# 20.1: I.1 Cylindrical coordinates

Let ψ be a scalar and $$\vec{u}$$ be a vector expressed as a linear combination of the cylindrical basis vectors: $$\vec{u} = u_r\hat{e}^{(r)} +u_\theta \hat{e}^{(θ)} +u_z\hat{e}^{(z)}$$.

$\vec{\nabla} \psi=\hat{e}^{(r)} \frac{\partial \psi}{\partial r}+\hat{e}^{(\theta)} \frac{1}{r} \frac{\partial \psi}{\partial \theta}+\hat{e}^{(z)} \frac{\partial \psi}{\partial z}\label{eqn:1}$

Divergence of a vector

$\vec{\nabla} \cdot \vec{u}=\frac{1}{r} \frac{\partial\left(r u_{r}\right)}{\partial r}+\frac{1}{r} \frac{\partial u_{\theta}}{\partial \theta}+\frac{\partial u_{z}}{\partial z}\label{eqn:2}$

Curl of a vector

$\vec{\nabla} \times \vec{u}=\hat{e}^{(r)}\left(\frac{1}{r} \frac{\partial u_{z}}{\partial \theta}-\frac{\partial u_{\theta}}{\partial z}\right)+\hat{e}^{(\theta)}\left(\frac{\partial u_{r}}{\partial z}-\frac{\partial u_{z}}{\partial r}\right)+\hat{e}^{(z)}\left(\frac{1}{r} \frac{\partial\left(r u_{\theta}\right)}{\partial r}-\frac{1}{r} \frac{\partial u_{r}}{\partial \theta}\right)\label{eqn:3}$

Laplacian of a scalar

$\nabla^{2} \psi=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial \psi}{\partial r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} \psi}{\partial \theta^{2}}+\frac{\partial^{2} \psi}{\partial z^{2}}\label{eqn:4}$

Laplacian of a vector

$\nabla^{2} \vec{u}=\hat{e}^{(r)}\left(\nabla^{2} u_{r}-\frac{u_{r}}{r^{2}}-\frac{2}{r^{2}} \frac{\partial u_{\theta}}{\partial \theta}\right)+\hat{e}^{(\theta)}\left(\nabla^{2} u_{\theta}+\frac{2}{r^{2}} \frac{\partial u_{r}}{\partial \theta}-\frac{u_{\theta}}{r^{2}}\right)+\hat{e}^{(z)} \nabla^{2} u_{z}\label{eqn:5}$

Material derivative

$\frac{D}{D t}=\frac{\partial}{\partial t}+u_{r} \frac{\partial}{\partial r}+\frac{u_{\theta}}{r} \frac{\partial}{\partial \theta}+u_{z} \frac{\partial}{\partial z}\label{eqn:6}$

The equations of motion are the same as those listed in section 6.8, except that the Navier-Stokes momentum equation Equation 6.8.2 has added terms on the right-hand side, representing the centrifugal force. They arise mathematically because in curvilinear coordinates, the directions of the basis vectors vary in space. Differentiating the velocity to get acceleration therefore involves differentiating the basis vectors as well.

In cylindrical coordinates, the basis vectors $$\hat{e}^{(r)}$$ and $$\hat{e}^{(\theta)}$$ vary in space but $$\hat{e}^{(z)}$$ does not. We can therefore consider the simpler case of polar coordinates $$\{r,\theta\}$$. Suppose a fluid particle at $$\vec{x}$$ has velocity

$\vec{u}=u_{r} \hat{e}^{(r)}+u_{\theta} \hat{e}^{(\theta)}.\label{eqn:7}$

Over a short time interval $$dt$$, this velocity carries the particle to a new location $$\vec{x}+d\vec{x}$$. In Figure $$\PageIndex{2}$$a, the basis vectors at the initial and final locations are color coded red and blue, respectively. Now how does the velocity change? Differentiating Equation $$\ref{eqn:7}$$ gives

$\frac{d}{d t} \vec{u}=\frac{d u_{r}}{d t} \hat{e}^{(r)}+u_{r} \frac{d \hat{e}^{(r)}}{d t}+\frac{d u_{\theta}}{d t} \hat{e}^{(\theta)}+u_{\theta} \frac{d \hat{e}^{(\theta)}}{d t}.\label{eqn:8}$

Figure $$\PageIndex{2}$$b shows the increments of the basis vectors, $$d\hat{e}^{(r)}$$ and $$d\hat{e}^{(θ)}$$. Because the basis vectors have length 1, the distance between their tips is approximated by the arc length $$d\theta$$ (accurate as $$|d\theta| \rightarrow 0$$). The direction of $$\hat{e}^{(r)}$$ is parallel to that of $$\hat{e}^{(\theta)}$$, while the direction of $$\hat{e}^{(\theta)}$$ is opposite to that of $$\hat{e}^{(r)}$$. Therefore:

$d \hat{e}^{(r)}=d \theta \hat{e}^{(\theta)} ; \quad d \hat{e}^{(\theta)}=-d \theta \hat{e}^{(r)}$

Dividing by $$dt$$ and taking the limit $$dt \rightarrow 0$$, we obtain the time derivatives of the basis vectors. We now substitute these into Equation $$\ref{eqn:8}$$ and find that

$\frac{d}{d t} \vec{u}=\frac{d u_{r}}{d t} \hat{e}^{(r)}+u_{r} \frac{d \theta}{d t} \hat{e}^{(\theta)}+\frac{d u_{\theta}}{d t} \hat{e}^{(\theta)}-u_{\theta} \frac{d \theta}{d t} \hat{e}^{(r)}.\label{eqn:9}$

Note finally that $$d\theta/dt = u_\theta /r$$, giving

$\frac{d}{d t} \vec{u}=\frac{d u_{r}}{d t} \hat{e}^{(r)}+\frac{u_{r} u_{\theta}}{r} \hat{e}^{(\theta)}+\frac{d u_{\theta}}{d t} \hat{e}^{(\theta)}-\frac{u_{\theta}^{2}}{r} \hat{e}^{(r)}.\label{eqn:10}$

The second and fourth terms on the right-hand side are the centripetal acceleration. If we write the material derivative as Equation $$\ref{eqn:6}$$, we must subtract these extra terms from the right hand side:

$\rho \frac{D \vec{u}}{D t}=\rho \vec{a}_{C}+\rho \vec{g}-\vec{\nabla} p+\mu \nabla^{2} \vec{u}+\mu \vec{\nabla}(\vec{\nabla} \cdot \vec{u}),\label{eqn:11}$

where

$\vec{a}_{C}=\frac{u_{\theta}^{2}}{r} \hat{e}^{(r)}-\frac{u_{r} u_{\theta}}{r} \hat{e}^{(\theta)}.\label{eqn:12}$