0
$\begingroup$

I am trying to understand in the original derivation of Brownain diffusion, where does the assumption of Markovian and Gaussian nature factor in.

In Albert Einstein's original work on Brownian motion in 1905, he derives the probability distribution density of particles using a random walk scheme as follows.

If the particle is at a position $x$ at at a time $t$ with the the probability density $p(x,t)$, then it makes a tiny jump of length $\Delta$ in a span of time $\tau$, where $\Delta$ is a random number with PDF $\phi(\Delta)$. The probabilty of finding the particle at any $x$ at a time $t+\tau$ can be given as,

$$ p(x,t+\tau)=\int_{-\infty}^{\infty} p(x-\Delta,t)\phi(\Delta)d\Delta $$ Question 1.

Does the above equation correspond to a Markovian approximation? I would assume so because the position of the particle at the next step depends only the previous step.

To proceed further, Einstein considers a Taylor expansion of $p(x,t)$ in both $x$ and $t$ about $\Delta$ and $\tau$ respectively. With this he obtains,

$$ \tau \frac{\partial p}{\partial t} + ... = \frac{1}{2}\frac{\partial^2 p}{\partial x^2}\left[ \int_0^\infty \Delta^2 \phi(\Delta) d\Delta \right] + ... $$

In these above steps, the first order term in \Delta vanishes for a symmetric distribution of jumps (i.e. $\phi(\Delta)$ is even).

Question 2.

Now, can we say that the truncation of the series to only upto the second term in $\Delta$ account to Gaussian approximation?

Lastly, it also seems like if we consider the complete expansion in $\Delta$, one can reproduce the Kramers-Moyal expansion. Is this essentially true?

$\endgroup$

1 Answer 1

0
$\begingroup$
  1. Does the above equation correspond to a Markovian approximation? The answer is yes. In reality, the motion of a diffusing particle is correlated over short times. This is known as velocity persistence and arises from the inertia of the particle that prevents the velocity from changing abruptly. Over time steps that represent a large enough umber of collisions the velocity becomes uncorrelated to its earlier state. In this limit the process may be described as Markov.

  2. Now, can we say that the truncation of the series to only up to the second term in Δ account to Gaussian approximation? The truncation of course is part of the Gaussian approximation but it is not the only necessary condition. To obtain the Gaussian solution we must also assume that $\Delta^2/\tau$ is constant. This is a good approximation for a dilute solute. In concentrated systems the diffusion coefficient may depend on concentration (in this case on $P(x)$ due to molecular interactions between solute molecules.

  3. The Kramers–Moyal expansion considers the general case where the walker can jump any distance and expands the master equation in terms of the moments of the jump distribution. If the jump length $\Delta$ is a Dirac delta, its moments are $\Delta^n$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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