# Physical meaning of the convection term in the momentum equation of acoustic wave

In deriving the acoustic wave equation, the momentum equation is used.

$$\frac{\partial \mathbf{u}}{\partial t}+ (\mathbf{u}\nabla)\mathbf{u}=-\frac{1}{\rho} \nabla p$$

Intuitively, the convection term $$(\mathbf{u}\nabla)\mathbf{u}$$ represents a component of acceleration, but how is this acceleration originate?

P.S. What is the difference between $$(\mathbf{u}\nabla)\mathbf{u}$$ and $$(\mathbf{u}\cdot\nabla)\mathbf{u}$$?

• They presumably mean the same thing, but the former is extremely confusing while the latter is perfectly clear. – G. Smith Aug 1 at 5:47
• Acoustic wave is linear, you should not need this term at all (unless there is some background flow velocity u$_0$). – Maxim Umansky Aug 1 at 6:25
• @G.Smith Is $(\mathbf{u}\cdot\nabla)\mathbf{u}=(\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z})\mathbf{u}? – ecook Aug 1 at 12:20 • You left out the final$ so the MathJax didn’t display as math. – G. Smith Aug 1 at 17:03
• $$(\mathbf{u}\cdot\nabla)\mathbf{u}=\left(u_x\frac{\partial}{\partial x}+u_y\frac{\partial}{\partial y}+u_z\frac{\partial}{\partial z}\right)\mathbf{u}$$ This is very different from what you had. The $\mathbf{u}\cdot\nabla$ is a scalar differential operator, and is not the divergence $\nabla\cdot\mathbf{u}$. – G. Smith Aug 1 at 17:11

The momentum equation written above is written for an incompressible fluid, otherwise the density $$\rho$$ would have to be written inside the partial derivative: $$\frac{\partial\rho\mathbf{u}}{\partial t}$$
• convection (transport) of momentum through the CV boundaries. This is the meaning of the term $$(\mathbf{u}\nabla)\mathbf{u}$$.