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1 files changed, 52 insertions, 51 deletions
diff --git a/source/know/concept/curvilinear-coordinates/index.md b/source/know/concept/curvilinear-coordinates/index.md index 3012ca6..cb22e43 100644 --- a/source/know/concept/curvilinear-coordinates/index.md +++ b/source/know/concept/curvilinear-coordinates/index.md @@ -14,15 +14,15 @@ is known as a **coordinate surface**, and the intersections of the surfaces of different coordinates are called **coordinate lines**. A **curvilinear** coordinate system is one where at least one of the coordinate surfaces is curved, -e.g. in cylindrical coordinates the line between $r$ and $z$ is a circle. +e.g. in cylindrical coordinates the line between $$r$$ and $$z$$ is a circle. If the coordinate surfaces are mutually perpendicular, it is an **orthogonal** system, which is generally desirable. -A useful attribute of a coordinate system is its **line element** $\dd{\ell}$, +A useful attribute of a coordinate system is its **line element** $$\dd{\ell}$$, which represents the differential element of a line in any direction. -For an orthogonal system, its square $\dd{\ell}^2$ is calculated -by taking the differential elements of the old Cartesian $(x, y, z)$ system -and writing them out in the new $(x_1, x_2, x_3)$ system. +For an orthogonal system, its square $$\dd{\ell}^2$$ is calculated +by taking the differential elements of the old Cartesian $$(x, y, z)$$ system +and writing them out in the new $$(x_1, x_2, x_3)$$ system. The resulting expression will be of the form: $$\begin{aligned} @@ -33,7 +33,7 @@ $$\begin{aligned} } \end{aligned}$$ -Where $h_1$, $h_2$, and $h_3$ are called **scale factors**, +Where $$h_1$$, $$h_2$$, and $$h_3$$ are called **scale factors**, and need not be constants. The equation above only contains quadratic terms because the coordinate system is orthogonal by assumption. @@ -45,20 +45,20 @@ and [cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coor In the following subsections, we derive general formulae to convert expressions -from Cartesian coordinates to the new orthogonal system $(x_1, x_2, x_3)$. +from Cartesian coordinates to the new orthogonal system $$(x_1, x_2, x_3)$$. ## Basis vectors -Consider the the vector form of the line element $\dd{\ell}$, -denoted by $\dd{\vu{\ell}}$ and expressed as: +Consider the the vector form of the line element $$\dd{\ell}$$, +denoted by $$\dd{\vu{\ell}}$$ and expressed as: $$\begin{aligned} \dd{\vu{\ell}} = \vu{e}_x \dd{x} + \vu{e}_y \dd{y} + \vu{e}_z \dd{z} \end{aligned}$$ -We can expand the Cartesian differential elements, e.g. $\dd{y}$, +We can expand the Cartesian differential elements, e.g. $$\dd{y}$$, in the new basis as follows: $$\begin{aligned} @@ -66,17 +66,17 @@ $$\begin{aligned} = \pdv{y}{x_1} \dd{x_1} + \pdv{y}{x_2} \dd{x_2} + \pdv{y}{x_3} \dd{x_3} \end{aligned}$$ -If we write this out for $\dd{x}$, $\dd{y}$ and $\dd{z}$, -and group the terms according to $\dd{x}_1$, $\dd{x}_2$ and $\dd{x}_3$, -we can compare it the alternative form of $\dd{\vu{\ell}}$: +If we write this out for $$\dd{x}$$, $$\dd{y}$$ and $$\dd{z}$$, +and group the terms according to $$\dd{x}_1$$, $$\dd{x}_2$$ and $$\dd{x}_3$$, +we can compare it the alternative form of $$\dd{\vu{\ell}}$$: $$\begin{aligned} \dd{\vu{\ell}} = \vu{e}_1 \:h_1 \dd{x_1} + \vu{e}_2 \:h_2 \dd{x_2} + \vu{e}_3 \:h_3 \dd{x_4} \end{aligned}$$ -From this, we can read off $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$. -Here we only give $\vu{e}_1$, since $\vu{e}_2$ and $\vu{e}_3$ are analogous: +From this, we can read off $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$. +Here we only give $$\vu{e}_1$$, since $$\vu{e}_2$$ and $$\vu{e}_3$$ are analogous: $$\begin{aligned} \boxed{ @@ -89,9 +89,9 @@ $$\begin{aligned} ## Gradient 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 $x_3$: +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 $$x_3$$: $$\begin{gathered} \boxed{ @@ -107,8 +107,8 @@ $$\begin{gathered} <label for="proof-grad">Proof</label> <div class="hidden" markdown="1"> <label for="proof-grad">Proof.