Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Many sources of practical interest (wires, coaxial cables, line charges, cylindrical capacitors) have an axis of symmetry. Describing them with $x$ and $y$ is clumsy, and it is much more natural to use the distance from the axis and the angle around it. Think of finding a seat in a round theatre: “row $ρ$ (how far from the stage centre), seat angle $ϕ$ , balcony level $z$ ” is far more convenient than two perpendicular distances.

Coordinates and coordinate surfaces

In circular cylindrical coordinates a point $P$ is described by (Figures 2.3.8 and 2.3.9):

$ρ$ , the perpendicular distance from the $z$ -axis, $0 \leq ρ < \infty$ ,
$ϕ$ , the angle measured from the $x z$ -half-plane ( $x > 0$ ) to the half-plane containing $P$ , counter-clockwise when seen from $+ z$ , $0 \leq ϕ < 2 π$ , and
$z$ , the same as in Cartesian coordinates, $- \infty < z < \infty$ .

The coordinate surfaces are a circular cylinder of radius $ρ$ around the $z$ -axis ( $ρ = const$ ), a half-plane hinged on the $z$ -axis ( $ϕ = const$ ), and a plane perpendicular to the $z$ -axis ( $z = const$ ). The point $P$ is the single point that lies on all three (Figure 2.3.8).

The three coordinate surfaces of cylindrical coordinates through a point \mathrm P: the cylinder \rho=\text{const} (blue), the half-plane \phi=\text{const} hinged on the z-axis (green) and the plane z=\text{const} (orange). Each pair meets in a coordinate line (red): a circle, a vertical line, a radial line. The unit vectors at \mathrm P are perpendicular to the three surfaces: \vec a_\rho points away from the axis, \vec a_\phi along the circle towards increasing \phi, \vec a_z upwards. — **Figure 2.3.8.** The three coordinate surfaces of cylindrical coordinates through a point $P$ : the cylinder $ρ = const$ (blue), the half-plane $ϕ = const$ hinged on the $z$ -axis (green) and the plane $z = const$ (orange). Each pair meets in a coordinate line (red): a circle, a vertical line, a radial line. The unit vectors at $P$ are perpendicular to the three surfaces: $a_{ρ}$ points away from the axis, $a_{ϕ}$ along the circle towards increasing $ϕ$ , $a_{z}$ upwards.

Projecting $P$ onto the $x y$ -plane we see from the right triangle in that plane (Figure 2.3.9) that

x = ρ cos ϕ, y = ρ sin ϕ, z = z, (2.3.15)

and conversely

ρ cos ϕ = x^{2} + y^{2}, = \frac{x}{ρ}, sin ϕ = \frac{y}{ρ}, z = z . (2.3.16)

The angle $ϕ$ is fixed by the pair of relations for $cos ϕ$ and $sin ϕ$ . Neither alone suffices on the whole range $0 \leq ϕ < 2 π$ : $arccos$ returns angles only in $[0, π]$ and $arcsin$ only in $[- π /2, π /2]$ , so $ϕ = arccos (x / ρ) = arcsin (y / ρ)$ holds only in the first quadrant. In general

ϕ = {arccos (x / ρ), 2 π - arccos (x / ρ), y \geq 0, y < 0, (2.3.17)

which is what the two-argument function $atan2 (y, x)$ of programming languages computes (up to adding $2 π$ to its negative values). For example the point $(x, y) = (- 1, - 1)$ has $ρ = 2$ and $ϕ = 5 π /4$ , not $3 π /4$ as $arccos (- 1/ 2)$ alone would suggest. On the $z$ -axis ( $ρ = 0$ ) the angle $ϕ$ is undefined.

Cylindrical coordinates of a point \mathrm P: \rho is the distance of the projection of \mathrm P on the xy-plane from the origin, \phi is the angle of that projection measured from the x-axis towards the y-axis, and z is the height. The small right-angle marks show that the projection on the x-axis is x=\rho\cos\phi and on the y-axis y=\rho\sin\phi. — **Figure 2.3.9.** Cylindrical coordinates of a point $P$ : $ρ$ is the distance of the projection of $P$ on the $x y$ -plane from the origin, $ϕ$ is the angle of that projection measured from the $x$ -axis towards the $y$ -axis, and $z$ is the height. The small right-angle marks show that the projection on the $x$ -axis is $x = ρ cos ϕ$ and on the $y$ -axis $y = ρ sin ϕ$ .

