# Navier-Stokes eqs. correspond to $F=m*a$

I have the (typical) Navier--Stokes system for incompressible fluid:

$$div(u)=0$$ $$\rho(u_t+u\cdot\nabla u)=-\nabla p+div(\nu\nabla u)+\rho g$$

In a paper that I'm reading says that the term

$$u_t+u\cdot\nabla u$$

is an acceleration. I can understand that $u_t$ (the derivative of the velocity $u$ with respect to time) is in fact an acceleration, but, why "$u_t+u\cdot\nabla u$" is also an acceleration?

My second question is: why the right side terms $$-\nabla p+div(\nu\nabla u)+\rho g$$ represent the sum forze?

• It would be useful for you to look at the Material Derivative – nluigi Jan 23 '18 at 9:23
• I find the integral formulations of conservation equations easier to understand than the differential formulations. You should find them in any good textbook on continuum mechanics and fluid dynamics. – Robin Jan 24 '18 at 9:25

$u_t$ on its own is the rate of change of velocity at a point. Fluid is flowing past that point, so the bit of fluid that the velocity is referring to is constantly changing. In order to apply $\vec F=m\vec a$, you need to think of a fluid parcel. You want to consider the acceleration of, and the forces acting on, a little box of fluid, the boundaries of which move along with the flow. For example, if you have steady flow in a pipe, and the pipe diameter decreases, the fluid speeds up as it squeezes into the smaller pipe. Another way of saying that is that as a bit of fluid comes along, it accelerates. However, $u_t$ is zero everywhere (it's steady flow). $u \cdot \nabla u$ is the part of the acceleration that the fluid experiences due to moving to a new location.