Proving $ \vec{V} \cdot (\vec{\nabla}\vec{V}) = (\vec{\nabla}\cdot\vec{V})\vec{V} $ using index notation

Question

In my fluid mechanics course we encounter a lot of vector calculus problems, one of which I have been struggling with for a while now. We must prove that $$ \vec{V} \cdot \left(\vec{\nabla}\vec{V}\right) = \left(\vec{\nabla}\cdot\vec{V}\right)\vec{V} $$ solely using summation/index notation. $\vec{\nabla}\vec{V}$ is a second order tensor which we denote by: $$\left(\sum\limits_{i}\hat{e}_{i}\frac{\partial}{\partial x_i}\right)\left(\sum\limits_{j}\hat{e}_{j}V_j\right).$$ I think my confusion lies in the use of $\frac{\partial}{\partial x_i}$ in a tensor since we haven't used tensors commonly before taking this course. Could someone maybe prove this and clarify how second order tensors work in general?

Answer 1

$\def\vv{{\bf v}} \def\del{\nabla} \def\o{\cdot} \def\pd{\partial}$Note that $$[\vv\o(\del\vv)]_j = v_i [\del\vv]_{ij} = v_i(\pd_i v_j)$$ and
$$[(\del\o\vv)\vv]_j = (\pd_i v_i)v_j.$$ But that $$v_i(\pd_i v_j)\ne (\pd_i v_i)v_j,$$ in general. (Repeated indices are to be summed over. This is Einstein's summation notation.)

For clarity let $\vv$ be two dimensional. For $j=1$ the claim is that $$v_1(\pd_1 v_1) + v_2(\pd_2 v_1) = (\pd_1 v_1+\pd_2 v_2)v_1.$$ Clearly this is false. For example, if $\vv=[x,y]^T$ this implies that $$x = 2x.$$

The intended claim is likely that $$\vv\o(\del\vv) = (\vv\o\del)\vv.$$ (Note that $\del\o\vv$ and $\vv\o\del$ are completely different objects. The first is a scalar. The second is a scalar differential operator.) This result can be easily proved, $$[\vv\o(\del\vv)]_j = v_i [\del\vv]_{ij} = v_i(\pd_i v_j) = (v_i\pd_i) v_j = [(\vv\o\del)\vv]_j.$$

Answer 2

The space that these vectors live in has a metric $g_{ij}$. For example, if you're in Euclidean space and you're using Cartesian coordinates, then $g_{ij}$ is equal to the Kronecker delta $\delta_{ij}$. If you're using non-Cartesian coordinates (e.g. polar coordinates) or you're working in non-Euclidean space, then your metric will be different. The index notation expression for a vector is $v^i$. The dot product between two vectors is represented by:

$$\vec{a}\cdot\vec{b}=a^ig_{ij}b^j=a^ib_i$$

This is, in fact, the definition of a metric - it tells you the "distance" between the tips of two vectors. The "lowered index" vector $b_i$ is defined straightforwardly, as long as you know what your metric is:

$$b_i=g_{ij}b^j$$

If you're in Euclidean space and you're using Cartesian coordinates, then we conveniently have that $b_i=b^i$, since $g_{ij}=\delta_{ij}$. With any other metric, this is not true. For example, in 2D polar coordinates (where $\vec{b}=b^r\hat{r}+b^\theta\hat{\theta}$), our metric is defined by $g_{rr}=1$ and $g_{\theta\theta}=r^2$, with the other two elements zero. In that case, we have that $b_r=g_{rr}b^r+g_{r\theta}b^{\theta}$, so $b_r=b^r$, but $b_\theta=g_{\theta r}b^r+g_{\theta \theta}b^{\theta}$, so $b_\theta=r^2b^{\theta}$. But as long as you know what your metric is, lowering the index of a vector should be straightforward.

For the rest of this discussion, let's assume that you're working in Euclidean space, since the differential geometry in non-Euclidean space gets complicated once you start taking derivatives.

The derivative operator $\vec{\nabla}$ is notated as $\partial^i$, which is shorthand for $\frac{\partial}{\partial x^i}$. The dyadic product $\vec{\nabla}\vec{V}$ is therefore notated as $\partial^j v^i$, which is shorthand for $\frac{\partial v^i}{\partial x^j}$.Putting all this together, the expression you need to prove is written as:

$$v^ig_{ij}\partial^kv^j=\partial^ig_{ij}v^jv^k$$

or, lowering the indices:

$$v_j\partial^kv^j=\partial_j v^jv^k$$

This should be enough information to start the proof.