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In a paper I have seen, somebody wrote down the Navier-Stokes as \begin{align} \rho\frac{\text{D}u}{\text{D}t} &= -\nabla p + \nabla \cdot \sigma + F\,,\\ \partial_t \rho + \nabla \cdot \left(\rho u\right) &= 0 \end{align} with the stress tensor $$ \sigma_{ij} = \mu \partial_{(i} u_{j)}\,. $$ ($\rho$: mass density, $u$: velocity field, $p$: pressure, $\sigma$: viscous stress tensor, $F$: external body force, $\mu$: viscosity; brackets in the indices denote symmetrization, of course)

It looks like the momentum balance equation for an incompressible fluid together with the mass balance for a compressible fluid. Is this just an imprecision (which I think is the case) or some common approximation that I happen to not know about? As far as I am aware, this would not be the Boussinesq approximation.

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The more general formula, $$\frac{\partial\rho}{\partial t}+\nabla\cdot\left(\rho\mathbf u\right)=0,\tag{1}$$ can be written in terms of the material derivative, $$\frac{D\rho}{Dt}+\rho\nabla\cdot\mathbf u=0,\tag{2}$$ simply by expanding the spatial derivative.

With an incompressible flow, the density is constant in time along the line of the flow, $$\frac{D\rho}{Dt}=0,$$ which requires $\nabla\cdot\mathbf u=0$ in Eq. (2). Hence, Eq. (1) can be used for incompressible flows, so long as the condition $\nabla\cdot\mathbf u=0$ is satisfied. So unless the authors neglected to mention incompressibility, I wouldn't call it "imprecise", though I guess your mileage may vary.

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  • $\begingroup$ Thanks for your answer. The divergence-free condition for the velocity field is not mentioned in the text. It goes like "the general Navier-Stokes equation is ...(momentum equation as above, explanation of symbols, stress tensor)... The continuity equation is ...(eqn as above)... Hereafter, only incompressible flows are considered." Which to me seems imprecise since the authors were either not aware that they were giving a non-general stress tensor (incompressibility already assumed) or they were adding unnecessary terms to the continuity equation. $\endgroup$
    – kricheli
    Commented Aug 18, 2022 at 18:47
  • $\begingroup$ As incompressibility necessitates divergence free, saying one implicitly demands the other. But like I said, I don't think that is "imprecise" but you are free to claim as much if you want more rigor (though (a) making claims like "they were unaware" is probably demeaning and ought to be avoided and (b) you're likely to find this level of rigor more frequently than not). $\endgroup$
    – Kyle Kanos
    Commented Aug 18, 2022 at 21:10
  • $\begingroup$ My thesaurus offers "precision" as a synonym for "rigour". So if there is a lack of rigour... ;) Anyway, we're agreed that rigor could be improved. Thank you. $\endgroup$
    – kricheli
    Commented Aug 19, 2022 at 6:25

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