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# The curl of a vector field
> *Definition*: the Levi-Civita permutation symbol is defined as
>
> $$
> e_{ijk} = \begin{cases} 0 &\text{ if $i,j,k$ are identical}, \\ 1 &\text{ if the permutation $(i,j,k)$ is even}, \\ -1 &\text{ if the permutation $(i,j,k)$ is odd}.\end{cases}
> $$
>
>
The curl of a vector field may describe the circulation of a vector field and is defined below.
> *Definition*: derivation and definition is missing for now.
Note that the "cross product " between the nabla operator and the vector field $\mathbf{v}$ does not imply anything and is only there for notational sake. An alternative to this notation is using $\text{rot } \mathbf{v}$ to denote the curl or rotation.
> *Theorem*: the curl of a vector field $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ for a curvilinear coordinate system is defined as
>
> $$
> \nabla \times \mathbf{v}(\mathbf{x}) = \frac{1}{\sqrt{g(\mathbf{x})}} e^{ijk} \partial_i \big(v_j(\mathbf{x}) \big) \mathbf{a}_k(\mathbf{x}),
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
??? note "*Proof*:"
Will be added later.
The curl of a vector field for a ortho-curvilinear coordinate system may also be derived and can be found below.
> *Corollary*: the curl of a vector field $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ for a ortho-curvilinear coordinate system is defined as
>
> $$
> \nabla \times \mathbf{v}(\mathbf{x}) = \frac{1}{h_1 h_2 h_3} e^{ijk} \partial_i \big(h_j v_{(j)}(\mathbf{x}) \big) h_k \mathbf{e}_{(k)},
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
??? note "*Proof*:"
Will be added later.
Please note that the scaling factors may also depend on the position $\mathbf{x} \in \mathbb{R}^3$ depending on the coordinate system.
> *Proposition*: let $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ be a vector field and $f: \mathbb{R}^3 \to \mathbb{R}$ a scalar field then we have
>
> $$
> \begin{align*}
> \nabla \cdot \big(\nabla \times \mathbf{v}(\mathbf{x}) \big) &= 0, \\
> \nabla \times \nabla f(\mathbf{x}) &= \mathbf{0},
> \end{align*}
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
??? note "*Proof*:"
Will be added later.
Similarly to the [divergence theorem](divergence.md#divergence-in-curvilinear-coordinates) for the divergence, the curl is related to Kelvin-Stokes theorem given below.
> *Theorem*: let $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ be a smooth vector field and $A \subset \mathbb{R}^3$ a closed surface with boundary curve $C \subset \mathbb{R}^3$ piecewise smooth we have that
>
> $$
> \oint_C \big\langle \mathbf{v}(\mathbf{x}), d\mathbf{x} \big\rangle = \int_A \big\langle \nabla \times \mathbf{v}(\mathbf{x}), d\mathbf{A} \big\rangle,
> $$
>
> is true.
??? note "*Proof*:"
Will be added later.

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# Curves
> *Definition*: a curve is a continuous vector-valued function of one real-valued parameter.
>
> * A closed curve $\mathbf{c}: \mathbb{R} \to \mathbb{R}^3$ is defined by $\mathbf{c}(a) = \mathbf{c}(b)$ with $a \in \mathbb{R}$ the begin point and $b \in \mathbb{R}$ the end point.
> * A simple curve has no crossings.
<br>
> *Definition*: let $\mathbf{c}: \mathbb{R} \to \mathbb{R}^3$ be a curve, the derivative of $\mathbf{c}$ is defined as the velocity of the curve $\mathbf{c}'$. The length of the velocity is defined as the speed of the curve $\|\mathbf{c}'\|$.
<br>
> *Proposition*: let $\mathbf{c}: \mathbb{R} \to \mathbb{R}^3$ be a curve, the velocity of the curve $\mathbf{c}'$ is tangential to the curve.
??? note "*Proof*:"
Will be added later.
<br>
> *Definition*: let $\mathbf{c}: \mathbb{R} \to \mathbb{R}^3$ be a differentiable curve, the infinitesimal arc length $ds: \mathbb{R} \to \mathbb{R}$ of the curve is defined as
>
> $$
> ds(t) := \|d \mathbf{c}(t)\| = \|\mathbf{c}'(t)\|dt
> $$
>
> for all $t \in \mathbb{R}$.
<br>
> *Theorem*: let $\mathbf{c}: \mathbb{R} \to \mathbb{R}^3$ be a differentiable curve, the arc length $s: \mathbb{R} \to \mathbb{R}$ of a section that start at $t_0 \in \mathbb{R}$ is given by
>
> $$
> s(t) = \int_{t_0}^t \|\mathbf{c}'(u)\|du,
> $$
>
> for all $t \in \mathbb{R}$.
??? note "*Proof*:"
Will be added later.
## Arc length parameterization
To obtain a speed of unity everywhere on the curve, or differently put equidistant arc lengths between each time step an arc length parameterization can be performed. It can be performed in 3 steps:
1. For a given curve determine the arc length function for a given start point.
2. Find the inverse of the arc length function if it exists.
3. Adopt the arc length as variable of the curve.
Obtaining a speed of unity on the entire defined curve.
For example consider a curve $\mathbf{c}: \mathbb{R} \to \mathbb{R}^3$ given in Cartesian coordinates by
$$
\mathbf{c}(\phi) = \begin{pmatrix} r \cos \phi \\ r \sin \phi \\ \rho r \phi\end{pmatrix},
$$
for all $\phi \in \mathbb{R}$ with $r, \rho \in \mathbb{R}^+$.
