So, I was trying to obtain the point form of the conservation of linear momentum equation in integral form, namely:

$\int_{\partial \Omega} \vec{V} \rho \vec{V} \cdot \vec{dS} + \int_{\partial \Omega} p \vec{dS} = 0$

According to the Gauss theorem for a closed surface $S$:

$\iint_S \vec{A} \cdot \vec{dS} = \iiint_V \nabla \cdot \vec{A} dV $

But if I apply that to the above equation I get

$\int_{\partial \Omega} \vec{V} \rho \vec{V} \cdot \vec{dS} = \int_{\Omega} \nabla \cdot (\vec{V} \rho \vec{V}) dV =\\ \quad \int_{\Omega} (\nabla \cdot \vec{V}) \rho \vec{V} dV $

Which can't be right, since for an incompressible flow $\nabla \cdot \vec{V} = 0$.

Isn't the dot product supposed to be commutative? What am I missing?

I apologize for any misuse of mathematical notation, let me know of any mistakes.

  • $\begingroup$ $\vec V\rho\vec V\cdot\overrightarrow{dS}$ is not of the form $\vec A\cdot\overrightarrow{dS}$. The former is a scalar multiple of $\vec V$, specifically $\rho\vec V\cdot\overrightarrow{dS}$ times $\vec V$—a vector—while the latter is the dot product of two vectors—a scalar. The term $\vec V\rho\vec V$ that you pulled out of the first expression is not a vector. Indeed, it’s nonsensical in isolation. $\endgroup$ – amd Mar 28 '18 at 8:10
  • $\begingroup$ @amd It's the dot product of a dyadic product of two vectors, with a vector, resulting in a vector. The term $\vec{V} \rho \vec{V}$ by itself is a 2nd order tensor. You can see that the same quantity is present in the answer below. :) $\endgroup$ – Jason Barmparesos Mar 29 '18 at 9:16

The correct identity is

$$\nabla\cdot (\rho \mathbf{v} \mathbf{v}) = \rho \mathbf{v} \cdot \nabla\mathbf{v} \, + \, (\nabla \cdot \mathbf{v}) \rho \mathbf{v},$$

where only the second term on the RHS vanishes in incompressible flow.

To check, this should all be consistent with the Navier-Stokes equations for steady flow (minus viscous terms that you have excluded):

$$\rho\mathbf{v} \cdot \nabla \mathbf{v} = - \nabla p,$$

which follows when $\nabla \cdot \mathbf{v} = 0$ from

$$\begin{align}0 &= \int_{\partial \Omega}\rho \mathbf{v} \mathbf{v} \cdot d\mathbf{S} + \int_{\partial \Omega} p \,d\mathbf{S} \\&= \int_{\Omega}\nabla \cdot (\rho \mathbf{v} \mathbf{v}) \, dV + \int_{\Omega} \nabla p \,dV\\ &= \int_{\Omega} \{ \rho \mathbf{v} \cdot \nabla \mathbf{v} + (\nabla \cdot \mathbf{v})\rho \mathbf{v} + \nabla p\} \, dV \\ &= \int_{\Omega} \{ \rho \mathbf{v} \cdot \nabla \mathbf{v} + \nabla p\} \, dV \end{align} $$

  • $\begingroup$ Ok, this makes more sense. Thanks! $\endgroup$ – Jason Barmparesos Mar 27 '18 at 23:01

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