Engineering at Alberta Courses » Vector Calculus in Cylindrical Coordinate Systems

Calculus: Vector Calculus in Cylindrical Coordinate Systems

Introduction

Polar Coordinate System

Consider the representation of a geometric plane using $\mathbb{R}^2$ with a chosen but arbitrary origin. The directions at every point in the plane are defined using the basis vectors $e_1=\{1,0\}$ and $e_2=\{0,1\}$ (Fig 1 left). In certain situations, it is more convenient to define directions or basis vectors at every point such that the first direction (first basis vector) points away from the origin (blue arrows in Fig 1 right) and the second direction is perpendicular to the first direction while maintaining the right hand rule (yellow arrows in Fig 1 right). Given that the coordinates of a point in the geometric plane are given by $x_1$ and $x_2$ , then, we can define at every point the following relations:

$\begin{split} x_1&=r\cos\theta\\ x_2&=r\sin\theta \end{split}$

with the inverse relationship:

$\begin{split} r&=\sqrt{x_1^2+x_2^2}\\ \theta&=\arctan[x_1,x_2] \end{split}$

taking into consideration the quadrant of $x_1$ and $x_2$ . It is important to note that the function $\arctan[x1,x2]$ is not defined at the origin. It is also important to note that $e_r$ and $e_\theta$ are themselves vector fields. $e_r$ represents a vector field of unit vectors that are pointing away from the origin and $e_\theta$ represents a vector field of unit vectors perpendicular to the vectors of the $e_r$ vector field maintaining the right hand orientation (Fig 1 left). Using a simple change of coordinates, the new basis set $B'=\{e_r,e_\theta\}$ at a point represented by the coordinates $x_1$ and $x_2$ (or the corresponding r and \theta) can be related to the Cartesian basis $e_1$ and $e_2$ using the relationships:

$\begin{split} e_r&=\cos(\theta)e_1+\sin(\theta)e_2\\ e_\theta&=-\sin(\theta)e_1+\cos(\theta)e_2 \end{split}$

Therefore, the coordinate transformation from the Cartesian basis to the polar coordinate system is described at every point using the matrix $Q$ :

$Q=\left(\begin{matrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{matrix}\right)$

As the vector fields $e_r$ and $e_\theta$ are functions of the two real numbers $r$ , and $\theta$ , then, we can find the derivatives of $e_r$ and $e_\theta$ as follows:

$e_r=\cos(\theta) e_1+\sin (\theta) e_2\Rightarrow \begin{cases}\frac{\partial e_r}{\partial r}=0\\ \frac{\partial e_r}{\partial \theta}=-\sin(\theta)e_1+\cos(\theta)e_2=e_\theta\end{cases}$

$e_\theta=-\sin(\theta) e_1+\cos (\theta) e_2\Rightarrow \begin{cases}\frac{\partial e_\theta}{\partial r}=0\\ \frac{\partial e_\theta}{\partial \theta}=-\cos(\theta)e_1-\sin(\theta)e_2=-e_r\end{cases}$

It is also useful to find the derivatives of the position vector of a point with respect to the independent variables $r$ and $\theta$ . The position $x\in\mathbb{R}^2$ of a point in the plane is given by:

$x=re_r(\theta)$

The derivatives of the position vector with respect to $r$ and $\theta$ are given by:

$\begin{split} \frac{\partial x}{\partial r}&=e_r(\theta)\\ \frac{\partial x}{\partial \theta}&=re_\theta(\theta) \end{split}$

These derivatives will be used to calculate the derivatives of other quantities in a polar coordinate system.

polar1 — Figure 1. Rectangular (left) vs. polar (right) coordinate systems in a plane

Cylindrical Coordinate System:

A cylindrical coordinate system is a system used for directions in \mathbb{R}^3 in which a polar coordinate system is used for the first plane (Fig 2 and Fig 3). The coordinate system directions can be viewed as three vector fields $e_r, e_\theta$ , and $e_z$ such that:

$\begin{split} e_r&=\cos(\theta)e_1+\sin(\theta)e_2\\ e_\theta&=-\sin(\theta)e_1+\cos(\theta)e_2\\ e_z&=e_3 \end{split}$

with $r$ and $\theta$ related to the coordinates $x_1$ and $x_2$ using the polar coordinate system relationships. The coordinate transformation from the Cartesian basis to the cylindrical coordinate system is described at every point using the matrix $Q$ :

$Q=\left(\begin{matrix}\cos(\theta)&\sin(\theta)&0\\-\sin(\theta)&\cos(\theta)&0\\0&0&1\end{matrix}\right)$

