Correct form of the convection term in the Navier-Stokes equation

Question

Usually the NS momentum equation for an compressible fluid is written in its convecting form, in the absence of external forces, as

$$ \rho\frac{\partial \vec u}{\partial t} +\rho(\vec u\cdot\nabla)\vec u+\nabla P = \mu\nabla^2\vec u +\frac{\mu}{3}\nabla(\nabla\cdot \vec u) $$

However, from the Wikipedia, this can be written in "conservation form", as

$$ \rho\frac{\partial \vec u}{\partial t} +\rho\nabla\cdot (\vec u\otimes\vec u)+\nabla P = \mu\nabla^2\vec u +\frac{\mu}{3}\nabla(\nabla\cdot \vec u) $$

I can't see how these two are consistent, the convection term in the lower one is

$$ \nabla_\alpha(u_\alpha u_\beta) = \vec u\nabla\cdot \vec u + (\vec u\cdot \nabla)\vec u $$

So there's an extra term $\vec u\nabla\cdot \vec u$ with respect to the first equation.

What am I not understanding here? I see that for incompressible fluids, there's a pressure equations that constraints the system to obey $\nabla\cdot \vec u=0$, but I'm still missing something about the general case.

Answer 1

You have changed the equation from what is in Wikipedia. You have $$ \rho\frac{\partial{\bf v }}{\partial t}+\rho \nabla\cdot ({\bf v}\otimes {\bf v})+\ldots $$ Wiki's conservation form has an expression with the $\rho$'s in the correct places $$ \frac{\partial{\rho \bf v }}{\partial t}+\nabla\cdot (\rho({\bf v}\otimes {\bf v}))+\ldots $$ The extra bit they add to the plain Euler equation is
$$ {\bf v}\left( \frac{\partial \rho }{\partial t} + \nabla \cdot (\rho {\bf v})\right) $$ which is zero by the continuity equation.