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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.

$\displaystyle r$ $\textstyle =$ $\displaystyle \sqrt{x^2 +y^2}$ (1)
$\displaystyle \theta$ $\textstyle =$ $\displaystyle \tan^{-1}\left({y\over x}\right)$ (2)
$\displaystyle z$ $\textstyle =$ $\displaystyle z,$ (3)

where $r\in [0,\infty)$, $\theta\in [0,2\pi)$, and $z\in (-\infty, \infty)$. In terms of $x$, $y$, and $z$
$\displaystyle x$ $\textstyle =$ $\displaystyle r\cos\theta$ (4)
$\displaystyle y$ $\textstyle =$ $\displaystyle r\sin\theta$ (5)
$\displaystyle z$ $\textstyle =$ $\displaystyle z.$ (6)

Morse and Feshbach (1953) define the cylindrical coordinates by
$\displaystyle x$ $\textstyle =$ $\displaystyle \xi_1\xi_2$ (7)
$\displaystyle y$ $\textstyle =$ $\displaystyle \xi_1\sqrt{1-{\xi_2}^2}$ (8)
$\displaystyle z$ $\textstyle =$ $\displaystyle \xi_3,$ (9)

where $\xi_1=r$ and $\xi_2=\cos\theta$. The Metric elements of the cylindrical coordinates are
$\displaystyle g_{rr}$ $\textstyle =$ $\displaystyle 1$ (10)
$\displaystyle g_{\theta\theta}$ $\textstyle =$ $\displaystyle r^2$ (11)
$\displaystyle g_{zz}$ $\textstyle =$ $\displaystyle 1,$ (12)

so the Scale Factors are
$\displaystyle g_r$ $\textstyle =$ $\displaystyle 1$ (13)
$\displaystyle g_\theta$ $\textstyle =$ $\displaystyle r$ (14)
$\displaystyle g_z$ $\textstyle =$ $\displaystyle 1.$ (15)

The Line Element is
d{\bf s} = dr \,\hat {\bf r} + r\,d\theta\, \hat {\boldsymbol{\theta}} + dz \,\hat {\bf z},
\end{displaymath} (16)

and the Volume Element is
dV = r dr\,d\theta\,dz.
\end{displaymath} (17)

The Jacobian is
\left\vert{\partial (x, y, z)\over \partial (r, \theta, z)}\right\vert = r.
\end{displaymath} (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].
\end{displaymath} (19)

To find the Unit Vectors,
$\displaystyle \hat {\bf r}$ $\textstyle \equiv$ $\displaystyle {{d{\bf r}\over dr}\over \left\vert\begin{array}{c}d{\bf r}\over ...
...t\vert} = \left[\begin{array}{c}\cos\theta\\  \sin\theta\\  0\end{array}\right]$ (20)
$\displaystyle \hat {\boldsymbol{\theta}}$ $\textstyle \equiv$ $\displaystyle {{d{\bf r}\over d\theta}\over\left\vert\begin{array}{c}d{\bf r}\o...
...\vert} = \left[\begin{array}{c}-\sin\theta\\  \cos\theta\\  0\end{array}\right]$ (21)
$\displaystyle \hat {\bf z}$ $\textstyle \equiv$ $\displaystyle {{d{\bf r}\over dz}\over \left\vert\begin{array}{c}d{\bf r}\over dz\end{array}\right\vert} = \left[\begin{array}{c}0\\  0\\  1\end{array}\right].$ (22)

