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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?

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  • $\begingroup$ Just to make it clear, have you proven that using normal vectors? Or are you trying to solve it using index notation directly? $\endgroup$
    – FGSUZ
    Jul 28 '20 at 19:11
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    $\begingroup$ I would recommend writing out the whole thing laboriously, just in order to learn, and then see how it can be captured in index notation. By "laboriously" I mean write out $\vec{\nabla} \vec{V}$ as a $3\times 3$ matrix (for the case of vectors in 3 dimensions). The first row is $(\partial V_x / \partial x, \; \partial V_y/\partial x, \partial V_z / \partial x)$. $\endgroup$ Jul 28 '20 at 19:25
  • $\begingroup$ @FGSUZ I think it has to be proven using normal vectors. The problem said "using summation" which is sometimes called index notation in the course. Thanks to the first answer I now know index notation to be something different than what is meant, though it is interesting to read about. $\endgroup$
    – Brentdb
    Jul 28 '20 at 19:26
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    $\begingroup$ The reason you can't prove it is that it is not true. $\endgroup$ Jul 28 '20 at 21:56
  • $\begingroup$ Isn't this only valid for irrotational flows and shouldn't the gradient of the vector be the transpose of the resulting rank-2 tensor? $\endgroup$ Feb 26 '21 at 23:32
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$\def\vv{{\bf v}} \def\del{\nabla} \def\o{\cdot} \def\pd{\partial}$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, $$[\vv\o(\del\vv)]_j = v_i [\del\vv]_{ij} = v_i(\pd_i v_j) = (v_i\pd_i) v_j = [(\vv\o\del)\vv]_j.$$

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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.

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  • $\begingroup$ Thanks for the elaborate answer, this way of solving it goes over my head though. Could you provide some reference on this notation and metrics? $\endgroup$
    – Brentdb
    Jul 28 '20 at 16:22
  • $\begingroup$ This reference assumes you're working in Cartesian coordinates: web.iitd.ac.in/~pmvs/courses/mcl702/notation.pdf $\endgroup$ Jul 28 '20 at 16:59
  • $\begingroup$ @Brentdb In order to make this any simpler, you have to make a choice of coordinate system. If you're allowed to do that in the proof, use the above reference instead. If instead you have to prove this for any choice of coordinate system, you have to allow for an arbitrary metric. $\endgroup$ Jul 28 '20 at 17:01
  • $\begingroup$ @probably_someone: Let ${\bf v} = [x,y]^T$. Then $v_j\partial^1 v^j = x$, $(\partial_j v^j)v^1 = 2x$, and $\partial_j(v^j v^1) = 3x$. $\endgroup$
    – user26872
    Jul 28 '20 at 19:42

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