# Is a vector times the divergence of the same vector equal to zero?

Is the following equal to zero because of the inner product?

$$\vec{U} \nabla \bullet \vec{U} = 0$$

I am confused because $\nabla \bullet \vec{U}$ is divergence of vector U which results in a scalar. And a scalar time a vector is a vector.

This came about because of some computational fluid dynamics mathematics. To get the "Total Derivative" as presented, the above would have to be zero from what I can tell.

Per Mr. Khor's request for additional information from the CFD book. Note that I am learning, and I figured out my own error as I was typing. Thanks Mr. Khor. :

I have the momentum equation in vector form $$\frac{\partial \rho \vec{U}}{\partial t} = - \nabla \bullet (\rho \vec{U} \otimes \vec{U}) - \nabla \bullet \tau - \nabla P + \rho \vec{g}$$

First, split the time and convection term by using the product rule

$$\frac{\partial \rho \vec{U}}{\partial t} = \rho \frac{\partial \vec{U}}{\partial t} + \vec{U} \frac{\partial \rho}{\partial t}$$

convection term

$$\nabla \bullet (\rho \vec{U} \otimes \vec{U}) = \rho \vec{U} \bullet \nabla \otimes \vec{U} + \vec{U} \nabla \bullet (\rho \vec{U})$$

Replace the values in the momentum equation with the new time and convection terms, cancel out the continuity part, and get $$\rho[\frac{\partial \vec{U}}{\partial t} + \vec{U} \bullet \nabla \otimes \vec{U}] + \vec{U}[\frac{\partial \rho}{\partial t} + \nabla \bullet (\rho \vec{U}] = - \nabla \bullet \tau - \nabla P + p \vec{g}$$

$$\rho[\frac{\partial \vec{U}}{\partial t} + \vec{U} \bullet \nabla \otimes \vec{U}] = - \nabla \bullet \tau - \nabla P + p \vec{g}$$

a non-conserved momentum total derivative form of

$$\rho \frac{D \vec{U}}{Dt} = -\nabla \bullet \tau - \nabla P + p \vec{g}$$.

I went on to evaluate $\vec{U} \bullet \nabla \otimes \vec{U}$ to see if I could get the total derivative in the form of $\frac{D \vec{U}}{Dt} = \frac{\partial \vec{U}}{\partial t} + \vec{U} \bullet \nabla \vec{U}$ from previous readings. Previous readings had also said that $\nabla \otimes \vec{U}$ could be written as $\nabla \vec{U}$ and I knew this but still had a brain malfunction.

From current evaluation, I discovered that $\vec{U} \bullet \nabla \otimes \vec{U}$ was already in the form needed for the total derivative.

Sorry for the confusion.

• well an incompressible fluid would certainly satisfy this. Otherwise, can you share what leads to this equation? Feb 23 '18 at 0:39
• Are you sure the equation isn't $u\cdot \nabla u$?
– user88319
Feb 23 '18 at 0:41
• I updated the information. Sorry for any confusion, but, upon another evaluation, I discovered my own error. A simple one at that. Feb 23 '18 at 2:57