Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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: $l^2/2\tau = \mu kT$?

share|cite|improve this question
up vote 0 down vote accepted

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.

share|cite|improve this answer

Your Answer


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.