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-rw-r--r-- | source/know/concept/capillary-action/index.md | 2 | ||||
-rw-r--r-- | source/know/concept/cylindrical-polar-coordinates/index.md | 273 | ||||
-rw-r--r-- | source/know/concept/hilbert-space/index.md | 55 | ||||
-rw-r--r-- | source/know/concept/material-derivative/index.md | 20 | ||||
-rw-r--r-- | source/know/concept/orthogonal-curvilinear-coordinates/index.md | 250 | ||||
-rw-r--r-- | source/know/concept/spherical-coordinates/index.md | 294 |
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diff --git a/source/know/concept/capillary-action/index.md b/source/know/concept/capillary-action/index.md index 4b9e76c..fea6ef8 100644 --- a/source/know/concept/capillary-action/index.md +++ b/source/know/concept/capillary-action/index.md @@ -58,7 +58,7 @@ $$\begin{aligned} } \end{aligned}$$ -The right-most side gives an alternative way of understanding $$\mathrm{Bo}$$: +The rightmost side gives an alternative way of understanding $$\mathrm{Bo}$$: $$m$$ is the mass of a cube with side $$L$$, such that the numerator is the weight force, and the denominator is the tension force of the surface. In any case, capillary action can be observed when $$\mathrm{Bo \ll 1}$$. diff --git a/source/know/concept/cylindrical-polar-coordinates/index.md b/source/know/concept/cylindrical-polar-coordinates/index.md index 43b4684..3c54ef8 100644 --- a/source/know/concept/cylindrical-polar-coordinates/index.md +++ b/source/know/concept/cylindrical-polar-coordinates/index.md @@ -8,11 +8,11 @@ categories: layout: "concept" --- -**Cylindrical polar coordinates** are an extension of polar coordinates to 3D, -which describes the location of a point in space -using the coordinates $$(r, \varphi, z)$$. -The $$z$$-axis is unchanged from Cartesian coordinates, -hence it is called a *cylindrical* system. +**Cylindrical polar coordinates** extend polar coordinates $$(r, \varphi)$$ to 3D, +by describing the location of a point in space +using the variables $$(r, \varphi, z)$$. +The $$z$$-axis is unchanged from the Cartesian system, +hence it is called *cylindrical*. Cartesian coordinates $$(x, y, z)$$ and the cylindrical system $$(r, \varphi, z)$$ are related by: @@ -20,78 +20,85 @@ and the cylindrical system $$(r, \varphi, z)$$ are related by: $$\begin{aligned} \boxed{ \begin{aligned} - x &= r \cos\varphi \\ - y &= r \sin\varphi \\ - z &= z + x + &= r \cos{\varphi} + \\ + y + &= r \sin{\varphi} + \\ + z + &= z \end{aligned} } \end{aligned}$$ Conversely, a point given in $$(x, y, z)$$ -can be converted to $$(r, \varphi, z)$$ -using these formulae: +can be converted to $$(r, \varphi, z)$$ using these formulae, +where $$\mathtt{atan2}$$ is the 2-argument arctangent, +which is needed to handle the signs correctly: $$\begin{aligned} \boxed{ - r = \sqrt{x^2 + y^2} - \qquad - \varphi = \mathtt{atan2}(y, x) - \qquad - z = z + \begin{aligned} + r + &= \sqrt{x^2 + y^2} + \\ + \varphi + &= \mathtt{atan2}(y, x) + \\ + z + &= z + \end{aligned} } \end{aligned}$$ -The cylindrical polar coordinates form +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: +whose **scale factors** $$h_r$$, $$h_\varphi$$ and $$h_z$$ we need. +To get those, we calculate the unnormalized local basis: $$\begin{aligned} - \dd{x} = \dd{r} \cos\varphi - \dd{\varphi} r \sin\varphi - \qquad - \dd{y} = \dd{r} \sin\varphi + \dd{\varphi} r \cos\varphi - \qquad - \dd{z} = \dd{z} -\end{aligned}$$ - -And then we calculate the line element $$\dd{\ell}^2$$, -skipping many terms thanks to orthogonality, - -$$\begin{aligned} - \dd{\ell}^2 - &= \dd{r}^2 \big( \cos^2(\varphi) + \sin^2(\varphi) \big) - + \dd{\varphi}^2 \big( r^2 \sin^2(\varphi) + r^2 \cos^2(\varphi) \big) - + \dd{z}^2 + h_r \vu{e}_r + &= \vu{e}_x \pdv{x}{r} + \vu{e}_y \pdv{y}{r} + \vu{e}_z \pdv{z}{r} + \\ + &= \vu{e}_x \cos{\varphi} + \vu{e}_y \sin{\varphi} + \\ + h_\varphi \vu{e}_\varphi + &= \vu{e}_x \pdv{x}{\varphi} + \vu{e}_y \pdv{y}{\varphi} + \vu{e}_z \pdv{z}{\varphi} \\ - &= \dd{r}^2 + r^2 \: \dd{\varphi}^2 + \dd{z}^2 + &= - \vu{e}_x \: r \sin{\varphi} + \vu{e}_y \: r \cos{\varphi} + \\ + h_z \vu{e}_z + &= \vu{e}_x \pdv{x}{z} + \vu{e}_y \pdv{y}{z} + \vu{e}_z \pdv{z}{z} + \\ + &= \vu{e}_z \end{aligned}$$ -Finally, we can simply read off -the squares of the desired scale factors -$$h_r^2$$, $$h_\varphi^2$$ and $$h_z^2$$: +By normalizing the **local basis vectors** +$$\vu{e}_r$$, $$\vu{e}_\varphi$$ and $$\vu{e}_z$$, +we arrive at these expressions: $$\begin{aligned} \boxed{ - h_r = 1 - \qquad - h_\varphi = r - \qquad - h_z = 1 + \begin{aligned} + h_r + &= 1 + \\ + h_\varphi + &= r + \\ + h_z + &= 1 + \end{aligned} } -\end{aligned}$$ - -With these factors, we can easily convert things from the Cartesian system -using the standard formulae for orthogonal curvilinear coordinates. -The basis vectors are: - -$$\begin{aligned} + \qquad\qquad \boxed{ \begin{aligned} \vu{e}_r - &= \cos\varphi \:\vu{e}_x + \sin\varphi \:\vu{e}_y + &= \vu{e}_x \cos{\varphi} + \vu{e}_y \sin{\varphi} \\ \vu{e}_\varphi - &= - \sin\varphi \:\vu{e}_x + \cos\varphi \:\vu{e}_y + &= - \vu{e}_x \sin{\varphi} + \vu{e}_y \cos{\varphi} \\ \vu{e}_z &= \vu{e}_z @@ -99,7 +106,52 @@ $$\begin{aligned} } \end{aligned}$$ -The basic vector operations (gradient, divergence, Laplacian and curl) are given by: +Thanks to these scale factors, we can easily convert calculus from the Cartesian system +using the standard formulae for orthogonal curvilinear coordinates. + + + +## Differential elements + +For line integrals, +the tangent vector element $$\dd{\vb{\ell}}$$ for a curve is as follows: + +$$\begin{aligned} + \boxed{ + \dd{\vb{\ell}} + = \vu{e}_r \dd{r} + + \: \vu{e}_\varphi \: r \dd{\varphi} + + \: \vu{e}_z \dd{z} + } +\end{aligned}$$ + +For surface integrals, +the normal vector element $$\dd{\vb{S}}$$ for a surface is given by: + +$$\begin{aligned} + \boxed{ + \dd{\vb{S}} + = \vu{e}_r \: r \dd{\varphi} \dd{z} + + \: \vu{e}_\varphi \dd{r} \dd{z} + + \: \vu{e}_z \: r \dd{r} \dd{\varphi} + } +\end{aligned}$$ + +And for volume integrals, +the infinitesimal volume $$\dd{V}$$ takes the following form: + +$$\begin{aligned} + \boxed{ + \dd{V} + = r \dd{r} \dd{\varphi} \dd{z} + } +\end{aligned}$$ + + + +## Common