# Chapter 1: Curvilinear Coordinates

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The material covered in this chapter is also presented in Boas Chapter 10, Sections 8 and 9.

## 1.1 Cartesian Coordinates

To locate a point $P$ in three-dimensional space requires the specification of three coordinates, and the simplest choise is usually to employ Cartesian coordinates $(x,y,z)$. The association between P and its coordinates $(x,y,z)$ depends on a choice of origin $O$ for the coordinate system, as well as a choice of orientation for the coordinate axes. These choices are arbitrary, and are usually made to simplify the mathematical formulation of a given problem.

The choice of coordinate system provides us with the three coordinate directions $\boldsymbol{\hat{x}}$, $\boldsymbol{\hat{y}}$, and $\boldsymbol{\hat{z}}$. These are unit vectors that point in the directions specified by the notation; for example, $\boldsymbol{\hat{x}}$ points in the direction of increasing $x$. The vectors are mutually orthogonal; for example $\boldsymbol{\hat{x}} \cdot \boldsymbol{\hat{y}} = 0$. (The direction vectors are sometimes denoted $\boldsymbol{\hat{\imath}}$, $\boldsymbol{\hat{\jmath}}$, and $\boldsymbol{\hat{k}}$. The notation used here is more direct and informative, and is compatible with the notation employed below to describe the direction vectors in curvilinear coordinates.)

A vector $\boldsymbol{A}$ can be decomposed in the vector basis provided by the direction vectors. We have $\boldsymbol{A}=A_x \, \boldsymbol{\hat{x}} + A_y \, \boldsymbol{\hat{y}} + A_z \, \boldsymbol{\hat{z}}$, where, for example, $A_x=\boldsymbol{A} \cdot \boldsymbol{\hat{x}}$ is the component of $\boldsymbol{A}$ along $\boldsymbol{\hat{x}}$. The position vector of the point P, denoted $\boldsymbol{r}$, is obviously decomposed as

$\boldsymbol{r}=x \, \boldsymbol{\hat{x}} + y \, \boldsymbol{\hat{x}} + z \, \boldsymbol{\hat{z}}. \tag{1.1}$

Pictorially, the vector $\boldsymbol{r}$ can be represented as an arrow that points from the origin $O$ to the point $P$. It is an example of a vector field, a vector that deponds on position in space. Other examples of vector fields are $\boldsymbol{A}=xy \, \boldsymbol{\hat{x}} + yz \, \boldsymbol{\hat{y}} + zx \, \boldsymbol{\hat{z}}$ and $\boldsymbol{B}= -y \, \boldsymbol{\hat{x}} + z \, \boldsymbol{\hat{y}}$.

The expression of Eq.(1.1) for the position vector leads us to a formal definition for the basis vectors. Suppose that we differentiate $\boldsymbol{r}$ with respect to $x$, keeping $y$ and $z$ fixed. This defines the partial derivative of $\boldsymbol{r}$ with respect to $x$, and the answer is clearly $\frac{\partial \boldsymbol{r}}{\partial x}=\boldsymbol{\hat{x}}$, because $\boldsymbol{\hat{x}}$, $\boldsymbol{\hat{y}}$, and $\boldsymbol{\hat{z}}$ are constant vectors that don't depend on $x$. Similarly, we can relate $\boldsymbol{\hat{y}}$ and $\boldsymbol{\hat{z}}$ to the partial derivatives of $\boldsymbol{r}$ with respect to $y$ and $z$, respectively. These results give us the definitions we were looking for,

$\boldsymbol{\hat{x}}=\frac{\partial \boldsymbol{r}}{\partial x} \qquad \boldsymbol{\hat{y}}=\frac{\partial \boldsymbol{r}}{\partial y} \qquad \boldsymbol{\hat{z}}=\frac{\partial \boldsymbol{r}}{\partial z}. \tag{1.2}$

Equations (1.2) tell us something we already know, for example, that $\boldsymbol{\hat{x}}$ points in the direction of increasing $x$. To see this, consider two points in three-dimensional space, $P$ situated at $(x, y, z)$, and $Q$ situated at $(x+dx, y, z)$; clearly, $Q$ is displaced by $dx$ relative to $P$. The position vector corresponding to $P$ is $\boldsymbol{r}(P) = x \,\boldsymbol{\hat{x}}+ y \,\boldsymbol{\hat{y}}+ z \,\boldsymbol{\hat{z}}$, while the position vector corresponding to $Q$ is $\boldsymbol{r}(Q) = (x+dx) \,\boldsymbol{\hat{x}} + y \,\boldsymbol{\hat{y}} + z\,\boldsymbol{\hat{z}}$ the displacement from $P$ to $Q$ is $d \boldsymbol{r}(PQ) = dx \, \boldsymbol{\hat{x}}$, and we see that \begin{equation} \boldsymbol{\hat{x}} = \frac{d\boldsymbol{r}(PQ)}{dx} \tag{1.3}\end{equation} is indeed a vector that points in the direction of increasing $x$, because the displacement vector $d\boldsymbol{r}(PQ)$ is constructed precisely to point in that direction; division by $dx$ ensures that $\boldsymbol{\hat{x}}$ is a unit vector. The definition of the partial derivative implies that $d\boldsymbol{r}(PQ)/dx \equiv \partial \boldsymbol{r}/\partial x$, because $y$ and $z$ were held fixed during the displacement, and we have reproduced the first of Eqs.(1.2).

We next consider a general displacement from the point $P$, described by $(x, y, z)$, to a new point $R$ described by $(x+dx, y+dy, z+dz)$; this time the displacement involves all three directions instead of just $x$. It should be clear that in this case, the displacement from $P$ to $R$ is described by \begin{equation} d\boldsymbol{r} = dx\, \boldsymbol{\hat{x}} + dy\, \boldsymbol{\hat{y}}+ dz\, \boldsymbol{\hat{z}} \tag{1.4}\end{equation} This is known as the infinitesimal displacement vector. The definitions introduced thus far imply that

\begin{equation} \boldsymbol{r}(x+dx, y+dy, z+dz) - \boldsymbol{r}(x, y, z) = d\boldsymbol{r},\tag{1.5} \end{equation}

with the first term on the left-hand side describing the position of $R$, and the second term giving the position of $P$.

The infinitesimal displacement vector $d\boldsymbol{r}$ is an exceedingly important construction that can be involved in many different calculations. It is, for example, featured in any line integral of the form $\int_C \boldsymbol{A} \cdot d \boldsymbol{r}$, where $C$ is a path in three-dimensional space, $d\boldsymbol{r}$ the displacement along this path, and $\boldsymbol{A}$ a vector field defined on the path. It also allows us to define surface and volume elements in Cartesian coordinates. Suppose, for example, that we wish to evaluate the surface integral $\int_S \boldsymbol{A} \cdot d\boldsymbol{a}$, where $S$ is a segment of the $x$-$y$ plane, and $d\boldsymbol{a}$ is the area element on this surface. Equation (1.4) informs us that the orthogonal displacements on the surface are $dx$ and $dy$ (because $dz = 0$), and since $d\boldsymbol{a}$ must point in the direction normal to the surface, it is given by $d\boldsymbol{a} = dxdy\, \boldsymbol{\hat{z}}$. As another example, suppose that we wish to obtain the volume element $dV$ in Cartesian coordinates. We turn to Eq(1.4), observe that the orthogonal displacements are $dx$, $dy$, and $dz$, and form the volume element by merging all these together,

\begin{equation} dV = dx dy dz. \tag{1.6} \end{equation}

The answer, of course, is well known. What matters here is that the path that led us to Eq.(1.6) originated with the $d\boldsymbol{r}$ of Eq.(1.4). Below we shall adapt the strategy to other coordinate systems.

We conclude this section with a refresher on the main ingredients of vector calculus, the gradient, divergence, and curl. In Cartesian coordinates, the gradient $\boldsymbol{\nabla} f$ of a scalar function $f(x,y,z)$ is given by

\begin{equation} \boldsymbol{\nabla}f = \frac{\partial f}{\partial x}\, \boldsymbol{\hat{x}} + \frac{\partial f}{\partial y}\, \boldsymbol{\hat{y}} + \frac{\partial f}{\partial z}\, \boldsymbol{\hat{z}}. \tag{1.7} \end{equation}

It is a vector with components given by all the partial derivatives of the function. The gradient is defined so that the change $df = f(x+dx,y+dy,z+dz) - f(x,y,z)$ in the function as we move from $P$ to the neighbouring point $R$ is given by

\begin{equation} df = \boldsymbol{\nabla} f \cdot d\boldsymbol{r}, \tag{1.8} \end{equation}

in terms of the gradient and the infinitesimal displacement vector. Working out the right-hand side gives

\begin{equation} df = \frac{\partial f}{\partial x}\, dx + \frac{\partial f}{\partial y}\, dy + \frac{\partial f}{\partial z}\, dz, \tag{1.9}\end{equation}

which is the statement of the fundamental theorem on partial derivatives. It should be noted that Eq.(1.8) is actually independent of the coordinates, an observation that will be important below.

In Cartesian coordinates, the divergence of a vector field $\boldsymbol{A}$ is given by

\begin{equation} \boldsymbol{\nabla} \cdot \boldsymbol{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}, \tag{1.10} \end{equation}

and its curl is given by

\begin{equation} \boldsymbol{\nabla} \times \boldsymbol{A} = \boldsymbol{\hat{x}} \bigg( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \bigg) + \boldsymbol{\hat{y}} \bigg( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \bigg) + \boldsymbol{\hat{z}} \bigg( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \bigg). \tag{1.11} \end{equation}

You may remember that the expression for the curl can be conveniently reproduced by constructing a determinant involving the basis vectors (in the first row), the partial derivatives (in the second row), and the components of $\boldsymbol{A}$ (in the third row).

