## continuity equation, cartisan to polar

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:

$\frac{1}{r}\frac{\partial}{\partial r}(r*V_{r}) +\frac{1}{r}\frac{\partial}{\partial \theta}(V_{\theta})=0$
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 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.