7.1 Scalars and Vectors

7.1.1 Introduction

Scalar Quantities. There are two types of quantities describing fluid flows: scalar and vector quantities. Scalar quantities are quantities that are described only by a magnitude. They do not have a direction of action. For fluids the relevant scalar quantities are

• temperature T , measured in K

• density ρ, measured in kg m^− 3

• pressure p, measured in Pa

In the following we will use the symbol ψ whenever referring to a scalar quantity.

Vector Quantities. Vector quantities have both magnitude and a direction of action. The relevant vector quantities of fluid flows are

• position $\overrightarrow{s}$, measured in m

• velocity $\overrightarrow{v}$, measured in m s^− 1

• acceleration $\overrightarrow{a}$, measured in m s^− 2

In the following we will use the symbol $\overrightarrow{F}$ whenever referring to a vector quantity.

Point Notation. In the following we will refer to a point in space using the symbol S. A point S is defined by the coordinates that will be written as

• S_x,y,z in Cartesian coordinates

• S_r,φ,z in cylindrical coordinates

• S_r,φ in polar coordinates

• S_r,φ,θ in spherical coordinates

Vector Notation. As stated, vectors will be written with an arrow, as in the case of $\overrightarrow{F}$. Vectors are always referred to by their respective coordinate system. In Cartesian coordinates, $\overrightarrow{s}$ would be written

$F=\left(\begin{array}{l}{F}_x\\ F_y\\ F_z\end{array}\right)=F_x{\overrightarrow{e}}_x+F_y{\overrightarrow{e}}_y+F_z{\overrightarrow{e}}_z$

Matrix Notation. We will refer to matrices as bold letters. The matrix A would be given as

$\mathrm{a}=\left(\begin{array}{cccc}\hfill a_{1,1}\hfill & \hfill a_{1,2}\hfill & \hfill \cdots \hfill & \hfill a_{1,n}\hfill \\ \hfill a_{2,1}\hfill & \hfill a_{2,2}\hfill & \hfill \cdots \hfill & \hfill a_{2,n}\hfill \\ \hfill ⋮\hfill & \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill \\ \hfill a_{m,1}\hfill & \hfill a_{m,2}\hfill & \hfill \cdots \hfill & \hfill a_{m,n}\hfill \end{array}\right)$

We will require matrices often, particularly when discussing numerical methods that use the orthogonal axis system ${\overrightarrow{e}}_x,{\overrightarrow{e}}_y,\text{and}{\overrightarrow{e}}_z$ of the Cartesian coordinate system. If other coordinate systems are used, different orthogonal axis vectors are used, e.g${\overrightarrow{e}}_{\phi }\mathrm{or}{\overrightarrow{e}}_r\text{.}$

7.1.2 Vector Operations

Flow fields cannot be described without the use of vectors. It therefore is essential to summarize some of the most important vector operations that we will use when discussing fluid mechanics.

7.1.2.4 Cross Product

The cross product is used to find the vector which forms the normal to both input vectors. The result therefore is obviously a vector. It is defined as

$\begin{array}{ll}\overrightarrow{F}\times \overrightarrow{G}\hfill & =\left(\begin{array}{l}{F}_1\\ F_2\\ F_n\end{array}\right)\times \left(\begin{array}{l}{G}_1\\ G_2\\ G_n\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{l}{F}_2{G}_3-F_3{G}_2\\ F_3{G}_1-F_1{G}_3\\ F_1{G}_2-F_2{G}_1\end{array}\right)\hfill \end{array}$

It may be diffcult to remember this formula. It is significantly easier if done graphically (see Fig. 7.1). For this, write the vector a second time just underneath the first vector. Now start with the left column, second line (F₂) and multiply it with the value one row farther down on the right side (G₃). Subtract from this the two values found when starting on the right column, second line (G₂) and moving to the left, and one line down to (F₃). You have effectively made a cross. This will give the first entry in the resulting vector. Proceed by repeating this procedure moving one line down to start (F₃) which will give you the second entry in the resulting vector. Repeat a third time, starting again one line farther down (F₁).

7.1.3 Operators

Fluid mechanics uses two operators in order to simplify notation, i.e., the nabla operator $\overrightarrow{\nabla }$ and the Laplace operator Δ. Both operators can be applied to scalars ψ and vectors $\overrightarrow{F}$ and we will briefly introduce the most important operations and their meaning.

7.2.1 Gauss’s Theorem

We begin with Gauss’s theorem named after German mathematician Carl Friedrich Gauß [3]. This theorem is often referred to as the divergence theorem. It is used in the derivation of some of the most elementary concepts in fluid mechanics and physics. Gauss’s theorem is used whenever the effect of a vector field across the boundary of a closed control volume is to be considered. Vector fields acting on the curved boundary of a control volume are notoriously difficult to describe analytically. Here, Gauss’s theorem is used, which allows a neat transformation trick.

