10.6. Tensor Calculus in Euclidean Spaces - cylindrical coordinates in \(E^3\)

10.6.1. Cylindrical coordiantes and cylindrical coordinates

Using cylindrical coordinates \((q^1, q^2, q^3) = (r, \theta, z)\) and cylindrical base vectors (uniform in space, so that their derivatives are zero), a point in Euclidean vector space \(E^3\) can be represented as

\[\vec{r} = r \cos \theta \, \hat{x} + r \sin \theta \, \hat{y} + z \, \hat{z} \ .\]

10.6.2. Natural basis, reciprocal basis, metric tensor, and Christoffel symbols

Natural basis

Natural basis reads

\[\begin{split}\begin{cases} \vec{b}_1 = \dfrac{\partial \vec{r}}{\partial q^1} = \dfrac{\partial \vec{r}}{\partial r } = \cos \theta \, \hat{x} + \sin \theta \, \hat{y} \\ \vec{b}_2 = \dfrac{\partial \vec{r}}{\partial q^2} = \dfrac{\partial \vec{r}}{\partial \theta} = - r \sin \theta \, \hat{x} + r \cos \theta \, \hat{y} \\ \vec{b}_3 = \dfrac{\partial \vec{r}}{\partial q^3} = \dfrac{\partial \vec{r}}{\partial z } = \hat{z} \\ \end{cases}\end{split}\]

Metric tensor

Covariant components of metric tensors,

\[g_{ab} = \vec{b}_a \cdot \vec{b}_b \ , \]

can be collected in the diagonal matrix

\[\begin{split}\left[ g_{ab} \right] = \begin{bmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & 1 \end{bmatrix} \ ,\end{split}\]

while its contra-variant components can be collected in the inverse matrix (easy to compute, since \(\left[ g_{ab} \right]\) is diagonal),

\[\begin{split}\left[ g^{ab} \right] = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{r^2} & 0 \\ 0 & 0 & 1 \end{bmatrix} \ .\end{split}\]

Reciprocal basis

Reciprocal basis is readily evaluated using \(\vec{b}^a = g^{ab} \, \vec{b}_b\),

\[\begin{split}\begin{cases} \vec{b}^1 = \cos \theta \, \hat{x} + \sin \theta \, \hat{y} \\ \vec{b}^2 = - \dfrac{1}{r} \sin \theta \, \hat{x} + \dfrac{1}{r} \cos \theta \, \hat{y} \\ \vec{b}^3 = \hat{z} \\ \end{cases}\end{split}\]

Physical basis

Since metric tensor is diagonal, the cylindrical coordinate system is orthogonal, and its natural and reciprocal basis are orthogonal. A unit orthogonal basis, usually named physical basis with unit vector with no physical dimension, is evalated by normalization process,

\[\begin{split}\begin{cases} \hat{r} = \hat{b}_1 = \dfrac{\vec{b}_1}{g_{11}} = \dfrac{\vec{b}^1}{g^{11}} = \cos \theta \, \hat{x} + \sin \theta \, \hat{y} \\ \hat{\theta} = \hat{b}_2 = \dfrac{\vec{b}_2}{g_{22}} = \dfrac{\vec{b}^2}{g^{22}} = -\sin \theta \, \hat{x} + \cos \theta \, \hat{y} \\ \hat{z} = \hat{b}_3 = \dfrac{\vec{b}_3}{g_{33}} = \dfrac{\vec{b}^3}{g^{33}} = \hat{z} \ . \end{cases}\end{split}\]

Derivatives of natural basis and Christoffel symbols

Derivatives of the natural basis read

\[\begin{split}\begin{aligned} \dfrac{\partial \vec{b}_1}{\partial q^1} & = \vec{0} \\ \dfrac{\partial \vec{b}_2}{\partial q^2} & = -r \cos \theta \, \hat{x} - r \sin \theta \, \hat{y} = - q^1 \, \vec{b}_1 \\ \dfrac{\partial \vec{b}_3}{\partial q^3} & = \vec{0} \\ \dfrac{\partial \vec{b}_2}{\partial q^1} = \dfrac{\partial \vec{b}_1}{\partial q^2} & = -\sin \theta \, \hat{x} + \cos \theta \hat{y} = \dfrac{1}{q^1} \, \vec{b}_2 \\ \dfrac{\partial \vec{b}_3}{\partial q^1} = \dfrac{\partial \vec{b}_1}{\partial q^3} & = \vec{0} \\ \dfrac{\partial \vec{b}_3}{\partial q^2} = \dfrac{\partial \vec{b}_2}{\partial q^3} & = \vec{0} \\ \end{aligned}\end{split}\]

so that non-zero Christoffel symbols of a cylindrical coordinate system are

\[\begin{split}\begin{aligned} & \Gamma_{12}^{2} = \Gamma_{21}^2 = \dfrac{1}{q^1} \\ & \Gamma_{22}^{1} = - q^1 \ . \end{aligned}\end{split}\]