The vector fields $e_r$ and $e_\theta$ are functions of $\theta$ and their derivatives with respect to $r$ and $\theta$ follow from the polar coordinate system. $e_z$ on the other hand is independent of $r$ and $\theta$ .

cylindrical2 — Figure 2. Rectangular (left) vs. cylindrical (right) local directions at a point in space

Fields in Cylindrical Coordinate System

Let $D$ be a subset of $\mathbb{R}^3$ . If $\phi:D\rightarrow \mathbb{R}$ , $u:D\rightarrow \mathbb{R}^3$ , and $T:D\rightarrow\mathbb{M}^3$ are smooth scalar, vector and second-order tensor fields, then they can be chosen to be functions of either the Cartesian coordinates $x_1, x_2$ , and $x_3$ , or the corresponding real numbers $r$ , $\theta$ , and $z$ . Also, $u$ and $T$ can have their components expressed in either the fixed orthonormal basis set $B=\{e_1,e_2,e_3\}$ , or can be expressed using the cylindrical coordinate system directions at every point. In other words, $u$ , as a vector field, can have the one of the following forms:

$\begin{split} u(x_1,x_2,x_3)&=u_1(x_1,x_2,x_3)e_1+u_2(x_1,x_2,x_3)e_2+u_3(x_1,x_2,x_3)e_3\\ &=u(r,\theta,z)\\ &=u_r(r,\theta,z)e_r(\theta)+u_\theta(r,\theta,z)e_\theta(\theta)+u_z(r,\theta,z)e_z \end{split}$

where $u_1,u_2,u_3,u_r,u_\theta$ , and $u_z$ are scalar fields. Similarly, $T$ , as a tensor field, can have one of the following forms:

$\begin{split} T(x_1,x_2,x_3)&=T_{11}(x_1,x_2,x_3)e_1\otimes e_1+T_{12}(x_1,x_2,x_3)e_1\otimes e_2+T_{13}(x_1,x_2,x_3)e_1\otimes e_3\\ & +T_{21}(x_1,x_2,x_3)e_2\otimes e_1+T_{22}(x_1,x_2,x_3)e_2\otimes e_2+T_{23}(x_1,x_2,x_3)e_2\otimes e_3\\ & +T_{31}(x_1,x_2,x_3)e_3\otimes e_1+T_{32}(x_1,x_2,x_3)e_3\otimes e_2+T_{33}(x_1,x_2,x_3)e_3\otimes e_3\\ &=T(r,\theta,z)\\ &=T_{rr}(r,\theta,z)e_r(\theta)\otimes e_r(\theta)+T_{r\theta}(r,\theta,z)e_r(\theta)\otimes e_\theta(\theta)+T_{rz}(r,\theta,z)e_r(\theta)\otimes e_z\\ & +T_{\theta r}(r,\theta,z)e_\theta(\theta)\otimes e_r(\theta)+T_{\theta \theta}(r,\theta,z)e_\theta(\theta)\otimes e_\theta(\theta)+T_{\theta z}(r,\theta,z)e_\theta(\theta)\otimes e_z\\ & +T_{zr}(r,\theta,z)e_z\otimes e_r(\theta)+T_{z\theta}(r,\theta,z)e_z\otimes e_\theta(\theta)+T_{zz}(r,\theta,z)e_z\otimes e_z \end{split}$

where $T_{11},T_{12},T_{13},T_{21},T_{22},T_{23},T_{31},T_{32},T_{33},T_{rr},T_{r\theta},T_{rz},T_{\theta r},T_{\theta \theta},T_{\theta z},T_{zr},T_{z\theta}$ , and $T_{zz}$ are scalar fields.

Field Derivatives in Cylindrical Coordinate Systems

Gradient of a Scalare Field

Let $\phi:D\subset \mathbb{R}^3\rightarrow \mathbb{R}$ be a scalar field such that $\phi=\phi(r,\theta,z)$ . The gradient of $\phi$ in a cylindrical coordinate system can be obtained using one of two ways. The first way is to find $\phi$ as a function of $x_1, x_2$ , and $x_3$ by simply replacing $r$ , $\theta$ , and $z$ . Then, finding the gradient of $\phi$ in the Cartesian coordinate system and then utilizing the relationship $\nabla \phi'=Q \cdot \nabla \phi$ . After that, the variables $x_1,x_2$ , and $x_3$ can be replaced with $r,\theta$ , and $z$ . Alternative, the gradient of $\phi$ can be obtained directly in the cylindrical coordinate system. In order to find the expression for the gradient, recall that a scalar field $\phi$ is differentiable if there exists $\nabla\phi$ such that $\forall x\in D$ , $\forall n\in\mathbb{R}^3$

