Cylindrical coordinate system

The cylindrical coordinate system extends planar polar coordinates by an axial coordinate: $(x,y,z)=(R\cos \varphi ,R\sin \varphi ,z)$.

For $x=r\cos \varphi$, $y=r\sin \varphi$ and unchanged $z$, the Jacobian determinant of Cartesian coordinates with respect to cylindrical coordinates is $r$. Hence a volume integral acquires the displayed factor for $r\ge 0$. The inverse determinant is $1/r$ where $r>0$; the axis is a coordinate degeneracy.

The azimuthal angle $\varphi$ in cylindrical coordinates is measured in the horizontal plane from a chosen reference axis. In the usual mathematical convention $x=s\cos \varphi$, $y=s\sin \varphi$, so positive $\varphi$ runs counterclockwise from positive $x$ when viewed from positive $z$. This same angular coordinate occurs in spherical coordinates. It is defined modulo $2\pi$ away from the polar axis and is nonunique on the axis.

The local orthonormal basis associated with cylindrical coordinates: ${\mathbf{e}}_r=(\cos \theta ,\sin \theta ,0)$, ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, and ${\mathbf{e}}_z=(0,0,1)$. The first two basis vectors depend on the angular coordinate and are undefined on the axis without a choice of angle.

The azimuthal unit vector is tangent to increasing azimuthal angle at fixed cylindrical radius and height. For $r>0$ it equals $(-y/r,x/r,0)$. A counterclockwise circle has tangent displacement $d\mathbf{r}=r\hat{\varphi }d\varphi$. Unlike a fixed Cartesian basis vector, its direction depends on position.

For a differentiable scalar field and cylindrical coordinates with $r>0$, the gradient has the displayed components in the local orthonormal basis. The chain rule applied to $r=\sqrt{x^2+y^2}$ and a local angle $\theta$ gives $r_x=\cos \theta$, $r_y=\sin \theta$, ${\theta }_x=-\sin \theta /r$, and ${\theta }_y=\cos \theta /r$. Substitution in $\nabla \varphi ={\varphi }_x{\mathbf{e}}_x+{\varphi }_y{\mathbf{e}}_y+{\varphi }_z{\mathbf{e}}_z$ proves the formula. The angular factor $1/r$ appears because an angular change $d\theta$ sweeps arc length $rd\theta$.

Differentiate the position vector $\mathbf{r}=r{\mathbf{e}}_r+z{\mathbf{e}}_z$ using ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$, ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$, and fixed ${\mathbf{e}}_z$. The first derivative is the velocity $\dot{r}{\mathbf{e}}_r+r\dot{\theta }{\mathbf{e}}_{\theta }+\dot{z}{\mathbf{e}}_z$; differentiating once more gives the displayed acceleration. In particular, $r$ times the azimuthal acceleration is $d(r^2\dot{\theta })/dt$.

If the cylindrical components of a vector field have dependence $e^{im\varphi }$, differentiation of the rotating basis adds the displayed cross product term. Scalar-component differentiation alone misses this contribution. For $\mathbf{B}=(J/2)s\hat{\mathbf{\varphi }}$, one also has $(\mathbf{v}\cdot \nabla )\mathbf{B}=(J/2)\hat{\mathbf{z}}\times \mathbf{v}$. These identities are useful in cylindrical magnetohydrodynamic waves.

The cylindrical radius $R=\sqrt{x^2+y^2}$ is the perpendicular distance from the symmetry axis.