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Mathematics
Linear and Abstract Algebra
Solving for the Nth divergence in any coordinate system
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[QUOTE="Vanilla Gorilla, post: 6869112, member: 682122"] OK, so I've taken a few days to mull this over, and I think this should be correct, but I'm not sure. My main concern is if ##m## is the appropriate number to sum to here $$\sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}}$$ However, I am unsure all around as to the correctness of my reasoning and conclusions. Any further criticism is appreciated! :) Let's construct an operator, such that $$\nabla^{n} =\sum_{i=1}^{m} \frac{\partial^n f}{\partial^n x_{i}}$$ in ##m## dimenions, when written in the Cartesian Basis. Then $$\nabla^n F = \frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_k\nabla^{n-1} F\right) = \frac{1}{\sqrt{\vert g\vert}}\partial_{i_1}(\sqrt{\vert g\vert} g^{i_1k_1}\partial_{k_1} \frac{1}{\sqrt{\vert g\vert}}\partial_{i_2}(\sqrt{\vert g\vert} g^{i_2k_2}\partial_{k_2} ... \frac{1}{\sqrt{\vert g\vert}}\partial_{i_n}(\sqrt{\vert g\vert} g^{i_nk_n}\partial_{k_n} F )...))$$ by [URL='https://www.physicsforums.com/threads/solving-for-the-nth-divergence-in-any-coordinate-system.1050912/post-6867737']This[/URL]. Let's convert this out of Einstein notation for our use in regular PDEs. $$\nabla^n F = \frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_k\nabla^{n-1} F\right) = \frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_k (\nabla^{n-1} F)\right) = \frac{\partial_i \left(\sqrt {\vert g \vert} g^{ik} \partial_k (\nabla^{n-1} F)\right)} {\sqrt{\vert g \vert}} =$$ $$\sum^{m}_{\begin{align} {i = 1} \nonumber \\ {k = 1} \nonumber \\ \end{align}} \frac{\partial_i \left(\sqrt {\vert g \vert} g^{ik} \partial_k (\nabla^{n-1} F)\right)} {\sqrt{\vert g \vert}} =$$ $$\sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-1 } F)\right ) } {\sqrt {\vert g \vert}}$$ Expanding this twice to get the gist of the pattern: $$\sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } ( \sum\limits^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-2 } F)\right ) } {\sqrt {\vert g \vert}} )\right ) } {\sqrt {\vert g \vert}} = $$ $$\sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } ( \sum\limits^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } ( \sum\limits^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-3 } F)\right ) } {\sqrt {\vert g \vert}} )\right ) } {\sqrt {\vert g \vert}} )\right ) } {\sqrt {\vert g \vert}}$$ However, this is highly cluttered, so let's replace it again with the following $$\sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-1 } F)\right ) } {\sqrt {\vert g \vert}}$$ For example, $$\nabla^3 F = \sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-1 } F)\right ) } {\sqrt {\vert g \vert}}$$ where $$\Delta F = \nabla^2 F = \nabla^{ n-1 } F = \frac{1} {\sqrt{\vert g \vert}} \partial_i \left( \sqrt { \vert g \vert } g^{ik} \partial_k F\right) = \frac{ \partial_{i_{1}} \left( \sqrt { \vert g \vert } g^{ {i_{1}} {i_{2}} } \partial_{i_{2}} F\right) } {\sqrt{\vert g \vert}} = \sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac{ \partial_{i_{1}} \left( \sqrt { \vert g \vert } g^{ {i_{1}} {i_{2}} } \partial_{i_{2}} F\right) } {\sqrt{\vert g \vert}} $$ [CENTER] [U]Summary[/U][/CENTER] So, in summary, my final results are that, 1, $$\nabla^3 F = \sum\limits^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } ( \sum\limits^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac{ \partial_{i_{1}} \left( \sqrt { \vert g \vert } g^{ {i_{1}} {i_{2}} } \partial_{i_{2}} F\right) } {\sqrt{\vert g \vert}} )\right ) } {\sqrt {\vert g \vert}}$$ in any general basis, while in the Cartesian Basis, it is $$\nabla^3 F = \sum_{i = 1}^{ m } \frac {\partial^3 F } {\partial x_i^3} $$ Likewise, $$\nabla^n F = \sum^{ m }_{\begin{align} { i_{1} = 1 } \nonumber \\ { i_{2} = 1 } \nonumber \\ \end{align}} \frac {\partial_{ i_{1} } \left( \sqrt {\vert g \vert} g^{ i_{1} i_{2} } \partial_{ i_{2} } (\nabla^{ n-1 } F)\right ) } {\sqrt {\vert g \vert}}$$ in any general basis, while in the Cartesian Basis, it is $$\nabla^{n} F = \sum_{i=1}^{m} \frac{\partial^n F}{\partial^n x_{i}}$$ [/QUOTE]
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