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Diffstat (limited to 'content/know/concept/curvilinear-coordinates/index.pdc')
| -rw-r--r-- | content/know/concept/curvilinear-coordinates/index.pdc | 154 |
1 files changed, 95 insertions, 59 deletions
diff --git a/content/know/concept/curvilinear-coordinates/index.pdc b/content/know/concept/curvilinear-coordinates/index.pdc index e1c0465..925eda3 100644 --- a/content/know/concept/curvilinear-coordinates/index.pdc +++ b/content/know/concept/curvilinear-coordinates/index.pdc @@ -50,7 +50,7 @@ and [parabolic cylindrical coordinates](/know/concept/parabolic-cylindrical-coor In the following subsections, we derive general formulae to convert expressions -from Cartesian coordinates in the new orthogonal system $(x_1, x_2, x_3)$. +from Cartesian coordinates to the new orthogonal system $(x_1, x_2, x_3)$. ## Basis vectors @@ -93,7 +93,26 @@ $$\begin{aligned} ## Gradient -For a given direction $\dd{\ell}$, we know that +In an orthogonal coordinate system, +the gradient $\nabla f$ of a scalar $f$ is as follows, +where $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$ +are the basis unit vectors respectively corresponding to $x_1$, $x_2$ and $x_3$: + +$$\begin{gathered} + \boxed{ + \nabla f + = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1} + + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2} + + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3} + } +\end{gathered}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-grad"/> +<label for="proof-grad">Proof</label> +<div class="hidden"> +<label for="proof-grad">Proof.</label> +For a direction $\dd{\ell}$, we know that $\dv*{f}{\ell}$ is the component of $\nabla f$ in that direction: $$\begin{aligned} @@ -104,7 +123,7 @@ $$\begin{aligned} \end{aligned}$$ Where $\vu{u}$ is simply a unit vector in the direction of $\dd{\ell}$. -We can thus find an expression for the gradient $\nabla f$ +We thus find the expression for the gradient $\nabla f$ by choosing $\dd{\ell}$ to be $h_1 \dd{x_1}$, $h_2 \dd{x_2}$ and $h_3 \dd{x_3}$ in turn: $$\begin{gathered} @@ -112,49 +131,59 @@ $$\begin{gathered} = \vu{e}_1 \dv{x_1}{\ell} \pdv{f}{x_1} + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2} + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3} - \\ - \boxed{ - \nabla f - = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1} - + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2} - + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3} - } \end{gathered}$$ - -Where $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$ -are the basis unit vectors respectively corresponding to $x_1$, $x_2$ and $x_3$. +</div> +</div> ## Divergence -Consider a vector $\vb{V}$ in the target coordinate system -with components $V_1$, $V_2$ and $V_3$: +The divergence of a vector $\vb{V} = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3$ +in an orthogonal system is given by: + +$$\begin{aligned} + \boxed{ + \nabla \cdot \vb{V} + = \frac{1}{h_1 h_2 h_3} + \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big) + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-div"/> +<label for="proof-div">Proof</label> +<div class="hidden"> +<label for="proof-div">Proof.</label> +As preparation, we rewrite $\vb{V}$ as follows +to introduce the scale factors: $$\begin{aligned} \vb{V} - &= \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3 - \\ &= \vu{e}_1 \frac{1}{h_2 h_3} (h_2 h_3 V_1) + \vu{e}_2 \frac{1}{h_1 h_3} (h_1 h_3 V_2) + \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3) \end{aligned}$$ -We take only the $\vu{e}_1$-component of this vector, -and expand its divergence using a vector identity, -where $f = h_2 h_3 V_1$ is a scalar -and $\vb{U} = \vu{e}_1 / (h_2 h_3)$ is a vector: +We start by taking only the $\vu{e}_1$-component of this vector, +and expand its divergence using the following vector identity: $$\begin{gathered} \nabla \cdot (\vb{U} \: f) = \vb{U} \cdot (\nabla f) + (\nabla \cdot \vb{U}) \: f - \\ +\end{gathered}$$ + +Inserting the scalar $f = h_2 h_3 V_1$ +the vector $\vb{U} = \vu{e}_1 / (h_2 h_3)$, +we arrive at: + +$$\begin{gathered} \nabla \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} (h_2 h_3 