Benutzer:Prolineserver/Vector fields in cylindrical and spherical coordinates

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) and z is the regular z-coordinate.

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

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

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.

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:

$\mathbf {\dot {A}} ={\dot {A}}_{x}\mathbf {\hat {x}} +{\dot {A}}_{y}\mathbf {\hat {y}} +{\dot {A}}_{z}\mathbf {\hat {z}}$

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:

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

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) and φ is the angle with the X-Z-plane (0 <= φ < 2π).

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

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

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.

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:

$\mathbf {\dot {A}} ={\dot {A}}_{x}\mathbf {\hat {x}} +{\dot {A}}_{y}\mathbf {\hat {y}} +{\dot {A}}_{z}\mathbf {\hat {z}}$

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:

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

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.