Common mistake

Computing $ϕ = arctan (y / x)$ with a calculator. The ratio $y / x$ is the same for $(1, 1)$ and $(- 1, - 1)$ , and $arctan$ returns angles between $- π /2$ and $π /2$ only, so points in the second and third quadrants come out wrong by $π$ . Always look at the signs of $x$ and $y$ (or use Eq. (2.3.17), or $atan2$ ) to place the point in the correct quadrant.

Quick check

Give the cylindrical coordinates of the Cartesian point $(0, - 2, 5)$ .

$ρ = 0 + 4 = 2$ . From $cos ϕ = 0$ and $sin ϕ = - 1$ we get $ϕ = 3 π /2$ (by Eq. (2.3.17): $2 π - arccos 0 = 2 π - π /2$ ), and $z = 5$ . Hence $(ρ, ϕ, z) = (2, 3 π /2, 5)$ .

Position vector, unit vectors and scale factors

Inserting Eq. (2.3.15) into the Cartesian position vector,

r = ρ cos ϕ a_{x} + ρ sin ϕ a_{y} + z a_{z} . (2.3.18)

Now we apply Eq. (2.3.5) to each coordinate in turn: differentiate $r$ , take the length of the result (the scale factor), and divide by it (the unit vector).

The coordinate ρ

\frac{\partial r}{\partial ρ} = cos ϕ a_{x} + sin ϕ a_{y} .

Its length is $cos^{2} ϕ + sin^{2} ϕ = 1$ , hence

h_{ρ} = 1, a_{ρ} = cos ϕ a_{x} + sin ϕ a_{y} . (2.3.19)

$a_{ρ}$ points horizontally away from the $z$ -axis, along the projection of $r$ on the $x y$ -plane. Since $ρ$ is itself a length, no scale factor is needed.

The coordinate φ

\frac{\partial r}{\partial ϕ} = - ρ sin ϕ a_{x} + ρ cos ϕ a_{y} = ρ (- sin ϕ a_{x} + cos ϕ a_{y}) .

The bracket has unit length, so

h_{ϕ} = ρ, a_{ϕ} = - sin ϕ a_{x} + cos ϕ a_{y} . (2.3.20)

$a_{ϕ}$ is horizontal, tangent to the circle of radius $ρ$ around the $z$ -axis, in the direction of increasing $ϕ$ . The scale factor $h_{ϕ} = ρ$ expresses the fact that an increase $d ϕ$ of the angle moves the point along an arc of length $ρ d ϕ$ : the farther from the axis, the longer the arc (Figure 2.3.10).

The coordinate z

\frac{\partial r}{\partial z} = a_{z}, h_{z} = 1.

Why h_\phi=\rho. The same change of angle d\phi moves a point near the axis (radius \rho_1) along a short arc \rho_1\,d\phi and a point far from it (radius \rho_2) along a long arc \rho_2\,d\phi: an angle is not a length, and the scale factor \rho converts it into one. A change d\rho (red) is already a length, so h_\rho=1. — **Figure 2.3.10.** Why $h_{ϕ} = ρ$ . The same change of angle $d ϕ$ moves a point near the axis (radius $ρ_{1}$ ) along a short arc $ρ_{1} d ϕ$ and a point far from it (radius $ρ_{2}$ ) along a long arc $ρ_{2} d ϕ$ : an angle is not a length, and the scale factor $ρ$ converts it into one. A change $d ρ$ (red) is already a length, so $h_{ρ} = 1$ .

One checks directly from Eqs. (2.3.19) and (2.3.20) that $a_{ρ} \cdot a_{ϕ} = - cos ϕ sin ϕ + sin ϕ cos ϕ = 0$ , and that both are perpendicular to $a_{z}$ : the system is orthogonal. It is also right-handed, the cyclic order being $ρ \to ϕ \to z \to ρ$ :

a_{ρ} \times a_{ϕ} a_{ϕ} \times a_{z} a_{z} \times a_{ρ} = a_{z}, = a_{ρ}, = a_{ϕ} . (2.3.21)

Finally, since $ρ cos ϕ a_{x} + ρ sin ϕ a_{y} = ρ a_{ρ}$ , the position vector (2.3.18) takes the compact form

r = ρ a_{ρ} + z a_{z} . (2.3.22)

There is no $a_{ϕ}$ term: the angle of $P$ is hidden inside $a_{ρ}$ , which is different for different $ϕ$ .

