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Define a dimensionless channel contrast $p=(b-a)/(a+b)$ and spacer ratio $s=r/(a+b)$. Positive $a,b$ give $|p|<1$, and substitution gives the displayed expression. Equal channel responses, $p=0$, eliminate the model's magnetic-configuration signal. Increasing the common spacer resistance at fixed $a,b$ increases the denominator and decreases $m$.

This is a controlled comparison within the model. In an actual multilayer, changing a spacer thickness can change the magnetic coupling and the interfaces as well as electrical transport. Spin-flip scattering also mixes the channels that the calculation kept independent.

When spin-flip scattering is weak, two spin populations have separately defined current responses and share a terminal electric field. Their electrical conductivities add: $\sigma ={\sigma }_{\uparrow }+{\sigma }_{\downarrow }$, so $\rho ={\rho }_{\uparrow }{\rho }_{\downarrow }/({\rho }_{\uparrow }+{\rho }_{\downarrow })$. Spin-dependent band structure and scattering set the channel parameters; being the majority population does not universally imply lower channel resistivity. The model is named for Nevill Francis Mott and is the independent-channel basis for many giant magnetoresistance models.

The Mott two-current model treats the two spin populations as parallel conduction channels when spin-flip scattering is weak. Thus $\sigma ={\sigma }_{\uparrow }+{\sigma }_{\downarrow }$ and ${\rho }^{-1}={\rho }_{\uparrow }^{-1}+{\rho }_{\downarrow }^{-1}$. Light, broad-band carriers conduct efficiently; scattering into narrow $d$-like states relaxes their current. In a simple spin-conserving $s$–$d$ picture,In copper, the filled $d$ bands supply little final-state phase space at $E_F$, so both spin channels have low resistivity. In cobalt, the nearly filled majority $d$ band suppresses majority-channel scattering, while minority $d$ states remain available. The stated iron reversal corresponds to stronger scattering-weighted majority-channel phase space: its majority $d$ band is not closed as in cobalt. Band filling, exchange splitting, velocities and scattering matrix elements set this scattering-weighted spin-channel asymmetry; majority population alone does not select the better conductor. The given iron ordering is a material/scattering input, not a universal consequence of the two-current postulate. Different impurities and interfaces can change the asymmetry.^[ref]

Spin relaxation is the approach of nonequilibrium spin populations and coherences toward their equilibrium values through coupling to other degrees of freedom or environmental fluctuations. A spin polarized away from equilibrium can lose its net polarization even when the total angular momentum of the larger closed system is conserved.

For a simple ensemble in a fixed magnetic field, longitudinal relaxation with time $T_1$ restores the population difference along the field, while transverse relaxation with time $T_2$ damps the rotating transverse spin polarization. Static field inhomogeneity can produce an additional reversible spread of precession phases, so an observed free-induction decay time need not equal the intrinsic $T_2$. Spin-flip scattering changes populations; phase-randomizing processes can also suppress transverse coherence without individual flips. A Hanle spin-precession measurement probes the competition of precession, transport and relaxation.

Consider two identical magnetic layers separated by a nonmagnetic layer, between reservoirs that impose the same terminal electric potential difference on both spin channels. Neglect spin-flip scattering and assume each layer obeys Ohm's law. In a magnetic layer, define positive channel electrical resistances $a,b>0$. Let the spacer contribute $r\ge 0$ to each channel. These are circuit parameters, not measured iron or chromium resistivities.

For parallel layer magnetizations, one channel meets $a,r,a$ and the other $b,r,b$. For opposite layer magnetizations, each meets one $a$, one $b$, and the spacer. The two channels share the same terminal electric potential difference and run in parallel. Writing $x\parallel y=xy/(x+y)$ givesThe inequality is algebra, given the specified network. It shows how a difference between spin-channel responses can yield a difference between magnetic configurations even though all the constituent resistors are ohmic.

For chosen $a=1\Omega$, $b=4\Omega$, $r=1\Omega$, the parallel channels have total electrical resistances $3$ and $9\Omega$, while the opposite-alignment channels each have $6\Omega$. Thus $R_P=2.25\Omega$ and $R_{AP}=3.00\Omega$. At a common terminal electric potential difference $1\mathrm{V}$, the total electric currents are $4/9\mathrm{A}$ and $1/3\mathrm{A}$, respectively.