$\lim_{h\rightarrow 0}\left|\frac{\phi(x+hn)-\phi(x)}{h}-\nabla\phi\cdot n\right|=0$

Using this definition, we will aim to find the representation of $\nabla \phi$ in a cylindrical coordinate system. Let $\nabla\phi=\nabla\phi_r e_r +\nabla\phi_\theta e_\theta+\nabla\phi_z e_z$ . We will strategically choose particular expressions for hn to obtain the required expression for the vector $\nabla\phi$ . First, we will choose hn such that it is equivalent to a path increment caused by a change \Delta r, i.e., hn=\Delta r \frac{\partial x}{\partial r}=\Delta r e_r (Fig. 4). Then:

$\lim_{\Delta r\rightarrow 0}\left|\frac{\phi(x+\Delta r e_r)-\phi(x)}{\Delta r}-\nabla\phi\cdot e_r\right|=0$

Therefore:

$\lim_{\Delta r\rightarrow 0}\left|\frac{\phi(r+\Delta r,\theta,z)-\phi(r,\theta,z)}{\Delta r}\right|=\nabla\phi_r$

Therefore:

$\nabla \phi_r=\frac{\partial \phi}{\partial r}$

Similarly, $hn$ can be chosen to be equivalent to a path increment caused by $\Delta \theta$ , i.e., $hn=\Delta \theta \frac{\partial x}{\partial \theta}=r\Delta \theta e_\theta$ (Fig. 4). Therefore:

$\lim_{\Delta \theta \rightarrow 0}\left|\frac{\phi(x+r\Delta \theta e_\theta)-\phi(x)}{\Delta \theta}-\nabla\phi\cdot re_\theta\right|=0$

Therefore:

$\lim_{\Delta \theta \rightarrow 0}\left|\frac{\phi(x+r\Delta \theta e_\theta)-\phi(x)}{\Delta \theta}\right|=\nabla\phi\cdot re_\theta=r\nabla\phi_\theta$

Therefore:

$\lim_{\Delta \theta \rightarrow 0}\left|\frac{\phi(r,\theta+\Delta\theta,z)-\phi(r,\theta,z)}{\Delta \theta}\right|=r\nabla\phi_\theta$

Therefore,

$\nabla\phi_\theta=\frac{\partial \phi}{r\partial\theta}$

The third component is straightforward and is equal to:

$\nabla\phi_z=\frac{\partial \phi}{\partial z}$

Therefore, the gradient of $\phi$ in a cylindrical coordinate system has the form:

$\nabla \phi = \frac{\partial \phi}{\partial r} e_r + \frac{\partial \phi}{r\partial \theta} e_\theta+ \frac{\partial \phi}{\partial z} e_z$

cylindrical3 — Figure 4. Variation of a scalar field along the directions of a polar coordinate system

Gradient of a Vector Field

Let $u:D\subset \mathbb{R}^3\rightarrow \mathbb{R}^3$ be a smooth vector field. The components of the tensor field $\nabla u$ in a cylindrical coordinate system can be obtained by a simple coordinate transformation using the components in the Cartesian coordinate system and the matrix of transformation $Q$ . I.e., $\nabla u'=Q\cdot\nabla u \cdot Q^T$ . Alternatively, if $u$ is already expressed in a cylindrical coordinate system, then, notice that the derivatives of $u$ with respect to $r$ , $\theta$ , and $z$ are given by:

$\begin{split} \frac{\partial u}{\partial r}&=\frac{\partial u_r}{\partial r}e_r+\frac{\partial u_\theta}{\partial r}e_\theta+\frac{\partial u_z}{\partial r}e_z\\ \frac{\partial u}{\partial \theta}&=\frac{\partial u_r}{\partial \theta}e_r+u_r e_\theta+\frac{\partial u_\theta}{\partial \theta}e_\theta-u_{\theta}e_r+\frac{\partial u_z}{\partial \theta}e_z\\ \frac{\partial u}{\partial z}&=\frac{\partial u_r}{\partial z}e_r+\frac{\partial u_\theta}{\partial z}e_\theta+\frac{\partial u_z}{\partial z}e_z \end{split}$

We can now assume assume that, in the cylindrical coordinate system, $\nabla u'$ has the following form:

$\nabla u'=\left(\begin{matrix}\nabla u_{rr} & \nabla u_{r\theta} & \nabla u_{rz}\\ \nabla u_{\theta r} & \nabla u_{\theta \theta} & \nabla u_{\theta z}\\\nabla u_{zr} & \nabla u_{z\theta} & \nabla u_{zz} \end{matrix}\right)$

Notice that:

$\begin{split} \nabla u_{rr}&=\nabla u e_r\cdot e_r\\ \nabla u_{r\theta}&=\nabla u e_\theta \cdot e_r\\ \nabla u_{rz}&=\nabla u e_z \cdot e_r\\ \nabla u_{\theta r}&=\nabla u e_r\cdot e_\theta \\ \nabla u_{\theta \theta}&=\nabla u e_\theta \cdot e_\theta \\ \nabla u_{\theta z}&=\nabla u e_z \cdot e_\theta \\ \nabla u_{z r}&=\nabla u e_r\cdot e_z \\ \nabla u_{z \theta}&=\nabla u e_\theta \cdot e_z \\ \nabla u_{z z}&=\nabla u e_z \cdot e_z \end{split}$

Recall that a vector field is differentiable if there exists a tensor field denoted $\nabla u$ such that $\forall x\in D,\forall n\in\mathbb{R}^3$ :

$\lim_{h\rightarrow 0}\left\|\frac{u(x+hn)-u(x)}{h}-\nabla u(n)\right\|=0$

Therefore, the component $\nabla u_{rr}$ can be obtained by setting $n=\Delta r e_r$ in the above relationship and taking the dot product with $e_r$ as follows:

$\lim_{\Delta r\rightarrow 0}\left(\frac{u(x+\Delta r e_r)-u(x)}{\Delta r}\right)\cdot e_r=\nabla u_{rr}$

I.e.

$\frac{\partial u}{\partial r}\cdot e_r=\nabla u_{rr}$

Therefore:

$\nabla u_{rr}=\frac{\partial u_r}{\partial r}$

The component $\nabla u_{r\theta}$ can be obtained by setting $n=r\Delta \theta e_\theta$ in the above relationship and taking the dot product with $e_r$ as follows:

$\lim_{\Delta \theta\rightarrow 0}\left(\frac{u(x+r\Delta \theta e_\theta)-u(x)}{\Delta \theta}\right)\cdot e_r=r\nabla u_{r\theta}$

I.e.

$\frac{\partial u}{\partial \theta}\cdot e_r=r\nabla u_{r\theta}$

Therefore:

$\nabla u_{r\theta}=\frac{\partial u}{r\partial \theta}\cdot e_r=\frac{\partial u_r}{r\partial \theta}-\frac{u_{\theta}}{r}$

Similarly,

$\begin{split} \nabla u_{\theta r}&=\frac{\partial u_\theta}{\partial r}\\ \nabla u_{\theta \theta}&=\frac{\partial u_\theta}{r\partial \theta}+\frac{u_r}{r}\\ \nabla u_{\theta z}&=\frac{\partial u_\theta}{\partial z}\\ \nabla u_{z \theta}&=\frac{\partial u_z}{r\partial \theta}\\ \nabla u_{rz}&=\frac{\partial u_r}{\partial z}\\ \nabla u_{z r}&=\frac{\partial u_z}{\partial r}\\ \nabla u_{zz}&=\frac{\partial u_z}{\partial z} \end{split}$

Therefore, $\nabla u'$ has the following form:

$\nabla u'=\left(\begin{matrix} \frac{\partial u_r}{\partial r} & \frac{\partial u_r}{r\partial \theta}-\frac{u_{\theta}}{r} &\frac{\partial u_r}{\partial z}\\ \frac{\partial u_\theta}{\partial r} & \frac{\partial u_\theta}{r\partial \theta}+\frac{u_r}{r} &\frac{\partial u_\theta}{\partial z}\\ \frac{\partial u_z}{\partial r} & \frac{\partial u_z}{r\partial \theta}& \frac{\partial u_z}{\partial z} \end{matrix}\right)$

Divergence of a Vector Field

If $u$ is given similar to the previous section, then, the divergence of $u$ in a cylindrical coordinate system is given by:

$\mathrm{div}(u)=\mathrm{Trace}(\nabla u)=\frac{\partial u_r}{\partial r}+\frac{\partial u_\theta}{r\partial \theta}+\frac{u_r}{r}+\frac{\partial u_z}{\partial z}$

Gradient of a Tensor Field

Let $T(r,\theta,z)$ be a tensor field with components $T_{\alpha\beta}(r,\theta,z)$ with $\alpha,\beta\in\{r,\theta,z\}$ . First, we use the tensor product representation of the tensor field to find the derivatives of $T$ with respect to $r,\theta$ , and $z$ . The derivatives with respect to $r$ and $z$ are straightforward since $e_r,e_\theta$ , and $e_z$ are independent of $r$ and $z$ :