Derivatives of unit Vectors with respect to the coordinates are
$\displaystyle {\partial \hat {\bf r}\over \partial r}$ $\textstyle =$ $\displaystyle {\bf0}$ (23)
$\displaystyle {\partial \hat{\bf r}\over \partial \theta}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}-\sin\theta\\  \cos\theta\\  0\end{array}\right] =\hat {\boldsymbol{\theta}}$ (24)
$\displaystyle {\partial \hat {\bf r}\over \partial z}$ $\textstyle =$ $\displaystyle {\bf0}$ (25)
$\displaystyle {\partial \hat{\boldsymbol{\theta}}\over\partial r}$ $\textstyle =$ $\displaystyle {\bf0}$ (26)
$\displaystyle {\partial \hat{\boldsymbol{\theta}}\over \partial \theta }$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}-\cos\theta\\  -\sin\theta\\  0\end{array}\right] = -\hat {\bf r}$ (27)
$\displaystyle {\partial \hat{\boldsymbol{\theta}}\over \partial z}$ $\textstyle =$ $\displaystyle {\bf0}$ (28)
$\displaystyle {\partial \hat {\bf z}\over \partial r}$ $\textstyle =$ $\displaystyle {\bf0}$ (29)
$\displaystyle {\partial \hat {\bf z}\over \partial \theta }$ $\textstyle =$ $\displaystyle {\bf0}$ (30)
$\displaystyle {\partial \hat {\bf z}\over \partial z}$ $\textstyle =$ $\displaystyle {\bf0}.$ (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},
\end{displaymath} (32)

so the Gradient components become
$\displaystyle \nabla_r\hat{\bf r}$ $\textstyle =$ $\displaystyle {\bf0}$ (33)
$\displaystyle \nabla_\theta \hat{\bf r}$ $\textstyle =$ $\displaystyle {1\over r}\hat{\boldsymbol{\theta}}$ (34)
$\displaystyle \nabla_z\hat{\bf r}$ $\textstyle =$ $\displaystyle {\bf0}$ (35)
$\displaystyle \nabla_r\hat{\boldsymbol{\theta}}$ $\textstyle =$ $\displaystyle {\bf0}$ (36)
$\displaystyle \nabla_\theta \hat{\boldsymbol{\theta}}$ $\textstyle =$ $\displaystyle -{1\over r}\hat{\bf r}$ (37)
$\displaystyle \nabla_z\hat{\boldsymbol{\theta}}$ $\textstyle =$ $\displaystyle {\bf0}$ (38)
$\displaystyle \nabla_r\hat{\bf z}$ $\textstyle =$ $\displaystyle {\bf0}$ (39)
$\displaystyle \nabla_\theta \hat{\bf z}$ $\textstyle =$ $\displaystyle {\bf0}$ (40)
$\displaystyle \nabla_z\hat{\bf z}$ $\textstyle =$ $\displaystyle {\bf0}.$ (41)

Now, since the Connection Coefficients are defined by
\Gamma^i_{jk}={\hat {\bf x}}_i\cdot(\nabla_k {\hat{\bf x}}_j),
\end{displaymath} (42)

$\displaystyle \Gamma^r$ $\textstyle =$ $\displaystyle \left[\begin{array}{ccc}0 & 0 & 0\\  0 & -{1\over r} & 0\\  0 & 0 & 0\end{array}\right]$ (43)
$\displaystyle \Gamma^\theta$ $\textstyle =$ $\displaystyle \left[\begin{array}{ccc}0 & {1\over r} & 0\\  0 & 0 & 0\\  0 & 0 & 0\end{array}\right]$ (44)
$\displaystyle \Gamma^z$ $\textstyle =$ $\displaystyle \left[\begin{array}{ccc}0 & 0 & 0\\  0 & 0 & 0\\  0 & 0 & 0\end{array}\right].$ (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,
\end{displaymath} (46)

