Why does the length of $\vec{e_\theta}=\frac{d}{d\theta}(\vec{e_r})$ matter?

Question

Define:

• $\vec{e_r}:=\vec{i}\cos\theta+\vec{j}\sin\theta$;
• $\vec{e_\theta}:=\frac{d}{d\theta}(\vec{e_r})=-\vec{i}\sin\theta+\vec{j}\cos\theta.$

Consider:

• a unit circle with centre $O$;
• a point $A_\theta$ with $\overrightarrow{OA_\theta}=\overrightarrow{e_r}$;
• a point $B_\theta$ with $\overrightarrow{A_\theta B_\theta}=\overrightarrow{e_\theta}$;
• a particle $P$ at $A_\theta$.

$P$ never reaches any given $B_\theta$ since $\overrightarrow{e_\theta}$ is continually deflected in a direction anti-parallel to $\overrightarrow{e_r}$. In fact, $P$ only ever travels a differential distance towards $B_\theta$. So what difference would it make if $\overrightarrow{e_\theta}$ had a different length?

Answer 1

This definition of $\vec{e}_r,\,\vec{e}_\theta$ in terms of $\vec{e}_i:=\vec{i},\,\vec{e}_j:=\vec{j}$ has the elegant consequence$$\vec{e}_k\cdot\vec{e}_l=\delta_{kl}$$whether we stick to the Cartesian or polar coordinate system, as well as$$\frac{d}{d\theta}\vec{e}_r=\vec{e}_\theta,\,\frac{d}{d\theta}\vec{e}_\theta=-\vec{e}_r.$$Thus$$\frac{\partial}{\partial\theta}(f(r,\,\theta)\vec{e}_r+g(r,\,\theta)\vec{e}_\theta)=\left(\frac{\partial f}{\partial\theta}-g\right)\vec{e}_r+\left(f+\frac{\partial g}{\partial\theta}\right)\vec{e}_\theta.$$If $\vec{e}_\theta$ weren't a unit vector, we'd have various coefficients complicating these display-line results, which would be especially confusing for orbits where $r$ varies. As it is, the position vector $\vec{r}=r\vec{e}_r$ satisfies $\frac{\partial}{\partial\theta}\vec{r}=r\vec{e}_\theta$ (not to mention $\frac{\partial}{\partial r}\vec{r}=\vec{e}_r$), which is orthogonal to $\vec{r}$ because $\vec{r}\cdot\vec{r}$ is $\theta$-independent. (If you're interested in why, for example, a physicist might find all this interesting, see here.) Meanwhile, the equations$$\vec{i}=\cos\theta\vec{e}_r-\sin\theta\vec{e}_\theta,\,\vec{j}=\sin\theta\vec{e}_r+\cos\theta\vec{e}_\theta$$help us prove$$\nabla r=\frac{1}{2r}\nabla\vec{r}\cdot\vec{r}=\vec{e}_r,\,\nabla\theta=\cos^2\theta\nabla\frac{y}{x}=\frac1r\vec{e}_\theta,$$a system of vectors orthonormal to $\partial_r\vec{r},\,\partial_\theta\vec{r}$ in the sense discussed here.

Answer 2

Circumferential and radial components vectors are differently sized. There is a fixed ratio between them.

If the logarithmic spiral angle is $\alpha $ then

$$\dfrac { r(\theta)\, d\theta } {d\,r(\theta)}=\tan \alpha = m $$

(here wlog $ \dfrac{d\theta}{dt}=\omega = 1 $ )

Integrate $$ r= r_i e^ {m\theta }$$

If $m$ is small we have a slow progressing spiral, never asymptotic to starting $r= r_i.$