Template:Use American English Template:Short description
Note: This page uses common physics notation for spherical coordinates, in which <math>\theta</math> is the angle between the z axis and the radius vector connecting the origin to the point in question, while <math>\phi</math> is the angle between the projection of the radius vector onto the x-y plane and the x axis. Several other definitions are in use, and so care must be taken in comparing different sources.<ref name="wolfram">Wolfram Mathworld, spherical coordinates</ref>
Cylindrical coordinate systemEdit
Vector fieldsEdit
Vectors are defined in cylindrical coordinates by (ρ, φ, z), where
- ρ is the length of the vector projected onto the xy-plane,
- φ is the angle between the projection of the vector onto the xy-plane (i.e. ρ) and the positive x-axis (0 ≤ φ < 2π),
- z is the regular z-coordinate.
(ρ, φ, z) is given in Cartesian coordinates by: <math display="block">\begin{bmatrix} \rho \\ \phi \\ z \end{bmatrix} = \begin{bmatrix} \sqrt{x^2 + y^2} \\ \operatorname{arctan}(y / x) \\ z \end{bmatrix},\ \ \ 0 \le \phi < 2\pi,
</math>
or inversely by: <math display="block">\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \rho\cos\phi \\ \rho\sin\phi \\ z \end{bmatrix}.</math>
Any vector field can be written in terms of the unit vectors as: <math display="block">\mathbf A = A_x \mathbf{\hat x} + A_y \mathbf{\hat y} + A_z \mathbf{\hat z} = A_\rho \mathbf{\hat \rho} + A_\phi \boldsymbol{\hat \phi} + A_z \mathbf{\hat z}</math> The cylindrical unit vectors are related to the Cartesian unit vectors by: <math display="block">\begin{bmatrix}\boldsymbol{\hat \rho} \\ \boldsymbol{\hat\phi} \\ \mathbf{\hat z}\end{bmatrix} = \begin{bmatrix}
\cos\phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi & 0 \\ 0 & 0 & 1
\end{bmatrix} \begin{bmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix}</math>
Note: the matrix is an orthogonal matrix, that is, its inverse is simply its transpose.
Time derivative of a vector fieldEdit
To find out how the vector field A changes in time, the time derivatives should be calculated. For this purpose Newton's notation will be used for the time derivative (<math>\dot{\mathbf{A}}</math>). In Cartesian coordinates this is simply: <math display="block">\dot{\mathbf{A}} = \dot{A}_x \hat{\mathbf{x}} + \dot{A}_y \hat{\mathbf{y}} + \dot{A}_z \hat{\mathbf{z}}</math>
However, in cylindrical coordinates this becomes: <math display="block">\dot{\mathbf{A}} = \dot{A}_\rho \hat{\boldsymbol{\rho}} + A_\rho \dot{\hat{\boldsymbol{\rho}}}
+ \dot{A}_\phi \hat{\boldsymbol{\phi}} + A_\phi \dot{\hat{\boldsymbol{\phi}}}
+ \dot{A}_z \hat{\boldsymbol{z}} + A_z \dot{\hat{\boldsymbol{z}}}</math>
The time derivatives of the unit vectors are needed. They are given by: <math display="block">\begin{align}
\dot{\hat{\boldsymbol{\rho}}} & = \dot{\phi} \hat{\boldsymbol{\phi}} \\
\dot{\hat{\boldsymbol{\phi}}} & = - \dot\phi \hat{\boldsymbol{\rho}} \\
\dot{\hat{\mathbf{z}}} & = 0
\end{align}</math>
So the time derivative simplifies to: <math display="block">\dot{\mathbf{A}} = \hat{\boldsymbol{\rho}} \left(\dot{A}_\rho - A_\phi \dot{\phi}\right)
+ \hat{\boldsymbol{\phi}} \left(\dot{A}_\phi + A_\rho \dot{\phi}\right)
+ \hat{\mathbf{z}} \dot{A}_z</math>
Second time derivative of a vector fieldEdit
The second time derivative is of interest in physics, as it is found in equations of motion for classical mechanical systems. The second time derivative of a vector field in cylindrical coordinates is given by: <math display="block">\ddot{\mathbf{A}} = \mathbf{\hat \rho} \left(\ddot A_\rho - A_\phi \ddot\phi - 2 \dot A_\phi \dot\phi - A_\rho \dot\phi^2\right)
+ \boldsymbol{\hat\phi} \left(\ddot A_\phi + A_\rho \ddot\phi + 2 \dot A_\rho \dot\phi - A_\phi \dot\phi^2\right)
+ \mathbf{\hat z} \ddot A_z</math>
To understand this expression, A is substituted for P, where P is the vector (ρ, φ, z).
This means that <math>\mathbf{A} = \mathbf{P} = \rho \mathbf{\hat \rho} + z \mathbf{\hat z}</math>.
After substituting, the result is given: <math display="block">\ddot\mathbf{P} = \mathbf{\hat \rho} \left(\ddot \rho - \rho \dot\phi^2\right)
+ \boldsymbol{\hat\phi} \left(\rho \ddot\phi + 2 \dot \rho \dot\phi\right)
+ \mathbf{\hat z} \ddot z</math>
In mechanics, the terms of this expression are called:
| <math> \ddot \rho \mathbf{\hat \rho} </math> | central outward acceleration |
| <math> -\rho \dot\phi^2 \mathbf{\hat \rho} </math> | centripetal acceleration |
| <math> \rho \ddot\phi \boldsymbol{\hat\phi} </math> | angular acceleration |
| <math> 2 \dot \rho \dot\phi \boldsymbol{\hat\phi} </math> | Coriolis effect |
| <math> \ddot z \mathbf{\hat z} </math> | Template:Mvar-acceleration |
Spherical coordinate systemEdit
Vector fieldsEdit
Vectors are defined in spherical coordinates by (r, θ, φ), where
- r is the length of the vector,
- θ is the angle between the positive Z-axis and the vector in question (0 ≤ θ ≤ π), and
- φ is the angle between the projection of the vector onto the xy-plane and the positive X-axis (0 ≤ φ < 2π).