The Laplacian of a scalar function $f$ is defined to be the divergence of the vector field $\boldsymbol{\nabla} f$,

\begin{equation} \nabla^2 f := \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} f) = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}. \tag{1.12} \end{equation}

## 1.2 Polar Coordinates

Cartesian coordinates are familiar and intuitive, but in some problems they are not necessarily the most convenient choice of coordinates. In many situations it is advantageous to give them up and adopt an alternative set of coordinates that is better suited to the given circumstances. To prepare the way for a general discussion of alternative coordinates in the next section, we examine first the familiar case of polar coordinates in the two-dimensional plane.

We work in the $x$-$y$ plane, and define the polar coordinates $(s,\phi)$ with the relations

\begin{equation} x = s\cos\phi, \qquad y = s\sin \phi.\tag{1.13} \end{equation}

The geometrical meaning of the coordinates is illustrated in Fig.1.1. The radial coordinate $s$ represents the distance of the point $P$ from the origin, and the angle $\phi$ refers to the $x$-axis. A point $P$ can be situated in the plane by providing values for both $s$ and $\phi$. It should be clear that curves of constant $s$ are circles of radius $s$ centred at the origin, while curves of constant $\phi$ are straight lines through the origin. It should also be clear that curves of constant $s$ are everywhere perpendicular to curves of constant $\phi$. The position vector of $P$ can be expressed as

\begin{equation} \boldsymbol{r}(s,\phi) = s\cos\phi\, \boldsymbol{\hat{x}} + s\sin\phi\, \boldsymbol{\hat{y}} \tag{1.14} \end{equation}

in terms of the new coordinates. Notice that $\boldsymbol{r}$ is still decomposed in terms of the old basis vectors. Figure 1.2: Basis vectors $\boldsymbol{\hat{s}}$ and $\boldsymbol{\hat{\phi}}$ associated with the polar coordinates. The vector $\boldsymbol{\hat{s}}$ always points away from the origin, in the direction of increasing $s$, and $\boldsymbol{\hat{\phi}}$ is always perpendicular to $\boldsymbol{\hat{s}}$, in the direction of increasing $\phi$.

We would like to introduce a new basis of vectors $\boldsymbol{\hat{s}}$ and $\boldsymbol{\hat{\phi}}$ adapted to the polar coordinates. We adopt a definition that is directly inspired from the definition of $\boldsymbol{\hat{x}}$ and $\boldsymbol{\hat{y}}$:

\begin{align} \boldsymbol{\hat{s}} &:= \text{unit vector pointing in the direction of increasing $s$}, \tag{1.15a}\\ \boldsymbol{\hat{\phi}} &:= \text{unit vector pointing in the direction of increasing $\phi$}. \tag{1.15b} \end{align}

The definition allows us to display the vectors pictorially in Fig.1.2. The most important observation is that while $\boldsymbol{\hat{s}}$ and $\boldsymbol{\hat{\phi}}$ are constant in length (because they are both unit vectors), they are not constant in direction, because the direction of increasing $s$ and the direction of increasing $\phi$ both depend on position in the plane. The new basis vectors are therefore vector fields that depend on position, and they are certainly not constant vectors like $\boldsymbol{\hat{x}}$ and $\boldsymbol{\hat{y}}$. This means that working with $\boldsymbol{\hat{s}}$ and $\boldsymbol{\hat{\phi}}$ requires some care.

To relate $\boldsymbol{\hat{s}}$ and $\boldsymbol{\hat{\phi}}$ to the old basis vectors, we simply follow the definitions. Let us begin with $\boldsymbol{\hat{s}}$, which we know must be pointing in the direction of increasing $s$. Let us, therefore, consider a displacement from the point $P$ at $(s,\phi)$ to a neighbouring point $Q$ at $(s+ds,\phi)$; the displacement is entirely in the $s$-direction, and $\phi$ is kept fixed. The displacement is described by the vector

\begin{equation} d\boldsymbol{r} = \boldsymbol{r}(s+ds, \phi) - \boldsymbol{r}(s, \phi),\tag{1.16} \end{equation}

and this gives us the correct direction for $\boldsymbol{\hat{s}}$. By the definition of the partial derivative we have that

\begin{equation} d\boldsymbol{r} =\frac{\partial \boldsymbol{r}}{\partial s}\, ds,\tag{1.17} \end{equation}

and we conclude that

\begin{equation} \boldsymbol{\hat{s}} \propto \frac{\partial \boldsymbol{r}}{\partial s}.\tag{1.18} \end{equation}

We have that $\boldsymbol{\hat{s}}$ points in the same direction as the partial derivative of $\boldsymbol{r}$ with respect to $s$, and working this out from Eq.(1.14) gives $\boldsymbol{\hat{s}} \propto \cos\phi\, \boldsymbol{\hat{x}} + \sin\phi\, \boldsymbol{\hat{y}}$. To determine the constant of proportionality we must calculate the length of the vector on the right-hand side, and if necessary, rescale that vector so that it acquires a unit length. In this case there is no need to rescale, because $\cos\phi\, \boldsymbol{\hat{x}} + \sin\phi\, \boldsymbol{\hat{y}}$ does have a unit length. We have obtained

\begin{equation} \boldsymbol{\hat{s}} = \frac{\partial \boldsymbol{r}}{\partial s} = \cos\phi\, \boldsymbol{\hat{x}} + \sin\phi\, \boldsymbol{\hat{y}}, \tag{1.19} \end{equation}

and have identified the first of our basis vectors. Notice that as expected, $\boldsymbol{\hat{s}}$ depends on $\phi$ and therefore on position in the plane

To obtain $\boldsymbol{\hat{\phi}}$ we follow a very similar route, this time considering a displacement in the $\phi$-direction, from $P$ at $(s,\phi)$ to a new point $R$ at $(s,\phi+d\phi)$. The displacement is now described by

\begin{equation} d\boldsymbol{r} = \boldsymbol{r}(s, \phi+d\phi) - \boldsymbol{r}(s, \phi),\tag{1.20} \end{equation}

or

\begin{equation} d\boldsymbol{r} =\frac{\partial \boldsymbol{r}}{\partial \phi}\, d\phi,\tag{1.21} \end{equation}

and this gives us the correct direction for $\boldsymbol{\hat{\phi}}$,

\begin{equation} \boldsymbol{\hat{\phi}} \propto \frac{\partial \boldsymbol{r}}{\partial \phi}. \tag{1.22}\end{equation}

The partial derivative of $\boldsymbol{r}$ with respect to $\phi$ is given by $-s\sin\phi\, \boldsymbol{\hat{x}} + s\cos\phi\, \boldsymbol{\hat{y}} = s (-\sin\phi\, \boldsymbol{\hat{x}}+\cos\phi\, \boldsymbol{\hat{y}})$, and we observe that the vector between brackets has a unit length. We conclude that the factor of proportionality must be $1/s$, to eliminate the factor of $s$ in $\partial \boldsymbol{r}/\partial \phi$. We have obtained

\begin{equation} \boldsymbol{\hat{\phi}} = \frac{1}{s} \frac{\partial \boldsymbol{r}}{\partial \phi} = -\sin\phi\, \boldsymbol{\hat{x}} + \cos\phi\, \boldsymbol{\hat{y}}, \tag{1.23} \end{equation}

the second of our basis vectors. Notice that like $\boldsymbol{\hat{s}}$, $\boldsymbol{\hat{\phi}}$ depends on $\phi$, and therefore on position in the plane.

Exercise 1:Check that $\boldsymbol{\hat{s}}$ and $\boldsymbol{\hat{\phi}}$ are everywhere orthogonal to each other.

We have seen that a vector $\boldsymbol{A}$ can be decomposed in the basis provided by $\boldsymbol{\hat{x}}$ and $\boldsymbol{\hat{y}}$. It can also be decomposed in the new basis, according to

\begin{equation} \boldsymbol{A} = A_s\, \boldsymbol{\hat{s}} + A_\phi\, \boldsymbol{\hat{\phi}},\tag{1.24} \end{equation}

where $A_s := \boldsymbol{A} \cdot \boldsymbol{\hat{s}}$ is the component of $\boldsymbol{A}$ along $\boldsymbol{\hat{s}}$, and $A_\phi := \boldsymbol{A} \cdot \boldsymbol{\hat{\phi}}$ its component along $\boldsymbol{\hat{\phi}}$. An example of such a decomposition is

\begin{equation} \boldsymbol{r} = s\, \boldsymbol{\hat{s}},\tag{1.25} \end{equation}

which gives the position vector in the new basis. In case you find this expression surprising, or expected the more symmetric (but wrong!) expression $\boldsymbol{r} = s\, \boldsymbol{\hat{s}} + \phi\, \boldsymbol{\hat{\phi}}$, take the time to reflect on the geometrical meaning of the vectors $\boldsymbol{r}$ and $\boldsymbol{\hat{s}}$.

It is sometimes useful to express the old basis vectors in terms of the new basis. It is easy to show that

\begin{align} \boldsymbol{\hat{x}} &= \cos\phi\, \boldsymbol{\hat{s}} - \sin\phi\, \boldsymbol{\hat{\phi}}, \tag{1.26a} \\ \boldsymbol{\hat{y}} &= \sin\phi\, \boldsymbol{\hat{s}} + \cos\phi\, \boldsymbol{\hat{\phi}}\tag{1.26b} \end{align} 4

provides the required relations.

Exercise 2:Verify these equations. This requires almost no work. Just remember, for example, that the coefficient in front of $\boldsymbol{\hat{s}}$ in the decomposition of $\boldsymbol{\hat{x}}$ is nothing more than $\boldsymbol{\hat{x}} \cdot \boldsymbol{\hat{s}} = \boldsymbol{\hat{s}} \cdot \boldsymbol{\hat{x}}$, which can be read off from Eq.(1.19).