</label> -For a direction $\dd{\ell}$, we know that -$\idv{f}{\ell}$ is the component of $\nabla f$ in that direction: +For a direction $$\dd{\ell}$$, we know that +$$\idv{f}{\ell}$$ is the component of $$\nabla f$$ in that direction: $$\begin{aligned} \dv{f}{\ell} @@ -117,9 +117,9 @@ $$\begin{aligned} = \nabla f \cdot \vu{u} \end{aligned}$$ -Where $\vu{u}$ is simply a unit vector in the direction of $\dd{\ell}$. -We thus find the expression for the gradient $\nabla f$ -by choosing $\dd{\ell}$ to be $h_1 \dd{x_1}$, $h_2 \dd{x_2}$ and $h_3 \dd{x_3}$ in turn: +Where $$\vu{u}$$ is simply a unit vector in the direction of $$\dd{\ell}$$. +We thus find the expression for the gradient $$\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} \nabla f @@ -127,13 +127,14 @@ $$\begin{gathered} + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2} + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3} \end{gathered}$$ + </div> </div> ## Divergence -The divergence of a vector $\vb{V} = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3$ +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: $$\begin{aligned} @@ -149,7 +150,7 @@ $$\begin{aligned} <label for="proof-div">Proof</label> <div class="hidden" markdown="1"> <label for="proof-div">Proof.</label> -As preparation, we rewrite $\vb{V}$ as follows +As preparation, we rewrite $$\vb{V}$$ as follows to introduce the scale factors: $$\begin{aligned} @@ -159,7 +160,7 @@ $$\begin{aligned} + \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3) \end{aligned}$$ -We start by taking only the $\vu{e}_1$-component of this vector, +We start by taking only the $$\vu{e}_1$$-component of this vector, and expand its divergence using the following vector identity: $$\begin{gathered} @@ -167,8 +168,8 @@ $$\begin{gathered} = \vb{U} \cdot (\nabla f) + (\nabla \cdot \vb{U}) f \end{gathered}$$ -Inserting the scalar $f = h_2 h_3 V_1$ -the vector $\vb{U} = \vu{e}_1 / (h_2 h_3)$, +Inserting the scalar $$f = h_2 h_3 V_1$$ +the vector $$\vb{U} = \vu{e}_1 / (h_2 h_3)$$, we arrive at: $$\begin{gathered} @@ -178,8 +179,8 @@ $$\begin{gathered} \end{gathered}$$ The first right-hand term is easy to calculate -thanks to our expression for the gradient $\nabla f$. -Only the $\vu{e}_1$-component survives due to the dot product: +thanks to our expression for the gradient $$\nabla f$$. +Only the $$\vu{e}_1$$-component survives due to the dot product: $$\begin{aligned} \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big) @@ -200,7 +201,7 @@ $$\begin{aligned} = \frac{\vu{e}_3}{h_3} \end{aligned}$$ -Because $\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$ in an orthogonal basis, +Because $$\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$$ in an orthogonal basis, these gradients can be used to express the vector whose divergence we want: $$\begin{aligned} @@ -210,7 +211,7 @@ $$\begin{aligned} \end{aligned}$$ We then apply the divergence and expand the expression using a vector identity. -In all cases, the curl of a gradient $\nabla \cross \nabla f$ is zero, so: +In all cases, the curl of a gradient $$\nabla \cross \nabla f$$ is zero, so: $$\begin{aligned} \nabla \cdot \frac{\vu{e}_1}{h_2 h_3} @@ -219,7 +220,7 @@ $$\begin{aligned} = 0 \end{aligned}$$ -After repeating this procedure for the other components of $\vb{V}$, +After repeating this procedure for the other components of $$\vb{V}$$, we get the desired general expression for the divergence. </div> </div> @@ -227,7 +228,7 @@ we get the desired general expression for the divergence. ## Laplacian -The Laplacian $\nabla^2 f$ is simply $\nabla \cdot \nabla f$, +The Laplacian $$\nabla^2 f$$ is simply $$\nabla \cdot \nabla f$$, so we can find the general formula by combining the two preceding results for the gradient and the divergence: @@ -247,7 +248,7 @@ $$\begin{aligned} ## Curl -The curl of a vector $\vb{V}$ is as follows +The curl of a vector $$\vb{V}$$ is as follows in a general orthogonal curvilinear system: $$\begin{aligned} @@ -269,22 +270,22 @@ $$\begin{aligned} <div class="hidden" markdown="1"> <label for="proof-curl">Proof.