$\begin{split} \frac{\partial T(r,\theta,z)}{\partial r}=&\frac{\partial T_{rr}}{\partial r}e_r\otimes e_r+\frac{\partial T_{r\theta}}{\partial r}e_r\otimes e_\theta+\frac{\partial T_{rz}}{\partial r}e_r\otimes e_z\\ & +\frac{\partial T_{\theta r}}{\partial r}e_\theta\otimes e_r+\frac{\partial T_{\theta\theta}}{\partial r}e_\theta\otimes e_\theta+\frac{\partial T_{\theta z}}{\partial r}e_\theta\otimes e_z\\ & +\frac{\partial T_{zr}}{\partial r}e_z\otimes e_r+\frac{\partial T_{z\theta}}{\partial r}e_z\otimes e_\theta+\frac{\partial T_{zz}}{\partial r}e_z\otimes e_z \end{split}$

Similarly,

$\begin{split} \frac{\partial T(r,\theta,z)}{\partial z}=&\frac{\partial T_{rr}}{\partial z}e_r\otimes e_r+\frac{\partial T_{r\theta}}{\partial z}e_r\otimes e_\theta+\frac{\partial T_{rz}}{\partial z}e_r\otimes e_z\\ & +\frac{\partial T_{\theta r}}{\partial z}e_\theta\otimes e_r+\frac{\partial T_{\theta\theta}}{\partial z}e_\theta\otimes e_\theta+\frac{\partial T_{\theta z}}{\partial z}e_\theta\otimes e_z\\ & +\frac{\partial T_{zr}}{\partial z}e_z\otimes e_r+\frac{\partial T_{z\theta}}{\partial z}e_z\otimes e_\theta+\frac{\partial T_{zz}}{\partial z}e_z\otimes e_z \end{split}$

However, the dependence of $e_r$ and $e_{\theta}$ on $\theta$ leads to the following derivative of $T$ with respect to $\theta$ :

$\begin{split} \frac{\partial T(r,\theta,z)}{\partial \theta}=&\frac{\partial T_{rr}}{\partial \theta}e_r\otimes e_r+\frac{\partial T_{r\theta}}{\partial \theta}e_r\otimes e_\theta+\frac{\partial T_{rz}}{\partial \theta}e_r\otimes e_z\\ & +\frac{\partial T_{\theta r}}{\partial \theta}e_\theta\otimes e_r+\frac{\partial T_{\theta\theta}}{\partial \theta}e_\theta\otimes e_\theta+\frac{\partial T_{\theta z}}{\partial \theta}e_\theta\otimes e_z\\ & +\frac{\partial T_{zr}}{\partial \theta}e_z\otimes e_r+\frac{\partial T_{z\theta}}{\partial \theta}e_z\otimes e_\theta+\frac{\partial T_{zz}}{\partial \theta}e_z\otimes e_z\\ &+T_{rr}e_r\otimes e_\theta+T_{rr}e_\theta\otimes e_r-T_{r\theta}e_r\otimes e_r+T_{r\theta}e_\theta\otimes e_\theta+T_{rz}e_\theta\otimes e_z\\ &+T_{\theta r}e_\theta\otimes e_\theta-T_{\theta r}e_r\otimes e_r-T_{\theta\theta}e_\theta\otimes e_r-T_{\theta\theta}e_r\otimes e_\theta-T_{\theta z}e_r\otimes e_z\\ &+T_{z r}e_z\otimes e_\theta-T_{z\theta}e_z\otimes e_r \end{split}$

Using the general definition of the gradient of a tensor field, the components of the gradient of $T$ denoted by $\nabla T$ in the cylindrical coordinate system can be obtained in a manner similar to the previous section. Let the components be $(\nabla T)_{\alpha\beta\gamma}$ with $\alpha,\beta,\gamma\in\{r,\theta,z\}$ . Then, it is straightforward (but with lots of details) to show that when $\gamma=r$ or $\gamma=z$ :

$(\nabla T)_{\alpha \beta \gamma}=\left(\frac{\partial T}{\partial \gamma}e_{\beta}\right)\cdot e_\alpha$

Therefore:

$\begin{split} (\nabla T)_{\alpha\beta r}=&\frac{\partial T_{\alpha \beta}}{\partial r}\\ (\nabla T)_{\alpha\beta z}=&\frac{\partial T_{\alpha \beta}}{\partial z} \end{split}$

However, the components when \gamma=\theta have the following form:

$\begin{split} (\nabla T)_{rr\theta}=&\frac{\partial T_{rr}}{r\partial \theta}-\frac{T_{r\theta}}{r}-\frac{T_{\theta r}}{r}\\ (\nabla T)_{r\theta\theta}=&\frac{\partial T_{r\theta}}{r\partial \theta}+\frac{T_{rr}}{r}-\frac{T_{\theta \theta}}{r}\\ (\nabla T)_{rz\theta}=&\frac{\partial T_{rz}}{r\partial \theta}-\frac{T_{\theta z}}{r}\\ (\nabla T)_{\theta r\theta}=&\frac{\partial T_{\theta r}}{r\partial \theta}+\frac{T_{rr}}{r}-\frac{T_{\theta \theta}}{r}\\ (\nabla T)_{\theta \theta \theta}=&\frac{\partial T_{\theta \theta}}{r\partial \theta}+\frac{T_{r\theta}}{r}+\frac{T_{\theta r}}{r}\\ (\nabla T)_{\theta z\theta}=&\frac{\partial T_{\theta z}}{r\partial \theta}+\frac{T_{rz}}{r}\\ (\nabla T)_{zr\theta}=&\frac{\partial T_{zr}}{r\partial \theta}-\frac{T_{z\theta}}{r}\\ (\nabla T)_{z \theta \theta}=&\frac{\partial T_{z\theta}}{r\partial \theta}+\frac{T_{zr}}{r}\\ (\nabla T)_{zz\theta}=&\frac{\partial T_{zz}}{r\partial \theta}\\ \end{split}$

Divergence of a Tensor Field

Let $T(r,\theta,z)$ be a tensor field with the cylindrical coordinate system components $T_{\alpha \beta}$ with $\alpha,\beta\in\{r,\theta,z\}$ .
Using the general definition of the divergence of a tensor field, the components of \mathrm{div}{(T)} in a cylindrical coordinate system can be obtained as follows:

$\begin{split} (\mathrm{div}(T))_r=&\mathrm{div}(Te_r)=\mathrm{Trace}(\nabla T e_r)\\ (\mathrm{div}(T))_\theta=&\mathrm{div}(Te_\theta)=\mathrm{Trace}(\nabla T e_\theta)\\ (\mathrm{div}(T))_z=&\mathrm{div}(Te_z)=\mathrm{Trace}(\nabla T e_z) \end{split}$

where $e_r, e_\theta$ , and $e_z$ are fixed in space at a particular point. I.e., the $\nabla$ operator only applies to $T$ . The procedure used in the gradient of a vector in a cylindrical coordinate system section combined with the derivatives of $T$ shown in the previous section can be used to reach the following formulas for the components of the divergence of $T$ in a cylindrical coordinate system:

$\begin{split} \mathrm{div}{(T)}&=\left(\begin{array}{c} \mathrm{Trace}(\nabla Te_r)\\\mathrm{Trace}(\nabla Te_\theta)\\\mathrm{Trace}(\nabla Te_z)\end{array}\right)\\ &=\left(\begin{array}{c} \left(\frac{\partial T}{\partial r}e_r \right)\cdot e_r + \left(\frac{\partial T}{r\partial \theta}e_r \right)\cdot e_\theta + \left(\frac{\partial T}{\partial z}e_r \right)\cdot e_z\\ \left(\frac{\partial T}{\partial r}e_\theta \right)\cdot e_r + \left(\frac{\partial T}{r\partial \theta}e_\theta \right)\cdot e_\theta + \left(\frac{\partial T}{\partial z}e_\theta \right)\cdot e_z\\ \left(\frac{\partial T}{\partial r}e_z \right)\cdot e_r + \left(\frac{\partial T}{r\partial \theta}e_z \right)\cdot e_\theta + \left(\frac{\partial T}{\partial z}e_z \right)\cdot e_z\end{array}\right)\\ &=\left(\begin{array}{c} \sum_{\alpha\in\{r,\theta,z\}}(\nabla T)_{\alpha r r}\\ \sum_{\alpha\in\{r,\theta,z\}}(\nabla T)_{\alpha \theta \theta} \\ \sum_{\alpha\in\{r,\theta,z\}}(\nabla T)_{\alpha z z}\end{array}\right) \end{split}$

Therefore:

$\mathrm{div}{(T)}=\left(\begin{array}{c} \frac{\partial T_{rr}}{\partial r}+\frac{\partial T_{r\theta}}{r\partial \theta}+\frac{T_{rr}-T_{\theta\theta}}{r}+\frac{\partial T_{rz}}{\partial z}\\ \frac{\partial T_{\theta r}}{\partial r}+\frac{\partial T_{\theta\theta}}{r\partial \theta}+\frac{T_{r\theta}+T_{\theta r}}{r}+\frac{\partial T_{\theta z}}{\partial z}\\ \frac{\partial T_{zr}}{\partial r}+\frac{\partial T_{z\theta}}{r\partial \theta}+\frac{T_{zr}}{r}+\frac{\partial T_{z z}}{\partial z} \end{array} \right)$

