Brownian Motion and the Correlation Function

Deriving the velocity correlation function ⟨v(0)v(t)⟩ from the Langevin equation — basically asking whether your current state still remembers where it started, gold-spoon style.

Ex 33.2

Derive the expression for the velocity correlation function < v(0)v(t) > in Brownian motion.

Equation from the original notes

If we

Equation from the original notes

read it like this, and take the ensemble average of it,

we can probably understand it as an index of “is there a relation between the initial condition and after some time has passed?” at about this level.

Hmm… so it’d be something like the “gold-spoon / dirt-spoon theory.”

Whether my current state has any connection to my initial condition (my parents’ wealth) or not….. putting it this way makes it more relatable, right???

So then, for Brownian motion,

let’s start from the Langevin equation

Equation from the original notes

Okay okay, here we go.

Equation from the original notes

I’ll express v_dot differently.

Equation from the original notes

Therefore

Equation from the original notes

As mentioned earlier, since F(t) is a haphazard force, a Random force,

the ensemble-averaged red term must come out to 0.

If it doesn’t come out to 0, then the meaning of “random force” is lost,

Equation from the original notes

And let’s send tau to the limit 0!!!!!

Equation from the original notes

Equation from the original notes

Yup!!! All we have to do is solve the diff-eq!!!~~

Equation from the original notes

When t=0,

Equation from the original notes

since that’s what it’ll be,

A is exactly that value.

Equation from the original notes

Looking at the equation above,

Equation from the original notes

this value converges to 0 as t→∞.

That is, we reach the conclusion that as time passes, the correlation with the initial condition gets fainter,

and we’ve derived the conclusion that this is the essence of Brownian Motion. hehehehehe hehe lololololol

Comments

Discussion happens via GitHub Discussions. You'll need a GitHub account to comment.