Vector fields in cylindrical and spherical coordinates
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In vector calculus and physics, a vector field is an assignment of a vector to each point in a space. When these spaces are in (typically) three dimensions, then the use of cylindrical or spherical coordinates to represent the position of objects in this space is useful in connection with objects and phenomena that have some rotational symmetry about the longitudinal axis, such as water flow in a straight pipe with round cross-section, heat distribution in a metal cylinder, electromagnetic fields produced by an electric current in a long, straight wire, accretion disks in astronomy, and so on. The mathematical properties of such vector fields are thus of interest to physicists and mathematicians alike, who study them to model systems arising in the natural world.
Note: This page uses common physics notation for spherical coordinates, in which <math>\theta</math> is the angle between the <math>z </math> axis and the radius vector <math>r </math> connecting the origin to the point in question, while <math>\phi</math> is the angle between the projection of the radius vector onto the <math>x - y </math> plane and the <math>x </math> 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 system
Vector fields
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 field
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 field
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 system
Vector fields
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 field
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 also
- Del in cylindrical and spherical coordinates for the specification of gradient, divergence, curl, and Laplacian in various coordinate systems.
References
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