Cylindrical Coordinates

Cylindrical coordinates are a generalization of 2-D Polar Coordinates to 3-D by superposing a height ($z$) axis. Unfortunately, there are a number of different notations used for the other two coordinates. Either $r$ or $\rho$ is used to refer to the radial coordinate and either $\phi$ or $\theta$ to the azimuthal coordinates. Arfken (1985), for instance, uses $(\rho, \phi, z)$, while Beyer (1987) uses $(r, \theta, z)$. In this work, the Notation $(r, \theta, z)$ is used.

$$r = \sqrt{x^2 +y^2}$$ (1)

$$\theta = \tan^{-1}\left({y\over x}\right)$$ (2)

$$z = z,$$ (3)

where $r\in [0,\infty)$, $\theta\in [0,2\pi)$, and $z\in (-\infty, \infty)$. In terms of $x$, $y$, and $z$

$$x = r\cos\theta$$ (4)

$$y = r\sin\theta$$ (5)

$$z = z.$$ (6)

Morse and Feshbach (1953) define the cylindrical coordinates by

$$x = \xi_1\xi_2$$ (7)

$$y = \xi_1\sqrt{1-{\xi_2}^2}$$ (8)

$$z = \xi_3,$$ (9)

where $\xi_1=r$ and $\xi_2=\cos\theta$. The Metric elements of the cylindrical coordinates are

$$g_{rr} = 1$$ (10)

$$g_{\theta\theta} = r^2$$ (11)

$$g_{zz} = 1,$$ (12)

so the Scale Factors are

$$g_r = 1$$ (13)

$$g_\theta = r$$ (14)

$$g_z = 1.$$ (15)

The Line Element is

$$d{\bf s} = dr \,\hat {\bf r} + r\,d\theta\, \hat {\boldsymbol{\theta}} + dz \,\hat {\bf z},$$ (16)

and the Volume Element is

$$dV = r dr\,d\theta\,dz.$$ (17)

The Jacobian is

$$\left\vert{\partial (x, y, z)\over \partial (r, \theta, z)}\right\vert = r.$$ (18)

A Cartesian Vector is given in Cylindrical Coordinates by

$${\bf r} = \left[{\matrix{r\cos \theta\cr r\sin \theta\cr z\cr}}\right].$$ (19)

To find the Unit Vectors,

$$\hat {\bf r} \equiv {{d{\bf r}\over dr}\over \left\vert\begin{array}{c}d{\bf r}\over ... ...t\vert} = \left[\begin{array}{c}\cos\theta\\ \sin\theta\\ 0\end{array}\right]$$ (20)

$$\hat {\boldsymbol{\theta}} \equiv {{d{\bf r}\over d\theta}\over\left\vert\begin{array}{c}d{\bf r}\o... ...\vert} = \left[\begin{array}{c}-\sin\theta\\ \cos\theta\\ 0\end{array}\right]$$ (21)

$$\hat {\bf z} \equiv {{d{\bf r}\over dz}\over \left\vert\begin{array}{c}d{\bf r}\over dz\end{array}\right\vert} = \left[\begin{array}{c}0\\ 0\\ 1\end{array}\right].$$ (22)

Derivatives of unit Vectors with respect to the coordinates are

$${\partial \hat {\bf r}\over \partial r} = {\bf0}$$ (23)

$${\partial \hat{\bf r}\over \partial \theta} = \left[\begin{array}{c}-\sin\theta\\ \cos\theta\\ 0\end{array}\right] =\hat {\boldsymbol{\theta}}$$ (24)

$${\partial \hat {\bf r}\over \partial z} = {\bf0}$$ (25)

$${\partial \hat{\boldsymbol{\theta}}\over\partial r} = {\bf0}$$ (26)

$${\partial \hat{\boldsymbol{\theta}}\over \partial \theta } = \left[\begin{array}{c}-\cos\theta\\ -\sin\theta\\ 0\end{array}\right] = -\hat {\bf r}$$ (27)

$${\partial \hat{\boldsymbol{\theta}}\over \partial z} = {\bf0}$$ (28)

$${\partial \hat {\bf z}\over \partial r} = {\bf0}$$ (29)

$${\partial \hat {\bf z}\over \partial \theta } = {\bf0}$$ (30)

$${\partial \hat {\bf z}\over \partial z} = {\bf0}.$$ (31)

