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What is the advantage of using a polar coordinate system with rotating unit vectors? Kleppner's and Kolenkow's An Introduction to Mechanics states that base vectors $\mathbf{ \hat{r}}$ and $\mathbf{\hat{\theta}}$ have a variable direction, such that for a Cartesian coordinates system's base vectors $\mathbf{ \hat{i}}$ and $\mathbf{ \hat{j}}$ we have

$$\mathbf{\hat{r}} = \cos \theta\ \mathbf{\hat{i}} + \sin \theta\ \mathbf{\hat{j}}$$

$$\mathbf{\hat{\theta}} = -\sin \theta\ \mathbf{\hat{i}} + \cos \theta\ \mathbf{\hat{j}}$$

Now, isn't counter-productive to define a coordinate system in terms of another? Why, at least in this book, we choose to use such a dependent coordinate system, instead of using a polar coordinate system employing a radius and the angle that this one forms with a polar axis, therefore independent of another coordinate system?

EDIT: Let me clarify that I'm not asking about the advantages of the polar coordinate system over the Cartesian one, but about the advantages of a polar coordinate system defined on rotating base vectors $\mathbf{ \hat{r}}$ and $\mathbf{\hat{\theta}}$ over another polar coordinate system where we employ a base vector $\mathbf{ \hat{r}}$ and the angle (hence a scalar and not a vector) that this one forms with a polar axis.

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    $\begingroup$ I think you may have misunderstood: the coordinate system described here is the plain polar coordinate system. The fact that the relationship between $(\mathbf{\hat{r}}, \mathbf{\hat{\theta}})$ and $(\mathbf{\hat{i}}, \mathbf{\hat{j}})$ is not constant is just a fact of life. $\endgroup$ – dmckee Jul 26 at 17:10
  • $\begingroup$ $\theta$ is a shorthand for $\tan^{-1}(y/x)$. This transform describes a polar coordinate system with the polar axis parallel to $\hat{\mathbf{i}}$, but one could set the polar axis to be anywhere else by using $\tan^{-1}(y/x) + \theta_0$ for some constant $\theta_0$ instead. $\endgroup$ – eyeballfrog Jul 26 at 17:37
  • $\begingroup$ Please check my edit. $\endgroup$ – torito verdejo Jul 26 at 17:49
  • $\begingroup$ There is no such thing as a polar coordinate system with constant basis vectors. The nature of the coordinate system is that the basis vectors depend on the location at which you evaluate them. $\endgroup$ – dmckee Jul 26 at 17:57
  • $\begingroup$ Thank you, @dmckee. After thinking about describing velocity and acceleration I've understood that a polar coordinate system without base vectors would be a headache when facing operations, derivation and integration among/on vectors. $\endgroup$ – torito verdejo Jul 26 at 18:04
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Some algebraic expressions become less complex (and others more complex).

Some expressions involving differentials, like integrals, can become solvable analytically, that are not solvable analytically in Cartesian coordinates.

Some symmetries, like solutions of differential equations in fewer dimensions, differ between coordinate systems. One dimensional solutions in Cartesian coordinates are similar to walls in three dimensions, whereas in polar coordinates are cylindrical solutions.

You have a lack of one degree of freedom in origo in polar coordinates.

Just because coordinate systems are transformable back and forth between each other doesn't mean that other derived effects are similar.

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