Determining the arc length function $s: \mathbb{R} \to \mathbb{R}$ of the curve
$$
\begin{align*}
s(\phi) &= \int_0^\phi \|\mathbf{c}'(u)\|du, \\
&= \int_0^\phi r \sqrt{1 + \rho^2}du, \\
&= \phi r \sqrt{1 + \rho^2},
\end{align*}
$$
for all $\phi \in \mathbb{R}$. It may be observed that $s$ is a bijective mapping.
The inverse of the arc length function $s^{-1}: \mathbb{R} \to \mathbb{R}$ is then given by
$$
s^{-1}(\phi) = \frac{\phi}{r\sqrt{a + \rho^2}},
$$
for all $\phi \in \mathbb{R}$.
The arc length parameterization $\mathbf{c}_s: \mathbb{R} \to \mathbb{R}^3$ of $\mathbf{c}$ is then given by
$$
\mathbf{c}_s(\phi) = \mathbf{c}(s^{-1}(\phi)) = \begin{pmatrix} r \cos (\phi / r\sqrt{a + \rho^2}) \\ r \sin (\phi / r\sqrt{a + \rho^2}) \\ \rho \phi / \sqrt{a + \rho^2}\end{pmatrix},
$$
for all $\phi \in \mathbb{R}$.

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# Curvilinear coordinate systems
In this section curvilinear coordinate systems will be presented, these are coordinate systems that are based on a set of basis vectors that are neither orthognal nor normalized.
> *Principle*: space can be equipped with a smooth and continuous coordinate net.
## Covariant basis
> *Definition*: consider a coordinate system $(x_1, x_2, x_3)$ that is mapped by $\mathbf{x}: \mathbb{R}^3 \to \mathbb{R}^3$ with respect to a reference coordinate system. Producing a position vector for every combination of coordinate values.
>
> * For two coordinates fixed, a coordinate curve is obtained.
> * For one coordinate fixed, a coordinate surface is obtained.
We will now use this coordinate system described as $\mathbf{x}$ to formulate a set of basis vectors.
> *Definition*: for a valid coordinate system $\mathbf{x}$ a set of linearly independent covariant (local) basis vectors can be described by
>
> $$
> \mathbf{a}_i(x_1, x_2, x_3) := \partial_i \mathbf{x}(x_1, x_2, x_3),
> $$
>
> for all $(x_1, x_2, x_3) \in \mathbb{R}^3$ and $i \in \{1, 2, 3\}$.
Obtaining basis vectors that are tangential to the corresponding coordinate curves. Therefore any vector $\mathbf{u} \in \mathbb{R}^3$ can be written in terms of its components with respect to this basis
$$
\mathbf{u} = \sum_{i=1}^3 u_i \mathbf{a}_i
$$
with $u_{1,2,3} \in \mathbb{R}$ the components.
> *Definition*: the Einstein summation convention omits the summation symbol and is defined by
>
> $$
> \mathbf{u} = \sum_{i=1}^3 u_i \mathbf{a}_i = u^i \mathbf{a}_i,
> $$
>
> with $u^{1,2,3} \in \mathbb{R}$ the contravariant components. The definition states that
>
> 1. When an index appears twice in a product, one as a subscript and once as a superscript, summation over that index is implied.
> 2. A superscript that appears in denominator counts as a subscript.
This convention makes writing summation a lot easier, though one may see it as a little unorthodox.
## The metric tensor
> *Definition*: for two vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^3$ that are represented in terms of a covariant basis, the scalar product is given by
>
> $$
> \langle \mathbf{u}, \mathbf{v} \rangle = u^i v^j \langle \mathbf{a}_i, \mathbf{a}_j \rangle = u^i v^j g_{ij},
> $$
>
> with $g_{ij}$ the components of a structure that is called the metric tensor given by
>
> $$
> (g_{ij}) := \begin{pmatrix} \langle \mathbf{a}_1, \mathbf{a}_1 \rangle & \langle \mathbf{a}_1, \mathbf{a}_2 \rangle & \langle \mathbf{a}_1, \mathbf{a}_3 \rangle \\ \langle \mathbf{a}_2, \mathbf{a}_1 \rangle & \langle \mathbf{a}_2, \mathbf{a}_2 \rangle & \langle \mathbf{a}_2, \mathbf{a}_3 \rangle \\ \langle \mathbf{a}_3, \mathbf{a}_1 \rangle & \langle \mathbf{a}_3, \mathbf{a}_2 \rangle & \langle \mathbf{a}_3, \mathbf{a}_3 \rangle \end{pmatrix}.
> $$
For the special case of an orthogonal set of basis vectors, all of-diagonal elements are zero and we have a metric tensor $g_{ij}$ given by
$$
(g_{ij}) = \begin{pmatrix} \langle \mathbf{a}_1, \mathbf{a}_1 \rangle & & \\ & \langle \mathbf{a}_2, \mathbf{a}_2 \rangle & \\ & & \langle \mathbf{a}_3, \mathbf{a}_3 \rangle\end{pmatrix} = \begin{pmatrix} h_1^2 & & \\ & h_2^2 & \\ & & h_3^2\end{pmatrix},
$$
with $h_i = \sqrt{\langle \mathbf{a}_i, \mathbf{a}_i \rangle} = \|\mathbf{a}_i\|$ the scale factors for $i \in \{1, 2, 3\}$.
