0
$\begingroup$

In the Navier-Stokes equations, there's a well-known convective term of the form: \begin{equation}(\mathbf{v}\cdot\nabla)\mathbf{v}\end{equation} I'm not able to understand it. As far as I know, the nabla operator is, as it name says, an operator, not simply an element of $\mathbb{R}^3$, it maps between function spaces. Therefore, notations for divergence and curl in terms of nabla are just useful, not rigurous.

So how in the world does this term make sense? Maybe there's a better definition for nabla, because, using the most popular one, this seems really strange. But, it's from Navier-Stokes, so, obviously, there's something I'm missing here.

Let it be noted, I'm not studying Navier-Stokes, so, physical interpretations and motivations for the term are not my main concern here. Actually, I just need to understand in order to lucidly prove this identity: \begin{equation}\nabla\times(\mathbf{v}\cdot\nabla)\mathbf{v}=-\nabla\times[\mathbf v\times(\nabla\times\mathbf v)] \end{equation} So, as though as I wish to understand why this term is written like that, it's not particularly important for me, right now, to dive too much into the hydrodynamics.

Any help will be appreciated.

$\endgroup$
1
  • $\begingroup$ What's not rigorous about divergence and curl in terms of nabla? Divergence and curl are operations. $\endgroup$
    – pwf
    Mar 20, 2023 at 16:23

1 Answer 1

0
$\begingroup$

You are correct that $\nabla$ can be ambiguous in coordinate systems that are not Cartesian

In Cartesian coordinates symbol $({\bf v}\cdot \nabla){\bf v}$ represents a vector field whose $j$'th cartesian component is $$ [({\bf v}\cdot \nabla){\bf v}]_j = (v_i \partial_i)v_j. $$ In general coordinates, or on Riemann manifolds, it will become $\nabla_{\bf v} {\bf v}$ where $\nabla_{\bf v}$ now represents a covariant directional derivative containing the connection induced by the metric

$\endgroup$
3
  • 2
    $\begingroup$ $\nabla$ is not ambiguous in non-Cartesian coordinate systems. You just have to be careful to remember that in curvilinear coordinates the coordinates depend on position. That does make it more awkward to use curvilinear coordinates for derivations like OP is looking for, and you're generally better off in Cartesian coordinates, but I wouldn't call it ambiguous. $\endgroup$
    – pwf
    Mar 20, 2023 at 15:57
  • $\begingroup$ @pwf I think that both the OP nad I mean that writing ${\rm div\,} {\bf v}\stackrel{?}{=} \nabla\cdot {\bf v}$ and $ {\rm curl \,} {\bf v}\stackrel{?}{=} \nabla\times {\bf v}$ is not strictly correct in non-cartesian coordinates. There is no simple object "$\nabla$" whose dot and cross product reproduce the LHS. $\endgroup$
    – mike stone
    Mar 20, 2023 at 18:57
  • 2
    $\begingroup$ Not to be argumentative, but unless I'm misunderstanding your point, I would still disagree. You can absolutely write down a "$\nabla$" whose dot and cross product with a vector field give the divergence and curl of the field, in any coordinate system. You just have to remember that in non-Cartesian systems, the unit vectors are locally defined, and therefore derivatives of the unit vectors are not trivial. On the other hand, if you just mean that "$\nabla$" is not a vector, then I would agree with you, but it's not a vector in Cartesian coordinates either. $\endgroup$
    – pwf
    Mar 20, 2023 at 19:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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