operations + +The basic vector operations (gradient, divergence, curl and Laplacian) are given by: $$\begin{aligned} \boxed{ @@ -113,7 +165,7 @@ $$\begin{aligned} $$\begin{aligned} \boxed{ \nabla \cdot \vb{V} - = \frac{1}{r} \pdv{(r V_r)}{r} + = \pdv{V_r}{r} + \frac{V_r}{r} + \frac{1}{r} \pdv{V_\varphi}{\varphi} + \pdv{V_z}{z} } @@ -121,81 +173,124 @@ $$\begin{aligned} $$\begin{aligned} \boxed{ + \begin{aligned} + \nabla \times \vb{V} + &= \quad \vu{e}_r \bigg( \frac{1}{r} \pdv{V_z}{\varphi} - \pdv{V_\varphi}{z} \bigg) + \\ + &\quad\: + \vu{e}_\varphi \bigg( \pdv{V_r}{z} - \pdv{V_z}{r} \bigg) + \\ + &\quad\: + \vu{e}_z \bigg( \pdv{V_\varphi}{r} + \frac{V_\varphi}{r} - \frac{1}{r} \pdv{V_r}{\varphi} \bigg) + \end{aligned} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ \nabla^2 f - = \frac{1}{r} \pdv{}{r}\Big( r \pdv{f}{r} \Big) + = \pdvn{2}{f}{r} + \frac{1}{r} \pdv{f}{r} + \frac{1}{r^2} \pdvn{2}{f}{\varphi} + \pdvn{2}{f}{z} } \end{aligned}$$ + + +## Uncommon operations + +Uncommon operations include: +the gradient of a divergence $$\nabla (\nabla \cdot \vb{V})$$, +the gradient of a vector $$\nabla \vb{V}$$, +the advection of a vector $$(\vb{U} \cdot \nabla) \vb{V}$$ with respect to $$\vb{U}$$, +the Laplacian of a vector $$\nabla^2 \vb{V}$$, +and the divergence of a 2nd-order tensor $$\nabla \cdot \overline{\overline{\vb{T}}}$$: + $$\begin{aligned} \boxed{ \begin{aligned} - \nabla \times \vb{V} - &= \vu{e}_r \Big( \frac{1}{r} \pdv{V_z}{\varphi} - \pdv{V_\varphi}{z} \Big) + \nabla (\nabla \cdot \vb{V}) + &= \quad \vu{e}_r \bigg( \pdvn{2}{V_r}{r} + \frac{1}{r} \mpdv{V_\varphi}{r}{\varphi} + \mpdv{V_z}{r}{z} + + \frac{1}{r} \pdv{V_r}{r} - \frac{1}{r^2} \pdv{V_\varphi}{\varphi} - \frac{V_r}{r^2} \bigg) \\ - &+ \vu{e}_\varphi \Big( \pdv{V_r}{z} - \pdv{V_z}{r} \Big) + &\quad\: + \vu{e}_\varphi \bigg( \frac{1}{r} \mpdv{V_r}{\varphi}{r} + \frac{1}{r^2} \pdvn{2}{V_\varphi}{\varphi} + + \frac{1}{r} \mpdv{V_z}{\varphi}{z} + \frac{1}{r^2} \pdv{V_r}{\varphi} \bigg) \\ - &+ \frac{\vu{e}_z}{r} \Big( \pdv{(r V_\varphi)}{r} - \pdv{V_r}{\varphi} \Big) + &\quad\: + \vu{e}_z \bigg( \mpdv{V_r}{z}{r} + \frac{1}{r} \mpdv{V_\varphi}{z}{\varphi} + \pdvn{2}{V_z}{z} + \frac{1}{r} \pdv{V_r}{z} \bigg) \end{aligned} } \end{aligned}$$ -The differential element of volume $$\dd{V}$$ -takes the following form: - $$\begin{aligned} \boxed{ - \dd{V} - = r \dd{r} \dd{\varphi} \dd{z} + \begin{aligned} + \nabla \vb{V} + &= \quad \vu{e}_r \vu{e}_r \pdv{V_r}{r} + + \vu{e}_r \vu{e}_\varphi \pdv{V_\varphi}{r} + + \vu{e}_r \vu{e}_z \pdv{V_z}{r} + \\ + &\quad\: + \vu{e}_\varphi \vu{e}_r \bigg( \frac{1}{r} \pdv{V_r}{\varphi} - \frac{V_\varphi}{r} \bigg) + + \vu{e}_\varphi \vu{e}_\varphi \bigg( \frac{1}{r} \pdv{V_\varphi}{\varphi} + \frac{V_r}{r} \bigg) + + \vu{e}_\varphi \vu{e}_z \frac{1}{r} \pdv{V_z}{\varphi} + \\ + &\quad\: + \vu{e}_z \vu{e}_r \pdv{V_r}{z} + + \vu{e}_z \vu{e}_\varphi \pdv{V_\varphi}{z} + + \vu{e}_z \vu{e}_z \pdv{V_z}{z} + \end{aligned} } \end{aligned}$$ -So, for example, an integral over all of space is converted like so: - -$$\begin{aligned} - \iiint_{-\infty}^\infty