> *Theorem*: the determinant of the metric tensor $g := \det(g_{ij})$ can be written as the square of the scalar triple product of the covariant basis vectors
>
> $$
> g = \langle \mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3 \rangle^2.
> $$
??? note "*Proof*:"
Will be added later.
<br>
> *Corollary*: consider a covariant basis and the infinitesimal coordinate transformations $(dx_1, dx_2, dx_3)$ spanned by the covariant basis then the volume defined by these infinitesimal coordinate transformations is given by
>
> $$
> \begin{align*}
> dV &= \langle dx_1 \mathbf{a}_2, dx_2 \mathbf{a}_1, dx_3 \mathbf{a}_3 \rangle, \\
> &= \sqrt{g} dx_1 dx_2 dx_3,
> \end{align*}
> $$
>
> by definition of the scalar triple product. For a function $f: \mathbb{R}^3 \to \mathbb{R}$ its integral in the domain $D \subseteq \mathbb{R}^3$ with $D = [a_1, b_1] \times [a_2, b_2] \times [a_3, b_3]$ and $a_i, b_i \in \mathbb{R}$ for $i \in \{1, 2, 3\}$ closed may be given by
>
> $$
> \int_D f(x_1, x_2, x_3)dV = \int_{a_1}^{b_1} \int_{a_2}^{b_2} \int_{a_3}^{b_3} f(x_1, x_2, x_3) \sqrt{g} dx_1 dx_2 dx_3.
> $$
??? note "*Proof*:"
Will be added later.
## Contravariant basis
The covariant basis vectors have been constructed as tangential vectors of the coordinate curves. An alternative basis can be constructed from vectors that are perpendicular to coordinate surfaces.
> *Definition*: for a valid set of covariant basis vectors the contravariant basis vectors may be defined, given by
>
> $$
> \begin{align*}
> \mathbf{a}^1 &:= \frac{1}{\sqrt{g}} (\mathbf{a}_2 \times \mathbf{a}_3), \\
> \mathbf{a}^2 &:= \frac{1}{\sqrt{g}} (\mathbf{a}_3 \times \mathbf{a}_1), \\
> \mathbf{a}^3 &:= \frac{1}{\sqrt{g}} (\mathbf{a}_1 \times \mathbf{a}_2)
> \end{align*}
> $$
From this definition it follows that $\langle \mathbf{a}^i, \mathbf{a}_j \rangle = \delta_j^i$, with $\delta_j^i$ the Kronecker delta defined by
> *Definition*: the Kronecker delta $\delta_{ij}$ is defined as
>
> $$
> \delta_{ij} = \begin{cases} 1 &\text{ if } i = j, \\ 0 &\text{ if } i \neq j.\end{cases}
> $$
A metric tensor for contravariant basis vectors may be defined. With which the relations between covariant and contravariant quantities can be found.
> *Definition*: the components of the metric tensor for contravariant basis vectors are defined as
>
> $$
> g^{ij} := \langle \mathbf{a}^i, \mathbf{a}^j \rangle,
> $$
>
> therefore the metric tensor for contravariant basis vectors is given by
>
> $$
> (g^{ij}) = \begin{pmatrix} \langle \mathbf{a}^1, \mathbf{a}^1 \rangle & \langle \mathbf{a}^1, \mathbf{a}^2 \rangle & \langle \mathbf{a}^1, \mathbf{a}^3 \rangle \\ \langle \mathbf{a}^2, \mathbf{a}^1 \rangle & \langle \mathbf{a}^2, \mathbf{a}^2 \rangle & \langle \mathbf{a}^2, \mathbf{a}^3 \rangle \\ \langle \mathbf{a}^3, \mathbf{a}^1 \rangle & \langle \mathbf{a}^3, \mathbf{a}^2 \rangle & \langle \mathbf{a}^3, \mathbf{a}^3 \rangle \end{pmatrix}.
> $$
These relations are stated in the proposition below.
> *Proposition*: considering the two ways of representing the vector $\mathbf{u} \in \mathbb{R}^3$ given by
>
> $$
> \mathbf{u} = u^i \mathbf{a}_i = u_i \mathbf{a}^i.
> $$
>
> From the definitions given above the relations between the covariant and contravariant quantities of the vector $\mathbf{u}$ have been found to be
>
> $$
> u_i = g_{ij} u^j, \qquad \mathbf{a}_i = g_{ij} \mathbf{a}^j,
> $$
>
> $$
> u^i = g^{ij} u_j, \qquad \mathbf{a}^i = g^{ij} \mathbf{a}_j.
> $$
??? note "*Proof*:"
Will be added later.
By combining the expressions for the components a relation can be established between $g_{ij}$ and $g^{ij}$.
> *Theorem*: the components of the metric tensor for covariant and contravariant basis vectors are related by
>
> $$
> g_{ij} g^{jk} = \delta_i^k.
> $$
??? note "*Proof*:"
Will be added later.
This is the index notation for $(g_{ij})(g^{ij}) = I$, with $I$ the identity matrix, therefore we have
$$
(g^{ij}) = (g_{ij})^{-1},
$$
concluding that both matrices are nonsingular.
> *Corollary*: let $\mathbf{u} \in \mathbb{R}^3$ be a vector, for orthogonal basis vectors it follows that the covariant and contravariant basis vectors are proportional by
>
> $$
> \mathbf{a}^i = \frac{1}{h_i^2} \mathbf{a}_i,
> $$
>
> and for the components of $\mathbf{u}$ we have
>
> $$
> u^i = \frac{1}{h_i^2} u_i,
> $$
>
> for all $i \in \{1, 2, 3\}$.
