Bend resistance ($R_{ij,kl}=(V_k-V_l)/I_{i\to j}$)

A four-terminal bend resistance uses adjacent current contacts and the other adjacent pair as voltage probes. It can be negative when a ballistic injector preferentially populates the opposite voltage probe. Its sign is a nonlocal routing property, not the sign of the total electrical dissipation.

For a balanced nonnegative transmission matrix and $I_i=g_0(N_i{V}_i-{\sum }_j{T}_{ij}{V}_j)$, electrical power obeyswith diagonal reflection included consistently or removed from both sides. A negative four-terminal voltage-to-current ratio is compatible with this positive total power because the measured voltage pair is not the current source's pair.

For left-deflected injection in the left 1,top 2,right 3,bottom 4 geometry, use 1 to 2 with $M$ modes, 2 to 3 with $M$, 3 to 4 with $N$, and 4 to 1 with $M$ plus 4 to 3 with $N-M$. The zero-current probes give $V_2=V_1$ and $V_3=M{V}_1/N$, hence $R_{14,23}=(h/e^2)(1/M-1/N)$. Reversing the transmission matrix gives equal probe potentials and zero bend resistance. Fixed-lead resistance need not be even in field; reciprocity also exchanges current and voltage pairs.

Number the leads left 1,top 2,right 3,bottom 4. Let narrow leads 1,2 have $M$ spin-resolved modes and wide leads 3,4 have $N>M$. Straight-through injection sends $M$ modes between 1 and 3 and between 2 and 4; each excess $N-M$ mode connects 3 and 4. With current 1 to 4, $V_4=0$ and zero current at 2,3, $V_2=0$, $V_3=M{V}_1/N$ and $I=(e^2/h)M(N-M)V_1/N$. Thus $R_{14,23}=-h/[e^2(N-M)]$. At $N=M$ this ideal network disconnects the current source from its drain.