We next turn to the infinitesimal displacement vector $d\boldsymbol{r}$, which was seen to play an important role in Sec.1.1. We consider a general displacement from $P$ at $(s,\phi)$ to a neighbouring point $S$ at $(s+ds, \phi+d\phi)$. The infinitesimal displacement between these points is

\begin{equation} d\boldsymbol{r} = \boldsymbol{r}(s+ds,\phi+d\phi) - \boldsymbol{r}(s,\phi), \tag{1.27} \end{equation}

and the fundamental theorem on partial derivatives allows us to express this as

\begin{equation} d\boldsymbol{r} = \frac{\partial \boldsymbol{r}}{\partial s}\, ds + \frac{\partial \boldsymbol{r}}{\partial \phi}\, d\phi. \tag{1.28}\end{equation}

Comparison with Eqs.(1.19) and (1.23) finally reveals that the infinitesimal displacement vector is given by

\begin{equation} d\boldsymbol{r} = ds\, \boldsymbol{\hat{s}} + sd\phi\, \boldsymbol{\hat{\phi}} \tag{1.29} \end{equation}

in polar coordinates. The vector is naturally decomposed in the basis provided by $\boldsymbol{\hat{s}}$ and $\boldsymbol{\hat{\phi}}$, and we observe that $ds$ and $sd\phi$ are revealed as the orthogonal displacements. These expressions are intuitive: $ds$ is easily seen to represent a displacement in the radial direction, and the factor $s$ converts the angle increment $d\phi$ to an actual displacement $s d\phi$ in the angular direction. From these orthogonal displacements we infer that $da = (ds)(sd\phi) = sdsd\phi$ is the area element in polar coordinates.

To conclude this section we note that it is trivial to extend the two-dimensional plane toward a third dimension by re-introducing the $z$ coordinate. If we join $z$ to the polar coordinates we obtain the set $(s,\phi,z)$ known as cylindrical coordinates. The results obtained in this section can then easily be generalized to the three-dimensional case. The position vector is now expressed as

\begin{equation} \boldsymbol{r}(s,\phi,z) = s\cos\phi\, \boldsymbol{\hat{x}} + s\sin\phi\, \boldsymbol{\hat{y}} + z\, \boldsymbol{\hat{z}},\tag{1.30} \end{equation}

the basis vectors are

\begin{align} \boldsymbol{\hat{s}} &= \frac{\partial \boldsymbol{r}}{\partial s} = \cos\phi\, \boldsymbol{\hat{x}} + \sin\phi\, \boldsymbol{\hat{y}}, \tag{1.31a} \\ \boldsymbol{\hat{\phi}} &= \frac{1}{s} \frac{\partial \boldsymbol{r}}{\partial \phi} = -\sin\phi\, \boldsymbol{\hat{x}} + \cos\phi\, \boldsymbol{\hat{y}}, \tag{1.31b}\\ \boldsymbol{\hat{z}} &= \frac{\partial \boldsymbol{r}}{\partial z} = \boldsymbol{\hat{z}}, \tag{1.31c} \end{align}

and the infinitesimal displacement vector is

\begin{equation} d\boldsymbol{r} = ds\, \boldsymbol{\hat{s}} + sd\phi\, \boldsymbol{\hat{\phi}} + dz\, \boldsymbol{\hat{z}}. \tag{1.32}\end{equation}

From this expression we deduce that the volume element is given by

\begin{equation} dV = (ds)(sd\phi)(dz) = sds\, d\phi\, dz \tag{1.33}\end{equation}

in cylindrical coordinates.

Exercise 3: Verify that the cylindrical coordinates $(s,\phi,z)$ define a right-handed system, in the sense that $\boldsymbol{\hat{s}} \times \boldsymbol{\hat{\phi}} = \boldsymbol{\hat{z}}$, $\boldsymbol{\hat{\phi}} \times \boldsymbol{\hat{z}} = \boldsymbol{\hat{s}}$, and $\boldsymbol{\hat{z}} \times \boldsymbol{\hat{s}} = \boldsymbol{\hat{\phi}}$.

## 1.3 General Orthogonal Coordinates

With the main ideas nicely illustrated in the specific cases of polar and cylindrical coordinates, we are now ready to formulate a general theory of curvilinear coordinates. We imagine that instead of the usual Cartesian coordinates $(x, y, z)$, we choose to work with an alternative set of coordinates $(q_1, q_2, q_3)$, which we collectively denote $q_j$. Obviously we need three independent coordinates to situate a point $P$ in three-dimensional space, and we shall assume that the $q_j$'s, whatever they are, do the job. We shall also assume that the $q_j$'s form an orthogonal coordinate system, which means that the curves $q_j = \text{constant}$ are assumed to cross each other at right angles. It is of course possible to introduce coordinates that don't possess this property, but such systems are typically not as useful or convenient as orthogonal coordinates. With the cylindrical coordinates considered previously, we would have $q_1 = s$, $q_2 = \phi$, and $q_3 = z$.

The new coordinates $q_j$ can be related to the old Cartesian coordinates by equations of the form

\begin{equation} x = x(q_1, q_2, q_3), \qquad y = y(q_1, q_2, q_3), \qquad z = z(q_1, q_2, q_3),\tag{1.34} \end{equation}

in which the Cartesian coordinates are expressed as functions of the new coordinates. In the specific example of cylindrical coordinates, the relations are $x = s\cos\phi$, $y = s\sin\phi$, and $z=z$. These relations can be used to express the position vector of a generic point $P$ as

\begin{equation} \boldsymbol{r}(q_j) = x(q_j)\, \boldsymbol{\hat{x}} + y(q_j)\, \boldsymbol{\hat{y}} + z(q_j)\, \boldsymbol{\hat{z}}, \tag{1.35} \end{equation}

with $q_j$ denoting the coordinates of $P$. The position vector is given as a function of $(q_1, q_2, q_3)$, but it is still decomposed in the old basis vectors $\boldsymbol{\hat{x}}$, $\boldsymbol{\hat{y}}$, and $\boldsymbol{\hat{z}}$.

We wish to construct a vector basis adapted to the new coordinates, consisting of $\boldsymbol{\hat{\text{q}}_1}$, $\boldsymbol{\hat{\text{q}}_2}$, and $\boldsymbol{\hat{\text{q}}_3}$, which we collectively denote $\boldsymbol{\hat{\text{q}}_j}$. With the cylindrical coordinates considered previously, we would have $\boldsymbol{\hat{\text{q}}_1} = \boldsymbol{\hat{s}}$, $\boldsymbol{\hat{\text{q}}_2 }= \boldsymbol{\hat{\phi}}$, and $\boldsymbol{\hat{\text{q}}_3} = \boldsymbol{\hat{z}}$. The definition of each basis vector is directly adapted from Eq.(1.15):

\begin{equation} \boldsymbol{\hat{\text{q}}_j} := \text{unit vector pointing in the direction of increasing $q_j$}. \tag{1.36} \end{equation}

Because the curves $q_j = \text{constant}$ are perpendicular to one another, these vectors are mutually orthogonal, so that $\boldsymbol{\hat{\text{q}}_j} \cdot \boldsymbol{\hat{\text{q}}_k} = 0$ when $j \neq k$. Of course we also have $\boldsymbol{\hat{\text{q}}_j} \cdot \boldsymbol{\hat{\text{q}}_k} = 1$ when $j = k$, and a convenient way to summarize all this is to write

\begin{equation} \boldsymbol{\hat{\text{q}}_j} \cdot \boldsymbol{\hat{\text{q}}_k} = \delta_{jk}, \tag{1.37}\end{equation}

where $\delta_{jk}$ is the Kronecker delta, equal to $1$ whenever $j = k$, and to $0$ whenever $j \neq k$. We assume that the coordinates $(q_1, q_2, q_3)$ form a right-handed system, in the sense that

\begin{equation} \boldsymbol{\hat{\text{q}}_1} \times \boldsymbol{\hat{\text{q}}_2} = \boldsymbol{\hat{\text{q}}_3}, \qquad \boldsymbol{\hat{\text{q}}_2} \times \boldsymbol{\hat{\text{q}}_3} = \boldsymbol{\hat{\text{q}}_1}, \qquad \boldsymbol{\hat{\text{q}}_3 }\times \boldsymbol{\hat{\text{q}}_1} = \boldsymbol{\hat{\text{q}}_2}. \tag{1.38} \end{equation}

Notice that in these equations, the indices are always put into the cyclic order $1 \to 2 \to 3 \to 1 \to \cdots$.

How do we construct the new basis vectors? Again we simply follow the definition. We know that $\boldsymbol{\hat{\text{q}}_j}$ points in the direction of increasing $q_j$, and we also know that $d\boldsymbol{r} = (\partial \boldsymbol{r}/\partial q_j)\, dq_j$ describes an infinitesimal displacement in that direction. These vectors must therefore be proportional to each other, and we write

\begin{equation} \boldsymbol{\hat{\text{q}}_j} = \frac{1}{h_j}\frac{\partial \boldsymbol{r}}{\partial q_j}, \tag{1.39} \end{equation}

where

\begin{equation} h_j := \sqrt{ \frac{\partial \boldsymbol{r}}{\partial q_j} \cdot \frac{\partial \boldsymbol{r}}{\partial q_j} } \tag{1.40} \end{equation}

is a scale factor that ensures that $\boldsymbol{\hat{\text{q}}_j}$ is properly normalized. We have seen that in cylindrical coordinates, $h_1 = 1$ (for $\boldsymbol{\hat{s}}$), $h_2 = s$ (for $\boldsymbol{\hat{\phi}}$), and $h_3 = 1$ (for $\boldsymbol{\hat{z}}$). The basis vectors and the scale factors can be calculated as soon as the relations $x(q_j)$, $y(q_j)$, and $z(q_j)$ are concretely specified.