??? note "*Proof*:"
Will be added later.
Therefore it also follows that for the special case of orthogonal basis vectors the metric tensor for contrariant basis vectors $(g^{ij})$ is given by
$$
(g^{ij}) = \begin{pmatrix} \langle \mathbf{a}^1, \mathbf{a}^1 \rangle & & \\ & \langle \mathbf{a}^2, \mathbf{a}^2 \rangle & \\ & & \langle \mathbf{a}^3, \mathbf{a}^3 \rangle\end{pmatrix} = \begin{pmatrix} \frac{1}{h_1^2} & & \\ & \frac{1}{h_2^2} & \\ & & \frac{1}{h_3^2}\end{pmatrix},
$$
with $h_i = \sqrt{\langle \mathbf{a}_i, \mathbf{a}_i \rangle} = \|\mathbf{a}_i\|$ the scale factors for $i \in \{1, 2, 3\}$.
## Phyiscal components
A third representation of vectors uses physical components and normalized basis vectors.
> *Definition*: from the above corollary the physical component representation for a vector $\mathbf{u} \in \mathbb{R}^3$ can be defined as
>
> $$
> \mathbf{e}_{(i)} := h_i \mathbf{a}^i = \frac{1}{h_i} \mathbf{a}_i,
> $$
>
> $$
> u_{(i)} := h_i u^i = \frac{1}{h_i} u_i,
> $$
>
> for all $i \in \{1, 2, 3\}$.
Contributing to the physical component representation given by
$$
\mathbf{u} = u^{(i)} \mathbf{e}_{(i)},
$$
for $i \in \{1, 2, 3\}$.
> *Proposition*: obtaining the properties
>
> $$
> \langle \mathbf{e}_{(i)}, \mathbf{e}_{(i)} \rangle = \frac{1}{h_i^2} \langle \mathbf{a}_i, \mathbf{a}_i \rangle = 1,
> $$
>
> and for vectors $\mathbf{u}, \mathbf{v} \in \mathbb{R}^3$ we have
>
> $$
> \langle \mathbf{u}, \mathbf{v} \rangle = u^{(i)} v_{(i)}.
> $$
??? note "*Proof*:"
Will be added later.
In particular the length of a vector $\mathbf{u} \in \mathbb{R}^3$ can then be determined by
$$
\|\mathbf{u}\| = \sqrt{u^{(i)} u_{(i)}}.
$$
We will discuss as an example the representations of the cartesian, cylindrical and spherical coordinate systems viewed from a cartesian perspective. This means that the coordinate maps are based on the cartesian interpretation of them. Every other interpretation could have been used, but our brains have a preference for cartesian it seems.
Let $\mathbf{x}: \mathbb{R}^3 \to \mathbb{R}^3$ map a cartesian coordinate system given by
$$
\mathbf{x}(x,y,z) = \begin{pmatrix} x \\ y \\ z\end{pmatrix},
$$
then we have the covariant basis vectors given by
$$
\mathbf{a}_i(x,y,z) = \partial_i \mathbf{x}(x,y,z),
$$
obtaining $\mathbf{a}_1 = \begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix}, \mathbf{a}_2 = \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}, \mathbf{a}_3 = \begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}$.
It may be observed that this set of basis vectors is orthogonal. Therefore the scaling factors are given by $h_1 = 1, h_2 = 1, h_3 = 1$ as to be expected for the reference.
<br>
Let $\mathbf{x}: \mathbb{R}^3 \to \mathbb{R}^3$ map a cylindrical coordinate system given by
$$
\mathbf{x}(r,\theta,z) = \begin{pmatrix} r \cos \theta \\ r \sin \theta \\ z\end{pmatrix},
$$
then we have the covariant basis vectors given by
$$
\mathbf{a}_i(r,\theta,z) = \partial_i \mathbf{x}(r,\theta,z),
$$
obtaining $\mathbf{a}_1(\theta) = \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0\end{pmatrix}, \mathbf{a}_2(r, \theta) = \begin{pmatrix} -r\sin \theta \\ r \cos \theta \\ 0\end{pmatrix}, \mathbf{a}_3 = \begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}$.
It may be observed that this set of basis vectors is orthogonal. Therefore the scaling factors are given by $h_1 = 1, h_2 = r, h_3 = 1$.
<br>
Let $\mathbf{x}: \mathbb{R}^3 \to \mathbb{R}^3$ map a spherical coordinate system given by
$$
\mathbf{x}(r,\theta,\varphi) = \begin{pmatrix}r \cos \theta \sin \varphi \\ r \sin \theta \sin \varphi \\ r \cos \varphi\end{pmatrix},
$$
using the mathematical convention, then we have the covariant basis vectors given by
$$
\mathbf{a}_i(r,\theta,\varphi) = \partial_i \mathbf{x}(r,\theta,\varphi),
$$
obtaining $\mathbf{a}_1(\theta, \varphi) = \begin{pmatrix} \cos \theta \sin \varphi \\ \sin \theta \sin \varphi\\ \cos \varphi\end{pmatrix}, \mathbf{a}_2(r, \theta, \varphi) = \begin{pmatrix} -r\sin \theta \sin \varphi \\ r \cos \theta \sin \varphi \\ 0\end{pmatrix}, \mathbf{a}_3 = \begin{pmatrix} r \cos \theta \cos \varphi \\ r \sin \theta \cos \varphi \\ - r \sin \varphi\end{pmatrix}$.