Curl of a Vector Field

Using the general definition of the Curl, if $u$ is a vector field given in terms of $r,\theta$ , and $z$ and represented in a cylindrical coodrinate system, then, the components of the curl of $u$ are given by:

$\begin{split} (\mbox{curl}(u))_r&=(\mbox{curl}(u))\cdot e_r=\mbox{div}(u\times e_r)=\mbox{Trace}(\nabla u\times e_r)\\ (\mbox{curl}(u))_\theta&=(\mbox{curl}(u))\cdot e_\theta=\mbox{div}(u\times e_\theta)=\mbox{Trace}(\nabla u\times e_\theta)\\ (\mbox{curl}(u))_z&=(\mbox{curl}(u))\cdot e_z=\mbox{div}(u\times e_z)=\mbox{Trace}(\nabla u\times e_z) \end{split}$

with the \nabla operator applied to u and not to the vectors $e_r, e_\theta$ , and $e_z$ . Therefore in a cylindrical coordinate system:

$\mbox{curl}(u)=\left(\begin{array}{c}\frac{\partial u_z}{r\partial \theta}-\frac{\partial u_{\theta}}{\partial z}\\ \frac{\partial u_r}{\partial z}-\frac{\partial u_z}{\partial r}\\ \frac{\partial u_\theta}{\partial r}+\frac{u_{\theta}}{r}-\frac{\partial u_r}{r\partial \theta} \end{array}\right)$

Laplacian of a Scalar Field

Using the general definition of the Laplacian, if $\phi$ is a scalar function given in terms of $r,\theta$ , and $z$ , then, the Laplacian of $\phi$ is given by:

$\begin{split} \nabla^2\phi&=\mathrm{div}(\nabla \phi)\\ &=\mathrm{div}\left(\frac{\partial \phi}{\partial r} e_r + \frac{\partial \phi}{r\partial \theta} e_\theta+ \frac{\partial \phi}{\partial z} e_z\right)\\ &=\frac{\partial^2 \phi}{\partial r^2}+\frac{\partial^2 \phi}{r^2\partial \theta^2}+\frac{\partial \phi}{r\partial r}+\frac{\partial^2 \phi}{\partial z^2} \end{split}$

Example

Let $u:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be given by $u=\{x_1^2+x_2^2,x_1x_2\}$ . Find the gradient of $u$ in the current coordinate system. Find the expression for $u$ and the gradient of $u$ in a polar coordinate system.

SOLUTION

In the Cartesian coordinate system, the gradient of $u$ has the form:

$\nabla u=\left(\begin{matrix}2x_1&2x_2\\x_2&x_1\end{matrix}\right)$

Note that $u$ and $\nabla u$ can be represented in the Cartesian coordinate system using $r$ and $\theta$ instead of $x_1$ and $x_2$ as follows:

$\begin{split} u=&\left(\begin{array}{c}r^2\\r^2\cos(\theta)\sin(\theta)\end{array}\right)\\ \nabla u=&\left(\begin{matrix}2r\cos(\theta)&2r\sin(\theta)\\r\sin(\theta)&r\cos(\theta)\end{matrix}\right) \end{split}$

It is important to note that this is a mere change of variables, but the components of $u$ and $\nabla u$ are still represented using the Cartesian coordinate system. The values themselves are just obtained using $r$ and $\theta$ , rather than $x_1$ and $x_2$ .

If $e_1$ , and $e_2$ are the basis vectors in the Cartesian coordinate system, and if $e_r$ and $e_\theta$ are the basis vectors in the cylindrical coordinate system, then the matrix of transformation from the Cartesian to the cylindrical coordinate system at a particular point depends on the value of theta at that particular point and is given by:

$Q=\left(\begin{matrix}\cos(\theta)&\sin(\theta)\\-\sin(\theta)&\cos(\theta)\end{matrix}\right)$

If $u'$ is the expression of $u$ in the polar coordinate system, it has the form:

$u'=Qu=\left(\begin{array}{c}u_r\\u_\theta\end{array}\right)=\left(\begin{array}{c}r^2\cos(\theta)\left(1+\sin^2(\theta)\right)\\-r^2\sin^3(\theta)\end{array}\right)$

The representation $\nabla u'$ in the cylindrical coordinate system can be obtained using the change of coordinates formula:

$\nabla u'=Q\nabla u Q^T=\left(\begin{matrix}-r\cos(\theta)(-3+\cos(2\theta))&r\cos(2\theta)\sin(\theta)\\-2r\sin^3(\theta)&r\cos(2\theta)\cos(\theta)\end{matrix}\right)$