V_1) \Big) = \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big) + \Big( \nabla \cdot \frac{\vu{e}_1}{h_2 h_3} \Big) (h_2 h_3 V_1) \end{gathered}$$ -The first term is straightforward to calculate -thanks to our preceding expression for the gradient. +The first right-hand term is easy to calculate +thanks to our expression for the gradient $\nabla f$. Only the $\vu{e}_1$-component survives due to the dot product: $$\begin{aligned} @@ -162,8 +191,8 @@ $$\begin{aligned} = \frac{\vu{e}_1}{h_1 h_2 h_3} \pdv{(h_2 h_3 V_1)}{x_1} \end{aligned}$$ -The second term is a bit more involved. -To begin with, we use the gradient formula to note that: +The second term is more involved. +First, we use the gradient formula to observe that: $$\begin{aligned} \nabla x_1 @@ -177,7 +206,7 @@ $$\begin{aligned} \end{aligned}$$ Because $\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$ in an orthogonal basis, -we can get the vector whose divergence we want: +these gradients can be used to express the vector whose divergence we want: $$\begin{aligned} \nabla x_2 \cross \nabla x_3 @@ -196,15 +225,9 @@ $$\begin{aligned} \end{aligned}$$ After repeating this procedure for the other components of $\vb{V}$, -we arrive at the following general expression for the divergence $\nabla \cdot \vb{V}$: - -$$\begin{aligned} - \boxed{ - \nabla \cdot \vb{V} - = \frac{1}{h_1 h_2 h_3} - \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big) - } -\end{aligned}$$ +we get the desired general expression for the divergence. +</div> +</div> ## Laplacian @@ -229,31 +252,55 @@ $$\begin{aligned} ## Curl -We find the curl in a similar way as the divergence. -Consider an arbitrary vector $\vb{V}$: +The curl of a vector $\vb{V}$ is as follows +in a general orthogonal curvilinear system: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \nabla \times \vb{V} + &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big) + \\ + &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big) + \\ + &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big) + \end{aligned} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-curl"/> +<label for="proof-curl">Proof</label> +<div class="hidden"> +<label for="proof-curl">Proof.</label> +The curl is found in a similar way as the divergence. +We rewrite $\vb{V}$ like so: $$\begin{aligned} \vb{V} - = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3 = \frac{\vu{e}_1}{h_1} (h_1 V_1) + \frac{\vu{e}_2}{h_2} (h_2 V_2) + \frac{\vu{e}_3}{h_3} (h_3 V_3) \end{aligned}$$ -We expand the curl of its $\vu{e}_1$-component using a vector identity, -where $f = h_1 V_1$ is a scalar and $\vb{U} = \vu{e}_1 / h_1$ is a vector: +We expand the curl of its $\vu{e}_1$-component using the following vector identity: $$\begin{gathered} \nabla \cross (\vb{U} \: f) = (\nabla \cross \vb{U}) \: f - \vb{U} \cross (\nabla f) - \\ +\end{gathered}$$ + +Inserting the scalar $f = h_1 V_1$ +and the vector $\vb{U} = \vu{e}_1 / h_1$, we arrive at: + +$$\begin{gathered} \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big) = \Big( \nabla \cross \frac{\vu{e}_1}{h_1} \Big) (h_1 V_1) - \frac{\vu{e}_1}{h_1} \cross \Big( \nabla (h_1 V_1) \Big) \end{gathered}$$ -Previously, when calculating the divergence, +Previously, when proving the divergence, we already showed that $\vu{e}_1 / h_1 = \nabla x_1$. Because the curl of a gradient is zero, -the first term thus disappears, leaving only the second, -which contains a gradient turning out to be: +the first term disappears, leaving only the second, +which contains a gradient that turns out to be: $$\begin{aligned} \nabla (h_1 V_1) @@ -273,20 +320,9 @@ $$\begin{aligned} \end{aligned}$$ If we go through the same process for the other components of $\vb{V}$ -and add the results together, we get the following expression for the curl $\nabla \cross \vb{V}$: - -$$\begin{aligned} - \boxed{ - \begin{aligned} - \nabla \times \vb{V} - &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big) - \\ - &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big) - \\ - &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big) - \end{aligned} - } -\end{aligned}$$ +and add up the results, we get the desired expression for the curl. +</div> +</div> ## Differential elements |