10.6.3. Differential operators

10.6.3.1. Gradient

Example 10.15 (Gradient of a scalar field)

\[\begin{split}\begin{aligned} \nabla F & = \vec{b}^a \dfrac{\partial F}{\partial q^a} = \\ & = \hat{b}_a \, g^{aa} \, \dfrac{\partial F}{\partial q^a} = \\ & = \hat{r} \, \dfrac{\partial F}{\partial r} + \hat{\theta} \, \dfrac{1}{r} \, \dfrac{\partial F}{\partial \theta} + \hat{z} \dfrac{\partial F}{\partial z} \ . \end{aligned}\end{split}\]

Example 10.16 (Gradient of a vector field)

\[\begin{split}\begin{aligned} \nabla F & = \vec{b}^a \otimes \vec{b}_b \left[ \dfrac{\partial F^b}{\partial q^a} + \Gamma_{ac}^b \, F^c \right] = \\ & = \dots = \\ & = \vec{b}^1 \otimes \vec{b}_1 \, \partial_1 F^1 && + \vec{b}^1 \otimes \vec{b}_2 \, \left[ \partial_1 F^2 + \Gamma_{12}^2 F^2 \right] && + \vec{b}^1 \otimes \vec{b}_3 \, \partial_1 F^3 \\ & + \vec{b}^2 \otimes \vec{b}_1 \, \left[ \partial_2 F^1 + \Gamma_{22}^1 F^2 \right] && + \vec{b}^2 \otimes \vec{b}_2 \, \left[ \partial_2 F^2 + \Gamma_{21}^2 F^1 \right] && + \vec{b}^2 \otimes \vec{b}_3 \, \partial_2 F^3 \\ & + \vec{b}^3 \otimes \vec{b}_1 \, \partial_3 F^1 && + \vec{b}^3 \otimes \vec{b}_2 \, \partial_3 F^2 && + \vec{b}^3 \otimes \vec{b}_3 \, \partial_3 F^3 \\ & = \hat{r } \otimes \hat{r } \, \partial_r F_r && + \hat{r } \otimes \hat{\theta} \, \frac{1}{r} \left[ \partial_r (r F_{\theta}) + F_{\theta} \right] && + \hat{r } \otimes \hat{z } \, \partial_r F_z \\ & + \hat{\theta} \otimes \hat{r } \, \frac{1}{r} \left[ \partial_\theta F_r - r F_{\theta} \right] && + \hat{\theta} \otimes \hat{\theta} \, \left[ \partial_\theta \left( \frac{F_\theta}{r} \right) + \frac{F_r}{r} \right] && + \hat{\theta} \otimes \hat{z } \, \frac{1}{r} \partial_{\theta} F_z \\ & + \hat{z } \otimes \hat{r } \, \partial_z F_x && + \hat{z } \otimes \hat{\theta} \, \frac{1}{r} \partial_\theta F_y && + \hat{z } \otimes \hat{z } \, \partial_z F_z \ . \end{aligned}\end{split}\]

Example 10.17 (Gradient of a \(2^{nd}\)-order tensor field)

10.6.3.2. Directional derivative

10.6.3.3. Divergence

Example 10.18 (Divergence of a vector field)

\[\begin{split}\begin{aligned} \nabla \cdot \vec{F} & = \dfrac{\partial F^a}{\partial q^a} + \Gamma_{ac}^a \, F^c = \\ & = \dfrac{\partial F_r}{\partial r} + \dfrac{\partial}{\partial \theta}\left( \frac{F_\theta}{r} \right) + \frac{F_\theta}{r} + \dfrac{\partial F_z}{\partial z} \ . \end{aligned}\end{split}\]

Divergence of a \(\ 2^{nd}\)-order tensor field

Using the general formula of the divergence of a \(2^{nd}\)-order tensor field (see Divergence)

\[\begin{split}\begin{aligned} \nabla \cdot \left( F^{bc} \vec{b}_b \otimes \vec{b}_c \right) & = C_{1}^{2} \left( \nabla F \right) = \\ & = \vec{b}_c \, \left[ \dfrac{\partial F^{ac}}{\partial q^a} + \Gamma_{ad}^a \, F^{dc} + \Gamma^{c}_{ad} \, F^{ad} \right] \end{aligned}\end{split}\]

the contravariant components in the natural basis induced by cylindrical coordinates of the divergence of a second order tensor reads

