# Viscous term in the Navier stoke equation

How exactly do you derive it without using tensors? You use Newton's frictional law on a 3d block of fluid, instead of just on fluid surfaces, but I can't 100% visualize how. So is there any simple derivation, or at least explanation?

$$\mu \nabla ^2 \vec{V}$$

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## Answers and Replies

arildno
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There are TWO critical steps here!
1. The divergence of a general STRESS tensor (which must be symmetrical, by arguments given at first by Cauchy and others) appears in the PRIMITIVE equation of motion for the continuum.
2. We may "always" extract an isotropic pressure element out of this, by algebraic shuffling, which yields the the gradient of the pressure as appearing in our eqs of motion. Then, for the rest, we must MODEL a relation between our stresses, and our strains or strain rates. Simple, NEWTONIAN fluids are those saying that there exists a constant of proportionality between the strain rate tensor and the stress tensor, a more complicated model would be that of an anisotropic Newtonian fluid, in which the single constant of proportionality is replaced with a suitable constant tensor of "proportionality. Other models will be for "non-Newtonian" fluids, for example that the viscosity proportionality factor depends on temperature and pressure, thereby linking the thermodynamics of the fluid with its overall behaviour of motion. Or, we might have fluids where the viscosity is itself dependent on the strain rates, introducing an additional non-linearity. Or, we might have fluids in which not only strain rates, but also strains themselves generates stress; this makes typically, the fluid exhibit elastic properties.
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The classical Navier-Stokes equations is gained as the simplest relationship, a single constant of proportionality between the stress tensor and the strain rate tensor is sufficient to describe the motion in the fluid.

Andy Resnick
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The stress tensor σ can, for Newtonian fluids, be decomposed into an isotropic (pressure) and velocity gradient tensor: σ= pI +μ(∇u+(∇u)$^{T}$). Using the identities div(pI) = grad p and div(grad u) = ∇$^{2}$u results int he usual N-S momentum equation.

arildno
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The stress tensor σ can, for Newtonian fluids, be decomposed into an isotropic (pressure) and velocity gradient tensor: σ= pI +μ(∇u+(∇u)$^{T}$). Using the identities div(pI) = grad p and div(grad u) = ∇$^{2}$u results int he usual N-S momentum equation.
You are correct in speaking that we also actually replace the strain rate tensor with the velocity gradient tensor; they differ, if I recall, by some extremely ugly non-linearities we ignore. I've seen the different expressions for the truly ugly deformation tensor relative to the displacement gradient tensor; the latter of which is typically the one kept for modelling an elastic medium. I'm sure the actual strain rate tensor is at least as ugly as the deformation tensor, but if I recall, in contrast to the deformation tensor, the full strain rate tensor is never used. Thus, I happened to identify the stain rate tensor with the velocity gradient tensor, which technically, was a mistake. I had forgotten about the conceptual difference. The velocity gradient tensor is the linearized strain rate tensor, and it is THAT simplified tensor we choose to keep in order to derive the N-S equations.

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Andy Resnick
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Replacing the stress tensor with a strain tensor requires a constitutive relation, Newtonian/inviscid/ideal fluids being the most simple. One could also use, for example, a Maxwell constitutive relation for a viscoelastic material, in which case the NS equation would appear very different. Another major class of 'misbehaving' materials are those with memory- the constitutive relation is a Volterra integral.

As you mention, the Newtonian constitutive relation is linear- nonlinear deformations are also very complex to deal with, but as I recall, the Cayley-Hamilton theroem for tensor invariants sets some bounds on what is allowed.