Alternatively, the gradient of u in the cylindrical coordinate system can be obtained directly using the components $u_r$ and $u_\theta$ :

$\nabla u'=\left(\begin{matrix} \frac{\partial u_r}{\partial r} & \frac{\partial u_r}{r\partial \theta}-\frac{u_{\theta}}{r} \\ \frac{\partial u_\theta}{\partial r} & \frac{\partial u_\theta}{r\partial \theta}+\frac{u_r}{r} \end{matrix}\right)= \left(\begin{matrix}2r\cos(\theta)(1+\sin^2(\theta))&r\cos(2\theta)\sin(\theta)\\-2r\sin^3(\theta)&r\cos(2\theta)\cos(\theta)\end{matrix}\right)$

Noting that $2r\cos(\theta)(1+\sin^2(\theta))=-r\cos(\theta)(-3+\cos(2\theta))$ , the two methods produce identical results for the components of the gradient in the polar coordinate system.
The following figure shows the directions at an arbitrary point $(x_1,x_2)$ in a Cartesian coordinate system (left) and the representation of $u$ and $\nabla u$ in that coordinate system. On the right, the polar coordinate system directions are shown along with the representation $u'$ and $\nabla u'$ .

polar2 — Figure 5. Components of the vector and its gradient in a Cartesian (left) and polar (right) coordinate systems

View Mathematica Code

Q={{Cos[th],Sin[th]},{-Sin[th],Cos[th]}};
u={x1^2+x2^2,x1*x2};
x={x1,x2}
rule={x1->r*Cos[th],x2->r*Sin[th]}
urth=FullSimplify[u/.rule]
Gu=Table[D[u[[i]],x[[j]]],{i,1,2},{j,1,2}]
Gurth=FullSimplify[Gu/.rule]
up=FullSimplify[Q.urth]
Gup=FullSimplify[Q.Gurth.Transpose[Q]]
Gup//MatrixForm
Guformula=FullSimplify[{{D[up[[1]],r],D[up[[1]],th]/r-up[[2]]/r},{D[up[[2]],r],up[[1]]/r+D[up[[2]],th]/r}}];
Guformula//MatrixForm
FullSimplify[Guformula-Gup]

View Python Code

import sympy as sp
from sympy import diff, simplify
from sympy.matrices import Matrix
sp.init_printing(use_latex = "mathjax")
x1,x2 = sp.symbols('x_1 x_2')
r, theta = sp.symbols('r \u03B8')
u1 = x1**2+x2**2
u2 = x1*x2
# cartesian coordinate system
u = Matrix([u1,u2])
display("u =",u)
x=Matrix([x1,x2])
# gradient
grad_u=Matrix([[diff(ui,xj) for xj in x] for ui in u])
display("\u2207u =",grad_u)
# Change of variables
display("The following is a simple change of variables but the components are still in the cartesian coordinate system")
urth = u.subs([(x1, r*sp.cos(theta)), (x2, r*sp.sin(theta))])
display("u_r_\u03B8 =",urth,simplify(urth))
# gradient
grad_u_rth = grad_u.subs([(x1, r*sp.cos(theta)), (x2, r*sp.sin(theta))])
display("\u2207u_r_\u03B8 =",grad_u_rth)
# rot matrix
Q = Matrix([[sp.cos(theta),sp.sin(theta)],[-sp.sin(theta),sp.cos(theta)]])
display("Q =",Q)
# Change of coordinates
display("The following representation is in the radial coordinate system")
upolar=simplify(Q*urth)
grad_u_polar=simplify(Q*grad_u_rth*Q.T)
display("u' = Q*u =",upolar)
display("\u2207u' = Q*\u2207u*Q^T =",grad_u_polar)
# Gradient formula in the polar coordinate system
display("The gradient can also be calculated using the gradient formula")
grad_u_formula=Matrix([[diff(upolar[0],r),diff(upolar[0],theta)/r-upolar[1]/r],
                [diff(upolar[1],r),upolar[0]/r+diff(upolar[1],theta)/r]])
grad_u_formula=simplify(grad_u_formula)
display(grad_u_formula)

Open Educational Resources

Calculus: Vector Calculus in Cylindrical Coordinate Systems

Introduction

Polar Coordinate System

Cylindrical Coordinate System:

Fields in Cylindrical Coordinate System

Field Derivatives in Cylindrical Coordinate Systems

Gradient of a Scalare Field

Gradient of a Vector Field

Divergence of a Vector Field

Gradient of a Tensor Field

Divergence of a Tensor Field

Curl of a Vector Field

Laplacian of a Scalar Field

Example

SOLUTION

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