f(x, y, z) \dd{V} - = \int_{-\infty}^{\infty} \int_0^{2\pi} \int_0^\infty f(r, \varphi, z) \: r \dd{r} \dd{\varphi} \dd{z} -\end{aligned}$$ - -The isosurface elements are as follows, where $$S_r$$ is a surface at constant $$r$$, etc.: - $$\begin{aligned} \boxed{ \begin{aligned} - \dd{S}_r = r \dd{\varphi} \dd{z} - \qquad - \dd{S}_\varphi = \dd{r} \dd{z} - \qquad - \dd{S}_z = r \dd{r} \dd{\varphi} + (\vb{U} \cdot \nabla) \vb{V} + &= \quad \vu{e}_r \bigg( U_r \pdv{V_r}{r} + \frac{U_\varphi}{r} \pdv{V_r}{\varphi} + U_z \pdv{V_r}{z} + - \frac{U_\varphi V_\varphi}{r} \bigg) + \\ + &\quad\: + \vu{e}_\varphi \bigg( U_r \pdv{V_\varphi}{r} + \frac{U_\varphi}{r} \pdv{V_\varphi}{\varphi} + U_z \pdv{V_\varphi}{z} + + \frac{U_\varphi V_r}{r} \bigg) + \\ + &\quad\: + \vu{e}_z \bigg( U_r \pdv{V_z}{r} + \frac{U_\varphi}{r} \pdv{V_z}{\varphi} + U_z \pdv{V_z}{z} \bigg) \end{aligned} } \end{aligned}$$ -Similarly, the normal vector element $$\dd{\vu{S}}$$ for an arbitrary surface is given by: - $$\begin{aligned} \boxed{ - \dd{\vu{S}} - = \vu{e}_r \: r \dd{\varphi} \dd{z} - + \vu{e}_\varphi \dd{r} \dd{z} - + \vu{e}_z \: r \dd{r} \dd{\varphi} + \begin{aligned} + \nabla^2 \vb{V} + &= \quad \vu{e}_r \bigg( \pdvn{2}{V_r}{r} + \frac{1}{r} \pdv{V_r}{r} + \frac{1}{r^2} \pdvn{2}{V_r}{\varphi} + + \pdvn{2}{V_r}{z} - \frac{2}{r^2} \pdv{V_\varphi}{\varphi} - \frac{V_r}{r^2} \bigg) + \\ + &\quad\: + \vu{e}_\varphi \bigg( \pdvn{2}{V_\varphi}{r} + \frac{1}{r} \pdv{V_\varphi}{r} + \frac{1}{r^2} \pdvn{2}{V_\varphi}{\varphi} + + \pdvn{2}{V_\varphi}{z} + \frac{2}{r^2} \pdv{V_r}{\varphi} - \frac{V_\varphi}{r^2} \bigg) + \\ + &\quad\: + \vu{e}_z \bigg( \pdvn{2}{V_z}{r} + \frac{1}{r} \pdv{V_z}{r} + + \frac{1}{r^2} \pdvn{2}{V_z}{\varphi} + \pdvn{2}{V_z}{z} \bigg) + \end{aligned} } \end{aligned}$$ -And finally, the tangent vector element $$\dd{\vu{\ell}}$$ of a given curve is as follows: - $$\begin{aligned} \boxed{ - \dd{\vu{\ell}} - = \vu{e}_r \dd{r} - + \vu{e}_\varphi \: r \dd{\varphi} - + \vu{e}_z \dd{z} + \begin{aligned} + \nabla \cdot \overline{\overline{\mathbf{T}}} + &= \quad \vu{e}_r \bigg( \pdv{T_{rr}}{r} + \frac{1}{r} \pdv{T_{\varphi r}}{\varphi} + \pdv{T_{zr}}{z} + + \frac{T_{rr}}{r} - \frac{T_{\varphi \varphi}}{r} \bigg) + \\ + &\quad\: + \vu{e}_\varphi \bigg( \pdv{T_{r \varphi}}{r} + \frac{1}{r} \pdv{T_{\varphi \varphi}}{\varphi} + \pdv{T_{z \varphi}}{z} + + \frac{T_{r \varphi}}{r} + \frac{T_{\varphi r}}{r} \bigg) + \\ + &\quad\: + \vu{e}_z \bigg( \pdv{T_{rz}}{r} + \frac{1}{r} \pdv{T_{\varphi z}}{\varphi} + \pdv{T_{zz}}{z} + + \frac{T_{rz}}{r} \bigg) + \end{aligned} } \end{aligned}$$ + ## References 1. M.L. Boas, *Mathematical methods in the physical sciences*, 2nd edition, Wiley. +2. B. Lautrup, + *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, + CRC Press. diff --git a/source/know/concept/hilbert-space/index.md b/source/know/concept/hilbert-space/index.md index 57926ce..42b9cb1 100644 --- a/source/know/concept/hilbert-space/index.md +++ b/source/know/concept/hilbert-space/index.md @@ -8,19 +8,20 @@ categories: layout: "concept" --- -A **Hilbert space**, also called an **inner product