I have been searching for this answer for awhile. I've read numerous texts and even watched some lectures online, but often times this is never explained and just given. The viscous stress term in the Navier-Stokes equations looks like

\begin{equation} \nabla \cdot \tau = \nabla \cdot \mu \left(\nabla\vec{u} + (\nabla\vec{u})^T\right) \end{equation}

Now the term $\nabla \cdot \mu \nabla\vec{u}$ is easy enough to understand as it is just velocity diffusion, but I have a hard time coming up with a physical interpretation of the term $\nabla \cdot \mu (\nabla\vec{u})^T$. After I expanded this term I ended up with

\begin{equation} \nabla \cdot \mu (\nabla\vec{u})^T = \begin{pmatrix} \frac{\partial}{\partial x} \nabla \cdot \vec{u} \\ \frac{\partial}{\partial y} \nabla \cdot \vec{u} \\ \frac{\partial}{\partial z} \nabla \cdot \vec{u} \end{pmatrix} \end{equation}

which seems to imply that this effect is not present in a divergence-free velocity field, but I still can't come up with or find any physical intuition about what this term actually means. Does anyone understand what this term physically represents?

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    $\begingroup$ Addition: you're correct that the term is absent in incompressible flow. It looks like it takes into account diffusion of momentum due to gradients in density. Two adjacent parcels of fluid can have the same velocity but different momentum, there's no shear stress between them but momentum will diffuse. $\endgroup$
    – Dan
    Apr 13, 2015 at 3:31
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    $\begingroup$ This question is on-topic for Engineering. I have removed several comments suggesting other sites for this question. In part because of asking for an applied understanding of the equation but also because this is a part of continuum mechanics. Please remember that it's okay to be a bit jealous of your site $\endgroup$
    – user16
    Apr 13, 2015 at 15:35
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    $\begingroup$ Related meta discussion: meta.engineering.stackexchange.com/questions/266/… $\endgroup$
    – user16
    Apr 14, 2015 at 21:04
  • $\begingroup$ The point about a momentum gradient being present because of a non-zero density gradient was a good one. Thank you everyone for your responses! $\endgroup$ Apr 17, 2015 at 15:08

2 Answers 2


You should not separate those two terms in search for physical interpretation. The term $\nabla\vec{u}+\left(\nabla\vec{u}\right)^T$ is the strain rate tensor $\dot{\gamma}$. The momentum flux (or stress) due to the fact we have a flowing fluid is accounted for by the whole term $\mu\left(\nabla\vec{u}+\left(\nabla\vec{u}\right)^T\right)$. In the NS equation both terms can be thought of as force densities (force per unit volume). You are right, that the second term is zero for incompressible flows (see here).

UPDATE: The complete derivation of the strain rate tensor is complex and it might be out of scope here. If you are interested I have found that a good resource is Introduction to Fluid Mechanics by Whitaker. Briefly, lets accept that the tensor $\nabla \vec{u}$ represents the strain rate and solid like rotational motion. Any tensor can be decomposed in the following way: $$\nabla \vec{u} = \frac{1}{2}\left(\nabla\vec{u}+\left(\nabla\vec{u}\right)^T\right)+\frac{1}{2}\left(\nabla\vec{u}-\left(\nabla\vec{u}\right)^T\right)$$ The first term is typically called the strain rate tensor, is symmetric, and it can be shown that it includes no rigid rotational motion. The second term is typically called the vorticity tensor, its is skew symmetric, and it can be shown that it does not contribute to the rate of strain and that it represents rigid like rotational motion.

  • $\begingroup$ This is what I found looking into it, but I was trying to find something like a derivation of the strain rate tensor before committing to an answer, to understand why it includes the regular and transpose matrix. $\endgroup$ Apr 16, 2015 at 2:51
  • $\begingroup$ Thank you, I went through the strain-rate tensor derivation from geometry as you suggested, and that helped me a lot. $\endgroup$ Apr 17, 2015 at 15:07

I agree with @sturgman one should not look at individual parts but try to understand it in ints context.

Looking at the very basic version of the Navier-Stokes-Equation (using Einstein-Notation):

$$\rho\frac{\mathrm{D}u_i}{\mathrm{D}t} = \rho k_i + \frac{\partial}{\partial x_i} \left( -p + \lambda^*\frac{\partial u_k}{\partial x_k}\right) + \underbrace{\frac{\partial}{\partial x_j} \left( \eta \left[ \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right] \right)}_{\nabla \,\cdot\, (\eta \, \left[ (\nabla\vec{u})+(\nabla\vec{u})^\mathrm{T}\right])} $$

The underbraced part in its original can be rewritten.

$$ \frac{\partial}{\partial x_j} \left( \eta \left[ \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right] \right) = \eta \left( \frac{\partial^2u_i}{\partial x_j \partial x_j} + \frac{\partial}{\partial x_i}\left[\frac{\partial u_k}{\partial{x_k}} \right]\right)$$

Which leads to:

$$\rho\frac{\mathrm{D}u_i}{\mathrm{D}t} = \underbrace{\rho k_i}_{\text{I}} - \underbrace{\frac{\partial p}{\partial x_i}}_{\text{II}} +\underbrace{(\lambda^* + \eta)\frac{\partial}{\partial x_i}\left[\frac{\partial u_k}{\partial x_k}\right]}_{\text{III}} + \underbrace{\eta \left[ \frac{\partial^2u_i}{\partial x_j \partial x_j}\right]}_{\text{IV}}$$

In symbolic notation this should look like:

$$\rho\frac{\mathrm{D}\vec{u}}{\mathrm{D}t} = \rho\vec{k} - \nabla p + (\lambda^* + \eta)\nabla(\nabla\cdot\vec{u}) + \eta\nabla\cdot\nabla\vec{u}$$

Part $\text{III}$ is not always shown like this depending on the way the Newtonian stress tensor was introduced. Since $\lambda^*$ is a fluid property which is very hard to measure but varies only little, Stokes Hypothesis sets it to $-2/3 \eta$ (Which is technically only true for monoatomic gases).

Part $\text{III}$ describes a feature of a fluid where the atomic structure of the fluid-molecule can absorb energy, it is sometimes referred to as pressure-viscosity. Whereas Part $\text{IV}$ describes the resistance of the flow when sheared, part $\text{III}$ describes the resistance of a fluid-volume when it is "isobarically" expanded or compressed.

  • $\begingroup$ I am sorry :-( It wasn't my intent. $\endgroup$
    – peterh
    Sep 24, 2015 at 20:50

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