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

I have earlier posted the same question here on `math stackexchange` but without any answer. As the question concerns tensors, I guess that I have come to the right place i.e. to physicists.

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 $\theta$.

I am thinking about `covariant derivatives` as the conversion process includes differentiation with respect to the bases. For example if $x=r \; \cos{\theta}$, then

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

So we have terms like $\frac{\partial{x}}{\partial{r}}$ and $\frac{\partial{x}}{\partial{\theta}}$ that concern basis vectors both in the `Cartesian` and in the `Polar` coordinate systems.

Hope that someone can shed some light on this.

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## 1 Answer

1) The first derivation is correct. When you make the second one for $\dot x$ written as $\dot x = \dfrac{d x}{d t}$ you make an error when using full derivatives instead of partial ones. In the latter case you would have obtained: $$\dot x=\dfrac{d x}{d t}=\dfrac{\partial x}{\partial \theta}\dfrac{d \theta}{d t} +\dfrac{\partial x}{\partial \phi}\dfrac{d \phi}{d t}+\dfrac{\partial x}{\partial r}\dfrac{d r}{d t}$$

Then, as $x=r \cos \theta$, you can insert $\dfrac{\partial x}{\partial \phi}=0, \dfrac{\partial x}{\partial \theta}=-r \sin \theta$, $\dfrac{\partial x}{\partial r}=\cos \theta$ into the last formula to obtain the same result you had before.

2) The derivatives you are using here are not covariant derivatives, they are always partial. The derivative $\dfrac{d}{d t}$ is expressed through partial derivatives, in contrast to full derivative $\dfrac{D}{D t}$, which is expressed through covariant derivatives. As you are actually not using covariant derivatives in your derivation, you don't have to worry about the basis vectors.

3) You could make you derivation alternatively by rewriting your original equations in a covariant form $M\dfrac{D^2 x^i}{D t^2}+C\dfrac{D x^i}{D t}+Kx^i=0$, where $x^i$ stands for $(x,y,z)$, stating that the equations of motion do not depend on the choice of coordinates, rewriting the equation in spherical coordinates as $M\dfrac{D^2 \tilde x^i}{D t^2}+C\dfrac{D \tilde x^i}{D t}+K \tilde x^i=0$, where $\tilde x = (r, \theta, \phi)$, and expanding all the derivatives $\dfrac{D}{D t}$ through partial derivatives and Christoffel symbols in spherical coordinates. You would have to take care of the basis, as the result would be in natural basis, whereas the equations of motion are usually written in orthonormal basis.

4) Another powerful way to transform coordinates would be by using lagrangian form of the equations of motion. This would be possible for the case of conservative systems only (no damping terms). In this case you could just rewrite the lagrangian of your system in spherical coordinates and write lagrangian equations from it in a more or less straightforward manner.

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1
As a coda to this, the critical place where covariant differentiation shows up irrevocably is in the case where you have spatial derivatives. – Jerry Schirmer Mar 13 '12 at 4:29
@Alexey Thank you. I am working in Polar coordinates with $r$ and $\theta$ and you have illustrated the case for spherical coordinates, thus the additional azimuth angle $\phi$ in you equations. 1 vote up. – yCalleecharan Mar 13 '12 at 8:09
@Jerry Schirmer Thanks. But we do have spatial derivatives of $x$ w.r.t. $r$ and so on as indicated in my post. 1 vote up. – yCalleecharan Mar 13 '12 at 8:11
@yCalleecharan: I mean, in the original Lagrangian. If you tried to convert something like the wave equation $\frac{1}{c^{2}}\frac{\partial^{2}\phi}{\partial t^{2}} - \left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right)\phi$ into polar coordinates, you'd find that the transformed equation would contain terms that only have first spatial derivatives of $\phi$. The coefficients of these terms are your nonzero $\Gamma_{ij}{}^{k}$ – Jerry Schirmer Mar 14 '12 at 5:25
@Alexey I have corrected my post and put the partial derivatives where necessary. – yCalleecharan Mar 17 '12 at 19:46
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