Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have two Brownian particles, $B_1$ and $B_2$ (with diffusion coefficients $D_1$ and $D_2$), at coordinates $P_1$ and $P_2$ in a three-dimensional fluid.

I let the system evolve for $t$ seconds. What is the variance and mean for the change in distance between $B_1$ and $B_2$? How does this change if we set $D_2 = 0$ $\mu^2 / sec$?

share|improve this question

1 Answer 1

up vote 0 down vote accepted

The relative motion $B_2-B_1$ is Brownian with diffusion coefficient $D=D_1+D_2$ and initial value $P_0=P_2-P_1$. Hence, at time $t$, $B_2-B_1=P_0+\sqrt{D}\cdot Z_t$, where $Z_t$ is a standard 3D Brownian particle.

The expression of the mean distance $\mathbb E(\|B_2-B_1\|)$ is complicated but the mean square distance is simply $\mathbb E(\|B_2-B_1\|^2)=\|P_0\|^2+D\cdot\mathbb E(\|Z_t\|^2)=\|P_0\|^2+\color{red}{2}\cdot D\cdot t$, where the coefficient $\color{red}{2}$ depends on the convention used and might be $1$ for a mathematician.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.