space**, is an -abstract **vector space** with a notion of length and angle. +A **Hilbert space**, also called an **inner product space**, +is an abstract **vector space** with a notion of length and angle. + ## Vector space -An abstract **vector space** $$\mathbb{V}$$ is a generalization of the -traditional concept of vectors as "arrows". It consists of a set of -objects called **vectors** which support the following (familiar) -operations: +An abstract **vector space** $$\mathbb{V}$$ is a generalization +of the traditional concept of vectors as "arrows". +It consists of a set of objects called **vectors** +which support the following (familiar) operations: -+ **Vector addition**: the sum of two vectors $$V$$ and $$W$$, denoted $$V + W$$. -+ **Scalar multiplication**: product of a vector $$V$$ with a scalar $$a$$, denoted $$a V$$. ++ **Vector addition**: the sum of two vectors $$V$$ and $$W$$, denoted by $$V + W$$. ++ **Scalar multiplication**: product of a vector $$V$$ with a scalar $$a$$, denoted by $$a V$$. In addition, for a given $$\mathbb{V}$$ to qualify as a proper vector space, these operations must obey the following axioms: @@ -34,24 +35,26 @@ space, these operations must obey the following axioms: + **Multiplication is distributive over scalars**: $$(a + b)V = aV + bV$$ + **Multiplication is distributive over vectors**: $$a (U + V) = a U + a V$$ -A set of $$N$$ vectors $$V_1, V_2, ..., V_N$$ is **linearly independent** if -the only way to satisfy the following relation is to set all the scalar coefficients $$a_n = 0$$: +A set of $$N$$ vectors $$V_1, V_2, ..., V_N$$ is **linearly independent** +if the only way to satisfy the following relation +is to set all the scalar coefficients $$a_n = 0$$: $$\begin{aligned} \mathbf{0} = \sum_{n = 1}^N a_n V_n \end{aligned}$$ -In other words, these vectors cannot be expressed in terms of each -other. Otherwise, they would be **linearly dependent**. +In other words, these vectors cannot be expressed in terms of each other. +Otherwise, they would be **linearly dependent**. -A vector space $$\mathbb{V}$$ has **dimension** $$N$$ if only up to $$N$$ of -its vectors can be linearly indepedent. All other vectors in -$$\mathbb{V}$$ can then be written as a **linear combination** of these $$N$$ **basis vectors**. +A vector space $$\mathbb{V}$$ has **dimension** $$N$$ +if only up to $$N$$ of its vectors can be linearly indepedent. +All other vectors in $$\mathbb{V}$$ can then be written +as a **linear combination** of these $$N$$ **basis vectors**. -Let $$\vu{e}_1, ..., \vu{e}_N$$ be the basis vectors, then any -vector $$V$$ in the same space can be **expanded** in the basis according to -the unique weights $$v_n$$, known as the **components** of $$V$$ -in that basis: +Let $$\vu{e}_1, ..., \vu{e}_N$$ be the basis vectors, +then any vector $$V$$ in the same space can be **expanded** +in the basis according to the unique weights $$v_n$$, +known as the **components** of $$V$$ in that basis: $$\begin{aligned} V = \sum_{n = 1}^N v_n \vu{e}_n @@ -71,19 +74,20 @@ $$\begin{gathered} \end{gathered}$$ + ## Inner product -A given