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In a proof that I'm working on the solution has that $$\hat{n}\cdot\nabla~\times~(\vec{V}~\times~\vec{C}) = (\hat{n}~\times~\nabla)~\times~\vec{V})\cdot\vec{C}$$

I'm wondering what the justification is for being able to flip $\hat{n}$ and $\vec{C}$ since one is in a cross product and one is in a dot product.

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  • $\begingroup$ 1. Use "\times" to get a cross product. 2. No doubt if you just do all the relevant algebra, it all works. In general, $a \cdot (b \times c)$ is the determinant of the $3 \times 3$ matrix whose rows are $a, b,$ and $c$, so you get things like $a \cdot (b \times c) = \pm (a \times b) \cdot c$, where you have to check the $\pm$ on a single example to get it right. Probably applying that rule a couple of times will do what you need. $\endgroup$ – John Hughes May 18 '18 at 1:30
  • $\begingroup$ Don't you have an extra set of parentheses in the left side (open right after the dot, end before the equal sign)? Otherwise $\hat n\cdot\nabla$ is not a vector $\endgroup$ – Andrei May 18 '18 at 1:34
  • $\begingroup$ @Andrei: I just assumed that "$\times$" had higher precedence than $\cdot$, but I think you must be right that the parentheses would surround the two cross-products. $\endgroup$ – John Hughes May 18 '18 at 1:44
  • $\begingroup$ I'm not sure this helps, but, $ \nabla \times (\vec{V} \times \vec{C}) = (\vec{C} \cdot \nabla ) \vec{V} -(\vec{V} \cdot \nabla ) \vec{C} +\vec{V}(\nabla \cdot \vec{C}) -\vec{C}(\nabla \cdot \vec{V}) $ $\endgroup$ – James S. Cook May 18 '18 at 1:48
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\vec{n}\cdot\nabla\times\pars{\vec{v}\times\vec{c}} = \bracks{\pars{\vec{n}\times\nabla}\times\vec{v}}\cdot\vec{c}:\ {\Large ?}.\qquad}$ Hereafter, $\ds{\epsilon_{\alpha\beta\gamma}}$ is the Levi-Civita Symbol.

\begin{align} \vec{n}\cdot\nabla\times\pars{\vec{v} \times \vec{c}} & = \sum_{i}n_{i}\bracks{\nabla\times\pars{\vec{v} \times \vec{c}}}_{i} = \sum_{i}n_{i}\sum_{jk}\epsilon_{ijk}\,\partiald{\pars{\vec{v} \times \vec{c}}_{k}}{x_{j}} \\[5mm] & = \sum_{ijk}n_{i}\,\epsilon_{ijk}\,\partiald{}{x_{j}} \sum_{\ell m}\epsilon_{k\ell m}\,v_{\ell}\,c_{m} = \sum_{ij\ell m}n_{i}\,\,\partiald{\pars{v_{\ell}\,c_{m}}}{x_{j}} \sum_{k}\epsilon_{kij}\epsilon_{k\ell m} \end{align}

However, $\ds{\sum_{k}\epsilon_{kij}\epsilon_{k\ell m} = \delta_{i\ell}\delta_{jm} - \delta_{im}\delta_{j\ell}}$.

Then, \begin{align} \vec{n}\cdot\nabla\times\pars{\vec{v} \times \vec{c}} & = \sum_{ij}n_{i}\,\,\partiald{\pars{v_{i}\,c_{j}}}{x_{j}} - \sum_{ij}n_{i}\,\,\partiald{\pars{v_{j}\,c_{i}}}{x_{j}} = \sum_{ij}\pars{n_{i}\,\partiald{}{x_{j}} - n_{j}\,\partiald{}{x_{i}}}v_{i}c_{j} \\[5mm] & = \sum_{j}\bracks{\sum_{i}\pars{n_{i}\,\partiald{}{x_{j}} - n_{j}\,\partiald{}{x_{i}}}v_{i}}c_{j}\label{1}\tag{1} \end{align}


