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Continuity equation, cartisan to polar

by arneolsen
Tags: cartisan, continuity, equation, polar
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Feb21-13, 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:
y = r*sin(\theta)[/itex]

I get:

\frac{dx}{dt} \\

V_{x} \\
cos(\theta) & -r*sin(\theta) \\
sin(\theta) & r*cos(\theta)\end{pmatrix}
V_{r} \\
, I have defined: [itex]
\frac{dr}{dt} \\
V_{r} \\

This gives:
[itex]V_{x}=cos(\theta)*V_{r}-r*sin(\theta)*V_{\theta}[/itex] and

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}

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
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Feb21-13, 04:35 PM
HW Helper
P: 1,008
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
\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, [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
\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
\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}[/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
\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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