\[\begin{split}\begin{aligned} \nabla \cdot \mathbb{F} & = \\ & = \vec{b}_1 \left[ \dfrac{\partial F^{11}}{\partial q^{1}} + \dfrac{\partial F^{21}}{\partial q^{2}} + \dfrac{\partial F^{31}}{\partial q^{3}} + \Gamma_{21}^{2} F^{11} + \Gamma_{22}^{1} F^{22} \right] + \\ & + \vec{b}_2 \left[ \dfrac{\partial F^{12}}{\partial q^{1}} + \dfrac{\partial F^{22}}{\partial q^{2}} + \dfrac{\partial F^{32}}{\partial q^{3}} + \Gamma_{21}^{2} F^{12} + \Gamma_{21}^{2} F^{12} + \Gamma_{21}^2 F^{21} \right] + \\ & + \vec{b}_3 \left[ \dfrac{\partial F^{13}}{\partial q^{1}} + \dfrac{\partial F^{23}}{\partial q^{2}} + \dfrac{\partial F^{33}}{\partial q^{3}} + \Gamma_{21}^{2} F^{13} \right] = \\ & = \vec{b}_1 \left[ \dfrac{\partial F^{11}}{\partial q^{1}} + \dfrac{\partial F^{21}}{\partial q^{2}} + \dfrac{\partial F^{31}}{\partial q^{3}} + \dfrac{1}{q^1} F^{11} - q^{1} F^{22} \right] + \\ & + \vec{b}_2 \left[ \dfrac{\partial F^{12}}{\partial q^{1}} + \dfrac{\partial F^{22}}{\partial q^{2}} + \dfrac{\partial F^{32}}{\partial q^{3}} + \dfrac{2}{q^1} F^{12} + \dfrac{1}{q^1} F^{21} \right] + \\ & + \vec{b}_3 \left[ \dfrac{\partial F^{13}}{\partial q^{1}} + \dfrac{\partial F^{23}}{\partial q^{2}} + \dfrac{\partial F^{33}}{\partial q^{3}} + \dfrac{1}{q^1} F^{13} \right] \ . \end{aligned}\end{split}\]

Next, it’s easy to exploit the definition of the coordinates \(\left( q^1, q^2, q^3 \right) = \left( r, \theta, z \right)\) and the relation between natural and physical basis and components to get

\[\begin{split}\begin{aligned} \nabla \cdot \mathbb{F} & = \\ & = \hat{r} \left[ \dfrac{\partial \left( F^{rr} \right)}{\partial r} + \dfrac{\partial \left( \frac{1}{r} F^{\theta r} \right)}{\partial \theta} + \dfrac{\partial F^{z r}}{\partial z} + \dfrac{1}{r} F^{rr} - r \dfrac{1}{r^2} F^{\theta \theta} \right] + \\ & + r \hat{\theta} \left[ \dfrac{\partial \left( \frac{1}{r} F^{r \theta} \right)}{\partial r} + \dfrac{1}{r} \dfrac{\partial \left(\frac{1}{r} F^{\theta \theta} \right)}{\partial \theta} + \dfrac{\partial F^{z \theta}}{\partial z} + \dfrac{2}{r} \frac{1}{r} F^{r \theta} + \dfrac{1}{r} \dfrac{1}{r} F^{\theta r} \right] + \\ & + \hat{z} \left[ \dfrac{\partial F^{r z}}{\partial r} + \dfrac{\partial \left( \frac{1}{r} F^{\theta z} \right)}{\partial \theta} + \dfrac{\partial F^{zz}}{\partial z} + \dfrac{1}{r} F^{rz} \right] \ . \end{aligned}\end{split}\]

or, after few algebraic manipulations,

\[\begin{split}\begin{aligned} \nabla \cdot \mathbb{F} & = \\ & = \hat{r} \left[ \dfrac{\partial F^{rr} }{\partial r} + \dfrac{1}{r} \dfrac{\partial F^{\theta r} }{\partial \theta} + \dfrac{\partial F^{z r}}{\partial z} + \dfrac{1}{r} F^{rr} - \dfrac{1}{r} F^{\theta \theta} \right] + \\ & + \hat{\theta} \left[ \dfrac{\partial F^{r \theta} }{\partial r} + \dfrac{1}{r} \dfrac{\partial F^{\theta \theta} }{\partial \theta} + \dfrac{\partial F^{z \theta}}{\partial z} + \dfrac{1}{r} F^{r \theta} + \dfrac{1}{r} F^{\theta r} \right] + \\ & + \hat{z} \left[ \dfrac{\partial F^{r z}}{\partial r} + \dfrac{1}{r} \dfrac{\partial F^{\theta z} }{\partial \theta} + \dfrac{\partial F^{zz}}{\partial z} + \dfrac{1}{r} F^{rz} \right] \ . \end{aligned}\end{split}\]

10.6.3.4. Laplacian

Example 10.19 (Laplacian of a scalar field)

Example 10.20 (Laplacian of a vector field)