The basis vectors $\boldsymbol{\hat{q}_j}$ can be used to decompose any vector $\boldsymbol{A}$. We have

\begin{equation} \boldsymbol{A} = A_1\, \boldsymbol{\hat{q}_1} + A_2\, \boldsymbol{\hat{q}_2} + A_3\, \boldsymbol{\hat{q}_3}, \tag{1.41}\end{equation}

where, for example, $A_2 = \boldsymbol{A} \cdot \boldsymbol{\hat{q}_2}$ is the component of $\boldsymbol{A}$ along $\boldsymbol{\hat{q}_2}$. In general we would write $A_j = \boldsymbol{A} \cdot \boldsymbol{\hat{q}_j}$ to represent a typical component of the vector. The position vector $\boldsymbol{r}$ can be decomposed in the new basis, but it is not possible to write a universal expression for it; the actual expression depends on the specific nature of the coordinates. (The position vector is NOT EQUAL to $q_1\, \boldsymbol{\hat{q}_1} + q_2\, \boldsymbol{\hat{q}_2} + q_3\, \boldsymbol{\hat{q}_3}$, however tempting it may be to believe that it is.)

The infinitesimal displacement $d\boldsymbol{r}$ does admit a universal expression. This, you will recall, describes the displacement from a point $P$ at $(q_1, q_2, q_3)$ to a neighbouring point $Q$ at $(q_1+dq_1, q_2+dq_2, q_3+dq_3)$. Its precise definition is

\begin{equation} d\boldsymbol{r} := \boldsymbol{r}(q_1+dq_1, q_2+dq_2, q_3+dq_3) - \boldsymbol{r}(q_1, q_2, q_3), \tag{1.42}\end{equation}

and the fundamental theorem on partial derivatives allows us to write this as

\begin{equation} d\boldsymbol{r} = \frac{\partial \boldsymbol{r}}{\partial q_1}\, dq_1 + \frac{\partial \boldsymbol{r}}{\partial q_2}\, dq_2 + \frac{\partial \boldsymbol{r}}{\partial q_3}\, dq_3. \tag{1.43}\end{equation}

Comparison with Eq.(1.39) brings this to its final form of

\begin{equation} d\boldsymbol{r} = h_1 dq_1\, \boldsymbol{\hat{q}}_1} + h_2 dq_2\, \boldsymbol{\hat{q}_2} + h_3 dq_3\, \boldsymbol{\hat{q}}_3}. \tag{1.44} \end{equation

This is a decomposition of $d\boldsymbol{r}$ in the new basis, and the components $dr_j = h_j\, dq_j$ represent the orthogonal displacements in the three directions specified by the new coordinates. The scale factor $h_j$ converts the increment $dq_j$ in the coordinate $q_j$ to an actual displacement in the direction of $\boldsymbol{\hat{q}_j}$. For example, we have seen that in cylindrical coordinates, $dr_\phi = s\, d\phi$, with the scale factor $s$ converting the angle increment $d\phi$ to an actual displacement $s\,d\phi$ in the direction of $\boldsymbol{\hat{\phi}}$.

As we have seen previously, the infinitesimal displacement vector of Eq.(1.44) is the source of a wealth of useful information. For example, it allows us to construct the volume element $dV$ in the coordinates $(q_1, q_2, q_3)$ by simply merging together the three orthogonal displacements,

\begin{equation} dV = (h_1\, dq_1) (h_2\, dq_2) (h_3\, dq_3) = h_1 h<p>_2 h_3\, dq_1 dq_2 dq_3. \tag{1.45} \end{equation}

The factor $h_1 h_2 h_3$ is known as the Jacobian of the transformation between the new coordinates $(q_1, q_2, q_3)$ and the old coordinates $(x, y, z)$. As another example, we can use Eq.(1.44) to form the area element $d\boldsymbol{a}$ on a segment of surface $q_2 = \text{constant}$. Because $q_2$ is constant, the relevant orthogonal displacements on the surface are $h_1\, dq_1$ and $h_3\, dq_3$. And because the direction normal to the surface is $\boldsymbol{\hat{q}_2}$, the area element is $d\boldsymbol{a} = (h_1\, dq_1)(h_3\, dq_3) \boldsymbol{\hat{q}_2}$. Analogous results hold for the area element on the surfaces $q_1 = \text{constant}$ and $q_3 = \text{constant}$.

The gradient $\boldsymbol{\nabla} f$ of a function $f$ is given by Eq.(1.7) in Cartesian coordinates. What is its expression in the new coordinates $(q_1, q_2, q_3)$? To answer the question we rely on the defining property of the gradient, as stated in Eq.(1.8), that it is related by

\begin{equation} df = \boldsymbol{\nabla} f \cdot d\boldsymbol{r} \tag{1.46}\end{equation}

to the change $df$ of the function $f$ induced by the infinitesimal displacement described by $d\boldsymbol{r}$. In this context we imagine that $f$ is a function of the coordinates $(q_1, q_2, q_3)$, and that $d\boldsymbol{r}$ is given by Eq.(1.44).

To begin we consider an infinitesimal displacement from a point $P$ at $(q_1, q_2, q_3)$ to a neighbouring point $Q$ at $(q_1 + dq_1, q_2, q_3)$. The displacement is in the direction of $q_1$, with $q_2$ and $q_3$ held fixed; it is described by the vector $d\boldsymbol{r} = h_1dq_1\, \boldsymbol{\hat{q}_1}$. The change in the function induced by the displacement is $df = f(q_1+dq_1, q_2, q_2) - f(q_1, q_2, q_3)$, and as we have seen, this is given by

\begin{equation} df = \boldsymbol{\nabla} f \cdot d\boldsymbol{r} = \boldsymbol{\nabla} f \cdot (h_1dq_1\, \boldsymbol{\hat{q}_1}) = h_1 (\boldsymbol{\nabla} f \cdot \boldsymbol{\hat{q}_1} )\, dq_1. \tag{1.47}\end{equation}

This implies that $\boldsymbol{\nabla} f \cdot \boldsymbol{\hat{q}_1} = h_1^{-1} (df/dq_1)$, and since $q_2$ and $q_3$ are held fixed during the displacement, the result is better expressed as

\begin{equation} \boldsymbol{\nabla} f \cdot \boldsymbol{\hat{q}_1} = \frac{1}{h_1} \frac{\partial f}{\partial q_1}, \tag{1.48} \end{equation}

in terms of a partial derivative with respect to $q_1$. We could, in a similar way, consider displacements in the other directions, $q_2$ and $q_3$, and obtain analogous results for these directions. All of these can be summarized by the single equation

\begin{equation} \boldsymbol{\nabla} f \cdot \boldsymbol{\hat{q}_j} = \frac{1}{h_j} \frac{\partial f}{\partial q_j}. \tag{1.49} \end{equation}

This is a key result that is just about to reveal the answer to our question.

The quantity $\boldsymbol{\nabla} f \cdot \boldsymbol{\hat{q}_j}$ is nothing but the component of the vector $\boldsymbol{\nabla} f$ in the direction of the basis vector $\boldsymbol{\hat{q}_j}$. The complete vector can be written as

\begin{equation} \boldsymbol{\nabla} f = (\boldsymbol{\nabla} f \cdot \boldsymbol{\hat{q}_1})\, \boldsymbol{\hat{q}_1} + (\boldsymbol{\nabla} f \cdot \boldsymbol{\hat{q}_2})\, \boldsymbol{\hat{q}_2} + (\boldsymbol{\nabla} f \cdot \boldsymbol{\hat{q}_3})\, \boldsymbol{\hat{q}_3}, \tag{1.50}\end{equation}

and substituting our previous result gives

\begin{equation} \boldsymbol{\nabla} f = \frac{1}{h_1} \frac{\partial f}{\partial q_1}\, \boldsymbol{\hat{q}_1} + \frac{1}{h_2} \frac{\partial f}{\partial q_2}\, \boldsymbol{\hat{q}_2} + \frac{1}{h_3} \frac{\partial f}{\partial q_3}\, \boldsymbol{\hat{q}_3}, \tag{1.51} \end{equation}

the desired expression for the gradient vector, decomposed in the new basis $\boldsymbol{\hat{q}_j}$

The expression of Eq.(1.51) is valid in any system of orthogonal coordinates. In Cartesian coordinates ($q_1 = x$, $q_2 = y$, $q_3 = z$, $h_1 = h_2 = h_3 = 1$), the expression obviously reduces to Eq.(1.7). In cylindrical coordinates ($q_1 = s$, $q_2 = \phi$, $q_3 = z$, $h_1 = h_3 = 1$, $h_2 = s$), it becomes

\begin{equation} \boldsymbol{\nabla} f = \frac{\partial f}{\partial s}\, \boldsymbol{\hat{s}} + \frac{1}{s} \frac{\partial f}{\partial \phi}\, \boldsymbol{\hat{\phi}} + \frac{\partial f}{\partial z}\, \boldsymbol{\hat{z}},\tag{1.52} \end{equation}

the standard expression that can be found in many textbooks.

