$$\frac{{\partial (\rho {v}_x})}{\partial t}+\nabla\cdot{(\rho {v}_x\mathbf{v}) = -(\nabla P)_x + \rho g_x+\nabla\cdot\mathbf{\tau_x}}$$

This is the $x$ component of the Cauchy momentum equation, where $\tau$ is the stress tensor and $\mathbf{\tau_x}$ is the vector form (i.e. transpose) of the first row of $\tau$. This is derived using standard assumptions (like the functions involved are continuously differentiable, pressure is isotropic, the momentum transfer occurs locally except for gravity). It can be derived by applying Newtons second law to the mass in an Eulerian control volume at an instant in time, then finding the vector components of all terms, then applying Gauss's theorem. More details on the derivation here. https://web.mit.edu/16.unified/www/FALL/fluids/Lectures/f07.pdf

In the case of a Newtonian fluid, we can apply the constitutive relation $\mathbf{\tau_x} = \mu \nabla v_x$. Now the RHS becomes $$ -(\nabla P)_x + \rho g_x+\mu \nabla^2 v_x$$ which is equal to the RHS of the standard Navier-Stokes equation. What about the LHS? It doesn't match up with the LHS of the N-S equation unless we cancel out all density variation terms. In that case, we are assuming the flow is incompressible, right? But then why do we have a compressible Navier Stokes equation. Is it because the effect of density variation in the LHS of the momentum equation is small and ignorable, but the effect is not ignorable in the mass continuity equation?


The continuity equation that expresses conservation of mass is

$$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbb{v}) = 0, $$

and this holds both for incompressible and compressible flow. (Of course, it reduces to the solenoidal condition $\nabla \cdot \mathbb{v} = 0$ when the flow is incompressible and $\rho$ is constant).

Applying the product rule for differentiation to the LHS, we get

$$\frac{\partial (\rho v_x)}{\partial t} + \nabla\cdot (\rho v_x\mathbf{v}) = v_x \frac{\partial \rho}{\partial t} + \rho \frac{\partial v_x}{\partial t}+ v_x \nabla \cdot (\rho \mathbb{v}) + \rho \mathbb{v} \cdot \nabla v_x \\ = v_x \underbrace{\left(\frac{\partial \rho}{\partial t}+ \nabla \cdot (\rho \mathbb{v})\right)}_{= 0}+ \rho \frac{\partial v_x}{\partial t}+ \rho \mathbb{v} \cdot \nabla v_x ,$$

Hence, even for compressible flow the LHS of the $v_x$-momentum eqation is $\displaystyle\rho \frac{\partial v_x}{\partial t}+ \rho \mathbb{v} \cdot \nabla v_x$.

This applies to the other components as well, and collecting together in vector form we get the usual LHS of the Navier-Stokes equations,

$$\rho \left(\frac{\partial \mathbb{v}}{\partial t}+\mathbb{v} \cdot \nabla \mathbb{v}\right)$$

  • $\begingroup$ facepalm of course. I'm shocked by how long that puzzled me for $\endgroup$
    – gdcpb
    Jun 19 '20 at 23:41
  • $\begingroup$ Just so you know this is the NS equations in what is called convective form as opposed to conservative form as you originally wrote. Now be warned that the stress term does not reduce to $\mu\nabla^2 \mathbb{v}$ in compressible flow -- there are some other terms that involve $\nabla \cdot \mathbb{v}$. See here $\endgroup$
    – RRL
    Jun 19 '20 at 23:44

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