From 0b6bada15afc0a3477316427e3fa145e78699d0c Mon Sep 17 00:00:00 2001 From: Prefetch Date: Mon, 5 Jun 2023 21:40:44 +0200 Subject: Improve knowledge base --- .../orthogonal-curvilinear-coordinates/index.md | 241 +++++++++++++++++---- 1 file changed, 197 insertions(+), 44 deletions(-) (limited to 'source/know') diff --git a/source/know/concept/orthogonal-curvilinear-coordinates/index.md b/source/know/concept/orthogonal-curvilinear-coordinates/index.md index 4fb45b4..675b83a 100644 --- a/source/know/concept/orthogonal-curvilinear-coordinates/index.md +++ b/source/know/concept/orthogonal-curvilinear-coordinates/index.md @@ -43,12 +43,12 @@ $$\begin{aligned} &= z(c_1, c_2, c_3) \end{aligned}$$ -A useful attribute of a coordinate system is its **line element** $$\dd{\vu{\ell}}$$, +A useful attribute of a coordinate system is its **line element** $$\dd{\vb{\ell}}$$, which represents the differential element of a line in any direction. Let $$\vu{e}_x$$, $$\vu{e}_y$$ and $$\vu{e}_z$$ be the Cartesian basis unit vectors: $$\begin{aligned} - \dd{\vu{\ell}} + \dd{\vb{\ell}} \equiv \vu{e}_x \dd{x} + \: \vu{e}_y \dd{y} + \: \vu{e}_z \dd{z} \end{aligned}$$ @@ -56,7 +56,7 @@ The Cartesian differential elements can be rewritten in $$(c_1, c_2, c_3)$$ with the chain rule: $$\begin{aligned} - \dd{\vu{\ell}} + \dd{\vb{\ell}} = \quad &\vu{e}_x \bigg( \pdv{x}{c_1} \dd{c_1} + \: \pdv{x}{c_2} \dd{c_2} + \: \pdv{x}{c_3} \dd{c_3} \!\bigg) \\ + \: &\vu{e}_y \bigg( \pdv{y}{c_1} \dd{c_1} + \: \pdv{y}{c_2} \dd{c_2} + \: \pdv{y}{c_3} \dd{c_3} \!\bigg) @@ -93,6 +93,7 @@ and orthogonal for any orthogonal curvilinear system. They are called *local* basis vectors because they generally depend on $$(c_1, c_2, c_3)$$, i.e. their directions vary from position to position. +Their definitions can also be inverted: $$\begin{aligned} \boxed{ @@ -109,7 +110,6 @@ $$\begin{aligned} } \end{aligned}$$ - In the following subsections, we use the scale factors $$h_1$$, $$h_2$$ and $$h_3$$ to derive general formulae for converting vector calculus from Cartesian coordinates to $$(c_1, c_2, c_3)$$. @@ -121,12 +121,12 @@ from Cartesian coordinates to $$(c_1, c_2, c_3)$$. The point of the scale factors $$h_1$$, $$h_2$$ and $$h_3$$, as can be seen from their derivation, is to correct for "distortions" of the coordinates compared to the Cartesian system, -such that the line element $$\dd{\vu{\ell}}$$ retains its length. +such that the line element $$\dd{\vb{\ell}}$$ retains its length. As was already established above: $$\begin{aligned} \boxed{ - \dd{\vu{\ell}} + \dd{\vb{\ell}} = \vu{e}_1 h_1 \dd{c_1} + \: \vu{e}_2 h_2 \dd{c_2} + \: \vu{e}_3 h_3 \dd{c_3} } \end{aligned}$$ @@ -148,7 +148,7 @@ can be expressed in terms of $$\dd{}_1\!\vb{x}$$, $$\dd{}_2\!\vb{x}$$ and $$\dd{ The differential normal vector element $$\dd{\vu{S}}$$ in a surface integral is hence given by: $$\begin{aligned} - \dd{\vu{S}} + \dd{\vb{S}} &= \dd{}_1\!