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diff --git a/source/know/concept/curvilinear-coordinates/index.md b/source/know/concept/curvilinear-coordinates/index.md deleted file mode 100644 index 48a5a72..0000000 --- a/source/know/concept/curvilinear-coordinates/index.md +++ /dev/null @@ -1,375 +0,0 @@ ---- -title: "Curvilinear coordinates" -sort_title: "Curvilinear coordinates" -date: 2021-03-03 -categories: -- Mathematics -- Physics -layout: "concept" ---- - -In a 3D coordinate system, the isosurface of a coordinate -(i.e. the surface where that coordinate is constant while the others vary) -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. -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** $$\dd{\ell}$$, -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. -The resulting expression will be of the form: - -$$\begin{aligned} - \boxed{ - \dd{\ell}^2 - = \dd{x}^2 + \dd{y}^2 + \dd{z}^2 - = h_1^2 \dd{x_1}^2 + h_2^2 \dd{x_2}^2 + h_3^2 \dd{x_3}^2 - } -\end{aligned}$$ - -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. - -Examples of orthogonal curvilinear coordinate systems include -[spherical coordinates](/know/concept/spherical-coordinates/), -[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/), -and [cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coordinates/). - -In the following subsections, -we derive general formulae to convert expressions -from Cartesian coordinates to the new orthogonal system $$(x_1, x_2, x_3)$$. - - - -## Basis vectors - -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. $$\dd{y}$$, -in the new basis as follows: - -$$\begin{aligned} - \dd{y} - = \pdv{y}{x_1} \dd{x_1} + \pdv{y}{x_2} \dd{x_2} + \pdv{y}{x_3} \dd{x_3} -\end{aligned}$$ - -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 $$\dd{\vu{\ell}}$$: - -$$\begin{aligned} - \dd{\vu{\ell}} - = \vu{e}_1 \:h_1 \dd{x_1} + \vu{e}_2 \:h_2 \dd{x_2} + \vu{e}_3 \:h_3 \dd{x_4} -\end{aligned}$$ - -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: - -$$\begin{aligned} - \boxed{ - h_1 \vu{e}_1 - = \vu{e}_x \pdv{x}{x_1} + \vu{e}_y \pdv{y}{x_1} + \vu{e}_z \pdv{y}{x_1} - } -\end{aligned}$$ - - - -## 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 $$x_3$$: - -$$\begin{gathered} - \boxed{ - \nabla f - = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1} - + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2} - + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3} - } -\end{gathered}$$ - - -{% include proof/start.html id="proof-grad" -%} -For a direction $$\dd{\ell}$$, we know that -$$\idv{f}{\ell}$$ is the component of $$\nabla f$$ in that direction: - -$$\begin{aligned} - \dv{f}{\ell} - = \pdv{f}{x} \dv{x}{\ell} + \pdv{f}{y} \dv{y}{\ell} + \pdv{f}{z} \dv{z}{\ell} - = \nabla f \cdot \bigg( \dv{x}{\ell}, \dv{y}{\ell}, \dv{z}{\ell} \bigg) - = \nabla f \cdot \vu{u} -\end{aligned}$$ - -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: - -$$\begin{gathered} - \nabla f - = \vu{e}_1 \dv{x_1}{\ell} \pdv{f}{x_1} - + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2} - + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3} -\end{gathered}$$ -{% include proof/end.html id="proof-grad" %} - - - -## Divergence - -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: - -$$\begin{aligned} - \boxed{ - \nabla \cdot \vb{V} - = \frac{1}{h_1 h_2 h_3} - \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big) - } -\end{aligned}$$ - - -{% include proof/start.html id="proof-div" -%} -As preparation, we rewrite $$\vb{V}$$ as follows -to introduce the scale factors: - -$$\begin{aligned} - \vb{V} - &= \vu{e}_1 \frac{1}{h_2 h_3} (h_2 h_3 V_1) - + \vu{e}_2 \frac{1}{h_1 h_3} (h_1 h_3 V_2) - + \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3) -\end{aligned}$$ - -We 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 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: - -$$\begin{aligned} - \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big) - = \frac{\vu{e}_1}{h_1 h_2 h_3} \pdv{(h_2 h_3 V_1)}{x_1} -\end{aligned}$$ - -The second term is more involved. -First, we use the gradient formula to observe that: - -$$\begin{aligned} - \nabla x_1 - = \frac{\vu{e}_1}{h_1} - \qquad \quad - \nabla x_2 - = \frac{\vu{e}_2}{h_2} - \qquad \quad - \nabla x_3 - = \frac{\vu{e}_3}{h_3} -\end{aligned}$$ - -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: - -$$\begin{aligned} - \nabla x_2 \cross \nabla x_3 - = \frac{\vu{e}_2}{h_2} \cross \frac{\vu{e}_3}{h_3} - = \frac{\vu{e}_1}{h_2 h_3} -\end{aligned}$$ - -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: - -$$\begin{aligned} - \nabla \cdot \frac{\vu{e}_1}{h_2 h_3} - = \nabla \cdot \big( \nabla x_2 \cross \nabla x_3 \big) - = \nabla x_3 \cdot (\nabla \cross \nabla x_2) - \nabla x_2 \cdot (\nabla \cross \nabla x_3) - = 0 -\end{aligned}$$ - -After repeating this procedure for the other components of $$\vb{V}$$, -we get the desired general expression for the divergence. -{% include proof/end.html id="proof-div" %} - - - -## Laplacian - -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: - -$$\begin{aligned} - \boxed{ - \nabla^2 f - = \frac{1}{h_1 h_2 h_3} - \bigg( - \pdv{}{x_1}\Big(\! \frac{h_2 h_3}{h_1} \pdv{f}{x_1} \!\Big) - + \pdv{}{x_2}\Big(\! \frac{h_1 h_3}{h_2} \pdv{f}{x_2} \!\Big) - + \pdv{}{x_3}\Big(\! \frac{h_1 h_2}{h_3} \pdv{f}{x_3} \!\Big) - \bigg) - } -\end{aligned}$$ - - - -## Curl - -The curl of a vector $$\vb{V}$$ is as follows -in a general orthogonal curvilinear system: - -$$\begin{aligned} - \boxed{ - \begin{aligned} - \nabla \times \vb{V} - &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big) - \\ - &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big) - \\ - &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big) - \end{aligned} - } -\end{aligned}$$ - - -{% include proof/start.html id="proof-curl" -%} -The curl is found in a similar way as the divergence. -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: - -$$\begin{gathered} - \nabla \cross (\vb{U} \: f) - = (\nabla \cross \vb{U}) f - \vb{U} \cross (\nabla f) -\end{gathered}$$ - -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 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 disappears, leaving only the second, -which contains a gradient that turns out to be: - -$$\begin{aligned} - \nabla (h_1 V_1) - = \vu{e}_1 \frac{1}{h_1} \pdv{(h_1 V_1)}{x_1} - + \vu{e}_2 \frac{1}{h_2} \pdv{(h_1 V_1)}{x_2} - + \vu{e}_3 \frac{1}{h_3} \pdv{(h_1 V_1)}{x_3} -\end{aligned}$$ - -Consequently, the curl of the first component of $$\vb{V}$$ is as follows, -using the fact that $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$ -are related to each other by cross products: - -$$\begin{aligned} - \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big) - = - \frac{\vu{e}_1}{h_1} \cross \Big( \nabla (h_1 V_1) \Big) - = - \frac{\vu{e}_3}{h_1 h_2} \pdv{(h_1 V_1)}{x_2} + \frac{\vu{e}_2}{h_1 h_3} \pdv{(h_1 V_1)}{x_3} -\end{aligned}$$ - -If we go through the same process for the other components of $$\vb{V}$$ -and add up the results, we get the desired expression for the curl. -{% include proof/end.html id="proof-curl" %} - - - -## Differential elements - -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 $$\dd{V}$$. - -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: - -$$\begin{aligned} - \boxed{ - \dd{V} - = \dd{x}\dd{y}\dd{z} - = h_1 h_2 h_3 \dd{x_1} \dd{x_2} \dd{x_3} - } -\end{aligned}$$ - -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: - -$$\begin{aligned} - \boxed{ - \begin{aligned} - \dd{S_1} &= h_2 h_3 \dd{x_2} \dd{x_3} - \\ - \dd{S_2} &= h_1 h_3 \dd{x_1} \dd{x_3} - \\ - \dd{S_3} &= h_1 h_2 \dd{x_1} \dd{x_2} - \end{aligned} - } -\end{aligned}$$ - -Using the same logic, the normal vector element $$\dd{\vu{S}}$$ -of an arbitrary surface is given by: - -$$\begin{aligned} - \boxed{ - \dd{\vu{S}} - = \vu{e}_1 h_2 h_3 \dd{x_2} \dd{x_3} + \vu{e}_2 h_1 h_3 \dd{x_1} \dd{x_3} + \vu{e}_3 h_1 h_2 \dd{x_1} \dd{x_2} - } -\end{aligned}$$ - -Finally, the tangent vector element $$\dd{\vu{\ell}}$$ takes the following form: - -$$\begin{aligned} - \boxed{ - \dd{\vu{\ell}} - = \vu{e}_1 h_1 \dd{x_1} + \vu{e}_2 h_2 \dd{x_2} + \vu{e}_3 h_3 \dd{x_3} - } -\end{aligned}$$ - - - -## References -1. M.L. Boas, - *Mathematical methods in the physical sciences*, 2nd edition, - Wiley. diff --git a/source/know/concept/cylindrical-parabolic-coordinates/index.md b/source/know/concept/cylindrical-parabolic-coordinates/index.md index c8e16da..766c9b6 100644 --- a/source/know/concept/cylindrical-parabolic-coordinates/index.md +++ b/source/know/concept/cylindrical-parabolic-coordinates/index.md @@ -37,8 +37,8 @@ $$\begin{aligned} = - \frac{y^2}{\tau^2} + \tau^2 \end{aligned}$$ -Cylindrical parabolic coordinates form an orthogonal -[curvilinear system](/know/concept/curvilinear-coordinates/), +Cylindrical parabolic coordinates form +an [orthogonal curvilinear system](/know/concept/orthogonal-curvilinear-coordinates/), so we would like to find its scale factors $$h_\sigma$$, $$h_\tau$$ and $$h_z$$. The differentials of the Cartesian coordinates are as follows: diff --git a/source/know/concept/cylindrical-polar-coordinates/index.md b/source/know/concept/cylindrical-polar-coordinates/index.md index 686a4ed..43b4684 100644 --- a/source/know/concept/cylindrical-polar-coordinates/index.md +++ b/source/know/concept/cylindrical-polar-coordinates/index.md @@ -41,8 +41,8 @@ $$\begin{aligned} } \end{aligned}$$ -The cylindrical polar coordinates form an orthogonal -[curvilinear system](/know/concept/curvilinear-coordinates/), +The cylindrical polar coordinates form +an [orthogonal curvilinear system](/know/concept/orthogonal-curvilinear-coordinates/), whose scale factors $$h_r$$, $$h_\varphi$$ and $$h_z$$ we want to find. To do so, we calculate the differentials of the Cartesian coordinates: diff --git a/source/know/concept/orthogonal-curvilinear-coordinates/index.md b/source/know/concept/orthogonal-curvilinear-coordinates/index.md new file mode 100644 index 0000000..4fb45b4 --- /dev/null +++ b/source/know/concept/orthogonal-curvilinear-coordinates/index.md @@ -0,0 +1,1027 @@ +--- +title: "Orthogonal curvilinear coordinates" +sort_title: "Orthogonal curvilinear coordinates" +date: 2023-05-29 # Originally 2021-03-03, major rewrite +categories: +- Mathematics +- Physics +layout: "concept" +--- + +In a 3D coordinate system, the isosurface of a coordinate +(i.e. the surface where that coordinate is constant while the others vary) +is known as a **coordinate surface**, and the intersection line +of two coordinates surfaces is called a **coordinate line**. + +A **curvilinear** coordinate system is one +where at least one of the coordinate surfaces is curved: +e.g. in cylindrical coordinates, the coordinate line of $$r$$ and $$z$$ is a circle. +Here we limit ourselves to **orthogonal** systems, +where the coordinate surfaces are always perpendicular. +Examples of such orthogonal curvilinear systems include +[spherical coordinates](/know/concept/spherical-coordinates/), +[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/), +and [cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coordinates/). + + + +## Scale factors and basis vectors + +Given such a system with coordinates $$(c_1, c_2, c_3)$$. +Their definition lets us convert all positions +to classic Cartesian coordinates $$(x, y, z)$$ +using functions $$x$$, $$y$$ and $$z$$: + +$$\begin{aligned} + x + &= x(c_1, c_2, c_3) + \\ + y + &= y(c_1, c_2, c_3) + \\ + z + &= z(c_1, c_2, c_3) +\end{aligned}$$ + +A useful attribute of a coordinate system is its **line element** $$\dd{\vu{\ell}}$$, +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: + +$$\begin{aligned} + \dd{\vu{\ell}} + \equiv \vu{e}_x \dd{x} + \: \vu{e}_y \dd{y} + \: \vu{e}_z \dd{z} +\end{aligned}$$ + +The Cartesian differential elements can be rewritten +in $$(c_1, c_2, c_3)$$ with the chain rule: + +$$\begin{aligned} + \dd{\vu{\ell}} + = \quad &\vu{e}_x \bigg( \pdv{x}{c_1} \dd{c_1} + \: \pdv{x}{c_2} \dd{c_2} + \: \pdv{x}{c_3} \dd{c_3} \!\bigg) + \\ + + \: &\vu{e}_y \bigg( \pdv{y}{c_1} \dd{c_1} + \: \pdv{y}{c_2} \dd{c_2} + \: \pdv{y}{c_3} \dd{c_3} \!\bigg) + \\ + + \: &\vu{e}_z \bigg( \pdv{z}{c_1} \dd{c_1} + \: \pdv{z}{c_2} \dd{c_2} + \: \pdv{z}{c_3} \dd{c_3} \!\bigg) + \\ + = \quad &\bigg( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \bigg) \dd{c_1} + \\ + + &\bigg( \pdv{x}{c_2} \vu{e}_x + \pdv{y}{c_2} \vu{e}_y + \pdv{z}{c_2} \vu{e}_z \bigg) \dd{c_2} + \\ + + &\bigg( \pdv{x}{c_3} \vu{e}_x + \pdv{y}{c_3} \vu{e}_y + \pdv{z}{c_3} \vu{e}_z \bigg) \dd{c_3} +\end{aligned}$$ + +From this we define the **scale factors** $$h_1$$, $$h_2$$ and $$h_3$$ +and **local basis vectors** $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + h_1 \vu{e}_1 + &\equiv \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z + \\ + h_2 \vu{e}_2 + &\equiv \pdv{x}{c_2} \vu{e}_x + \pdv{y}{c_2} \vu{e}_y + \pdv{z}{c_2} \vu{e}_z + \\ + h_3 \vu{e}_3 + &\equiv \pdv{x}{c_3} \vu{e}_x + \pdv{y}{c_3} \vu{e}_y + \pdv{z}{c_3} \vu{e}_z + \end{aligned} + } +\end{aligned}$$ + +Where $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$ are normalized, +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. + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \vu{e}_x + &\equiv \pdv{c_1}{x} h_1 \vu{e}_1 + \pdv{c_2}{x} h_2 \vu{e}_2 + \pdv{c_3}{x} h_3 \vu{e}_3 + \\ + \vu{e}_y + &\equiv \pdv{c_1}{y} h_1 \vu{e}_1 + \pdv{c_2}{y} h_2 \vu{e}_2 + \pdv{c_3}{y} h_3 \vu{e}_3 + \\ + \vu{e}_z + &\equiv \pdv{c_1}{z} h_1 \vu{e}_1 + \pdv{c_2}{z} h_2 \vu{e}_2 + \pdv{c_3}{z} h_3 \vu{e}_3 + \end{aligned} + } +\end{aligned}$$ + + +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 $$(c_1, c_2, c_3)$$. + + + +## Differential elements + +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. +As was already established above: + +$$\begin{aligned} + \boxed{ + \dd{\vu{\ell}} + = \vu{e}_1 h_1 \dd{c_1} + \: \vu{e}_2 h_2 \dd{c_2} + \: \vu{e}_3 h_3 \dd{c_3} + } +\end{aligned}$$ + +These terms are the differentials along each of the local basis vectors. +Let us now introduce the following notation, e.g. for $$c_1$$: + +$$\begin{aligned} + \dd{}_1\!\vb{x} + \equiv \pdv{\vb{x}}{c_1} \dd{c_1} + = \Big( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \Big) \dd{c_1} + = \vu{e}_1 h_1 \dd{c_1} +\end{aligned}$$ + +And likewise we define $$\dd{}_2\!\vb{x}$$ and $$\dd{}_3\!