$\displaystyle A_{r;r}$ $\textstyle =$ $\displaystyle {\partial A_r\over\partial r}-\Gamma^i_{rr}A_i = {\partial A_r\over\partial r}$ (47)
$\displaystyle A_{r;\theta}$ $\textstyle =$ $\displaystyle {1\over r}{\partial A_r\over\partial\theta}-\Gamma^i_{r\theta}A_i = {1\over r}{\partial A_\theta\over\partial r}-\Gamma^\theta_{r\theta}A_\theta$  
  $\textstyle =$ $\displaystyle {1\over r}{\partial A_r\over\partial\theta}-{A_\theta\over r}$ (48)
$\displaystyle A_{r;z}$ $\textstyle =$ $\displaystyle {\partial A_r\over\partial z}-\Gamma^i_{rz}A_i = {\partial A_r\over\partial z}$ (49)
$\displaystyle A_{\theta;r}$ $\textstyle =$ $\displaystyle {\partial A_\theta\over\partial r}\Gamma^i_{\theta r}A_i = {\partial A_\theta\over\partial r}$ (50)
$\displaystyle A_{\theta;\theta}$ $\textstyle =$ $\displaystyle {1\over r}{\partial A_\theta\over\partial\theta}-\Gamma^i_{\theta...
...i = {1\over r}{\partial A_\theta\over\partial\theta}-\Gamma^r_{\theta\theta}A_r$  
  $\textstyle =$ $\displaystyle {1\over r}{\partial A_\theta\over\partial\theta} +{A_r\over r}$ (51)
$\displaystyle A_{\theta;z}$ $\textstyle =$ $\displaystyle {\partial A_\theta\over\partial z}-\Gamma^i_{\theta z}A_i = {\partial A_\theta\over\partial z}$ (52)
$\displaystyle A_{z;r}$ $\textstyle =$ $\displaystyle {\partial A_z\over\partial r}-\Gamma^i_{zr}A_i={\partial A_z\over\partial r}$ (53)
$\displaystyle A_{z;\theta}$ $\textstyle =$ $\displaystyle {1\over r}{\partial A_z\over\partial\theta}-\Gamma^i_{z\theta}A_i ={1\over r}{\partial A_z\over\partial\theta}$ (54)
$\displaystyle A_{z;z}$ $\textstyle =$ $\displaystyle {\partial A_z\over\partial z}-\Gamma^i_{zz}A_i={\partial A_z\over\partial z}.$ (55)

Cross Products of the coordinate axes are
$\displaystyle \hat{\bf r} \times\hat{\bf z}$ $\textstyle =$ $\displaystyle -\hat{\boldsymbol{\theta}}$ (56)
$\displaystyle \hat{\boldsymbol{\theta}}\times\hat{\bf z}$ $\textstyle =$ $\displaystyle \hat {\bf r}$ (57)
$\displaystyle \hat{\bf r} \times\hat{\boldsymbol{\theta}}$ $\textstyle =$ $\displaystyle \hat{\bf z}.$ (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,
\end{displaymath} (59)

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

so $c_{rr}^\alpha = c_{\theta\theta}^\alpha=c_{\phi\phi}^\alpha=0$, where $\alpha=r,\theta,\phi$. Also
\begin{displaymath}[\hat {\bf r},\hat {\boldsymbol{\theta}}]= -[{\hat {\boldsymb...
...t{\boldsymbol{\theta}} = -{1\over r}\hat{\boldsymbol{\theta}},
\end{displaymath} (61)

so $c_{r\theta}^\theta =-c_{\theta r}^\theta =-{1\over r}$, $c_{r\theta}^r=c_{r\theta }^\phi = 0$. Finally,
\begin{displaymath}[\hat {\bf r},\hat {\boldsymbol{\phi}}]= [\hat {\boldsymbol{\theta}},\hat {\boldsymbol{\phi}}] = 0.
\end{displaymath} (62)

$\displaystyle c^r$ $\textstyle =$ $\displaystyle \left[\begin{array}{ccc}0 & 0 & 0\\  0 & 0 & 0\\  0 & 0 & 0\end{array}\right]$ (63)
$\displaystyle c^\theta$ $\textstyle =$ $\displaystyle \left[\begin{array}{ccc}0 & -{1\over r} & 0\\  {1\over r} & 0 & 0\\  0 & 0 & 0\end{array}\right]$ (64)
$\displaystyle c^\phi$ $\textstyle =$ $\displaystyle \left[\begin{array}{ccc}0 & 0 & 0\\  0 & 0 & 0\\  0 & 0 & 0\end{array}\right].$ (65)

