# How to deal with motion on a 2-D lattice in terms of dimension?

I am reading a paper titled: Random walks of molecular motors arising from diffusional encounters with immobilized filaments. There the authors consider the molecular motor moving on a 1-D protein filament in a 2-D unbounded media. They have used the framework of 2-D lattice for there analysis. The probability of jumping from the filament to the unbounded medium is $$\epsilon/2$$, the probability of retaining on a certain location on filament $$\gamma$$, the probability of moving backward $$0.5\delta$$ and then the authors say that the average velocity of moving on the filament is $$v=1-\gamma-\delta-0.5\epsilon$$ I have two questions regarding this argument:

Suppose I have a point $$d$$ units away from the origin then what should be the dimension of $$d/v$$, I mean $$v$$ is not in m/s. I can assume probabilities to be in $$s^{-1}$$ but that does not lead to $$\text{dim}\{d/v\}=s$$. Is there some concept of dimensionless being involved in this? How can change the parameters to get proper dimensionality? ## 1 Answer

I think that the $$v$$ they means is in Lattice units/Time step. Probabilities are dimensionless usually (not a decay probability, that is $$s^{-1}$$) and indicate the jump probability during an unit of "time step" in your lattice simulation. How long to take this time step, you can choose to match the real situation. In this case, how much time typically take your molecule to jump from one site to another.

Put lattice distance $$l= 1$$ Ångstrom, as distance between lattice points, and time step lasting $$\tau =$$ 1 nanosecond; you should get a dimensional speed $$v_d= v*(l/\tau)$$.

Working on lattice you don't care about actual dimensions, all is in simulation units, thus $$d/v$$ has dimension of time units, how they are called in your paper, $$n$$ or $$t$$. This is not physical time, just the number of time steps; to get physical time, you should multiply that by the corresponding physical duration of the time step $$\tau$$.