It may be observed that this set of basis vectors is orthogonal. Therefore the scaling factors are given by $h_1 = 1, h_2 = r \sin \varphi, h_3 = r$.

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# The divergence of a vector field
## Flux densities
Considering a medium with a mass density $\rho: \mathbb{R}^4 \to \mathbb{R}$ and a velocity field $\mathbf{v}: \mathbb{R}^4 \to \mathbb{R}^3$ consisting of a orientable finite sized surface element $d\mathbf{A} \in \mathbb{R}^3$.
> *Definition*: a surface must be orientable for the surface integral to exist. It must be able to move along the surface continuously without ending up on the "other side".
We then have a volume $dV \in \mathbb{R}$ defined by the parallelepiped formed by $dV = \langle d\mathbf{x}, d\mathbf{A} \rangle$ with the vector $d\mathbf{x} = \mathbf{v} dt$, for a time interval $dt \in \mathbb{R}$. The mass flux $d\Phi$ per unit of time through the surface element $d\mathbf{A}$ may then be given by
$$
d \Phi = \rho \langle \mathbf{v}, d\mathbf{A} \rangle.
$$
The mass flux $\Phi: \mathbb{R} \to \mathbb{R}$ through a orientable finite sized surface $A \subseteq \mathbb{R}^3$ is then given by
$$
\Phi(t) = \int_A \Big\langle \rho(\mathbf{x}, t) \mathbf{v}(\mathbf{x}, t), d\mathbf{A} \Big\rangle,
$$
for all $t \in \mathbb{R}$.
> *Definition*: let $\mathbf{\Gamma}: \mathbb{R}^4 \to \mathbb{R}^3$ be the (mass) flux density given by
>
> $$
> \mathbf{\Gamma}(\mathbf{x},t) := \rho(\mathbf{x},t) \mathbf{v}(\mathbf{x},t),
> $$
>
> for all $(\mathbf{x},t) \in \mathbb{R}^4$.
The (mass) flux density is a vector-valued function of position and time that expresses the rate of transport of a quantity per unit of time of area perpendicular to its direction.
The mass flux $\Phi$ through $A$ may then be given by
$$
\Phi(t) = \int_A \Big\langle \mathbf{\Gamma}(\mathbf{x},t), d\mathbf{A} \Big\rangle,
$$
for all $t \in \mathbb{R}$.
## Definition of the divergence
> *Definition*: the divergence of a flux density $\mathbf{\Gamma}: \mathbb{R}^4 \to \mathbb{R}^3$ is given by
>
> $$
>\nabla \cdot \mathbf{\Gamma}(\mathbf{x}, t) = \lim_{V \to 0} \frac{1}{V} \int_A \Big\langle \mathbf{\Gamma}(\mathbf{x}, t), d\mathbf{A} \Big\rangle,
> $$
>
> for all $(\mathbf{x}, t) \in \mathbb{R}^4$ for a volume $V \subset \mathbb{R}^3$ with closed orientable boundary surface $A \subset V$.
Note that this "dot product" between the nabla operator and the flux density $\mathbf{\Gamma}$ does not imply anything and is only there for notational sake. An alternative to this notation is using $\text{div } \mathbf{\Gamma}$ to denote the divergence.
The definition of the divergence can be interpreted with the particle mass balance for a medium with a particle density $n: \mathbb{R}^4 \to \mathbb{R}$ and a velocity field $\mathbf{v}: \mathbb{R}^4 \to \mathbb{R}^3$. Furthermore we have that the particles are produced at a rate $S: \mathbb{R}^4 \to \mathbb{R}^3$.
We then have the particle flux density $\mathbf{\Gamma}: \mathbb{R}^4 \to \mathbb{R}^3$ given by
$$
\mathbf{\Gamma}(\mathbf{x},t) = n(\mathbf{x},t) \mathbf{v}(\mathbf{x},t),
$$
for all $(\mathbf{x},t) \in \mathbb{R}^4$.
For a volume $V \subseteq \mathbb{R}^3$ with a closed orientable boundary surface $A \subseteq \mathbb{R}^3$ we have that the amount of particles inside this volume for a specific time is given by
$$
\int_V n(\mathbf{x}, t) dV,
$$
for all $t \in \mathbb{R}$. We have that the particle flux through $A$ is given by
$$
\int_A \Big\langle \mathbf{\Gamma}(\mathbf{x},t), d\mathbf{A} \Big\rangle,
$$
for all $t \in \mathbb{R}$ and we have that the particle production rate in this volume $V$ is given by
$$
\int_V S(\mathbf{x}, t)dV,
$$
for all $t \in \mathbb{R}$. We conclude that the sum of the particle flux through $A$ and the time derivative of the particles inside the volume $V$ must be equal to the production rate inside this volume $V$. Therefore we have
$$
d_t \int_V n(\mathbf{x}, t) dV + \int_A \Big\langle \mathbf{\Gamma}(\mathbf{x},t), d\mathbf{A} \Big\rangle = \int_V S(\mathbf{x}, t)dV,
$$
for all $t \in \mathbb{R}$.
Assuming the system is stationary the time derivative of the particles inside the volume $V$ must vanish. The divergence is then defined to be the total production for a position $\mathbf{x} \in V$.
