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In an unbiased random walk in one dimension, the coefficient of diffusion is $D = l^2/2\tau$, where $l$ is the size of the jump and $\tau$ is time taken for that jump.

In simple Brownian motion, Einstein's relation says that $D = \mu kT$ where $\mu$ is the mobility, $k$ is the Boltzmann constant and $T$ is temperature.

In both cases, it turns out: $$ \langle x^2 \rangle = 2Dt$$ So my question is, can we equate the two constants? So, can we say that: $l^2/2\tau = \mu kT$?

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Yes. $D$ refers to the same quantity in both expressions. In fact, equating them is one way to derive the Einstein-Smoluchowski relation. See the last section of these notes for more details.

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