# Vector fields in cylindrical and spherical coordinates

 Contents

## Vector fields in cylindrical coordinates

Vectors are defined in cylindrical coordinates by (ρ,φ,z), where

• ρ is the length of the vector projected onto the X-Y-plane,
• φ is the angle of the projected vector with the positive X-axis (0 <= φ < 2π),
• z is the regular z-coordinate.

(ρ,φ,z) is given in cartesian coordinates by:

[itex]\left[\begin{matrix}
   \rho & = & \sqrt{x^2 + y^2} \\
\phi & = & \operatorname{atan2}(y, x) \\
z & = & z \end{matrix}\right.[itex]


or inversely by:

[itex]\left[\begin{matrix}
   x & = & \rho\cos\phi \\
y & = & \rho\sin\phi \\
z & = & z \end{matrix}\right.[itex]


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

[itex]\mathbf A = A_x\mathbf{\hat x} + A_y\mathbf{\hat y} + A_z\mathbf{\hat z}
                = A_\rho\boldsymbol{\hat \rho} + A_\phi\boldsymbol{\hat \phi} + A_z\boldsymbol{\hat z}[itex]


The cylindrical unit vectors are related to the cartesian unit vectors by:

[itex]\begin{bmatrix}\boldsymbol\hat\rho \\ \boldsymbol\hat\phi \\ \boldsymbol\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}[itex]


## Time derivative of a vector field in cylindrical coordinates

To find out how the vector field A changes in time we calculate the time derivatives. In cartesian coordinates this is simply:

[itex]\mathbf\dot A = \dot A_x \mathbf\hat x + \dot A_y \mathbf\hat y + \dot A_z \mathbf\hat z[itex]

However, in cylindrical coordinates this becomes:

[itex]\mathbf\dot A = \dot A_\rho \boldsymbol\hat\rho + A_\rho \boldsymbol\dot\hat\rho
 + \dot A_\phi \boldsymbol\hat\phi + A_\phi \boldsymbol\dot\hat\phi
+ \dot A_z \boldsymbol\hat z + A_z \boldsymbol\dot\hat z[itex]


We need the time derivatives of the unit vectors. They are given by:

[itex]\left[\begin{matrix}
 \boldsymbol\dot\hat\rho & = & \dot\phi \boldsymbol\hat\phi \\
\boldsymbol\dot\hat\phi & = & - \dot\phi \boldsymbol\hat\rho \\
\boldsymbol\dot\hat z   & = & 0 \end{matrix}\right.[itex]


So the time derivative simplifies to:

[itex]\mathbf\dot A = \boldsymbol\hat\rho (\dot A_\rho - A_\phi \dot\phi)
 + \boldsymbol\hat\phi (\dot A_\phi + A_\rho \dot\phi)
+ \boldsymbol\hat z \dot A_z[itex]


## Gradient, divergence, curl, and laplacian in cylindrical coordinates

The specification of gradient, divergence, curl, and laplacian in cylindrical coordinates can be found in the article Nabla in cylindrical and spherical coordinates.

## Vector fields in spherical coordinates

Vectors are defined in spherical coordinates by (r,θ,φ), where

• r is the length of the vector,
• θ is the angle with the positive Z-axis (0 <= θ <= π),
• φ is the angle with the X-Z-plane (0 <= φ < 2π).

(r,θ,φ) is given in cartesian coordinates by:

[itex]\left[\begin{matrix}
   r & = & \sqrt{x^2 + y^2 + z^2} \\
\theta & = & \arccos(z / r) \\
\phi & = & \operatorname{atan2}(y, x) \end{matrix}\right.[itex]


or inversely by:

[itex]\left[\begin{matrix}
   x & = & r\sin\theta\cos\phi \\
y & = & r\sin\theta\sin\phi \\
z & = & r\cos\theta \end{matrix}\right.[itex]


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

[itex]\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}[itex]


The spherical unit vectors are related to the cartesian unit vectors by:

[itex]\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}[itex]


## Time derivative of a vector field in spherical coordinates

To find out how the vector field A changes in time we calculate the time derivatives. In cartesian coordinates this is simply:

[itex]\mathbf\dot A = \dot A_x \mathbf\hat x + \dot A_y \mathbf\hat y + \dot A_z \mathbf\hat z[itex]

However, in spherical coordinates this becomes:

[itex]\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[itex]


We need the time derivatives of the unit vectors. They are given by:

[itex]\begin{bmatrix}\boldsymbol\dot\hat r \\ \boldsymbol\dot\hat\theta \\ \boldsymbol\dot\hat\phi \end{bmatrix}
 = \begin{bmatrix} 0           & \dot\theta & \dot\phi \sin\theta \\
-\dot\theta & 0          & \dot\phi \cos\theta \\
-\dot\phi \sin\theta & -\dot\phi \cos\theta & 0 \end{bmatrix}
\begin{bmatrix} \boldsymbol\hat r \\ \boldsymbol\hat\theta \\ \boldsymbol\hat\phi \end{bmatrix}[itex]


So the time derivative becomes:

[itex]\mathbf\dot A = \boldsymbol\hat r (\dot A_r - A_\theta \dot\theta - A_\phi \dot\phi \sin\theta)
 + \boldsymbol\hat\theta (\dot A_\theta + A_r \dot\theta - A_\phi \dot\phi \cos\theta)
+ \boldsymbol\hat\phi (\dot A_\phi + A_r \dot\phi \sin\theta + A_\phi \dot\phi \cos\theta)[itex]


## Gradient, divergence, curl, and laplacian in spherical coordinates

The specification of gradient, divergence, curl, and laplacian in spherical coordinates can be found in the article Nabla in cylindrical and spherical coordinates.

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