Timeline for What is the correct way to calculate deviatoric stress tensor in lattice Boltzmann method?

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Dec 30, 2019 at 19:10	comment	added	Mithridates the Great		Dimensionless dynamic and kinematic viscosities are equal: $$\hat{\mu} = \frac{\mu \Delta t}{\rho_{f} \Delta x^{2}} = \frac{\nu \Delta t} {\Delta x^{2}} = \hat{\nu}$$ where kinematic viscosity is defined as $\nu = \frac{\mu}{\rho_{f}}$.
Dec 30, 2019 at 16:13	comment	added	nluigi		@AloneProgrammer - I dont understand your point that dimensionless dynamic and kinematic viscosities are equal. I think you have some confusion about dimensions in LB; you are mixing dimensionless variables with variables in lattice units. You should be consistent in any unit system. Most likely i think RDj is correct and you are simply confusing dynamic and kinematic viscosity.
Dec 29, 2019 at 13:06	history	edited	Anton Menshov♦	CC BY-SA 4.0	added 1 character in body
Dec 28, 2019 at 18:33	comment	added	Mithridates the Great		Sorry, but no, it's not an answer to my question. The dimensionless dynamics viscosity is defined as: $$\hat{\mu} = \frac{\mu\Delta t}{\rho_{f} \Delta x^{2}}$$ Where $\rho_{f}$ is constant density of fluid. You see that dimensionless dynamic and kinematic viscosities are indeed equal: $$\hat{\mu} = \hat{\nu}$$ So it doesn't matter here. The main idea here is that how close instantaneous and constant fluid densities are or in another how small $Mach$ is.
Dec 28, 2019 at 13:45	review	First posts
Dec 29, 2019 at 13:07
Dec 28, 2019 at 13:42	history	answered	RDj	CC BY-SA 4.0