Vector fields in cylindrical and spherical coordinates



Note: This page uses common physics notation for spherical coordinates, in which $$\theta$$ is the angle between the z axis and the radius vector connecting the origin to the point in question, while $$\phi$$ 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.

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: $$\begin{bmatrix} \rho \\ \phi \\ z \end{bmatrix} = \begin{bmatrix} \sqrt{x^2 + y^2} \\ \operatorname{arctan}(y / x) \\ z \end{bmatrix},\ \ \ 0 \le \phi < 2\pi, $$ or inversely by: $$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \rho\cos\phi \\ \rho\sin\phi \\ z \end{bmatrix}.$$

Any vector field can be written in terms of the unit vectors as: $$\mathbf A = A_x \mathbf{\hat x} + A_y \mathbf{\hat y} + A_z \mathbf{\hat z} = A_\rho \mathbf{\hat \rho} + A_\phi \boldsymbol{\hat \phi} + A_z \mathbf{\hat z}$$ The cylindrical unit vectors are related to the Cartesian unit vectors by: $$\begin{bmatrix}\mathbf{\hat \rho} \\ \boldsymbol{\hat\phi} \\ \mathbf{\hat z}\end{bmatrix} = \begin{bmatrix} \cos\phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix}$$

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 ($$\dot{\mathbf{A}}$$). In Cartesian coordinates this is simply: $$\dot{\mathbf{A}} = \dot{A}_x \hat{\mathbf{x}} + \dot{A}_y \hat{\mathbf{y}} + \dot{A}_z \hat{\mathbf{z}}$$

However, in cylindrical coordinates this becomes: $$\dot{\mathbf{A}} = \dot{A}_\rho \hat{\boldsymbol{\rho}} + A_\rho \dot{\hat{\boldsymbol{\rho}}} + \dot{A}_\phi \hat{\boldsymbol{\phi}} + A_\phi \dot{\hat{\boldsymbol{\phi}}} + \dot{A}_z \hat{\boldsymbol{z}} + A_z \dot{\hat{\boldsymbol{z}}}$$

The time derivatives of the unit vectors are needed. They are given by: $$\begin{align} \dot{\hat{\mathbf{\rho}}} & = \dot{\phi} \hat{\boldsymbol{\phi}} \\ \dot{\hat{\boldsymbol{\phi}}} & = - \dot\phi \hat{\mathbf{\rho}} \\ \dot{\hat{\mathbf{z}}} & = 0 \end{align}$$

So the time derivative simplifies to: $$\dot{\mathbf{A}} = \hat{\boldsymbol{\rho}} \left(\dot{A}_\rho - A_\phi \dot{\phi}\right) + \hat{\boldsymbol{\phi}} \left(\dot{A}_\phi + A_\rho \dot{\phi}\right) + \hat{\mathbf{z}} \dot{A}_z$$

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: $$\mathbf{\ddot A} = \mathbf{\hat \rho} \left(\ddot A_\rho - A_\phi \ddot\phi - 2 \dot A_\phi \dot\phi - A_\rho \dot\phi^2\right) + \boldsymbol{\hat\phi} \left(\ddot A_\phi + A_\rho \ddot\phi + 2 \dot A_\rho \dot\phi - A_\phi \dot\phi^2\right) + \mathbf{\hat z} \ddot A_z$$

To understand this expression, A is substituted for P, where P is the vector (ρ, φ, z).

This means that $$\mathbf{A} = \mathbf{P} = \rho \mathbf{\hat \rho} + z \mathbf{\hat z}$$.

After substituting, the result is given: $$\ddot\mathbf{P} = \mathbf{\hat \rho} \left(\ddot \rho - \rho \dot\phi^2\right) + \boldsymbol{\hat\phi} \left(\rho \ddot\phi + 2 \dot \rho \dot\phi\right) + \mathbf{\hat z} \ddot z$$

In mechanics, the terms of this expression are called:

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: $$\begin{bmatrix}r \\ \theta \\ \phi \end{bmatrix} = \begin{bmatrix} \sqrt{x^2 + y^2 + z^2} \\ \arccos(z / \sqrt{x^2 + y^2 + z^2}) \\ \arctan(y / x) \end{bmatrix},\ \ \ 0 \le \theta \le \pi,\ \ \ 0 \le \phi < 2\pi, $$ or inversely by: $$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} r\sin\theta\cos\phi \\ r\sin\theta\sin\phi \\ r\cos\theta\end{bmatrix}.$$

Any vector field can be written in terms of the unit vectors as: $$\mathbf A = A_x\mathbf{\hat x} + A_y\mathbf{\hat y} + A_z\mathbf{\hat z} = A_r\boldsymbol{\hat r} + A_\theta\boldsymbol{\hat \theta} + A_\phi\boldsymbol{\hat \phi}$$ The spherical unit vectors are related to the Cartesian unit vectors by: $$\begin{bmatrix}\boldsymbol{\hat{r}} \\ \boldsymbol{\hat\theta} \\ \boldsymbol{\hat\phi} \end{bmatrix} = \begin{bmatrix} \sin\theta\cos\phi & \sin\theta\sin\phi & \cos\theta \\ \cos\theta\cos\phi & \cos\theta\sin\phi & -\sin\theta \\ -\sin\phi         & \cos\phi           & 0 \end{bmatrix} \begin{bmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix}$$

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: $$\begin{bmatrix}\mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix} = \begin{bmatrix} \sin\theta\cos\phi & \cos\theta\cos\phi & -\sin\phi \\ \sin\theta\sin\phi & \cos\theta\sin\phi & \cos\phi \\ \cos\theta        & -\sin\theta        & 0 \end{bmatrix} \begin{bmatrix} \boldsymbol{\hat{r}} \\ \boldsymbol{\hat\theta} \\ \boldsymbol{\hat\phi} \end{bmatrix}$$

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: $$\mathbf{\dot A} = \dot A_x \mathbf{\hat x} + \dot A_y \mathbf{\hat y} + \dot A_z \mathbf{\hat z}$$ However, in spherical coordinates this becomes: $$\mathbf{\dot A} = \dot A_r \boldsymbol{\hat r} + A_r \boldsymbol{\dot{\hat r}} + \dot A_\theta \boldsymbol{\hat\theta} + A_\theta \boldsymbol{\dot{\hat\theta}} + \dot A_\phi \boldsymbol{\hat\phi} + A_\phi \boldsymbol{\dot{\hat\phi}}$$ The time derivatives of the unit vectors are needed. They are given by: $$\begin{align} \boldsymbol{\dot{\hat r}} &= \dot\theta \boldsymbol{\hat\theta} + \dot\phi\sin\theta \boldsymbol{\hat\phi} \\ \boldsymbol{\dot{\hat\theta}} &= - \dot\theta \boldsymbol{\hat r} + \dot\phi\cos\theta \boldsymbol{\hat\phi} \\ \boldsymbol{\dot{\hat\phi}} &= - \dot\phi\sin\theta \boldsymbol{\hat{r}} - \dot\phi\cos\theta \boldsymbol{\hat\theta} \end{align}$$ Thus the time derivative becomes: $$\mathbf{\dot A} = \boldsymbol{\hat r} \left(\dot A_r - A_\theta \dot\theta - A_\phi \dot\phi \sin\theta \right) + \boldsymbol{\hat\theta} \left(\dot A_\theta + A_r \dot\theta - A_\phi \dot\phi \cos\theta\right) + \boldsymbol{\hat\phi} \left(\dot A_\phi + A_r \dot\phi \sin\theta + A_\theta \dot\phi \cos\theta\right)$$