Commonly used orthogonal coordinate systems

Cartesian coordinate

$\begin{align*} &r = (x,y,z) \\ &\vec h_x = \frac{\partial r}{\partial x} = (1,0,0) \\ &\vec h_y = \frac{\partial r}{\partial y} = (0,1,0) \\ &\vec h_z = \frac{\partial r}{\partial z} = (0,0,1) \\ &h_x=1,\ e_x=(1,0,0)\\ &h_y=1,\ e_y=(0,1,0)\\ &h_z=1,\ e_z=(0,0,1)\\ &r = (x,y,z)\cdot (h_x, h_y, h_z) = xe_x+ye_y+ze_z \\ &{\rm d}V=||J|| {\rm d}x {\rm d}y {\rm d}z = {\rm d}x {\rm d}y {\rm d}z \end{align*}$

orthogonal, no singularity point

Cylindrical polar coordinates

$\begin{align*} &\vec r = (x,y,z) = (\rho \cos\theta,\rho \sin \theta, z)\\ &\vec h_\rho = \frac{\partial \vec r}{\partial \rho} = (\cos \theta ,\sin \theta,0) \\ &\vec h_\theta = \frac{\partial \vec r}{\partial \theta} = (- \rho \sin \theta, \rho \cos \theta,0) \\ &\vec h_z = \frac{\partial \vec r}{\partial z} = (0,0,1) \\ &h_\rho=1,\ e_\rho=(\cos \theta, \sin \theta,0)\\ &h_\theta=\rho,\ e_\theta=(- \sin \theta, \cos \theta,0)\\ &h_z=1,\ e_z=(0,0,1)\\ &r = (\rho, \theta,z)\cdot (h_\rho, h_\theta, h_z) = \rho e_\rho+\theta \rho e_\theta+ze_z \\ &{\rm d}V=||J|| {\rm d}\rho {\rm d}\theta {\rm d}z = \rho {\rm d}\rho {\rm d}\theta {\rm d}z &\end{align*}$

orthogonal, singularity at

Sperical polar coordinates

$\begin{align*} &\vec r = (x,y,z) = (\rho \sin\theta \cos \phi,\rho \cos \theta \sin \phi, \rho \cos \theta)\\ &\vec h_\rho = \frac{\partial \vec r}{\partial \rho} = ( \sin\theta \cos \phi, \cos \theta \sin \phi, \cos \theta) \\ &\vec h_\theta = \frac{\partial \vec r}{\partial \theta} = (\rho \cos \theta \cos \phi, - \rho \sin \theta \sin \phi, - \rho \sin \theta) \\ &\vec h_\phi = \frac{\partial \vec r}{\partial z} = (- \rho \sin \theta \sin \phi, \rho \sin \theta \cos \phi,0) \\ &h_\rho=1,\ e_\rho=(\sin \theta \cos \phi, \cos \theta \sin \phi, \cos \theta)\\ &h_\theta=\rho,\ e_\theta=(\rho \cos \theta \cos \phi, \rho \cos \theta \sin \phi, - \rho \sin \theta)\\ &h_\phi= \rho \sin \theta ,\ e_\phi=(- \rho \sin \theta \sin \phi,\rho \sin \theta \cos \phi ,0)\\ &r = (\rho, \theta, \phi) \cdot (h_\rho, h_\theta, h_\phi) = \rho e_\rho+\theta \rho e_\theta+\phi \rho e_z \\ &{\rm d}V=||J|| {\rm d}\rho {\rm d}\theta {\rm \phi}z = \rho^2 \sin \theta {\rm d}\rho {\rm d}\theta {\rm d}\phi &\end{align*}$

orthogonal, singularity at or
To verify orthogonality:

, which is the are contravariant derivatives, usually written as
If given , , which are the covariant derivatives, usually written as
The concept of a metric tensor:

the metric tensor being diagonal is equivalent to the unit vectors being orthogonal, also

Vector calculus in orthogonal coordinates

Notice that
both by the vector identity
We finally retrieve the grad and curl of vector fields in curvilinear coordinates
$\begin{align*} &\vec \nabla \cdot \vec F = \frac{1}{h_1h_2h_3}\sum _{i = 1, j \neq i, k \neq i, i \neq j}^{i=3}\frac{\partial }{\partial q_i}(h_jh_kF_i) \\ &\vec \nabla \times \vec F = \frac{1}{h_1h_2h_3} \left | \begin{array}{cccc} h_1e_1 & h_2e_2 & h_3e_3 \\ \frac{\partial}{\partial q_1} & \frac{\partial}{\partial q_2} & \frac{\partial}{\partial q_3} \\ h_1F_1 & h_2F_2 & h_3F_3 \end{array} \right | \end{align*}$
To calculate the Laplacian of a vector field, the difficulty lies in the non-constant unit vectors in general coordinates, please resolve to the following vector identity.

2018-10-17

Commonly used orthogonal coordinate systems

Cartesian coordinate

Cylindrical polar coordinates

Sperical polar coordinates

Vector calculus in orthogonal coordinates

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