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The friction term in Navier-Stokes equation assumes that the viscosity coefficients are the same for the longitudinal and transverse directions. This doesn't seem intuitive, because the former is essentially a bulk modulus while the latter doesn't involve any compression of the fluid. How is the assumption justified?

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I don't know the answer, but good question ;-) – David Z Jul 31 '11 at 19:39
up vote 10 down vote accepted

Actually, there are two different viscosity coefficients. You can see this from the stress tensor $$ \sigma_{ij} = -p_0 \delta_{ij} + \eta \left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} - \frac{2}{3} \delta_{ij} \frac{\partial v_k}{\partial x_k} \right) + \zeta \delta_{ij} \frac{\partial v_k}{\partial x_k} $$ which has the two coefficients of viscosity $\eta$ and $\zeta$ (see Landau & Lifshitz, Fluid Mechanics, for example). The pressure $p_0$ is given by the thermodynamic equation of state, but this is not the whole pressure $p$. The latter is given by the mean normal stress $$ p = - \frac{1}{3} \sigma_{ii} = p_0 - \zeta \frac{\partial v_k}{\partial x_k} $$ so that the stress tensor is $$ \sigma_{ij} = -p \delta_{ij} + \eta \left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} - \frac{2}{3} \delta_{ij} \frac{\partial v_k}{\partial x_k} \right) .$$

That's why sometimes you don't see the coefficient $\zeta$ (often called second viscosity) in the Navier-Stokes equation. It is hidden in the pressure, but it's there.

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Thanks! I indeed thought there should be something like the second viscosity. Is the second viscosity usually larger or smaller than the first viscosity? – felix Aug 2 '11 at 20:27
@felix: I don't know. For water, $\zeta$ ~ 3 cP at 15 $^\circ$C, according to a Wikipedia entry ( At this temperature, water has $\eta$ ~ 1 cP. – Carlos Aug 3 '11 at 0:54
felix, for gases it can be assumed to be zero. In assumptions of usual kinetic theory of gases (pair collisions, etc) it can be calculated to be zero. – Yrogirg Aug 15 '12 at 13:09

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