The unit vectors depend on position

Equations (2.3.19) and (2.3.20) contain $ϕ$ : the directions of $a_{ρ}$ and $a_{ϕ}$ change from point to point (Figure 2.3.11), while $a_{z}$ is constant. Differentiating,

\frac{\partial a _{ρ}}{\partial ϕ} \frac{\partial a _{ρ}}{\partial ρ} = a_{ϕ}, \frac{\partial a _{ϕ}}{\partial ϕ} = - a_{ρ}, = \frac{\partial a _{ϕ}}{\partial ρ} = \frac{\partial a _{ρ}}{\partial z} = \frac{\partial a _{ϕ}}{\partial z} = 0 . (2.3.23)

(For example $\partial a_{ρ} / \partial ϕ = - sin ϕ a_{x} + cos ϕ a_{y} = a_{ϕ}$ .) This has two practical consequences that we shall meet again and again.

Components of vectors at different points cannot simply be added. For example the unit vector $a_{ρ}$ at $ϕ = 0$ is $a_{x}$ , and at $ϕ = π$ it is $- a_{x}$ , so their sum is $0$ , not “ $2 a_{ρ}$ ”. When the contributions of a source distribution are summed by an integral, $a_{ρ}$ and $a_{ϕ}$ must be expressed through the constant Cartesian unit vectors before they are taken out of the integral.
Derivatives of a vector field act on the unit vectors too. This is why the divergence and curl in cylindrical coordinates are not obtained by merely replacing $x, y, z$ by $ρ, ϕ, z$ in Eqs. (2.2.6) and (2.2.8). The correct forms are derived in Section 2.3.10.

Top view (looking down the z-axis) of the cylindrical unit vectors at four points on the circle \rho=\text{const}. \vec a_\rho (red) points away from the axis and \vec a_\phi (blue) is tangent to the circle, towards increasing \phi. Both rotate with the point, unlike \vec a_x and \vec a_y. At every point \vec a_\rho\times\vec a_\phi=\vec a_z, which points out of the page. — **Figure 2.3.11.** Top view (looking down the $z$ -axis) of the cylindrical unit vectors at four points on the circle $ρ = const$ . $a_{ρ}$ (red) points away from the axis and $a_{ϕ}$ (blue) is tangent to the circle, towards increasing $ϕ$ . Both rotate with the point, unlike $a_{x}$ and $a_{y}$ . At every point $a_{ρ} \times a_{ϕ} = a_{z}$ , which points out of the page.

Change $ϕ$ in the explorer below and watch $a_{ρ}$ and $a_{ϕ}$ rotate with the point while $a_{z}$ stays fixed. The three coordinate surfaces through the point are a cylinder, a half-plane and a plane.

InteractiveLoading simulation…

Common mistake

Adding unit vectors that belong to different points. For a ring of charge around the $z$ -axis, each small piece at angle $ϕ^{'}$ produces at a point on the axis a field whose horizontal part points along $- a_{ρ}$ evaluated at $ϕ^{'}$ (from the piece towards the axis). These are different vectors for different pieces (for two opposite pieces, at $ϕ^{'} = 0$ and $ϕ^{'} = π$ , they are $- a_{x}$ and $+ a_{x}$ ), so the horizontal parts cancel instead of adding up to a large multiple of “ $- a_{ρ}$ ”. Before adding or integrating, write $a_{ρ}$ and $a_{ϕ}$ with Eqs. (2.3.19) and (2.3.20) in terms of $a_{x}$ and $a_{y}$ .

Quick check

What are $a_{ρ}$ and $a_{ϕ}$ at a point with $ϕ = π /2$ (on the positive $y$ -axis)?

