# Del operator and divergence

1. Jan 11, 2012

### Hypatio

I'm trying to understand why the del operator is working a certain way.

So in my literature there is a term:

$\nabla \cdot \rho_a \mathbf{v}$

but then after saying that

$\rho_a=w_a\rho$

the term can somehow become

$\rho (\mathbf{v}\cdot \nabla w_a)$

I do not understand how nabla and the velocity, v, get flipped.. Is there some assumption that needs to be made for this to be true?

2. Jan 11, 2012

### Elliptic

What literature is that?
You must know something about nabla and on what operates nabla

3. Jan 11, 2012

### Hypatio

It's literature on fluid dynamics and chemical reactions. The equation I am looking at gives

$\frac{\partial \rho_a}{\partial t}+(\nabla \cdot \rho_a \mathbf{v})=0$

and then after defining that rho_a=rho*w_a they rearrange the equation to give

$\rho \left ( \frac{\partial w_a}{\partial t} + \mathbf{v}\cdot \nabla w_a\right )=0$

so rho is a density, t is time, w is a mass fraction, and v is velocity.

I'm just not understanding what rule you follow, if any, to flip $\nabla \cdot \mathbf{v}$ without it meaning something completely different.

Does this make sense?

Last edited: Jan 11, 2012
4. Jan 11, 2012

### Elliptic

here is something in attachment

#### Attached Files:

• ###### Eqn3.gif
File size:
2.6 KB
Views:
143
5. Jan 11, 2012

### Hypatio

Thanks I think my misunderstanding is resolved. They are the same because in the product rule the divergence term drops when incompressibility is assumed. Thank you.