vector fields in cylindrical and spherical coordinates

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:

$$\backslash begin\{bmatrix\}\; \backslash rho\; \backslash \backslash \; \backslash phi\; \backslash \backslash \; z\; \backslash end\{bmatrix\}\; =$$

\begin{bmatrix}

\sqrt{x^2 + y^2} \\ \operatorname{arctan}(y / x) \\ z

\end{bmatrix},\ \ \ 0 \le \phi < 2\pi,

thumb

or inversely by:

$$\backslash begin\{bmatrix\}\; x\; \backslash \backslash \; y\; \backslash \backslash \; z\; \backslash end\{bmatrix\}\; =$$

\begin{bmatrix} \rho\cos\phi \\ \rho\sin\phi \\ z \end{bmatrix}.

Any vector field can be written in terms of the unit vectors as:

$$\backslash mathbf\; A$$

= 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:

$$\backslash begin\{bmatrix\}\backslash boldsymbol\{\backslash hat\; \backslash rho\}\; \backslash \backslash \; \backslash boldsymbol\{\backslash hat\backslash phi\}\; \backslash \backslash \; \backslash mathbf\{\backslash hat\; z\}\backslash end\{bmatrix\}$$

= \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.

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 ($\backslash dot\{\backslash mathbf\{A\}\}$).

In Cartesian coordinates this is simply:

$$\backslash dot\{\backslash mathbf\{A\}\}\; =\; \backslash dot\{A\}\_x\; \backslash hat\{\backslash mathbf\{x\}\}\; +\; \backslash dot\{A\}\_y\; \backslash hat\{\backslash mathbf\{y\}\}\; +\; \backslash dot\{A\}\_z\; \backslash hat\{\backslash mathbf\{z\}\}$$

However, in cylindrical coordinates this becomes:

$$\backslash dot\{\backslash mathbf\{A\}\}\; =\; \backslash dot\{A\}\_\backslash rho\; \backslash hat\{\backslash boldsymbol\{\backslash rho\}\}\; +\; A\_\backslash rho\; \backslash dot\{\backslash hat\{\backslash boldsymbol\{\backslash rho\}\}\}$$

+ \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:

$$\backslash begin\{align\}$$

\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}

So the time derivative simplifies to:

$$\backslash dot\{\backslash mathbf\{A\}\}$$

= \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

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:

$$\backslash ddot\{\backslash mathbf\{A\}\}$$

= \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

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

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

After substituting, the result is given:

$$\backslash ddot\backslash mathbf\{P\}$$

= \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:

{{See also|Centripetal force|Angular acceleration|Coriolis effect}}

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:

$$\backslash begin\{bmatrix\}r\; \backslash \backslash \; \backslash theta\; \backslash \backslash \; \backslash phi\; \backslash end\{bmatrix\}\; =$$

\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:

$$\backslash begin\{bmatrix\}\; x\; \backslash \backslash \; y\; \backslash \backslash \; z\; \backslash end\{bmatrix\}\; =$$

\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:

$$\backslash mathbf\; A$$

= 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 basis vectors are related to the Cartesian basis vectors by the Jacobian matrix:

$$\backslash begin\{bmatrix\}\backslash boldsymbol\{\backslash hat\{r\}\}\; \backslash \backslash \; \backslash boldsymbol\{\backslash hat\backslash theta\}\; \backslash \backslash \; \backslash boldsymbol\{\backslash hat\backslash phi\}\; \backslash end\{bmatrix\}$$

= \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}

Normalizing the Jacobian matrix so that the spherical basis vectors have unit length we get:

$$\backslash begin\{bmatrix\}\backslash boldsymbol\{\backslash hat\{r\}\}\; \backslash \backslash \; \backslash boldsymbol\{\backslash hat\backslash theta\}\; \backslash \backslash \; \backslash boldsymbol\{\backslash hat\backslash phi\}\; \backslash end\{bmatrix\}$$

= \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:

$$\backslash begin\{bmatrix\}\backslash mathbf\{\backslash hat\; x\}\; \backslash \backslash \; \backslash mathbf\{\backslash hat\; y\}\; \backslash \backslash \; \backslash mathbf\{\backslash hat\; z\}\; \backslash end\{bmatrix\}$$

= \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}

To find out how the vector field A changes in time, the time derivatives should be calculated.

In Cartesian coordinates this is simply:

$$\backslash mathbf\{\backslash dot\; A\}\; =\; \backslash dot\; A\_x\; \backslash mathbf\{\backslash hat\; x\}\; +\; \backslash dot\; A\_y\; \backslash mathbf\{\backslash hat\; y\}\; +\; \backslash dot\; A\_z\; \backslash mathbf\{\backslash hat\; z\}$$

However, in spherical coordinates this becomes:

$$\backslash mathbf\{\backslash dot\; A\}\; =\; \backslash dot\; A\_r\; \backslash boldsymbol\{\backslash hat\; r\}\; +\; A\_r\; \backslash boldsymbol\{\backslash dot\{\backslash hat\; r\}\}$$

+ \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:

$$\backslash begin\{align\}$$

\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:

$$\backslash mathbf\{\backslash dot\; A\}$$

= \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)

• Del in cylindrical and spherical coordinates for the specification of gradient, divergence, curl, and Laplacian in various coordinate systems.

{{DEFAULTSORT:Vector Fields In Cylindrical And Spherical Coordinates}}

Category:Vector calculus

Category:Coordinate systems