## 1.5 Divergence

We now wish to express the divergence $\boldsymbol{\nabla} \cdot \boldsymbol{A}$ of a vector field $\boldsymbol{A}$ in the coordinates $(q_1, q_2, q_3)$. Recalling that the vector is decomposed as

\begin{equation} \boldsymbol{A} =A_1\, \boldsymbol{\hat{q}_1} + A_2\, \boldsymbol{\hat{q}_2} + A_3\, \boldsymbol{\hat{q}_3},\tag{1.53} \end{equation}

we will show that the divergence is given by

\begin{equation} \boldsymbol{\nabla} \cdot \boldsymbol{A} = \frac{1}{h_1 h_2 h_3} \bigg[ \frac{\partial}{\partial q_1} \big( h_2 h_3 A_1 \big) + \frac{\partial}{\partial q_2} \big( h_1 h_3 A_2 \big) + \frac{\partial}{\partial q_3} \big( h_1 h_2 A_3 \big) \bigg]. \tag{1.54} \end{equation}

In Cartesian coordinates this obviously reduces to the expression of Eq.(1.10). In cylindrical coordinates, we have that Eq.(1.54) becomes

\begin{equation} \boldsymbol{\nabla} \cdot \boldsymbol{A} = \frac{1}{s} \frac{\partial}{\partial s} ( s A_s ) + \frac{1}{s} \frac{\partial A_\phi}{\partial \phi} + \frac{\partial A_z}{\partial z}, \tag{1.55}\end{equation}

which is the standard expression.

Exercise 4:Verify that the preceding result follows from Eq.(1.54).

To prove Eq.(1.54) we rely on the identities

\begin{equation} \boldsymbol{\nabla} \cdot \bigg( \frac{\boldsymbol{\hat{q}_1}}{h_2 h_3} \bigg) = 0, \qquad \boldsymbol{\nabla} \cdot \bigg( \frac{\boldsymbol{\hat{q}_2}}{h_1 h_3} \bigg) = 0, \qquad \boldsymbol{\nabla} \cdot \bigg( \frac{\boldsymbol{\hat{q}_3}}{h_1 h_2} \bigg) = 0, \tag{1.56} \end{equation}

to be established below, as well as the standard identity of vector calculus,

\begin{equation} \boldsymbol{\nabla} \cdot (f \boldsymbol{v}) = \boldsymbol{v} \cdot \boldsymbol{\nabla} f + f \boldsymbol{\nabla} \cdot \boldsymbol{v}, \tag{1.57} \end{equation}

in which $f$ is any function and $\boldsymbol{v}$ any vector field.

Exercise 5:Verify the identity of Eq.(1.57) by working out both sides of the equation.

We begin by writing

\begin{align} \boldsymbol{\nabla} \cdot \boldsymbol{A} &= \boldsymbol{\nabla} \cdot \big( A_1\, \boldsymbol{\hat{q}_1} + A_2\, \boldsymbol{\hat{q}_2} + A_3\, \boldsymbol{\hat{q}_3} \big) \nonumber \\ &= \boldsymbol{\nabla} \cdot \bigg( h_2 h_3 A_1\, \frac{\boldsymbol{\hat{q}_1}}{h_2 h_3} + h_1 h_3 A_2\, \frac{\boldsymbol{\hat{q}_2}}{h_1 h_3} + h_1 h_2 A_3\, \frac{\boldsymbol{\hat{q}_3}}{h_1 h_2} \bigg) \nonumber \\ &= \boldsymbol{\nabla} \cdot \bigg( h_2 h_3 A_1\, \frac{\boldsymbol{\hat{q}_1}}{h_2 h_3} \bigg) + \boldsymbol{\nabla} \cdot \bigg( h_1 h_3 A_2\, \frac{\boldsymbol{\hat{q}_2}}{h_1 h_3} \bigg) + \boldsymbol{\nabla} \cdot \bigg( h_1 h_2 A_3\, \frac{\boldsymbol{\hat{q}_3}}{h_1 h_2} \bigg).\tag{1.58} \end{align}

Let us focus on the first term of the right-hand side, and set $f := h_2 h_3 A_1$, $\boldsymbol{v} := \boldsymbol{\hat{q}_1}/(h_2 h_3)$. This term has the form of $\boldsymbol{\nabla} \cdot (f \boldsymbol{v})$, and using the identity of Eq.(1.57), this is $\boldsymbol{v} \cdot \boldsymbol{\nabla} f + f \boldsymbol{\nabla} \cdot \boldsymbol{v}$. But $\boldsymbol{\nabla} \cdot \boldsymbol{v} = 0$ by virtue of the first of Eqs.(1.56), and we have found that

\begin{equation} \boldsymbol{\nabla} \cdot \bigg( h_2 h_3 A_1\, \frac{\boldsymbol{\hat{q}_1}}{h_2 h_3} \bigg) = \frac{\boldsymbol{\hat{q}_1}}{h_2 h_3} \cdot \boldsymbol{\nabla} (h_2 h_3 A_1). \tag{1.59}\end{equation}

The right-hand side of the equation is proportional to $\boldsymbol{\hat{q}_1} \cdot \boldsymbol{\nabla} f$, and this, according to Eq.(1.49), is simply $h_1^{-1} \partial f/\partial q_1$. What we have, therefore, is

\begin{equation} \boldsymbol{\nabla} \cdot \bigg( h_2 h_3 A_1\, \frac{\boldsymbol{\hat{q}_1}}{h_2 h_3} \bigg) = \frac{1}{h_1 h_2 h_3} \frac{\partial}{\partial q_1} (h_2 h_3 A_1)\tag{1.60} \end{equation}

for the first term in our expression for $\boldsymbol{\nabla} \cdot \boldsymbol{A}$. With similar manipulations for the remaining two terms, we finally arrive at Eq.(1.54).

Exercise 6:Complete the derivation of Eq.(1.54).

We have yet to establish the identities of Eq.(1.56). For this we need two more vector-calculus identities,

\begin{align} \boldsymbol{\nabla} \times (\boldsymbol{\nabla} f) &= 0, \tag{1.61a} \\ \boldsymbol{\nabla} \cdot ( \boldsymbol{u} \times \boldsymbol{v} \bigr) &= \boldsymbol{v} \cdot ( \boldsymbol{\nabla} \times \boldsymbol{u} ) - \boldsymbol{u} \cdot ( \boldsymbol{\nabla} \times \boldsymbol{v} ), \tag{1.61b} \end{align}

in which $f$ is any function, while $\boldsymbol{u}$ and $\boldsymbol{v}$ are arbitrary vector fields.

Exercise 7:Verify Eqs.(1.61).

We begin with Eq.(1.51) for $\boldsymbol{\nabla} f$, which we apply to the specific function $f = q_1$. Because $q_1$ is independent of both $q_2$ and $q_3$, the formula returns $\boldsymbol{\nabla} q_1 = \boldsymbol{\hat{q}_1}/h_1$. We obtain analogous results when we calculate $\boldsymbol{\nabla} q_2$ and $\boldsymbol{\nabla} q_3$, and the summary of all these is

\begin{equation} \boldsymbol{\nabla} q_j = \frac{1}{h_j}\, \boldsymbol{\hat{q}_j}. \tag{1.62} \end{equation}

These results imply that $\boldsymbol{\nabla}q_1 \times \boldsymbol{\nabla} q_2 = \boldsymbol{\hat{q}_1} \times \boldsymbol{\hat{q}_2}/(h_1 h_2) = \boldsymbol{\hat{q}_3}/(h_1 h_2)$, where we used Eq.(1.38) to re-express the cross product. We next take the divergence of this equation,

\begin{equation} \boldsymbol{\nabla} \cdot \bigg( \frac{\boldsymbol{\hat{q}_3}}{h_1 h_2} \bigg) = \boldsymbol{\nabla} \cdot \big( \boldsymbol{\nabla} q_1 \times \boldsymbol{\nabla} q_2 \big), \tag{1.63}\end{equation}

and apply Eq.(1.61b) with $\boldsymbol{u} = \boldsymbol{\nabla} q_1$ and $\boldsymbol{v} = \boldsymbol{\nabla} q_2$. The result involves $\boldsymbol{\nabla} \times \boldsymbol{u} = \boldsymbol{\nabla} \times (\boldsymbol{\nabla} q_1)$ and $\boldsymbol{\nabla} \times \boldsymbol{v} = \boldsymbol{\nabla} \times (\boldsymbol{\nabla} q_2)$, which are both zero by virtue of Eq.(1.61a) applied to $f = q_1$ and $f = q_2$, respectively. We have obtained

\begin{equation} \boldsymbol{\nabla} \cdot \bigg( \frac{\boldsymbol{\hat{q}_3}}{h_1 h_2} \bigg) = 0, \tag{1.64}\end{equation}

which is the third of Eqs.(1.56). We can prove the remaining two statements by following similar steps, beginning this time with either $\boldsymbol{\nabla} q_2 \times \boldsymbol{\nabla} q_3 = \boldsymbol{\hat{q}_1}/(h_2 h_3)$ or $\boldsymbol{\nabla} q_3 \times \boldsymbol{\nabla} q_1 = \boldsymbol{\hat{q}_2}/(h_1 h_3)$.

Exercise 8: Complete the derivation of Eqs.(1.56).

