Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I know that $\vec \nabla \cdot \vec v=0$, can I say that:

$$( \vec v \cdot \vec \nabla )\vec v=\underbrace{(\vec \nabla \cdot \vec v)}_{=0} \vec v=0 $$ ?

Note: this is a question I asked in Physics StackExchange but as it is mainly mathematical I thought it could be relevant to post it here too.

share|improve this question
    
What is $\bar{v}$? Is it like $(v_1,v_2,v_3,...)$ or $v$ is a multivariable function. I am asking this because $\nabla v$ is defined when $v$ is of the latter ones. –  Babak S. Jun 11 '13 at 9:24
    
@BabakS. In fluids, $v\cdot \nabla$ is an operator on a vector. –  user71815 Jun 11 '13 at 9:25
    
@user71815: oh i see now –  Babak S. Jun 11 '13 at 9:25
    
Hi, I added a parenthesis in the convective term in your question to avoid some confusion in notation. You could use tensor notation , but this way of writing is more mathematically sound for it can be interpreted as directional derivative. –  Shuhao Cao Jun 11 '13 at 20:19

1 Answer 1

up vote 4 down vote accepted

No. Try $v := (x,-y)$. Then $\nabla \cdot v = 0$ but $(v\cdot \nabla) v = (x,y)$. Problem is $(v \cdot \nabla) v \neq (\nabla \cdot v) v$ in general.

Edit: There seems to be some confusion as to what $v \cdot \nabla$ means, especially from the comments of the OP who seem to suggest that he/she treats $v$ and $\nabla$ as vectors which could be freely multiplied together in any order.

In the above example, $v\cdot\nabla$ is the operator $x \frac{\partial}{\partial x} - y \frac{\partial}{\partial y}$ which, being an operator, does not make sense unless you apply it to something. On the other hand, $\nabla \cdot v$ means apply the operator "$\nabla \cdot$" to $v$, resulting in the scalar function $\frac{\partial x}{\partial x} + \frac{\partial (-y)}{\partial y} = 0$.

share|improve this answer
    
But $v \cdot \nabla = \nabla \cdot v$ in general, right ? This is what troubles me. –  snickers Jun 11 '13 at 9:30
    
@snickers What are you talking about?? $v\cdot \nabla$ is an operator, so to say anything about it you have to apply it to some vector/scalar. On the other hand, $\nabla \cdot v$ is a scalar function, and makes sense on its own. They are completely different things! –  user71815 Jun 11 '13 at 9:32
    
Oh yes, I was a bit confused sorry. Now I understand, thank you for your answer. –  snickers Jun 11 '13 at 9:37
    
@snickers See the edit –  user71815 Jun 11 '13 at 9:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.