {\mathbf{H}_\text{m}}+ \mathbf{H}_\text{a}\quad,$$ (1.68)

where the first two terms take into account the exchange and anisotropy interjections. In other words, these interactions effectively act on the magnetization as they were suitable fields:

$$\mathbf{H}_ =\frac{2}{\mu_0 M_s}\nabla\cdot(A\nabla\textbf{{m}}) \quad,$$ (1.69)

$$\mathbf{H}_ =\frac{1}{\mu_0 M_s}\frac{\partial f_\text{an}}{\partial \textbf{{m}}} \quad.$$ (1.70)

Eqs. (1.67) can be rewritten as

$$\left\{\begin{aligned} &\mu_0 M_s \textbf{{m}}\times \textbf... ...\bigg\vert _{\partial \Omega} =\mathbf{0} \end{aligned} \right. .$$

The Brown's equations allow one to find the equilibrium configuration of the magnetization within the body. The first equation states that the torque exerted on magnetization by the effective field must vanish at the equilibrium. It is important to notice that Eqs. (1.71) are nonlinear, since the effective field (1.68) has a functional dependance on the whole vector field $\textbf{{m}}(\cdot)$. As we will discuss later, the existence of exact analytical solutions is subject to appropriate simplifying assumptions. For this reason, in most cases numerical solution of Eqs. (1.71) is required. In addition, as mentioned in section 1.1.2, the model must be completed with a dynamic equation to properly describe the evolution of the system. This will be done in the following section.