Let an arbitrary vector field $\overrightarrow{F}$ act across a closed volume V with the closed boundary or surface $A=\partial V$ (see Fig. 7.2a). The action of the vector field across the boundary $\partial V$ can be represented as the action of the vector field across the infinitesimal surface area dA. This surface area can be described by means of its normal vector $\overrightarrow{n}$. If the vector field acts on dA under an angle (ϕ and θ), only the projection of $\overrightarrow{F}$ on $\overrightarrow{n}$ must be considered. Thus the sum of the effect of the vector field on the surface is described by the integral of $\overrightarrow{F}$ over $\partial V$ as

${\displaystyle \oint \overrightarrow{F}\cdot \overrightarrow{n}dA}$

Here, the symbol ${\displaystyle \oint }$ indicates that the integral is to be taken over a closed surface, i.e., the start and the end point of the integration are identical (see section 7.2.5). The integral is the surface integral of the normal component of the vector field $\overrightarrow{F}$ acting on the infinitesimal surface $dA=\mathit{dxdy}\text{.}$. Now these integrals are very difficult to calculate as the normal vector $\overrightarrow{n}$ is a function of the position and thus not constant in space. This is where the Gauss’s theorem is used. We will not detail the mathematical basics of the theorem here. The theorem states that for closed volumes, this surface integral can be written as a volume integral

${\displaystyle \oint \overrightarrow{F}\cdot \overrightarrow{n}dA={\displaystyle \int \overrightarrow{\nabla }\cdot \overrightarrow{F}dV}}$

So in simple words: The action of a vector field across the boundary of a volume must equal the divergence (see section 7.1.3.3) inside of the volume. If we take mass transport as an example of a vector field this can be interpreted as follows: If mass flows across the boundary of a control volume, the sum of mass lost from the system is equal to the divergence of the flow inside of the system. So, if the mass inside of the system primarily diverges, i.e., it flows in outward direction it must naturally cross the boundary of the system, i.e., the surface $A=\partial V\text{.}$ So if we want to know the total amount of mass flowing across the boundary of the system, we do not need to evaluate the surface integral ${\displaystyle \oint \overrightarrow{F}\cdot \overrightarrow{n}dA}\text{.}$ It is enough to know the divergence of the mass flow inside of the volume to come up with the same result.

In many cases in fluid physics, it is significantly more difficult to calculate the surface integral of a vector field than to calculate the integral over the divergence of the vector field. In these cases, Eq. 7.10 makes calculating complicated surface integrals significantly simpler.

7.2.2 Stoke’s Theorem

The next theorem that we will discuss is named after Irish mathematician and physicist George Stokes [4]. Compared to the Gauss’s theorem, which allows substituting the surface integral of a vector field by the volume integral of the vector field’s divergence, Stoke’s theorem is used to simplify line integrals of vector fields on surfaces. As we will see, this theorem allows substituting a line integral of a vector field on a surface by the surface integral of the curl or rotation (see section 7.1.3.4) of the vector field on a surface. This is why it is sometimes referred to as the flux theorem.

Let us consider Fig. 7.2b-i. A surface A has a closed boundary $\partial A$ which is oriented such that the integration variable $d\overrightarrow{r}$ progresses along the boundary in counterclockwise fashion. In this case, the normal of the surface is (per definition) directed outward. This surface is under the influence of a vector field $\overrightarrow{F}$. The vector field may be defined in the whole volume, but we will only consider it on the surface as this is the only region we are interested in. Now let us consider the line integral

${\oint }_{\partial A}\overrightarrow{F}\cdot d\overrightarrow{r}$

This integral sums of the effect of the vector field $\overrightarrow{F}$ on the surface as we move along $d\overrightarrow{r}$. Moving from 1 to 2, the vector field $\overrightarrow{F}$ is directed in the same direction as $d\overrightarrow{r}$. Thus the product $\overrightarrow{F}\cdot d\overrightarrow{r}$ will result in a positive contribution to the line integral in this region. From 2 to 3, $d\overrightarrow{r}$ and $\overrightarrow{F}$ are perpendicular. Therefore we will not have a contribution to the overall line integral. Going from 3 to 4, $d\overrightarrow{r}$ and $\overrightarrow{F}$ point in opposite directions and thus, there will be a negative contribution to the line integral which is equal (in magnitude) to the line integral obtained for the section from 1 to 2. In the section from 4 to 1, $d\overrightarrow{r}$ and $\overrightarrow{F}$ are again perpendicular, therefore there is no contribution to the line integral. In this case we can see, that the overall line integral would be pretty much 0, as the contributions along the section 1 to 2 and 3 to 4 cancel each other.

Now let us consider the case displayed in Fig. 7.2b-ii. Here, we have a vector field which is not uniform across the surface A. It is rotating, i.e., it has a curl. Repeating the exercise we will see, that in this region, all sections of the circumference will contribute to the overall line integral and we will obtain a value bigger than zero. The difference compared to the case displayed in Fig. 7.2b-i is the curl of the vector field. So we already see that the overall curl of the vector field will have an influence on the magnitude of the line integral. This is exactly what the Stoke’s theorem states

${\oint }_{\partial A}\overrightarrow{F}\cdot d\overrightarrow{r}={\displaystyle \int \overrightarrow{\nabla }\times \overrightarrow{F}dA}$

As we can see, the theorem allows us to transform a line integral over a vector field $\overrightarrow{F}$ into a surface integral over the curl of the vector field $\overrightarrow{F}$. This is just as the Gauss’s theorem allowed us to transform a surface integral $\overrightarrow{F}$ into a volume integral over the divergence of the vector field $\overrightarrow{F}$. In many cases, these two theorems allow replacing a difficult integral with an integral which is easier to solve.

