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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.

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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. – S. Snape 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 – S. Snape 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
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 $\frac{\partial x}{\partial x} + \frac{\partial (-y)}{\partial y} = 0$.

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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

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