Laboratoire de Physique de l'Ecole Normale Supérieure, PSL University, CNRS, Sorbonne Universités, 24 rue Lhomond, 75231 Paris, France.
LPTMS, CNRS, Université Paris-Sud, Université Paris-Saclay, 91405 Orsay, France.
Phys Rev E. 2019 Jul;100(1-1):012113. doi: 10.1103/PhysRevE.100.012113.
We study active particles performing independent run-and-tumble motion on an infinite line with velocities v_{0}σ(t), where σ(t)=±1 is a dichotomous telegraphic noise with constant flipping rate γ. We first consider a single particle in the presence of an absorbing wall at x=0 and calculate the probability that it has survived up to time t and is at position x at time t. We then consider two particles with independent telegraphic noises and compute exactly the probability that they do not cross up to time t. In contrast with the case of passive (Brownian) particles this problem of two run-and-tumble particles (RTPs) cannot be reduced to a single RTP with an absorbing wall. Nevertheless, we are able to compute exactly the probability of no crossing of two independent RTPs up to time t and find that it decays at large time as t^{-1/2} with an amplitude that depends on the initial condition. We show that this amplitude vanishes if one extrapolates the starting distance between the two particles to a negative value-similarly to the Milne extrapolation length in neutron scattering.
我们研究在具有速度 v_{0}σ(t) 的无限线上独立进行跑-跌运动的活性粒子,其中 σ(t)=±1 是具有恒定翻转率 γ 的二分电报噪声。我们首先考虑一个在 x=0 处存在吸收壁的单个粒子,并计算该粒子在时间 t 内幸存且在时间 t 时处于位置 x 的概率。然后,我们考虑具有独立电报噪声的两个粒子,并精确计算它们在时间 t 内不相交的概率。与被动(布朗)粒子的情况不同,这个具有两个跑-跌粒子(RTP)的问题不能简化为具有吸收壁的单个 RTP。尽管如此,我们能够精确地计算出两个独立的 RTP 在时间 t 内不相交的概率,并发现它在大时间上按 t^{-1/2}衰减,其幅度取决于初始条件。我们表明,如果将两个粒子之间的起始距离外推到负值(类似于中子散射中的米尔恩外推长度),则该幅度会消失。