## 1.6 Curl

We intend to show that the curl of a vector field $\boldsymbol{A}$ is given by

\begin{align} \boldsymbol{\nabla} \times \boldsymbol{A} &= \frac{\boldsymbol{\hat{q}_1}}{h_2 h_3} \bigg[ \frac{\partial}{\partial q_2} (h_3 A_3) - \frac{\partial}{\partial q_3}(h_2 A_2) \bigg] + \frac{\boldsymbol{\hat{q}_2}}{h_1 h_3} \bigg[ \frac{\partial}{\partial q_3} (h_1 A_1) - \frac{\partial}{\partial q_1}(h_3 A_3) \bigg] \nonumber \\ & \quad \text{} + \frac{\boldsymbol{\hat{q}_3}}{h_1 h_2} \bigg[ \frac{\partial}{\partial q_1} (h_2 A_2) - \frac{\partial}{\partial q_2}(h_1 A_1) \bigg] \tag{1.65}\end{align}

in the coordinates $(q_1, q_2, q_3)$. The expression is a bit formidable, but you will recognize a clear pattern if you examine it closely; you will notice, in particular, the role of the cyclic order $1 \to 2 \to 3 \to 1 \to \cdots$. In the usual Cartesian coordinates, the general expression of Eq.(1.65) clearly reduces to Eq.(1.11). In cylindrical coordinates ($q_1 = s$, $q_2 = \phi$, $q_3 = z$ with $h_1 = 1$, $h_2 = s$, $h_3 = 1$), we have that

\begin{align} \boldsymbol{\nabla} \times \boldsymbol{A} &= \boldsymbol{\hat{s}} \bigg[ \frac{1}{s} \frac{\partial A_z}{\partial \phi} - \frac{\partial A_\phi}{\partial z} \bigg] + \boldsymbol{\hat{\phi}} \bigg[ \frac{\partial A_s}{\partial z} - \frac{\partial A_z}{\partial s} \bigg] + \boldsymbol{\hat{z}} \biggl[ \frac{1}{s} \frac{\partial}{\partial s}(s A_\phi) - \frac{1}{s} \frac{\partial A_s}{\partial \phi} \biggr].\tag{1.66} \end{align}

Exercise 9: Verify that $\boldsymbol{\nabla}\times \boldsymbol{A}$, as given previously in cylindrical coordinates, follows from Eq.(1.65).

To establish Eq.(1.65) we will need the vector-calculus identity

\begin{equation} \boldsymbol{\nabla} \times (f \boldsymbol{v}) = -\boldsymbol{v} \times \boldsymbol{\nabla} f + f \boldsymbol{\nabla} \times \boldsymbol{v}, \tag{1.67} \end{equation}

where $f$ is any scalar function and $\boldsymbol{v}$ any vector field, as well as

\begin{equation} \boldsymbol{\nabla} \times \bigg( \frac{\boldsymbol{\hat{q}_j}}{h_j} \bigg) = \boldsymbol{0}, \tag{1.68} \end{equation}

which follows directly from Eqs.(1.61a) and (1.62).

Exercise 10: Verify the identities of Eqs.(1.67) and (1.68).

We begin by calculating the curl of the vector field $\boldsymbol{A}$, decomposed in the basis provided by the unit vectors $\boldsymbol{\hat{q}_j}$,

\begin{align} \boldsymbol{\nabla} \times \boldsymbol{A} &= \boldsymbol{\nabla} \times \big( A_1\, \boldsymbol{\hat{q}_1} + A_2\, \boldsymbol{\hat{q}_2} + A_3\, \boldsymbol{\hat{q}_3} \big) \nonumber \\ &= \boldsymbol{\nabla} \times \bigg( h_1 A_1\, \frac{\boldsymbol{\hat{q}_1}}{h_1} + h_2 A_2\, \frac{\boldsymbol{\hat{q}_2}}{h_2} + h_3 A_3\, \frac{\boldsymbol{\hat{q}_3}}{h_3} \bigg) \nonumber \\ &= \boldsymbol{\nabla} \times \bigg( h_1 A_1\, \frac{\boldsymbol{\hat{q}_1}}{h_1} \bigg) + \boldsymbol{\nabla} \times \bigg( h_2 A_2\, \frac{\boldsymbol{\hat{q}_2}}{h_2} \bigg) + \boldsymbol{\nabla} \times \bigg( h_3 A_3\, \frac{\boldsymbol{\hat{q}_3}}{h_3} \bigg). \tag{1.69}\end{align}

Each term in this expression is of the form given by Eq.(1.67). For example, if we single out the first term, we recognize $h_1 A_1$ as the function $f$, and $\boldsymbol{\hat{q}_1}/h_1$ as the vector $\boldsymbol{v}$. Because $\boldsymbol{v}$ has a vanishing curl by virtue of Eq.(1.68), we have that the first term can be written as $-\boldsymbol{v} \times \boldsymbol{\nabla} f$, or $-(\boldsymbol{\hat{q}_1}/h_1) \times \boldsymbol{\nabla}(h_1 A_1)$. Repeating the same steps for the remaining two terms, we have that

\begin{equation} \boldsymbol{\nabla} \times \boldsymbol{A} = -\frac{\boldsymbol{\hat{q}_1}}{h_1} \times \boldsymbol{\nabla} (h_1 A_1) - \frac{\boldsymbol{\hat{q}_2}}{h_2} \times \boldsymbol{\nabla} (h_2 A_2) - \frac{\boldsymbol{\hat{q}_3}}{h_3} \times \boldsymbol{\nabla} (h_3 A_3).\tag{1.70} \end{equation}

At this stage we substitute Eq.(1.51) for the gradient of a function $f$, and obtain the very long expression

\begin{align} \boldsymbol{\nabla} \times \boldsymbol{A} &= -\frac{\boldsymbol{\hat{q}_1}}{h_1} \times \bigg[ \frac{1}{h_1} \frac{\partial}{\partial q_1} (h_1 A_1)\, \boldsymbol{\hat{q}_1} + \frac{1}{h_2} \frac{\partial}{\partial q_2} (h_1 A_1)\, \boldsymbol{\hat{q}_2} + \frac{1}{h_3} \frac{\partial}{\partial q_3} (h_1 A_1)\, \boldsymbol{\hat{q}_3} \bigg] \nonumber \\ & \quad \mbox{} -\frac{\boldsymbol{\hat{\text{q}}_2}}{h_2} \times \bigg[ \frac{1}{h_1} \frac{\partial}{\partial q_1} (h_2 A_2)\, \boldsymbol{\hat{q}_1} + \frac{1}{h_2} \frac{\partial}{\partial q_2} (h_2 A_2)\, \boldsymbol{\hat{q}_2} + \frac{1}{h_3} \frac{\partial}{\partial q_3} (h_2 A_2)\, \boldsymbol{\hat{q}_3} \bigg] \nonumber \\ & \quad \mbox{} -\frac{\boldsymbol{\hat{q}_3}}{h_3} \times \bigg[ \frac{1}{h_1} \frac{\partial}{\partial q_1} (h_3 A_3)\, \boldsymbol{\hat{q}_1} + \frac{1}{h_2} \frac{\partial}{\partial q_2} (h_3 A_3)\, \boldsymbol{\hat{q}_2} + \frac{1}{h_3} \frac{\partial}{\partial q_3} (h_3 A_3)\, \boldsymbol{\hat{q}_3} \bigg]. \tag{1.71}\end{align}

Fortunately this can be simplified dramatically by invoking the relations of Eq.(1.38). We get

\begin{align} \boldsymbol{\nabla} \times \boldsymbol{A} &= -\frac{(+\boldsymbol{\hat{q}_3})}{h_1 h_2} \frac{\partial}{\partial q_2}(h_1 A_1)\, - \frac{(-\boldsymbol{\hat{q}_2})}{h_1 h_3} \frac{\partial}{\partial q_3}(h_1 A_1)\, \nonumber \\ & \quad \mbox{} - \frac{(-\boldsymbol{\hat{q}_3})}{h_1 h_2} \frac{\partial}{\partial q_1}(h_2 A_2)\, - \frac{(+\boldsymbol{\hat{q}_1})}{h_2 h_3} \frac{\partial}{\partial q_3}(h_2 A_2)\, \nonumber \\ & \quad \mbox{} - \frac{(+\boldsymbol{\hat{q}_2})}{h_1 h_3} \frac{\partial}{\partial q_1}(h_3 A_3)\, - \frac{(-\boldsymbol{\hat{q}_1})}{h_3 h_3} \frac{\partial}{\partial q_2}(h_3 A_3),\tag{1.72} \end{align}

which is the same statement as in Eq.(1.65).

Exercise 11: Be sure to reproduce all the steps that led us to Eq.(1.65).

## 1.7 Laplacian

The Laplacian $\nabla^2 f$ of a scalar function $f$ is defined as the divergence of the vector field $\boldsymbol{\nabla} f$,

\begin{equation} \nabla^2 f := \boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} f), \tag{1.73}\end{equation}

and this definition is independent of the choice of coordinates. With the gradient given by Eq.(1.51) and the divergence given by Eq.(1.54), we have that the Laplacian is

\begin{equation} \nabla^2 f = \frac{1}{h_1 h_2 h_3} \bigg[ \frac{\partial}{\partial q_1} \bigg( \frac{h_2 h_3}{h_1} \frac{\partial f}{\partial q_1} \bigg) + \frac{\partial}{\partial q_2} \bigg( \frac{h_1 h_3}{h_2} \frac{\partial f}{\partial q_2} \bigg) + \frac{\partial}{\partial q_3} \bigg( \frac{h_1 h_2}{h_3} \frac{\partial f}{\partial q_3} \bigg) \bigg] \tag{1.74}\end{equation}

in the coordinates $(q_1, q_2, q_3)$. The expression clearly reduces to Eq.(1.12) in Cartesian coordinates. The general expression also implies that the Laplacian is given by

\begin{equation} \nabla^2 f = \frac{1}{s} \frac{\partial}{\partial s} \bigg( s \frac{\partial f}{\partial s} \bigg) + \frac{1}{s^2} \frac{\partial^2 f}{\partial \phi^2} + \frac{\partial^2 f}{\partial z^2}\tag{1.75} \end{equation}

in cylindrical coordinates

Exercise 12: Verify the general expression of Eq.(1.74), and reproduce its specific form in cylindrical coordinates.

## 1.8 Spherical Coordinates

In the preceding sections we developed a general theory of orthogonal coordinates $(q_1, q_2, q_3)$, and derived expressions for the gradient, divergence, curl, and Laplacian operators in this general coordinate system. Throughout the discussion we illustrated these results with the specific case of cylindrical coordinates $(s, \phi, z)$. In this section we provide further illustration with another familiar coordinate system, the spherical coordinates $(r, \theta, \phi)$.