Moreover, \begin{align} \bracks{\pars{\vec{n}\times\nabla}\times\vec{v}}_{j} & = \sum_{k\ell}\epsilon_{jk\ell}\pars{\vec{n}\times\nabla}_{k}\, v_{\ell} = \sum_{k\ell}\epsilon_{jk\ell}\pars{\sum_{mp}\epsilon_{kmp}\, n_{m}\,\partiald{}{x_{p}}}v_{\ell} \\[5mm] & = \sum_{\ell mp}\pars{\sum_{k}\epsilon_{k\ell j}\,\epsilon_{kmp}} n_{m}\,\partiald{}{x_{p}}v_{\ell} = \sum_{\ell mp}\pars{\delta_{\ell m}\delta_{jp} - \delta_{\ell p}\delta_{jm}} n_{m}\,\partiald{}{x_{p}}v_{\ell} \\[5mm] & = \sum_{\ell}\pars{n_{\ell}\,\partiald{}{x_{j}}v_{\ell} - n_{j}\,\partiald{}{x_{\ell}}v_{\ell}} = \sum_{i}\pars{n_{i}\,\partiald{}{x_{j}} - n_{j}\,\partiald{}{x_{i}}}v_{i}\label{2}\tag{2} \end{align}

Compare \eqref{1} and \eqref{2}:

$$ \vec{n}\cdot\nabla\times\pars{\vec{v} \times \vec{c}} = \sum_{j}\bracks{\pars{\vec{n}\times\nabla}\times\vec{v}}_{j}c_{j} = \bbx{\bracks{\pars{\vec{n}\times\nabla}\times\vec{v}}\cdot\vec{c}} $$

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  • $\begingroup$ There is no need for the double-epsilon identity in this problem. I love that identity, but I must admit, you could just as well match-up the levi-civita expressions directly. $\endgroup$ – James S. Cook May 18 '18 at 14:12
  • $\begingroup$ @JamesS.Cook Thanks. I agree with your point of view. However, sometimes is better to provide additional details. $\endgroup$ – Felix Marin May 18 '18 at 18:59
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Maybe the easiest way to see this is using tensor notation and the Levi-Civita symbol $\epsilon_{abc}$. In this notation the dot-product of two vectors $A_i$ and $B_i$ is written $A_iB_i$ with the sum over $i$ being implicit (note that I'm using all indices lowered, which is a bit different from typical Einstein summation convention, but that's moot here since everything is Euclidean); the cross-product of two vectors $A\times B$ is then $(A\times B)_i=\epsilon_{ijk}A_jB_k$ (note that this hides a double summation, but almost all the terms are zero).

Written this way, we have $(V\times C)_i=\epsilon_{ijk} V_jC_k$, so $(\nabla\times(V\times C))_i = \epsilon_{ijk}\nabla_j\epsilon_{klm}V_lC_m$, and $n\cdot(\nabla\times(V\times C)) = n_i\epsilon_{ijk}\nabla_j\epsilon_{klm}V_lC_m$ (note that every index appears exactly twice, so this is a scalar). Similarly, $((n\times \nabla)\times V)\cdot C = \epsilon_{ijk}\epsilon_{jlm}n_l\nabla_mV_kC_i$.

From here, the two can be shown the same by some rearrangement and renaming of indices; the first is $\epsilon_{ijk}\epsilon_{klm}n_i\nabla_jV_lC_m$, and by renaming $l\mapsto i, m\mapsto j, j\mapsto k, k\mapsto l, i\mapsto m$ we can write the second expression as $\epsilon_{mkl}\epsilon_{kij}n_i\nabla_jV_lC_m$. But $\epsilon_{kij}=-\epsilon_{ijk}$ and $\epsilon_{mkl}=-\epsilon_{klm}$, so these two expressions are equal.

(There's a little bit of subtlety since we can't just move $\nabla$ around willy-nilly, but note that we actually keep all of the 'vectors' $n, \nabla, V, C$ themselves here in the same order; hopefully you can convince yourself that it always applies to the same terms.)

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  • $\begingroup$ this is a much better path than the one I was about to go down... $\endgroup$ – James S. Cook May 18 '18 at 1:50
  • $\begingroup$ @JamesS.Cook I had the Einstein summation convention and index notation drilled into me by some general relativity courses I took in college and it really makes these questions infinitely easier (once you prove or convince yourself that all the manipulations 'go through' correctly, of course). $\endgroup$ – Steven Stadnicki May 18 '18 at 1:51
  • $\begingroup$ The sad thing is I know Einstein notation and the levi-civita stuff, I just got this idea to use that identity which is silly since it does not nicely involve dot and cross products in the needed way. $\endgroup$ – James S. Cook May 18 '18 at 14:03

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