# Conversion of motion equation from Cartesian to Polar coordinates: Is covariant differentiation necessary?

Say I have the following equation of motion in the Cartesian coordinate system for a typical mass spring damper system:

$$M \; \ddot{x} + C \; \dot{x} + K \; x = 0$$

where the dot $^\dot{}$ represents differentiation with respect to time.

Now I would like to convert this equation to Polar coordinates. So I introduce

$$x=r \; \cos{\theta}$$ to obtain

$$\dot{x}=\dot{r} \; \cos{\theta} - r \; \dot{\theta} \sin{\theta}$$

and $$\ddot{x}=\ddot{r} \; \cos{\theta}-2 \; \dot{r} \; \dot{\theta} \; \sin{\theta}-r \; \dot{\theta}^2 \; \cos{\theta}- r \; \ddot{\theta} \; \sin{\theta}$$

I can insert $x, \; \dot{x} \; \text{and} \; \ddot{x}$ in my original equation in the Cartesian coordinate system to yield

$$M \; (\ddot{r} \; \cos{\theta}-2 \; \dot{r} \; \dot{\theta} \; \sin{\theta}-r \; \dot{\theta}^2 \; \cos{\theta}- r \; \ddot{\theta} \; \sin{\theta}) + C \; (\dot{r} \; \cos{\theta} - r \; \dot{\theta} \sin{\theta}) + K \; (r \; \cos{\theta}) = 0$$

Note: I am just showing the equation and derivatives in the x-direction. But the full system has both $x$ and $y$ components.

I wonder if the above way of thinking is right. I am very new to tensors and I after reading about covariant derivatives, I am now thinking that one should include consider the basis vectors of the Polarcoordinate system (a non-Cartesiancoordinate system) also since unlike the basis vectors of the Cartesian coordinate system which do not change direction in the 2D space, Polar coordinate basis vectors change direction depending on the angle $\theta$.

Hope that someone can shed some light on this.

Thanks a lot...

Update

Note: The conversion process includes differentiation with respect to the bases. For example if $x=r \; \cos{\theta}$, then

$$\dot{x}=\frac{dx}{dt}=\frac{dx}{dr} \cdot \frac{dr}{dt} + \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$$

So we have terms like $\frac{dx}{dr}$ and $\frac{dx}{d\theta}$ that concern basis vectors both in the Cartesian and in the Polar coordinate systems.

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You've already accounted for the moving basis when you do the substitutions. Moreover, since you're not differentiating w.r.t. the spatial coordinates, you don't need to worry about covariant derivatives anyway. – Zhen Lin Mar 8 '12 at 13:56
@Zhen Lin Thanks. I have updated my post. As you can see we have differentiation with respect to the spatial coordinates. 1 vote up. – yCalleecharan Mar 8 '12 at 14:04
Yes, of course there's differentiation w.r.t. spatial coordinates when you're computing the substitutions. But your differential equations don't involve spatial coordinates. – Zhen Lin Mar 8 '12 at 17:14
@zhen Lin Yes not explicitly perhaps when just looking at the equation of motion but we are definitely using the spatial coordinates derivatives to arrive at the new equation of motion. – yCalleecharan Mar 8 '12 at 18:17
crossposted to Physics.SE: physics.stackexchange.com/q/22288/2451 – Qmechanic Mar 12 '12 at 21:06