The coordinates are defined by their relation to the Cartesian coordinates $(x, y, z)$, given by

\begin{equation} x = r\sin\theta\cos\phi, \qquad y = r\sin\theta\sin\phi, \qquad z = r\cos\theta, \tag{1.76}\end{equation}

and their geometrical meaning is displayed in Fig.1.3. We adopt the notation $q_1 \equiv r$, $q_2 \equiv \theta$, and $q_3 \equiv \phi$. The position vector $\boldsymbol{r}$ of a generic point $P$ in three-dimensional space can be expressed as

\begin{equation} \boldsymbol{r}(r,\theta,\phi) = r\sin\theta\cos\phi\, \boldsymbol{\hat{x}} + r\sin\theta\sin\phi\, \boldsymbol{\hat{y}} + r\cos\theta\, \boldsymbol{\hat{z}} \tag{1.77} \end{equation}

in spherical coordinates. You can verify that the length of the position vector is given by $|\boldsymbol{r}| = r$. According to the general theory formulated in Sec.1.3, the basis vectors $\boldsymbol{\hat{r}}$, $\boldsymbol{\hat{\theta}}$, and $\boldsymbol{\hat{\phi}}$ are proportional to $\partial \boldsymbol{r}/\partial r$, $\partial \boldsymbol{r}/\partial \theta$, and $\partial \boldsymbol{r}/\partial \phi$, respectively. Working out the partial derivatives yields

\begin{align} \frac{\partial \boldsymbol{r}}{\partial r} &= \big( \sin\theta\cos\phi\, \boldsymbol{\hat{x}} + \sin\theta\sin\phi\, \boldsymbol{\hat{y}} + \cos\theta\, \boldsymbol{\hat{z}} \big),\tag{1.78a} \\ \frac{\partial \boldsymbol{r}}{\partial \theta} &= r \big( \cos\theta\cos\phi\, \boldsymbol{\hat{x}} + \cos\theta\sin\phi\, \boldsymbol{\hat{y}} - \sin\theta\, \boldsymbol{\hat{z}} \big),\tag{1.78b} \\ \frac{\partial \boldsymbol{r}}{\partial \phi} &= r\sin\theta \big( -\sin\phi\, \boldsymbol{\hat{x}} + \cos\phi\, \boldsymbol{\hat{y}} \big),\tag{1.78c} \end{align}

and we observe that all vectors within round brackets possess a unit length. The calculation therefore identifies the basis vectors attached to the spherical coordinates, along with the scale factors $h_j$. We have

\begin{align} \boldsymbol{\hat{r}} &= \frac{\partial \boldsymbol{r}}{\partial r} = \sin\theta\cos\phi\, \boldsymbol{\hat{x}} + \sin\theta\sin\phi\, \boldsymbol{\hat{y}} + \cos\theta\, \boldsymbol{\hat{z}}, \tag{1.79a}\\ \boldsymbol{\hat{\theta}}&=\frac{1}{r} \frac{\partial \boldsymbol{r}}{\partial \theta} = \cos\theta\cos\phi\, \boldsymbol{\hat{x}} + \cos\theta\sin\phi\, \boldsymbol{\hat{y}} - \sin\theta\, \boldsymbol{\hat{z}},\tag{1.79b} \\ \boldsymbol{\hat{\phi}} &= \frac{1}{r\sin\theta} \frac{\partial \boldsymbol{r}}{\partial \phi} = -\sin\phi\, \boldsymbol{\hat{x}} + \cos\phi\, \boldsymbol{\hat{y}}, \tag{1.79c}\end{align}

as well as

\begin{equation} h_r = 1, \qquad h_\theta = r, \qquad h_\phi = r\sin\theta. \tag{1.80}\end{equation}

You will observe that when decomposed in the new basis, the position vector is given simply by $\boldsymbol{r} = r \boldsymbol{\hat{r}}$.

Exercise 13:Verify that the basis vectors $\boldsymbol{\hat{r}}$, $\boldsymbol{\hat{\theta}}$, and $\boldsymbol{\hat{\phi}}$ are all mutually orthogonal, and that they satisfy the right-handed conditions $\boldsymbol{\hat{r}} \times \boldsymbol{\hat{\theta}} = \boldsymbol{\hat{\phi}}$, $\boldsymbol{\hat{\theta}} \times \boldsymbol{\hat{\phi}} = \boldsymbol{\hat{r}}$, and $\boldsymbol{\hat{\phi}} \times \boldsymbol{\hat{r}} = \boldsymbol{\hat{\theta}}$.

Exercise 14:Decompose the Cartesian basis vectors $\boldsymbol{\hat{x}}$, $\boldsymbol{\hat{y}}$, and $\boldsymbol{\hat{z}}$ in the spherical basis $\boldsymbol{\hat{r}}$, $\boldsymbol{\hat{\theta}}$, and $\boldsymbol{\hat{\phi}}$. That is, invert the relations of Eqs.(1.79). Recall from Sec.1.2 that this requires very little work.

According to the general expression of Eq.(1.44), the infinitesimal displacement vector is given by

\begin{equation} d\boldsymbol{r} = dr\, \boldsymbol{\hat{r}} + rd\theta\, \boldsymbol{\hat{\theta}}+ r\sin\theta d\phi\, \boldsymbol{\hat{\phi}} \tag{1.81} \end{equation}

in spherical coordinates. This identifies the orthogonal displacements as $dr$, $r\, d\theta$, and $r\sin\theta\, d\phi$. Joining them together forms the volume element

\begin{equation} dV = (dr)(r\, d\theta)(r\sin\theta\, d\phi) = r^2dr\, \sin\theta d\theta\, d\phi. \tag{1.82} \end{equation}

Area elements can be formed in a similar way. For example, the area element on a surface $\theta = \text{constant}$ is given by $d\boldsymbol{a} = (dr)(r\sin\theta\, d\phi) \boldsymbol{\hat{\theta}}$.

The general expression of Eq.(1.51) becomes

\begin{equation} \boldsymbol{\nabla} f = \frac{\partial f}{\partial r}\, \boldsymbol{\hat{r}} + \frac{1}{r} \frac{\partial f}{\partial \theta}\, \boldsymbol{\hat{\theta}} + \frac{1}{r\sin\theta} \frac{\partial f}{\partial \phi}\, \boldsymbol{\hat{\phi}} \tag{1.83} \end{equation}

in spherical coordinates. From Eqs.(1.51) and (1.65) we get

\begin{equation} \boldsymbol{\nabla} \cdot \boldsymbol{A} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 A_r) + \frac{1}{r\sin\theta} \frac{\partial}{\partial\theta} (\sin\theta A_\theta) + \frac{1}{r\sin\theta} \frac{\partial A_\phi}{\partial \phi} \tag{1.84} \end{equation}

and

\begin{align} \boldsymbol{\nabla} \times \boldsymbol{A} &= \frac{\boldsymbol{\hat{r}}}{r\sin\theta} \bigg[ \frac{\partial}{\partial\theta} (\sin\theta A_\phi) - \frac{\partial A_\theta}{\partial \phi} \bigg] + \frac{\boldsymbol{\hat{\theta}}}{r} \bigg[ \frac{1}{\sin\theta} \frac{\partial A_r}{\partial \phi} - \frac{\partial}{\partial r}(r A_\phi) \bigg] \nonumber \\ & \quad \text{} + \frac{\boldsymbol{\hat{\phi}}}{r} \bigg[ \frac{\partial}{\partial r} (r A_\theta) - \frac{\partial A_r}{\partial \theta} \bigg], \tag{1.85} \end{align}

respectively, where the vector field $\boldsymbol{A}$ is decomposed as $\boldsymbol{A} = A_r\, \boldsymbol{\hat{r}} + A_\theta\, \boldsymbol{\hat{\theta}} + A_\phi\, \boldsymbol{\hat{\phi}}$. And according to Eq.(1.74), the Laplacian operator is given by

\begin{equation} \nabla^2 f = \frac{1}{r^2} \frac{\partial}{\partial r} \bigg( r^2 \frac{\partial f}{\partial r} \bigg) + \frac{1}{r^2\sin\theta} \frac{\partial}{\partial \theta} \bigg( \sin\theta \frac{\partial f}{\partial \theta} \bigg) + \frac{1}{r^2\sin^2\theta} \frac{\partial^2 f}{\partial \phi^2} \tag{1.86} \end{equation}

in spherical coordinates.

Exercise 15: Verify the foregoing expressions for the gradient, divergence, curl, and Laplacian operators in spherical coordinates.

## 1.9 Parabolic Coordinates

To conclude the chapter we examine another system of orthogonal coordinates that is less familiar than the cylindrical and spherical coordinates considered previously. This system, denoted $(u, v, z)$ and known as parabolic coordinates, is defined by

\begin{equation} x = uv, \qquad y = \frac{1}{2}(v^2 - u^2), \qquad z = z. \tag{1.87}\end{equation}

The geometrical meaning of the coordinates can be deduced by examining the curves $u = \text{constant}$ and $v = \text{constant}$ in the $x$-$y$ plane. If we eliminate $v$ from the preceding equations, we find that the curves $u = \text{constant}$ are described by

\begin{equation} y = \frac{x^2}{2u^2} - \frac{1}{2} u^2; \tag{1.88}\end{equation}

the equation reveals that these curves are parabolas. On the other hand, eliminating $u$ from the equations reveals that the curves $v = \text{constant}$ are described by

\begin{equation} y = \frac{1}{2} v^2 - \frac{x^2}{2v^2}, \tag{1.89}\end{equation}

and these are inverted parabolas. The coordinate system is illustrated in Fig.1.4. Figure 1.4: Parabolic coordinates. Curves of constant $u$ are shown in blue, while curves of constant $v$ are shown in green.

