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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    $\begingroup$ I don't know the answer, but good question ;-) $\endgroup$
    – David Z
    Jul 31, 2011 at 19:39

1 Answer 1


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.

  • $\begingroup$ Thanks! I indeed thought there should be something like the second viscosity. Is the second viscosity usually larger or smaller than the first viscosity? $\endgroup$
    – felix
    Aug 2, 2011 at 20:27
  • $\begingroup$ @felix: I don't know. For water, $\zeta$ ~ 3 cP at 15 $^\circ$C, according to a Wikipedia entry (en.wikipedia.org/wiki/Volume_viscosity). At this temperature, water has $\eta$ ~ 1 cP. $\endgroup$
    – Carlos
    Aug 3, 2011 at 0:54
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    $\begingroup$ 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. $\endgroup$
    – Yrogirg
    Aug 15, 2012 at 13:09

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