7.2.3 Green’s Theorem

We will also briefly introduce the third theorem that is often stated in combination with Gauss’s and Stoke’s theorem: the theorem of Green. It was originally described by and thus named after British mathematician George Green [1]. Green’s theorem is a special case of Stoke’s theorem (see section 7.2.2). It refers to the case where a planar surface A, which is defined in the x/y-plane, is under the influence of a vector field $\overrightarrow{F}$ that is only a function of x and y, but not of z (see Fig. 7.2c). So, this vector field does not change as we move along the z-direction. We can write this vector field as

$\overrightarrow{F}=\left(\begin{array}{l}{F}_x\left(x,y\right)\\ F_y\left(x,y\right)\\ \hfill 0\hfill \end{array}\right)$

So the vector field is planar and it is described by its components F_x and F_y that are only functions of x and y. Now Stoke’s theorem can be used to evaluate the line integral ${\displaystyle {\oint }_{\partial A}\overrightarrow{F}\cdot d\overrightarrow{r}}$ of this vector field and the plane A according to Eq. 7.11. For the vector field $\overrightarrow{F}$ defined by Eq. 7.12, the curl can be calculated using Eq. 7.5 as

$\overrightarrow{\nabla }\times \overrightarrow{F}=\left(\begin{array}{l}\frac{\partial F_z}{\partial y}-\frac{\partial F_y}{\partial z}\\ \frac{\partial F_x}{\partial z}-\frac{\partial F_z}{\partial x}\\ \frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\end{array}\right)=\left(\begin{array}{l}\hfill 0\hfill \\ \hfill 0\hfill \\ \frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\end{array}\right)$

So using this, we can reformulate Eq. 7.11 for this specific case to be

${\oint }_{\partial A}\overrightarrow{F}\cdot d\overrightarrow{r}={\int }_A\left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)dA$

Eq. 7.13 is more commonly known as the theorem of Green.

7.2.4 Reynolds’ Transport Theorem

There is a fourth theorem you may potentially come across which is the Reynolds transport theorem. The theorem was originally described by Reynolds¹[6] and is often considered the three-dimensional equivalent of the Leibniz integral rule (see section 5.2.4). Compared to the “big three” theorems we have just described, the Reynolds transport theorem may be the least known.

For the Reynolds transport theorem we look at a space-fixed control volume Ω through which fluid (or any likewise field) passes (see Fig. 7.3). You can think of the control volume Ω as being a pinhole through which we look onto the flow. We don’t see all of the flow, but our field of view is sufficient to serve as control volume.

Now let us suppose we are interested in a property of the flow B which could be, e.g., the mass m. We use two different notations of this property. B is the standard notation of the property given in its physical units. The second notation is the specific property b, which is simply given by dividing B by the mass. The unit of b is the physical unit of B divided by mass. Now the total amount of property within Ω will be given by

$B={\int }_{\Omega }\mathit{\rho bdV}$

You may wonder why the density ρ of the fluid has to be brought in here. If you consider the units, you will see that this equation is correct only if bringing in the density as

$\begin{array}{ll}\left[\rho \right]\hfill & =\mathrm{kg}{\mathrm{m}}^{-3}\hfill \\ \left[b\right]\hfill & =\left[B\right]{\mathrm{kg}}^{-1}\hfill \\ \left[bdV\right]\hfill & =\left[B\right]{\mathrm{m}}^3{\mathrm{kg}}^{-1}\hfill \\ \left[\mathit{\rho bdV}\right]\hfill & =\left[B\right]{\mathrm{k}\mathrm{g}\mathrm{m}}^3{\mathrm{kg}}^{-1}{\mathrm{m}}^{-3}=\left[B\right]\hfill \end{array}$

We are interested in which way the amount of B changes over time inside of Ω, i.e., we seek to find $\frac{\mathit{dB}}{dt}\text{.}$ Intuitively we know that the changes are due to two mechanisms

1. influx ${\stackrel{.}{B}}_{\mathrm{in}}$ and outflux ${\stackrel{.}{B}}_{\mathrm{out}}$ into and out of the control volume

2. generation and consumption of B over time inside of the control volume given by $\frac{\partial B}{\partial t}$

In- and Outflux. Calculating the in- and outflux is straightforward using Gauss’s theorem which tells us that

$\frac{\partial B_{\mathrm{in}‐/\mathrm{outflux}}}{dt}={\stackrel{.}{B}}_{\mathrm{in}}-{\stackrel{.}{B}}_{\mathrm{out}}=\underset{\partial \Omega }{\oint }B\cdot \overrightarrow{n}dA$

which means that the change of quantity inside of Ω can be tracked by summing up all fluxes across the surface of the control volume. Please note that this notation is also true for any dimension in which case B would be a vector.