The Gradient of a Vector Field in cylindrical coordinates is given by

$$\nabla\equiv\hat{\bf r}{\partial\over\partial r}+\hat{\bolds... ...ial\over\partial\theta}+\hat {\bf z}{\partial\over\partial z},$$ (32)

so the Gradient components become

$$\nabla_r\hat{\bf r} = {\bf0}$$ (33)

$$\nabla_\theta \hat{\bf r} = {1\over r}\hat{\boldsymbol{\theta}}$$ (34)

$$\nabla_z\hat{\bf r} = {\bf0}$$ (35)

$$\nabla_r\hat{\boldsymbol{\theta}} = {\bf0}$$ (36)

$$\nabla_\theta \hat{\boldsymbol{\theta}} = -{1\over r}\hat{\bf r}$$ (37)

$$\nabla_z\hat{\boldsymbol{\theta}} = {\bf0}$$ (38)

$$\nabla_r\hat{\bf z} = {\bf0}$$ (39)

$$\nabla_\theta \hat{\bf z} = {\bf0}$$ (40)

$$\nabla_z\hat{\bf z} = {\bf0}.$$ (41)

Now, since the Connection Coefficients are defined by

$$\Gamma^i_{jk}={\hat {\bf x}}_i\cdot(\nabla_k {\hat{\bf x}}_j),$$ (42)

$$\Gamma^r = \left[\begin{array}{ccc}0 & 0 & 0\\ 0 & -{1\over r} & 0\\ 0 & 0 & 0\end{array}\right]$$ (43)

$$\Gamma^\theta = \left[\begin{array}{ccc}0 & {1\over r} & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{array}\right]$$ (44)

$$\Gamma^z = \left[\begin{array}{ccc}0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{array}\right].$$ (45)

The Covariant Derivatives, given by

$$A_{j;k}={1\over g^{kk}}{\partial A_j\over \partial x_k}-\Gamma^i_{jk}A_i,$$ (46)

are

$$A_{r;r} = {\partial A_r\over\partial r}-\Gamma^i_{rr}A_i = {\partial A_r\over\partial r}$$ (47)

$$A_{r;\theta} = {1\over r}{\partial A_r\over\partial\theta}-\Gamma^i_{r\theta}A_i = {1\over r}{\partial A_\theta\over\partial r}-\Gamma^\theta_{r\theta}A_\theta$$

$$= {1\over r}{\partial A_r\over\partial\theta}-{A_\theta\over r}$$ (48)

$$A_{r;z} = {\partial A_r\over\partial z}-\Gamma^i_{rz}A_i = {\partial A_r\over\partial z}$$ (49)

$$A_{\theta;r} = {\partial A_\theta\over\partial r}\Gamma^i_{\theta r}A_i = {\partial A_\theta\over\partial r}$$ (50)

$$A_{\theta;\theta} = {1\over r}{\partial A_\theta\over\partial\theta}-\Gamma^i_{\theta... ...i = {1\over r}{\partial A_\theta\over\partial\theta}-\Gamma^r_{\theta\theta}A_r$$

$$= {1\over r}{\partial A_\theta\over\partial\theta} +{A_r\over r}$$ (51)

$$A_{\theta;z} = {\partial A_\theta\over\partial z}-\Gamma^i_{\theta z}A_i = {\partial A_\theta\over\partial z}$$ (52)

$$A_{z;r} = {\partial A_z\over\partial r}-\Gamma^i_{zr}A_i={\partial A_z\over\partial r}$$ (53)

$$A_{z;\theta} = {1\over r}{\partial A_z\over\partial\theta}-\Gamma^i_{z\theta}A_i ={1\over r}{\partial A_z\over\partial\theta}$$ (54)

$$A_{z;z} = {\partial A_z\over\partial z}-\Gamma^i_{zz}A_i={\partial A_z\over\partial z}.$$ (55)

