Physical Meaning of Divergence of Convective Velocity Term

When taking the divergence of the convective velocity term, I get the following: \begin{align} \nabla\cdot\left[\mathbf u\cdot\nabla\mathbf u\right]&=\frac{\partial}{\partial x_i}\left[u_j\frac{\partial u_i}{\partial x_j}\right]\\ &=\frac{\partial u_j}{\partial x_i}\frac{\partial u_i}{\partial x_i}+u_j\frac{\partial^2u_i}{\partial x_j\partial x_i} \\ &=\mathbf u\cdot\nabla q+\left(\nabla\mathbf u\right)\cdot\left(\nabla\mathbf u\right)^T \end{align} where $q=\nabla\cdot\mathbf u$.

I know the first term on the right hand side represents the convective term for the dilatation component of the velocity field (from Helmholtz decomposition), but I can't quite get the physical meaning of the second term. The gradient of velocity is a 2nd order tensor, but what is the physical meaning of the product of a second order tensor with its transpose? Is there a way to manipulate it to get a better physical meaning out of it?

• Generally I find it's easier to get meaning from the summation notation than from the vector notation. Also note -- you flipped the order of the terms from the second-to-last part of the expression to the last part of the expression. That said, I'm playing with it trying to figure out what the term means... – tpg2114 Sep 30 '14 at 23:34
• Under what situations are you interested in taking the divergence of the convective velocity? Is there a use-case here or a reference that you're trying to follow along? I don't know that I've come across this before but it looks kind of familiar all the same. – tpg2114 Sep 30 '14 at 23:37
• I'm looking at the Helmholtz Decomposition of the Navier Stokes equations. Basically you can decompose the equation into a solenoidal (vorticity) and irrotational (dilatation) part. The above is a piece of the dilatation form. In regards to the flipping the order - does it really matter since it's a dot product? – Kimusubi Oct 1 '14 at 2:29
• Unfortunately, my books on compressible turbulence that talk about the splitting of the fields is in the lab so I can't look it up right now, but I thought those looked familiar. And what I meant by the terms flipping -- the first term after the second $=$ is the 2nd term after the last $=$. In other words, $\frac{\partial u_j}{\partial x_i}\frac{\partial u_i}{\partial x_j} = (\nabla \vec{u})\cdot(\nabla \vec{u})^T$. I just wanted to make sure everybody was clear that the order of the terms changed from the second-to-last to the last step. – tpg2114 Oct 1 '14 at 2:53
• Shouldn't the last term be the double-dot product? The first term, $\mathbf{u} \cdot \nabla$q is a scalar (rank zero tensor) but currently the second term would be a rank one tensor. So I think it should be $\nabla \mathbf{u} \ : \ (\nabla \mathbf{u})^{T}$, right? – honeste_vivere Oct 31 '14 at 13:54

The term in equation is:

$$\frac{\partial u_i}{\partial x_j}\frac{\partial u_j}{\partial x_i}$$

So let's take a step back and think about what kinds of terms can appear in conservation equations. There can be a production term, a transport term, and a dissipation term. The transport term is the $\vec{u}\cdot\nabla q$ term that you noted. When you look at the full coupled set of equations (vorticity and dilatation conservation equations), there are some production and dissipation terms that transfer dilational velocity into vorticity and vice-versa.

Now, I'm unfamiliar with the decomposition here specifically. However, looking at some other equations which I am familiar with (turbulent kinetic energy), I will go out on a limb and say that that term is a dissipation term. In all the conservation laws I have seen, terms that look like the term in question are dissipation terms -- this goes to answer your question about how to think about terms like this in general.

This hypothesis seems to be backed up by a few papers I've found and scanned quickly, and this thesis in Eq 2.14d which lumps the term in question into a viscous dissipation term.

My vote -- it's a dissipation of dilatation.