\vb{x} \cross \dd{}_2\!\vb{x} + \dd{}_2\!\vb{x} \cross \dd{}_3\!\vb{x} + \dd{}_3\!\vb{x} \cross \dd{}_1\!\vb{x} \\ &= (\vu{e}_1 \cross \vu{e}_2) \: h_1 h_2 \dd{c_1} \dd{c_2} @@ -163,7 +163,7 @@ $$\vu{e}_3 \cross \vu{e}_1 = \vu{e}_2$$, so: $$\begin{aligned} \boxed{ - \dd{\vu{S}} + \dd{\vb{S}} = \vu{e}_1 \: h_2 h_3 \dd{c_2} \dd{c_3} + \: \vu{e}_2 \: h_1 h_3 \dd{c_1} \dd{c_3} + \: \vu{e}_3 \: h_1 h_2 \dd{c_1} \dd{c_2} } \end{aligned}$$ @@ -357,8 +357,8 @@ because $$\ipdv{\vu{e}_1}{c_1}$$ must be orthogonal to $$\vu{e}_1$$. ## Gradient of a scalar -In $$(c_1, c_2, c_3)$$, the gradient $$\nabla f$$ of a scalar field $$f$$ -has the following components: +The gradient $$\nabla f$$ of a scalar field $$f$$ +has the following components in $$(c_1, c_2, c_3)$$: $$\begin{aligned} \boxed{ @@ -483,13 +483,13 @@ Now, to proceed, it is easiest to just write out the index notation: $$\begin{aligned} \nabla \cdot \vb{V} - &= \quad\: \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j} + \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} + &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j} + \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} \\ - &= \quad\: \frac{1}{h_1} \pdv{V_1}{c_1} + \frac{V_1}{h_1 h_2} \pdv{h_2}{c_1} + \frac{V_1}{h_1 h_3} \pdv{h_3}{c_1} + &= \quad \frac{1}{h_1} \pdv{V_1}{c_1} + \frac{V_1}{h_1 h_2} \pdv{h_2}{c_1} + \frac{V_1}{h_1 h_3} \pdv{h_3}{c_1} \\ - &\quad\:\: + \frac{V_2}{h_1 h_2} \pdv{h_1}{c_2} + \frac{1}{h_2} \pdv{V_2}{c_2} + \frac{V_2}{h_2 h_3} \pdv{h_3}{c_2} + &\quad\: + \frac{V_2}{h_1 h_2} \pdv{h_1}{c_2} + \frac{1}{h_2} \pdv{V_2}{c_2} + \frac{V_2}{h_2 h_3} \pdv{h_3}{c_2} \\ - &\quad\:\: + \frac{V_3}{h_1 h_3} \pdv{h_1}{c_3} + \frac{V_3}{h_2 h_3} \pdv{h_2}{c_3} + \frac{1}{h_3} \pdv{V_3}{c_3} + &\quad\: + \frac{V_3}{h_1 h_3} \pdv{h_1}{c_3} + \frac{V_3}{h_2 h_3} \pdv{h_2}{c_3} + \frac{1}{h_3} \pdv{V_3}{c_3} \\ &= \frac{1}{h_1 h_2 h_3} \bigg( h_2 h_3 \pdv{V_1}{c_1} + h_3 V_1 \pdv{h_2}{c_1} + h_2 V_1 \pdv{h_3}{c_1} \\ @@ -550,36 +550,6 @@ and simply add up the results to get the desired formula. -## Laplacian of a scalar - -The Laplacian $$\nabla^2 f$$ of a scalar field $$f$$ -is calculated as follows in $$(c_1, c_2, c_3)$$: - -$$\begin{aligned} - \boxed{ - \nabla^2 f - = \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{f}{c_j} \bigg) - } -\end{aligned}$$ - -Where $$H \equiv h_1 h_2 h_3$$. -When this index notation is written out in full, it becomes: - -$$\begin{aligned} - \nabla^2 f - = \frac{1}{h_1 h_2 h_3} - \bigg( - \pdv{}{c_1}\Big(\! \frac{h_2 h_3}{h_1} \pdv{f}{c_1} \!\Big) - + \pdv{}{c_2}\Big(\! \frac{h_1 h_3}{h_2} \pdv{f}{c_2} \!\Big) - + \pdv{}{c_3}\Big(\! \frac{h_1 h_2}{h_3} \pdv{f}{c_3} \!