From Eqs. (2.3.19) and (2.3.20) with $cos ϕ = 0$ , $sin ϕ = 1$ : $a_{ρ} = a_{y}$ (away from the axis, along $+ y$ ) and $a_{ϕ} = - a_{x}$ (counter-clockwise seen from above). Check: $a_{ρ} \times a_{ϕ} = a_{y} \times (- a_{x}) = a_{z}$ .

Differential elements

With $h_{ρ} = 1$ , $h_{ϕ} = ρ$ , $h_{z} = 1$ , the general results (2.3.6) to (2.3.8) give

Differential elements in cylindrical coordinates

d ℓ = d ρ a_{ρ} + ρ d ϕ a_{ϕ} + d z a_{z}, (2.3.24)

d S_{ρ} d S_{ϕ} d S_{z} = a_{ρ} ρ d ϕ d z, = a_{ϕ} d ρ d z, = a_{z} ρ d ρ d ϕ, (2.3.25)

d V = ρ d ρ d ϕ d z . (2.3.26)

Figure 2.3.12 shows the volume element. It is bounded by the cylinders $ρ$ and $ρ + d ρ$ , the half-planes $ϕ$ and $ϕ + d ϕ$ , and the planes $z$ and $z + d z$ . Its edges have lengths $d ρ$ (radial), $ρ d ϕ$ (an arc of radius $ρ$ ) and $d z$ (vertical). To first order it is a rectangular box, and its volume is the product of the three edges. Each face is the product of two edges: the curved face on the cylinder $ρ$ has area $ρ d ϕ d z$ , the flat face on a half-plane $ϕ = const$ has area $d ρ d z$ , and the top and bottom faces have area $ρ d ρ d ϕ$ . A useful check of any cylindrical expression is dimensional: $d ϕ$ is dimensionless, so it must always be accompanied by a length $ρ$ to form an arc, an area or a volume with the right units.

The differential volume in cylindrical coordinates, bounded by the surfaces \rho and \rho+d\rho, \phi and \phi+d\phi, z and z+dz. Its three edges have lengths d\rho, \rho\,d\phi and dz, so dV=\rho\,d\rho\,d\phi\,dz. The blue outline in the xy-plane is its projection, the area element \rho\,d\rho\,d\phi. — **Figure 2.3.12.** The differential volume in cylindrical coordinates, bounded by the surfaces $ρ$ and $ρ + d ρ$ , $ϕ$ and $ϕ + d ϕ$ , $z$ and $z + d z$ . Its three edges have lengths $d ρ$ , $ρ d ϕ$ and $d z$ , so $d V = ρ d ρ d ϕ d z$ . The blue outline in the $x y$ -plane is its projection, the area element $ρ d ρ d ϕ$ .

In plain words

In cylindrical coordinates every step in $ϕ$ must be multiplied by $ρ$ to become a length: $ρ d ϕ$ . That single factor is the origin of the $ρ$ in $d S_{ρ}$ , in $d S_{z}$ and in $d V$ . Patches with the same $d ρ$ and $d ϕ$ are bigger far from the axis than near it.

Common mistake

Forgetting the Jacobian $ρ$ : writing $d V = d ρ d ϕ d z$ or $d S_{z} = d ρ d ϕ$ . The result then has the wrong units (one power of length is missing) and the wrong value: for a disc of radius $b$ one would get $2 π b$ instead of $π b^{2}$ (Example 2.3.12).

Quick check

What is the length of a quarter of the circle $ρ = 2$ (in the plane $z = 0$ ), and what is the area of the quarter disc $0 \leq ρ \leq 2$ , $0 \leq ϕ \leq π /2$ ?

Length: $\int_{0}^{π /2} ρ d ϕ = 2 \cdot π /2 = π$ . Area: $\int_{0}^{π /2} \int_{0}^{2} ρ d ρ d ϕ = \frac{2 ^{2}}{2} \cdot \frac{π}{2} = π$ . (The two numbers are equal only because $ρ = 2$ , and their units are different: $m$ and $m^{2}$ .)

2.3.4 Cylindrical coordinates

Coordinates and coordinate surfaces

Position vector, unit vectors and scale factors

The coordinate ρ

The coordinate φ

The coordinate z

The unit vectors depend on position

Differential elements