## Divergence in curvilinear coordinates
> *Theorem*: the divergence of a flux density $\mathbf{\Gamma}: \mathbb{R}^4 \to \mathbb{R}^3$ for a curvilinear coordinate system is given by
>
> $$
> \nabla \cdot \mathbf{\Gamma}(\mathbf{x}, t) = \frac{1}{\sqrt{g(\mathbf{x})}} \partial_i \Big(\Gamma^i(\mathbf{x},t) \sqrt{g(\mathbf{x})} \Big)
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$ and $i \in \{1, 2, 3\}$.
??? note "*Proof*:"
Will be added later.
We may also give the divergence for ortho-curvilinear coordinate systems.
> *Corollary*: the divergence of a flux density $\mathbf{\Gamma}: \mathbb{R}^3 \to \mathbb{R}^3$ for a ortho-curvilinear coordinate system is given by
>
> $$
> \nabla \cdot \mathbf{\Gamma}(\mathbf{x}, t) = \frac{1}{h_1h_2h_3} \partial_i \Big(\Gamma^{(i)}(\mathbf{x},t)\frac{1}{h_i} h_1 h_2 h_3 \Big)
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$ and $i \in \{1, 2, 3\}$.
??? note "*Proof*:"
Will be added later.
Please note that the scaling factors may also depend on the position $\mathbf{x} \in \mathbb{R}^3$ depending on the coordinate system.
It has been found that the volume integral over the divergence of a vector field is equal to the integral of the vector field itself over the surface that bounds the volume. It is known as the divergence theorem and is given below.
> *Theorem*: for a volume $V \subset \mathbb{R}^3$ with a closed and orientable boundary surface $A \subset V$ with a continuously differentiable flux density $\mathbf{\Gamma}: \mathbb{R}^4 \to \mathbb{R}^3$ we have that
>
> $$
> \int_A \Big\langle \mathbf{\Gamma}(\mathbf{x}, t), d\mathbf{A} \Big\rangle = \int_V \nabla \cdot \mathbf{\Gamma}(\mathbf{x}, t) dV,
> $$
>
> for all $t \in \mathbb{R}$.
??? note "*Proof*:"
Will be added later.

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# The gradient of a scalar field
Considering a scalar field $f: \mathbb{R}^3 \to \mathbb{R}$, if the field is continuously differentiable we have
$$
df(\mathbf{x}) = \partial_i f(\mathbf{x}) dx_i,
$$
for all $x \in \mathbb{R}^4$ and $i \in \{1,2,3\}$. We may rewrite this in terms of a displacement vector $d\mathbf{x} = \mathbf{a}_i dx^i$ into
$$
\begin{align*}
df &= \partial_i f(\mathbf{x}) \delta^i_j dx^j, \\
&= \partial_i f(\mathbf{x}) \langle \mathbf{a}^i, \mathbf{a}_j \rangle dx^j, \\
&= \partial_i f(\mathbf{x})\langle \mathbf{a}^i, d\mathbf{x} \rangle.
\end{align*}
$$
> *Definition*: the gradient of a scalar field $f: \mathbb{R}^3 \to \mathbb{R}$ for a curvilinear coordinate system is defined as
>
> $$
> \nabla f(\mathbf{x}) := \partial_i f(\mathbf{x}) \mathbf{a}^i,
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
Note that in the differentation section of multivariable calculus the definition of the gradient explicitly for Cartesian coordinate systems was given. This definition is rather general for all coordinate systems, although is limited to only 3 dimensions.
> *Proposition*: let $f: \mathbb{R}^3 \to \mathbb{R}$ be a scalar field, the gradient of $f$ points in the direction for which $f$ has the greatest increase.
??? note "*Proof*:"
Will be added later.
The following definition introduces the material derivative, it may appear to be a little unorthodox.
> *Definition*: let $f: \mathbb{R}^4 \to \mathbb{R}$ be a scalar field and $\mathbf{x}: \mathbb{R} \to \mathbb{R}^3$ a vector field. The material derivative of $f$ is defined as
>
> $$
> f'(\mathbf{x}(t), t) := \big\langle \nabla f(\mathbf{x}, t), \mathbf{x}'(t) \big\rangle + \partial_t f(\mathbf{x}, t),
> $$
>
> for all $t \in \mathbb{R}$. Note that the gradient in the scalar product is only taken for $\mathbf{x}$.
The following definition introduces the directional derivative.
> *Definition*: let $f: \mathbb{R}^3 \to \mathbb{R}$ be a scalar field and $\mathbf{v} \in \mathbb{R}^3$ a normalised vector such that $\|\mathbf{v}\| = 1$. The directional derivative of $f$ in the direction of $\mathbf{v}$ is defined as
>
> $$
> D_{\mathbf{v}} f(\mathbf{x}) := \big\langle \mathbf{v}, \nabla f(\mathbf{x}) \big\rangle,
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
Consider a vector field $\mathbf{u}: \mathbb{R}^3 \to \mathbb{R}^3$, the integral of $\mathbf{u}$ along a curve $C \subset \mathbb{R}^3$ is given by
$$
\int_C \big\langle \mathbf{u}(\mathbf{x}), d\mathbf{x} \big\rangle.
$$
> *Theorem*: let $f: \mathbb{R}^3 \to \mathbb{R}$ be a scalar field and consider a curve $C \subset \mathbb{R}^3$ then we have
>
> $$
> \int_C \big\langle \nabla f(\mathbf{x}), d\mathbf{x} \big\rangle = \big[f(\mathbf{x}) \big]_C.
> $$
??? note "*Proof*:"
Will be added later.