</label> The curl is found in a similar way as the divergence. -We rewrite $\vb{V}$ like so: +We rewrite $$\vb{V}$$ like so: $$\begin{aligned} \vb{V} = \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}$$ -We expand the curl of its $\vu{e}_1$-component using the following vector identity: +We expand the curl of its $$\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}$$ -Inserting the scalar $f = h_1 V_1$ -and the vector $\vb{U} = \vu{e}_1 / h_1$, we arrive at: +Inserting the scalar $$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) @@ -292,7 +293,7 @@ $$\begin{gathered} \end{gathered}$$ Previously, when proving the divergence, -we already showed that $\vu{e}_1 / h_1 = \nabla x_1$. +we already showed that $$\vu{e}_1 / h_1 = \nabla x_1$$. Because the curl of a gradient is zero, the first term disappears, leaving only the second, which contains a gradient that turns out to be: @@ -304,8 +305,8 @@ $$\begin{aligned} + \vu{e}_3 \frac{1}{h_3} \pdv{(h_1 V_1)}{x_3} \end{aligned}$$ -Consequently, the curl of the first component of $\vb{V}$ is as follows, -using the fact that $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$ +Consequently, the curl of the first component of $$\vb{V}$$ is as follows, +using the fact that $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$ are related to each other by cross products: $$\begin{aligned} @@ -314,7 +315,7 @@ $$\begin{aligned} = - \frac{\vu{e}_3}{h_1 h_2} \pdv{(h_1 V_1)}{x_2} + \frac{\vu{e}_2}{h_1 h_3} \pdv{(h_1 V_1)}{x_3} \end{aligned}$$ -If we go through the same process for the other components of $\vb{V}$ +If we go through the same process for the other components of $$\vb{V}$$ and add up the results, we get the desired expression for the curl. </div> </div> @@ -322,14 +323,14 @@ and add up the results, we get the desired expression for the curl. ## Differential elements -The point of the scale factors $h_1$, $h_2$ and $h_3$, as can seen from their derivation, +The point of the scale factors $$h_1$$, $$h_2$$ and $$h_3$$, as can seen from their derivation, is to correct for "distortions" of the coordinates compared to the Cartesian system, -such that the line element $\dd{\ell}$ retains its length. -This property extends to the surface $\dd{S}$ and volume $\dd{V}$. +such that the line element $$\dd{\ell}$$ retains its length. +This property extends to the surface $$\dd{S}$$ and volume $$\dd{V}$$. When handling a differential volume in curvilinear coordinates, e.g. for a volume integral, -the size of the box $\dd{V}$ must be corrected by the scale factors: +the size of the box $$\dd{V}$$ must be corrected by the scale factors: $$\begin{aligned} \boxed{ @@ -339,8 +340,8 @@ $$\begin{aligned} } \end{aligned}$$ -The same is true for the isosurfaces $\dd{S_1}$, $\dd{S_2}$ and $\dd{S_3}$ -where the coordinates $x_1$, $x_2$ and $x_3$ are respectively kept constant: +The same is true for the isosurfaces $$\dd{S_1}$$, $$\dd{S_2}$$ and $$\dd{S_3}$$ +where the coordinates $$x_1$$, $$x_2$$ and $$x_3$$ are respectively kept constant: $$\begin{aligned} \boxed{ @@ -354,7 +355,7 @@ $$\begin{aligned} } \end{aligned}$$ -Using the same logic, the normal vector element $\dd{\vu{S}}$ +Using the same logic, the normal vector element $$\dd{\vu{S}}$$ of an arbitrary surface is given by: $$\begin{aligned} @@ -364,7 +365,7 @@ $$\begin{aligned} } \end{aligned}$$ -Finally, the tangent vector element $\dd{\vu{\ell}}$ takes the following form: +Finally, the tangent vector element $$\dd{\vu{\ell}}$$ takes the following form: $$\begin{aligned} \boxed{ |