I am looking for information on the coefficient $\lambda$ in the following formulation of the barotropic compressible Navier-Stokes system:

\begin{align} & \partial_t \rho + \text{div}(\rho u) = 0,\\ & \rho \frac{D}{Dt} u = - \nabla p + \mu \Delta u + \lambda \nabla (\text{div} u), \end{align}

where $p := P(\rho)$ for some $P$ sufficiently smooth, $\rho, u$ are the density, velocity respectively, and the coefficients $\mu, \lambda$ satisy:

\begin{align} \mu \geq 0, \ 2\mu + 3\lambda \geq 0. \end{align}

This particular formulation is taken from P.G. Lemarié-Rieusset's 'The Navier-Stokes Problem in the 21st Century,' and I understand that it is somewhat simplified compared to more physics-oriented formulations like, say, Landau and Lifschitz's 'Fluid Mechanics'.

I am particularly interested in whether or not the mathematically useful assumption $\lambda = 0$ (without necessarily also forcing $\mu =0$) has any physical basis behind it. I have so far only found mention of Stokes's hypothesis, which wouldn't eliminate the $\nabla (\text{div} u )$ term entirely. Again, using Lemarié-Rieusset's notation, the Stokes hypothesis entails

\begin{equation} 2\mu + 3\lambda = 0, \end{equation}

which is not what I'm looking for.

Any keywords or names for the $\lambda = 0, \ \mu \neq 0$ assumption that I'm looking for, or sources with info on this matter would be greatly appreciated. In short, I'd like to know if there's any real-life scenario where this assumption makes sense, or if it's a mere "toy model" for convenience in maths.


2 Answers 2


See Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 1. They indicate that, to linear terms in the velocity components, the portion of the stress tensor over and above the isotropic pressure is proportional to the "deviatoric" rate of deformation tensor, plus a "dilatational viscosity" term. For monoatomic gases, kinetic theory of gases indicates that the dilatational viscosity is zero, and Bird et al indicate that for most other gases it is negligible. Under these circumstances $\lambda=-\frac{2}{3}\mu$.

  • $\begingroup$ I have read this book, and I understand that this is just saying that Stokes's hypothesis holds for monatomic gases. I also see that, under Stokes's hypothesis, my desired setting $\lambda =0$ would also set $\mu=0$, and we would just have Euler's equation. But is that to say that there is never a situation where only $\lambda =0$? When we are dealing with polyatomic gases, for example? $\endgroup$
    – gbnhgbnhg
    Commented Feb 2, 2022 at 6:25

Indeed, physicists prefer to use $\eta=\mu$ (shear viscosity) and $\zeta = \lambda+(2/3)\mu$ (bulk viscosity).

The second law implies that the rate of entropy production is positive, and therefore $\eta,\zeta\geq0$.

It is generally believed (but has not been rigorously proven) that the shear viscosity of a real fluid (a fluid made of atoms or elementary particles) is strictly positive, $\eta>0$. However, the laws of physics allow for scale invariant fluids with $\zeta=0$.

A mono-atomic gas is not, strictly speaking, scale invariant, and it does have a non-zero bulk viscosity. However, it can be modeled to a good approximation as point-like particles that only suffer elastic two-body collisions. This is a mathematical idealization that corresponds to zero bulk viscosity. The physical bulk viscosity of a mono-atomic gas is indeed small.

The bulk viscosity of di-atomic gases is comparable to their shear viscosity.

The case $\lambda=0$ corresponds to $\zeta=(2/3)\eta$. This is not forbidden, but there is no symmetry that would lead to this result, and it would therefore require fine-tuning. For example, $\eta$ and $\zeta$ typically differ in their dependence on temperature, so there may be a special temperature $T_0$ so that $\zeta(T_0)=(2/3)\eta(T_0)$ and $\lambda(T_0)=0$.

There is no simple mathematical idealization that leads to $\zeta=(2/3)\eta$.