vector space $$\mathbb{V}$$ can be promoted to a **Hilbert space** -or **inner product space** if it supports an operation $$\Inprod{U}{V}$$ -called the **inner product**, which takes two vectors and returns a -scalar, and has the following properties: +A given vector space $$\mathbb{V}$$ can be promoted to a **Hilbert space** or **inner product space** +if it supports an operation $$\Inprod{U}{V}$$ called the **inner product**, +which takes two vectors and returns a scalar, +and has the following properties: + **Skew symmetry**: $$\Inprod{U}{V} = (\Inprod{V}{U})^*$$, where $${}^*$$ is the complex conjugate. + **Positive semidefiniteness**: $$\Inprod{V}{V} \ge 0$$, and $$\Inprod{V}{V} = 0$$ if $$V = \mathbf{0}$$. + **Linearity in second operand**: $$\Inprod{U}{(a V + b W)} = a \Inprod{U}{V} + b \Inprod{U}{W}$$. -The inner product describes the lengths and angles of vectors, and in -Euclidean space it is implemented by the dot product. +The inner product describes the lengths and angles of vectors, +and in Euclidean space it is implemented by the dot product. The **magnitude** or **norm** $$|V|$$ of a vector $$V$$ is given by $$|V| = \sqrt{\Inprod{V}{V}}$$ and represents the real positive length of $$V$$. @@ -123,6 +127,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Infinite dimensions As the dimensionality $$N$$ tends to infinity, things may or may not diff --git a/source/know/concept/material-derivative/index.md b/source/know/concept/material-derivative/index.md index 7225053..d11287d 100644 --- a/source/know/concept/material-derivative/index.md +++ b/source/know/concept/material-derivative/index.md @@ -88,26 +88,6 @@ $$\begin{aligned} } \end{aligned}$$ -Where the advective term is to be evaluated in the following way in Cartesian coordinates: - -$$\begin{aligned} - (\va{v} \cdot \nabla) \va{U} - = - \begin{bmatrix} v_x \\ v_y \\ v_z \end{bmatrix} - \cdot - \begin{bmatrix} - \displaystyle\pdv{U_x}{x} & \displaystyle\pdv{U_x}{y} & \displaystyle\pdv{U_x}{z} \\ - \displaystyle\pdv{U_y}{x} & \displaystyle\pdv{U_y}{y} & \displaystyle\pdv{U_y}{z} \\ - \displaystyle\pdv{U_z}{x} & \displaystyle\pdv{U_z}{y} & \displaystyle\pdv{U_z}{z} - \end{bmatrix} - = - \begin{bmatrix} - v_x \displaystyle\pdv{U_x}{x} & v_y \displaystyle\pdv{U_x}{y} & v_z \displaystyle\pdv{U_x}{z} \\ - v_x \displaystyle\pdv{U_y}{x} & v_y \displaystyle\pdv{U_y}{y} & v_z \displaystyle\pdv{U_y}{z} \\ - v_x \displaystyle\pdv{U_z}{x} & v_y \displaystyle\pdv{U_z}{y} & v_z \displaystyle\pdv{U_z}{z} - \end{bmatrix} -\end{aligned}$$ - ## References 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, diff --git a/source/know/concept/spherical-coordinates/index.md b/source/know/concept/spherical-coordinates/index.md index f037182..01c5a61 100644 --- a/source/know/concept/spherical-coordinates/index.md +++ b/source/know/concept/spherical-coordinates/index.md @@ -8,9 +8,9 @@ categories: layout: "concept" --- -**Spherical coordinates** are an extension of polar coordinates to 3D. +**Spherical coordinates** are an extension of polar coordinates $$(r, \varphi)$$ to 3D. The position of a given point in