How to prove that $(\vec n\cdot\nabla)\vec n$ is orthogonal to $\vec n$ for unit vector $\vec n$?

Question

I'm trying to prove or disprove that if $\vec n(x,y,z)$ is a unit vector, then $(\vec n\cdot\nabla)\vec n$ is orthogonal to $\vec n$. For this I first tried to compute $\vec n\cdot((\vec n\cdot\nabla)\vec n)$ and show that it's zero. I wrote it out in components, but the resulting expression contained only additions, no subtractions at all, so I couldn't simply cancel out something.

Then I wrote $\vec n$ as

$$\vec n=\vec n_0/|\vec n_0|$$

and tried to do the same with its components. But the expression appeared too large to handle manually. I then used Wolfram Mathematica to check that $\vec n\cdot((\vec n\cdot\nabla)\vec n)=0$ symbolically, and it confirmed it.

But I still wonder, are there any ways of proving this without needing to handle large intermediate expressions?

Answer 1

A demonstration sans coordinates:

We have

$\langle \vec n, \vec n \rangle = 1; \tag{1}$

thus, for any vector field $\vec v$,

$(\vec v \cdot \nabla) \langle \vec n, \vec n\ \rangle = 0; \tag{2}$

but

$(\vec v \cdot \nabla) \langle \vec n, \vec n \rangle = \langle (\vec v \cdot \nabla) \vec n, \vec n\ \rangle + \langle \vec n,(\vec v \cdot \nabla)\vec n \rangle$ $= 2\langle \vec n,(\vec v \cdot \nabla) \vec n \rangle; \tag{3}$

combining (2) with (3) yields

$\langle \vec n,(\vec v \cdot \nabla)\vec n \rangle = 0; \tag{4}$

we now take $\vec v = \vec n$, whence

$\langle \vec n,(\vec n \cdot \nabla)\vec n \rangle = 0, \tag{5}$

showing, as required, that $\vec n$ is orthogonal to $(\vec n \cdot \nabla) \vec n$.

Note we have in fact established the much more general (4); that is, $\vec n$ is orthogonal to $(\vec v \cdot \nabla)\vec n$ for any $\vec v$.

Answer 2

Let $(n\cdot\nabla) n=(n_1 \partial_1+n_2\partial_2 + n_3\partial_3) n$. Then $n\cdot ((n\cdot\nabla)n)$ equals $\sum_{i,j} n_i (n_j \partial_j) n_i=\sum_{i,j} n_i n_j (\partial_j n_i)$.

Applying the gradient to $n\cdot n=const$ gives $\sum_j (\partial_i n_j) n_j=0$ for all $i$. Multiplying the last vector by $n$, we get that $$ \sum_{i,j} n_i (\partial_i n_j) n_j=0. $$