\Big) - \bigg) -\end{aligned}$$ - -This is trivial to prove: $$\nabla^2 f = \nabla \cdot (\nabla f)$$, -so combining our previous formulas is enough. - - - ## Curl of a vector The curl of a vector field $$\vb{V}$$ has the following components in $$(c_1, c_2, c_3)$$, @@ -703,6 +673,56 @@ and simply add up the results to get the desired formula. +## Laplacian of a scalar + +The Laplacian $$\nabla^2 f$$ of a scalar field $$f$$ +is calculated as follows in $$(c_1, c_2, c_3)$$: + +$$\begin{aligned} + \boxed{ + \nabla^2 f + = \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{f}{c_j} \bigg) + } +\end{aligned}$$ + +Where $$H \equiv h_1 h_2 h_3$$. +When this index notation is written out in full, it becomes: + +$$\begin{aligned} + \nabla^2 f + = \frac{1}{h_1 h_2 h_3} + \bigg( + \pdv{}{c_1}\Big(\! \frac{h_2 h_3}{h_1} \pdv{f}{c_1} \!\Big) + + \pdv{}{c_2}\Big(\! \frac{h_1 h_3}{h_2} \pdv{f}{c_2} \!\Big) + + \pdv{}{c_3}\Big(\! \frac{h_1 h_2}{h_3} \pdv{f}{c_3} \!\Big) + \bigg) +\end{aligned}$$ + +This is trivial to prove: $$\nabla^2 f = \nabla \cdot (\nabla f)$$, +so combining our previous formulae is enough. + + + +## Gradient of a divergence + +The gradient of a divergence $$\nabla (\nabla \cdot \vb{V})$$ +has the following components in $$(c_1, c_2, c_3)$$: + +$$\begin{aligned} + \boxed{ + \big( \nabla (\nabla \cdot \vb{V}) \big)_j + = \frac{1}{h_j} \pdv{}{c_j} \bigg( \sum_{k} \frac{1}{H} \pdv{}{c_k} \Big( \frac{H V_k}{h_k} \Big) \bigg) + } +\end{aligned}$$ + +Where $$H \equiv h_1 h_2 h_3$$. +This is trivial to prove: $$\nabla \cdot \vb{V}$$ is a scalar, +which we insert into our gradient formula. +We no longer write out the index notation, +as the formulae become quite long. + + + ## Gradient of a vector It also possible to take the gradient of a vector @@ -1015,6 +1035,139 @@ so we can sum over $$j \neq m$$ instead. +## Laplacian of a vector + +The Laplacian $$\nabla^2 \vb{V}$$ of a vector $$\vb{V}$$ +has the following components in $$(c_1, c_2, c_3)$$: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + (\nabla^2 \vb{V})_j + &= \sum_{k} \frac{1}{H} \pdv{}{c_k} \bigg( \frac{H}{h_k^2} \pdv{V_j}{c_k} \bigg) + \\ + &\quad\: + \sum_{k \neq j} \frac{1}{H} \bigg( \pdv{}{c_j} \Big( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \Big) + - \pdv{}{c_k} \Big( \frac{H V_k}{h_j h_k^2} \pdv{h_k}{c_j} \Big) \bigg) + \\ + &\quad\: + \sum_{k \neq j} \frac{1}{h_j h_k} \bigg( \frac{1}{h_j} \pdv{V_k}{c_j} \pdv{h_j}{c_k} + - \frac{1}{h_k} \pdv{V_k}{c_k} \pdv{h_k}{c_j} \bigg) + \\ + &\quad\:- \sum_{k \neq j} \frac{1}{h_j h_k} \bigg( \frac{V_j}{h_j h_k} \Big( \pdv{h_j}{c_k} \Big)^2 + + \sum_{l \neq k} \frac{V_l}{h_k h_l} \pdv{h_k}{c_l} \pdv{h_k}{c_j} \bigg) + \end{aligned} + } +\end{aligned}$$ + + +{% include proof/start.html id="proof-lap-vector" -%} +We already know how to calculate the Laplacian $$\nabla^2 f$$ of a scalar. +From that, we read out the $$\nabla^2$$-operator +and apply it to a vector $$\vb{V}$$ instead: + +$$\begin{aligned} + \nabla^2 \vb{V} + &= \bigg( \sum_{j} \frac{1}{H} \pdv{}{c_j} \Big( \frac{H}{h_j^2} \pdv{}{c_j} \Big) \bigg) + \bigg( \sum_{k} V_k \vu{e}_k \bigg) + \\ + &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{}{c_j} (V_k \vu{e}_k) \bigg) + \\ + &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \vu{e}_k + \frac{H V_k}{h_j^2} \pdv{\vu{e}_k}{c_j} \bigg) + \\ + &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \vu{e}_k + + \frac{H V_k}{h_j^2} \Big( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_k}{c_l} \vu{e}_l \Big) \bigg) + \\ + &= \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \sum_{k} \frac{H}{h_j^2} \pdv{V_k}{c_j} \vu{e}_k + + \sum_{k} \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \vu{e}_j + - \sum_{l} \frac{H V_j}{h_j^2 h_l} \pdv{h_j}{c_l} \vu{e}_l \bigg) + \\ + &= \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \sum_{k} \frac{H}{h_j^2} \pdv{V_k}{c_j} \vu{e}_k + + \sum_{k \neq j} \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \vu{e}_j + - \sum_{k \neq j} \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \vu{e}_k \bigg) +\end{aligned}$$ + +Where we have noticed that the latter two terms cancel out if $$j = k$$. +We expand according to the product rule of differentiation: + +$$\begin{aligned} + \nabla^2 \vb{V} + &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \bigg) \vu{e}_k + + \sum_{jk} \frac{1}{H} \frac{H}{h_j^2} \pdv{V_k}{c_j} \pdv{\vu{e}_k}{c_j} + \\ + &\quad\: + \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_j + + \sum_{j} \sum_{k \neq j} \frac{1}{H} \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \pdv{\vu{e}_j}{c_j} + \\ + &\quad\: - \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_k + - \sum_{j} \sum_{k \neq j} \frac{1}{H} \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \pdv{\vu{e}_k}{c_j} +\end{aligned}$$ + +Substituting our expression for the derivatives of the local basis vectors, we find: + +$$\begin{aligned} + \nabla^2 \vb{V} + &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \bigg) \vu{e}_k + + \sum_{jk} \frac{1}{h_j^2} \pdv{V_k}{c_j} + \bigg( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_k}{c_l} \vu{e}_l \bigg) + \\ + &\quad\: + \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_j + - \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg( \sum_{l \neq j} \frac{1}{h_l} \pdv{h_j}{c_l} \vu{e}_l \bigg) + \\ + &\quad\: - \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_k + - \sum_{j} \sum_{k \neq j} \frac{V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j \bigg) + \\ + &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \bigg) \vu{e}_k + + \sum_{jk} \frac{1}{h_j^2 h_k} \pdv{V_k}{c_j} \pdv{h_j}{c_k} \vu{e}_j + - \sum_{jl} \frac{1}{h_j^2 h_l} \pdv{V_j}{c_j} \pdv{h_j}{c_l} \vu{e}_l + \\ + &\quad\: + \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_j + - \sum_{j} \sum_{k \neq j} \sum_{l \neq j} \frac{V_k}{h_j^2 h_k h_l} \pdv{h_j}{c_k} \pdv{h_j}{c_l} \vu{e}_l + \\ + &\quad\: - \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_k + - \sum_{j} \sum_{k \neq j} \frac{V_j}{h_j^2 h_k^2} \bigg( \pdv{h_j}{c_k} \bigg)^2 \vu{e}_j + \\ + &= \sum_{jk} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \bigg) \vu{e}_k + + \sum_{j} \sum_{k \neq j} \frac{1}{h_j^2 h_k} \pdv{V_k}{c_j} \pdv{h_j}{c_k} \vu{e}_j + - \sum_{j} \sum_{k \neq j} \frac{1}{h_j^2 h_k} \pdv{V_j}{c_j} \pdv{h_j}{c_k} \vu{e}_k + \\ + &\quad\: + \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_j + - \sum_{j} \sum_{k \neq j} \sum_{l \neq j} \frac{V_k}{h_j^2 h_k h_l} \pdv{h_j}{c_k} \pdv{h_j}{c_l} \vu{e}_l + \\ + &\quad\: - \sum_{j} \sum_{k \neq j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_k + - \sum_{j} \sum_{k \neq j} \frac{V_j}{h_j^2 h_k^2} \bigg( \pdv{h_j}{c_k} \bigg)^2 \vu{e}_j +\end{aligned}$$ + +Where we have once again noticed that terms #2 and #3 cancel out if $$j = k$$. +Next, we isolate the $$c_m$$-component by dot-multiplying with $$\vu{e}_m$$: + +$$\begin{aligned} + (\nabla^2 \vb{V})_m + &= (\nabla^2 \vb{V}) \cdot \vu{e}_m + \\ + &= \sum_{jk} \frac{\delta_{km}}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_k}{c_j} \bigg) + + \sum_{j} \sum_{k \neq j} \frac{\delta_{jm}}{h_j^2 h_k} \pdv{V_k}{c_j} \pdv{h_j}{c_k} + - \sum_{j} \sum_{k \neq j} \frac{\delta_{km}}{h_j^2 h_k} \pdv{V_j}{c_j} \pdv{h_j}{c_k} + \\ + &\quad\: + \sum_{j} \sum_{k \neq j} \delta_{jm} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_k}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) + - \sum_{j} \sum_{k \neq j} \sum_{l \neq j} \delta_{lm} \frac{V_k}{h_j^2 h_k h_l} \pdv{h_j}{c_k} \pdv{h_j}{c_l} + \\ + &\quad\: - \sum_{j} \sum_{k \neq j} \delta_{km} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_k} \pdv{h_j}{c_k} \bigg) + - \sum_{j} \sum_{k \neq j} \delta_{jm} \frac{V_j}{h_j^2 h_k^2} \bigg( \pdv{h_j}{c_k} \bigg)^2 + \\ + &= \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H}{h_j^2} \pdv{V_m}{c_j} \bigg) + + \sum_{k \neq m} \frac{1}{h_k h_m^2} \pdv{V_k}{c_m} \pdv{h_m}{c_k} + - \sum_{j \neq m} \frac{1}{h_j^2 h_m} \pdv{V_j}{c_j} \pdv{h_j}{c_m} + \\ + &\quad\: + \sum_{k \neq m} \frac{1}{H} \pdv{}{c_m} \bigg( \frac{H V_k}{h_m^2 h_k} \pdv{h_m}{c_k} \bigg) + - \sum_{j \neq m} \sum_{k \neq j} \frac{V_k}{h_j^2 h_k h_m} \pdv{h_j}{c_k} \pdv{h_j}{c_m} + \\ + &\quad\: - \sum_{j \neq m} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j^2 h_m} \pdv{h_j}{c_m} \bigg) + - \sum_{k \neq m} \frac{V_m}{h_m^2 h_k^2} \bigg( \pdv{h_m}{c_k} \bigg)^2 +\end{aligned}$$ + +Which gives the desired formula after some simple index renaming and rearranging. +{% include proof/end.html id="proof-lap-vector" %} + + + ## References 1. M.L. Boas, *Mathematical methods in the physical sciences*, 2nd edition, -- cgit v1.2.3