(r, θ, φ) is given in Cartesian coordinates by: <math display="block">\begin{bmatrix}r \\ \theta \\ \phi \end{bmatrix} = \begin{bmatrix} \sqrt{x^2 + y^2 + z^2} \\ \arccos(z / \sqrt{x^2 + y^2 + z^2}) \\ \arctan(y / x) \end{bmatrix},\ \ \ 0 \le \theta \le \pi,\ \ \ 0 \le \phi < 2\pi, </math> or inversely by: <math display="block">\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} r\sin\theta\cos\phi \\ r\sin\theta\sin\phi \\ r\cos\theta\end{bmatrix}.</math>
Any vector field can be written in terms of the unit vectors as: <math display="block">\mathbf A
= A_x\mathbf{\hat x} + A_y\mathbf{\hat y} + A_z\mathbf{\hat z}
= A_r\boldsymbol{\hat r} + A_\theta\boldsymbol{\hat \theta} + A_\phi\boldsymbol{\hat \phi}</math>
The spherical basis vectors are related to the Cartesian basis vectors by the Jacobian matrix:
<math display="block">\begin{bmatrix}\boldsymbol{\hat{r}} \\ \boldsymbol{\hat\theta} \\ \boldsymbol{\hat\phi} \end{bmatrix}
= \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} & \frac{\partial z}{\partial r} \\
\frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} & \frac{\partial z}{\partial \theta} \\
\frac{\partial x}{\partial \phi} & \frac{\partial y}{\partial \phi} & \frac{\partial z}{\partial \phi} \end{bmatrix}
\begin{bmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix}</math>
Normalizing the Jacobian matrix so that the spherical basis vectors have unit length we get:
<math display="block">\begin{bmatrix}\boldsymbol{\hat{r}} \\ \boldsymbol{\hat\theta} \\ \boldsymbol{\hat\phi} \end{bmatrix}
= \begin{bmatrix} \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\
\cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\
-\sin\phi & \cos\phi & 0 \end{bmatrix}
\begin{bmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix}</math>
Note: the matrix is an orthogonal matrix, that is, its inverse is simply its transpose.
The Cartesian unit vectors are thus related to the spherical unit vectors by:
<math display="block">\begin{bmatrix}\mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix}
= \begin{bmatrix} \sin\theta\cos\phi & \cos\theta\cos\phi & -\sin\phi \\
\sin\theta\sin\phi & \cos\theta\sin\phi & \cos\phi \\
\cos\theta & -\sin\theta & 0 \end{bmatrix}
\begin{bmatrix} \boldsymbol{\hat{r}} \\ \boldsymbol{\hat\theta} \\ \boldsymbol{\hat\phi} \end{bmatrix}</math>
Time derivative of a vector fieldEdit
To find out how the vector field A changes in time, the time derivatives should be calculated. In Cartesian coordinates this is simply: <math display="block">\mathbf{\dot A} = \dot A_x \mathbf{\hat x} + \dot A_y \mathbf{\hat y} + \dot A_z \mathbf{\hat z}</math> However, in spherical coordinates this becomes: <math display="block">\mathbf{\dot A} = \dot A_r \boldsymbol{\hat r} + A_r \boldsymbol{\dot{\hat r}}
+ \dot A_\theta \boldsymbol{\hat\theta} + A_\theta \boldsymbol{\dot{\hat\theta}}
+ \dot A_\phi \boldsymbol{\hat\phi} + A_\phi \boldsymbol{\dot{\hat\phi}}</math>
The time derivatives of the unit vectors are needed. They are given by: <math display="block">\begin{align}
\boldsymbol{\dot{\hat r}} &= \dot\theta \boldsymbol{\hat\theta} + \dot\phi\sin\theta \boldsymbol{\hat\phi} \\
\boldsymbol{\dot{\hat\theta}} &= - \dot\theta \boldsymbol{\hat r} + \dot\phi\cos\theta \boldsymbol{\hat\phi} \\
\boldsymbol{\dot{\hat\phi}} &= - \dot\phi\sin\theta \boldsymbol{\hat{r}} - \dot\phi\cos\theta \boldsymbol{\hat\theta}
\end{align}</math> Thus the time derivative becomes: <math display="block">\mathbf{\dot A} = \boldsymbol{\hat r} \left(\dot A_r - A_\theta \dot\theta - A_\phi \dot\phi \sin\theta \right)
+ \boldsymbol{\hat\theta} \left(\dot A_\theta + A_r \dot\theta - A_\phi \dot\phi \cos\theta\right)
+ \boldsymbol{\hat\phi} \left(\dot A_\phi + A_r \dot\phi \sin\theta + A_\theta \dot\phi \cos\theta\right)</math>
See alsoEdit
- Del in cylindrical and spherical coordinates for the specification of gradient, divergence, curl, and Laplacian in various coordinate systems.
ReferencesEdit
<references/>