Time Derivatives of the Vector are

$\displaystyle \dot {\bf r}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}\cos\theta\,\dot r-r\sin\theta\,\dot \theta...
...\,\hat {\bf r} + r\dot\theta\,\hat {\boldsymbol{\theta}} + \dot z\,\hat {\bf z}$ (66)
$\displaystyle \ddot {\bf r}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}-\sin\theta\,\dot r\dot\theta+\cos\theta\,\...
...n\theta\, {\dot\theta}^2+r\cos\theta\,\ddot \theta\\  \ddot z\end{array}\right]$  
  $\textstyle =$ $\displaystyle \left[\begin{array}{c}-2\sin\theta\,\dot r\dot\theta+\cos\theta\,...
...n\theta \,{\dot\theta}^2+r\cos\theta\,\ddot \theta\\  \ddot z\end{array}\right]$  
  $\textstyle =$ $\displaystyle (\ddot r-r{\dot\theta}^2)\hat{\bf r}+(2\dot r\dot\theta+r\ddot\theta)\hat{\boldsymbol{\theta}}+\ddot z\,\hat{\bf z}.$ (67)

Speed is given by

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

Time derivatives of the unit Vectors are
$\displaystyle \dot{\hat{\bf r}}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}-\sin\theta\,\dot\theta\\
\cos\theta\,\dot\theta\\  0\end{array}\right]=\dot\theta\,\hat{\boldsymbol{\theta}}$  
$\displaystyle \dot{\hat{\boldsymbol{\theta}}}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}-\cos\theta\,\dot\theta\\
-\sin\theta\,\dot\theta\\  0\end{array}\right]=-\dot\theta\,\hat{\bf r}$  
$\displaystyle \dot{\hat{\bf z}}$ $\textstyle =$ $\displaystyle \left[\begin{array}{c}0\\  0\\  0\end{array}\right]={\bf0}.$ (69)

Cross Products of the axes are
$\displaystyle \hat{\bf r} \times\hat{\bf z}$ $\textstyle =$ $\displaystyle -\hat{\boldsymbol{\theta}}$ (70)
$\displaystyle \hat{\boldsymbol{\theta}}\times\hat{\bf z}$ $\textstyle =$ $\displaystyle \hat {\bf r}$ (71)
$\displaystyle \hat{\bf r} \times\hat{\boldsymbol{\theta}}$ $\textstyle =$ $\displaystyle \hat{\bf z}.$ (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}.
\end{displaymath} (73)

To rewrite this, use the identity
\nabla({\bf A}\cdot{\bf B})={\bf A}\times (\nabla\times {\bf... A})+({\bf A}\cdot\nabla){\bf B}+({\bf B}\cdot\nabla){\bf A}
\end{displaymath} (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},
\end{displaymath} (75)

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

{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).
\end{displaymath} (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},
\end{displaymath} (78)


-\dot{\bf r}\times (\nabla\times\dot{\bf r}) = -2\dot\theta ...
...\dot\theta\hat{\boldsymbol{\theta}}-2r\dot\theta^2\hat{\bf r}.
\end{displaymath} (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.
\end{displaymath} (80)

Examining this term by term,
$\displaystyle \dot r\nabla\dot r$ $\textstyle =$ $\displaystyle \dot r{\partial \over \partial t}\nabla r = \dot r {\partial \ove...
... {\bf r} =\dot r\dot {\hat{\bf r}}=\dot r\dot \theta \hat {\boldsymbol{\theta}}$ (81)
$\displaystyle r\dot \theta \nabla(r\dot \theta )$ $\textstyle =$ $\displaystyle r\dot \theta \left[{r{\partial \over \partial t}\nabla\theta +\do...
...({{1\over r}\hat {\boldsymbol{\theta}}}\right)+\dot \theta \hat {\bf r}}\right]$  
  $\textstyle =$ $\displaystyle r\dot\theta\left[{r\left({-{1\over r^2}\dot r\hat{\boldsymbol{\th...
...1\over r}\dot{\hat{\boldsymbol{\theta}}}}\right)+\dot\theta\hat {\bf r}}\right]$  
  $\textstyle =$ $\displaystyle -\dot \theta \dot r\hat {\boldsymbol{\theta}}+r\dot \theta (-\dot...
...f r}) +r\dot \theta^2\hat {\bf r}=-\dot \theta \dot r\hat {\boldsymbol{\theta}}$ (82)
$\displaystyle \dot z\nabla\dot z$ $\textstyle =$ $\displaystyle \dot z{\partial \over \partial t}\nabla z =\dot z {\partial\over \partial t}\hat {\bf z} = \dot z\dot {\hat {\bf z}} = {\bf0},$ (83)

so, as expected,
\nabla ({\textstyle{1\over 2}}\dot {\bf r}^2)={\bf0}.
\end{displaymath} (84)

