Note: This page uses common physics notation for spherical coordinates, in which ${\displaystyle \theta }$ is the angle between the z axis and the radius vector connecting the origin to the point in question, while ${\displaystyle \phi }$ is the angle between the projection of the radius vector onto the x-y plane and the x axis. Several other definitions are in use, and so care must be taken in comparing different sources.^[1]

Cylindrical coordinate system

Vector fields

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

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

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

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

or inversely by:

$${\displaystyle {\begin{bmatrix}x\\y\\z\end{bmatrix))={\begin{bmatrix}\rho \cos \phi \\\rho \sin \phi \\z\end{bmatrix)).}$$

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

$${\displaystyle \mathbf {A} =A_{x}\mathbf {\hat {x)) +A_{y}\mathbf {\hat {y)) +A_{z}\mathbf {\hat {z)) =A_{\rho }\mathbf {\hat {\rho )) +A_{\phi }{\boldsymbol {\hat {\phi ))}+A_{z}\mathbf {\hat {z)) }$$

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

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

Time derivative of a vector field

To find out how the vector field A changes in time, the time derivatives should be calculated. For this purpose Newton's notation will be used 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:

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

The time derivatives of the unit vectors are needed. They are given by:

$${\displaystyle {\begin{aligned}{\dot {\hat {\mathbf {\rho } ))}&={\dot {\phi )){\hat {\boldsymbol {\phi ))}\\{\dot {\hat {\boldsymbol {\phi ))))&=-{\dot {\phi )){\hat {\mathbf {\rho } ))\\{\dot {\hat {\mathbf {z} ))}&=0\end{aligned))}$$

So the time derivative simplifies to:

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

Second time derivative of a vector field

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

$${\displaystyle \mathbf {\ddot {A)) =\mathbf {\hat {\rho )) \left({\ddot {A))_{\rho }-A_{\phi }{\ddot {\phi ))-2{\dot {A))_{\phi }{\dot {\phi ))-A_{\rho }{\dot {\phi ))^{2}\right)+{\boldsymbol {\hat {\phi ))}\left({\ddot {A))_{\phi }+A_{\rho }{\ddot {\phi ))+2{\dot {A))_{\rho }{\dot {\phi ))-A_{\phi }{\dot {\phi ))^{2}\right)+\mathbf {\hat {z)) {\ddot {A))_{z))$$

To understand this expression, A is substituted for P, where P is the vector (ρ, φ, z).

This means that ${\displaystyle \mathbf {A} =\mathbf {P} =\rho \mathbf {\hat {\rho )) +z\mathbf {\hat {z)) }$.

After substituting, the result is given:

$${\displaystyle {\ddot {\mathbf {P} ))=\mathbf {\hat {\rho )) \left({\ddot {\rho ))-\rho {\dot {\phi ))^{2}\right)+{\boldsymbol {\hat {\phi ))}\left(\rho {\ddot {\phi ))+2{\dot {\rho )){\dot {\phi ))\right)+\mathbf {\hat {z)) {\ddot {z))}$$

In mechanics, the terms of this expression are called:

$${\displaystyle {\begin{aligned}{\ddot {\rho ))\mathbf {\hat {\rho )) &={\text{central outward acceleration))\\-\rho {\dot {\phi ))^{2}\mathbf {\hat {\rho )) &={\text{centripetal acceleration))\\\rho {\ddot {\phi )){\boldsymbol {\hat {\phi ))}&={\text{angular acceleration))\\2{\dot {\rho )){\dot {\phi )){\boldsymbol {\hat {\phi ))}&={\text{Coriolis effect))\\{\ddot {z))\mathbf {\hat {z)) &={\text{z-acceleration))\end{aligned))}$$

Spherical coordinate system

Vector fields

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

• r is the length of the vector,
• θ is the angle between the positive Z-axis and the vector in question (0 ≤ θ ≤ π), and
• φ is the angle between the projection of the vector onto the xy-plane and the positive X-axis (0 ≤ φ < 2π).

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

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

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

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

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

$${\displaystyle {\begin{bmatrix}{\boldsymbol {\hat {r))}\\{\boldsymbol {\hat {\theta ))}\\{\boldsymbol {\hat {\phi ))}\end{bmatrix))={\begin{bmatrix}\sin \theta \cos \phi &\sin \theta \sin \phi &\cos \theta \\\cos \theta \cos \phi &\cos \theta \sin \phi &-\sin \theta \\-\sin \phi &\cos \phi &0\end{bmatrix)){\begin{bmatrix}\mathbf {\hat {x)) \\\mathbf {\hat {y)) \\\mathbf {\hat {z)) \end{bmatrix))}$$

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

The Cartesian unit vectors are thus related to the spherical unit vectors by:

$${\displaystyle {\begin{bmatrix}\mathbf {\hat {x)) \\\mathbf {\hat {y)) \\\mathbf {\hat {z)) \end{bmatrix))={\begin{bmatrix}\sin \theta \cos \phi &\cos \theta \cos \phi &-\sin \phi \\\sin \theta \sin \phi &\cos \theta \sin \phi &\cos \phi \\\cos \theta &-\sin \theta &0\end{bmatrix)){\begin{bmatrix}{\boldsymbol {\hat {r))}\\{\boldsymbol {\hat {\theta ))}\\{\boldsymbol {\hat {\phi ))}\end{bmatrix))}$$

Time derivative of a vector field

To find out how the vector field A changes in time, the time derivatives should be calculated. In Cartesian coordinates this is simply:

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

However, in spherical coordinates this becomes:

$${\displaystyle \mathbf {\dot {A)) ={\dot {A))_{r}{\boldsymbol {\hat {r))}+A_{r}{\boldsymbol {\dot {\hat {r))))+{\dot {A))_{\theta }{\boldsymbol {\hat {\theta ))}+A_{\theta }{\boldsymbol {\dot {\hat {\theta ))))+{\dot {A))_{\phi }{\boldsymbol {\hat {\phi ))}+A_{\phi }{\boldsymbol {\dot {\hat {\phi ))))}$$

The time derivatives of the unit vectors are needed. They are given by:

$${\displaystyle {\begin{aligned}{\boldsymbol {\dot {\hat {r))))&={\dot {\theta )){\boldsymbol {\hat {\theta ))}+{\dot {\phi ))\sin \theta {\boldsymbol {\hat {\phi ))}\\{\boldsymbol {\dot {\hat {\theta ))))&=-{\dot {\theta )){\boldsymbol {\hat {r))}+{\dot {\phi ))\cos \theta {\boldsymbol {\hat {\phi ))}\\{\boldsymbol {\dot {\hat {\phi ))))&=-{\dot {\phi ))\sin \theta {\boldsymbol {\hat {r))}-{\dot {\phi ))\cos \theta {\boldsymbol {\hat {\theta ))}\end{aligned))}$$

Thus the time derivative becomes:

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