Generation and Consumption. The generation and consumption of B inside of Ω can be described using Eq. 7.14 as

$\frac{\partial B_{\mathrm{generation}/\mathrm{consumption}}}{dt}=\frac{\partial }{\partial t}{\int }_{\Omega }\mathit{\rho bdV}$

Final Form. Using Eq. 7.15 and Eq. 7.16 we can now formulate the final form of the theorem as

$\begin{array}{ll}\frac{\partial B}{\partial t}\hfill & =\frac{\partial B_{\mathrm{in}‐/\mathrm{outflux}}}{dt}+\frac{\partial B_{\mathrm{generation}/\mathrm{consumption}}}{dt}\hfill \\ \hfill & =\frac{\partial }{\partial t}{\int }_{\Omega }\mathit{\rho bdV}+{\oint }_{\partial \Omega }B\cdot \overrightarrow{n}dA\hfill \\ \hfill & ={\int }_{\Omega }\frac{\partial \rho b}{\partial t}dV+{\oint }_{\partial \Omega }B\cdot \overrightarrow{n}dA\hfill \end{array}$

where Eq. 7.17 gives the theorem in its most commonly encountered form.

7.2.5 A Note on Notations

Single or Multiple Integral Signs. In many textbooks, the surface or volume integrals of Gauss’s, Stoke’s, and Green’s theorem are given with multiple integral. As an example, you may find Gauss’s theorem in the following (equivalent) forms

$\begin{array}{ll}{\displaystyle ∯\overrightarrow{F}\cdot \overrightarrow{n}dA}\hfill & ={\displaystyle {\displaystyle \iiint \overrightarrow{\nabla }\cdot \overrightarrow{F}dV}}\hfill \\ {\displaystyle \oint }\overrightarrow{F}\cdot \overrightarrow{n}dA\hfill & ={\displaystyle \int }\overrightarrow{\nabla }\cdot \overrightarrow{F}dV\hfill \end{array}$

As stated, they represent the same information. The first version highlights that the integral dA is a surface integral as dA = dx dy from which we can derive that we (usually) have to perform two integrations (unless x and y are somehow interdependent). Identically, the triple integral ∫∫∫ indicated that dV is a triple integral, most likely a volume integral. The second form of Gauss’s theorem says exactly the same. However, it implies that the reader knows that dA is a surface integral and dV a volume integral. In most cases this is trivial, as the variables are setup this way. Just as dA = dx dy the volume integral would read dV = dx dy dz. If this is clear from the context, the simple integrals will do just fine. In general, single integral notation is used in this book.

Closed Path, Surface, and Volume Integrals. There is a distinct difference between the simple integral ∫ and the close integral . The regular integral assumes a lower and an upper boundary for the integration both must be known in order to determine the integration constant correctly. A closed integral will assume the lower and the upper boundary to be the same value. In other words, the integration will proceed along a path, or a surface and terminate at the same point at which it started. This is why it is referred to as a closed integral. This distinction must be made.

Indicating the Integration Region. Usually the region on which the integral is applied will be indicated next to the integration symbol if the integral is not specific about this. As an example, consider Eq. 7.13. The information that the integral is to be taken on the boundary $\partial A$ is indicated by ${\oint }_{\partial A}$ because the integration variable $d\overrightarrow{r}$ is also defined inside of the area (not just on the boundary). Therefore this additional information clarifies the integral. Usually if the integral is taken over a surface dA, this information could also be applied to the integration sign ${\int }_A$, but it is redundant.

7.3 Coordinate System Transformation

7.3.1 Introduction

We will primarily use four coordinate systems: Cartesian, cylindrical, polar, and spherical coordinates (see Fig. 7.4).

Cartesian Coordinate System. The Cartesian coordinate system (see Fig. 7.4a) uses the orthogonal axis vectors ${\overrightarrow{e}}_x,{\overrightarrow{e}}_y,\mathrm{and}{\overrightarrow{e}}_z\text{.}$ This coordinate system is used primarily for systems with rectangular shape, e.g., microfluidic channels with rectangular cross-section.

Cylindrical Coordinate System. The cylindrical coordinate system (see Fig. 7.4b) uses the orthogonal axis vectors ${\overrightarrow{e}}_r$ (radius), ${\overrightarrow{e}}_{\phi }$ (rotation around ${\overrightarrow{e}}_z$, azimuthal angle), and the ${\overrightarrow{e}}_z$ axis. This coordinate system is used primarily for systems with a tubular shape, e.g., capillaries.

Polar Coordinate System. The polar coordinate system (see Fig. 7.4c) is the planar, i.e., two-dimensional version of the cylindrical coordinate system. It uses the orthogonal axis vectors ${\overrightarrow{e}}_r$ (radius) and ${\overrightarrow{e}}_{\phi }$ (rotation around the origin). It is mainly used to describe rotationally-symmetric planar flow problems.

Spherical Coordinate System. The spherical coordinate system (see Fig. 7.4d) uses the orthogonal axis vectors ${\overrightarrow{e}}_{\rho }$ (radius in space; ρ is used instead of r to reflect the fact that the radius is out-of-plane), ${\overrightarrow{e}}_{\theta }$ axis (zenith angle, rotation about the normal of the plane defined by the rotation about φ in the origin) and ${\overrightarrow{e}}_{\phi }$ (rotation around ${\overrightarrow{e}}_z$, which is not used in the spherical coordinate system, azimuthal angle). This coordinate system is used rarely, but lends itself well, e.g., for the fluid mechanics of circular particles in flow fields. Please note that in order for the system to form a right-hand system, the order of the axes is r, θ, and ϕ.

Conversion. In many cases it is necessary to transform one coordinate system into another, i.e., to express a vector which was defined in one coordinate system into a different coordinate system. In order to do so, the respective axis vectors of the coordinate systems into which the transformation is to be made must be expressed by the coordinate system from which the transformation is to be made. In the next section, we will study the most important conversions.

