Cylindrical Coordinates

In the cylindrical coordinate system, $u_1=r$ , $u_2=\theta$ , and $u_3=z$ , where $r=\sqrt{x^{\,2}+y^{\,2}}$ , $\theta=\tan^{-1}(y/x)$ , and $x$ , $y$ , $z$ are standard Cartesian coordinates. Thus, $r$ is the perpendicular distance from the $z$ -axis, and $\theta$ the angle subtended between the projection of the radius vector (i.e., the vector connecting the origin to a general point in space) onto the $x$ -$y$ plane and the $x$ -axis. (See Figure C.1.)

Figure C.1: Cylindrical coordinates.

A general vector ${\bf A}$ is written

$${\bf A} = A_r\,{\bf e}_r+ A_\theta\,{\bf e}_\theta + A_z\,{\bf e}_z,$$ (C.29)

where ${\bf e}_r=\nabla r/\vert\nabla r\vert$ , ${\bf e}_\theta = \nabla\theta/\vert\nabla\theta\vert$ , and ${\bf e}_z=\nabla z/\vert\nabla z\vert$ . Of course, the unit basis vectors ${\bf e}_r$ , ${\bf e}_\theta$ , and ${\bf e}_z$ are mutually orthogonal, so $A_r = {\bf A}\cdot {\bf e}_r$ , et cetera.

As is easily demonstrated, an element of length (squared) in the cylindrical coordinate system takes the form

$$d{\bf x}\cdot d{\bf x} = dr^{\,2} + r^{\,2}\,d\theta^{\,2} + dz^{\,2}.$$ (C.30)

Hence, comparison with Equation (C.6) reveals that the scale factors for this system are

$$h_r = 1,$$ (C.31)

$$h_\theta = r,$$ (C.32)

$$h_z = 1.$$ (C.33)

Thus, surface elements normal to ${\bf e}_r$ , ${\bf e}_\theta$ , and ${\bf e}_z$ are written

$$dS_r = r\,d\theta\,dz,$$ (C.34)

$$dS_\theta = dr\,dz,$$ (C.35)

$$dS_z = r\,dr\,d\theta,$$ (C.36)

respectively, whereas a volume element takes the form

$$dV = r\,dr\,d\theta\,dz.$$ (C.37)

According to Equations (C.13), (C.15), and (C.18), gradient, divergence, and curl in the cylindrical coordinate system are written

$$\nabla \psi = \frac{\partial \psi}{\partial r}\,{\bf e}_r + \frac{1}{r}\frac{... ...{\partial\theta}\,{\bf e}_\theta + \frac{\partial \psi}{\partial z}\,{\bf e}_z,$$ (C.38)

$$\nabla\cdot{\bf A} =\frac{1}{r}\,\frac{\partial}{\partial r}\,(r\,A_r) + \frac{1}{r}\,\frac{\partial A_\theta}{\partial\theta} + \frac{\partial A_z}{\partial z},$$ (C.39)

$$\nabla\times {\bf A} = \left(\frac{1}{r}\,\frac{\partial A_z}{\partial \theta}-\frac{\... ...\,A_\theta) - \frac{1}{r}\,\frac{\partial A_r}{\partial\theta}\right){\bf e}_z,$$ (C.40)

respectively. Here, $\psi({\bf r})$ is a general scalar field, and ${\bf A}({\bf r})$ a general vector field.

According to Equation (C.19), when expressed in cylindrical coordinates, the Laplacian of a scalar field becomes

$$\nabla^{\,2} \psi = \frac{1}{r}\,\frac{\partial}{\partial r}\left... ...,2} \psi}{\partial\theta^{\,2}} + \frac{\partial^{\,2} \psi}{\partial z^{\,2}}.$$ (C.41)

Moreover, from Equation (C.23), the components of $({\bf A}\cdot\nabla)\,{\bf A}$ in the cylindrical coordinate system are

$$[({\bf A}\cdot\nabla)\,{\bf A}]_r ={\bf A}\cdot\nabla A_r - \frac{A_\theta^{\,2}}{r},$$ (C.42)

$$[({\bf A}\cdot\nabla)\,{\bf A}]_\theta = {\bf A}\cdot\nabla A_\theta + \frac{A_r\,A_\theta}{r},$$ (C.43)

$$[({\bf A}\cdot\nabla)\,{\bf A}]_z = {\bf A}\cdot\nabla A_z.$$ (C.44)

Let us define the symmetric gradient tensor

$$\widetilde{\nabla {\bf A}} = \frac{1}{2}\left[\nabla {\bf A} + (\nabla{\bf A})^T\right].$$ (C.45)

Here, the superscript $T$ denotes a transpose. Thus, if the $ij$ element of some second-order tensor ${\bf S}$ is $S_{ij}$ then the corresponding element of ${\bf S}^T$ is $S_{ji}$ . According to Equation (C.26), the components of $\widetilde{\nabla {\bf A}}$ in the cylindrical coordinate system are

$$(\widetilde{\nabla {\bf A}})_{rr} = \frac{\partial A_r}{\partial r},$$ (C.46)

$$(\widetilde{\nabla {\bf A}})_{\theta\theta} =\frac{1}{r} \frac{\partial A_\theta}{\partial \theta}+ \frac{A_r}{r},$$ (C.47)

$$(\widetilde{\nabla {\bf A}})_{zz} = \frac{\partial A_z}{\partial z},$$ (C.48)

$$(\widetilde{\nabla {\bf A}})_{r\theta}=(\widetilde{\nabla {\bf A}})_{\theta r} = \frac{1}{2}\left(\frac{1}{r}\,\frac{\partial A_r}{\partial\theta} + \frac{\partial A_\theta}{\partial r} - \frac{A_\theta}{r}\right),$$ (C.49)

$$(\widetilde{\nabla {\bf A}})_{rz}=(\widetilde{\nabla {\bf A}})_{zr} =\frac{1}{2}\left(\frac{\partial A_r}{\partial z} + \frac{\partial A_z}{\partial r}\right),$$ (C.50)

$$(\widetilde{\nabla {\bf A}})_{\theta z}=(\widetilde{\nabla {\bf A}})_{z\theta} =\frac{1}{2}\left(\frac{\partial A_\theta}{\partial z} + \frac{1}{r}\frac{\partial A_z}{\partial \theta}\right).$$ (C.51)

Finally, from Equation (C.28), the components of $\nabla^{\,2}{\bf A}$ in the cylindrical coordinate system are

$$(\nabla^{\,2}{\bf A})_r =\nabla^{\,2} A_r - \frac{A_r}{r^{\,2}} - \frac{2}{r^{\,2}}\,\frac{\partial A_\theta}{\partial\theta},$$ (C.52)

$$(\nabla^{\,2}{\bf A})_\theta = \nabla^{\,2} A_\theta + \frac{2}{r^{\,2}}\,\frac{\partial A_r}{\partial \theta}-\frac{A_\theta}{r^{\,2}},$$ (C.53)

$$(\nabla^{\,2}{\bf A})_z =\nabla^{\,2} A_z.$$ (C.54)