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# Vector operators
## Properties of the gradient, divergence and curl
> *Proposition*: let $a,b \in \mathbb{R}$, $f,g: \mathbb{R}^3 \to \mathbb{R}$ be differentiable scalar fields and $\mathbf{u}, \mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ be differentiable vector fields. Then we have the following identities:
>
> **Linearity:**
>
> $$
> \begin{align*}
> \nabla (af + bg) &= a \nabla f + b \nabla g, \\
> \nabla \cdot (a\mathbf{u} + b \mathbf{v}) &= a (\nabla \cdot \mathbf{u}) + b (\nabla \cdot \mathbf{v}), \\
> \nabla \times (a\mathbf{u} + b \mathbf{v}) &= a (\nabla \times \mathbf{u}) + b (\nabla \times\mathbf{v}).
> \end{align*}
> $$
>
> **Multiplication rules:**
>
> $$
> \begin{align*}
> \nabla (fg) &= f \nabla g+ g \nabla f, \\
> \nabla \cdot (f \mathbf{u}) &= f (\nabla \cdot \mathbf{u}) + \langle \nabla f, \mathbf{u} \rangle, \\
> \nabla \cdot (\mathbf{u} \times \mathbf{v}) &= \langle \nabla \times \mathbf{u}, \mathbf{v} \rangle - \langle \mathbf{u}, \nabla \times \mathbf{v} \rangle, \\
> \nabla \times (f\mathbf{u}) &= f (\nabla \times \mathbf{u}) + \nabla f \times \mathbf{u}.
> \end{align*}
> $$
??? note "*Proof*:"
Will be added later.
## The laplacian
> *Definition*: the laplacian of a differentiable scalar field $f: \mathbb{R}^3 \to \mathbb{R}$ is defined as
>
> $$
> \nabla^2 f(\mathbf{x}) := \nabla \cdot \nabla f(\mathbf{x}),
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
The notation may be unorthodox for some. An alternative notatation for the laplacian is $\Delta f$, though generally deprecated.
We can also rewrite the laplacian for curvilinear coordinate systems as has been done below.
> *Theorem*: the laplacian of a differentiable scalar field $f: \mathbb{R}^3 \to \mathbb{R}$ for a curvilinear coordinate system is given by
>
> $$
> \nabla^2 f(\mathbf{x}) = \frac{1}{g(\mathbf{x})} \partial_i \Big(\sqrt{g(\mathbf{x})} g^{ij}(\mathbf{x}) \partial_j f(\mathbf{x}) \Big),
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
??? note "*Proof*:"
Will be added later.
The laplacian for a ortho-curvilinear coordinate system may also be derived and can be found below.
> *Corollary*: the laplacian of a differentiable scalar field $f: \mathbb{R}^3 \to \mathbb{R}$ for a ortho-curvilinear coordinate system is given by
>
> $$
> \nabla^2 f(\mathbf{x}) = \frac{1}{h_1 h_2 h_3} \bigg(\partial_1 \Big(\frac{h_2 h_3}{h_1} \partial_1 f(\mathbf{x}) \Big) + \partial_2 \Big(\frac{h_1 h_3}{h_2} \partial_2 f(\mathbf{x}) \Big) + \partial_3 \Big(\frac{h_1 h_2}{h_3} \partial_3 f(\mathbf{x}) \Big) \bigg),
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
??? note "*Proof*:"
Will be added later.
Please note that the scaling factors may also depend on the position $\mathbf{x} \in \mathbb{R}^3$ depending on the coordinate system.
> *Proposition*: the laplacian of a differentiable vector field $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ is given by
>
> $$
> \nabla^2 \mathbf{v}(\mathbf{x}) = \nabla \big(\nabla \cdot \mathbf{v}(\mathbf{x})\big) - \nabla \times \big(\nabla \times \mathbf{v}(\mathbf{x})\big),
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
??? note "*Proof*:"
Will be added much later.
## Potentials
> *Definition*: a vector field $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ is irrotational or curl free if
>
> $$
> \nabla \times \mathbf{v}(\mathbf{x}) = \mathbf{0},
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
If $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ is the gradient of some scalar field $\Phi: \mathbb{R}^3 \to \mathbb{R}$ it is irrotational since
$$
\nabla \times\big (\nabla \Phi(\mathbf{x})\big) = \mathbf{0},
$$
for all $\mathbf{x} \in \mathbb{R}^3$.
> *Proposition*: an irrotational vector field $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ has a scalar potential $\Phi: \mathbb{R}^3 \to \mathbb{R}$ such that
>
> $$
> \mathbf{v}(\mathbf{x}) = \nabla \Phi(\mathbf{x}),
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
??? note "*Proof*:"
Will be added later.
In physics the scalar potential is generally given by the negative of the gradient, both are correct but one is more stupid than the other.
> *Definition*: a vector field $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ is solenoidal or divergence-free if
>
> $$
> \nabla \cdot \mathbf{v}(\mathbf{x}) = 0,
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
If $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ is the curl of some vector field $\mathbf{u}: \mathbb{R}^3 \to \mathbb{R}^3$ it is solenoidal since
$$
\nabla \cdot \big(\nabla \times \mathbf{u}(\mathbf{x}) \big) = 0,
$$
for all $\mathbf{x} \in \mathbb{R}^3$.
> *Proposition*: a solenoidal vector field $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ has a vector potential $\mathbf{u}: \mathbb{R}^3 \to \mathbb{R}^3$ such that
>
> $$
> \mathbf{v}(\mathbf{x}) = \nabla \times \mathbf{u}(\mathbf{x}),
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
??? note "*Proof*:"
Will be added later.
