Can Vector Calculus Verify This Identity?

[nsiderbam]

Homework Statement

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

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

Homework Equations

Divergence product rule
<br /> \nabla \cdot (\vec{F} \phi) = \nabla (\phi) \cdot \vec{F} + \phi (\nabla \cdot \vec{F})<br />

The Attempt at a Solution

By the product rule,

<br /> \nabla \cdot (\vec{\Omega} n) = \nabla n \cdot \vec{\Omega} + n (\nabla \cdot \vec{\Omega})<br />

Therefore,

<br /> \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})<br />

and

<br /> 0 = n (\nabla \cdot \vec{\Omega})<br />

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
<br /> \vec{\Omega}<br />
and
<br /> n<br />
but in that case n is no trivial function
<br /> n=f(r,E,\vec{\Omega},t)<br />
and
<br /> \vec{\Omega}<br /> is the neutron direction unit vector.

 

[LCKurtz]

Homework Statement

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

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

Homework Equations

Divergence product rule
<br /> \nabla \cdot (\vec{F} \phi) = \nabla (\phi) \cdot \vec{F} + \phi (\nabla \cdot \vec{F})<br />

The Attempt at a Solution

By the product rule,

<br /> \nabla \cdot (\vec{\Omega} n) = \nabla n \cdot \vec{\Omega} + n (\nabla \cdot \vec{\Omega})<br />

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