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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 $D_2 = 0$ $\mu^2 / sec$? |
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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. |
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