2018-10-17

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 $\rho=0$

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 $r=0$ or $\theta=0,\pi$
To verify orthogonality:

$\vec h_i \cdot \vec h_j = 0\ \forall i \neq j$ , which is the are contravariant derivatives, usually written as $\vec v = v^i e_i$
If given $q(x,y,z)$ , $\vec \nabla q_i \cdot \vec \nabla q_j = 0 \ \forall i \neq j$ , which are the covariant derivatives, usually written as $\vec v = v_ie^i$
The concept of a metric tensor:
$g_{ij} =\frac{\partial x_k}{\partial q_i} \frac{\partial x_k}{\partial q_j}$
the metric tensor being diagonal is equivalent to the unit vectors being orthogonal, also
$h_1h_2h_3 = \sqrt {{\rm det} g}$
${\rm d}s^2 = g_{ij}{\rm d}q_i {\rm d}q_j$

Vector calculus in orthogonal coordinates

$\begin{align*} &{\rm d}\vec r = {\rm d}q_i \vec h_i \\ &\vec{\nabla}\phi = \frac{e_i}{h_i} \frac{\partial \phi}{\partial q_i} \end{align*}$

Notice that
$\begin{align*} &\vec \nabla (\frac{e_1}{h_2h_3}) = \vec \nabla \times (q_2\vec \nabla q_3) = 0 \\ &\vec \nabla \times (\frac{e_1}{h_1}) = \vec \nabla \times (\vec \nabla q_1) = 0 \end{align*}$
both by the vector identity $\vec \nabla \cdot (\vec \nabla \times \vec F) = 0$
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.
$\vec \nabla ^2 \vec F = \vec \nabla(\vec \nabla \cdot \vec F) - \vec \nabla \times(\vec \nabla \times \vec F)$

最后编辑于：2018.10.20 00:43:41