1.2.1.1 Exchange

Let us take the first-order variation of Eq. (1.38):

$$\delta F_ =F_ (\textbf{{m}}+\delta\textbf{{m}})-F_ (\textbf{{m}})=\int_\Omega 2A\nabla\textbf{{m}}\cdot\nabla\delta\textbf{{m}} dV \quad,$$ (1.54)

where $\nabla\textbf{{m}}\cdot\nabla\delta\textbf{{m}}$ is a compact notation for $\nabla m_x\cdot\nabla\delta m_x+\nabla m_y\cdot\nabla\delta m_y +\nabla m_z\cdot\nabla\delta m_z$. Now we proceed in the derivation for the $x$ component, the remaining $y,z$ can be treated analogously. By applying the vector identity

$$\textbf{v}\cdot\nabla f=\nabla\cdot(f\textbf{v})-f\nabla\cdot\textbf{v}\quad,$$ (1.55)

in which we put $f=\delta m_x$ and $\textbf{v}=\nabla m_x$, one obtains:

$$\int_\Omega \nabla m_x\cdot\nabla\delta m_x dV=\int_\Omega \bi... ...delta m_x A \nabla m_x) - \delta m_x \nabla\cdot(A\nabla m_x) \big] dV \quad.$$ (1.56)

By using the divergence theorem, the first term can be written as surface integral over the boundary $\partial \Omega$

$$\int_\Omega \nabla m_x\cdot\nabla\delta m_x dV=\int_{\partial\... ...\textbf{n}} dS - \int_\Omega \delta m_x \nabla\cdot(A\nabla m_x) dV \quad.$$ (1.57)

By substituting the latter equation and the analogous for the $y,z$ components into Eq. (1.54), one ends up with:

$$\delta F_ =-\int_\Omega \Big[ 2\nabla\cdot(A\nabla\textbf{{m}})\cdot\delta... ... \textbf{{m}}}{\partial \textbf{n}}\cdot\delta \textbf{{m}}\right] dS \quad,$$ (1.58)

which is the exchange contribution to the first-order variation of the free energy functional.