the velocity feedback term which we are considering here as a drive (which one might as well call the extra damping on the particle at x$_{1}$ than that in the position x$_{2}$ ) is in effect changing the local temperature of the particle at x$_{1}$ than that at x$_{2}$ resulting in a nonequilibrium situation. This nonequilibrium situation would sustain all over the time because within the scope of the model the Gaussian noise strength and the so called damping on the particles are held constant. Thus, effectively in this model the local environment of the two interacting particles are kept fixed at two different temperatures and that keeps the nonequilibrium situation prevailing for all the time. The above reasoning makes it clear why, in the present context, we prefer calling the feedback term as a drive than considering it merely a part of the damping on the particle at x$_{1}$ . To analyze this model further, let us move to the center of mass (CM) X = ( x$_{1}$ + x$_{2}$ ) / 2 and the internal gap Z = ( x$_{1}$ − x$_{2}$ ) coordinates. The model now reduces to the form ˙ Z = $_{−}$α (2 − β ) 1 − β Z + ξ$_{Z}$ ˙ X = − αβ 2(1 − β ) Z + ξ$_{X}$ (4) where, < ξ$_{i}$ > = 0, < ξ$_{Z}$ ( t ) ξ$_{Z}$ ( t ′ ) > = 2 T D$_{1}$δ ( t − t ′ ), < ξ$_{X}$ ( t ) ξ$_{X}$ ( t ′ ) > = TD$_{1}$ 2 δ ( t − t ′ ) and D$_{1}$ = (1 + 1 (1 $_{−}$β ) $_{2}$). Eq.2 clearly shows that the internal coordinate Z will have an equilibrium distribution P$_{eq}$ ( Z ) = exp ( − α (2 − β ) 2 T D$_{1}$ (1 − β ) Z $^{2}$) . (5) Since, the average velocity of the CM < ˙ X > is proportional to the < Z > , it can be easily calculated. Note that, although Z has a probability distribution with zero mean the volume exclusion will prevent Z being less than zero and thus we will actually have a nonzero < Z > by simply integrating one one half of the distribution it has. This is exactly where we explicitly see the role of volume exclusion in producing directional transport. Thus, we arrive at < Z > = √ 2 T D$_{1}$ (1 − β ) πα (2 − β ) < ˙ X > = − αβ 2(1 − β ) < Z > = − αβ 2(1 − β ) × √ 2 T D$_{1}$ (1 − β ) πα (2 − β ) . (6) Note that, at β = 0 the average velocity of the system vanishes identically. This is in keeping with the fact that one cannot expect directed transport in equilibrium. Interestingly, in the 4