7.3.2 Cartesian/Cylindrical Coordinates Conversion

Transforming the Independent Variables. As an example, expressing the position vector $\overrightarrow{s}$

$\overrightarrow{s}\left(r,y,z\right)=x{\overrightarrow{e}}_x+y{\overrightarrow{e}}_y+z{\overrightarrow{e}}_z$

defined in Cartesian coordinates in cylindrical coordinates (see Fig. 7.4b) where we can express the independent variables x, y, and z as the dependent variables r, φ, and z by noting

$x=rcos\phi$

$y=rsin\phi$

$z=z$

which in turn gives us the vector $\overrightarrow{s}$ as function of r,φ, and z

$\overrightarrow{s}\left(r,\phi ,z\right)=rcos\phi {\overrightarrow{e}}_x+rsin\phi {\overrightarrow{e}}_y+z{\overrightarrow{e}}_z$

We have now transformed the vector $\overrightarrow{F}$ into a function of r, φ, and z but we still use the Cartesian coordinate system. In the next step, we need to transform the basis vectors as well.

Transforming the Basis Vectors. Transforming the basis vectors is simply a question of forming the partial derivatives of the position vector $\overrightarrow{s}\left(x,\phi ,z\right)$ givn by Eq. 7.22 along the desired coordinate axis. Once these derivatives have been found, we only need to normalize the vectors, i.e., divide them by their respective norm. For ${\overrightarrow{e}}_r$ we find

$\begin{array}{ll}\frac{\partial \overrightarrow{s}\left(r,\phi ,z\right)}{\partial r}\hfill & =\frac{\partial \left(rcos\phi \right)}{\partial r}{\overrightarrow{e}}_x+\frac{\partial \left(rsin\phi \right)}{\partial r}{\overrightarrow{e}}_y+\frac{\partial z}{\partial r}{\overrightarrow{e}}_z\hfill \\ \hfill & =cos\phi {\overrightarrow{e}}_x+sin\phi {\overrightarrow{e}}_y\hfill \\ ‖\frac{\partial \overrightarrow{s}\left(r,\phi ,z\right)}{\partial r}‖\hfill & =\sqrt{{cos}^2\phi +{sin}^2\phi }=1\hfill \\ {\overrightarrow{e}}_r\hfill & =\frac{\frac{\partial \overrightarrow{s}\left(r,\phi ,z\right)}{\partial r}}{‖\frac{\partial \overrightarrow{s}\left(r,\phi ,z\right)}{\partial r}‖}=cos\phi {\overrightarrow{e}}_x+sin\phi {\overrightarrow{e}}_y\hfill \end{array}$

where we observe that the new basis vector in cylindrical coordinates ${\overrightarrow{e}}_r$ depends on two basis vectors from Cartesian coordinates, i.e., ${\overrightarrow{e}}_x$ and ${\overrightarrow{e}}_y$.

We proceed likewise with ${\overrightarrow{e}}_{\phi }$ to find

$\begin{array}{ll}\frac{\partial \overrightarrow{s}\left(r,\phi ,z\right)}{\partial \phi }\hfill & =\frac{\partial \left(rcos\phi \right)}{\partial \phi }{\overrightarrow{e}}_x+\frac{\partial \left(rsin\phi \right)}{\partial \phi }{\overrightarrow{e}}_y+\frac{\partial z}{\partial \phi }{\overrightarrow{e}}_z\hfill \\ \hfill & =-rsin\phi {\overrightarrow{e}}_x+rcos\phi {\overrightarrow{e}}_y\hfill \\ ‖\frac{\partial \overrightarrow{s}\left(r,\phi ,z\right)}{\partial \phi }‖\hfill & =\sqrt{r^2{sin}^2\phi +r^2{cos}^2\phi }=r\hfill \\ {\overrightarrow{e}}_{\phi }\hfill & =\frac{\frac{\partial \overrightarrow{s}\left(r,\phi ,z\right)}{\partial \phi }}{‖\frac{\partial \overrightarrow{s}\left(r,\phi ,z\right)}{\partial \phi }‖}=-sin\phi {\overrightarrow{e}}_x+cos\phi {\overrightarrow{e}}_y\hfill \end{array}$

where we again observe that the basis vector in cylindrical coordinates is dependent on the two basis vectors ${\overrightarrow{e}}_x$ and ${\overrightarrow{e}}_y$ in Cartesian coordinates. Also the axis vectors depend on the same variable (in this case φ) which makes for interesting derivatives as we will see in a moment.

