Suppose $v$ is a vector valued function. Then $\nabla v$ is a rank 2 tensor, i.e. a matrix. In the Navier-Stokes equations we see a term of the form $$(v \cdot \nabla) v.$$ At first I thought that the dot would indicate matrix multiplication, but as it's written, this does not work due to dimensions. How is one supposed to interpret/evaluate this term?

Also, what does the divergence of this term look like, i.e. $\nabla\cdot\big((v \cdot \nabla) v\big)$? Does it follow the usual product rule?

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  • $\begingroup$ As $v\cdot\nabla$ is defined for scalar functions, presumably $(v\cdot\nabla)v$ is just $v\cdot\nabla$ applied to each coordinate of $v$, yielding a vector function. $\endgroup$
    – anon
    Aug 13, 2022 at 23:31
  • $\begingroup$ @PrincessEev Thanks, the answer in that question helps me in evaluating the dot product, but it now leads me to wonder where that expression comes from? Given that we are taking a linear combination of the columns of (using their notation) $\nabla b$ with the $i$th coefficient being the $i$th value of $v$, why isn't this commonly written as $(\nabla b) \cdot a$, as one would expect from normal matrix-vector multiplication? $\endgroup$
    – CBBAM
    Aug 13, 2022 at 23:36

1 Answer 1


Think about partial derivatives. The derivatives $\partial/\partial x_1$ and $\partial/\partial x_2$, for example, are derivative in the $x_1$ direction and in the $x_2$ direction, and I assume you have no problem with an expression like $\partial v/\partial x_1$. But now we see that $\partial/\partial x_2 = e_1\cdot\nabla$ and $\partial/\partial x_2 = e_2\cdot\nabla$, where we are simply forming differential operators and not differentiating anything in these expressions. (Of course, here $e_1, e_2$ are part of the standard basis.) This line of reasoning gives you $v\cdot\nabla$ as the derivative in the $v/|v|$ direction scaled by $|v|$. I would really think of this as the coordinate-free version of partial derivatives. In fact, for any totally-differentiable function $F(x)$, it's total derivative at $x$ can be expressed as the linear map $$ DF_x(v) = (v\cdot\nabla)F(x), $$ and the matrix of this map in the standard basis is the Jacobian matrix.

The same way $\partial v/\partial x = (e_1\cdot\nabla)v$ is valid, so is $(v\cdot\nabla)v$; this is "the derivative of $v$ in the $v$ direction (scaled by $|v|$)". It's important to note here that I am fairly certain the intent is not to differentiate the $v$ in $v\cdot\nabla$; doing so is valid, but is harder to interpret.

And that brings us to your last question. These sorts of expressions are much easier to reason about when you're explicit about what's being differentiated. For example, I would write $(v\cdot\dot\nabla)\dot v$ to indicate that the right-hand $v$ is the only one being differentiated. This makes it easy to evaluate your divergence expression: $$\begin{aligned} \nabla\cdot((v\cdot\nabla)v) &= \dot\nabla\cdot((\dot v\cdot\hat\nabla)\hat v) + \dot\nabla\cdot((v\cdot\hat\nabla)\dot{\hat v}) \\ &= (\dot v\cdot\hat\nabla)(\dot\nabla\cdot\hat v) + (v\cdot\hat\nabla)(\dot\nabla\cdot\dot{\hat v}) \\ &= (\dot v\cdot\hat\nabla)(\dot\nabla\cdot\hat v) + (v\cdot\nabla)(\nabla\cdot v), \end{aligned}$$ where the hats and dots indicate which derivative is acting on what. In the last line, I've made the differentiation implicit in the right-hand term since it follows usual conventions. This term is also easy to interpret: it's the $v$-directional derivative of the divergence of $v$. The left-hand term is stranger, a sort of "cross-divergence"; in coordinates $$ (\dot v\cdot\hat\nabla)(\dot\nabla\cdot\hat v) = \sum_{i,j}\frac{\partial v_i}{\partial x_j}\frac{\partial v_j}{\partial x_i}. $$

  • $\begingroup$ I realized I didn't make it explicit, but this was part of the initial question: the dot in $v\cdot\nabla$ is the dot product/inner product. $\endgroup$ Aug 14, 2022 at 2:02
  • $\begingroup$ This was extremely helpful, thank you! Do you have any resources for learning more "generalized" multivariable/vector calculus? $\endgroup$
    – CBBAM
    Aug 14, 2022 at 3:51
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    $\begingroup$ @CBBAM Glad to help! These sorts of manipulations I got from studying geometric calculus, specifically what's called the vector derivative. I would recommend Geometric Algebra for Physicists (2003) by Chris Doran and Anthony Lasenby; however, this would entail learning about geometric/Clifford algebras. (Though I think that is very worthwhile!) I've not seen these sorts of manipulations in standard vector calculus, so unfortunately I don't have a reference like that. $\endgroup$ Aug 14, 2022 at 7:20

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