Cross Products of the coordinate axes are

$$\hat{\bf r} \times\hat{\bf z} = -\hat{\boldsymbol{\theta}}$$ (56)

$$\hat{\boldsymbol{\theta}}\times\hat{\bf z} = \hat {\bf r}$$ (57)

$$\hat{\bf r} \times\hat{\boldsymbol{\theta}} = \hat{\bf z}.$$ (58)

The Commutation Coefficients are given by

$$c_{\alpha\beta}^\mu \vec e_\mu = [\vec e_\alpha, \vec e_\beta] = \nabla_\alpha\vec e_\beta-\nabla_\beta \vec e_\alpha,$$ (59)

But

$$[\hat {\bf r},\hat {\bf r}]= [{\hat {\boldsymbol{\theta}}},{\... ...hat {\boldsymbol{\phi}}},{\hat {\boldsymbol{\phi}}}] = {\bf0},$$ (60)

so $c_{rr}^\alpha = c_{\theta\theta}^\alpha=c_{\phi\phi}^\alpha=0$, where $\alpha=r,\theta,\phi$. Also

$$[\hat {\bf r},\hat {\boldsymbol{\theta}}]= -[{\hat {\boldsymb... ...t{\boldsymbol{\theta}} = -{1\over r}\hat{\boldsymbol{\theta}},$$ (61)

so $c_{r\theta}^\theta =-c_{\theta r}^\theta =-{1\over r}$, $c_{r\theta}^r=c_{r\theta }^\phi = 0$. Finally,

$$[\hat {\bf r},\hat {\boldsymbol{\phi}}]= [\hat {\boldsymbol{\theta}},\hat {\boldsymbol{\phi}}] = 0.$$ (62)

Summarizing,

$$c^r = \left[\begin{array}{ccc}0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{array}\right]$$ (63)

$$c^\theta = \left[\begin{array}{ccc}0 & -{1\over r} & 0\\ {1\over r} & 0 & 0\\ 0 & 0 & 0\end{array}\right]$$ (64)

$$c^\phi = \left[\begin{array}{ccc}0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\end{array}\right].$$ (65)

Time Derivatives of the Vector are

$$\dot {\bf r} = \left[\begin{array}{c}\cos\theta\,\dot r-r\sin\theta\,\dot \theta... ...\,\hat {\bf r} + r\dot\theta\,\hat {\boldsymbol{\theta}} + \dot z\,\hat {\bf z}$$ (66)

$$\ddot {\bf r} = \left[\begin{array}{c}-\sin\theta\,\dot r\dot\theta+\cos\theta\,\... ...n\theta\, {\dot\theta}^2+r\cos\theta\,\ddot \theta\\ \ddot z\end{array}\right]$$

$$= \left[\begin{array}{c}-2\sin\theta\,\dot r\dot\theta+\cos\theta\,... ...n\theta \,{\dot\theta}^2+r\cos\theta\,\ddot \theta\\ \ddot z\end{array}\right]$$

$$= (\ddot r-r{\dot\theta}^2)\hat{\bf r}+(2\dot r\dot\theta+r\ddot\theta)\hat{\boldsymbol{\theta}}+\ddot z\,\hat{\bf z}.$$ (67)

Speed is given by

$$v \equiv \vert\dot{\bf r}\vert = \sqrt{\dot r^2 +r^2\dot\theta^2+\dot z^2}.$$ (68)

Time derivatives of the unit Vectors are

$$\dot{\hat{\bf r}} = \left[\begin{array}{c}-\sin\theta\,\dot\theta\\ \cos\theta\,\dot\theta\\ 0\end{array}\right]=\dot\theta\,\hat{\boldsymbol{\theta}}$$

$$\dot{\hat{\boldsymbol{\theta}}} = \left[\begin{array}{c}-\cos\theta\,\dot\theta\\ -\sin\theta\,\dot\theta\\ 0\end{array}\right]=-\dot\theta\,\hat{\bf r}$$

$$\dot{\hat{\bf z}} = \left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]={\bf0}.$$ (69)