space is described by -three coordinates $$(r, \theta, \varphi)$$, defined as: +three variables $$(r, \theta, \varphi)$$, defined as: * $$r$$: the **radius** or **radial distance**: distance to the origin. * $$\theta$$: the **elevation**, **polar angle** or **colatitude**: @@ -18,6 +18,10 @@ three coordinates $$(r, \theta, \varphi)$$, defined as: * $$\varphi$$: the **azimuth**, **azimuthal angle** or **longitude**: angle from the positive $$x$$-axis, typically in the counter-clockwise sense. +Note that this is the standard notation among physicists, +but mathematicians often switch the definitions of $$\theta$$ and $$\varphi$$, +while still writing $$(r, \theta, \varphi)$$. + Cartesian coordinates $$(x, y, z)$$ and the spherical system $$(r, \theta, \varphi)$$ are related by: @@ -32,104 +36,142 @@ $$\begin{aligned} \end{aligned}$$ Conversely, a point given in $$(x, y, z)$$ -can be converted to $$(r, \theta, \varphi)$$ -using these formulae: +can be converted to $$(r, \theta, \varphi)$$ using these formulae, +where $$\mathtt{atan2}$$ is the 2-argument arctangent, +which is needed to handle the signs correctly: $$\begin{aligned} \boxed{ - r = \sqrt{x^2 + y^2 + z^2} - \qquad - \theta = \arccos(z / r) - \qquad - \varphi = \mathtt{atan2}(y, x) + \begin{aligned} + r + &= \sqrt{x^2 + y^2 + z^2} + \\ + \theta + &= \arccos(z / r) + \\ + \varphi + &= \mathtt{atan2}(y, x) + \end{aligned} } \end{aligned}$$ -The spherical coordinate system is +Spherical coordinates form an [orthogonal curvilinear system](/know/concept/orthogonal-curvilinear-coordinates/), -whose scale factors $$h_r$$, $$h_\theta$$ and $$h_\varphi$$ we want to find. -To do so, we calculate the differentials of the Cartesian coordinates: +whose **scale factors** $$h_r$$, $$h_\theta$$ and $$h_\varphi$$ we need. +To get those, we calculate the unnormalized local basis: $$\begin{aligned} - \dd{x} &= \dd{r} \sin\theta \cos\varphi + \dd{\theta} r \cos\theta \cos\varphi - \dd{\varphi} r \sin\theta \sin\varphi + h_r \vu{e}_r + &= \vu{e}_x \pdv{x}{r} + \vu{e}_y \pdv{y}{r} + \vu{e}_z \pdv{z}{r} \\ - \dd{y} &= \dd{r} \sin\theta \sin\varphi + \dd{\theta} r \cos\theta \sin\varphi + \dd{\varphi} r \sin\theta \cos\varphi + &= \vu{e}_x \sin{\theta} \cos{\varphi} + \vu{e}_y \sin{\theta} \sin{\varphi} + \vu{e}_z \cos{\theta} \\ - \dd{z} &= \dd{r} \cos\theta - \dd{\theta} r \sin\theta -\end{aligned}$$ - -And then we calculate the line element $$\dd{\ell}^2$$, -skipping many terms thanks to orthogonality: - -$$\begin{aligned} - \dd{\ell}^2 - &= \:\:\:\: \dd{r}^2 \big( \sin^2(\theta) \cos^2(\varphi) + \sin^2(\theta) \sin^2(\varphi) + \cos^2(\theta) \big) + h_\theta \vu{e}_\theta + &= \vu{e}_x \pdv{x}{\theta} + \vu{e}_y \pdv{y}{\theta} + \vu{e}_z \pdv{z}{\theta} \\ - &\quad + \dd{\theta}^2 \big( r^2 \cos^2(\theta) \cos^2(\varphi) + r^2 \cos^2(\theta) \sin^2(\varphi) + r^2 \sin^2(\theta) \big) + &= \vu{e}_x \: r \cos{\theta} \cos{\varphi} + \vu{e}_y \: r \cos{\theta} \sin{\varphi} - \vu{e}_z \: r \sin{\theta} \\ - &\quad + \dd{\varphi}^2 \big( r^2 \sin^2(\theta) \sin^2(\varphi) + r^2 \sin^2(\theta) \cos^2(\varphi) \big) + |