From 3138ead6bfd6e88e8cdbf9e4c32df64e18bc4595 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Fri, 9 Jun 2023 19:52:54 +0200 Subject: Improve knowledge base --- .../orthogonal-curvilinear-coordinates/index.md | 250 ++++++++++----------- 1 file changed, 125 insertions(+), 125 deletions(-) (limited to 'source/know/concept/orthogonal-curvilinear-coordinates/index.md') diff --git a/source/know/concept/orthogonal-curvilinear-coordinates/index.md b/source/know/concept/orthogonal-curvilinear-coordinates/index.md index 675b83a..c7299ee 100644 --- a/source/know/concept/orthogonal-curvilinear-coordinates/index.md +++ b/source/know/concept/orthogonal-curvilinear-coordinates/index.md @@ -910,131 +910,6 @@ Dot-multiplying by $$\vu{e}_j$$ isolates the $$c_j$$-component and gives the des -## Divergence of a tensor - -It also possible to take the divergence of a 2nd-order tensor $$\overline{\overline{\mathbf{T}}}$$, -yielding a vector with these components in $$(c_1, c_2, c_3)$$: - -$$\begin{aligned} - \boxed{ - (\nabla \cdot \overline{\overline{\mathbf{T}}})_j - = \sum_{k} \frac{1}{h_k} \pdv{T_{kj}}{c_k} - + \sum_{k \neq j} \frac{T_{jk}}{h_j h_k} \pdv{h_j}{c_k} - - \sum_{k \neq j} \frac{T_{kk}}{h_j h_k} \pdv{h_k}{c_j} - + \sum_{k} \sum_{l \neq k} \frac{T_{lj}}{h_k h_l} \pdv{h_k}{c_l} - } -\end{aligned}$$ - -{% include proof/start.html id="proof-div-tensor" -%} -From our earlier calculation of $$\nabla f$$, -we know how to express the del $$\nabla$$ in $$(c_1, c_2, c_3)$$. -Now we simply take the dot product and evaluate: - -$$\begin{aligned} - \nabla \cdot \overline{\overline{\mathbf{T}}} - &= \bigg( \vu{e}_1 \frac{1}{h_1} \pdv{}{c_1} + \vu{e}_2 \frac{1}{h_2} \pdv{}{c_2} + \vu{e}_3 \frac{1}{h_3} \pdv{}{c_3} \bigg) - \\ - &\quad\:\:\: \cdot \Big( T_{11} \vu{e}_1 \vu{e}_1 + T_{12} \vu{e}_1 \vu{e}_2 + T_{13} \vu{e}_1 \vu{e}_3 - \\ - &\qquad + T_{21} \vu{e}_2 \vu{e}_1 + T_{22} \vu{e}_2 \vu{e}_2 + T_{23} \vu{e}_2 \vu{e}_3 - \\ - &\qquad + T_{31} \vu{e}_3 \vu{e}_1 + T_{32} \vu{e}_3 \vu{e}_2 + T_{33} \vu{e}_3 \vu{e}_3 \Big) - \\ - &= \bigg( \sum_{j} \vu{e}_j \frac{1}{h_j} \pdv{}{c_j} \bigg) \cdot \bigg( \sum_{kl} T_{kl} \vu{e}_k \vu{e}_l \bigg) - \\ - &= \sum_{jkl} \vu{e}_j \cdot \frac{1}{h_j} \pdv{}{c_j} (T_{kl} \vu{e}_k \vu{e}_l) -\end{aligned}$$ - -We apply the product rule of differentiation -and use that $$\vb{c} \cdot (\vb{a} \vb{b}) = (\vb{c} \cdot \vb{a}) \vb{b}$$: - -$$\begin{aligned} - \nabla \cdot \overline{\overline{\mathbf{T}}} - &= \sum_{jkl} \bigg( (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{T_{kl}}{c_j} \vu{e}_l - + (\vu{e}_j \cdot \vu{e}_k) \frac{T_{kl}}{h_j} \pdv{\vu{e}_l}{c_j} - + \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg) - \\ - &= \sum_{jkl} \bigg( \delta_{jk} \frac{1}{h_j} \pdv{T_{kl}}{c_j} \vu{e}_l + \delta_{jk} \frac{T_{kl}}{h_j} \pdv{\vu{e}_l}{c_j} - + \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg) - \\ - &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j} - + \sum_{k} \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg) -\end{aligned}$$ - -Inserting our expressions for the derivatives of the basis vectors -in the last term, we find: - -$$\begin{aligned} - \nabla \cdot \overline{\overline{\mathbf{T}}} - &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j} - + \sum_{k} \vu{e}_j \cdot - \Big( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{m} \frac{1}{h_m} \pdv{h_k}{c_m} \vu{e}_m \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg) - \\ - &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j} - + \sum_{k} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l - - \sum_{m} (\vu{e}_j \cdot \vu{e}_m) \frac{T_{jl}}{h_j h_m} \pdv{h_j}{c_m} \vu{e}_l \bigg) - \\ - &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j} - + \sum_{k} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l - \frac{T_{jl}}{h_j h_j} \pdv{h_j}{c_j} \vu{e}_l \bigg) - \\ - &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j} - + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l \bigg) -\end{aligned}$$ - -Where we noticed that the latter two terms cancel out if $$k = j$$. -Next, rewriting $$\ipdv{\vu{e}_l}{c_j}$$: - -$$\begin{aligned} - \nabla \cdot \overline{\overline{\mathbf{T}}} - &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l - + \frac{T_{jl}}{h_j} \Big( \frac{1}{h_l} \pdv{h_j}{c_l} \vu{e}_j - \delta_{jl} \sum_{m} \frac{1}{h_m} \pdv{h_l}{c_m} \vu{e}_m \Big) - + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l \bigg) - \\ - &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l - + \frac{T_{jl}}{h_j h_l} \pdv{h_j}{c_l} \vu{e}_j - \delta_{jl} \sum_{m} \frac{T_{jl}}{h_j h_m} \pdv{h_l}{c_m} \vu{e}_m - + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l \bigg) - \\ - &= \sum_{jl} \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l - + \sum_{jl} \frac{T_{jl}}{h_j h_l} \pdv{h_j}{c_l} \vu{e}_j - - \sum_{jm} \frac{T_{jj}}{h_j h_m} \pdv{h_j}{c_m} \vu{e}_m - + \sum_{jl} \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l -\end{aligned}$$ - -Renaming the indices such that each term contains $$\vu{e}_l$$, -we arrive at the full result: - -$$\begin{aligned} - \nabla \cdot \overline{\overline{\mathbf{T}}} - &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} - + \frac{T_{lj}}{h_j h_l} \pdv{h_l}{c_j} - - \frac{T_{jj}}{h_j h_l} \pdv{h_j}{c_l} - + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_l -\end{aligned}$$ - -To isolate the $$c_m$$-component, we dot-multiply by $$\vu{e}_m$$ -and resolve the Kronecker delta $$\delta_{lm}$$: - -$$\begin{aligned} - (\nabla \cdot \overline{\overline{\mathbf{T}}})_m - &= (\nabla \cdot \overline{\overline{\mathbf{T}}}) \cdot \vu{e}_m - \\ - &= \sum_{jl} \delta_{lm} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} - + \frac{T_{lj}}{h_j h_l} \pdv{h_l}{c_j} - - \frac{T_{jj}}{h_j h_l} \pdv{h_j}{c_l} - + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \bigg) - \\ - &= \sum_{j} \frac{1}{h_j} \pdv{T_{jm}}{c_j} - + \sum_{j} \frac{T_{mj}}{h_j h_m} \pdv{h_m}{c_j} - - \sum_{j} \frac{T_{jj}}{h_j h_m} \pdv{h_j}{c_m} - + \sum_{j} \sum_{k \neq j} \frac{T_{km}}{h_j h_k} \pdv{h_j}{c_k} -\end{aligned}$$ - -The second and third terms cancel out for $$j = m$$, -so we can sum over $$j \neq m$$ instead. -{% include proof/end.html id="proof-div-tensor" %} - - - ## Laplacian of a vector The Laplacian $$\nabla^2 \vb{V}$$ of a vector $$\vb{V}$$ @@ -1168,6 +1043,131 @@ Which gives the desired formula after some simple index renaming and rearranging +## Divergence of a tensor + +It also possible to take the divergence of a 2nd-order tensor $$\overline{\overline{\mathbf{T}}}$$, +yielding a vector with these components in $$(c_1, c_2, c_3)$$: + +$$\begin{aligned} + \boxed{ + (\nabla \cdot \overline{\overline{\mathbf{T}}})_j + = \sum_{k} \frac{1}{h_k} \pdv{T_{kj}}{c_k} + + \sum_{k \neq j} \frac{T_{jk}}{h_j h_k} \pdv{h_j}{c_k} + - \sum_{k \neq j} \frac{T_{kk}}{h_j h_k} \pdv{h_k}{c_j} + + \sum_{k} \sum_{l \neq k} \frac{T_{lj}}{h_k h_l} \pdv{h_k}{c_l} + } +\end{aligned}$$ + +{% include proof/start.html id="proof-div-tensor" -%} +From our earlier calculation of $$\nabla f$$, +we know how to express the del $$\nabla$$ in $$(c_1, c_2, c_3)$$. +Now we simply take the dot product and evaluate: + +$$\begin{aligned} + \nabla \cdot \overline{\overline{\mathbf{T}}} + &= \bigg( \vu{e}_1 \frac{1}{h_1} \pdv{}{c_1} + \vu{e}_2 \frac{1}{h_2} \pdv{}{c_2} + \vu{e}_3 \frac{1}{h_3} \pdv{}{c_3} \bigg) + \\ + &\quad\:\:\: \cdot \Big( T_{11} \vu{e}_1 \vu{e}_1 + T_{12} \vu{e}_1 \vu{e}_2 + T_{13} \vu{e}_1 \vu{e}_3 + \\ + &\qquad + T_{21} \vu{e}_2 \vu{e}_1 + T_{22} \vu{e}_2 \vu{e}_2 + T_{23} \vu{e}_2 \vu{e}_3 + \\ + &\qquad + T_{31} \vu{e}_3 \vu{e}_1 + T_{32} \vu{e}_3 \vu{e}_2 + T_{33} \vu{e}_3 \vu{e}_3 \Big) + \\ + &= \bigg( \sum_{j} \vu{e}_j \frac{1}{h_j} \pdv{}{c_j} \bigg) \cdot \bigg( \sum_{kl} T_{kl} \vu{e}_k \vu{e}_l \bigg) + \\ + &= \sum_{jkl} \vu{e}_j \cdot \frac{1}{h_j} \pdv{}{c_j} (T_{kl} \vu{e}_k \vu{e}_l) +\end{aligned}$$ + +We apply the product rule of differentiation +and use that $$\vb{c} \cdot (\vb{a} \vb{b}) = (\vb{c} \cdot \vb{a}) \vb{b}$$: + +$$\begin{aligned} + \nabla \cdot \overline{\overline{\mathbf{T}}} + &= \sum_{jkl} \bigg( (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{T_{kl}}{c_j} \vu{e}_l + + (\vu{e}_j \cdot \vu{e}_k) \frac{T_{kl}}{h_j} \pdv{\vu{e}_l}{c_j} + + \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg) + \\ + &= \sum_{jkl} \bigg( \delta_{jk} \frac{1}{h_j} \pdv{T_{kl}}{c_j} \vu{e}_l + \delta_{jk} \frac{T_{kl}}{h_j} \pdv{\vu{e}_l}{c_j} + + \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg) + \\ + &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j} + + \sum_{k} \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg) +\end{aligned}$$ + +Inserting our expressions for the derivatives of the basis vectors +in the last term, we find: + +$$\begin{aligned} + \nabla \cdot \overline{\overline{\mathbf{T}}} + &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j} + + \sum_{k} \vu{e}_j \cdot + \Big( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{m} \frac{1}{h_m} \pdv{h_k}{c_m} \vu{e}_m \Big) \frac{T_{kl}}{h_j} \vu{e}_l \bigg) + \\ + &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j} + + \sum_{k} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l + - \sum_{m} (\vu{e}_j \cdot \vu{e}_m) \frac{T_{jl}}{h_j h_m} \pdv{h_j}{c_m} \vu{e}_l \bigg) + \\ + &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j} + + \sum_{k} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l - \frac{T_{jl}}{h_j h_j} \pdv{h_j}{c_j} \vu{e}_l \bigg) + \\ + &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + \frac{T_{jl}}{h_j} \pdv{\vu{e}_l}{c_j} + + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l \bigg) +\end{aligned}$$ + +Where we noticed that the latter two terms cancel out if $$k = j$$. +Next, rewriting $$\ipdv{\vu{e}_l}{c_j}$$: + +$$\begin{aligned} + \nabla \cdot \overline{\overline{\mathbf{T}}} + &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + + \frac{T_{jl}}{h_j} \Big( \frac{1}{h_l} \pdv{h_j}{c_l} \vu{e}_j - \delta_{jl} \sum_{m} \frac{1}{h_m} \pdv{h_l}{c_m} \vu{e}_m \Big) + + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l \bigg) + \\ + &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + + \frac{T_{jl}}{h_j h_l} \pdv{h_j}{c_l} \vu{e}_j - \delta_{jl} \sum_{m} \frac{T_{jl}}{h_j h_m} \pdv{h_l}{c_m} \vu{e}_m + + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l \bigg) + \\ + &= \sum_{jl} \frac{1}{h_j} \pdv{T_{jl}}{c_j} \vu{e}_l + + \sum_{jl} \frac{T_{jl}}{h_j h_l} \pdv{h_j}{c_l} \vu{e}_j + - \sum_{jm} \frac{T_{jj}}{h_j h_m} \pdv{h_j}{c_m} \vu{e}_m + + \sum_{jl} \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \vu{e}_l +\end{aligned}$$ + +Renaming the indices such that each term contains $$\vu{e}_l$$, +we arrive at the full result: + +$$\begin{aligned} + \nabla \cdot \overline{\overline{\mathbf{T}}} + &= \sum_{jl} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} + + \frac{T_{lj}}{h_j h_l} \pdv{h_l}{c_j} + - \frac{T_{jj}}{h_j h_l} \pdv{h_j}{c_l} + + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \bigg) \vu{e}_l +\end{aligned}$$ + +To isolate the $$c_m$$-component, we dot-multiply by $$\vu{e}_m$$ +and resolve the Kronecker delta $$\delta_{lm}$$: + +$$\begin{aligned} + (\nabla \cdot \overline{\overline{\mathbf{T}}})_m + &= (\nabla \cdot \overline{\overline{\mathbf{T}}}) \cdot \vu{e}_m + \\ + &= \sum_{jl} \delta_{lm} \bigg( \frac{1}{h_j} \pdv{T_{jl}}{c_j} + + \frac{T_{lj}}{h_j h_l} \pdv{h_l}{c_j} + - \frac{T_{jj}}{h_j h_l} \pdv{h_j}{c_l} + + \sum_{k \neq j} \frac{T_{kl}}{h_j h_k} \pdv{h_j}{c_k} \bigg) + \\ + &= \sum_{j} \frac{1}{h_j} \pdv{T_{jm}}{c_j} + + \sum_{j} \frac{T_{mj}}{h_j h_m} \pdv{h_m}{c_j} + - \sum_{j} \frac{T_{jj}}{h_j h_m} \pdv{h_j}{c_m} + + \sum_{j} \sum_{k \neq j} \frac{T_{km}}{h_j h_k} \pdv{h_j}{c_k} +\end{aligned}$$ + +The second and third terms cancel out for $$j = m$$, +so we can sum over $$j \neq m$$ instead. +{% include proof/end.html id="proof-div-tensor" %} + + + ## References 1. M.L. Boas, *Mathematical methods in the physical sciences*, 2nd edition, -- cgit v1.2.3