Cross Products of the axes are

$$\hat{\bf r} \times\hat{\bf z} = -\hat{\boldsymbol{\theta}}$$ (70)

$$\hat{\boldsymbol{\theta}}\times\hat{\bf z} = \hat {\bf r}$$ (71)

$$\hat{\bf r} \times\hat{\boldsymbol{\theta}} = \hat{\bf z}.$$ (72)

The Convective Derivative is

$${D\dot {\bf r}\over Dt} \equiv \left({{\partial \over \parti... ... {\bf r}\over \partial t}+\dot {\bf r}\cdot\nabla\dot {\bf r}.$$ (73)

To rewrite this, use the identity

$$\nabla({\bf A}\cdot{\bf B})={\bf A}\times (\nabla\times {\bf... ...bf A})+({\bf A}\cdot\nabla){\bf B}+({\bf B}\cdot\nabla){\bf A}$$ (74)

and set ${\bf A}={\bf B}$, to obtain

$$\nabla({\bf A}\cdot{\bf A}) = 2{\bf A}\times (\nabla\times {\bf A})+2({\bf A}\cdot \nabla) {\bf A},$$ (75)

so

$$({\bf A}\cdot \nabla){\bf A} = \nabla({\textstyle{1\over 2}}{\bf A}^2)-{\bf A}\times (\nabla\times {\bf A}).$$ (76)

Then

$${D\dot {\bf r}\over Dt}=\ddot{\bf r}+\nabla({\textstyle{1\ov... ...s \dot {\bf r} + \nabla({\textstyle{1\over 2}}\dot {\bf r}^2).$$ (77)

The Curl in the above expression gives

$$\nabla\times \dot {\bf r}={1\over r}{\partial \over \partial r}(r^2\dot \theta)\hat {\bf z} = 2\dot \theta \hat {\bf z},$$ (78)

so

$$-\dot{\bf r}\times (\nabla\times\dot{\bf r}) = -2\dot\theta ... ...\dot\theta\hat{\boldsymbol{\theta}}-2r\dot\theta^2\hat{\bf r}.$$ (79)

We expect the gradient term to vanish since Speed does not depend on position. Check this using the identity $\nabla(f^2)=2f\nabla f$,

$$\nabla ({\textstyle{1\over 2}}\dot {\bf r}^2) = {\textstyle{... ...a\dot r+r\dot \theta \nabla(r\dot \theta )+\dot z\nabla\dot z.$$ (80)

Examining this term by term,

$$\dot r\nabla\dot r = \dot r{\partial \over \partial t}\nabla r = \dot r {\partial \ove... ... {\bf r} =\dot r\dot {\hat{\bf r}}=\dot r\dot \theta \hat {\boldsymbol{\theta}}$$ (81)

$$r\dot \theta \nabla(r\dot \theta ) = r\dot \theta \left[{r{\partial \over \partial t}\nabla\theta +\do... ...({{1\over r}\hat {\boldsymbol{\theta}}}\right)+\dot \theta \hat {\bf r}}\right]$$

$$= r\dot\theta\left[{r\left({-{1\over r^2}\dot r\hat{\boldsymbol{\th... ...1\over r}\dot{\hat{\boldsymbol{\theta}}}}\right)+\dot\theta\hat {\bf r}}\right]$$

$$= -\dot \theta \dot r\hat {\boldsymbol{\theta}}+r\dot \theta (-\dot... ...f r}) +r\dot \theta^2\hat {\bf r}=-\dot \theta \dot r\hat {\boldsymbol{\theta}}$$ (82)

$$\dot z\nabla\dot z = \dot z{\partial \over \partial t}\nabla z =\dot z {\partial\over \partial t}\hat {\bf z} = \dot z\dot {\hat {\bf z}} = {\bf0},$$ (83)

so, as expected,

$$\nabla ({\textstyle{1\over 2}}\dot {\bf r}^2)={\bf0}.$$ (84)

