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Is there an easy way to show that

$$\vec u\times(\nabla \times \vec v)+\vec v\times (\nabla\times \vec u)+ (\vec v\cdot \nabla)\vec u+(\vec u \cdot \nabla )\vec v$$

is equal to

$$(\vec u \times \nabla)\times\vec v+(\vec v\times \nabla)\times \vec u+\vec u (\nabla \cdot v)+\vec v(\nabla \cdot u)$$

other than straight forward computation using components? I know that both are ways to find $\nabla(\vec u \cdot \vec v)$ but I am guessing there must be a way to use identities so that all the computational work is avoided.

DMH16
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  • Applying the $bac-cab$ rule is the standard way of doing such computations to get rid of all of the "cross products" – Ninad Munshi Apr 07 '20 at 01:05
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    @NinadMunshi The "bac cab" rule works for a cross product of 3 vectors, but that doesn't seem to help directly with $(u \times \nabla) \times v$ – Ben Grossmann Apr 07 '20 at 01:12
  • @NinadMunshi wouldn't del create a problem when applying such rule? – DMH16 Apr 07 '20 at 01:13
  • @Omnomnomnom thats why I have cross products in quotes, some of the terms aren't real cross products – Ninad Munshi Apr 07 '20 at 01:16
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    @Ninad I am not aware of a "bac cab" rule for expressions like $(u \times \nabla) \times v$. If such a rule is well known, perhaps you could point to a link or reference. – Ben Grossmann Apr 07 '20 at 01:18
  • @Omnomnomnom $(u\times \nabla) \times v = \nabla(u\cdot v) - (v\cdot \nabla) u$ where mnemonically you can use the intermediate step $(u \times \nabla) \times v = - v \times (u\times \nabla)$ to get that the original expression is equal to $cab-bac$. The purpose of the $bac-cab$ rule in this case is to get the correct ordering of the dot products on the inside. The ordering of the outside is negotiable depending on what doesn't leave an operator hanging without an argument. – Ninad Munshi Apr 07 '20 at 01:22
  • @NinadMunshi Could you point to a reference or link for it? The fact that it follows a mnemonic does not make it correct – Ben Grossmann Apr 07 '20 at 01:27
  • @Omnomnomnom This follows from advice I got from a (math) professor long ago, and it more or less follows from just deriving the $bac-cab$ rule from scratch, but without assuming that the dot products commute. I was never made aware of a reference to look at. Maybe a physics book would have one, but I am not sure. – Ninad Munshi Apr 07 '20 at 01:29
  • @NinadMunshi Interesting, hopefully somebody will find something or maybe write out a justification. Thanks for sharing – Ben Grossmann Apr 07 '20 at 01:57

1 Answers1

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Applying the ideas below to $u \times (\nabla \times v)$ and $(u \times \nabla) \times v$ (as is done on my post here for the second term) yields the identities $$ u \times (\nabla \times v) = u \cdot (\nabla v) - (u \cdot \nabla) v,\\ (u \times \nabla) \times v = u \cdot (\nabla v) - u (\nabla \cdot v). $$ With that, we can simplify your first expression as $$ u\times(\nabla \times v)+v\times (\nabla\times u)+ (v\cdot \nabla) u+( u \cdot \nabla ) v =\\ [u\times(\nabla \times v) + (u \cdot \nabla )v]+ [v\times (\nabla\times u) + (v \cdot \nabla)u]= \\u \cdot (\nabla v) + v \cdot (\nabla u). $$ We can simplify the second expression similarly.

Partial answer:

Write $v = (v_1,v_2,v_3)$, $f = (f_1,f_2,f_3)$, and take $e_1,e_2,e_3$ to be the canonical basis vectors (that is, $i,j,k$). We can write $f = \sum_{j=1}^3 f_j e_j$, and $$ \nabla \cdot f = \sum_{i=1}^3 e_i \cdot \frac{\partial f}{\partial x_i}, \qquad \nabla \times f = \sum_{i=1}^3 e_i \times \frac{\partial f}{\partial x_i}, \\ (v \cdot \nabla)f_j = \sum_{i=1}^3 (v \cdot e_i) \frac{\partial f_j}{\partial x_i}, \qquad (v \times \nabla)f_j = \sum_{i=1}^3 (v \times e_i) \frac{\partial f_j}{\partial x_i},\\ (v \times \nabla) \times f = \sum_{i=1}^3 \sum_{j=1}^3 [(v \times e_i) \times e_j] \frac{\partial f_j}{\partial x_i} $$ With that said, we can write the first expression as $$ \sum_{i=1}^3\sum_{j=1}^3 \left[u \times (e_i \times e_j)\frac{\partial v_j}{\partial x_i} + v \times (e_i \times e_j)\frac{\partial u_j}{\partial x_i} + \left(v_i \frac{\partial u_j}{\partial x_i} + u_i \frac{\partial v_j}{\partial x_i}\right)e_j \right], $$ and the second expression as $$ \sum_{i=1}^3 \sum_{j=1}^3 \left[(u \times e_i) \times e_j\frac{\partial v_j}{\partial x_i} + (v \times e_i) \times e_j\frac{\partial u_j}{\partial x_i} + \left( u_i \frac{\partial v_j}{\partial x_j} + v_i \frac{\partial u_j}{\partial x_j}\right) e_i \right]. $$ The rest can be resolved using vector algebra identities.

Ben Grossmann
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