
#1
Feb2113, 03:12 PM

P: 1

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 [itex]\frac{\partial V_{x}}{\partial x}+\frac{\partial V_{y}}{\partial y}=0[/itex] by using: [itex]x=r*cos(\theta)\\ y = r*sin(\theta)[/itex] I get: [itex] \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} [/itex], I have defined: [itex] \begin{pmatrix} \frac{dr}{dt} \\ \frac{d\theta}{dt}\end{pmatrix} = \begin{pmatrix} V_{r} \\ V_{\theta}\end{pmatrix} [/itex] This gives: [itex]V_{x}=cos(\theta)*V_{r}r*sin(\theta)*V_{\theta}[/itex] and [itex]V_{y}=sin(\theta)*V_{r}+r*cos(\theta)*V_{\theta}[/itex] Now my problem arises, I do not see how I am supposed to calculate: [itex]\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} [/itex] Can you guys help me how to end this? It is supposed to be at the end: [itex] \frac{1}{r}\frac{\partial}{\partial r}(r*V_{r}) +\frac{1}{r}\frac{\partial}{\partial \theta}(V_{\theta})=0 [/itex] 



#2
Feb2113, 04:35 PM

HW Helper
P: 774

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 [itex]\nabla[/itex] or you will get things wrong.
The continuity equation is [tex]\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \mathbf{v}) = 0[/tex] which for a constant density [itex]\rho[/itex] reduces to [tex]\nabla \cdot \mathbf{v} = 0.[/tex] In cartesian coordinates, [tex]\nabla = \mathbf{e}_x \frac{\partial}{\partial x} + \mathbf{e}_y \frac{\partial}{\partial y}[/tex] where the unit vectors [itex]\mathbf{e}_x[/itex] and [itex]\mathbf{e}_y[/itex] are constant, so that if [itex]\mathbf{v} = v_x \mathbf{e}_x + v_y \mathbf{e}_y[/itex] then [tex] \nabla \cdot \mathbf{v} = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} [/tex] The unit vectors in polar coordinates are, in terms of the cartesian unit vectors, [itex]\mathbf{e}_r = \cos\theta \mathbf{e}_x + \sin\theta \mathbf{e}_y[/itex] and [itex]\mathbf{e}_\theta = \sin\theta \mathbf{e}_x + \cos\theta \mathbf{e}_y[/itex]. These are not constant but depend on [itex]\theta[/itex]. Thus, if [itex]\mathbf{v} = v_r\mathbf{e}_r + v_\theta \mathbf{e}_\theta[/itex] then [tex] \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 [/tex] and [tex] \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 [/tex] It is fairly easy to show that, in polar coordinates, [tex] \nabla = \mathbf{e}_r\frac{\partial }{\partial r} + \mathbf{e}_\theta \frac1r \frac{\partial}{\partial \theta}[/tex] (the best way is to express the derivatives with respect to [itex]r[/itex] and [itex]\theta[/itex] in terms of those with respect to [itex]x[/itex] and [itex]y[/itex] using the chain rule, and then invert a 2x2 matrix to obtain the derivatives with respect to [itex]x[/itex] and [itex]y[/itex] in terms of those with respect to [itex]r[/itex] and [itex]\theta[/itex], and substitute those expressions into the cartesian version of [itex]\nabla[/itex]) and it is straightforward to compute [tex] \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)[/tex] to get the correct expression. 


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