We do not require a transformation on the last basis vector ${\overrightarrow{e}}_z$ which remains unchanged. We therefore find

${\overrightarrow{e}}_r=cos\phi {\overrightarrow{e}}_x+sin\phi {\overrightarrow{e}}_y$

${\overrightarrow{e}}_{\phi }=-sin\phi {\overrightarrow{e}}_x+cos\phi {\overrightarrow{e}}_y$

${\overrightarrow{e}}_z={\overrightarrow{e}}_z$

We can now solve Eq. 7.23 and Eq. 7.24 for ${\overrightarrow{e}}_x$ and ${\overrightarrow{e}}_y$ to find

$\begin{array}{l}sin\phi {\overrightarrow{e}}_r=sin\phi cos{\overrightarrow{e}}_x+{sin}^2\phi {\overrightarrow{e}}_y\\ cos\phi {\overrightarrow{e}}_{\phi }=-sin\phi cos\phi {\overrightarrow{e}}_x+{cos}^2\phi {\overrightarrow{e}}_y\end{array}$

which after adding up results in

$\begin{array}{ll}cos\phi {\overrightarrow{e}}_{\phi }+sin\phi {\overrightarrow{e}}_r\hfill & =\left({sin}^2\phi +{cos}^2\phi \right){\overrightarrow{e}}_y\hfill \\ {\overrightarrow{e}}_y\hfill & =cos\phi {\overrightarrow{e}}_{\phi }+sin\phi {\overrightarrow{e}}_r\hfill \end{array}$

where we have used the fact that ${sin}^2\phi +{cos}^2\phi =1$ (see Eq. 3.22). If we reinsert Eq. 7.26 into Eq. 7.23 we find

$\begin{array}{ll}{\overrightarrow{e}}_r\hfill & =cos\phi {\overrightarrow{e}}_x+sin\phi \left(cos\phi {\overrightarrow{e}}_{\phi }+sin\phi {\overrightarrow{e}}_r\right)\hfill \\ \hfill & =cos\phi {\overrightarrow{e}}_x+sin\phi cos\phi {\overrightarrow{e}}_{\phi }+{sin}^2\phi {\overrightarrow{e}}_r\hfill \\ cos\phi {\overrightarrow{e}}_x\hfill & =sin\phi cos\phi {\overrightarrow{e}}_{\phi }+\left({sin}^2\phi -1\right){\overrightarrow{e}}_r\hfill \\ \hfill & =sin\phi cos\phi {\overrightarrow{e}}_{\phi }+\left({sin}^2\phi -\left(si{n}^2\phi +co{s}^2\phi \right)\right){\overrightarrow{e}}_r\hfill \\ \hfill & =sin\phi cos\phi {\overrightarrow{e}}_{\phi }-{cos}^2\phi {\overrightarrow{e}}_r\hfill \\ {\overrightarrow{e}}_x\hfill & =cos\phi {\overrightarrow{e}}_r-sin\phi {\overrightarrow{e}}_{\phi }\hfill \end{array}$

which in summary gives us

${\overrightarrow{e}}_x=cos\phi {\overrightarrow{e}}_r-sin\phi {\overrightarrow{e}}_{\phi }$

${\overrightarrow{e}}_y=sin\phi {\overrightarrow{e}}_r-cos\phi {\overrightarrow{e}}_{\phi }$

${\overrightarrow{e}}_z={\overrightarrow{e}}_z$

We now have all the equations to express the Cartesian basis vectors in terms of the cylindrical unit vectors (see Eq. 7.23, Eq. 7.24, and Eq. 7.25) and v.v. (see Eq. 7.28, Eq. 7.29, and Eq. 7.30). In order to convert a vector $\overrightarrow{F}$ from Cartesian to cylindrical coordinates we use

$\begin{array}{ll}\left(\begin{array}{l}{F}_x\\ F_y\\ F_z\end{array}\right)\hfill & =F_x{\overrightarrow{e}}_x+F_y{\overrightarrow{e}}_y+F_z{\overrightarrow{e}}_z\hfill \\ \hfill & =F_x\left(cos\phi {\overrightarrow{e}}_r-sin\phi {\overrightarrow{e}}_{\phi }\right){\overrightarrow{e}}_r+F_y\left(cos\phi {\overrightarrow{e}}_{\phi }-sin\phi {\overrightarrow{e}}_r\right)+F_z{\overrightarrow{e}}_z\hfill \\ \hfill & =\left(F_{x}cos\phi +F_{y}sin\phi \right){\overrightarrow{e}}_r+\left(F_{x}sin\phi +F_{y}cos\phi \right){\overrightarrow{e}}_{\phi }+F_z{\overrightarrow{e}}_z\hfill \\ \hfill & =F_r{\overrightarrow{e}}_r+F_{\phi }{\overrightarrow{e}}_{\phi }+F_z{\overrightarrow{e}}_z\hfill \\ \hfill & =\left(\begin{array}{l}{F}_r\\ F_{\phi }\\ F_z\end{array}\right)\hfill \end{array}$

while in order to convert a vector $\overrightarrow{F}$ from cylindrical to Cartesian coordinates we find

$\begin{array}{ll}\left(\begin{array}{l}{F}_r\\ F_{\phi }\\ F_z\end{array}\right)\hfill & =F_r{\overrightarrow{e}}_r+F_{\phi }{\overrightarrow{e}}_{\phi }+F_z{\overrightarrow{e}}_z\hfill \\ \hfill & =F_r\left(cos\phi {\overrightarrow{e}}_x+sin\phi {\overrightarrow{e}}_y\right)+F_{\phi }\left(-sin\phi {\overrightarrow{e}}_x+cos\phi {\overrightarrow{e}}_y\right)+F_z{\overrightarrow{e}}_z\hfill \\ \hfill & =\left(F_{r}cos\phi -F_{\phi }sin\phi \right){\overrightarrow{e}}_x+\left(F_{r}sin\phi +F_{\phi }cos\phi \right){\overrightarrow{e}}_y+F_z{\overrightarrow{e}}_z\hfill \\ \hfill & =F_x{\overrightarrow{e}}_x+F_y{\overrightarrow{e}}_y+F_z{\overrightarrow{e}}_z\hfill \\ \hfill & =\left(\begin{array}{l}{F}_x\\ F_y\\ F_z\end{array}\right)\hfill \end{array}$