Finally, whereas $\zeta=0$ has specific physical consequences for the behavior of the fluid, there is nothing special that happens for $\lambda=0$. In particular, if $\zeta=0$ then a monopole oscillation is complete undamped, and the isotropic expansion of a gas produces no entropy.

Postscript: Physicists like to write the Navier-Stokes equation in "conservative" form $$ \partial_t \pi_i + \nabla_j \Pi_{ij} = 0 $$ where $\pi_i=\rho u_i$ is the density of momentum ($\rho$ is the mass density, and $u_i$ is the fluid velocity), and $\Pi_{ij}$ is a symmetric tensor, called the stress tensor (engineers add the word Cauchy). This equation expresses conservation of momentum.

The stress tensor can be expanded in derivatives of the fluid variables. At zero'th order we obtain the ideal stress tensor $$ \Pi_{ij}^{(0)} = P \delta_{ij} + \rho u_i u_j. $$ The corresponding conservation law is the Euler equation. At first order we add $$ \Pi_{ij}^{(1)} = -\eta\sigma_{ij} - \zeta \delta_{ij} (\nabla\cdot u) $$ where $\sigma_{ij}$ is the symmetric traceless tensor $$ \sigma_{ij} = \nabla_i u_j + \nabla_j u_i -\frac{2}{3}\delta_{ij} (\nabla \cdot u). $$ This corresponds to the Navier-Stokes equation.

Note that what we have done is to decompose $\Pi_{ij}^{(1)}$ into irreducible tensors that transform under the rotation group, a scalar (the bulk term), and a traceless rank two tensor (the shear term). This explains why the two terms contribute separately to the rate of entropy production, and the 2nd law constrains $\eta,\zeta$, not $\lambda,\mu$.

  • $\begingroup$ This is a good answer, and conforms with everything I've read so far. One point I'd like to ask about though is the idea that nothing special happens for $\lambda =0$. Under this assumption, the $\nabla \text{div}$ term disappears, and we're left with quite a different equation. I'm not familiar enough with the physics to say what exactly that term was doing but surely the fluid must be behaving in some particular way when following this equation? $\endgroup$
    – gbnhgbnhg
    Commented Feb 4, 2022 at 8:33
  • $\begingroup$ Another question, actually, does the coefficient $\lambda$ on its own not represent any physical property? Is it only $\zeta = \lambda + 2\mu /3$ that corresponds easily with a property? $\lambda$ doesn't get its own name in Lemarié-Rieusset or Landau & Lifschitz as far as I can tell, and the wikipedia article for the Navier-Stokes equation (erroneously?) calls $\lambda$ the $\textit{bulk viscosity}$... $\endgroup$
    – gbnhgbnhg
    Commented Feb 4, 2022 at 8:36
  • 1
    $\begingroup$ @DavidHughes I added a postscript. Yes, the wikipedia article calls $\zeta$ the "second vsicosity". A physicist will know what you mean by 2nd viscosity (indeed, Landau uses the term), but consider the words "bulk viscosity" and "2nd viscosity" synonymous. $\endgroup$
    – Thomas
    Commented Feb 4, 2022 at 14:43
  • 1
    $\begingroup$ Both fluid dynamics (2nd law constrains $\zeta$, not $\lambda$) and kinetic theory ($\zeta$ vanishes for elastic two-body collisions, not $\lambda$) indicate that $\zeta$ is the relevant physical object. Of course, in the end $(\eta,\zeta)$ are just linear combinations of $(\eta,\zeta)$, so clearly $\lambda$ affects fluid flow. There just isn't an obvious limit in which $\lambda$ vanishes, or any symmetry of the laws of fluid dynamics that emerges in that case. $\endgroup$
    – Thomas
    Commented Feb 4, 2022 at 14:45
  • 1
    $\begingroup$ .. linear combinations of $(\mu,\lambda)$ … $\endgroup$
    – Thomas
    Commented Feb 4, 2022 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.