The position vector of a point $P$ can be expressed as

\begin{equation} \boldsymbol{r}(u,v,z) = uv\, \boldsymbol{\hat{x}} + \frac{1}{2}(v^2 - u^2)\, \boldsymbol{\hat{y}} + z\, \boldsymbol{\hat{z}}\tag{1.90} \end{equation}

in terms of the parabolic coordinates $q_1 \equiv u$, $q_2 \equiv v$, and $q_3 \equiv z$. The basis vectors $\boldsymbol{\hat{u}}$ and $\boldsymbol{\hat{v}}$, defined to be unit vectors pointing in the directions of increasing $u$ and $v$, respectively, are easily shown to be given by

\begin{align} \boldsymbol{\hat{u}} &= \frac{1}{\sqrt{u^2 + v^2}} \frac{\partial \boldsymbol{r}}{\partial u} = \frac{v}{\sqrt{u^2 + v^2}}\, \boldsymbol{\hat{x}} - \frac{u}{\sqrt{u^2 + v^2}}\, \boldsymbol{\hat{y}}, \tag{1.91a}\\ \boldsymbol{\hat{v}} &= \frac{1}{\sqrt{u^2 + v^2}} \frac{\partial \boldsymbol{r}}{\partial v} = \frac{u}{\sqrt{u^2 + v^2}}\, \boldsymbol{\hat{x}} + \frac{v}{\sqrt{u^2 + v^2}}\, \boldsymbol{\hat{y}}. \tag{1.91b} \end{align}

The scale factors are obviously

\begin{equation} h_u = \sqrt{u^2 + v^2}, \qquad h_v = \sqrt{u^2 + v^2}, \qquad h_z = 1. \tag{1.92} \end{equation}

Exercise 16: Verify that the basis vectors $\boldsymbol{\hat{u}}$ and $\boldsymbol{\hat{v}}$ are given by the preceding expressions. Then show that these vectors are orthogonal to each other. Finally, verify the right-handed conditions $\boldsymbol{\hat{u}} \times \boldsymbol{\hat{v}} = \boldsymbol{\hat{z}}$, $\boldsymbol{\hat{v}} \times \boldsymbol{\hat{z}} = \boldsymbol{\hat{u}}$, and $\boldsymbol{\hat{z}} \times \boldsymbol{\hat{u}} = \boldsymbol{\hat{v}}$.

Exercise 17: Decompose the Cartesian basis vectors $\boldsymbol{\hat{x}}$ and $\boldsymbol{\hat{y}}$ in the parabolic basis $\boldsymbol{\hat{u}}$ and $\boldsymbol{\hat{v}}$.

Exercise 18: Decompose the position vector $\boldsymbol{r}$ in the parabolic basis $\boldsymbol{\hat{u}}$, $\boldsymbol{\hat{v}}$, and $\boldsymbol{\hat{z}}$.

The general expression of Eq.(1.44) implies that the infinitesimal displacement vector is given by

\begin{equation} d\boldsymbol{r} = \sqrt{u^2 + v^2}\, du\, \boldsymbol{\hat{u}} + \sqrt{u^2 + v^2}\, dv\, \boldsymbol{\hat{v}} + dz\, \boldsymbol{\hat{z}} \end{equation}

in parabolic coordinates. It follows that the volume element is $dV = (u^2+v^2)\, dudvdz$ in this coordinate system.

Exercise 19: Derive expressions for $\boldsymbol{\nabla} f$, $\boldsymbol{\nabla} \cdot \boldsymbol{A}$, $\boldsymbol{\nabla} \times \boldsymbol{A}$ in parabolic coordinates.

Exercise 20: Show that the Laplacian operator is given by $\nabla^2 f = \frac{1}{u^2+v^2} \bigg( \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2} \bigg) + \frac{\partial^2 f}{\partial z^2}$ in parabolic coordinates.

## 1.10 Practice Problems

1. (Boas Chapter 10, Section 8, Problem 4) Express the vector field $\boldsymbol{v} = y\, \boldsymbol{\hat{x}} - x\, \boldsymbol{\hat{y}} + \boldsymbol{\hat{z}}$ in cylindrical coordinates. The vector must be decomposed in the cylindrical basis $\boldsymbol{\hat{s}}$, $\boldsymbol{\hat{\phi}}$, and $\boldsymbol{\hat{z}}$, and its components must be expressed as functions of $s$, $\phi$, and $z$.
2. Express the vector $\boldsymbol{v}$ of the preceding problem in spherical coordinates.

3. (Boas Chapter 10, Section 8, Problem 7) Find the basis vectors $\boldsymbol{\hat{u}}$, $\boldsymbol{\hat{v}}$ and the scale factors $h_u$, $h_v$ for the elliptic cylindrical coordinates $(u, v)$ defined by $x = a\cosh u\, \cos v$, $y = a\sinh u\, \sin v$, where $a$ is a constant.

4. (Boas Chapter 10, Section 9, Problem 16) Calculate $\boldsymbol{\nabla} \cdot \boldsymbol{\hat{s}}$, $\boldsymbol{\nabla} \cdot \boldsymbol{\hat{\phi}}$, $\boldsymbol{\nabla} \times \boldsymbol{\hat{s}}$, and $\boldsymbol{\nabla} \times \boldsymbol{\hat{\phi}}$.

5. (Boas Chapter 10, Section 9, Problem 17) Calculate $\boldsymbol{\nabla} \cdot \boldsymbol{\hat{r}}$, $\boldsymbol{\nabla} \cdot \boldsymbol{\hat{\theta}}$, $\boldsymbol{\nabla} \times \boldsymbol{\hat{\theta}}$, and $\boldsymbol{\nabla} \times \boldsymbol{\hat{\phi}}$.

6. (Boas Chapter 10, Section 9, Problem 18) Calculate $\boldsymbol{\nabla} \times (\ln s\, \boldsymbol{\hat{z}})$, $\boldsymbol{\nabla} \ln s$, and $\boldsymbol{\nabla} \cdot (s\, \boldsymbol{\hat{s}} + z\, \boldsymbol{\hat{z}})$.

7. (Boas Chapter 10, Section 9, Problem 19) Calculate $\boldsymbol{\nabla} \times (r\, \boldsymbol{\hat{\theta}})$, $\boldsymbol{\nabla} (r\cos\theta)$, and $\boldsymbol{\nabla} \cdot \boldsymbol{r}$.

8. (Boas Chapter 10, Section 9, Problem 20) Calculate $\nabla^2 s$, $\nabla^2 s^{-1}$, and $\nabla^2 \ln s$.

9. (Boas Chapter 10, Section 9, Problem 21) Calculate $\nabla^2 r$, $\nabla^2 r^2$, $\nabla^2 r^{-2}$, and $\nabla^2 e^{ikr\cos\theta}$, where $i$ is the imaginary number and $k$ is a constant.

10. Express the Laplacian $\nabla^2 f$ of a function $f$ in the elliptical cylindrical coordinates introduced in problem 3.

## 1.11 Challenge Problems

1. A vector field $\boldsymbol{v}$ is given by

$\boldsymbol{v} = -\frac{y}{x^2+y^2}\, \boldsymbol{\hat{x}} + \frac{x}{x^2+y^2}\, \boldsymbol{\hat{y}}$

a) Express the vector field in cylindrical coordinates. Make sure to decompose it in the vector basis of this coordinate system.

b) Calculate $\boldsymbol{\nabla} \times \boldsymbol{v}$.

c) Calculate the line integral $\oint_C \boldsymbol{v} \cdot d\boldsymbol{r}$ when the closed path $C$ is a circle of radius $s$ in the $x$-$y$ plane.

d) Discuss the previous results in light of the curl integral theorem. How do you resolve the apparent contradiction?

2. Calculate the partial derivatives of $(\boldsymbol{\hat{r}}, \boldsymbol{\hat{\theta}}, \boldsymbol{\hat{\phi}})$ with respect to $(r, \theta, \phi)$. There is a total of 9 quantities to calculate. Make  sure to express your results in terms of the vector basis of spherical coordinates.

3. A particle moves in the $x$-$y$ plane, and its position is described as a function of time by the spherical coordinates $r = r(t)$, $\theta = \pi/2$, and $\phi = \phi(t)$. Calculate the components of the particle's velocity vector $\boldsymbol{v} = d\boldsymbol{r}/dt$ and acceleration vector $\boldsymbol{a} = d\boldsymbol{v}/dt$ in spherical coordinates.

4. A coordinate system $(\mu,\theta,\phi)$ is defined by its relationship

$x = \cosh \mu\, \sin\theta\, \cos\phi, \qquad y = \cosh \mu\, \sin\theta\, \sin\phi, \qquad z = \sinh \mu\, \cos\theta$

to the Cartesian coordinates $(x, y, z)$. The new coordinates satisfy $\mu \geq 0$, $0 \leq \theta \leq \pi$, and $0 \leq \phi < 2\pi$.

a) Working in the $x$-$z$ plane ($\phi = 0$) for simplicity, describe the family of curves $\mu = \text{constant}$ and the family of curves $\theta = \text{constant}$. Be mathematically precise, and give as many interesting details as you can.

b) Still working in the $x$-$z$ plane, provide a parametric plot of this family of curves. Your plot should display a good number of these curves. Keep $\mu < 0.5$ to produce a nice plot, and make sure that the aspect ratio is equal to 1, so that a circle would appear round.

c) Calculate the basis vectors $\boldsymbol{\hat{\mu}}$, $\boldsymbol{\hat{\theta}}$, and $\boldsymbol{\hat{\phi}}$. Simplify your expressions as much as possible by exploiting the identities satisfied by trigonometric and hyperbolic functions. Verify that these vectors are all mutually orthogonal.

d) Calculate the scaling factors $h_\mu$, $h_\theta$, and $h_\phi$. Again, simplify your expressions as much as possible.

e) Work out the Laplacian $\nabla^2 f$ in this coordinate system.