Orthogonality. We will shortly verify that the found axis vectors for the cylindrical coordinate systems are effectively orthogonal. For this we will calculate the cross products of the three vectors

$\begin{array}{lll}{\overrightarrow{e}}_r\times {\overrightarrow{e}}_{\phi }\hfill & =\left(\begin{array}{ccc}\hfill \hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill \hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill \hfill & \hfill {cos}^2\phi +{sin}^2\phi \hfill & \hfill \hfill \end{array}\right)=\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)\hfill & ={\overrightarrow{e}}_z\hfill \\ {\overrightarrow{e}}_r\times {\overrightarrow{e}}_z\hfill & =\left(\begin{array}{ccc}\hfill \hfill & \hfill sin\phi \hfill & \hfill \hfill \\ \hfill \hfill & \hfill -cos\phi \hfill & \hfill \hfill \\ \hfill \hfill & \hfill 0\hfill & \hfill \hfill \end{array}\right)\hfill & ={\overrightarrow{e}}_{\phi }\hfill \\ {\overrightarrow{e}}_{\phi }\times {\overrightarrow{e}}_z\hfill & =\left(\begin{array}{ccc}\hfill \hfill & \hfill cos\phi \hfill & \hfill \hfill \\ \hfill \hfill & \hfill sin\phi \hfill & \hfill \hfill \\ \hfill \hfill & \hfill 0\hfill & \hfill \hfill \end{array}\right)\hfill & ={\overrightarrow{e}}_r\hfill \end{array}$

As can be seen, the respective cross products of two of the three axis vectors are identical to the third respective unit vector. This proofs that we have indeed an orthogonal coordinate system.

7.3.3 Cartesian/Polar Coordinates Conversion

The conversion from Cartesian to polar coordinates (and v.v.) is simply a special case of the conversion from Cartesian to cylindrical coordinates while keeping z = 0.

Transforming the Independent Variables. For the conversion of the independent variables x and y we therefore find in analogy to Eq. 7.19 and Eq. 7.20

$x=rcos\phi$

$y=rsin\phi$

Transforming the Basis Vectors. Consequently we find the conversion of the basis vectors in analogy to Eq. 7.23, Eq. 7.24, Eq. 7.28, and Eq. 7.29 to be

${\overrightarrow{e}}_r=cos\phi {\overrightarrow{e}}_x+sin\phi {\overrightarrow{e}}_y$

${\overrightarrow{e}}_{\phi }=-sin\phi {\overrightarrow{e}}_x+cos\phi {\overrightarrow{e}}_y$

and

${\overrightarrow{e}}_x=cos\phi {\overrightarrow{e}}_r-sin\phi {\overrightarrow{e}}_{\phi }$

${\overrightarrow{e}}_y=sin\phi {\overrightarrow{e}}_r-cos\phi {\overrightarrow{e}}_{\phi }$

which allows us to convert the vector $\overrightarrow{F}$ from Cartesian to polar coordinates using

$\begin{array}{ll}\left(\begin{array}{l}{F}_x\\ F_y\end{array}\right)\hfill & =F_x{\overrightarrow{e}}_x+F_y{\overrightarrow{e}}_y\hfill \\ \hfill & =F_x\left(cos\phi {\overrightarrow{e}}_r-sin\phi {\overrightarrow{e}}_{\phi }\right)+F_y\left(sin\phi {\overrightarrow{e}}_r+cos\phi {\overrightarrow{e}}_{\phi }\right)\hfill \\ \hfill & =\left(F_{x}cos\phi -F_{y}sin\phi \right){\overrightarrow{e}}_r+\left(F_{x}sin\phi +F_{y}cos\phi \right){\overrightarrow{e}}_{\phi }\hfill \\ \hfill & =F_r{\overrightarrow{e}}_r+F_{\phi }{\overrightarrow{e}}_{\phi }\hfill \\ \hfill & =\left(\begin{array}{l}{F}_r\\ F_{\phi }\end{array}\right)\hfill \end{array}$

while the inverse conversion of a vector $\overrightarrow{F}$ from cylindrical to Cartesian coordinates is given by

$\begin{array}{ll}\left(\begin{array}{l}{F}_r\\ F_{\phi }\end{array}\right)\hfill & =F_r{\overrightarrow{e}}_r+F_{\phi }{\overrightarrow{e}}_{\phi }\hfill \\ \hfill & =F_r\left(cos\phi {\overrightarrow{e}}_x+sin\phi {\overrightarrow{e}}_y\right)+F_{\phi }\left(-sin\phi {\overrightarrow{e}}_x+cos\phi {\overrightarrow{e}}_y\right)\hfill \\ \hfill & =\left(F_{r}cos\phi -F_{\phi }sin\phi \right){\overrightarrow{e}}_x+\left(F_{r}sin\phi +F_{\phi }cos\phi \right){\overrightarrow{e}}_y\hfill \\ \hfill & =F_x{\overrightarrow{e}}_x+F_y{\overrightarrow{e}}_y\hfill \\ \hfill & =\left(\begin{array}{l}{F}_x\\ F_y\end{array}\right)\hfill \end{array}$

Orthogonality. We can skip the proof of the orthogonality of the polar coordinate system as this proof has already been done implicitly while proofing the orthogonality of the cylindrical coordinate system.

