Question

How to convert cylindrical to spherical coordinates matrix?

Answer 1

Vectors are defined in cylindrical coordinatesby (r, θ, z), where

r is the length of the vector projected onto the xy-plane,θ is the angle between the projection of the vector onto the xy-plane (i.e. r) and the positive x-axis (0 ≤ θ < 2π),z is the regular z-coordinate.

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

{\displaystyle {\begin{bmatrix}r\\\theta \\z\end{bmatrix}}={\begin{bmatrix}{\sqrt {x^{2}+y^{2}}}\\\operatorname {arctan} (y/x)\\z\end{bmatrix}},\ \ \ 0\leq \theta <2\pi ,}



or inversely by:

{\displaystyle {\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}r\cos \theta \\r\sin \theta \\z\end{bmatrix}}.}

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

{\displaystyle \mathbf {A} =A_{x}\mathbf {\hat {x}} +A_{y}\mathbf {\hat {y}} +A_{z}\mathbf {\hat {z}} =A_{r}\mathbf {\hat {r}} +A_{\theta }{\boldsymbol {\hat {\theta }}}+A_{z}\mathbf {\hat {z}} }

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

{\displaystyle {\begin{bmatrix}\mathbf {\hat {r}} \\{\boldsymbol {\hat {\theta }}}\\\mathbf {\hat {z}} \end{bmatrix}}={\begin{bmatrix}\cos \theta &\sin \theta &0\\-\sin \theta &\cos \theta &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 fieldEdit

To find out how the vector field A changes in time we calculate the time derivatives. For this purpose we use Newton's notation for the time derivative ({\displaystyle {\dot {\mathbf {A} }}}). In cartesian coordinates this is simply:

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

However, in cylindrical coordinates this becomes:

{\displaystyle {\dot {\mathbf {A} }}={\dot {A}}_{r}{\hat {\boldsymbol {r}}}+A_{r}{\dot {\hat {\boldsymbol {r}}}}+{\dot {A}}_{\theta }{\hat {\boldsymbol {\theta }}}+A_{\theta }{\dot {\hat {\boldsymbol {\theta }}}}+{\dot {A}}_{z}{\hat {\boldsymbol {z}}}+A_{z}{\dot {\hat {\boldsymbol {z}}}}}

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

{\displaystyle {\begin{aligned}{\dot {\hat {\mathbf {r} }}}&={\dot {\theta }}{\hat {\boldsymbol {\theta }}}\\{\dot {\hat {\boldsymbol {\theta }}}}&=-{\dot {\theta }}{\hat {\mathbf {r} }}\\{\dot {\hat {\mathbf {z} }}}&=0\end{aligned}}}

So the time derivative simplifies to:

{\displaystyle {\dot {\mathbf {A} }}={\hat {\boldsymbol {r}}}({\dot {A}}_{r}-A_{\theta }{\dot {\theta }})+{\hat {\boldsymbol {\theta }}}({\dot {A}}_{\theta }+A_{r}{\dot {\theta }})+{\hat {\mathbf {z} }}{\dot {A}}_{z}}

Second time derivative of a vector fieldEdit

The second time derivative is of interest in physics, as it is found in equations of motionfor classical mechanical systems. The second time derivative of a vector field in cylindrical coordinates is given by:

{\displaystyle \mathbf {\ddot {A}} =\mathbf {\hat {r}} ({\ddot {A}}_{r}-A_{\theta }{\ddot {\theta }}-2{\dot {A}}_{\theta }{\dot {\theta }}-A_{r}{\dot {\theta }}^{2})+{\boldsymbol {\hat {\theta }}}({\ddot {A}}_{\theta }+A_{r}{\ddot {\theta }}+2{\dot {A}}_{r}{\dot {\theta }}-A_{\theta }{\dot {\theta }}^{2})+\mathbf {\hat {z}} {\ddot {A}}_{z}}

To understand this expression, we substitute A = P, where p is the vector (r, θ, z).

This means that {\displaystyle \mathbf {A} =\mathbf {P} =r\mathbf {\hat {r}} +z\mathbf {\hat {z}} }.

After substituting we get:

{\displaystyle {\ddot {\mathbf {P} }}=\mathbf {\hat {r}} ({\ddot {r}}-r{\dot {\theta }}^{2})+{\boldsymbol {\hat {\theta }}}(r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }})+\mathbf {\hat {z}} {\ddot {z}}}

In mechanics, the terms of this expression are called:

{\displaystyle {\begin{aligned}{\ddot {r}}\mathbf {\hat {r}} &={\mbox{central outward acceleration}}\\-r{\dot {\theta }}^{2}\mathbf {\hat {r}} &={\mbox{centripetal acceleration}}\\r{\ddot {\theta }}{\boldsymbol {\hat {\theta }}}&={\mbox{angular acceleration}}\\2{\dot {r}}{\dot {\theta }}{\boldsymbol {\hat {\theta }}}&={\mbox{Coriolis effect}}\\{\ddot {z}}\mathbf {\hat {z}} &={\mbox{z-acceleration}}\end{aligned}}}