We have already computed $\ddot{\bf r}$, so combining all three pieces gives
$\displaystyle {D\dot {\bf r}\over Dt}$ $\textstyle =$ $\displaystyle (\ddot r-r\dot \theta^2-2r\dot \theta^2)\hat {\bf r} +(2\dot r\do...
...+2\dot r\dot \theta+r\ddot\theta)\hat {\boldsymbol{\theta}}+\ddot z\hat {\bf z}$  
  $\textstyle =$ $\displaystyle (\ddot r-3r\dot \theta^2)\hat {\bf r} +(4\dot r\dot \theta +r\ddot\theta)\hat {\boldsymbol{\theta}}+\ddot z\hat {\bf z}.$ (85)

The Divergence is
$\displaystyle \nabla\cdot A$ $\textstyle =$ $\displaystyle A^r_{;r} = A^r_{,r}+\left({\Gamma^r_{rr}A^t+\Gamma^r_{\theta r}A^\theta+\Gamma^r_{z r}A^z}\right)+A^\theta_{,\theta}$  
  $\textstyle \phantom{=}$ $\displaystyle \mathop{+}\left({\Gamma^\theta_{r\theta}A^r +\Gamma^\theta _{\theta\theta}A^\theta +\Gamma^\theta_{z\theta}A^z}\right)$  
  $\textstyle \phantom{=}$ $\displaystyle \mathop{+} A^z_{,z} +\left({\Gamma^z_{rz}A^r +\Gamma^z_{\theta z}A^\theta +\Gamma^z_{zz}A^z}\right)$  
  $\textstyle =$ $\displaystyle A^r_{,r}+ A^\theta_{,\theta}+A^z_{,z} +\left({0+0+0}\right)+\left({{1\over r}+0+0}\right)$  
  $\textstyle \phantom{=}$ $\displaystyle +\left({0+0+0}\right)$  
  $\textstyle =$ $\displaystyle {1\over g_r}{\partial\over\partial r}A^r+{1\over g_\theta}{\parti...
...\partial\theta}A^\theta +{1\over g_z}{\partial\over\partial z}A^z+{1\over r}A^r$  
  $\textstyle =$ $\displaystyle \left({{\partial\over\partial r}+{1\over r}}\right)A^r+{1\over r}{\partial\over\partial\theta}A^\theta +{\partial\over\partial z}A^z,$ (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}.
\end{displaymath} (87)

The Cross Product is
$\displaystyle \nabla \times {\bf F}$ $\textstyle =$ $\displaystyle \left({{1\over r} {\partial F_z\over\partial\theta} - {\partial F...
...ver\partial z} - {\partial F_z\over\partial r}}\right)\hat{\boldsymbol{\theta}}$  
  $\textstyle \phantom{=}$ $\displaystyle + {1\over r}\left[{{\partial\over\partial r}(rF_\theta) - {\partial F_r\over\partial\theta}}\right]\hat {\bf z},$ (88)

and the Laplacian is
$\displaystyle \nabla^2 f$ $\textstyle \equiv$ $\displaystyle {1\over r} {\partial \over \partial r} \left({r {\partial f\over ...
... } {\partial ^2 f\over\partial \theta ^2 } + {\partial ^2 f\over \partial z^2 }$  
  $\textstyle =$ $\displaystyle {\partial ^2 f\over \partial r^2 } + {1\over r} {\partial f\over ...
...} {\partial ^2 f\over \partial \theta ^2 } + {\partial ^2 f\over\partial z^2 }.$ (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
\end{displaymath} (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


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.

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© 1996-9 Eric W. Weisstein