# Derivatives of polar coordiantes?

I'm a undergraduate and I was reading about the polar coordinate system specifically this paper. I don't understand the term: $$\frac{de_r}{d\theta} = e_\theta \text{, and } \frac{de_\theta}{d\theta} = -e_r$$
I don't see how you can have a the derivative of $$de_r$$ over $$d\theta$$ since they are not the same variable, if that makes any sence.

In reference to the diagram, ask yourself what happens to the unit vector $$\hat{e}_r$$ as you increase the angle $$\varphi$$ (as measured counterclockwise from the rightward pointing horizontal axis) by an infinitesimal amount $$d\varphi$$. Note that they use $$\varphi$$ in this diagram instead of $$\theta$$.

In words, the tail of the vector $$\hat{e}_r$$ moves counterclockwise along the circumference of the circle, with its tip always pointing radially outwards from the centre.

The difference between the transformed $$\hat{e}_r$$ (call it $$\hat{e}_{r1}$$), and the original $$\hat{e}_r$$ (call it $$\hat{e}_{r0}$$), is another vector (call it $$d\hat{e}_r$$), which points from the tip of $$\hat{e}_{r0}$$ to the tip of $$\hat{e}_{r1}.$$ That is, $$d\hat{e}_r=\hat{e}_{r1}-\hat{e}_{r0}.$$

In the limit as $$d\varphi\rightarrow 0,$$ that is, as $$\hat{e}_{r1}$$ and $$\hat{e}_{r0}$$ become closer and closer together, $$d\hat{e}_r$$ becomes closer and closer to a vector pointing perpendicular to $$\hat{e}_{r0}$$, i.e. it becomes close and closer to the vector $$\hat{e}_\varphi$$. Draw some pictures to convince yourself of this

In other words, as $$d\varphi\rightarrow 0,$$ $$d\hat{e}_r\rightarrow \hat{e}_\varphi,$$ which is to say that $$\frac{d\hat{e}_r}{d\varphi} = \hat{e}_\varphi.$$

Reference for the diagram

Edit:

To address @R.W.Bird's comment. I'll try to show that the magnitude of $$\frac{d\hat{e}_r}{d\varphi} = \hat{e}_\varphi$$ indeed is 1.

The magnitude of $$d\hat{e}_r$$ is given by

$$\left|{d\hat{e}_r}\right|=\left|\hat{e}_{r1}-\hat{e}_{r0}\right|=\sqrt{\hat{e}_{r1}^2-2\hat{e}_{r1}\cdot\hat{e}_{r0}+\hat{e}_{r0}^2}.$$

If we use the fact that by definition $$\left|\hat{e}_{r}\right|=1$$, then

$$\left|{d\hat{e}_r}\right|=\sqrt{1-2\cos(d\varphi)+1}=\sqrt{2-2\cos(d\varphi)},$$

since the angle between $$\hat{e}_{r0}$$ and $$\hat{e}_{r1}$$ is $$d\varphi$$. Now,

$$\left|\frac{{\left|d\hat{e}_r\right|}}{d\varphi}\right|=\left|\frac{d}{d\varphi}\sqrt{2-2\cos(d\varphi)}\right|=\left|\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|.$$

As $$d\varphi\rightarrow 0,$$

$$\lim_{d\varphi\rightarrow 0}\left|\frac{{\left|d\hat{e}_r\right|}}{d\varphi}\right|=\lim_{d\varphi\rightarrow 0}\left|\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|$$

$$=\lim_{d\varphi\rightarrow 0}\left|\frac{d}{d\varphi}\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|$$

$$=\lim_{d\varphi\rightarrow 0}\left|\frac{\cos (d\varphi)}{\sqrt{2-2 \cos (d\varphi)}}-\frac{\sin ^{2}(d\varphi)}{(2-2 \cos (d\varphi))^{3 / 2}}\right|$$

$$=1$$

where from the first to second line we have used l'Hopital's rule. This argument is a little hand-wavy, but it gets us the right answer... sorry about that.

• Your explanations of the directions are clear enough. It is not so obvious that the magnitude of the derivative should be 1. Oct 31, 2019 at 17:43
• Thanks alot, I understand alot better now, Thanks! Nov 1, 2019 at 16:34

$$\hat{e}_r$$ is a radial unit vector. It obviously points in different directions for different values of $$\theta$$. So it has a derivative expressing that rate of change.