# Bounce-back rule

Hi Dear All,
I have a question about bounce-back rule if some one like to answer.
We all know that LBM is second order accurate and bounce-back is first order accurate. My question is in case of hitting the particle to the wall before collision then it’s goes to it’s previous position after collision. What’s the reason behind this and what’s physically it’s shows to us.
This issue is disscus in some paper but i want to the actual physical meaning for more understanding if some friends explain this.
Thanks
Yuanchaomei

Bounce-back is second order accurate if the wall is midway between two lattice nodes.
Be also aware that there is no collision rule for the particles on the wall. In the fluid, the populations usually experience collision and the propagation. On the wall, there is first bounce-back and then propagation, i.e., collision is replaced by bounce-back on the wall nodes. The physical meaning of this is that the molecules are reflected at the walls. But we know from macroscopic hydrodynamics that there is a no-slip condition at the walls. To account for this, the molecules do not undergo specular reflection but a full bounce-back.
I hope this helps a bit.

Timm

A little bit more comments on the Bounce-Back nodes - the location of the boundary is viscosity dependent on the parameter \omega. For some cases, as Poiseuille flow, the location of the boundary can be calculated analytically, i.e. \Delta = 16/3*(1/tau-1/2)^2 - where Delta is the distance from the bounce back node.

Therefore, when we talk about the second order convergence we actually talk about the convergence of the error in location of the boundary, which we think is on the midway, but is actually somewhere in another place. Probably, the good way is to use other solid boundary conditions giving you precise locations instead of the bounce-back.

Hopefully, it’s clear,
Alex

alex, can you tell us where your formula for delta comes from? i thought that at tau=1 the wall is half-way between the nodes, whereas with your formula you get delta(tau=1) = 4/3.