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Lattice Boltzmann "collision" step

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Old   January 24, 2006, 18:22
Default Lattice Boltzmann "collision" step
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Joseph Bergevin
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I've been working with BGK Lattice Boltzmann models, and have come to wonder what the utility of the collision step might be. From what I gather, its incorporation only serves to slow the time required for a given configuration to reach its ultimate "equilibrium" state. This could be a factor in highly dynamic situations, but in a constant flow with a static set of boundaries there doesn't seem to be a need for the collide step. Given a relaxation time of 1, the step reduces to a simple assignment of the PDF. Am I missing something? I'm not very advanced in my maths, but fortunately LBMs are quite intuitive (simple conservations and redistributions).
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Old   January 25, 2006, 15:56
Default Re: Lattice Boltzmann "collision" step
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w
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first, if you don't have a collision operator/step at all, you do not have dynamics in your system, i.e., your initial velocity profile will remain the same for ever. Second, the LBGK eq. is

f_new = (1/t)*feq + (1-1/t)*f_old

with t the relaxation time, and feq the equilibrium distributions. If yoy have t=1, then:

f_new = feq

and since feq=feq(rho,velocity) (function of density and velocity at each point), again, after you put an initial velocity field, you do not have any evolution any more.

Besides, remember that the value of t determines your fluid viscosity, so you are not completly free to choose its value if you need a certain Reynolds number in your flow.

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