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I'm trying to get an intuitive understanding of the Incompressible Navier Stokes equation (as for me thats the only way I can use it effectively and avoid the rote learning method):

$\rho(\frac{\partial \bf u}{\partial t} + \sum_i u_i\frac{\partial \bf u}{\partial x_i}) = - \nabla P + \nu \nabla^2 \bf u$

So far I get the idea that its an applied Newtons Second Law, $F = ma$.

With (volume specific) mass times acceleration on the LHS and the forces from the pressure gradient field, and fluid friction forces on the RHS. (I've omitted the other force term for simplicity).

The term that I'm struggling to fit into this picture is the '$\rho\sum_i u_i\frac{\partial \bf u}{\partial x_i}$' term.

So far my best attempt to explain it is that each directional momentum component is multiplied with the change (pd) of the velocity field relative to that direction. This gives an idea of the potential change in the direction of momentum with change in positional coordinate, which in turn describes the turbulence of the system. - Personally I feel this is a poor explanation that doesn't quite ring true and I'm don't feel like I have understood it properly. Could someone offer a better explanation of the physical intuition of this term in simple language?

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    $\begingroup$ This might help: en.wikipedia.org/wiki/… $\endgroup$ – Robin Ekman Mar 29 '18 at 9:26
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    $\begingroup$ You already have your answer by now but I'll just point out a big picture detail that you might already know. Newtonian mechanics is based on $\vec{F} = m\vec{a}$. Fluid mechanics also relies on principles of newtonian mechanics. In one class, everything is discrete. In the other, everything is continuous. And in one class, since everything is discrete, you take the Lagrangian approach (you follow discrete particles around). In the other class since everything is continuous, it's often easier to take the Eulerian approach. Between the 2 approaches, the way you interpret functions is completely $\endgroup$ – DWade64 Mar 29 '18 at 12:27
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The term combines with the $\partial \mathbf{u} / \partial t$ term to form a convective derivative, which has a simple physical interpretation.

Suppose you're standing by a highway and a red car going 80 mph passes by. Half a second later a blue car going 10 mph passes by. Your reaction wouldn't be "wow, that guy in the red car must have really slammed the brakes, since 10 is much less than 80!", because these are not the speeds of the same object. The convective derivative fixes this mistake.

Mathematically, if you want to write Newton's second law for a particle, the acceleration is simple $d \mathbf{u} / d t$ where $\mathbf{u}(t)$ is the particle's velocity at time $t$. For a fluid, $\mathbf{u}$ is now a function $\mathbf{u}(\mathbf{r}, t)$ which describes the local velocity of the fluid at position $\mathbf{r}$ at time $t$. But $\partial \mathbf{u} / \partial t$ doesn't describe the acceleration of any physical object. If you compare $\mathbf{u}(\mathbf{r}, t)$ and $\mathbf{u}(\mathbf{r}, t + \Delta t)$ you are comparing the velocities of two different pieces of the fluid, not computing the acceleration of one piece.

To fix this, let's instead consider the velocity of a single, fixed piece of fluid, $$\mathbf{v}(t) = \mathbf{u}(\mathbf{r}(t), t)$$ where $\mathbf{r}(t)$ is the position of that piece of fluid at time $t$. The acceleration of that piece is $$\mathbf{a}_{\text{piece}} = \frac{d \mathbf{v}}{dt} = \frac{\partial \mathbf{u}}{\partial \mathbf{r}} \cdot \frac{d \mathbf{r}}{dt} + \frac{\partial \mathbf{u}}{\partial t}$$ by the chain rule. But by definition $d\mathbf{r}/dt = \mathbf{u}$, so $$\mathbf{a}_{\text{piece}} = \frac{\partial \mathbf{u}}{\partial r_i} u_i + \frac{\partial \mathbf{u}}{\partial t}$$ which is exactly what appears in the equation.

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  • $\begingroup$ That's given me a really clear explanation. Just wanted to say that you're identification of me considering the position as the point of focus rather than a discrete piece of fluid was spot on. Quick question since I'm new here: Why is it considered a conventional faux pas on Stack Exchange to say thanks in response to a good answer? (Thanks!) $\endgroup$ – Ben Mar 29 '18 at 9:40
  • $\begingroup$ @Ben It's for posterity; somebody just like you who comes here in a year wants to see the question and answer, not "hi, my name is..." and "thanks", etc. $\endgroup$ – knzhou Mar 29 '18 at 9:45
  • $\begingroup$ @Ben But, you're welcome! $\endgroup$ – knzhou Mar 29 '18 at 9:45

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