Constant circular motion: understanding $\underline{e_{\theta}}=\frac{d(\underline{e_r})}{d\theta}$

Question

Context: 1st year BSc Mathematics, Vectors and Mechanics module, constant circular motion.

This may be trivial, but can someone tell me what's wrong with the following reasoning?

$$\underline{e_r}=\underline{i}\cos\theta+\underline{j}\sin\theta=(1,\theta) \;\;(1);$$ $$\underline{e_\theta}=\frac{d(\underline{e_r})}{d\theta}=-\underline{i}\sin\theta+\underline{j}\cos\theta \;\;(2);$$ so $$(1),(2):\;\; \underline{e_\theta}=\frac{d}{d\theta}((1,\theta))=(1,\frac{d\theta}{d\theta})=(1,1) \;\;(3),$$ so $$(2),(3): \;\; (1,1)=-\underline{i}\sin\theta+\underline{j}\cos\theta \;\; (4),$$ an undesirable conclusion.

Answer 1

The equation $\displaystyle \frac{d}{d\theta}(r,\theta)=\left(\frac{dr}{d\theta},\frac{d\theta}{d\theta}\right)$ is not valid if $(r,\theta)$ is not Cartesian.

Note $(r,\theta)$ is supposed to represent the position $\vec{r}$, which is a function of the angle $\theta$ from some chosen origin and positive axis in a 2D plane. The derivative $d\vec{r}/d\theta$ is supposed to be interpreted as

$$ \frac{d\vec{r}}{d\theta}= \lim_{\Delta\theta\to0}\frac{\Delta\vec{r}}{\Delta\theta}. \tag{$\ast$}$$

Note $\Delta\vec{r}$ is vector displacement, and the vector displacement between $(r_1,\theta_1)$ and $(r_2,\theta_2)$ is not given in polar form by the formula $(r_2-r_1,\theta_2-\theta_1)$. If $r_1=r_2$ then obviously the "rotational displacement" i.e. the amount of rotation to go from one to the other is $\theta_2-\theta_1$, but the derivative $(\ast)$ describes the linear displacement in polar coordinates. For instance, if $\theta=t$ (so a point particle is going around the origin at constant unit speed), then the derivative is just the velocity vector, $d\vec{r}/d\theta=d\vec{r}/dt=\vec{v}$.

The angular component of $\vec{v}$ does not describe how $\theta$ is changing (i.e. it is not $d\theta/d\theta$), it says which direction the particle is moving. Similarly, radial component of $\vec{v}$ describes the speed, which is different from $dr/d\theta$ since the particle may have a non-radial component to its movement. The radical component of $\vec{v}$ would only match $dr/dt$ if it is (at a given moment) moving entirely radially, in which case $d\theta=0$ anyway. Also in this case, if $\theta$ is constant, the direction of movement is the same as the direction of position, so the angular component of $\vec{v}$ is also $\theta$, not $0$.

Answer 2

Unfortunately, most non-Cartesian coordinates don't let us just differentiate the coordinates characterizing a unit vector, whether we try all of them or a plausible subset. For a generally viable treatment, it helps to treat the $\theta$ subscript in $\underline{e}_\theta$ as a parameter and write the vector as a rotation of the $\theta=0$ case, say $\underline{e}_\theta=R_\theta\underline{e}_0$, so$$\frac{d}{d\theta}\underline{e}_\theta=\lim_{h\to0}\frac{\underline{e}_{\theta+h}-\underline{e}_\theta}{h}=R_\theta\lim_{h\to0}\frac{\underline{e}_{h}-\underline{e}_0}{h},$$a calculation I've made easier to read by placing the polar subscript outside the underline. To prove $\frac{d}{d\theta}\underline{e}_\theta=\underline{e}_{\theta-\pi/2}$, it suffices to check the case $\theta=0$, which can be done with a diagram instead of Cartesian coordinates.