\vb{x}$$. +All differential elements (as found in e.g. integrals) +can be expressed in terms of $$\dd{}_1\!\vb{x}$$, $$\dd{}_2\!\vb{x}$$ and $$\dd{}_3\!\vb{x}$$. + +The differential normal vector element $$\dd{\vu{S}}$$ in a surface integral is hence given by: + +$$\begin{aligned} + \dd{\vu{S}} + &= \dd{}_1\!\vb{x} \cross \dd{}_2\!\vb{x} + \dd{}_2\!\vb{x} \cross \dd{}_3\!\vb{x} + \dd{}_3\!\vb{x} \cross \dd{}_1\!\vb{x} + \\ + &= (\vu{e}_1 \cross \vu{e}_2) \: h_1 h_2 \dd{c_1} \dd{c_2} + + \: (\vu{e}_2 \cross \vu{e}_3) \: h_2 h_3 \dd{c_2} \dd{c_3} + + \: (\vu{e}_3 \cross \vu{e}_1) \: h_1 h_3 \dd{c_1} \dd{c_3} +\end{aligned}$$ + +In an orthonormal basis we have +$$\vu{e}_1 \cross \vu{e}_2 = \vu{e}_3$$, +$$\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$$ and +$$\vu{e}_3 \cross \vu{e}_1 = \vu{e}_2$$, so: + +$$\begin{aligned} + \boxed{ + \dd{\vu{S}} + = \vu{e}_1 \: h_2 h_3 \dd{c_2} \dd{c_3} + \: \vu{e}_2 \: h_1 h_3 \dd{c_1} \dd{c_3} + \: \vu{e}_3 \: h_1 h_2 \dd{c_1} \dd{c_2} + } +\end{aligned}$$ + +Next, the differential volume $$\dd{V}$$ +must also be corrected by the scale factors: + +$$\begin{aligned} + \dd{V} + = \dd{}_1\!\vb{x} \cross \dd{}_2\!\vb{x} \cdot \dd{}_3\!\vb{x} + = (\vu{e}_1 \cross \vu{e}_2 \cdot \vu{e}_3) \: h_1 h_2 h_3 \dd{c_1} \dd{c_2} \dd{c_3} +\end{aligned}$$ + +Once again $$\vu{e}_1 \cross \vu{e}_2 = \vu{e}_3$$, +so the vectors disappear from the expression, leaving: + +$$\begin{aligned} + \boxed{ + \dd{V} + = h_1 h_2 h_3 \dd{c_1} \dd{c_2} \dd{c_3} + } +\end{aligned}$$ + + + +## Basis vector derivatives + +Orthonormality tells us that $$\vu{e}_j \cdot \vu{e}_j = 1$$ for $$j = 1,2,3$$. +Differentiating with respect to $$c_k$$: + +$$\begin{aligned} + \pdv{}{c_k} (\vu{e}_j \cdot \vu{e}_j) + = 2 \pdv{\vu{e}_j}{c_k} \cdot \vu{e}_j + = \pdv{}{c_k} 1 + = 0 +\end{aligned}$$ + +This means that the $$c_k$$-derivative of $$\vu{e}_j$$ +will always be orthogonal to $$\vu{e}_j$$, for all $$j$$ and $$k$$. +Indeed, the general expression for the derivative of a local basis vector is: + +$$\begin{aligned} + \boxed{ + \pdv{\vu{e}_j}{c_k} + = \frac{1}{h_j} \pdv{h_k}{c_j} \vu{e}_k - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_j}{c_l} \vu{e}_l + } +\end{aligned}$$ + +Where $$\delta_{jk}$$ is the Kronecker delta. +For example, if $$j = 1$$, writing this out gives: + +$$\begin{aligned} + \pdv{\vu{e}_1}{c_1} + &= - \frac{1}{h_2} \pdv{h_1}{c_2} \vu{e}_2 - \frac{1}{h_3} \pdv{h_1}{c_3} \vu{e}_3 + \\ + \pdv{\vu{e}_1}{c_2} + &= \frac{1}{h_1} \pdv{h_2}{c_1} \vu{e}_2 + \\ + \pdv{\vu{e}_1}{c_3} + &= \frac{1}{h_1} \pdv{h_3}{c_1} \vu{e}_3 +\end{aligned}$$ + + +{% include proof/start.html id="proof-deriv-basis" -%} +In this proof we set $$j = 1$$ and $$k = 2$$ for clarity, +but the approach is valid for any $$j \neq k$$. +We know the definitions of $$h_1 \vu{e}_1$$ and $$h_2 \vu{e}_2$$, +and that differentiations can be reordered: + +$$\begin{aligned} + \pdv{}{c_2} (h_1 \vu{e}_1) + &= \pdv{}{c_2} \pdv{}{c_1} \big( x \vu{e}_x + y \vu{e}_y + z \vu{e}_z \big) + = \pdv{}{c_1} (h_2 \vu{e}_2) +\end{aligned}$$ + +Expanding this according to the product rule of differentiation: + +$$\begin{aligned} + \pdv{h_1}{c_2} \vu{e}_1 + h_1 \pdv{\vu{e}_1}{c_2} + = \pdv{h_2}{c_1} \vu{e}_2 + h_2 \pdv{\vu{e}_2}{c_1} +\end{aligned}$$ + +We rearrange this in two different ways. +Indeed, these two equations are identical: + +$$\begin{aligned} + h_1 \pdv{\vu{e}_1}{c_2} + &= \pdv{h_2}{c_1} \vu{e}_2 + \Big( h_2 \pdv{\vu{e}_2}{c_1} - \pdv{h_1}{c_2} \vu{e}_1 \Big) + \\ + h_2 \pdv{\vu{e}_2}{c_1} + &= \pdv{h_1}{c_2} \vu{e}_1 + \Big( h_1 \pdv{\vu{e}_1}{c_2} - \pdv{h_2}{c_1} \vu{e}_2 \Big) +\end{aligned}$$ + +Recall that all derivatives of $$\vu{e}_j$$ are orthogonal to $$\vu{e}_j$$. +Therefore, the first equation's right-hand side must be orthogonal to $$\vu{e}_1$$, +and the second's to $$\vu{e}_2$$. +We deduce that the parenthesized expressions +are proportional to $$\vu{e}_3$$, +and call the proportionality factors $$\lambda_{123}$$ and $$\lambda_{213}$$: + +$$\begin{aligned} + h_1 \pdv{\vu{e}_1}{c_2} + &= \pdv{h_2}{c_1} \vu{e}_2 + \lambda_{213} \vu{e}_3 + \\ + h_2 \pdv{\vu{e}_2}{c_1} + &= \pdv{h_1}{c_2} \vu{e}_1 + \lambda_{123} \vu{e}_3 +\end{aligned}$$ + +Since these equations are identical, +by comparing the definition of $$\lambda_{123}$$ to the other side of the equation, +we see that $$\lambda_{123} = \lambda_{213}$$: + +$$\begin{aligned} + \lambda_{123} \vu{e}_3 + &= h_1 \pdv{\vu{e}_1}{c_2} - \pdv{h_2}{c_1} \vu{e}_2 + = h_2 \pdv{\vu{e}_2}{c_1} - \pdv{h_1}{c_2} \vu{e}_1 + = \lambda_{213} \vu{e}_3 +\end{aligned}$$ + +In general, $$\lambda_{jkl} = \lambda_{kjl}$$ for $$j \neq k \neq l$$. +Next, we dot-multiply $$\lambda_{123}$$'s equation by $$\vu{e}_3$$, +using that $$\vu{e}_2 \cdot \vu{e}_3 = 0$$ +and consequently $$\ipdv{(\vu{e}_2 \cdot \vu{e}_3)}{c_1} = 0$$: + +$$\begin{aligned} + \lambda_{123} + &= h_2 \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_3 - \pdv{h_1}{c_2} \vu{e}_1 \cdot \vu{e}_3 + = h_2 \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_3 + = - h_2 \frac{h_3}{h_3} \pdv{\vu{e}_3}{c_1} \cdot \vu{e}_2 + = - \frac{h_2}{h_3} \lambda_{132} +\end{aligned}$$ + +In general, $$\lambda_{jkl} = - h_k \lambda_{jlk} / h_l$$ for $$j \neq k \neq l$$. +Combining this fact with $$\lambda_{jkl} = \lambda_{kjl}$$ gives: + +$$\begin{aligned} + \lambda_{jkl} + = - \frac{h_k}{h_l} \lambda_{jlk} + = - \frac{h_k}{h_l} \lambda_{ljk} + = \frac{h_k}{h_l} \frac{h_j}{h_k} \lambda_{lkj} + = \frac{h_j}{h_l} \lambda_{klj} + = - \frac{h_j}{h_l} \frac{h_l}{h_j} \lambda_{kjl} + = - \lambda_{jkl} +\end{aligned}$$ + +But $$\lambda_{jkl} = -\lambda_{jkl}$$ is only possible if $$\lambda_{jkl}$$ is zero. +Thus $$\lambda_{123}$$'s equation reduces to: + +$$\begin{aligned} + h_2 \pdv{\vu{e}_2}{c_1} + &= \pdv{h_1}{c_2} \vu{e}_1 + \qquad \implies \qquad + \pdv{\vu{e}_2}{c_1} + = \frac{1}{h_2} \pdv{h_1}{c_2} \vu{e}_1 +\end{aligned}$$ + +This gives us the general expression for $$\ipdv{\vu{e}_j}{c_k}$$ when $$j \neq k$$, +but what about $$j = k$$? +Well, from orthogonality we know: + +$$\begin{aligned} + 0 + = \vu{e}_2 \cdot \vu{e}_1 + = \pdv{}{c_1} (\vu{e}_2 \cdot \vu{e}_1) + = \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_1 + \vu{e}_2 \cdot \pdv{\vu{e}_1}{c_1} +\end{aligned}$$ + +We just calculated one of those terms, so this equation gives us the other: + +$$\begin{aligned} + \vu{e}_2 \cdot \pdv{\vu{e}_1}{c_1} + = - \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_1 + = - \frac{1}{h_2} \pdv{h_1}{c_2} +\end{aligned}$$ + +Now we have the $$\vu{e}_2$$-component of $$\ipdv{\vu{e}_1}{c_1}$$, +and can find the $$\vu{e}_3$$-component in the same way: + +$$\begin{aligned} + \vu{e}_3 \cdot \pdv{\vu{e}_1}{c_1} + = - \pdv{\vu{e}_3}{c_1} \cdot \vu{e}_1 + = - \frac{1}{h_3} \pdv{h_1}{c_3} +\end{aligned}$$ + +Adding up the $$\vu{e}_2$$- and $$\vu{e}_3$$-components gives the desired formula. +There is no $$\vu{e}_1$$-component +because $$\ipdv{\vu{e}_1}{c_1}$$ must be orthogonal to $$\vu{e}_1$$. +{% include proof/end.html id="proof-deriv-basis" -%} + + + +## Gradient of a scalar + +In $$(c_1, c_2, c_3)$$, the gradient $$\nabla f$$ of a scalar field $$f$$ +has the following components: + +$$\begin{aligned} + \boxed{ + (\nabla f)_j + = \frac{1}{h_j} \pdv{f}{c_j} + } +\end{aligned}$$ + +When this index notation is written out in full, +the gradient $$\nabla f$$ becomes: + +$$\begin{aligned} + \nabla f + = \frac{1}{h_1} \pdv{f}{c_1} \vu{e}_1 + + \frac{1}{h_2} \pdv{f}{c_2} \vu{e}_2 + + \frac{1}{h_3} \pdv{f}{c_3} \vu{e}_3 +\end{aligned}$$ + + +{% include proof/start.html id="proof-grad-scalar" -%} +For any unit vector $$\vu{u}$$, we can project $$\nabla f$$ onto it +to get the component of $$\nabla f$$ along $$\vu{u}$$. +Let us choose $$\vu{u} = \vu{e}_1$$, then such a projection gives: + +$$\begin{aligned} + \nabla f \cdot \vu{e}_1 + &= \bigg( \pdv{f}{x} \vu{e}_x + \pdv{f}{y} \vu{e}_y + \pdv{f}{z} \vu{e}_z \bigg) + \cdot \frac{1}{h_1} \bigg( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \bigg) + \\ + &= \frac{1}{h_1} \bigg( \pdv{f}{x} \pdv{x}{c_1} + \pdv{f}{y} \pdv{y}{c_1} + \pdv{f}{z} \pdv{z}{c_1} \bigg) + \\ + &= \frac{1}{h_1} \pdv{f}{c_1} +\end{aligned}$$ + +And we can do the same for $$\vu{e}_2$$ and $$\vu{e}_3$$, +yielding analogous results: + +$$\begin{aligned} + \nabla f \cdot \vu{e}_2 + = \frac{1}{h_2} \pdv{f}{c_2} + \qquad \qquad + \nabla f \cdot \vu{e}_3 + = \frac{1}{h_3} \pdv{f}{c_3} +\end{aligned}$$ + +Finally, to express $$\nabla f$$ in the new coordinate system $$(c_1, c_2, c_3)$$, +we simply combine these projections for all the basis vectors: + +$$\begin{aligned} + \nabla f + = (\nabla f \cdot \vu{e}_1) \vu{e}_1 + + (\nabla f \cdot \vu{e}_2) \vu{e}_2 + + (\nabla f \cdot \vu{e}_3) \vu{e}_3 +\end{aligned}$$ +{% include proof/end.html id="proof-grad-scalar" %} + + + +## Divergence of a vector + +The divergence of a vector field $$\vb{V} = V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3$$ +is given in $$(c_1, c_2, c_3)$$ by: + +$$\begin{aligned} + \boxed{ + \nabla \cdot \vb{V} + = \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j} \bigg) + } +\end{aligned}$$ + +Where $$H \equiv h_1 h_2 h_3$$. +When this index notation is written out in full, it becomes: + +$$\begin{aligned} + \nabla \cdot \vb{V} + = \frac{1}{h_1 h_2 h_3} + \bigg( \pdv{}{c_1} (h_2 h_3 V_1) + \pdv{}{c_2} (h_1 h_3 V_2) + \pdv{}{c_3} (h_1 h_2 V_3) \bigg) +\end{aligned}$$ + + +{% include proof/start.html id="proof-div-vector-1" label="Proof 1" -%} +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 of $$\nabla$$ and $$\vb{V}$$: + +$$\begin{aligned} + \nabla \cdot \vb{V} + &= \bigg( \vu{e}_1 \frac{1}{h_1} \pdv{}{c_1} + \vu{e}_2 \frac{1}{h_2} \pdv{}{c_2} + \vu{e}_3 \frac{1}{h_3} \pdv{}{c_3} \bigg) + \cdot \bigg( V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3 \bigg) + \\ + &= \bigg( \sum_{j} \vu{e}_j \frac{1}{h_j} \pdv{}{c_j} \bigg) \cdot \bigg( \sum_{k} V_k \vu{e}_k \bigg) + \\ + &= \sum_{jk} \vu{e}_j \cdot \frac{1}{h_j} \pdv{}{c_j} (V_k \vu{e}_k) + \\ + &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j} + + \sum_{jk} \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{V_k}{h_j} +\end{aligned}$$ + +Substituting our expression for the derivatives of the local basis vectors, we find: + +$$\begin{aligned} + \nabla \cdot \vb{V} + &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j} + + \sum_{jk} \vu{e}_j + \cdot \bigg( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_k}{c_l} \vu{e}_l \bigg) \frac{V_k}{h_j} + \\ + &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j} + + \sum_{jk} (\vu{e}_j \cdot \vu{e}_j) \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} + - \sum_{jl} (\vu{e}_j \cdot \vu{e}_l) \frac{V_j}{h_j h_l} \pdv{h_j}{c_l} + \\ + &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j} + + \sum_{jk} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} + - \sum_{j} \frac{V_j}{h_j h_j} \pdv{h_j}{c_j} + \\ + &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j} + + \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} + \\ +\end{aligned}$$ + +Where we noticed that the latter two terms cancel out if $$k = j$$. +Now, to proceed, it is easiest to just write out the index notation: + +$$\begin{aligned} + \nabla \cdot \vb{V} + &= \quad\: \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j} + \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k} + \\ + &= \quad\: \frac{1}{h_1} \pdv{V_1}{c_1} + \frac{V_1}{h_1 h_2} \pdv{h_2}{c_1} + \frac{V_1}{h_1 h_3} \pdv{h_3}{c_1} + \\ + &\quad\:\: + \frac{V_2}{h_1 h_2} \pdv{h_1}{c_2} + \frac{1}{h_2} \pdv{V_2}{c_2} + \frac{V_2}{h_2 h_3} \pdv{h_3}{c_2} + \\ + &\quad\:\: + \frac{V_3}{h_1 h_3} \pdv{h_1}{c_3} + \frac{V_3}{h_2 h_3} \pdv{h_2}{c_3} + \frac{1}{h_3} \pdv{V_3}{c_3} + \\ + &= \frac{1}{h_1 h_2 h_3} \bigg( h_2 h_3 \pdv{V_1}{c_1} + h_3 V_1 \pdv{h_2}{c_1} + h_2 V_1 \pdv{h_3}{c_1} + \\ + &\qquad\qquad + h_3 V_2 \pdv{h_1}{c_2} + h_1 h_3 \pdv{V_2}{c_2} + h_1 V_2 \pdv{h_3}{c_2} + \\ + &\qquad\qquad + h_2 V_3 \pdv{h_1}{c_3} + h_1 V_3 \pdv{h_2}{c_3} + h_1 h_2 \pdv{V_3}{c_3} \bigg) +\end{aligned}$$ + +Which can clearly be rewritten with the product rule, +leading to the desired formula. +{% include proof/end.html id="proof-div-vector-1" label="Proof 1" %} + + +Boas gives an alternative proof, which is shorter but more specialized: + + +{% include proof/start.html id="proof-div-vector-2" label="Proof 2" -%} +We take the divergence of the $$c_1$$-component of $$\vb{V}$$ and expand it: + +$$\begin{aligned} + \nabla \cdot (V_1 \vu{e}_1) + &= \nabla \cdot \bigg( \Big( h_2 h_3 V_1 \Big) \Big( \frac{\vu{e}_1}{h_2 h_3} \Big) \bigg) + \\ + &= \nabla (h_2 h_3 V_1) \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} \Big) + + (h_2 h_3 V_1) \bigg( \nabla \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} \Big) \bigg) +\end{aligned}$$ + +The latter term is zero, because +in any orthogonal basis $$\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$$, +and according to our gradient formula we have $$\nabla c_2 = \vu{e}_2 / h_2$$ etc., so: + +$$\begin{aligned} + \nabla \cdot \bigg( \frac{\vu{e}_1}{h_2 h_3} \bigg) + &= \nabla \cdot \bigg( \frac{\vu{e}_2}{h_2} \cross \frac{\vu{e}_3}{h_3} \bigg) + \\ + &= \nabla \cdot \big( \nabla c_2 \cross \nabla c_3 \big) + \\ + &= \nabla c_3 \cdot (\nabla \cross \nabla c_2) - \nabla c_2 \cdot (\nabla \cross \nabla c_3) + \\ + &= 0 +\end{aligned}$$ + +Where we used a vector identity and the fact that the curl of a gradient must vanish. +We are thus left with the former term, +to which we apply our gradient formula again, +where only the $$\vu{e}_1$$-term survives due to the dot product and orthogonality: + +$$\begin{aligned} + \nabla \cdot (V_1 \vu{e}_1) + &= \nabla (h_2 h_3 V_1) \cdot \frac{\vu{e}_1}{h_2 h_3} + \\ + &= \frac{1}{h_1 h_2 h_3} \pdv{}{c_1} (h_2 h_3 V_1) +\end{aligned}$$ + +We then repeat this procedure for $$\vb{V}$$'s other components, +and simply add up the results to get the desired formula. +{% include proof/end.html id="proof-div-vector-2" label="Proof 2" %} + + + +## 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: + +$$\begin{aligned} + \nabla^2 f + = \frac{1}{h_1 h_2 h_3} + \bigg( + \pdv{}{c_1}\Big(\! \frac{h_2 h_3}{h_1} \pdv{f}{c_1} \!\Big) + + \pdv{}{c_2}\Big(\! \frac{h_1 h_3}{h_2} \pdv{f}{c_2} \!\Big) + + \pdv{}{c_3}\Big(\! \frac{h_1 h_2}{h_3} \pdv{f}{c_3} \!\Big) + \bigg) +\end{aligned}$$ |