We have already computed $\ddot{\bf r}$, so combining all three pieces gives

$${D\dot {\bf r}\over Dt} = (\ddot r-r\dot \theta^2-2r\dot \theta^2)\hat {\bf r} +(2\dot r\do... ...+2\dot r\dot \theta+r\ddot\theta)\hat {\boldsymbol{\theta}}+\ddot z\hat {\bf z}$$

$$= (\ddot r-3r\dot \theta^2)\hat {\bf r} +(4\dot r\dot \theta +r\ddot\theta)\hat {\boldsymbol{\theta}}+\ddot z\hat {\bf z}.$$ (85)

The Divergence is

$$\nabla\cdot A = A^r_{;r} = A^r_{,r}+\left({\Gamma^r_{rr}A^t+\Gamma^r_{\theta r}A^\theta+\Gamma^r_{z r}A^z}\right)+A^\theta_{,\theta}$$

$$\phantom{=} \mathop{+}\left({\Gamma^\theta_{r\theta}A^r +\Gamma^\theta _{\theta\theta}A^\theta +\Gamma^\theta_{z\theta}A^z}\right)$$

$$\phantom{=} \mathop{+} A^z_{,z} +\left({\Gamma^z_{rz}A^r +\Gamma^z_{\theta z}A^\theta +\Gamma^z_{zz}A^z}\right)$$

$$= A^r_{,r}+ A^\theta_{,\theta}+A^z_{,z} +\left({0+0+0}\right)+\left({{1\over r}+0+0}\right)$$

$$\phantom{=} +\left({0+0+0}\right)$$

$$= {1\over g_r}{\partial\over\partial r}A^r+{1\over g_\theta}{\parti... ...\partial\theta}A^\theta +{1\over g_z}{\partial\over\partial z}A^z+{1\over r}A^r$$

$$= \left({{\partial\over\partial r}+{1\over r}}\right)A^r+{1\over r}{\partial\over\partial\theta}A^\theta +{\partial\over\partial z}A^z,$$ (86)

or, in Vector notation

$$\nabla\cdot{\bf F} = {1\over r} {\partial\over\partial r} (r... ... F_\theta\over\partial\theta} + {\partial F_z\over\partial z}.$$ (87)

The Cross Product is

$$\nabla \times {\bf F} = \left({{1\over r} {\partial F_z\over\partial\theta} - {\partial F... ...ver\partial z} - {\partial F_z\over\partial r}}\right)\hat{\boldsymbol{\theta}}$$

$$\phantom{=} + {1\over r}\left[{{\partial\over\partial r}(rF_\theta) - {\partial F_r\over\partial\theta}}\right]\hat {\bf z},$$ (88)

and the Laplacian is

$$\nabla^2 f \equiv {1\over r} {\partial \over \partial r} \left({r {\partial f\over ... ... } {\partial ^2 f\over\partial \theta ^2 } + {\partial ^2 f\over \partial z^2 }$$

$$= {\partial ^2 f\over \partial r^2 } + {1\over r} {\partial f\over ... ...} {\partial ^2 f\over \partial \theta ^2 } + {\partial ^2 f\over\partial z^2 }.$$ (89)

The vector Laplacian is

$$\nabla ^2 {\bf v} =\left[{\matrix{{\partial^2 v_r\over \part... ..._z\over \partial r}\cr}}\right]. \hrule width 0pt height 7.5pt$$ (90)

The Helmholtz Differential Equation is separable in cylindrical coordinates and has Stäckel Determinant $S=1$ (for $r$, $\theta$, $z$) or $S=1/(1-{\xi_2}^2)$ (for Morse and Feshbach's $\xi_1$, $\xi_2$, $\xi_3$).

See also Elliptic Cylindrical Coordinates, Helmholtz Differential Equation--Circular Cylindrical Coordinates, Polar Coordinates, Spherical Coordinates

References

Arfken, G. ``Circular Cylindrical Coordinates.'' §2.4 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 95-101, 1985.

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 212, 1987.

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, p. 657, 1953.