Brownian Motion - "no inertia"

[SchroedingersLion]

Greetings,

I currently work my way through Langevin Dynamics which, in a certain limit, becomes Brownian Motion.
I refer to this brief article on Wikipedia: https://en.wikipedia.org/wiki/Brownian_dynamics

I understand the general LD equation given there. In order to obtain Brownian Dynamics, one sets the net acceleration (to be precise, its average) to 0 and reorganizes the equations. I don't really see how this is an assumption of "no inertia". If inertia is the resistance of mass against acceleration, than zero acceleration (independently of the applied force) should correspond to infinite inertia, should it not?

 

[Dale]

0 average acceleration is not the same as 0 acceleration.

 

[SchroedingersLion]

I know, but how does this answer my question?

 

[Dale]

0 acceleration would indeed imply infinite mass, but it is not 0 acceleration. It is 0 average acceleration, which in no way implies infinite mass.

 

[SchroedingersLion]

Fair enough, so infinite inertia would be false. However, I still don't see why it is called "no inertia". On the same note, I don't see why it is called the "high friction" limit. Why is $\gamma \rightarrow \infty$ the same as setting the average acceleration to zero?

 

[vanhees71]

It's just neglecting the term $m \ddot{x}$ against the other terms in the equation. What I don't understand is, why they call this "Brownian motion". For me Brownian motion is the motion described by the full Langevin equation (usually without external forces).

 

[radium]

This is usually called the overdamped limit. In this case γ is large enough so that the velocity equilibrates very rapidly so you can set dv/dt=0 to find the behavior at long times. The joint pdf p(x,v) then factorizes into a stationary Maxwellian distribution for p(v) times a time dependent distribution p(x(t)) approaching a diffusion process.