# Continuity equation, cartisan to polar

1. Feb 21, 2013

### arneolsen

Hello, I've allways wondered how to get to polar coordinates from cartisan coordinates. I took a course in fluid mechanics but we never learned how to get the continuity equation from cartisan to polar. I know you can use physics to derive the polar equation, but I want to do it just by using mathematics and the cartisan equation.

In cartisan the equation is
$\frac{\partial V_{x}}{\partial x}+\frac{\partial V_{y}}{\partial y}=0$

by using:
$x=r*cos(\theta)\\ y = r*sin(\theta)$

I get:

$\begin{pmatrix} \frac{dx}{dt} \\ \frac{dy}{dt}\end{pmatrix} = \begin{pmatrix} V_{x} \\ V_{y}\end{pmatrix} = \begin{pmatrix} cos(\theta) & -r*sin(\theta) \\ sin(\theta) & r*cos(\theta)\end{pmatrix} * \begin{pmatrix} V_{r} \\ V_{\theta}\end{pmatrix}$
, I have defined: $\begin{pmatrix} \frac{dr}{dt} \\ \frac{d\theta}{dt}\end{pmatrix} = \begin{pmatrix} V_{r} \\ V_{\theta}\end{pmatrix}$

This gives:
$V_{x}=cos(\theta)*V_{r}-r*sin(\theta)*V_{\theta}$ and
$V_{y}=sin(\theta)*V_{r}+r*cos(\theta)*V_{\theta}$

Now my problem arises, I do not see how I am supposed to calculate:
$\frac{\partial V_{x}}{\partial x}+\frac{\partial V_{y}}{\partial y}= \frac{\partial (cos(\theta)*V_{r}-r*sin(\theta)*V_{\theta})}{\partial x} + \frac{ \partial (sin(\theta)*V_{r}+r*cos(\theta)*V_{\theta}) }{\partial y}$

Can you guys help me how to end this? It is supposed to be at the end:

$\frac{1}{r}\frac{\partial}{\partial r}(r*V_{r}) +\frac{1}{r}\frac{\partial}{\partial \theta}(V_{\theta})=0$

Last edited: Feb 21, 2013
2. Feb 21, 2013

### pasmith

Unfortunately vector calculus doesn't work that way. If you're trying to convert from cartesian coordinates then you need to work in terms of the vector derivative operator $\nabla$ or you will get things wrong.

The continuity equation is
$$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) = 0$$
which for a constant density $\rho$ reduces to
$$\nabla \cdot \mathbf{v} = 0.$$

In cartesian coordinates,
$$\nabla = \mathbf{e}_x \frac{\partial}{\partial x} + \mathbf{e}_y \frac{\partial}{\partial y}$$
where the unit vectors $\mathbf{e}_x$ and $\mathbf{e}_y$ are constant, so that if $\mathbf{v} = v_x \mathbf{e}_x + v_y \mathbf{e}_y$ then
$$\nabla \cdot \mathbf{v} = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y}$$

The unit vectors in polar coordinates are, in terms of the cartesian unit vectors, $\mathbf{e}_r = \cos\theta \mathbf{e}_x + \sin\theta \mathbf{e}_y$ and $\mathbf{e}_\theta = -\sin\theta \mathbf{e}_x + \cos\theta \mathbf{e}_y$. These are not constant but depend on $\theta$. Thus, if $\mathbf{v} = v_r\mathbf{e}_r + v_\theta \mathbf{e}_\theta$ then
$$\frac{\partial}{\partial \theta} (v_r\mathbf{e}_r ) = \frac{\partial v_r}{\partial \theta} \mathbf{e}_r + v_r \frac{\partial}{\partial \theta} \mathbf{e}_r = \frac{\partial v_r}{\partial \theta} \mathbf{e}_r + v_r \mathbf{e}_\theta$$
and
$$\frac{\partial}{\partial \theta} (v_\theta\mathbf{e}_\theta ) = \frac{\partial v_\theta}{\partial \theta} \mathbf{e}_\theta + v_\theta \frac{\partial}{\partial \theta} \mathbf{e}_\theta = \frac{\partial v_\theta}{\partial \theta} \mathbf{e}_\theta - v_\theta \mathbf{e}_r$$

It is fairly easy to show that, in polar coordinates,
$$\nabla = \mathbf{e}_r\frac{\partial }{\partial r} + \mathbf{e}_\theta \frac1r \frac{\partial}{\partial \theta}$$
(the best way is to express the derivatives with respect to $r$ and $\theta$ in terms of those with respect to $x$ and $y$ using the chain rule, and then invert a 2x2 matrix to obtain the derivatives with respect to $x$ and $y$ in terms of those with respect to $r$ and $\theta$, and substitute those expressions into the cartesian version of $\nabla$) and it is straightforward to compute
$$\nabla \cdot \mathbf{v} = \left(\mathbf{e}_r\frac{\partial }{\partial r} + \mathbf{e}_\theta \frac1r \frac{\partial}{\partial \theta}\right)\cdot \left(v_r\mathbf{e}_r + v_\theta \mathbf{e}_\theta\right)$$
to get the correct expression.