CIP thickness-averaged two-current model

In a model of carriers sampling alternating magnetic and nonmagnetic layers, assume spin conservation, comparable carrier transport weights and negligible extra interface scattering. Averaging relaxation rates by the time spent in each material then gives a channel electrical resistivity $(t_1{\rho }_{\mathrm{FM},\mathrm{s}}+t_2{\rho }_{\mathrm{NM}})/(t_1+t_2)$. Here all constituent resistivities are per spin channel. A long mean free path compared with a period motivates interlayer sampling; it does not by itself prove the other assumptions. This is a homogenized current-in-plane model, rather than an independent parallel-layer limit or a current-perpendicular-to-plane resistor chain.

From the four-layer CIP sheet resistance, $R_P=ab/[2D^2(a+b)]$ and $R_{AP}=(a+b)/(8D^2)$. Subtraction gives $R_{AP}-R_P=(a-b{)}^2/[8D^2(a+b)]\ge 0$ for positive channels. The ratio in the title normalizes by the low-resistance state; normalizing by the high-resistance state instead gives $(a-b{)}^2/(a+b{)}^2$. Common spacer scattering adds equally to $a,b$, diluting the contrast. The result assumes identical magnetic layers; arbitrary materials need not have the same sign of spin asymmetry.

Let $D=t_1+t_2$, $a=t_1{\rho }_L+t_2{\rho }_N$ and $b=t_1{\rho }_H+t_2{\rho }_N$. A square of four layers has total thickness $2D$. For parallel moments its channel sheet resistances are $a/(2D^2)$ and $b/(2D^2)$; for antiparallel moments both are $(a+b)/(4D^2)$. Adding parallel spin conductances gives the title. The factor of four-layer thickness is essential; changing the normalization changes the prefactors. The nonmagnetic bulk resistivity would be ${\rho }_N/2$ when ${\rho }_N$ is this model's per-spin value.