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 Polar
coordinate system (a non-Cartesian
coordinate 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.