7.3.4 Cartesian/Spherical Coordinates Conversion

Transforming the Independent Variables. In an analogy we will now transform Cartesian to spherical coordinates (see Fig. 7.4d) where we note the conversion of the independent variables x, y, and z to the independent variables r, θ, and φ as

$\begin{array}{l}x=rsin\theta cos\phi \\ y=rsin\theta cos\phi \\ z=rcos\theta \end{array}$

which in turn gives us the vector $\overrightarrow{s}$ as function of ρ, θ , and φ

$\overrightarrow{s}\left(\rho ,\theta ,\phi \right)=\mathit{\rho sin\theta }cos\phi {\overrightarrow{e}}_x+\rho sin\theta sin\phi {\overrightarrow{e}}_y+\rho cos\theta {\overrightarrow{e}}_z$

Transforming the Basis Vectors. We now form the partial derivatives along the new independent variables. Once we have these vectors, we divide them by their norm to find the new unit vectors. For ${\overrightarrow{e}}_{\rho }$ we find

$\begin{array}{ll}\frac{\partial \overrightarrow{s}\left(\rho ,\theta ,\phi \right)}{\partial \rho }\hfill & =\frac{\partial \left(\rho sin\theta cos\phi \right)}{\partial \rho }{\overrightarrow{e}}_x+\frac{\partial \left(\rho sin\theta sin\phi \right)}{\partial \rho }{\overrightarrow{e}}_y+\frac{\partial \left(\rho cos\theta \right)}{\partial \rho }{\overrightarrow{e}}_z\hfill \\ \hfill & =sin\theta cos\phi {\overrightarrow{e}}_x+sin\theta sin\phi {\overrightarrow{e}}_y+cos\theta {\overrightarrow{e}}_z\hfill \\ ‖\frac{\partial \overrightarrow{s}\left(\rho ,\theta ,\phi \right)}{\partial \rho }‖\hfill & =\sqrt{{\left(sin\theta cos\phi \right)}^2+{\left(sin\theta sin\phi \right)}^2+{cos}^2\theta }\hfill \\ \hfill & =\sqrt{sin{\theta }^2\left({cos}^2\phi +{sin}^2\phi \right)+{cos}^2\theta }=\sqrt{sin{\theta }^2+cos{\theta }^2}=1\hfill \\ {\overrightarrow{e}}_{\rho }\hfill & =\frac{\frac{\partial \overrightarrow{s}\left(\rho ,\phi ,\theta \right)}{\partial \rho }}{‖\frac{\partial \overrightarrow{s}\left(\rho ,\phi ,\theta \right)}{\partial \rho }‖}\hfill \\ \hfill & =sin\theta cos\phi {\overrightarrow{e}}_x+sin\theta sin\phi {\overrightarrow{e}}_y+cos\theta {\overrightarrow{e}}_z\hfill \end{array}$

Again, we see that the unit vectors in spherical coordinates depend on the unit vectors of the Cartesian coordinate system.

We now proceed likewise for ${\overrightarrow{e}}_{\theta }$ for which we find

$\begin{array}{ll}\frac{\partial \overrightarrow{s}\left(\rho ,\theta ,\phi \right)}{\partial \theta }\hfill & =\frac{\partial \left(\rho sin\theta cos\phi \right)}{\partial \theta }{\overrightarrow{e}}_x+\frac{\partial \left(\rho sin\theta sin\phi \right)}{\partial \theta }{\overrightarrow{e}}_y+\frac{\partial \left(\rho cos\theta \right)}{\partial \theta }{\overrightarrow{e}}_z\hfill \\ \hfill & =\rho cos\theta cos\phi {\overrightarrow{e}}_x+\rho cos\theta sin\phi {\overrightarrow{e}}_y+-\rho sin\theta {\overrightarrow{e}}_z\hfill \\ ‖\frac{\partial \overrightarrow{s}\left(\rho ,\theta ,\phi \right)}{\partial \theta }‖\hfill & =\sqrt{{\left(\rho cos\theta cos\phi \right)}^2+{\left(\rho cos\theta sin\phi \right)}^2+{\rho }^2{sin}^2\theta }\hfill \\ \hfill & =\sqrt{{\rho }^2{cos}^2\theta \left(cos{\phi }^2+{sin}^2\phi \right)+{\rho }^2{sin}^2\theta }=\sqrt{sin{\theta }^2+cos{\theta }^2}=\rho \hfill \\ {\overrightarrow{e}}_{\theta }\hfill & =\frac{\frac{\partial \overrightarrow{s}\left(\rho ,\phi ,\theta \right)}{\partial \theta }}{‖\frac{\partial \overrightarrow{s}\left(\rho ,\phi ,\theta \right)}{\partial \theta }‖}\hfill \\ \hfill & =cos\theta cos\phi {\overrightarrow{e}}_x+sin\theta sin\phi {\overrightarrow{e}}_y-sin\theta {\overrightarrow{e}}_z\hfill \end{array}$