The theorem below is the Helmholtz decomposition theorem and states that every vector field can be written in terms of two potentials.
> *Theorem*: every vector field $\mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3$ can be written in terms of a scalar $\Phi: \mathbb{R}^3 \to \mathbb{R}$ and a vector $\mathbf{u}: \mathbb{R}^3 \to \mathbb{R}^3$ potential as
>
> $$
> \mathbf{v}(\mathbf{x}) = \nabla \Phi(\mathbf{x}) + \nabla \times \mathbf{u}(\mathbf{x}),
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
??? note "*Proof*:"
Will be added later.
It then follows that the scalar and vector potentials can be determined for a volume $V \subset \mathbb{R}^3$ with a boundary surface $A \subset \mathbb{R}^3$ that encloses the domain $V$.
> *Corollary*: the scalar $\Phi: \mathbb{R}^3 \to \mathbb{R}$ and vector $\mathbf{u}: \mathbb{R}^3 \to \mathbb{R}^3$ potentials for a volume $V \subset \mathbb{R}^3$ with a boundary surface $A \subset \mathbb{R}^3$ that encloses the domain $V$ are given by
>
> $$
> \begin{align*}
> \Phi(\mathbf{x}) &= \frac{1}{4\pi} \int_V \frac{\nabla \cdot \mathbf{v}(\mathbf{r})}{\|\mathbf{x} - \mathbf{r}\|}dV - \frac{1}{4\pi} \oint_A \bigg\langle \frac{1}{\|\mathbf{x} - \mathbf{r}\|} \mathbf{v}(\mathbf{r}), d\mathbf{A} \bigg\rangle, \\
> \\
> \mathbf{u}(\mathbf{x}) &= \frac{1}{4\pi} \int_V \frac{\nabla \times \mathbf{v}(\mathbf{r})}{\|\mathbf{x} - \mathbf{r}\|}dV - \frac{1}{4\pi} \oint_A \frac{1}{\|\mathbf{x} - \mathbf{r}\|} \mathbf{v}(\mathbf{r}) \times d\mathbf{A},
> \end{align*}
> $$
>
> for all $\mathbf{x} \in \mathbb{R}^3$.
??? note "*Proof*:"
Will be added later.

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# Vectors and geometry
## Axiomatic geometry
The defining property of axiomatic geometry is that it can be introduced without any reference to a coordinate system. The 5 postulates of classical geometry are listed below.
1. A straight line segment can be drawn between any pair of two points.
2. A straight line segment can be extended indefinitely into a straight line.
3. A line segment is the radius of a circle with one of the end points as its center.
4. All right angles are congruent.
The fifth postulate as formulated below is only valid for Euclidean geometry; flat space informally.
<ol start="5">
<li>Given in a plane, a line and a point not on that line there is only one line through that point that does not intersect with the other line.</li>
</ol>
## Vectors
Referring to linear algebra section [vector spaces](../../../mathematics/linear-algebra/vector-spaces.md) for the axioms of the Euclidean vector space and its vector definitions. Some vector products in 3 dimensional Euclidean space are defined below
> *Definition*: the Euclidean scalar product of $\mathbf{u}, \mathbf{v} \in \mathbb{R}^3$ is given by
>
> $$
> \langle \mathbf{u}, \mathbf{v} \rangle := \|\mathbf{u}\| \|\mathbf{v}\| \cos \varphi,
> $$
>
> with $\|\mathbf{u}\|$ and $\|\mathbf{v}\|$ the length of $\mathbf{u}$ and $\mathbf{v}$ and the $\varphi$ the angle between $\mathbf{u}$ and $\mathbf{v}$.
It follows than that for $\mathbf{v} = \mathbf{u}$ we have
$$
\|\mathbf{u}\| = \langle \mathbf{u}, \mathbf{u} \rangle.
$$
> *Definition*: the Euclidean cross product of $\mathbf{u}, \mathbf{v} \in \mathbb{R}^3$ is given by
>
> $$
> \|\mathbf{u} \times \mathbf{v}\| := \|\mathbf{u}\| \|\mathbf{v}\| \sin \varphi,
> $$
>
> with $\|\mathbf{u}\|$ and $\|\mathbf{v}\|$ the length of $\mathbf{u}$ and $\mathbf{v}$ and the $\varphi$ the angle between $\mathbf{u}$ and $\mathbf{v}$. Defining the area of a parallelogram span by $\mathbf{u}$ and $\mathbf{v}$. The normal direction of the surface is obtained by not taking the length of the cross product.
The scalar and cross product can be combined obtaining a parallelepiped spanned by three 3-dimensional vectors.
> *Definition*: the Euclidean scalar triple of $\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbb{R}^3$ is given by
>
> $$
> \langle \mathbf{u}, \mathbf{v}, \mathbf{w} \rangle := \langle \mathbf{u}, \mathbf{v} \times \mathbf{w} \rangle,
> $$
>
> defining the volume of a parallelepiped spanned by $\mathbf{u}, \mathbf{v}$ and $\mathbf{w}$.
Let $J$ be a $3 \times 3$ matrix given by $J = (\mathbf{u}^T, \mathbf{v}^T, \mathbf{w}^T)$, the Euclidean scalar product may also be defined as
$$
\langle \mathbf{u}, \mathbf{v}, \mathbf{w} \rangle = \det (J),
$$
with $\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbb{R}^3$. We also have that
$$
\langle \mathbf{u}, \mathbf{v}, \mathbf{w} \rangle^2 = \det (J^TJ).
$$