Can Vector Calculus Verify This Identity?

[nsiderbam]

Homework Statement

Use your knowledge of vector algebra to verify the following identity:

$$\vec{\Omega} \cdot \nabla n = \nabla \cdot \vec{\Omega} n<br />$$

Homework Equations

Divergence product rule
$$\nabla \cdot (\vec{F} \phi) = \nabla (\phi) \cdot \vec{F} + \phi (\nabla \cdot \vec{F})$$

The Attempt at a Solution

By the product rule,

$$\nabla \cdot (\vec{\Omega} n) = \nabla n \cdot \vec{\Omega} + n (\nabla \cdot \vec{\Omega})$$

Therefore,

$$\vec{\Omega} \cdot \nabla n = \nabla \cdot \vec{\Omega} n = \nabla \cdot (\vec{\Omega} n) = \nabla n \cdot \vec{\Omega} + n (\nabla \cdot \vec{\Omega})$$

and

$$0 = n (\nabla \cdot \vec{\Omega})$$

I'm not quite sure what I'm doing wrong. Maybe it's a grouping thing. Any help would be appreciated.

 

[HallsofIvy]

It would help us if you would explain your notation. What is $\vec{\Omega}$? Is it a general vector function? And what is n?

 

[nsiderbam]

edit: Confirmed that omega is an arbitrary vector and n is a scalar.

Sorry about that. I believe omega is a general vector function and n is simply a scalar -- it is not specified but imo the problem statement phrases it as like it is a general identity. This is for a reactor physics course and which features
$$\vec{\Omega}$$
and
$$n$$
but in that case n is no trivial function
$$n=f(r,E,\vec{\Omega},t)$$
and
$$\vec{\Omega}$$ is the neutron direction unit vector.

 

[LCKurtz]

Homework Statement

Use your knowledge of vector algebra to verify the following identity:

$$\vec{\Omega} \cdot \nabla n = \nabla \cdot \vec{\Omega} n<br />$$

Homework Equations

Divergence product rule
$$\nabla \cdot (\vec{F} \phi) = \nabla (\phi) \cdot \vec{F} + \phi (\nabla \cdot \vec{F})$$

The Attempt at a Solution

By the product rule,

$$\nabla \cdot (\vec{\Omega} n) = \nabla n \cdot \vec{\Omega} + n (\nabla \cdot \vec{\Omega})$$

That would give your result if $\nabla\cdot\Omega = 0$. Is there some reason from the physical situation for why that would be true?

 

[HallsofIvy]

$\vec{\Omega}\cdot\nabla n= \nabla\cdot \left(\vec{\Omega}n\right)$
is certainly NOT true in general.
For example, if $\vec{\Omega}= x\vec{i}+ y\vec{j}+ z\vec{k}$ and $n= x^2$ then $\vec{\Omega}n= x^3\vec{i}+ x^2y\vec{j}+ x^2z\vec{k}$ and $\nabla\cdot \left(\vec{\Omega}n\right)= 3x^2+ x^2+ x^2= 4x^2$ while $\vec{\Omega}\cdot\nabla n= (x\vec{i}+ y\vec{j}+ z\vec{k})\cdot(2x\vec{i})= 2x^2$