Symmetry and dynamics (Video)

Moving domain wall

Consider two magnetic domains oppositely magnetised, e.g., $\bm{m}=(0,0,1)$ and $\bm{m}=(0,0,-1)$ respectively. Expanding the one domain is tantamount to moving the domain wall to the direction of the other domain.

J. A. Fernandez-Roldan et al, Sci. Rep. 9, 5130 (2019).

J. A. Fernandez-Roldan et al, Sci. Rep. 9, 5130 (2019).

Remark. A traveling domain wall changes the balance between neighbouring (oppositely magnetised) domains. It changes the magnetisation of the system.

Conservation of the total magnetisation

Consider the conservative model with exchange and uniaxial anisotropy

$$ \dot{\bm{m}} = -\bm{m}\times (\bm{m}'' + k^2 m_3\bm{\hat{e}}_3). $$

Define the total magnetisation

$$ \mathcal{M} = \int m_3\, dx $$

and calculate its time dependence

$$ \frac{d\mathcal{M}}{dt} = \int \dot{m}_3\, dx = -\int [\bm{m}\times (\bm{m}'' + k^2 m_3\bm{\hat{e}}_3)]_3\, dx. $$

The second term has no $z$ (third) component because $\bm{m}\times \bm{\hat{e}}_3 \perp \bm{\hat{e}}_3$, and

$$ \int \bm{m}\times \bm{m}'' dx = \int (\bm{m}\times\bm{m}')' dx = [\bm{m}\times\bm{m}']_{-\infty}^\infty = 0. $$

[We assume $\bm{m}'=0$ at the boundaries of the system.]

The result is

$$ \frac{d\mathcal{M}}{dt} = 0 \Rightarrow \mathcal{M}: \text{const.} $$

The total magnetisation is a conserved quantity for the above conservative model.