• That seems interesting that a dissipation term would arise from the inertial terms without accounting for specific fluid properties like the viscosity (i.e. not accounting for any sort of molecular transport of momentum, etc. across the fluid surfaces). It sort of makes sense simply considering that the two tensors are relative gradients of the velocity w.r.t. each other, but I would like to look further into it. – Kimusubi Oct 1 '14 at 3:30
• @Kimusubi Yeah, I'm just as reluctant to call it a viscous dissipation but it does get lumped in to those terms. Since we're talking about an "artificial" thing here, it may just be that convection of dilatation results in a reduction of dilatation in addition to transport. In a sense this would mean no flows can stay irrotational forever without a production term, which I could kind of believe as plausible. It's certainly interesting to think about though. – tpg2114 Oct 1 '14 at 3:43
• That's interesting to think of it like that. But wouldn't that defy Helmholtz persistence theorem - in the absence of viscosity, irrotational flow must stay irrotational (ignoring for now the deviatoric stress tensor)? I'm specifically referring to where you said that an irrotational flow wouldn't remain irrotational forever. – Kimusubi Oct 1 '14 at 4:01
• @Kimusubi It, it may not. I wonder (I don't know for sure) what this dissipation term looks like in an irrotational flow. It may be 0 in the case. So maybe this term only is non-zero when there is a rotational component to the flow. Again, I don't know for sure and am merely speculating since I haven't worked with the decomposed fields before. – tpg2114 Oct 1 '14 at 4:15
• Could we describe divergence with respect a group of people running down an alleyway, and into a wider road. Do they "diverge" at the transition between the narrow alley and wider road? – iantresman Dec 14 '14 at 13:34

I will start off by apologizing for being unfamiliar with the specifics of this problem, and the notation commonly used. I will therefore use notation and terminology that I am used to; we can decompose the gradient of the velocity field as $$\partial_ju_i = \omega_{ij} + \sigma_{ij} + \frac{1}{3}\delta_{ij}\theta,$$ where $\omega_{ij} = \partial_{[j}u_{i]}$ is the anti-symmetric part (vorticity), $\sigma_{ij} = \partial_{(j}u_{i)} - \delta_{ij}\theta/3$ is the symmetric and trace-free part (shear), and $\theta = \partial_iu_i$ is the trace/divergence (expansion parameter, $q$ in your notation). Then the term $$\partial_iu_j\partial_ju_i = \sigma^2 + \frac{1}{3}\theta^2 - \omega^2,$$ where $\sigma^2 = \sigma_{ij}\sigma_{ij}$, and $\omega^2 = \omega_{ij}\omega_{ij}$.

Since the other term gives the change of the expansion parameter along the fluid flow, we can make the interpretation (by moving the squared terms to the other side of the equation) that non-zero shear and expansion serves to reduce expansion along the fluid flow, while non-zero vorticity serves to increase expansion along the fluid flow.

We may be able to find the meaning of this term by considering an easier problem, an incompressible fluid. Taking the divergence of the Navier-Stokes equation in this case yields:

$$\nabla^2p=-\rho\left(\nabla\mathbf u\right)\cdot\left(\nabla\mathbf u\right)^T$$

We can see how the term in question is directly related to the Laplacian of the pressure field. Since this term exists in an incompressible flow, we can say it has physical meaning beyond or even different than dilation.

If we think of a Stokes flow, where this term is considered negligible due to the dominance of viscous terms, the nullity of this term serves to highlight the fact that there is no source or sink of convective acceleration due to pressure.

In a case where this is non-zero like in a turbulent flow (high Reynolds number), first: this tells us about the non-locality property of the pressure field (which you could see if you think of integrating the equality shown above). Second: It tells us about how pressure acts as a source or sink of of a fluid not by dilating or contracting it but by the purely non-linear nature of turbulence. The reason behind that is the fact that the term in question comes in by taking the divergence of the convective term of acceleration in the NS equation. The divergence of the acceleration field at a point, if non-zero, indicated the presence of a source or sink of acceleration.

• Your ideas around this term are confused: The term has nothing to do, per se, with turbulence (it's just as important in laminar flows), and I don't understand your musings regarding a "source or sink of convective acceleration due to pressure". – Pirx Feb 14 '17 at 1:31
• @Pirx Thank you for pointing out my mistake about laminar flows, it actually should be stokes flows where Re numbers are so low that the non-linear term $\left(\mathbf u \cdot \nabla\right)\left(\mathbf u\right)$ cancel in front of the viscous term, in which case you get a linearly varying pressure field along the direction of motion (by solving the resulting Laplace equation. As to your second comment, my thinking is that the term in question comes in by taking the divergence of convective term of acceleration in NS equation, which if non zero, can be interpreted as either source or sink. – Assaad Mrad Feb 14 '17 at 4:07