Vector fields in cylindrical and spherical coordinates

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1 Cylindrical coordinate system
- 1.1 Vector fields
- 1.2 Time derivative of a vector field
2 Spherical coordinate system
- 2.1 Vector fields
- 2.2 Time derivative of a vector field
3 See also

[edit] Cylindrical coordinate system

[edit] Vector fields

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:

$\left[\begin{matrix} \rho & = & \sqrt{x^2 + y^2} \\ \phi & = & \operatorname{arctan}(y / x), & 0 \le \phi < 2\pi \\ z & = & z \end{matrix}\right.$

or inversely by:

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

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

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

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

$\begin{bmatrix}\boldsymbol{\hat\rho} \\ \boldsymbol{\hat\phi} \\ \boldsymbol{\hat z}\end{bmatrix} = \begin{bmatrix} \cos\phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \mathbf{\hat x} \\ \mathbf{\hat y} \\ \mathbf{\hat z} \end{bmatrix}$

Note: the matrix is an orthogonal matrix, that is, its inverse is simply its transpose.

[edit] Time derivative of a vector field

To find out how the vector field A changes in time we calculate the time derivatives. 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 cylindrical coordinates this becomes:

$\mathbf{\dot A} = \dot A_\rho \boldsymbol{\hat\rho} + A_\rho \boldsymbol{\dot{\hat\rho}} + \dot A_\phi \boldsymbol{\hat\phi} + A_\phi \boldsymbol{\dot{\hat\phi}} + \dot A_z \boldsymbol{\hat z} + A_z \boldsymbol{\dot{\hat z}}$

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

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

So the time derivative simplifies to:

$\mathbf{\dot A} = \boldsymbol{\hat\rho} (\dot A_\rho - A_\phi \dot\phi) + \boldsymbol{\hat\phi} (\dot A_\phi + A_\rho \dot\phi) + \boldsymbol{\hat z} \dot A_z$

[edit] Spherical coordinate system

[edit] Vector fields

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:

$\left[\begin{matrix} r & = & \sqrt{x^2 + y^2 + z^2} \\ \theta & = & \arccos\left( z / r\right), & 0 \le \theta \le \pi \\ \phi & = & \operatorname{arctan}(y / x), & 0 \le \phi < 2\pi \end{matrix}\right.$

or inversely by:

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

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.

[edit] Time derivative of a vector field

To find out how the vector field A changes in time we calculate the time derivatives. 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:

$\mathbf{\dot A} = \dot A_r \boldsymbol{\hat r} + A_r \boldsymbol{\dot{\hat r}} + \dot A_\theta \boldsymbol{\hat\theta} + A_\theta \boldsymbol{\dot{\hat\theta}} + \dot A_\phi \boldsymbol{\hat\phi} + A_\phi \boldsymbol{\dot{\hat\phi}}$

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

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

So the time derivative becomes:

$\mathbf{\dot A} = \boldsymbol{\hat r} (\dot A_r - A_\theta \dot\theta - A_\phi \dot\phi \sin\theta) + \boldsymbol{\hat\theta} (\dot A_\theta + A_r \dot\theta - A_\phi \dot\phi \cos\theta) + \boldsymbol{\hat\phi} (\dot A_\phi + A_r \dot\phi \sin\theta + A_\theta \dot\phi \cos\theta)$