Vector fields in cylindrical and spherical coordinates

Vector fields in cylindrical coordinates

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

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

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

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}

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

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

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

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 )